ChemWiki - User contributions [en]

MRD:MolecularReactionDynamicsas11815

2018-05-25T11:01:53Z

As11815: /* Polanyi’s empirical rules */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂i=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center>∆ =((∂2z)/∂x∂y))2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and (∂2z)/(∂x2) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From Figure 3 it can be seen that MEP the line is not wavy because it represents the gradient of the surface at a point. However, for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy indicating that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during the first 2-3 s and then reaches plateau once the bond is completely formed. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially. (It appears linear as it is a very small range plotted on graph)

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C is reversible with A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
{|
|[[File:as11815Fig19.png|300px|thumb|''Figure 19:'' Contour plot]]
|[[File:as11815Fig20.png|300px|thumb|''Figure 20:'' Internuclear momenta VS time plot]]
|}

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -5, produces a reactive reaction (Figure 21), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (Figure 22). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

{|
|[[File:as11815Fig21.png|300px|thumb|''Figure 21: Reactive reaction p1 = 2.0, p2 = -5]]
|[[File:as11815Fig22.png|300px|thumb|''Figure 22: Unreactive reaction p1 = 6, p2 = -2]]
|}

The HF+H is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. Thus, the exact opposite would be observed compared with F+H2

===References ===

MRD:MolecularReactionDynamicsas11815

2018-05-25T11:01:38Z

As11815: /* References */

MRD:MolecularReactionDynamicsas11815

2018-05-25T11:01:21Z

As11815: /* Polanyi’s empirical rules */

MRD:MolecularReactionDynamicsas11815

2018-05-25T11:01:00Z

As11815: /* References */

MRD:MolecularReactionDynamicsas11815

2018-05-25T11:00:50Z

As11815: /* Polanyi’s empirical rules */

MRD:MolecularReactionDynamicsas11815

2018-05-25T11:00:14Z

As11815: /* References */

MRD:MolecularReactionDynamicsas11815

2018-05-25T11:00:06Z

As11815: /* Polanyi’s empirical rules */

MRD:MolecularReactionDynamicsas11815

2018-05-25T10:59:43Z

As11815: /* Polanyi’s empirical rules */

MRD:MolecularReactionDynamicsas11815

2018-05-25T10:59:22Z

As11815: /* Polanyi’s empirical rules */

MRD:MolecularReactionDynamicsas11815

2018-05-25T10:59:00Z

As11815: /* Polanyi’s empirical rules */

MRD:MolecularReactionDynamicsas11815

2018-05-25T10:58:09Z

As11815: /* Polanyi’s empirical rules */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂i=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center>∆ =((∂2z)/∂x∂y))2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and (∂2z)/(∂x2) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From Figure 3 it can be seen that MEP the line is not wavy because it represents the gradient of the surface at a point. However, for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy indicating that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during the first 2-3 s and then reaches plateau once the bond is completely formed. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially. (It appears linear as it is a very small range plotted on graph)

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C is reversible with A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
{|
|[[File:as11815Fig19.png|300px|thumb|''Figure 19:'' Contour plot]]
|[[File:as11815Fig20.png|300px|thumb|''Figure 20:'' Internuclear momenta VS time plot]]
|}

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -5, produces a reactive reaction (Figure 21), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (Figure 22). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

{|
|[[File:as11815Fig21.png|300px|thumb|''Figure 21: "Reactive reaction p1 = 2.0, p2 = -5]]
|[[File:as11815Fig22.png|300px|thumb|''Figure 22: "Unreactive reaction p1 = 6, p2 = -2]]
|}

The HF+H is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. Thus, the exact opposite would be observed compared with F+H2.

MRD:MolecularReactionDynamicsas11815

2018-05-25T10:57:43Z

As11815: /* Polanyi’s empirical rules */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂i=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center>∆ =((∂2z)/∂x∂y))2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and (∂2z)/(∂x2) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From Figure 3 it can be seen that MEP the line is not wavy because it represents the gradient of the surface at a point. However, for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy indicating that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during the first 2-3 s and then reaches plateau once the bond is completely formed. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially. (It appears linear as it is a very small range plotted on graph)

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C is reversible with A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
{|
|[[File:as11815Fig19.png|300px|thumb|''Figure 19:'' Contour plot]]
|[[File:as11815Fig20.png|300px|thumb|''Figure 20:'' Internuclear momenta VS time plot]]
|}

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -5, produces a reactive reaction (Figure 21), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (Figure 22). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

{|
|[[File:as11815Fig21.png|300px|thumb|''Figure 21: "Reactive reaction p1 = 2.0, p2 = -5]]
|[[File:as11815Fig22.png|300px|thumb|''Figure 22: "Unreactive reaction p1 = -5, p1 = 2.0]]
|}

The HF+H is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. Thus, the exact opposite would be observed compared with F+H2.

MRD:MolecularReactionDynamicsas11815

2018-05-25T10:56:10Z

As11815: /* Polanyi’s empirical rules */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂i=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center>∆ =((∂2z)/∂x∂y))2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and (∂2z)/(∂x2) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From Figure 3 it can be seen that MEP the line is not wavy because it represents the gradient of the surface at a point. However, for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy indicating that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during the first 2-3 s and then reaches plateau once the bond is completely formed. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially. (It appears linear as it is a very small range plotted on graph)

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C is reversible with A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
{|
|[[File:as11815Fig19.png|300px|thumb|''Figure 19:'' Contour plot]]
|[[File:as11815Fig20.png|300px|thumb|''Figure 20:'' Internuclear momenta VS time plot]]
|}

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -5, produces a reactive reaction (Figure 21), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (Figure 22). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

{|
|[[File:as11815Fig21.png|300px|thumb|''Figure 21]]
|[[File:as11815Fig22.png|300px|thumb|''Figure 22]]
|}

The HF+H is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. Thus, the exact opposite would be observed compared with F+H2.

MRD:MolecularReactionDynamicsas11815

2018-05-25T10:55:56Z

As11815: /* Reaction dynamics */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂i=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center>∆ =((∂2z)/∂x∂y))2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and (∂2z)/(∂x2) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From Figure 3 it can be seen that MEP the line is not wavy because it represents the gradient of the surface at a point. However, for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy indicating that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during the first 2-3 s and then reaches plateau once the bond is completely formed. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially. (It appears linear as it is a very small range plotted on graph)

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C is reversible with A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
{|
|[[File:as11815Fig19.png|300px|thumb|''Figure 19:'' Contour plot]]
|[[File:as11815Fig20.png|300px|thumb|''Figure 20:'' Internuclear momenta VS time plot]]
|}

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -5, produces a reactive reaction (Figure 21), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (Figure 22). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

{|
|[[File:as11815Fig21.png|300px|thumb|centre''Figure 21]]
|[[File:as11815Fig22.png|300px|thumb|centre''Figure 22]]
|}

The HF+H is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. Thus, the exact opposite would be observed compared with F+H2.

