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MRD:jhl416

2018-05-25T16:29:21Z

Jhl416: /* Distribution of energy */

== Molecular Reaction Dynamics: Applications to Triatomic systems ==

=== EXERCISE 1: H + H2 system ===

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

''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.''

With reference to the diagram provided (Gradients at the transition state, as well as in the reactants and product regions.) in the experimental script, q1 and q2 are defined as a different set of coordinates different from the coordinates of r1 and r2. q2 is along the direction of the reaction pathway and q1 is orthogonal to the direction of the reaction pathway.

Both the transition structures and minima have a 0 gradient where the first derivative would both be 0, i.e. ∂Ve(q1)/∂q1 * ∂Ve(q2)/∂q2 = 0

Their second derivatives need to be used in order to distinguish between the two.

'''For transition structures:'''

As the transition state is at the maximum along the reaction pathway, the second derivative of the transition state is < 0 along the q2 direction, i.e. ∂Ve2(q2)/∂q22 < 0

While the second derivative along the q1 direction is > 0 as it is a minimum along any other directions except from the direction of the reaction pathway, i.e. ∂Ve2(q1)/∂q12 > 0.

'''For minima:'''

Minima are at minimum along all directions, including the direction of the reaction pathway. Thus, both of the second derivatives (in q1 and q2 directions) are > 0, i.e. ∂Ve2(q2)/∂q22 > 0 and ∂Ve2(q1)/∂q12 > 0.

==== Trajectories from r1 = r2: locating the transition state ====

''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.''

At the transition state position, there is 0 gradient. Thus, without initial momentum, the kinetic energy of the system should be 0 where there is no oscillation present. Thus, distanceA-B and distanceB-C should be constant in the system.

[[File: Jhl416_ts_position.png |thumb|center|upright=2|Figure 1 - A plot of Internuclear Distance vs Time when r1 = r2 = 0.90777 Å | 600px]]

By testing different rts values, the best estimation of the transition state position is found to be 0.90777 Å. As seen from figure 1 above, when rts = 0.90777 Å, distanceA-B and distanceB-C are seen to be constant with no oscillation, which aligns with the above.

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

The reaction path calculated using different methods of mep and dynamics are presented below in figure 2 and 3.

[[File: jhl416_mep_surface.png |thumb|center|upright=2|Figure 2: mep surface plot at r1 = 0.91777, r2 = 0.90777, p1 = p2 = 0 ]]

From figure 2, it is seen that the trajectory calculated using mep appears to be a straight line. This is due to the fact that mep shows an infinitely slow motion of reaction path. As velocity is set to zero at every time step, the momentum and thus kinetic energy is also zero at each time step. This results in no oscillation motion and hence a straight line of the trajectory.

[[File: jhl416_dynamics_surface.png |thumb|center|upright=2|Figure 3: Dynamics surface plot at r1 = 0.91777, r2 = 0.90777, p1 = p2 = 0 ]]

From figure 3, the trajectory appears to be more wavy when calculated using the dynamics method. This gives a more realistic picture of the motion of atoms where masses of atoms are taken into account. Thus, the oscillation motion of the atoms can then be observed.

==== Reactive and unreactive trajectories ====

{| class="wikitable" border=1
! p1 !! p2 !! Reactive/Unreactive !! Description !! Image
|-
| -1.25 || -2.5 || Reactive || Along the trajectory, Hb and Hc are kept at a constant distance with no oscillation observed in the entrance channel. As Ha approaches and collision occurs, kinetic energy is enough for reaction to occur. The bond between Hb and Hc breaks and the bond between Ha and Hb is formed while Hc leaves in an opposite direction. Translational energy is converted into vibrational energy such that oscillation is observed in the exit channel between Ha and Hb. ||
[[File:jhl416_plot1.png |thumb|center|upright=1.5| Figure 4]]
|-
| -1.5 || -2.0 || Unreactive || Comparing with the first momenta combination, AB momentum is lower in this momenta combination, resulting in insufficient kinetic energy to overcome the activation barrier that the trajectory is unreactive. || [[File:jhl416_plot2.png|thumb|center|upright=1.5|Figure 5]]
|-
| -1.5 || -2.5 || Reactive || Similar to the first momenta combination, this momenta combination results in a similar trajectory. The major difference is observed at the entrance channel where oscillation between Hb and Hc before the collision with Ha. This can be explained by the increased magnitude of p1 (which corresponds to BC momentum) from 1.25 to 1.5. This also means an increase in kinetic energy of atoms such that oscillation can be observed. || [[File:jhl416_plot3.png|thumb|center|upright=1.5|Figure 6]]
|-
| -2.5 || -5.0 || Unreactive || As Ha collides with the bonded Hb and Hc, the bond breaks and the the Ha-Hb bond is formed. Yet, due to the high momentum of atoms, the newly-formed bond breaks again that the Hb-Hc bond is formed again while Ha leaves in the opposite direction in a reduced speed. Trajectory is unreactive.|| [[File:jhl416_plot4.png|thumb|center|upright=1.5|Figure 7]]
|-
| -2.5 || -5.2 || Reactive || Comparing with the previous case in figure 7, the magnitude of p2 (which corresponds to AB momentum) is increased from 5.0 to 5.2. Atoms behave similarly with that in figure 7. However, due to the high momentum, after the reformed Hb-Hc bond, it breaks again to form a Ha-Hb bond, leaving Hc to leave in an opposite direction. || [[File:jhl416_plot5.png|thumb|center|upright=1.5|Figure 8]]
|}

==== Assumptions of Transition State Theory ====

The following are the main assumptions of the Transition State Theory<ref name=tst/>
<references> :
== References ==
<ref name=tst> J. W. Moore, R. G. Pearson, Kinetics and Mechanism, '''1981''', 166
</ref>
</references>

1. Atoms behave in classical mechanics that a classical trajectory can be an accurate description of the motion.

2. There is an equilibrium between reactants and transition states.

3. Barrier recrossing does not occur.

As the Transition State Theory assumes no barrier recrossing occurs, its predicted values for reaction rate is likely to be overestimated in comparison with the experimental values. Under the assumption of the Transition State Theory, all systems will continue to form products once the energy barrier is crossed for one time. However, as demonstrated in figure 7, systems do recross the barrier under certain conditions. This indicates that some of the reactive reactions predictions under the Transition State Theory might not be true. Thus, the reaction rate in reality should be lower, leading to lower experimental values in comparison with the predictions under Transition State Theory.

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

==== PES inspection ====

=====Endothermic or exothermic reactions=====

[[File: jhl416_exo_fh2.png |thumb|center|upright=2|Figure 9: Surface plot of the F + H2 reaction ]]

As shown in figure 9, the F + H2 reaction is exothermic as the energy level of the exit channel (product: H-F) is lower than that of the entrance channel (reactant: H2). An exothermic reaction indicates a negative energy change. It means more energy is released in the bond formation (formation of H-F bond) than the energy is taken in during the bond breaking (breaking of H-H bond). Thus, the bond strength of H-F is stronger than that of H-H.

[[File: jhl416_endo_hhf.png |thumb|center|upright=2|Figure 10: Surface plot of the H + HF reaction ]]

As shown in figure 10, the H + HF reaction is endothermic as the energy level of the exit channel (product: H2) is higher than that of the entrance channel (reactant: H-F). An endothermic reaction indicates a positive energy change. It means more energy is taken in during the bond breaking (breaking of H-F bond) than released in the bond formation (formation of H-H bond), once again indicating the stronger bond strength of H-F than H-H.

=====Location of transition state=====

In the exothermic reaction, the activation energy is too small that it is difficult to locate the transition state. According to the Hammond postulate, the transition state resembles the structure of the reactants more than that of the products as the energy gap between the transition state and reactants is smaller than that between the transition state and products. Using this as a guide for the location, the transition state is located at where rHF = 1.8113 Å and rHH = 0.7445 Å.

[[File: Jhl416_tts_location.png |thumb|center|upright=2|Figure 11: Plot showing internuclear distance vs time when rHF = 1.8113 Å and rHH = 0.7445 Å ]]

Figure 11 above shows that when rHF = 1.8113 Å and rHH = 0.7445 Å, all distances between atoms are at constant, indicating that it is the point of transition state.

=====Activation energy for both reactions=====

====== F + H2 reaction ======

[[File: Jhl416_ea_fh2_surface.png |thumb|center|upright=2|Figure 12: MEP surface plot when rHF = 1.8213 Å and rHH = 0.7445 Å ]]

[[File: Jhl416_ea_fh2_energy.png |thumb|center|upright=2|Figure 13: Plot showing energy vs time when rHF = 1.8213 Å and rHH = 0.7445 Å ]]

By carrying out mep calculations from a structure neighbouring the transition state, figures 12 and 13 shows the energy difference between the transition state and reactants, i.e. the activation energy, for the F + H2 reaction. An activation energy of 0.243 kcal/mol.

