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MRD:00948108ab6114

2017-05-26T15:29:57Z

Ab6114: Created page with "=='''Molecular Reaction Dynamics'''== ==Exercise 1: H + H2 system== ==='''Question(1)'''=== What value does the total gradient of the potential energy surface have at a mi..."

=='''Molecular Reaction Dynamics'''==

==Exercise 1: H + H2 system==

==='''Question(1)'''===

What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

'''Answer:'''

Analysis of the PES curvature shows a maxima between two minimum points, all being stationary points with a gradient of zero. The first derivative of the curve, where the rate of change is zero, is found to confirm this. As Transition structure's inherently exist at the maximum point, it can be inferred that the transition structure lies exists at this point. In order to differentiate between the transition state and the minima, second derivative calculations can be performed, producing a negative value for a maxima and a positive value for a minima.<ref name="transition" />

----

==='''Question(2)'''===

Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.

'''Answer:'''

The approximate transition state position (r(ts)) was determined through analysis of the surface plot. This was confirmed by examining the internuclear distance vs time plot, to identify the point of intersection between the maxima and the minima - r(ts) is shown in the table below:

{| class="wikitable" border="1"
|+ Table 1
! r1 !! r2 !! r(ts)
|-
| 1.20 || 1.20 || 0.9075
|-
| 0.91 || 0.91 || 0.9075
|}

[[File:ab6114Figure1.PNG|680px|thumb|centre|Figure 1 - Internuclear distance vs Time plot of transition state position]]

A transition state is a point where the internuclear distance between all bonds remain constant and thus do not oscillate. Figure 1 shows that the internuclear distance does not oscillate at 0.9075 Å - hence this is the transition state position.

----

==='''Question(3)'''===

Comment on how the mep and the trajectory you just calculated differ.

'''Answer:'''

Dynamic calculations and mep are distinct for two main reasons. Calculations of mep span only the distance engaged by the transition state, while the dynamic calculation incorporates more of the reaction pathway - the dynamic calculation also takes mass into account, thus resulting in an oscillating reaction pathway line. No mass is included in the mep calculation, so no oscillation is observed.

----

==='''Question(4)'''===

Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.

'''Answer:'''

The effect of momentum variation on whether a trajectory is reactive or unreactive (for the initial positions r1 = 0.74 and r2 = 2.0):

{| class="wikitable" border="1"
|+ Table 2
! p1 !! p2 !! Trajectory !! Figure No.
|-
| -1.25 || -2.5 || Reactive || 2
|-
| -1.5 || -2.0 || Unreactive || 3
|-
| -1.5 || -2.5 || Reactive || 4
|-
| -2.5 || -5.0 || Unreactive || 5
|-
| -2.5 || -5.2 || Reactive || 6
|}

'''Figure 2'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 2. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc while at the same time the B-C bond distance remains relatively constant. This line continues through the transition state (r(A-B)=r(B-C)) and r(A-B) reaches a minimum and oscillates thus indicating that the Ha-Hb bond has formed. Simultaneously r(B-C) increases, confirming that the Hb-Hc bond has broken and Hc now exists as a separate hydrogen atom.

[[File:Ab6114_figure2.PNG|680px|thumb|centre|Figure 2 - The reactive trajectory associated with the specific parameters for Figure 2]]

'''Figure 3'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 3. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc. Since Ha has a momentum insufficient to overcome the activation energy barrier, the trajectory fails to reach the transition state - this is shown by the plot through r(A-B) returning to a larger distance without reaching transition state, and r(B-C) only oscillating due to vibrational fluctuations within the molecule.

[[File:Ab6114_figure3test2.PNG|680px|thumb|centre|Figure 3 - The unreactive trajectory associated with the specific parameters for Figure 3]]

'''Figure 4'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 4. The trajectory line illustrates a reaction where Ha had a momentum sufficient to overcome the activation energy barrier required to successfully interact with Hb-Hc, pass through the transition state, and form the molecule Ha-Hb.

[[File:Ab6114_figure4.PNG|680px|thumb|centre|Figure 4 - The reactive trajectory associated with the specific parameters for Figure 4]]

'''Figure 5'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 5. The trajectory line for these set parameters, initially show Ha to approach Hb-Hc with sufficient momentum to overcome the activation energy barrier and reach the transition state. However, instead of forming the product Ha-Hb, the trajectory line reverses and returns back to the reactants, with Hb-Hc exhibiting large vibrational energy fluctuations.

[[File:Ab6114_figure5.PNG|680px|thumb|centre|Figure 5 - The unreactive trajectory associated with the specific parameters for Figure 5]]

'''Figure 6'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 6. The trajectory line illustrates that Ha has sufficient momentum to overcome the activation energy barrier. However after reaching the transition state, the trajectory line goes back towards the reactants which is shown by r(B-C) decreasing and r(A-B) increasing in figure 6. The r(A-B) rose to a maxima before decreasing and oscillating, thus illustrating the formation of Ha-Hb. This reaction trajectory suggests that potentially an alternative reaction pathway was taken, one which does not involve a 'saddle point' on the PE plot corresponding with the transition states that normally produce the desired Ha and Hb-Hc.