MRD:MolecularReactionDynamicsas11815

2018-05-25T10:55:08Z

As11815: /* Polanyi’s empirical rules */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂i=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center>∆ =((∂2z)/∂x∂y))2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and (∂2z)/(∂x2) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From Figure 3 it can be seen that MEP the line is not wavy because it represents the gradient of the surface at a point. However, for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy indicating that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during the first 2-3 s and then reaches plateau once the bond is completely formed. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially. (It appears linear as it is a very small range plotted on graph)

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C is reversible with A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
{|
|[[File:as11815Fig19.png|300px|thumb|centre''Figure 19:'' Contour plot]]
|[[File:as11815Fig20.png|300px|thumb|centre''Figure 20:'' Internuclear momenta VS time plot]]
|}

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -5, produces a reactive reaction (Figure 21), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (Figure 22). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

{|
|[[File:as11815Fig21.png|300px|thumb|centre''Figure 21]]
|[[File:as11815Fig22.png|300px|thumb|centre''Figure 22]]
|}

The HF+H is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. Thus, the exact opposite would be observed compared with F+H2.

MRD:MolecularReactionDynamicsas11815

2018-05-25T10:54:32Z

As11815: /* Polanyi’s empirical rules */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂i=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center>∆ =((∂2z)/∂x∂y))2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and (∂2z)/(∂x2) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From Figure 3 it can be seen that MEP the line is not wavy because it represents the gradient of the surface at a point. However, for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy indicating that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during the first 2-3 s and then reaches plateau once the bond is completely formed. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially. (It appears linear as it is a very small range plotted on graph)

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C is reversible with A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
{|
|[[File:as11815Fig19.png|300px|thumb|centre''Figure 19:'' Contour plot]]
|[[File:as11815Fig20.png|300px|thumb|centre''Figure 20:'' Internuclear momenta VS time plot]]
|}

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -5, produces a reactive reaction (Figure 21), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (Figure 22). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

{|
|[[File:as11815Fig21.png|300px|thumb|centre ''Figure 21"]]
|[[File:as11815Fig22.png|300px|thumb|centre ''Figure 22"]]
|}

The HF+H is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. Thus, the exact opposite would be observed compared with F+H2.

MRD:MolecularReactionDynamicsas11815

2018-05-25T10:54:17Z

As11815: /* Polanyi’s empirical rules */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂i=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center>∆ =((∂2z)/∂x∂y))2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and (∂2z)/(∂x2) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From Figure 3 it can be seen that MEP the line is not wavy because it represents the gradient of the surface at a point. However, for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy indicating that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during the first 2-3 s and then reaches plateau once the bond is completely formed. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially. (It appears linear as it is a very small range plotted on graph)

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C is reversible with A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
{|
|[[File:as11815Fig19.png|300px|thumb|centre''Figure 19:'' Contour plot]]
|[[File:as11815Fig20.png|300px|thumb|centre''Figure 20:'' Internuclear momenta VS time plot]]
|}

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -5, produces a reactive reaction (Figure 21), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (Figure 22). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

{|
|[[File:as11815Fig21.png|300px|thumb|centre: ''Figure 21]]
|[[File:as11815Fig22.png|300px|thumb|centre: ''Figure 22]]
|}

The HF+H is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. Thus, the exact opposite would be observed compared with F+H2.

MRD:MolecularReactionDynamicsas11815

2018-05-25T10:53:08Z

As11815: /* EXERCISE 2: F-H-H and H-H-F systems */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂i=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center>∆ =((∂2z)/∂x∂y))2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and (∂2z)/(∂x2) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From Figure 3 it can be seen that MEP the line is not wavy because it represents the gradient of the surface at a point. However, for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy indicating that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during the first 2-3 s and then reaches plateau once the bond is completely formed. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially. (It appears linear as it is a very small range plotted on graph)

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C is reversible with A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
{|
|[[File:as11815Fig19.png|300px|thumb|centre''Figure 19:'' Contour plot]]
|[[File:as11815Fig20.png|300px|thumb|centre''Figure 20:'' Internuclear momenta VS time plot]]
|}

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -5, produces a reactive reaction (Figure 21), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (Figure 22). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

{|
|[[File:as11815Fig21.png|300px|thumb|centre''Figure 21]]
|[[File:as11815Fig22.png|300px|thumb|centre''Figure 22]]
|}

The HF+H is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. Thus, the exact opposite would be observed compared with F+H2.

MRD:MolecularReactionDynamicsas11815

2018-05-25T10:47:54Z

As11815: /* MEP and Dynamic Pathways: Calculating the reaction path */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂i=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center>∆ =((∂2z)/∂x∂y))2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and (∂2z)/(∂x2) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From Figure 3 it can be seen that MEP the line is not wavy because it represents the gradient of the surface at a point. However, for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy indicating that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during the first 2-3 s and then reaches plateau once the bond is completely formed. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially. (It appears linear as it is a very small range plotted on graph)

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C is reversible with A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

(You must insert a figure to show this)

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
{|
|[[File:as11815Fig19.png|300px|thumb|centre''Figure 19:'' Contour plot]]
|[[File:as11815Fig20.png|300px|thumb|centre''Figure 20:'' Internuclear momenta VS time plot]]
|}

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -5, produces a reactive reaction (Figure 21), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (Figure 22). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

{|
|[[File:as11815Fig21.png|300px|thumb|centre''Figure 21]]
|[[File:as11815Fig22.png|300px|thumb|centre''Figure 22]]
|}

The HF+H is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. Thus, the exact opposite would be observed compared with F+H2.

MRD:MolecularReactionDynamicsas11815

2018-05-25T10:46:27Z

As11815: /* MEP and Dynamic Pathways: Calculating the reaction path */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂i=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center>∆ =((∂2z)/∂x∂y))2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and (∂2z)/(∂x2) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From Figure 3 it can be seen that MEP the line is not wavy because it represents the gradient of the surface at a point. However, for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy indicating that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during the first 2-3 s and then reaches plateau once the bond is completely formed. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially.