====== H + HF reaction ======

[[File: Jhl416_ea_hhf_surface.png |thumb|center|upright=2|Figure 14: MEP surface plot when rHF = 1.8113 Å and rHH = 0.7345 Å ]]

[[File: Jhl416_ea_hhf_energy.png |thumb|center|upright=2|Figure 15: Plot showing energy vs time when rHF = 1.8113 Å and rHH = 0.7345 Å ]]

By carrying out mep calculations from a structure neighbouring the transition state, figures 14 and 15 shows the energy difference between the transition state and reactants, i.e. the activation energy, for the F + H2 reaction. An activation energy of 30.075 kcal/mol.

===== Reaction dynamics =====

====== Mechanism of release of reaction energy ======

[[File: Jhl416_rd_surface.png |thumb|center|upright=2|Figure 16: Surface plot when r1 = 2.3, r2 = 0.74, p1 = -2.5 and p2 = -1.5 ]]

[[File: Jhl416_red_m.png |thumb|center|upright=2|Figure 17: Plot showing internuclear momenta vs time when r1 = 2.3, r2 = 0.74, p1 = -2.5 and p2 = -1.5 ]]

With the initial conditions of r1 = 2.3, r2 = 0.74, p1 = -2.5, p2 = -1.5, it is shown in figure 16 that the trajectory is reactive.

By looking at the animation of the reaction, it can be observed that H2 approaches F atom with little oscillation. As they collides, H-H bond breaks and H-F bond is formed. However, the H-F bonds breaks again right afterwards and H-H bond is reformed. The oscillation of H-H bond becomes stronger and collides with F atom again that H-F bond is eventually formed with stronger oscillation while H atoms leaves in an opposite direction.

From the internuclear momenta vs time plot in figure 17, it can be seen that oscillation is initially absent between atoms A and B (i.e. atoms F and H) while present between atoms B and C (i.e. the two H atoms). Yet, after the reaction, strong oscillation is present between atoms A and B while absent between atoms B and C. In comparison, the oscillation in the initial H-H molecule is much smaller than that in the final H-F molecule. This indicates an increase of vibrational energy in the system, which is transferred from translational and potential energy.

In light of the conservation of energy, when potential energy is transferred to vibrational energy that is released in the form of heat, the temperature of the reaction will increase as a result. Thus, the mechanism of release of reaction energy can be confirmed by monitoring the change in temperature using calorimetry.

====== Distribution of energy ======

The H + HF reaction is an endothermic reaction that it has a late transition state, i.e. a late-barrier. According to Polanyi's empirical rules, vibrational energy should be more efficient than translational energy to promote this type of reactions. <ref name=per/><references>

== References ==
<ref name=per> Z. Zhang, Y. Zhou, D. H. Zhang, J. Phys. Chem. Lett. '''2012''', 3, 3416
</ref>
</references>

In the following cases, initial positions are set at the bottom of the entry channel where rHF = 0.93.

[[File: Jhl416_lowt_highv_reactive.png |thumb|center|upright=2|Figure 18: Contour plot when r1 = 2.3, r2 = 0.93, p1 = -1.5 and p2 = 7 ]]

In figure 18,
The initial conditions applied are of high vibrational energy with low translational energy, such that the the trajectory is expected to be reactive according to Polanyi's empirical rules. The experimental result is shown to be aligned with the theory.

[[File: Jhl416_hight_lowv_unreactive.png |thumb|center|upright=2|Figure 19: Contour plot when r1 = 2.3, r2 = 0.93, p1 = -5 and p2 = 0.5 ]]

In figure 19,
The initial conditions applied are of low vibrational energy with high translational energy, such that the the trajectory is expected to be unreactive according to Polanyi's empirical rules. The unreactive trajectory also aligns with the prediction.

However, there are also conditions found where the Polanyi's empirical rules is not obeyed.

[[File: Jhl416_lowt_highv_unreactive.png |thumb|center|upright=2|Figure 20: Contour plot when r1 = 2.3, r2 = 0.93, p1 = -1 and p2 = 8 ]]

In figure 20,
The initial conditions applied are of high vibrational energy with low translational energy, such that the the trajectory is expected to be reactive according to Polanyi's empirical rules. However, the trajectory is shown to be unreactive, disagreeing with the theory.

[[File: Jhl416_hight_lowv_reactive.png |thumb|center|upright=2|Figure 21: Contour plot when r1 = 2.3, r2 = 0.93, p1 = -10 and p2 = 0.5 ]]

In figure 21,
The initial conditions applied are of low vibrational energy with high translational energy, such that the the trajectory is expected to be unreactive according to Polanyi's empirical rules. Yet, a reactive trajectory is shown that it disagrees with the theory once again.

The above cases show that the Polanyi's empirical rules do not hold true at all times.

MRD:jhl416

2018-05-25T16:28:47Z

Jhl416: /* Distribution of energy */

== Molecular Reaction Dynamics: Applications to Triatomic systems ==

=== EXERCISE 1: H + H2 system ===

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

''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.''

With reference to the diagram provided (Gradients at the transition state, as well as in the reactants and product regions.) in the experimental script, q1 and q2 are defined as a different set of coordinates different from the coordinates of r1 and r2. q2 is along the direction of the reaction pathway and q1 is orthogonal to the direction of the reaction pathway.

Both the transition structures and minima have a 0 gradient where the first derivative would both be 0, i.e. ∂Ve(q1)/∂q1 * ∂Ve(q2)/∂q2 = 0

Their second derivatives need to be used in order to distinguish between the two.

'''For transition structures:'''

As the transition state is at the maximum along the reaction pathway, the second derivative of the transition state is < 0 along the q2 direction, i.e. ∂Ve2(q2)/∂q22 < 0

While the second derivative along the q1 direction is > 0 as it is a minimum along any other directions except from the direction of the reaction pathway, i.e. ∂Ve2(q1)/∂q12 > 0.

'''For minima:'''

Minima are at minimum along all directions, including the direction of the reaction pathway. Thus, both of the second derivatives (in q1 and q2 directions) are > 0, i.e. ∂Ve2(q2)/∂q22 > 0 and ∂Ve2(q1)/∂q12 > 0.

==== Trajectories from r1 = r2: locating the transition state ====

''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.''

At the transition state position, there is 0 gradient. Thus, without initial momentum, the kinetic energy of the system should be 0 where there is no oscillation present. Thus, distanceA-B and distanceB-C should be constant in the system.

[[File: Jhl416_ts_position.png |thumb|center|upright=2|Figure 1 - A plot of Internuclear Distance vs Time when r1 = r2 = 0.90777 Å | 600px]]

By testing different rts values, the best estimation of the transition state position is found to be 0.90777 Å. As seen from figure 1 above, when rts = 0.90777 Å, distanceA-B and distanceB-C are seen to be constant with no oscillation, which aligns with the above.

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

The reaction path calculated using different methods of mep and dynamics are presented below in figure 2 and 3.

[[File: jhl416_mep_surface.png |thumb|center|upright=2|Figure 2: mep surface plot at r1 = 0.91777, r2 = 0.90777, p1 = p2 = 0 ]]

From figure 2, it is seen that the trajectory calculated using mep appears to be a straight line. This is due to the fact that mep shows an infinitely slow motion of reaction path. As velocity is set to zero at every time step, the momentum and thus kinetic energy is also zero at each time step. This results in no oscillation motion and hence a straight line of the trajectory.

[[File: jhl416_dynamics_surface.png |thumb|center|upright=2|Figure 3: Dynamics surface plot at r1 = 0.91777, r2 = 0.90777, p1 = p2 = 0 ]]

From figure 3, the trajectory appears to be more wavy when calculated using the dynamics method. This gives a more realistic picture of the motion of atoms where masses of atoms are taken into account. Thus, the oscillation motion of the atoms can then be observed.

==== Reactive and unreactive trajectories ====

{| class="wikitable" border=1
! p1 !! p2 !! Reactive/Unreactive !! Description !! Image
|-
| -1.25 || -2.5 || Reactive || Along the trajectory, Hb and Hc are kept at a constant distance with no oscillation observed in the entrance channel. As Ha approaches and collision occurs, kinetic energy is enough for reaction to occur. The bond between Hb and Hc breaks and the bond between Ha and Hb is formed while Hc leaves in an opposite direction. Translational energy is converted into vibrational energy such that oscillation is observed in the exit channel between Ha and Hb. ||
[[File:jhl416_plot1.png |thumb|center|upright=1.5| Figure 4]]
|-
| -1.5 || -2.0 || Unreactive || Comparing with the first momenta combination, AB momentum is lower in this momenta combination, resulting in insufficient kinetic energy to overcome the activation barrier that the trajectory is unreactive. || [[File:jhl416_plot2.png|thumb|center|upright=1.5|Figure 5]]
|-
| -1.5 || -2.5 || Reactive || Similar to the first momenta combination, this momenta combination results in a similar trajectory. The major difference is observed at the entrance channel where oscillation between Hb and Hc before the collision with Ha. This can be explained by the increased magnitude of p1 (which corresponds to BC momentum) from 1.25 to 1.5. This also means an increase in kinetic energy of atoms such that oscillation can be observed. || [[File:jhl416_plot3.png|thumb|center|upright=1.5|Figure 6]]
|-
| -2.5 || -5.0 || Unreactive || As Ha collides with the bonded Hb and Hc, the bond breaks and the the Ha-Hb bond is formed. Yet, due to the high momentum of atoms, the newly-formed bond breaks again that the Hb-Hc bond is formed again while Ha leaves in the opposite direction in a reduced speed. Trajectory is unreactive.|| [[File:jhl416_plot4.png|thumb|center|upright=1.5|Figure 7]]
|-
| -2.5 || -5.2 || Reactive || Comparing with the previous case in figure 7, the magnitude of p2 (which corresponds to AB momentum) is increased from 5.0 to 5.2. Atoms behave similarly with that in figure 7. However, due to the high momentum, after the reformed Hb-Hc bond, it breaks again to form a Ha-Hb bond, leaving Hc to leave in an opposite direction. || [[File:jhl416_plot5.png|thumb|center|upright=1.5|Figure 8]]
|}