[[File:Ab6114_figure6.PNG|680px|thumb|centre|Figure 6 - The reactive trajectory associated with the specific parameters for Figure 6]]

----

==='''Question(5)'''===

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

'''Answer:'''

The Transition State Theory's main assumption is that the reactants and activated transition states are at quasi-equilibrium. A quasi-equilibrium utilises thermodynamic molecular driving forces that modify the systems state being very small and slow, allowing for the existence of an internal equilibrium that undergoes only small deviations from this system. Classical Transition State Theory explains that an atom with a momentum satisfying the activation energy barrier requirements, will react upon collision with a given molecule and synthesize products. As this theory however ignores potential quantum mechanical interactions, reaction rates calculated will be slower than what is found in experimental data. Quantum mechanical contributing calculations allow for the tunneling behaviour of particles to be factored into an alternative reaction pathway. When an atom and a molecule lack the sufficient activation energy to overcome this barrier, the atom is able to 'tunnel' through this barrier, due to the particle-wave duality observed in quantum-scale objects.<ref name="thermodynamics" />

----

==Exercise 2: F - H - H system==

==='''Question(6)'''===

Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

'''Answer:'''

Analysis:

- H-F Bond Energy = 565 KJ/mol<ref name="bond energies" />

- H-H Bond Energy = 432 KJ/mol<ref name="bond energies" />

- H-F is a stronger bond than H-H

- ΔE = E(bonds cleaved)-E(bonds formed)

'''F + H2''' --> ΔE = 432 - 565 = -133 KJ/mol '''THUS EXOTHERMIC'''

'''H + H-F''' --> ΔE = 565 - 432 = 133 KJ/mol '''THUS ENDOTHERMIC'''

----

==='''Question(7)'''===

Locate the approximate position of the transition state.

'''Answer:'''

In order to locate the position of the transition state, the radii corresponding to a stationary total system energy (gradient=0) were identified. The positions of the radii have a zero net force acting upon them, so plotting the internuclear distance vs time results in a non-oscillating constant internuclear separation - see the internuclear separations below and the corresponding plot vs time in figure 7:

- r(H-F) = 1.815Å

- r(H-H) = 0.74Å

[[File:ab6114 figure7.PNG|680px|thumb|centre|Figure 7 - Internuclear distance vs Time plot for the transition state position]]

----

==='''Question(8)'''===

Report the activation energy for both reactions.

'''Answer:'''

Determining the activation energy for both reactions, required chemical intuition and trial and error in order to identify an adequate reaction trajectory. The activation energy is equal to the energy difference between the reactants and the transition state - activation energy values calculated in absence of mathematical analysis, thus all energy values involved would actually have +/- values.

'''Reaction 1: F + H2 --> HF + H'''

- Ea = -103.7 - (-103.9) = 0.2 kcal/mol

Energy values used for this calculation were obtained from Figure 8 (reactants) and Figure 9 (transition state) below:

[[File:ab6114 figure8.PNG|470px|thumb|centre|Figure 8 - Surface plot illustrating energy of the reactants for F + H2 --> HF + H]]

[[File:ab6114 figure9.PNG|470px|thumb|centre|Figure 9 - Surface plot illustrating the energy of the transition state]]

'''Reaction 2: H + HF --> F + H2'''

- Ea = -103.7 - (-133.9) = 30.2 kcal/mol

Energy values used for this calculation were obtained from Figure 10 (reactants) and Figure 9 (transition state) below:

[[File:ab6114 figure9.PNG|470px|thumb|centre|Figure 9 - Surface plot illustrating the energy of the transition state]]

[[File:ab6114 figure10.PNG|470px|thumb|centre|Figure 10 - Surface plot showing energy of the reactants for H + HF --> F + H2]]

----

==='''Question(9)'''===

In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

'''Answer:'''

The Conservation of Energy Law <ref name="conservationofE" /> dictates that the total energy of an isolated system must be conserved, as energy cannot be created or destroyed, and is thus just converted from one form to another. For this reaction, the total energy is the sum of the kinetic and potential energies of each reactant involved in the reaction. For the exothermic reaction F + H2, the potential energy of the products is less than the reactants thus meaning that the kinetic energy of the products must be higher - this is illustrated by identifying a successful reaction trajectory and plotting a kinetic energy vs time graph as shown below in figure 11:

[[File:ab6114 figure11.PNG|470px|thumb|centre|Figure 11 - A successful reaction pathway determined through trial and error]]

[[File:ab6114 figure12.PNG|470px|thumb|centre|Figure 12 - Kinetic energy vs Time plot of the successful reaction pathway]]

As the kinetic energy increases, products from the reaction can be analysed using infrared spectroscopy - oscillations of the H-F bond exhibit vibrational energy that is illustrated by a peak on an IR spectrum, while the H-H reactant peak will be missing. Calorimetry could also be used for confirmation. As its an exothermic reaction, the heat released can be measured (at constant pressure), the enthalpy change equates to the net energy change, thus the reactions net free energy change can be found.