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C is reversible with A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

(You must insert a figure to show this)

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
{|
|[[File:as11815Fig19.png|300px|thumb|centre''Figure 19:'' Contour plot]]
|[[File:as11815Fig20.png|300px|thumb|centre''Figure 20:'' Internuclear momenta VS time plot]]
|}

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -5, produces a reactive reaction (Figure 21), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (Figure 22). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

{|
|[[File:as11815Fig21.png|300px|thumb|centre''Figure 21]]
|[[File:as11815Fig22.png|300px|thumb|centre''Figure 22]]
|}

The HF+H is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. Thus, the exact opposite would be observed compared with F+H2.

MRD:MolecularReactionDynamicsas11815

2018-05-25T10:44:39Z

As11815: /* MEP and Dynamic Pathways: Calculating the reaction path */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂i=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center>∆ =((∂2z)/∂x∂y))2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and (∂2z)/(∂x2) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From Figure 3 it can be seen that MEP the line is not wavy because it represents the gradient of the surface at a point. However, for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy indicating that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during the first 0.5s and then reaches plateau (still oscillating due to vibrational energy) once the bond is completely formed. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially.

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C is reversible with A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

(You must insert a figure to show this)

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
{|
|[[File:as11815Fig19.png|300px|thumb|centre''Figure 19:'' Contour plot]]
|[[File:as11815Fig20.png|300px|thumb|centre''Figure 20:'' Internuclear momenta VS time plot]]
|}

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -5, produces a reactive reaction (Figure 21), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (Figure 22). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

{|
|[[File:as11815Fig21.png|300px|thumb|centre''Figure 21]]
|[[File:as11815Fig22.png|300px|thumb|centre''Figure 22]]
|}

The HF+H is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. Thus, the exact opposite would be observed compared with F+H2.

MRD:MolecularReactionDynamicsas11815

2018-05-25T10:42:16Z

As11815: /* MEP and Dynamic Pathways: Calculating the reaction path */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂i=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center>∆ =((∂2z)/∂x∂y))2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and (∂2z)/(∂x2) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From Figure 3 it can be seen that MEP the line is not wavy because it represents the gradient of the surface at a point. However, for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy indicating that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during (how many) s and then reaches plateaus once the bond is completely formed and is static. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially.

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C is reversible with A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

(You must insert a figure to show this)

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
{|
|[[File:as11815Fig19.png|300px|thumb|centre''Figure 19:'' Contour plot]]
|[[File:as11815Fig20.png|300px|thumb|centre''Figure 20:'' Internuclear momenta VS time plot]]
|}

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -5, produces a reactive reaction (Figure 21), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (Figure 22). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

{|
|[[File:as11815Fig21.png|300px|thumb|centre''Figure 21]]
|[[File:as11815Fig22.png|300px|thumb|centre''Figure 22]]
|}

The HF+H is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. Thus, the exact opposite would be observed compared with F+H2.

MRD:MolecularReactionDynamicsas11815

2018-05-25T10:41:07Z

As11815: /* Locating the transition state */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂i=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center>∆ =((∂2z)/∂x∂y))2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and (∂2z)/(∂x2) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From figure 3 it can be seen that MEP the line is not wavy because we are looking at the gradient of the surface at a point. However for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy which indicated that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during (how many) s and then reaches plateaus once the bond is completely formed and is static. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially.

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C is reversible with A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

(You must insert a figure to show this)

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
{|
|[[File:as11815Fig19.png|300px|thumb|centre''Figure 19:'' Contour plot]]
|[[File:as11815Fig20.png|300px|thumb|centre''Figure 20:'' Internuclear momenta VS time plot]]
|}

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -5, produces a reactive reaction (Figure 21), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (Figure 22). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

{|
|[[File:as11815Fig21.png|300px|thumb|centre''Figure 21]]
|[[File:as11815Fig22.png|300px|thumb|centre''Figure 22]]
|}

The HF+H is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. Thus, the exact opposite would be observed compared with F+H2.

MRD:MolecularReactionDynamicsas11815

2018-05-25T10:40:17Z

As11815: /* Dynamics from the transition state region */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂i=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center>∆ =((∂2z)/∂x∂y))2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and (∂2z)/(∂x2) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From figure 3 it can be seen that MEP the line is not wavy because we are looking at the gradient of the surface at a point. However for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy which indicated that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during (how many) s and then reaches plateaus once the bond is completely formed and is static. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially.

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C is reversible with A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

(You must insert a figure to show this)

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
{|
|[[File:as11815Fig19.png|300px|thumb|centre''Figure 19:'' Contour plot]]
|[[File:as11815Fig20.png|300px|thumb|centre''Figure 20:'' Internuclear momenta VS time plot]]
|}

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -5, produces a reactive reaction (Figure 21), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (Figure 22). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

{|
|[[File:as11815Fig21.png|300px|thumb|centre''Figure 21]]
|[[File:as11815Fig22.png|300px|thumb|centre''Figure 22]]
|}

The HF+H is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. Thus, the exact opposite would be observed compared with F+H2.

MRD:MolecularReactionDynamicsas11815

2018-05-25T10:39:31Z

As11815: /* Dynamics from the transition state region */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂ri=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center>∆ =((∂2z)/∂x∂y))2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and (∂2z)/(∂x2) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From figure 3 it can be seen that MEP the line is not wavy because we are looking at the gradient of the surface at a point. However for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy which indicated that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during (how many) s and then reaches plateaus once the bond is completely formed and is static. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially.

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C is reversible with A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

(You must insert a figure to show this)

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
{|
|[[File:as11815Fig19.png|300px|thumb|centre''Figure 19:'' Contour plot]]
|[[File:as11815Fig20.png|300px|thumb|centre''Figure 20:'' Internuclear momenta VS time plot]]
|}

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -5, produces a reactive reaction (Figure 21), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (Figure 22). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

{|
|[[File:as11815Fig21.png|300px|thumb|centre''Figure 21]]
|[[File:as11815Fig22.png|300px|thumb|centre''Figure 22]]
|}

The HF+H is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. Thus, the exact opposite would be observed compared with F+H2.

MRD:MolecularReactionDynamicsas11815

2018-05-25T10:39:17Z

As11815: /* H + H2 system Analysis */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂ri=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center>∆ =((∂2z)/∂x∂y〗2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and (∂2z)/(∂x2) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From figure 3 it can be seen that MEP the line is not wavy because we are looking at the gradient of the surface at a point. However for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy which indicated that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during (how many) s and then reaches plateaus once the bond is completely formed and is static. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially.