==== Assumptions of Transition State Theory ====

The following are the main assumptions of the Transition State Theory<ref name=tst/>
<references> :
== References ==
<ref name=tst> J. W. Moore, R. G. Pearson, Kinetics and Mechanism, '''1981''', 166
</ref>
</references>

1. Atoms behave in classical mechanics that a classical trajectory can be an accurate description of the motion.

2. There is an equilibrium between reactants and transition states.

3. Barrier recrossing does not occur.

As the Transition State Theory assumes no barrier recrossing occurs, its predicted values for reaction rate is likely to be overestimated in comparison with the experimental values. Under the assumption of the Transition State Theory, all systems will continue to form products once the energy barrier is crossed for one time. However, as demonstrated in figure 7, systems do recross the barrier under certain conditions. This indicates that some of the reactive reactions predictions under the Transition State Theory might not be true. Thus, the reaction rate in reality should be lower, leading to lower experimental values in comparison with the predictions under Transition State Theory.

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

==== PES inspection ====

=====Endothermic or exothermic reactions=====

[[File: jhl416_exo_fh2.png |thumb|center|upright=2|Figure 9: Surface plot of the F + H2 reaction ]]

As shown in figure 9, the F + H2 reaction is exothermic as the energy level of the exit channel (product: H-F) is lower than that of the entrance channel (reactant: H2). An exothermic reaction indicates a negative energy change. It means more energy is released in the bond formation (formation of H-F bond) than the energy is taken in during the bond breaking (breaking of H-H bond). Thus, the bond strength of H-F is stronger than that of H-H.

[[File: jhl416_endo_hhf.png |thumb|center|upright=2|Figure 10: Surface plot of the H + HF reaction ]]

As shown in figure 10, the H + HF reaction is endothermic as the energy level of the exit channel (product: H2) is higher than that of the entrance channel (reactant: H-F). An endothermic reaction indicates a positive energy change. It means more energy is taken in during the bond breaking (breaking of H-F bond) than released in the bond formation (formation of H-H bond), once again indicating the stronger bond strength of H-F than H-H.

=====Location of transition state=====

In the exothermic reaction, the activation energy is too small that it is difficult to locate the transition state. According to the Hammond postulate, the transition state resembles the structure of the reactants more than that of the products as the energy gap between the transition state and reactants is smaller than that between the transition state and products. Using this as a guide for the location, the transition state is located at where rHF = 1.8113 Å and rHH = 0.7445 Å.

[[File: Jhl416_tts_location.png |thumb|center|upright=2|Figure 11: Plot showing internuclear distance vs time when rHF = 1.8113 Å and rHH = 0.7445 Å ]]

Figure 11 above shows that when rHF = 1.8113 Å and rHH = 0.7445 Å, all distances between atoms are at constant, indicating that it is the point of transition state.

=====Activation energy for both reactions=====

====== F + H2 reaction ======

[[File: Jhl416_ea_fh2_surface.png |thumb|center|upright=2|Figure 12: MEP surface plot when rHF = 1.8213 Å and rHH = 0.7445 Å ]]

[[File: Jhl416_ea_fh2_energy.png |thumb|center|upright=2|Figure 13: Plot showing energy vs time when rHF = 1.8213 Å and rHH = 0.7445 Å ]]

By carrying out mep calculations from a structure neighbouring the transition state, figures 12 and 13 shows the energy difference between the transition state and reactants, i.e. the activation energy, for the F + H2 reaction. An activation energy of 0.243 kcal/mol.

====== H + HF reaction ======

[[File: Jhl416_ea_hhf_surface.png |thumb|center|upright=2|Figure 14: MEP surface plot when rHF = 1.8113 Å and rHH = 0.7345 Å ]]

[[File: Jhl416_ea_hhf_energy.png |thumb|center|upright=2|Figure 15: Plot showing energy vs time when rHF = 1.8113 Å and rHH = 0.7345 Å ]]

By carrying out mep calculations from a structure neighbouring the transition state, figures 14 and 15 shows the energy difference between the transition state and reactants, i.e. the activation energy, for the F + H2 reaction. An activation energy of 30.075 kcal/mol.

===== Reaction dynamics =====

====== Mechanism of release of reaction energy ======

[[File: Jhl416_rd_surface.png |thumb|center|upright=2|Figure 16: Surface plot when r1 = 2.3, r2 = 0.74, p1 = -2.5 and p2 = -1.5 ]]

[[File: Jhl416_red_m.png |thumb|center|upright=2|Figure 17: Plot showing internuclear momenta vs time when r1 = 2.3, r2 = 0.74, p1 = -2.5 and p2 = -1.5 ]]

With the initial conditions of r1 = 2.3, r2 = 0.74, p1 = -2.5, p2 = -1.5, it is shown in figure 16 that the trajectory is reactive.

By looking at the animation of the reaction, it can be observed that H2 approaches F atom with little oscillation. As they collides, H-H bond breaks and H-F bond is formed. However, the H-F bonds breaks again right afterwards and H-H bond is reformed. The oscillation of H-H bond becomes stronger and collides with F atom again that H-F bond is eventually formed with stronger oscillation while H atoms leaves in an opposite direction.

From the internuclear momenta vs time plot in figure 17, it can be seen that oscillation is initially absent between atoms A and B (i.e. atoms F and H) while present between atoms B and C (i.e. the two H atoms). Yet, after the reaction, strong oscillation is present between atoms A and B while absent between atoms B and C. In comparison, the oscillation in the initial H-H molecule is much smaller than that in the final H-F molecule. This indicates an increase of vibrational energy in the system, which is transferred from translational and potential energy.

In light of the conservation of energy, when potential energy is transferred to vibrational energy that is released in the form of heat, the temperature of the reaction will increase as a result. Thus, the mechanism of release of reaction energy can be confirmed by monitoring the change in temperature using calorimetry.

====== Distribution of energy ======

The H + HF reaction is an endothermic reaction that it has a late-barrier. According to Polanyi's empirical rules, vibrational energy should be more efficient than translational energy to promote this type of reactions. <ref name=per/><references>

== References ==
<ref name=per> Z. Zhang, Y. Zhou, D. H. Zhang, J. Phys. Chem. Lett. '''2012''', 3, 3416
</ref>
</references>

In the following cases, initial positions are set at the bottom of the entry channel where rHF = 0.93.

[[File: Jhl416_lowt_highv_reactive.png |thumb|center|upright=2|Figure 18: Contour plot when r1 = 2.3, r2 = 0.93, p1 = -1.5 and p2 = 7 ]]

In figure 18,
The initial conditions applied are of high vibrational energy with low translational energy, such that the the trajectory is expected to be reactive according to Polanyi's empirical rules. The experimental result is shown to be aligned with the theory.

[[File: Jhl416_hight_lowv_unreactive.png |thumb|center|upright=2|Figure 19: Contour plot when r1 = 2.3, r2 = 0.93, p1 = -5 and p2 = 0.5 ]]

In figure 19,
The initial conditions applied are of low vibrational energy with high translational energy, such that the the trajectory is expected to be unreactive according to Polanyi's empirical rules. The unreactive trajectory also aligns with the prediction.

However, there are also conditions found where the Polanyi's empirical rules is not obeyed.

[[File: Jhl416_lowt_highv_unreactive.png |thumb|center|upright=2|Figure 20: Contour plot when r1 = 2.3, r2 = 0.93, p1 = -1 and p2 = 8 ]]

In figure 20,
The initial conditions applied are of high vibrational energy with low translational energy, such that the the trajectory is expected to be reactive according to Polanyi's empirical rules. However, the trajectory is shown to be unreactive, disagreeing with the theory.

[[File: Jhl416_hight_lowv_reactive.png |thumb|center|upright=2|Figure 21: Contour plot when r1 = 2.3, r2 = 0.93, p1 = -10 and p2 = 0.5 ]]

In figure 21,
The initial conditions applied are of low vibrational energy with high translational energy, such that the the trajectory is expected to be unreactive according to Polanyi's empirical rules. Yet, a reactive trajectory is shown that it disagrees with the theory once again.

The above cases show that the Polanyi's empirical rules do not hold true at all times.

MRD:jhl416

2018-05-25T16:25:01Z

Jhl416: /* Distribution of energy */

== Molecular Reaction Dynamics: Applications to Triatomic systems ==

=== EXERCISE 1: H + H2 system ===

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

''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.''