----

==References==

<references>
<ref name="transition">P. Jonathan P. K. Doye, David J. Wales, Journal of mathematical physics, 1996, '''48''', 843 - 846.</ref>
<ref name="thermodynamics">Chem Libre texts - Wave-particle duality (2015) [online] Available at: https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Quantum_Mechanics/09._The_Hydrogen_Atom/Atomic_Theory/Electrons_in_Atoms/Wave-Particle_Duality [Assessed 28 Sep 2015].</ref>
<ref name="bond energies">The Department of Chemistry Illinois (2017). Average Bond Energies [online] Available at: http://butane.chem.uiuc.edu/cyerkes/Chem104ACSpring2009/Genchemref/bondenergies.html [Accessed 9 Feb 2017].</ref>
<ref name="conservationofE">Hiebert, E.N. (1981). Historical Roots of the Principle of Conservation of Energy. Madison, Wis.: Ayer Co Pub.</ref>

MRD:ab6114

2017-05-26T15:27:22Z

Ab6114:

File:Ab6114 figure12.PNG

2017-05-26T15:08:39Z

Ab6114:

File:Ab6114 figure11.PNG

2017-05-26T15:07:16Z

Ab6114:

MRD:ab6114

2017-05-26T15:01:58Z

Ab6114:

=='''Molecular Reaction Dynamics'''==

==Exercise 1: H + H2 system==

==='''Question(1)'''===

What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

'''Answer:'''

Analysis of the PES curvature shows a maxima between two minimum points, all being stationary points with a gradient of zero. The first derivative of the curve, where the rate of change is zero, is found to confirm this. As Transition structure's inherently exist at the maximum point, it can be inferred that the transition structure lies exists at this point. In order to differentiate between the transition state and the minima, second derivative calculations can be performed, producing a negative value for a maxima and a positive value for a minima.<ref name="transition" />

----

==='''Question(2)'''===

Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.

'''Answer:'''

The approximate transition state position (r(ts)) was determined through analysis of the surface plot. This was confirmed by examining the internuclear distance vs time plot, to identify the point of intersection between the maxima and the minima - r(ts) is shown in the table below:

{| class="wikitable" border="1"
|+ Table 1
! r1 !! r2 !! r(ts)
|-
| 1.20 || 1.20 || 0.9075
|-
| 0.91 || 0.91 || 0.9075
|}

[[File:ab6114Figure1.PNG|680px|thumb|centre|Figure 1 - Internuclear distance vs Time plot of transition state position]]

A transition state is a point where the internuclear distance between all bonds remain constant and thus do not oscillate. Figure 1 shows that the internuclear distance does not oscillate at 0.9075 Å - hence this is the transition state position.

----

==='''Question(3)'''===

Comment on how the mep and the trajectory you just calculated differ.

'''Answer:'''

Dynamic calculations and mep are distinct for two main reasons. Calculations of mep span only the distance engaged by the transition state, while the dynamic calculation incorporates more of the reaction pathway - the dynamic calculation also takes mass into account, thus resulting in an oscillating reaction pathway line. No mass is included in the mep calculation, so no oscillation is observed.

----

==='''Question(4)'''===

Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.

'''Answer:'''

The effect of momentum variation on whether a trajectory is reactive or unreactive (for the initial positions r1 = 0.74 and r2 = 2.0):

{| class="wikitable" border="1"
|+ Table 2
! p1 !! p2 !! Trajectory !! Figure No.
|-
| -1.25 || -2.5 || Reactive || 2
|-
| -1.5 || -2.0 || Unreactive || 3
|-
| -1.5 || -2.5 || Reactive || 4
|-
| -2.5 || -5.0 || Unreactive || 5
|-
| -2.5 || -5.2 || Reactive || 6
|}

'''Figure 2'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 2. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc while at the same time the B-C bond distance remains relatively constant. This line continues through the transition state (r(A-B)=r(B-C)) and r(A-B) reaches a minimum and oscillates thus indicating that the Ha-Hb bond has formed. Simultaneously r(B-C) increases, confirming that the Hb-Hc bond has broken and Hc now exists as a separate hydrogen atom.

[[File:Ab6114_figure2.PNG|680px|thumb|centre|Figure 2 - The reactive trajectory associated with the specific parameters for Figure 2]]

'''Figure 3'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 3. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc. Since Ha has a momentum insufficient to overcome the activation energy barrier, the trajectory fails to reach the transition state - this is shown by the plot through r(A-B) returning to a larger distance without reaching transition state, and r(B-C) only oscillating due to vibrational fluctuations within the molecule.