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C is reversible with A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

(You must insert a figure to show this)

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
{|
|[[File:as11815Fig19.png|300px|thumb|centre''Figure 19:'' Contour plot]]
|[[File:as11815Fig20.png|300px|thumb|centre''Figure 20:'' Internuclear momenta VS time plot]]
|}

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -5, produces a reactive reaction (Figure 21), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (Figure 22). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

{|
|[[File:as11815Fig21.png|300px|thumb|centre''Figure 21]]
|[[File:as11815Fig22.png|300px|thumb|centre''Figure 22]]
|}

The HF+H is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. Thus, the exact opposite would be observed compared with F+H2.

MRD:MolecularReactionDynamicsas11815

2018-05-25T10:37:51Z

As11815: /* Dynamics from the transition state region */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂ri=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center> ∆ =((∂2z)/∂x∂y)2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and ((∂2z)/(∂x2)) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From figure 3 it can be seen that MEP the line is not wavy because we are looking at the gradient of the surface at a point. However for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy which indicated that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during (how many) s and then reaches plateaus once the bond is completely formed and is static. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially.

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C is reversible with A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

(You must insert a figure to show this)

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
{|
|[[File:as11815Fig19.png|300px|thumb|centre''Figure 19:'' Contour plot]]
|[[File:as11815Fig20.png|300px|thumb|centre''Figure 20:'' Internuclear momenta VS time plot]]
|}

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -5, produces a reactive reaction (Figure 21), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (Figure 22). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

{|
|[[File:as11815Fig21.png|300px|thumb|centre''Figure 21]]
|[[File:as11815Fig22.png|300px|thumb|centre''Figure 22]]
|}

The HF+H is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. Thus, the exact opposite would be observed compared with F+H2.

MRD:MolecularReactionDynamicsas11815

2018-05-22T22:22:51Z

As11815: /* Polanyi’s empirical rules */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂ri=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center> ∆ =〖((∂2z)/∂x∂y)〗2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and ((∂2z)/(∂x2)) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From figure 3 it can be seen that MEP the line is not wavy because we are looking at the gradient of the surface at a point. However for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy which indicated that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during (how many) s and then reaches plateaus once the bond is completely formed and is static. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially.

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C is reversible with A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

(You must insert a figure to show this)

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
{|
|[[File:as11815Fig19.png|300px|thumb|centre''Figure 19:'' Contour plot]]
|[[File:as11815Fig20.png|300px|thumb|centre''Figure 20:'' Internuclear momenta VS time plot]]
|}

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -5, produces a reactive reaction (Figure 21), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (Figure 22). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

{|
|[[File:as11815Fig21.png|300px|thumb|centre''Figure 21]]
|[[File:as11815Fig22.png|300px|thumb|centre''Figure 22]]
|}

The HF+H is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. Thus, the exact opposite would be observed compared with F+H2.

MRD:MolecularReactionDynamicsas11815

2018-05-22T22:21:33Z

As11815: /* MEP and Dynamic Pathways: Calculating the reaction path */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂ri=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center> ∆ =〖((∂2z)/∂x∂y)〗2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and ((∂2z)/(∂x2)) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From figure 3 it can be seen that MEP the line is not wavy because we are looking at the gradient of the surface at a point. However for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy which indicated that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during (how many) s and then reaches plateaus once the bond is completely formed and is static. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially.

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C is reversible with A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

(You must insert a figure to show this)

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
{|
|[[File:as11815Fig19.png|300px|thumb|centre''Figure 19:'' Contour plot]]
|[[File:as11815Fig20.png|300px|thumb|centre''Figure 20:'' Internuclear momenta VS time plot]]
|}

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -5, produces a reactive reaction (Figure 21), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (figure 22). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

{|
|[[File:as11815Fig21.png|300px|thumb|centre''Figure 21]]
|[[File:as11815Fig22.png|300px|thumb|centre''Figure 22]]
|}

The HF+H is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. Thus, the exact opposite would be observed compared with F+H2.

File:As11815Fig22.png

2018-05-22T22:20:23Z

As11815:

File:As11815Fig21.png

2018-05-22T22:19:54Z

As11815:

MRD:MolecularReactionDynamicsas11815

2018-05-22T22:19:41Z

As11815: /* Polanyi’s empirical rules */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂ri=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center> ∆ =〖((∂2z)/∂x∂y)〗2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and ((∂2z)/(∂x2)) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From figure 3 it can be seen that MEP the line is not wavy because we are looking at the gradient of the surface at a point. However for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy which indicated that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during (how many) s and then reaches plateaus once the bond is completely formed and is static. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially.

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C (reversible sign) A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

(You must insert a figure to show this)

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
{|
|[[File:as11815Fig19.png|300px|thumb|centre''Figure 19:'' Contour plot]]
|[[File:as11815Fig20.png|300px|thumb|centre''Figure 20:'' Internuclear momenta VS time plot]]
|}

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -5, produces a reactive reaction (Figure 21), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (figure 22). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

{|
|[[File:as11815Fig21.png|300px|thumb|centre''Figure 21]]
|[[File:as11815Fig22.png|300px|thumb|centre''Figure 22]]
|}

The HF+H is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. Thus, the exact opposite would be observed compared with F+H2.

MRD:MolecularReactionDynamicsas11815

2018-05-22T22:18:58Z

As11815: /* Polanyi’s empirical rules */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂ri=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center> ∆ =〖((∂2z)/∂x∂y)〗2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and ((∂2z)/(∂x2)) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From figure 3 it can be seen that MEP the line is not wavy because we are looking at the gradient of the surface at a point. However for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy which indicated that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during (how many) s and then reaches plateaus once the bond is completely formed and is static. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially.

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C (reversible sign) A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

(You must insert a figure to show this)

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
{|
|[[File:as11815Fig19.png|300px|thumb|centre''Figure 19:'' Contour plot]]
|[[File:as11815Fig20.png|300px|thumb|centre''Figure 20:'' Internuclear momenta VS time plot]]
|}

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -5, produces a reactive reaction (Figure 21), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (figure 22). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

{|
|[[File:as11815Fig21.png|300px|thumb|centre''Figure 21]]
|[[File:as11815Fig22.png|300px|thumb|centre''Figure 22]]
|}

The F+H2 is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. Thus, the exact opposite would be observed for F+H2 compared with

MRD:MolecularReactionDynamicsas11815

2018-05-22T22:18:10Z

As11815: /* Polanyi’s empirical rules */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂ri=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center> ∆ =〖((∂2z)/∂x∂y)〗2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and ((∂2z)/(∂x2)) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From figure 3 it can be seen that MEP the line is not wavy because we are looking at the gradient of the surface at a point. However for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy which indicated that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during (how many) s and then reaches plateaus once the bond is completely formed and is static. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially.