With reference to the diagram provided (Gradients at the transition state, as well as in the reactants and product regions.) in the experimental script, q1 and q2 are defined as a different set of coordinates different from the coordinates of r1 and r2. q2 is along the direction of the reaction pathway and q1 is orthogonal to the direction of the reaction pathway.

Both the transition structures and minima have a 0 gradient where the first derivative would both be 0, i.e. ∂Ve(q1)/∂q1 * ∂Ve(q2)/∂q2 = 0

Their second derivatives need to be used in order to distinguish between the two.

'''For transition structures:'''

As the transition state is at the maximum along the reaction pathway, the second derivative of the transition state is < 0 along the q2 direction, i.e. ∂Ve2(q2)/∂q22 < 0

While the second derivative along the q1 direction is > 0 as it is a minimum along any other directions except from the direction of the reaction pathway, i.e. ∂Ve2(q1)/∂q12 > 0.

'''For minima:'''

Minima are at minimum along all directions, including the direction of the reaction pathway. Thus, both of the second derivatives (in q1 and q2 directions) are > 0, i.e. ∂Ve2(q2)/∂q22 > 0 and ∂Ve2(q1)/∂q12 > 0.

==== Trajectories from r1 = r2: locating the transition state ====

''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.''

At the transition state position, there is 0 gradient. Thus, without initial momentum, the kinetic energy of the system should be 0 where there is no oscillation present. Thus, distanceA-B and distanceB-C should be constant in the system.

[[File: Jhl416_ts_position.png |thumb|center|upright=2|Figure 1 - A plot of Internuclear Distance vs Time when r1 = r2 = 0.90777 Å | 600px]]

By testing different rts values, the best estimation of the transition state position is found to be 0.90777 Å. As seen from figure 1 above, when rts = 0.90777 Å, distanceA-B and distanceB-C are seen to be constant with no oscillation, which aligns with the above.

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

The reaction path calculated using different methods of mep and dynamics are presented below in figure 2 and 3.

[[File: jhl416_mep_surface.png |thumb|center|upright=2|Figure 2: mep surface plot at r1 = 0.91777, r2 = 0.90777, p1 = p2 = 0 ]]

From figure 2, it is seen that the trajectory calculated using mep appears to be a straight line. This is due to the fact that mep shows an infinitely slow motion of reaction path. As velocity is set to zero at every time step, the momentum and thus kinetic energy is also zero at each time step. This results in no oscillation motion and hence a straight line of the trajectory.

[[File: jhl416_dynamics_surface.png |thumb|center|upright=2|Figure 3: Dynamics surface plot at r1 = 0.91777, r2 = 0.90777, p1 = p2 = 0 ]]

From figure 3, the trajectory appears to be more wavy when calculated using the dynamics method. This gives a more realistic picture of the motion of atoms where masses of atoms are taken into account. Thus, the oscillation motion of the atoms can then be observed.

==== Reactive and unreactive trajectories ====

{| class="wikitable" border=1
! p1 !! p2 !! Reactive/Unreactive !! Description !! Image
|-
| -1.25 || -2.5 || Reactive || Along the trajectory, Hb and Hc are kept at a constant distance with no oscillation observed in the entrance channel. As Ha approaches and collision occurs, kinetic energy is enough for reaction to occur. The bond between Hb and Hc breaks and the bond between Ha and Hb is formed while Hc leaves in an opposite direction. Translational energy is converted into vibrational energy such that oscillation is observed in the exit channel between Ha and Hb. ||
[[File:jhl416_plot1.png |thumb|center|upright=1.5| Figure 4]]
|-
| -1.5 || -2.0 || Unreactive || Comparing with the first momenta combination, AB momentum is lower in this momenta combination, resulting in insufficient kinetic energy to overcome the activation barrier that the trajectory is unreactive. || [[File:jhl416_plot2.png|thumb|center|upright=1.5|Figure 5]]
|-
| -1.5 || -2.5 || Reactive || Similar to the first momenta combination, this momenta combination results in a similar trajectory. The major difference is observed at the entrance channel where oscillation between Hb and Hc before the collision with Ha. This can be explained by the increased magnitude of p1 (which corresponds to BC momentum) from 1.25 to 1.5. This also means an increase in kinetic energy of atoms such that oscillation can be observed. || [[File:jhl416_plot3.png|thumb|center|upright=1.5|Figure 6]]
|-
| -2.5 || -5.0 || Unreactive || As Ha collides with the bonded Hb and Hc, the bond breaks and the the Ha-Hb bond is formed. Yet, due to the high momentum of atoms, the newly-formed bond breaks again that the Hb-Hc bond is formed again while Ha leaves in the opposite direction in a reduced speed. Trajectory is unreactive.|| [[File:jhl416_plot4.png|thumb|center|upright=1.5|Figure 7]]
|-
| -2.5 || -5.2 || Reactive || Comparing with the previous case in figure 7, the magnitude of p2 (which corresponds to AB momentum) is increased from 5.0 to 5.2. Atoms behave similarly with that in figure 7. However, due to the high momentum, after the reformed Hb-Hc bond, it breaks again to form a Ha-Hb bond, leaving Hc to leave in an opposite direction. || [[File:jhl416_plot5.png|thumb|center|upright=1.5|Figure 8]]
|}

==== Assumptions of Transition State Theory ====

The following are the main assumptions of the Transition State Theory<ref name=tst/>
<references> :
== References ==
<ref name=tst> J. W. Moore, R. G. Pearson, Kinetics and Mechanism, '''1981''', 166
</ref>
</references>

1. Atoms behave in classical mechanics that a classical trajectory can be an accurate description of the motion.

2. There is an equilibrium between reactants and transition states.

3. Barrier recrossing does not occur.

As the Transition State Theory assumes no barrier recrossing occurs, its predicted values for reaction rate is likely to be overestimated in comparison with the experimental values. Under the assumption of the Transition State Theory, all systems will continue to form products once the energy barrier is crossed for one time. However, as demonstrated in figure 7, systems do recross the barrier under certain conditions. This indicates that some of the reactive reactions predictions under the Transition State Theory might not be true. Thus, the reaction rate in reality should be lower, leading to lower experimental values in comparison with the predictions under Transition State Theory.

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

==== PES inspection ====

=====Endothermic or exothermic reactions=====

[[File: jhl416_exo_fh2.png |thumb|center|upright=2|Figure 9: Surface plot of the F + H2 reaction ]]

As shown in figure 9, the F + H2 reaction is exothermic as the energy level of the exit channel (product: H-F) is lower than that of the entrance channel (reactant: H2). An exothermic reaction indicates a negative energy change. It means more energy is released in the bond formation (formation of H-F bond) than the energy is taken in during the bond breaking (breaking of H-H bond). Thus, the bond strength of H-F is stronger than that of H-H.

[[File: jhl416_endo_hhf.png |thumb|center|upright=2|Figure 10: Surface plot of the H + HF reaction ]]

As shown in figure 10, the H + HF reaction is endothermic as the energy level of the exit channel (product: H2) is higher than that of the entrance channel (reactant: H-F). An endothermic reaction indicates a positive energy change. It means more energy is taken in during the bond breaking (breaking of H-F bond) than released in the bond formation (formation of H-H bond), once again indicating the stronger bond strength of H-F than H-H.

=====Location of transition state=====

In the exothermic reaction, the activation energy is too small that it is difficult to locate the transition state. According to the Hammond postulate, the transition state resembles the structure of the reactants more than that of the products as the energy gap between the transition state and reactants is smaller than that between the transition state and products. Using this as a guide for the location, the transition state is located at where rHF = 1.8113 Å and rHH = 0.7445 Å.

[[File: Jhl416_tts_location.png |thumb|center|upright=2|Figure 11: Plot showing internuclear distance vs time when rHF = 1.8113 Å and rHH = 0.7445 Å ]]

Figure 11 above shows that when rHF = 1.8113 Å and rHH = 0.7445 Å, all distances between atoms are at constant, indicating that it is the point of transition state.

=====Activation energy for both reactions=====

====== F + H2 reaction ======

[[File: Jhl416_ea_fh2_surface.png |thumb|center|upright=2|Figure 12: MEP surface plot when rHF = 1.8213 Å and rHH = 0.7445 Å ]]

[[File: Jhl416_ea_fh2_energy.png |thumb|center|upright=2|Figure 13: Plot showing energy vs time when rHF = 1.8213 Å and rHH = 0.7445 Å ]]

By carrying out mep calculations from a structure neighbouring the transition state, figures 12 and 13 shows the energy difference between the transition state and reactants, i.e. the activation energy, for the F + H2 reaction. An activation energy of 0.243 kcal/mol.

====== H + HF reaction ======

[[File: Jhl416_ea_hhf_surface.png |thumb|center|upright=2|Figure 14: MEP surface plot when rHF = 1.8113 Å and rHH = 0.7345 Å ]]

[[File: Jhl416_ea_hhf_energy.png |thumb|center|upright=2|Figure 15: Plot showing energy vs time when rHF = 1.8113 Å and rHH = 0.7345 Å ]]

By carrying out mep calculations from a structure neighbouring the transition state, figures 14 and 15 shows the energy difference between the transition state and reactants, i.e. the activation energy, for the F + H2 reaction. An activation energy of 30.075 kcal/mol.