[[File:Ab6114_figure3test2.PNG|680px|thumb|centre|Figure 3 - The unreactive trajectory associated with the specific parameters for Figure 3]]

'''Figure 4'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 4. The trajectory line illustrates a reaction where Ha had a momentum sufficient to overcome the activation energy barrier required to successfully interact with Hb-Hc, pass through the transition state, and form the molecule Ha-Hb.

[[File:Ab6114_figure4.PNG|680px|thumb|centre|Figure 4 - The reactive trajectory associated with the specific parameters for Figure 4]]

'''Figure 5'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 5. The trajectory line for these set parameters, initially show Ha to approach Hb-Hc with sufficient momentum to overcome the activation energy barrier and reach the transition state. However, instead of forming the product Ha-Hb, the trajectory line reverses and returns back to the reactants, with Hb-Hc exhibiting large vibrational energy fluctuations.

[[File:Ab6114_figure5.PNG|680px|thumb|centre|Figure 5 - The unreactive trajectory associated with the specific parameters for Figure 5]]

'''Figure 6'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 6. The trajectory line illustrates that Ha has sufficient momentum to overcome the activation energy barrier. However after reaching the transition state, the trajectory line goes back towards the reactants which is shown by r(B-C) decreasing and r(A-B) increasing in figure 6. The r(A-B) rose to a maxima before decreasing and oscillating, thus illustrating the formation of Ha-Hb. This reaction trajectory suggests that potentially an alternative reaction pathway was taken, one which does not involve a 'saddle point' on the PE plot corresponding with the transition states that normally produce the desired Ha and Hb-Hc.

[[File:Ab6114_figure6.PNG|680px|thumb|centre|Figure 6 - The reactive trajectory associated with the specific parameters for Figure 6]]

----

==='''Question(5)'''===

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

'''Answer:'''

The Transition State Theory's main assumption is that the reactants and activated transition states are at quasi-equilibrium. A quasi-equilibrium utilises thermodynamic molecular driving forces that modify the systems state being very small and slow, allowing for the existence of an internal equilibrium that undergoes only small deviations from this system. Classical Transition State Theory explains that an atom with a momentum satisfying the activation energy barrier requirements, will react upon collision with a given molecule and synthesize products. As this theory however ignores potential quantum mechanical interactions, reaction rates calculated will be slower than what is found in experimental data. Quantum mechanical contributing calculations allow for the tunneling behaviour of particles to be factored into an alternative reaction pathway. When an atom and a molecule lack the sufficient activation energy to overcome this barrier, the atom is able to 'tunnel' through this barrier, due to the particle-wave duality observed in quantum-scale objects.<ref name="thermodynamics" />

----

==Exercise 2: F - H - H system==

==='''Question(6)'''===

Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

'''Answer:'''

Analysis:

- H-F Bond Energy = 565 KJ/mol<ref name="bond energies" />

- H-H Bond Energy = 432 KJ/mol<ref name="bond energies" />

- H-F is a stronger bond than H-H

- ΔE = E(bonds cleaved)-E(bonds formed)

'''F + H2''' --> ΔE = 432 - 565 = -133 KJ/mol '''THUS EXOTHERMIC'''

'''H + H-F''' --> ΔE = 565 - 432 = 133 KJ/mol '''THUS ENDOTHERMIC'''

----

==='''Question(7)'''===

Locate the approximate position of the transition state.

'''Answer:'''

In order to locate the position of the transition state, the radii corresponding to a stationary total system energy (gradient=0) were identified. The positions of the radii have a zero net force acting upon them, so plotting the internuclear distance vs time results in a non-oscillating constant internuclear separation - see the internuclear separations below and the corresponding plot vs time in figure 7:

- r(H-F) = 1.815Å

- r(H-H) = 0.74Å

[[File:ab6114 figure7.PNG|680px|thumb|centre|Figure 7 - Internuclear distance vs Time plot for the transition state position]]

----

==='''Question(8)'''===

Report the activation energy for both reactions.

'''Answer:'''

Determining the activation energy for both reactions, required chemical intuition and trial and error in order to identify an adequate reaction trajectory. The activation energy is equal to the energy difference between the reactants and the transition state - activation energy values calculated in absence of mathematical analysis, thus all energy values involved would actually have +/- values.