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C (reversible sign) A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

(You must insert a figure to show this)

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
{|
|[[File:as11815Fig19.png|300px|thumb|centre''Figure 19:'' Contour plot]]
|[[File:as11815Fig20.png|300px|thumb|centre''Figure 20:'' Internuclear momenta VS time plot]]
|}

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -5, produces a reactive reaction (Figure 21), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (figure 22). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

The F+H2 is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. Thus, the exact opposite would be observed for F+H2 compared with

MRD:MolecularReactionDynamicsas11815

2018-05-22T22:15:40Z

As11815: /* Polanyi’s empirical rules */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂ri=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center> ∆ =〖((∂2z)/∂x∂y)〗2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and ((∂2z)/(∂x2)) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From figure 3 it can be seen that MEP the line is not wavy because we are looking at the gradient of the surface at a point. However for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy which indicated that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during (how many) s and then reaches plateaus once the bond is completely formed and is static. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially.

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C (reversible sign) A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

(You must insert a figure to show this)

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
{|
|[[File:as11815Fig19.png|300px|thumb|centre''Figure 19:'' Contour plot]]
|[[File:as11815Fig20.png|300px|thumb|centre''Figure 20:'' Internuclear momenta VS time plot]]
|}

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -5, produces a reactive reaction (Figure 21), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (figure 22). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

The F+H2 is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. Thus, the surface plot obeys Polanyi’s rules as p1 = vibrational energy (high) = 6.0, and p2 = translational energy (low) = -2, produces a reactive reaction (Figure A), whereas as p1 = vibrational energy (low) = 2, and p2 = translational energy (high) = -6, produces an unreactive reaction (figure B). Thus, it is confirmed that high vibrational energies result in successful endothermic reactions.

MRD:MolecularReactionDynamicsas11815

2018-05-22T22:08:19Z

As11815: /* Reaction dynamics */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂ri=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center> ∆ =〖((∂2z)/∂x∂y)〗2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and ((∂2z)/(∂x2)) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From figure 3 it can be seen that MEP the line is not wavy because we are looking at the gradient of the surface at a point. However for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy which indicated that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during (how many) s and then reaches plateaus once the bond is completely formed and is static. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially.

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C (reversible sign) A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

(You must insert a figure to show this)

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
{|
|[[File:as11815Fig19.png|300px|thumb|centre''Figure 19:'' Contour plot]]
|[[File:as11815Fig20.png|300px|thumb|centre''Figure 20:'' Internuclear momenta VS time plot]]
|}

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -6, produces a reactive reaction (Figure A), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (figure B). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

The F+H2 is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (high) = 6.0, and p2 = translational energy (low) = -2, produces a reactive reaction (Figure A), whereas as p1 = vibrational energy (low) = 2, and p2 = translational energy (high) = -6, produces an unreactive reaction (figure B). Thus, it is confirmed that high vibrational energies result in successful endothermic reactions.

File:As11815Fig20.png

2018-05-22T22:07:46Z

As11815:

MRD:MolecularReactionDynamicsas11815

2018-05-22T22:07:35Z

As11815: /* Reaction dynamics */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂ri=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center> ∆ =〖((∂2z)/∂x∂y)〗2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and ((∂2z)/(∂x2)) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From figure 3 it can be seen that MEP the line is not wavy because we are looking at the gradient of the surface at a point. However for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy which indicated that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during (how many) s and then reaches plateaus once the bond is completely formed and is static. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially.

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C (reversible sign) A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

(You must insert a figure to show this)

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.

[[File:as11815Fig19.png|300px|thumb|centre''Figure 19:'' Contour plot]]
[[File:as11815Fig20.png|300px|thumb|centre''Figure 19:'' Internuclear momenta VS time plot]]

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -6, produces a reactive reaction (Figure A), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (figure B). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

The F+H2 is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (high) = 6.0, and p2 = translational energy (low) = -2, produces a reactive reaction (Figure A), whereas as p1 = vibrational energy (low) = 2, and p2 = translational energy (high) = -6, produces an unreactive reaction (figure B). Thus, it is confirmed that high vibrational energies result in successful endothermic reactions.

File:As11815Fig19.png

2018-05-22T22:06:16Z

As11815: As11815 uploaded a new version of File:As11815Fig19.png

File:As11815Fig19.png

2018-05-22T22:05:41Z

As11815:

MRD:MolecularReactionDynamicsas11815

2018-05-22T22:05:23Z

As11815: /* EXERCISE 2: F-H-H and H-H-F systems */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂ri=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center> ∆ =〖((∂2z)/∂x∂y)〗2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and ((∂2z)/(∂x2)) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From figure 3 it can be seen that MEP the line is not wavy because we are looking at the gradient of the surface at a point. However for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy which indicated that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during (how many) s and then reaches plateaus once the bond is completely formed and is static. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially.

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C (reversible sign) A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

(You must insert a figure to show this)

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.

[[File:as11815Fig19.png|300px|thumb|centre''Figure 19:'' Contour plot]]

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -6, produces a reactive reaction (Figure A), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (figure B). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

The F+H2 is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (high) = 6.0, and p2 = translational energy (low) = -2, produces a reactive reaction (Figure A), whereas as p1 = vibrational energy (low) = 2, and p2 = translational energy (high) = -6, produces an unreactive reaction (figure B). Thus, it is confirmed that high vibrational energies result in successful endothermic reactions.

MRD:MolecularReactionDynamicsas11815

2018-05-22T22:04:45Z

As11815: /* Reaction dynamics */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂ri=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center> ∆ =〖((∂2z)/∂x∂y)〗2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and ((∂2z)/(∂x2)) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From figure 3 it can be seen that MEP the line is not wavy because we are looking at the gradient of the surface at a point. However for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy which indicated that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during (how many) s and then reaches plateaus once the bond is completely formed and is static. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially.

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C (reversible sign) A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

(You must insert a figure to show this)

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.1, p1 = -1.2, p2 = -1.6.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.

[[File:as11815Fig19.png|300px|thumb|centre

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -6, produces a reactive reaction (Figure A), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (figure B). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

The F+H2 is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (high) = 6.0, and p2 = translational energy (low) = -2, produces a reactive reaction (Figure A), whereas as p1 = vibrational energy (low) = 2, and p2 = translational energy (high) = -6, produces an unreactive reaction (figure B). Thus, it is confirmed that high vibrational energies result in successful endothermic reactions.