===== Reaction dynamics =====

====== Mechanism of release of reaction energy ======

[[File: Jhl416_rd_surface.png |thumb|center|upright=2|Figure 16: Surface plot when r1 = 2.3, r2 = 0.74, p1 = -2.5 and p2 = -1.5 ]]

[[File: Jhl416_red_m.png |thumb|center|upright=2|Figure 17: Plot showing internuclear momenta vs time when r1 = 2.3, r2 = 0.74, p1 = -2.5 and p2 = -1.5 ]]

With the initial conditions of r1 = 2.3, r2 = 0.74, p1 = -2.5, p2 = -1.5, it is shown in figure 16 that the trajectory is reactive.

By looking at the animation of the reaction, it can be observed that H2 approaches F atom with little oscillation. As they collides, H-H bond breaks and H-F bond is formed. However, the H-F bonds breaks again right afterwards and H-H bond is reformed. The oscillation of H-H bond becomes stronger and collides with F atom again that H-F bond is eventually formed with stronger oscillation while H atoms leaves in an opposite direction.

From the internuclear momenta vs time plot in figure 17, it can be seen that oscillation is initially absent between atoms A and B (i.e. atoms F and H) while present between atoms B and C (i.e. the two H atoms). Yet, after the reaction, strong oscillation is present between atoms A and B while absent between atoms B and C. In comparison, the oscillation in the initial H-H molecule is much smaller than that in the final H-F molecule. This indicates an increase of vibrational energy in the system, which is transferred from translational and potential energy.

In light of the conservation of energy, when potential energy is transferred to vibrational energy that is released in the form of heat, the temperature of the reaction will increase as a result. Thus, the mechanism of release of reaction energy can be confirmed by monitoring the change in temperature using calorimetry.

====== Distribution of energy ======

The H + HF reaction is an endothermic reaction that it has a late-barrier. According to Polanyi's empirical rules, vibrational energy should be more efficient than translational energy to promote this type of reactions. <ref name=per/><references>

== References ==
<ref name=per> Z. Zhang, Y. Zhou, D. H. Zhang, J. Phys. Chem. Lett. '''2012''', 3, 3416
</ref>
</references>

In the following cases, initial positions are set at the bottom of the entry channel where rHF = 0.93.

In figure,
The initial conditions applied are of high vibrational energy with low translational energy, such that the the trajectory is expected to be reactive according to Polanyi's empirical rules. The experimental result is shown to be aligned with the theory.

In figure,
The initial conditions applied are of low vibrational energy with high translational energy, such that the the trajectory is expected to be unreactive according to Polanyi's empirical rules. The unreactive trajectory also aligns with the prediction.

However, there are also conditions found where the Polanyi's empirical rules is not obeyed.

In figure,
The initial conditions applied are of high vibrational energy with low translational energy, such that the the trajectory is expected to be reactive according to Polanyi's empirical rules. However, the trajectory is shown to be unreactive, disagreeing with the theory.

In figure,
The initial conditions applied are of low vibrational energy with high translational energy, such that the the trajectory is expected to be unreactive according to Polanyi's empirical rules. Yet, a reactive trajectory is shown that it disagrees with the theory once again.

The above cases show that the Polanyi's empirical rules do not hold true at all times.

MRD:jhl416

2018-05-25T16:23:31Z

MRD:jhl416

2018-05-25T15:20:24Z

Jhl416: /* Distribution of energy */

2018-05-25T15:18:02Z

Jhl416:

MRD:jhl416

2018-05-25T15:15:19Z

Jhl416: /* Mechanism of release of reaction energy */

== Molecular Reaction Dynamics: Applications to Triatomic systems ==

=== EXERCISE 1: H + H2 system ===

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

''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.''

With reference to the diagram provided (Gradients at the transition state, as well as in the reactants and product regions.) in the experimental script, q1 and q2 are defined as a different set of coordinates different from the coordinates of r1 and r2. q2 is along the direction of the reaction pathway and q1 is orthogonal to the direction of the reaction pathway.

Both the transition structures and minima have a 0 gradient where the first derivative would both be 0, i.e. ∂Ve(q1)/∂q1 * ∂Ve(q2)/∂q2 = 0

Their second derivatives need to be used in order to distinguish between the two.

'''For transition structures:'''

As the transition state is at the maximum along the reaction pathway, the second derivative of the transition state is < 0 along the q2 direction, i.e. ∂Ve2(q2)/∂q22 < 0

While the second derivative along the q1 direction is > 0 as it is a minimum along any other directions except from the direction of the reaction pathway, i.e. ∂Ve2(q1)/∂q12 > 0.

'''For minima:'''

Minima are at minimum along all directions, including the direction of the reaction pathway. Thus, both of the second derivatives (in q1 and q2 directions) are > 0, i.e. ∂Ve2(q2)/∂q22 > 0 and ∂Ve2(q1)/∂q12 > 0.

==== Trajectories from r1 = r2: locating the transition state ====

''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.''

At the transition state position, there is 0 gradient. Thus, without initial momentum, the kinetic energy of the system should be 0 where there is no oscillation present. Thus, distanceA-B and distanceB-C should be constant in the system.

[[File: Jhl416_ts_position.png |thumb|center|upright=2|Figure 1 - A plot of Internuclear Distance vs Time when r1 = r2 = 0.90777 Å | 600px]]

By testing different rts values, the best estimation of the transition state position is found to be 0.90777 Å. As seen from figure 1 above, when rts = 0.90777 Å, distanceA-B and distanceB-C are seen to be constant with no oscillation, which aligns with the above.

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

The reaction path calculated using different methods of mep and dynamics are presented below in figure 2 and 3.

[[File: jhl416_mep_surface.png |thumb|center|upright=2|Figure 2: mep surface plot at r1 = 0.91777, r2 = 0.90777, p1 = p2 = 0 ]]

From figure 2, it is seen that the trajectory calculated using mep appears to be a straight line. This is due to the fact that mep shows an infinitely slow motion of reaction path. As velocity is set to zero at every time step, the momentum and thus kinetic energy is also zero at each time step. This results in no oscillation motion and hence a straight line of the trajectory.

[[File: jhl416_dynamics_surface.png |thumb|center|upright=2|Figure 3: Dynamics surface plot at r1 = 0.91777, r2 = 0.90777, p1 = p2 = 0 ]]

From figure 3, the trajectory appears to be more wavy when calculated using the dynamics method. This gives a more realistic picture of the motion of atoms where masses of atoms are taken into account. Thus, the oscillation motion of the atoms can then be observed.

==== Reactive and unreactive trajectories ====

{| class="wikitable" border=1
! p1 !! p2 !! Reactive/Unreactive !! Description !! Image
|-
| -1.25 || -2.5 || Reactive || Along the trajectory, Hb and Hc are kept at a constant distance with no oscillation observed in the entrance channel. As Ha approaches and collision occurs, kinetic energy is enough for reaction to occur. The bond between Hb and Hc breaks and the bond between Ha and Hb is formed while Hc leaves in an opposite direction. Translational energy is converted into vibrational energy such that oscillation is observed in the exit channel between Ha and Hb. ||
[[File:jhl416_plot1.png |thumb|center|upright=1.5| Figure 4]]
|-
| -1.5 || -2.0 || Unreactive || Comparing with the first momenta combination, AB momentum is lower in this momenta combination, resulting in insufficient kinetic energy to overcome the activation barrier that the trajectory is unreactive. || [[File:jhl416_plot2.png|thumb|center|upright=1.5|Figure 5]]
|-
| -1.5 || -2.5 || Reactive || Similar to the first momenta combination, this momenta combination results in a similar trajectory. The major difference is observed at the entrance channel where oscillation between Hb and Hc before the collision with Ha. This can be explained by the increased magnitude of p1 (which corresponds to BC momentum) from 1.25 to 1.5. This also means an increase in kinetic energy of atoms such that oscillation can be observed. || [[File:jhl416_plot3.png|thumb|center|upright=1.5|Figure 6]]
|-
| -2.5 || -5.0 || Unreactive || As Ha collides with the bonded Hb and Hc, the bond breaks and the the Ha-Hb bond is formed. Yet, due to the high momentum of atoms, the newly-formed bond breaks again that the Hb-Hc bond is formed again while Ha leaves in the opposite direction in a reduced speed. Trajectory is unreactive.|| [[File:jhl416_plot4.png|thumb|center|upright=1.5|Figure 7]]
|-
| -2.5 || -5.2 || Reactive || Comparing with the previous case in figure 7, the magnitude of p2 (which corresponds to AB momentum) is increased from 5.0 to 5.2. Atoms behave similarly with that in figure 7. However, due to the high momentum, after the reformed Hb-Hc bond, it breaks again to form a Ha-Hb bond, leaving Hc to leave in an opposite direction. || [[File:jhl416_plot5.png|thumb|center|upright=1.5|Figure 8]]
|}

==== Assumptions of Transition State Theory ====

The following are the main assumptions of the Transition State Theory<ref name=tst/>
<references> :
== References ==
<ref name=tst> J. W. Moore, R. G. Pearson, Kinetics and Mechanism, '''1981''', 166
</ref>
</references>

1. Atoms behave in classical mechanics that a classical trajectory can be an accurate description of the motion.

2. There is an equilibrium between reactants and transition states.

3. Barrier recrossing does not occur.

As the Transition State Theory assumes no barrier recrossing occurs, its predicted values for reaction rate is likely to be overestimated in comparison with the experimental values. Under the assumption of the Transition State Theory, all systems will continue to form products once the energy barrier is crossed for one time. However, as demonstrated in figure 7, systems do recross the barrier under certain conditions. This indicates that some of the reactive reactions predictions under the Transition State Theory might not be true. Thus, the reaction rate in reality should be lower, leading to lower experimental values in comparison with the predictions under Transition State Theory.