'''Reaction 1: F + H2 --> HF + H'''

- Ea = -103.7 - (-103.9) = 0.2 kcal/mol

Energy values used for this calculation were obtained from Figure 8 (reactants) and Figure 9 (transition state) below:

[[File:ab6114 figure8.PNG|470px|thumb|centre|Figure 8 - Surface plot illustrating energy of the reactants for F + H2 --> HF + H]]

[[File:ab6114 figure9.PNG|470px|thumb|centre|Figure 9 - Surface plot illustrating the energy of the transition state]]

'''Reaction 2: H + HF --> F + H2'''

- Ea = -103.7 - (-133.9) = 30.2 kcal/mol

Energy values used for this calculation were obtained from Figure 10 (reactants) and Figure 9 (transition state) below:

[[File:ab6114 figure9.PNG|470px|thumb|centre|Figure 9 - Surface plot illustrating the energy of the transition state]]

[[File:ab6114 figure10.PNG|470px|thumb|centre|Figure 10 - Surface plot showing energy of the reactants for H + HF --> F + H2]]

----

==='''Question(9)'''===

In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

'''Answer:'''

The Conservation of Energy Law <ref name="conservationofE" /> dictates that the total energy of an isolated system must be conserved, as energy cannot be created or destroyed, and is thus just converted from one form to another. For this reaction, the total energy is the sum of the kinetic and potential energies of each reactant involved in the reaction. For the exothermic reaction F + H2, the potential energy of the products is less than the reactants thus meaning that the kinetic energy of the products must be higher - this is illustrated by identifying a successful reaction trajectory and plotting a kinetic energy vs time graph as shown below in figure 11:

[[File:ab6114 figure11.PNG|470px|thumb|centre|Figure 11 - A successful reaction pathway determined through trial and error]]

[[File:ab6114 figure12.PNG|470px|thumb|centre|Figure 12 - Kinetic energy vs Time plot of the successful reaction pathway]]

The increased kinetic energy allows the products to be detected by IR analysis. Since the H-F bond is oscillating, it will release energy in the form of vibrational energy, resulting in a peak on the spectrum. The H-H vibrational peak present on the reactant spectrum will now be absent. Another possible experimental technique is calorimetry. Since the reaction is exothermic, the amount of heat released can be measured and since, at constant pressure, the enthalpy change is equal to the net change in energy, the net free energy change of the reaction can be determined.

----

==='''Q(10)'''===

Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

'''A:'''

Using trial and error, a successful reaction trajectory was found for both reactions shown below. The parameters used can be seen within the figures.

'''Reaction 1: F + H2 --> HF + H'''

'''Lower momentum'''

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 13 - An ultimately successful reaction pathway at high momentum indicating the preference of an early transition state]]

'''Higher momentum'''

[[File:HZ4315 Figure12.PNG|500px|thumb|centre|Figure 14 - The reaction pathway at an even larger momentum to confirm the theory]]

Reaction 1, which is an exothermic reaction, exhibits an increased translational energy in response to increasing the momentum of the reactants. According to Polanyi's rules, higher translational energies are more efficient at activating an early transition state (ie. an early barrier) than vibrational energy.<ref name="polanyi" /> Since early transition states resemble the reactants (as stated by Hammond's Postulate)<ref name="hammond" />, the reaction trajectory may be expected to bounce back towards the reactants. Figure 12 confirms this as an increased momentum (and therefore increased translational energy) sees the trajectory bounce back several times before eventually proceeding to the form the products. Figures 11 and 12 are an example of a set of inefficient reaction trajectories as the barrier is recrossed several times which is not ideal in the formation of the product.

The exothermic nature of the reaction also alters the position of the transition state away from the lowest 'saddle point'. This goes against the assumptions of the Transition State Theory which states that reaction trajectories should pass through this. As temperature increases, a greater proportion of excited vibrational states become occupied, leading to transition states occuring at different energies. As a result, the associated reaction trajectory at higher temperature travels through different activated complexes, making the reaction harder to model and any associated results more difficult to compare.

'''Reaction 2: H + HF --> F + H2'''

Reaction 2, which is the reverse reaction of Reaction 1, is endothermic with the energy of the products being higher than the reactants. As a result, according the Hammond's Postulate, the position of the transition state is later in the reaction trajectory and more closely resembles the products. Contrary to the preferred conditions for exothermic reactions, a larger amount of vibrational energy is needed. To achieve this, lower momentum would be adopted. Providing this is achieved, the set of reaction trajectories would be an example of an efficient reaction as the reaction trajectories would pass through the transition state once and to the products.

----

==References==

<references>
<ref name="transition">P. Jonathan P. K. Doye, David J. Wales, Journal of mathematical physics, 1996, '''48''', 843 - 846.</ref>
<ref name="thermodynamics">Chem Libre texts - Wave-particle duality (2015) [online] Available at: https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Quantum_Mechanics/09._The_Hydrogen_Atom/Atomic_Theory/Electrons_in_Atoms/Wave-Particle_Duality [Assessed 28 Sep 2015].</ref>
<ref name="bond energies">The Department of Chemistry Illinois (2017). Average Bond Energies [online] Available at: http://butane.chem.uiuc.edu/cyerkes/Chem104ACSpring2009/Genchemref/bondenergies.html [Accessed 9 Feb 2017].</ref>
<ref name="conservationofE">Hiebert, E.N. (1981). Historical Roots of the Principle of Conservation of Energy. Madison, Wis.: Ayer Co Pub.</ref>
<ref name="polanyi">Yan, S., Y.-T. Wu, and K. Liu. "Tracking The Energy Flow Along The Reaction Path". Proceedings of the National Academy of Sciences 105.35 (2008): 12667-12672.</ref>
<ref name="hammond">Chemistry LibreTexts. (2017). Hammond’s Postulate. [online] Available at: https://chem.libretexts.org/Reference/Organic_Chemistry_Glossary/Hammond’s_Postulate [Accessed 11 May 2017].</ref>
<references/>