MRD:MolecularReactionDynamicsas11815

2018-05-22T22:00:18Z

As11815: /* EXERCISE 2: F-H-H and H-H-F systems */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂ri=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center> ∆ =〖((∂2z)/∂x∂y)〗2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and ((∂2z)/(∂x2)) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From figure 3 it can be seen that MEP the line is not wavy because we are looking at the gradient of the surface at a point. However for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy which indicated that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during (how many) s and then reaches plateaus once the bond is completely formed and is static. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially.

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C (reversible sign) A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

(You must insert a figure to show this)

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|centre|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
(I already chose different initial conditions)
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.2, p1 = -1.0, p2 = -1.7.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -6, produces a reactive reaction (Figure A), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (figure B). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

The F+H2 is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (high) = 6.0, and p2 = translational energy (low) = -2, produces a reactive reaction (Figure A), whereas as p1 = vibrational energy (low) = 2, and p2 = translational energy (high) = -6, produces an unreactive reaction (figure B). Thus, it is confirmed that high vibrational energies result in successful endothermic reactions.

MRD:MolecularReactionDynamicsas11815

2018-05-22T21:59:32Z

As11815: /* EXERCISE 2: F-H-H and H-H-F systems */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂ri=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center> ∆ =〖((∂2z)/∂x∂y)〗2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and ((∂2z)/(∂x2)) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From figure 3 it can be seen that MEP the line is not wavy because we are looking at the gradient of the surface at a point. However for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy which indicated that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during (how many) s and then reaches plateaus once the bond is completely formed and is static. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially.

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C (reversible sign) A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

(You must insert a figure to show this)

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at -103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants

F+H2 reaction: Activation energy = -103.7 Kcal/mol – (-104) = 0.3 Kcal/mol

HF+H reaction: Activation energy = -103.7 Kcal/mol – (-136) = 32.3 Kcal/mol

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
(I already chose different initial conditions)
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.2, p1 = -1.0, p2 = -1.7.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.

===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -6, produces a reactive reaction (Figure A), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (figure B). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

The F+H2 is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (high) = 6.0, and p2 = translational energy (low) = -2, produces a reactive reaction (Figure A), whereas as p1 = vibrational energy (low) = 2, and p2 = translational energy (high) = -6, produces an unreactive reaction (figure B). Thus, it is confirmed that high vibrational energies result in successful endothermic reactions.

File:As11815Fig18.png

2018-05-22T21:56:41Z

As11815: As11815 uploaded a new version of File:As11815Fig18.png

File:As11815Fig18.png

2018-05-22T21:53:47Z

As11815:

MRD:MolecularReactionDynamicsas11815

2018-05-22T21:53:23Z

As11815: /* EXERCISE 2: F-H-H and H-H-F systems */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂ri=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center> ∆ =〖((∂2z)/∂x∂y)〗2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and ((∂2z)/(∂x2)) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From figure 3 it can be seen that MEP the line is not wavy because we are looking at the gradient of the surface at a point. However for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy which indicated that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during (how many) s and then reaches plateaus once the bond is completely formed and is static. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially.

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C (reversible sign) A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

(You must insert a figure to show this)

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below, H-F = 0.75 Å and H-H = 1.71 Å
[[File:as11815Fig18.png|300px|thumb|''Figure 18:'' Energy plot for H-F = 0.75 Å and H-H = 1.71 Å]]
The transition state was found to occur at '''-103.7 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants
F+H2 reaction: Activation energy = '''-103.7 Kcal/mo'''l – (-104)
HF+H reaction: Activation energy = '''-103.7 Kcal/mol''' – (-136)

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
(I already chose different initial conditions)
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.2, p1 = -1.0, p2 = -1.7.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -6, produces a reactive reaction (Figure A), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (figure B). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

The F+H2 is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (high) = 6.0, and p2 = translational energy (low) = -2, produces a reactive reaction (Figure A), whereas as p1 = vibrational energy (low) = 2, and p2 = translational energy (high) = -6, produces an unreactive reaction (figure B). Thus, it is confirmed that high vibrational energies result in successful endothermic reactions.

MRD:MolecularReactionDynamicsas11815

2018-05-22T21:48:07Z

As11815: /* PES inspection */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂ri=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center> ∆ =〖((∂2z)/∂x∂y)〗2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and ((∂2z)/(∂x2)) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From figure 3 it can be seen that MEP the line is not wavy because we are looking at the gradient of the surface at a point. However for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy which indicated that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during (how many) s and then reaches plateaus once the bond is completely formed and is static. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially.

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C (reversible sign) A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

(You must insert a figure to show this)

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. In order to find the maximum energy change, the reactant energies were deviated slightly from the transition state bond lengths so that MEP calculations could be carried out to find the activation energy. For the plots shown below,
The transition state was found to occur at '''-103.75 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants
F+H2 reaction: Activation energy = '''-103.75 Kcal/mo'''l – (-104)
HF+H reaction: Activation energy = '''-103.75 Kcal/mol''' – (-136)

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
(I already chose different initial conditions)
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.2, p1 = -1.0, p2 = -1.7.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -6, produces a reactive reaction (Figure A), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (figure B). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

The F+H2 is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (high) = 6.0, and p2 = translational energy (low) = -2, produces a reactive reaction (Figure A), whereas as p1 = vibrational energy (low) = 2, and p2 = translational energy (high) = -6, produces an unreactive reaction (figure B). Thus, it is confirmed that high vibrational energies result in successful endothermic reactions.

MRD:MolecularReactionDynamicsas11815

2018-05-22T21:37:23Z

As11815:

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂ri=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center> ∆ =〖((∂2z)/∂x∂y)〗2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and ((∂2z)/(∂x2)) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From figure 3 it can be seen that MEP the line is not wavy because we are looking at the gradient of the surface at a point. However for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy which indicated that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during (how many) s and then reaches plateaus once the bond is completely formed and is static. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially.

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C (reversible sign) A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

(You must insert a figure to show this)

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants.
The transition state was found to occur at '''-103.75 Kcal/mol'''

Activation energy = energy of transition state – energy of reactants
F+H2 reaction: Activation energy = '''-103.75 Kcal/mo'''l – (-104)
HF+H reaction: Activation energy = '''-103.75 Kcal/mol''' – (-136)

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
(I already chose different initial conditions)
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.2, p1 = -1.0, p2 = -1.7.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -6, produces a reactive reaction (Figure A), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (figure B). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

The F+H2 is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (high) = 6.0, and p2 = translational energy (low) = -2, produces a reactive reaction (Figure A), whereas as p1 = vibrational energy (low) = 2, and p2 = translational energy (high) = -6, produces an unreactive reaction (figure B). Thus, it is confirmed that high vibrational energies result in successful endothermic reactions.