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

==== PES inspection ====

=====Endothermic or exothermic reactions=====

[[File: jhl416_exo_fh2.png |thumb|center|upright=2|Figure 9: Surface plot of the F + H2 reaction ]]

As shown in figure 9, the F + H2 reaction is exothermic as the energy level of the exit channel (product: H-F) is lower than that of the entrance channel (reactant: H2). An exothermic reaction indicates a negative energy change. It means more energy is released in the bond formation (formation of H-F bond) than the energy is taken in during the bond breaking (breaking of H-H bond). Thus, the bond strength of H-F is stronger than that of H-H.

[[File: jhl416_endo_hhf.png |thumb|center|upright=2|Figure 10: Surface plot of the H + HF reaction ]]

As shown in figure 10, the H + HF reaction is endothermic as the energy level of the exit channel (product: H2) is higher than that of the entrance channel (reactant: H-F). An endothermic reaction indicates a positive energy change. It means more energy is taken in during the bond breaking (breaking of H-F bond) than released in the bond formation (formation of H-H bond), once again indicating the stronger bond strength of H-F than H-H.

=====Location of transition state=====

In the exothermic reaction, the activation energy is too small that it is difficult to locate the transition state. According to the Hammond postulate, the transition state resembles the structure of the reactants more than that of the products as the energy gap between the transition state and reactants is smaller than that between the transition state and products. Using this as a guide for the location, the transition state is located at where rHF = 1.8113 Å and rHH = 0.7445 Å.

[[File: Jhl416_tts_location.png |thumb|center|upright=2|Figure 11: Plot showing internuclear distance vs time when rHF = 1.8113 Å and rHH = 0.7445 Å ]]

Figure 11 above shows that when rHF = 1.8113 Å and rHH = 0.7445 Å, all distances between atoms are at constant, indicating that it is the point of transition state.

=====Activation energy for both reactions=====

====== F + H2 reaction ======

[[File: Jhl416_ea_fh2_surface.png |thumb|center|upright=2|Figure 12: MEP surface plot when rHF = 1.8213 Å and rHH = 0.7445 Å ]]

[[File: Jhl416_ea_fh2_energy.png |thumb|center|upright=2|Figure 13: Plot showing energy vs time when rHF = 1.8213 Å and rHH = 0.7445 Å ]]

By carrying out mep calculations from a structure neighbouring the transition state, figures 12 and 13 shows the energy difference between the transition state and reactants, i.e. the activation energy, for the F + H2 reaction. An activation energy of 0.243 kcal/mol.

====== H + HF reaction ======

[[File: Jhl416_ea_hhf_surface.png |thumb|center|upright=2|Figure 14: MEP surface plot when rHF = 1.8113 Å and rHH = 0.7345 Å ]]

[[File: Jhl416_ea_hhf_energy.png |thumb|center|upright=2|Figure 15: Plot showing energy vs time when rHF = 1.8113 Å and rHH = 0.7345 Å ]]

By carrying out mep calculations from a structure neighbouring the transition state, figures 14 and 15 shows the energy difference between the transition state and reactants, i.e. the activation energy, for the F + H2 reaction. An activation energy of 30.075 kcal/mol.

===== Mechanism of release of reaction energy =====

[[File: Jhl416_rd_surface.png |thumb|center|upright=2|Figure 16: Surface plot when r1 = 2.3, r2 = 0.74, p1 = -2.5 and p2 = -1.5 ]]

[[File: Jhl416_rd_momentum.png |thumb|center|upright=2|Figure 17: Plot showing internuclear momenta vs time when r1 = 2.3, r2 = 0.74, p1 = -2.5 and p2 = -1.5 ]]

With the initial conditions of r1 = 2.3, r2 = 0.74, p1 = -2.5, p2 = -1.5, it is shown in figure 16 that the trajectory is reactive.

By looking at the animation of the reaction, it can be observed that H2 approaches F atom with little oscillation. As they collides, H-H bond breaks and H-F bond is formed. However, the H-F bonds breaks again right afterwards and H-H bond is reformed. The oscillation of H-H bond becomes stronger and collides with F atom again that H-F bond is eventually formed with stronger oscillation while H atoms leaves in an opposite direction.

From the internuclear momenta vs time plot in figure 17, it can be seen that oscillation is initially absent between atoms A and B (i.e. atoms F and H) while present between atoms B and C (i.e. the two H atoms). Yet, after the reaction, strong oscillation is present between atoms A and B while absent between atoms B and C. In comparison, the oscillation in the initial H-H molecule is much smaller than that in the final H-F molecule. This indicates an increase of vibrational energy in the system, which is transferred from translational and potential energy.

In light of the conservation of energy, when potential energy is transferred to vibrational energy that is released in the form of heat, the temperature of the reaction will increase as a result. Thus, the mechanism of release of reaction energy can be confirmed by monitoring the change in temperature using calorimetry.

MRD:jhl416

2018-05-25T15:13:03Z

Jhl416: /* Mechanism of release of reaction energy */

== Molecular Reaction Dynamics: Applications to Triatomic systems ==

=== EXERCISE 1: H + H2 system ===

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

''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.''

With reference to the diagram provided (Gradients at the transition state, as well as in the reactants and product regions.) in the experimental script, q1 and q2 are defined as a different set of coordinates different from the coordinates of r1 and r2. q2 is along the direction of the reaction pathway and q1 is orthogonal to the direction of the reaction pathway.

Both the transition structures and minima have a 0 gradient where the first derivative would both be 0, i.e. ∂Ve(q1)/∂q1 * ∂Ve(q2)/∂q2 = 0

Their second derivatives need to be used in order to distinguish between the two.

'''For transition structures:'''

As the transition state is at the maximum along the reaction pathway, the second derivative of the transition state is < 0 along the q2 direction, i.e. ∂Ve2(q2)/∂q22 < 0

While the second derivative along the q1 direction is > 0 as it is a minimum along any other directions except from the direction of the reaction pathway, i.e. ∂Ve2(q1)/∂q12 > 0.

'''For minima:'''

Minima are at minimum along all directions, including the direction of the reaction pathway. Thus, both of the second derivatives (in q1 and q2 directions) are > 0, i.e. ∂Ve2(q2)/∂q22 > 0 and ∂Ve2(q1)/∂q12 > 0.

==== Trajectories from r1 = r2: locating the transition state ====

''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.''

At the transition state position, there is 0 gradient. Thus, without initial momentum, the kinetic energy of the system should be 0 where there is no oscillation present. Thus, distanceA-B and distanceB-C should be constant in the system.

[[File: Jhl416_ts_position.png |thumb|center|upright=2|Figure 1 - A plot of Internuclear Distance vs Time when r1 = r2 = 0.90777 Å | 600px]]

By testing different rts values, the best estimation of the transition state position is found to be 0.90777 Å. As seen from figure 1 above, when rts = 0.90777 Å, distanceA-B and distanceB-C are seen to be constant with no oscillation, which aligns with the above.

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

The reaction path calculated using different methods of mep and dynamics are presented below in figure 2 and 3.

[[File: jhl416_mep_surface.png |thumb|center|upright=2|Figure 2: mep surface plot at r1 = 0.91777, r2 = 0.90777, p1 = p2 = 0 ]]

From figure 2, it is seen that the trajectory calculated using mep appears to be a straight line. This is due to the fact that mep shows an infinitely slow motion of reaction path. As velocity is set to zero at every time step, the momentum and thus kinetic energy is also zero at each time step. This results in no oscillation motion and hence a straight line of the trajectory.

[[File: jhl416_dynamics_surface.png |thumb|center|upright=2|Figure 3: Dynamics surface plot at r1 = 0.91777, r2 = 0.90777, p1 = p2 = 0 ]]

From figure 3, the trajectory appears to be more wavy when calculated using the dynamics method. This gives a more realistic picture of the motion of atoms where masses of atoms are taken into account. Thus, the oscillation motion of the atoms can then be observed.