File:Ab6114 figure10.PNG

2017-05-26T14:30:46Z

Ab6114:

MRD:ab6114

2017-05-26T14:30:18Z

Ab6114:

=='''Molecular Reaction Dynamics'''==

==Exercise 1: H + H2 system==

==='''Question(1)'''===

What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

'''Answer:'''

Analysis of the PES curvature shows a maxima between two minimum points, all being stationary points with a gradient of zero. The first derivative of the curve, where the rate of change is zero, is found to confirm this. As Transition structure's inherently exist at the maximum point, it can be inferred that the transition structure lies exists at this point. In order to differentiate between the transition state and the minima, second derivative calculations can be performed, producing a negative value for a maxima and a positive value for a minima.<ref name="transition" />

----

==='''Question(2)'''===

Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.

'''Answer:'''

The approximate transition state position (r(ts)) was determined through analysis of the surface plot. This was confirmed by examining the internuclear distance vs time plot, to identify the point of intersection between the maxima and the minima - r(ts) is shown in the table below:

{| class="wikitable" border="1"
|+ Table 1
! r1 !! r2 !! r(ts)
|-
| 1.20 || 1.20 || 0.9075
|-
| 0.91 || 0.91 || 0.9075
|}

[[File:ab6114Figure1.PNG|680px|thumb|centre|Figure 1 - Internuclear distance vs Time plot of transition state position]]

A transition state is a point where the internuclear distance between all bonds remain constant and thus do not oscillate. Figure 1 shows that the internuclear distance does not oscillate at 0.9075 Å - hence this is the transition state position.

----

==='''Question(3)'''===

Comment on how the mep and the trajectory you just calculated differ.

'''Answer:'''

Dynamic calculations and mep are distinct for two main reasons. Calculations of mep span only the distance engaged by the transition state, while the dynamic calculation incorporates more of the reaction pathway - the dynamic calculation also takes mass into account, thus resulting in an oscillating reaction pathway line. No mass is included in the mep calculation, so no oscillation is observed.

----

==='''Question(4)'''===

Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.

'''Answer:'''

The effect of momentum variation on whether a trajectory is reactive or unreactive (for the initial positions r1 = 0.74 and r2 = 2.0):

{| class="wikitable" border="1"
|+ Table 2
! p1 !! p2 !! Trajectory !! Figure No.
|-
| -1.25 || -2.5 || Reactive || 2
|-
| -1.5 || -2.0 || Unreactive || 3
|-
| -1.5 || -2.5 || Reactive || 4
|-
| -2.5 || -5.0 || Unreactive || 5
|-
| -2.5 || -5.2 || Reactive || 6
|}

'''Figure 2'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 2. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc while at the same time the B-C bond distance remains relatively constant. This line continues through the transition state (r(A-B)=r(B-C)) and r(A-B) reaches a minimum and oscillates thus indicating that the Ha-Hb bond has formed. Simultaneously r(B-C) increases, confirming that the Hb-Hc bond has broken and Hc now exists as a separate hydrogen atom.

[[File:Ab6114_figure2.PNG|680px|thumb|centre|Figure 2 - The reactive trajectory associated with the specific parameters for Figure 2]]

'''Figure 3'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 3. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc. Since Ha has a momentum insufficient to overcome the activation energy barrier, the trajectory fails to reach the transition state - this is shown by the plot through r(A-B) returning to a larger distance without reaching transition state, and r(B-C) only oscillating due to vibrational fluctuations within the molecule.

[[File:Ab6114_figure3test2.PNG|680px|thumb|centre|Figure 3 - The unreactive trajectory associated with the specific parameters for Figure 3]]

'''Figure 4'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 4. The trajectory line illustrates a reaction where Ha had a momentum sufficient to overcome the activation energy barrier required to successfully interact with Hb-Hc, pass through the transition state, and form the molecule Ha-Hb.

[[File:Ab6114_figure4.PNG|680px|thumb|centre|Figure 4 - The reactive trajectory associated with the specific parameters for Figure 4]]

'''Figure 5'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 5. The trajectory line for these set parameters, initially show Ha to approach Hb-Hc with sufficient momentum to overcome the activation energy barrier and reach the transition state. However, instead of forming the product Ha-Hb, the trajectory line reverses and returns back to the reactants, with Hb-Hc exhibiting large vibrational energy fluctuations.