MRD:MolecularReactionDynamicsas11815

2018-05-22T21:36:24Z

As11815: /* PES inspection */

==H + H2 system Analysis==

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

The potential energy surface plot of a reaction is a function of the potential energy surface as a function of the reaction coordinate. Any minimum or at any maximum - the transition state; also known as the saddle point, that corresponds to the transient state of a reaction coordinate, at this point, both derivatives with respect to the transition state will be zero. In this reaction, ∂V(ri)/∂ri=0 was used to find the minima and transition state. The two can be differentiated from the second derivative.

<center> ∆ =〖((∂2z)/∂x∂y)〗2 - ((∂2z)/(∂x2))((∂2z)/(∂y2))</center>

where x and y = critical points and z = the derivative of the critical points.

If, ∆ > 0 then the point will be a saddle point. If ∆ < 0 and ((∂2z)/(∂x2)) > 0 then the point will be a local minima. <ref> C. M. Bender and S. A. Orszag, Advanced Mathematical Methods for Scientists and Engineers I: Asymptotic Methods and Perturbation Theory, Springer New York, 2013.</ref>

=== Locating the transition state ===
The gradient of the potential energy surface at the transition state is zero. This means the net force acting on the hydrogen molecule must also be zero. The system has no kinetic energy and all kinetic energy is converted to potential energy – thus we are looking at the highest potential energy point. Additionally, no net force means the momentum is zero. For the H+ H2 symmetric system, the transition state will occur when the r1 = r2. The value of this was found using GUI software by testing different constant bond distances and initial momentum of zero, so the molecule has no energy to escape the minimum potential well, to find that r1 = r2 = 0.907 Å was the best approximate match for a flat distance vs time plot (Figure 1). When there are no oscillations, it is confirmed that the molecule does not move and this must be the transition state.

[[File:MRDas11815Figure1.png|800px|thumb|centre|''Figure 1:'' Internuclear distance vs time plot for r1= r2=0.907]]

=== MEP and Dynamic Pathways: Calculating the reaction path ===
The minimum energy path (MEP) is the lowest energy pathway from reactants to products. This is the most likely mechanism to occur. In this part, the GUI is set as r1=rts+0.01=0.9175, r2= rts=0.9075, zero momenta and it is found that depending on the calculation used – MEP or Dynamics, two different trajectories are obtained.

Countour plots: From figure 3 it can be seen that MEP the line is not wavy because we are looking at the gradient of the surface at a point. However for the Dynamics calculation (Figure 2), the reaction path for the plot is wavy which indicated that the vibrations of the diatomic molecule match the inertial trajectory in which the molecule does oscillate. As a result of this, in the MEP plot the reaction doesn’t go to completion whereas in the Dynamics plot it does.

{|
|[[File:as11815Fig2.png|350px|thumb|centre|''Figure 2:'' Dynamic Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig3.png|350px|thumb|centre|''Figure 3:'' MEP Contour plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

Internuclear distance vs time plots: In the MEP plot (Figure 5), all the bond distances are initially constant. The internuclear distance however, decreases as the A-B bond forms during (how many) s and then reaches plateaus once the bond is completely formed and is static. Likewise, B-C and A-C internuclear distances gradually increase as these bonds break. For the dynamic plot (Figure 4) it can be seen that for a constant A-B, the B-C and A-C internuclear distances increase exponentially.

{|
|[[File:as11815Fig4.png|350px|thumb|centre|''Figure 4:'' Dynamic Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|[[File:as11815Fig5.png|350px|thumb|centre|''Figure 5:'' MEP Internuclear Distance VS Time plot for r1= rts +0.01 =0.9175, r2= rts]]
|}

If the initial conditions are changed to r1 = rts and r2 = rts +0.01, the reverse reaction pathway is calculated as the internuclear distance vs time (Figure 6) and momenta vs time (Figure 7) graphs reverse in sign. Figure 8 shows the internuclear momenta vs time with initial conditions. These are the final positions of the trajectories for which A collides with B-C to form A-B + C but because the internuclear distance and momenta vs time graphs oscillate it shows A-B + C (reversible sign) A + B-C. So the forward and reverse reactions happen sequentially.

{|
|[[File:as11815Fig6.png|350px|thumb|centre|''Figure 6:'' Internuclear Distance VS Time plot for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig7.png|350px|thumb|centre|''Figure 7:'' Internuclear Momenta vs Time for r1= rts, r2= rts +0.01 = 0.9175]]
|[[File:as11815Fig8.png|350px|thumb|centre|''Figure 8:'' Internuclear Momenta vs Time for r1= rts +0.01 =0.9175, r2= rts]]
|}

=== Reactive and Unreactive Trajectories ===

When a system has enough energy to overcome the activation barrier towards the transition state, the reaction is considered to go to completion. With constant initial positions of r1 = 0.745, r2 = 1.805 and varying momenta, the dependence of the reactivity of reaction pathway on momenta values are investigated.

{| class="wikitable" border="1"
|+ Momentum parameters for the computed trajectories
! p1 !! p2 !! Total energy (kcal/mol) !! Contour plot !! Reactivity
|-
| -1.25 || -2.5 || -99.018 || [[File:as11815Fig9.png|300px]] || Reactive: The trajectory is reactive. The reaction path starts at the reactants and channels towards the products, where it ends. Initially, the H2 molecule oscillates much less than the new A-B molecule formed. This suggests a lot of H-H bond potential energy is converted to kinetic energy of H-F during the reaction.
|-
| -1.5 || -2.0 || -100.456 ||[[File:as11815Fig10.png|300px]] ||Unreactive: The reaction path begins at the reactants but then oscillates back to the initial point before it can surpass the transition state. The reaction path corresponds to vibrating atom A approaching the lower vibrational B-C, which begins to vibrate as well. During this vibrational energy transfer the reactants interact such as to reach an energy near the A—B—C transition state but as there is not enough energy to form the transition state completely, so the complex reverses back to A and B-C reactants.
|-
| -1.5 || -2.5 || -98.956 ||[[File:as11815Fig11.png|300px]] ||Reactive: This is reactive. The vibrations of both initial A and B-C are more visible than the first example. This is due to the higher momentum B-C is given so it the initial oscillation of B, as well as the total vibrational energy during the reaction, is higher.
|-
| -2.5 || -5.0 || -84.956 ||[[File:as11815Fig12.png|300px]] ||Unreactive: Compared with the second example, the higher momentum means there is a higher vibrational energy - clearly indicated by the increased oscillations. However, although the vibrational energy may be enough to overcome the activation barrier of the transition state, the reaction pathway recrosses the transition state towards the reactants. This is because there is not enough translational energy and there needs to be a combination of both vibrational and translational energy for the energy barrier to be overcome.
|-
| -2.5 || -5.2 || -83.416 ||[[File:as11815Fig13.png|300px]] ||Reactive: This is similar to the first and third example, there is enough vibrational and translational energy to overcome the transition state.
|}

The main assumptions of transition state theory are :

1) The activated complex is in quasi-equilibrium with the reactants only and not the products.