==== Reactive and unreactive trajectories ====

{| class="wikitable" border=1
! p1 !! p2 !! Reactive/Unreactive !! Description !! Image
|-
| -1.25 || -2.5 || Reactive || Along the trajectory, Hb and Hc are kept at a constant distance with no oscillation observed in the entrance channel. As Ha approaches and collision occurs, kinetic energy is enough for reaction to occur. The bond between Hb and Hc breaks and the bond between Ha and Hb is formed while Hc leaves in an opposite direction. Translational energy is converted into vibrational energy such that oscillation is observed in the exit channel between Ha and Hb. ||
[[File:jhl416_plot1.png |thumb|center|upright=1.5| Figure 4]]
|-
| -1.5 || -2.0 || Unreactive || Comparing with the first momenta combination, AB momentum is lower in this momenta combination, resulting in insufficient kinetic energy to overcome the activation barrier that the trajectory is unreactive. || [[File:jhl416_plot2.png|thumb|center|upright=1.5|Figure 5]]
|-
| -1.5 || -2.5 || Reactive || Similar to the first momenta combination, this momenta combination results in a similar trajectory. The major difference is observed at the entrance channel where oscillation between Hb and Hc before the collision with Ha. This can be explained by the increased magnitude of p1 (which corresponds to BC momentum) from 1.25 to 1.5. This also means an increase in kinetic energy of atoms such that oscillation can be observed. || [[File:jhl416_plot3.png|thumb|center|upright=1.5|Figure 6]]
|-
| -2.5 || -5.0 || Unreactive || As Ha collides with the bonded Hb and Hc, the bond breaks and the the Ha-Hb bond is formed. Yet, due to the high momentum of atoms, the newly-formed bond breaks again that the Hb-Hc bond is formed again while Ha leaves in the opposite direction in a reduced speed. Trajectory is unreactive.|| [[File:jhl416_plot4.png|thumb|center|upright=1.5|Figure 7]]
|-
| -2.5 || -5.2 || Reactive || Comparing with the previous case in figure 7, the magnitude of p2 (which corresponds to AB momentum) is increased from 5.0 to 5.2. Atoms behave similarly with that in figure 7. However, due to the high momentum, after the reformed Hb-Hc bond, it breaks again to form a Ha-Hb bond, leaving Hc to leave in an opposite direction. || [[File:jhl416_plot5.png|thumb|center|upright=1.5|Figure 8]]
|}

==== Assumptions of Transition State Theory ====

The following are the main assumptions of the Transition State Theory<ref name=tst/>
<references> :
== References ==
<ref name=tst> J. W. Moore, R. G. Pearson, Kinetics and Mechanism, '''1981''', 166
</ref>
</references>

1. Atoms behave in classical mechanics that a classical trajectory can be an accurate description of the motion.

2. There is an equilibrium between reactants and transition states.

3. Barrier recrossing does not occur.

As the Transition State Theory assumes no barrier recrossing occurs, its predicted values for reaction rate is likely to be overestimated in comparison with the experimental values. Under the assumption of the Transition State Theory, all systems will continue to form products once the energy barrier is crossed for one time. However, as demonstrated in figure 7, systems do recross the barrier under certain conditions. This indicates that some of the reactive reactions predictions under the Transition State Theory might not be true. Thus, the reaction rate in reality should be lower, leading to lower experimental values in comparison with the predictions under Transition State Theory.

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

==== PES inspection ====

=====Endothermic or exothermic reactions=====

[[File: jhl416_exo_fh2.png |thumb|center|upright=2|Figure 9: Surface plot of the F + H2 reaction ]]

As shown in figure 9, the F + H2 reaction is exothermic as the energy level of the exit channel (product: H-F) is lower than that of the entrance channel (reactant: H2). An exothermic reaction indicates a negative energy change. It means more energy is released in the bond formation (formation of H-F bond) than the energy is taken in during the bond breaking (breaking of H-H bond). Thus, the bond strength of H-F is stronger than that of H-H.

[[File: jhl416_endo_hhf.png |thumb|center|upright=2|Figure 10: Surface plot of the H + HF reaction ]]

As shown in figure 10, the H + HF reaction is endothermic as the energy level of the exit channel (product: H2) is higher than that of the entrance channel (reactant: H-F). An endothermic reaction indicates a positive energy change. It means more energy is taken in during the bond breaking (breaking of H-F bond) than released in the bond formation (formation of H-H bond), once again indicating the stronger bond strength of H-F than H-H.

=====Location of transition state=====

In the exothermic reaction, the activation energy is too small that it is difficult to locate the transition state. According to the Hammond postulate, the transition state resembles the structure of the reactants more than that of the products as the energy gap between the transition state and reactants is smaller than that between the transition state and products. Using this as a guide for the location, the transition state is located at where rHF = 1.8113 Å and rHH = 0.7445 Å.

[[File: Jhl416_tts_location.png |thumb|center|upright=2|Figure 11: Plot showing internuclear distance vs time when rHF = 1.8113 Å and rHH = 0.7445 Å ]]

Figure 11 above shows that when rHF = 1.8113 Å and rHH = 0.7445 Å, all distances between atoms are at constant, indicating that it is the point of transition state.

=====Activation energy for both reactions=====

====== F + H2 reaction ======

[[File: Jhl416_ea_fh2_surface.png |thumb|center|upright=2|Figure 12: MEP surface plot when rHF = 1.8213 Å and rHH = 0.7445 Å ]]

[[File: Jhl416_ea_fh2_energy.png |thumb|center|upright=2|Figure 13: Plot showing energy vs time when rHF = 1.8213 Å and rHH = 0.7445 Å ]]

By carrying out mep calculations from a structure neighbouring the transition state, figures 12 and 13 shows the energy difference between the transition state and reactants, i.e. the activation energy, for the F + H2 reaction. An activation energy of 0.243 kcal/mol.

====== H + HF reaction ======

[[File: Jhl416_ea_hhf_surface.png |thumb|center|upright=2|Figure 14: MEP surface plot when rHF = 1.8113 Å and rHH = 0.7345 Å ]]

[[File: Jhl416_ea_hhf_energy.png |thumb|center|upright=2|Figure 15: Plot showing energy vs time when rHF = 1.8113 Å and rHH = 0.7345 Å ]]

By carrying out mep calculations from a structure neighbouring the transition state, figures 14 and 15 shows the energy difference between the transition state and reactants, i.e. the activation energy, for the F + H2 reaction. An activation energy of 30.075 kcal/mol.

===== Mechanism of release of reaction energy =====

[[File: Jhl416_rd_surface.png |thumb|center|upright=2|Figure 16: Surface plot when r1 = 2.3, r2 = 0.74, p1 = -2.5 and p2 = -1.5 ]]

[[File: Jhl416_momentum.png |thumb|center|upright=2|Figure 17: Plot showing internuclear momenta vs time when r1 = 2.3, r2 = 0.74, p1 = -2.5 and p2 = -1.5 ]]

With the initial conditions of r1 = 2.3, r2 = 0.74, p1 = -2.5, p2 = -1.5, it is shown in figure 16 that the trajectory is reactive.

By looking at the animation of the reaction, it can be observed that H2 approaches F atom with little oscillation. As they collides, H-H bond breaks and H-F bond is formed. However, the H-F bonds breaks again right afterwards and H-H bond is reformed. The oscillation of H-H bond becomes stronger and collides with F atom again that H-F bond is eventually formed with stronger oscillation while H atoms leaves in an opposite direction.

From the internuclear momenta vs time plot in figure 17, it can be seen that oscillation is initially absent between atoms A and B (i.e. atoms F and H) while present between atoms B and C (i.e. the two H atoms). Yet, after the reaction, strong oscillation is present between atoms A and B while absent between atoms B and C. In comparison, the oscillation in the initial H-H molecule is much smaller than that in the final H-F molecule. This indicates an increase of vibrational energy in the system, which is transferred from translational and potential energy.

In light of the conservation of energy, when potential energy is transferred to vibrational energy that is released in the form of heat, the temperature of the reaction will increase as a result. Thus, the mechanism of release of reaction energy can be confirmed by monitoring the change in temperature using calorimetry.

File:Jhl416 rd surface.png

2018-05-25T15:11:02Z

Jhl416:

MRD:jhl416

2018-05-25T15:09:40Z

Jhl416: /* Mechanism of release of reaction energy */

== Molecular Reaction Dynamics: Applications to Triatomic systems ==

=== EXERCISE 1: H + H2 system ===

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

''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.''

With reference to the diagram provided (Gradients at the transition state, as well as in the reactants and product regions.) in the experimental script, q1 and q2 are defined as a different set of coordinates different from the coordinates of r1 and r2. q2 is along the direction of the reaction pathway and q1 is orthogonal to the direction of the reaction pathway.

Both the transition structures and minima have a 0 gradient where the first derivative would both be 0, i.e. ∂Ve(q1)/∂q1 * ∂Ve(q2)/∂q2 = 0

Their second derivatives need to be used in order to distinguish between the two.

'''For transition structures:'''

As the transition state is at the maximum along the reaction pathway, the second derivative of the transition state is < 0 along the q2 direction, i.e. ∂Ve2(q2)/∂q22 < 0

While the second derivative along the q1 direction is > 0 as it is a minimum along any other directions except from the direction of the reaction pathway, i.e. ∂Ve2(q1)/∂q12 > 0.

'''For minima:'''

Minima are at minimum along all directions, including the direction of the reaction pathway. Thus, both of the second derivatives (in q1 and q2 directions) are > 0, i.e. ∂Ve2(q2)/∂q22 > 0 and ∂Ve2(q1)/∂q12 > 0.

==== Trajectories from r1 = r2: locating the transition state ====

''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.''

At the transition state position, there is 0 gradient. Thus, without initial momentum, the kinetic energy of the system should be 0 where there is no oscillation present. Thus, distanceA-B and distanceB-C should be constant in the system.