[[File:Ab6114_figure5.PNG|680px|thumb|centre|Figure 5 - The unreactive trajectory associated with the specific parameters for Figure 5]]

'''Figure 6'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 6. The trajectory line illustrates that Ha has sufficient momentum to overcome the activation energy barrier. However after reaching the transition state, the trajectory line goes back towards the reactants which is shown by r(B-C) decreasing and r(A-B) increasing in figure 6. The r(A-B) rose to a maxima before decreasing and oscillating, thus illustrating the formation of Ha-Hb. This reaction trajectory suggests that potentially an alternative reaction pathway was taken, one which does not involve a 'saddle point' on the PE plot corresponding with the transition states that normally produce the desired Ha and Hb-Hc.

[[File:Ab6114_figure6.PNG|680px|thumb|centre|Figure 6 - The reactive trajectory associated with the specific parameters for Figure 6]]

----

==='''Question(5)'''===

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

'''Answer:'''

The Transition State Theory's main assumption is that the reactants and activated transition states are at quasi-equilibrium. A quasi-equilibrium utilises thermodynamic molecular driving forces that modify the systems state being very small and slow, allowing for the existence of an internal equilibrium that undergoes only small deviations from this system. Classical Transition State Theory explains that an atom with a momentum satisfying the activation energy barrier requirements, will react upon collision with a given molecule and synthesize products. As this theory however ignores potential quantum mechanical interactions, reaction rates calculated will be slower than what is found in experimental data. Quantum mechanical contributing calculations allow for the tunneling behaviour of particles to be factored into an alternative reaction pathway. When an atom and a molecule lack the sufficient activation energy to overcome this barrier, the atom is able to 'tunnel' through this barrier, due to the particle-wave duality observed in quantum-scale objects.<ref name="thermodynamics" />

----

==Exercise 2: F - H - H system==

==='''Question(6)'''===

Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

'''Answer:'''

Analysis:

- H-F Bond Energy = 565 KJ/mol<ref name="bond energies" />

- H-H Bond Energy = 432 KJ/mol<ref name="bond energies" />

- H-F is a stronger bond than H-H

- ΔE = E(bonds cleaved)-E(bonds formed)

'''F + H2''' --> ΔE = 432 - 565 = -133 KJ/mol '''THUS EXOTHERMIC'''

'''H + H-F''' --> ΔE = 565 - 432 = 133 KJ/mol '''THUS ENDOTHERMIC'''

----

==='''Question(7)'''===

Locate the approximate position of the transition state.

'''Answer:'''

In order to locate the position of the transition state, the radii corresponding to a stationary total system energy (gradient=0) were identified. The positions of the radii have a zero net force acting upon them, so plotting the internuclear distance vs time results in a non-oscillating constant internuclear separation - see the internuclear separations below and the corresponding plot vs time in figure 7:

- r(H-F) = 1.815Å

- r(H-H) = 0.74Å

[[File:ab6114 figure7.PNG|680px|thumb|centre|Figure 7 - Internuclear distance vs Time plot for the transition state position]]

----

==='''Question(8)'''===

Report the activation energy for both reactions.

'''Answer:'''

Determining the activation energy for both reactions, required chemical intuition and trial and error in order to identify an adequate reaction trajectory. The activation energy is equal to the energy difference between the reactants and the transition state - activation energy values calculated in absence of mathematical analysis, thus all energy values involved would actually have +/- values.

'''Reaction 1: F + H2 --> HF + H'''

- Ea = -103.7 - (-103.9) = 0.2 kcal/mol

Energy values used for this calculation were obtained from Figure 8 (reactants) and Figure 9 (transition state) below:

[[File:ab6114 figure8.PNG|470px|thumb|centre|Figure 8 - Surface plot illustrating energy of the reactants]]

[[File:ab6114 figure9.PNG|470px|thumb|centre|Figure 9 - Surface plot illustrating the energy of the transition state]]

'''Reaction 2: H + HF --> F + H2'''

- Ea = -103.7 - (-133.9) = 30.2 kcal/mol

Energy values used for this calculation were obtained from Figure 10 (reactants) and Figure 9 (transition state) below:

[[File:ab6114 figure9.PNG|470px|thumb|centre|Figure 9 - Surface plot illustrating the energy of the transition state]]

[[File:ab6114 figure10.PNG|470px|thumb|centre|Figure 10 - Surface plot showing energy of the reactants for H + HF --> F + H2]]

----

==='''Q(9)'''===

In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

'''A:'''

According to the Law of Conservation of Energy, the total energy of an isolated system must be conserved as energy is neither created nor destroyed, just converted from one form to another.<ref name="lawE" /> For this specific reaction system, the total energy is the sum of the kinetic and potential energies of each substrate involved in the reaction. For the exothermic reaction F + H2, the potential energy of the products is lower than the reactants hence, the kinetic energy of the products must be higher. This is illustrated by determining a successful reaction trajectory and plotting a kinetic energy vs time graph as shown below:

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 11 - A reaction pathway for an ultimately successful reaction (parameters determined using trial and error)]]

[[File:KEvsTIME.PNG|500px|thumb|centre|Figure 12 - Kinetic energy vs Time plot of the successful reaction - the small oscillations on the left correspond to the reactants which are clearly converted to products as the kinetic energy increases significantly]]

The increased kinetic energy allows the products to be detected by IR analysis. Since the H-F bond is oscillating, it will release energy in the form of vibrational energy, resulting in a peak on the spectrum. The H-H vibrational peak present on the reactant spectrum will now be absent. Another possible experimental technique is calorimetry. Since the reaction is exothermic, the amount of heat released can be measured and since, at constant pressure, the enthalpy change is equal to the net change in energy, the net free energy change of the reaction can be determined.