2) The reactant nuclei are assumed to obey classical mechanics. This comes from the Born-Oppenheimer approximation of separating nuclei and electron motion due to nuclei being much heavier. As quantum mechanical pathways such as the lower energy tunneling is ignored, so transition state theory may overestimate the activation barrier, and falsely claim that a reaction is unreactive.

3) The reaction pathway must pass through the transition state on a potential energy diagram. This assumes that the reactant molecules follow Boltzmann distribution, which assumes that within the time of the reaction the system has reached thermal equilibrium. For high temperatures, population of higher energy states may be possible for which the transition state theory would also overestimate the activation barrier. <ref> P. Atkins, J. de Paula, Atkins' Physical Chemistry, Oxford University Press, 2014, 894-902</ref>

==EXERCISE 2: F-H-H and H-H-F systems==
<div style="text-align: justify;">

===PES inspection===

(You must insert a figure to show this)

The endothermic/exothermic nature of the the following reactions are direct consequences of:
H-H bond enthalpy = 104 kcal/mol
H-F bond enthalpy = 136 kcal/mol
<ref>http://www.cchem.berkeley.edu/rsgrp/groupsheets/bondenergies.pdf</ref>

The high electronegativity difference gives H-F a partial ionic character, making it stronger and more energetically stable.

F+H2 are found to react exothermically: ''The reactants are in a higher energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of formation of the more stable and thus, more negative value of H-F compared with the more positive dissociation energy associated with H-H leads to an overall negative energy change. Energy flows out of the system.''

H and HF react endothermically: ''The reactants are in a lower energy state than the products. This is a direct consequence of the more energetically stable bond of H-F in comparison with H-H. The enthalpy of more positive enthalpy of dissociation of the stronger H-F bond is not compensated for by the smaller negative value formation of H-H. Energy flows out of the system.''

The approximate position of the transition state can be located using Hammond’s postulate which states that the transition state of a reaction will resemble either the reactants or the products depending on which one it is energetically closer to. Thus, for exothermic reactions (F+H2) it will resemble the reactants and for endothermic reactions (H+HF) it will resemble the products. In the case of these reactions, the transition state corresponds to H-F-H complex which will always be closer to the H-H bond as it is higher in energy. Bond distances close to this energy were tested so that the contour plot showed no oscillations (Figure 14) and the energy plot (Figure 15) showed relative kinetic energy of zero compared with the total energy of the system because the system will be at its maximum potential energy. The transitions state values that best match this were found to be H-F = 0.74631 Å and H-H = 1.81009 Å. he transition state for the H + HF would also be the same as both reactions form the same [H-H-F]‡ complex. Thus it is expected that H-H = 0.74631 Å and H-F = 1.81009 Å also correspond to the transition state for the F+H2 reaction and this is confirmed from the plots. (Figure 16) and (Figure 17).

{|
||[[File:as11815Fig14.png|300px|thumb|''Figure 14:'' Contour plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|[[File:as11815Fig15.png|300px|thumb|''Figure 15:'' Energy plot for H-F = 0.74631 Å and H-H = 1.81009 Å]]
|-
|[[File:as11815Fig16.png|300px|thumb|''Figure 16:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|[[File:as11815Fig17.png|300px|thumb|''Figure 17:'' Contour plot for H-H = 0.74631 Å and H-F = 1.81009 Å]]
|}

The activation energy is the difference in energy between the maximum of the transition state and the minimum of the reactants. The transition state was found to occur at '''-103.75 Kcal/mol'''

Acitvation energy = energy of transition state – energy of reactants
F+H2 reaction: Activation energy = '''-103.75 Kcal/mo'''l – (-104)
HF+H reaction: Activation energy = '''-103.75 Kcal/mol''' – (-136)

<ref>“Organic Chemistry”, Clayden, Greeves, Warren and Wothers, (Oxford University Press, 2001)</ref>

===Reaction dynamics===
(I already chose different initial conditions)
The initial conditions that result in a reactive trajectory for the F+H2 were found to be: r1 =0.76, r2 = 2.2, p1 = -1.0, p2 = -1.7.
At the start of the reaction, the H-H molecule, (represented by B-C on the plot), oscillates. After collision, the reaction occurs and the H-F bond that is formed (represented by A-B on the plot) oscillates continuously. By the law of conservation of energy, the energy released in the exothermic reaction is converted from potential energy to kinetic energy. This can further be seen on the internuclear momenta vs time graph where the product H-F molecule’s oscillations are of a higher amplitude than H-H. The higher vibrational motion of H-F can be experimentally using several methods. IR spectroscopy can be used to check that H-F shows overtone bands due to the higher vibrational energy. Calorimetry can be used to calculate changes in enthalpy as a function of reaction coordinate. Emission spectroscopy to determine the photons released by vibrational molecules could further be used to determine vibrational energies.
===Polanyi’s empirical rules ===
In order for a chemical reaction to occur, the reactant must overcome the activation barrier of the transition state to form the products using its translational and vibrational energies. Polanyi’s Empirical Rules state that translational energy is more effective than vibrational when promoting an early transition barrier, whilst vibrational energy is more effective than translational energy when promoting a late transitional barrier. <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

The F+H2 is exothermic and found to have an early transition state by Hammond’s postulate, so an increase in translational energy would be expected to have a larger effect than vibrational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (low) = 2.0, and p2 = translational energy (high) = -6, produces a reactive reaction (Figure A), whereas as p1 = vibrational energy (high) = 6, and p2 = translational energy (low) = -2, produces an unreactive reaction (figure B). Thus, it is confirmed that high translational energies result in successful exothermic reactions.

The F+H2 is endothermic and found to have a late transition state by Hammond’s postulate, so an increase in vibrational energy would be expected to have a larger effect than translational energy on the reactivity. The surface plot obeys Polanyi’s rules as p1 = vibrational energy (high) = 6.0, and p2 = translational energy (low) = -2, produces a reactive reaction (Figure A), whereas as p1 = vibrational energy (low) = 2, and p2 = translational energy (high) = -6, produces an unreactive reaction (figure B). Thus, it is confirmed that high vibrational energies result in successful endothermic reactions.