[[File: Jhl416_ts_position.png |thumb|center|upright=2|Figure 1 - A plot of Internuclear Distance vs Time when r1 = r2 = 0.90777 Å | 600px]]

By testing different rts values, the best estimation of the transition state position is found to be 0.90777 Å. As seen from figure 1 above, when rts = 0.90777 Å, distanceA-B and distanceB-C are seen to be constant with no oscillation, which aligns with the above.

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

The reaction path calculated using different methods of mep and dynamics are presented below in figure 2 and 3.

[[File: jhl416_mep_surface.png |thumb|center|upright=2|Figure 2: mep surface plot at r1 = 0.91777, r2 = 0.90777, p1 = p2 = 0 ]]

From figure 2, it is seen that the trajectory calculated using mep appears to be a straight line. This is due to the fact that mep shows an infinitely slow motion of reaction path. As velocity is set to zero at every time step, the momentum and thus kinetic energy is also zero at each time step. This results in no oscillation motion and hence a straight line of the trajectory.

[[File: jhl416_dynamics_surface.png |thumb|center|upright=2|Figure 3: Dynamics surface plot at r1 = 0.91777, r2 = 0.90777, p1 = p2 = 0 ]]

From figure 3, the trajectory appears to be more wavy when calculated using the dynamics method. This gives a more realistic picture of the motion of atoms where masses of atoms are taken into account. Thus, the oscillation motion of the atoms can then be observed.

==== Reactive and unreactive trajectories ====

{| class="wikitable" border=1
! p1 !! p2 !! Reactive/Unreactive !! Description !! Image
|-
| -1.25 || -2.5 || Reactive || Along the trajectory, Hb and Hc are kept at a constant distance with no oscillation observed in the entrance channel. As Ha approaches and collision occurs, kinetic energy is enough for reaction to occur. The bond between Hb and Hc breaks and the bond between Ha and Hb is formed while Hc leaves in an opposite direction. Translational energy is converted into vibrational energy such that oscillation is observed in the exit channel between Ha and Hb. ||
[[File:jhl416_plot1.png |thumb|center|upright=1.5| Figure 4]]
|-
| -1.5 || -2.0 || Unreactive || Comparing with the first momenta combination, AB momentum is lower in this momenta combination, resulting in insufficient kinetic energy to overcome the activation barrier that the trajectory is unreactive. || [[File:jhl416_plot2.png|thumb|center|upright=1.5|Figure 5]]
|-
| -1.5 || -2.5 || Reactive || Similar to the first momenta combination, this momenta combination results in a similar trajectory. The major difference is observed at the entrance channel where oscillation between Hb and Hc before the collision with Ha. This can be explained by the increased magnitude of p1 (which corresponds to BC momentum) from 1.25 to 1.5. This also means an increase in kinetic energy of atoms such that oscillation can be observed. || [[File:jhl416_plot3.png|thumb|center|upright=1.5|Figure 6]]
|-
| -2.5 || -5.0 || Unreactive || As Ha collides with the bonded Hb and Hc, the bond breaks and the the Ha-Hb bond is formed. Yet, due to the high momentum of atoms, the newly-formed bond breaks again that the Hb-Hc bond is formed again while Ha leaves in the opposite direction in a reduced speed. Trajectory is unreactive.|| [[File:jhl416_plot4.png|thumb|center|upright=1.5|Figure 7]]
|-
| -2.5 || -5.2 || Reactive || Comparing with the previous case in figure 7, the magnitude of p2 (which corresponds to AB momentum) is increased from 5.0 to 5.2. Atoms behave similarly with that in figure 7. However, due to the high momentum, after the reformed Hb-Hc bond, it breaks again to form a Ha-Hb bond, leaving Hc to leave in an opposite direction. || [[File:jhl416_plot5.png|thumb|center|upright=1.5|Figure 8]]
|}

==== Assumptions of Transition State Theory ====

The following are the main assumptions of the Transition State Theory<ref name=tst/>
<references> :
== References ==
<ref name=tst> J. W. Moore, R. G. Pearson, Kinetics and Mechanism, '''1981''', 166
</ref>
</references>

1. Atoms behave in classical mechanics that a classical trajectory can be an accurate description of the motion.

2. There is an equilibrium between reactants and transition states.

3. Barrier recrossing does not occur.

As the Transition State Theory assumes no barrier recrossing occurs, its predicted values for reaction rate is likely to be overestimated in comparison with the experimental values. Under the assumption of the Transition State Theory, all systems will continue to form products once the energy barrier is crossed for one time. However, as demonstrated in figure 7, systems do recross the barrier under certain conditions. This indicates that some of the reactive reactions predictions under the Transition State Theory might not be true. Thus, the reaction rate in reality should be lower, leading to lower experimental values in comparison with the predictions under Transition State Theory.

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

==== PES inspection ====

=====Endothermic or exothermic reactions=====

[[File: jhl416_exo_fh2.png |thumb|center|upright=2|Figure 9: Surface plot of the F + H2 reaction ]]

As shown in figure 9, the F + H2 reaction is exothermic as the energy level of the exit channel (product: H-F) is lower than that of the entrance channel (reactant: H2). An exothermic reaction indicates a negative energy change. It means more energy is released in the bond formation (formation of H-F bond) than the energy is taken in during the bond breaking (breaking of H-H bond). Thus, the bond strength of H-F is stronger than that of H-H.

[[File: jhl416_endo_hhf.png |thumb|center|upright=2|Figure 10: Surface plot of the H + HF reaction ]]

As shown in figure 10, the H + HF reaction is endothermic as the energy level of the exit channel (product: H2) is higher than that of the entrance channel (reactant: H-F). An endothermic reaction indicates a positive energy change. It means more energy is taken in during the bond breaking (breaking of H-F bond) than released in the bond formation (formation of H-H bond), once again indicating the stronger bond strength of H-F than H-H.

=====Location of transition state=====

In the exothermic reaction, the activation energy is too small that it is difficult to locate the transition state. According to the Hammond postulate, the transition state resembles the structure of the reactants more than that of the products as the energy gap between the transition state and reactants is smaller than that between the transition state and products. Using this as a guide for the location, the transition state is located at where rHF = 1.8113 Å and rHH = 0.7445 Å.

[[File: Jhl416_tts_location.png |thumb|center|upright=2|Figure 11: Plot showing internuclear distance vs time when rHF = 1.8113 Å and rHH = 0.7445 Å ]]

Figure 11 above shows that when rHF = 1.8113 Å and rHH = 0.7445 Å, all distances between atoms are at constant, indicating that it is the point of transition state.

=====Activation energy for both reactions=====

====== F + H2 reaction ======

[[File: Jhl416_ea_fh2_surface.png |thumb|center|upright=2|Figure 12: MEP surface plot when rHF = 1.8213 Å and rHH = 0.7445 Å ]]

[[File: Jhl416_ea_fh2_energy.png |thumb|center|upright=2|Figure 13: Plot showing energy vs time when rHF = 1.8213 Å and rHH = 0.7445 Å ]]

By carrying out mep calculations from a structure neighbouring the transition state, figures 12 and 13 shows the energy difference between the transition state and reactants, i.e. the activation energy, for the F + H2 reaction. An activation energy of 0.243 kcal/mol.

====== H + HF reaction ======

[[File: Jhl416_ea_hhf_surface.png |thumb|center|upright=2|Figure 14: MEP surface plot when rHF = 1.8113 Å and rHH = 0.7345 Å ]]

[[File: Jhl416_ea_hhf_energy.png |thumb|center|upright=2|Figure 15: Plot showing energy vs time when rHF = 1.8113 Å and rHH = 0.7345 Å ]]

By carrying out mep calculations from a structure neighbouring the transition state, figures 14 and 15 shows the energy difference between the transition state and reactants, i.e. the activation energy, for the F + H2 reaction. An activation energy of 30.075 kcal/mol.

===== Mechanism of release of reaction energy =====

With the initial conditions of r1 = 2.3, r2 = 0.74, p1 = -2.5, p2 = -1.5, it is shown in figure that the trajectory is reactive.

By looking at the animation of the reaction, it can be observed that H2 approaches F atom with little oscillation. As they collides, H-H bond breaks and H-F bond is formed. However, the H-F bonds breaks again right afterwards and H-H bond is reformed. The oscillation of H-H bond becomes stronger and collides with F atom again that H-F bond is eventually formed with stronger oscillation while H atoms leaves in an opposite direction.

From the internuclear momenta vs time plot in figure, it can be seen that oscillation is initially absent between atoms A and B (i.e. atoms F and H) while present between atoms B and C (i.e. the two H atoms). Yet, after the reaction, strong oscillation is present between atoms A and B while absent between atoms B and C. In comparison, the oscillation in the initial H-H molecule is much smaller than that in the final H-F molecule. This indicates an increase of vibrational energy in the system, which is transferred from translational and potential energy.

In light of the conservation of energy, when potential energy is transferred to vibrational energy that is released in the form of heat, the temperature of the reaction will increase as a result. Thus, the mechanism of release of reaction energy can be confirmed by monitoring the change in temperature using calorimetry.

MRD:jhl416