----

==='''Q(10)'''===

Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

'''A:'''

Using trial and error, a successful reaction trajectory was found for both reactions shown below. The parameters used can be seen within the figures.

'''Reaction 1: F + H2 --> HF + H'''

'''Lower momentum'''

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 13 - An ultimately successful reaction pathway at high momentum indicating the preference of an early transition state]]

'''Higher momentum'''

[[File:HZ4315 Figure12.PNG|500px|thumb|centre|Figure 14 - The reaction pathway at an even larger momentum to confirm the theory]]

Reaction 1, which is an exothermic reaction, exhibits an increased translational energy in response to increasing the momentum of the reactants. According to Polanyi's rules, higher translational energies are more efficient at activating an early transition state (ie. an early barrier) than vibrational energy.<ref name="polanyi" /> Since early transition states resemble the reactants (as stated by Hammond's Postulate)<ref name="hammond" />, the reaction trajectory may be expected to bounce back towards the reactants. Figure 12 confirms this as an increased momentum (and therefore increased translational energy) sees the trajectory bounce back several times before eventually proceeding to the form the products. Figures 11 and 12 are an example of a set of inefficient reaction trajectories as the barrier is recrossed several times which is not ideal in the formation of the product.

The exothermic nature of the reaction also alters the position of the transition state away from the lowest 'saddle point'. This goes against the assumptions of the Transition State Theory which states that reaction trajectories should pass through this. As temperature increases, a greater proportion of excited vibrational states become occupied, leading to transition states occuring at different energies. As a result, the associated reaction trajectory at higher temperature travels through different activated complexes, making the reaction harder to model and any associated results more difficult to compare.

'''Reaction 2: H + HF --> F + H2'''

Reaction 2, which is the reverse reaction of Reaction 1, is endothermic with the energy of the products being higher than the reactants. As a result, according the Hammond's Postulate, the position of the transition state is later in the reaction trajectory and more closely resembles the products. Contrary to the preferred conditions for exothermic reactions, a larger amount of vibrational energy is needed. To achieve this, lower momentum would be adopted. Providing this is achieved, the set of reaction trajectories would be an example of an efficient reaction as the reaction trajectories would pass through the transition state once and to the products.

----

==References==

<references>
<ref name="transition">P. Jonathan P. K. Doye, David J. Wales, Journal of mathematical physics, 1996, '''48''', 843 - 846.</ref>
<ref name="thermodynamics">Chem Libre texts - Wave-particle duality (2015) [online] Available at: https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Quantum_Mechanics/09._The_Hydrogen_Atom/Atomic_Theory/Electrons_in_Atoms/Wave-Particle_Duality [Assessed 28 Sep 2015].</ref>
<ref name="bond energies">The Department of Chemistry Illinois (2017). Average Bond Energies [online] Available at: http://butane.chem.uiuc.edu/cyerkes/Chem104ACSpring2009/Genchemref/bondenergies.html [Accessed 9 Feb 2017].</ref>
<ref name="lawE">Mach, E. (1872). History and Root of the Principles of the Conservation of Energy. Open Court Pub. Co., Illinois.</ref>
<ref name="polanyi">Yan, S., Y.-T. Wu, and K. Liu. "Tracking The Energy Flow Along The Reaction Path". Proceedings of the National Academy of Sciences 105.35 (2008): 12667-12672.</ref>
<ref name="hammond">Chemistry LibreTexts. (2017). Hammond’s Postulate. [online] Available at: https://chem.libretexts.org/Reference/Organic_Chemistry_Glossary/Hammond’s_Postulate [Accessed 11 May 2017].</ref>
<references/>

MRD:ab6114

2017-05-26T14:22:46Z

Ab6114: /* Question(8) */

=='''Molecular Reaction Dynamics'''==

==Exercise 1: H + H2 system==

==='''Question(1)'''===

What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

'''Answer:'''

Analysis of the PES curvature shows a maxima between two minimum points, all being stationary points with a gradient of zero. The first derivative of the curve, where the rate of change is zero, is found to confirm this. As Transition structure's inherently exist at the maximum point, it can be inferred that the transition structure lies exists at this point. In order to differentiate between the transition state and the minima, second derivative calculations can be performed, producing a negative value for a maxima and a positive value for a minima.<ref name="transition" />

----

==='''Question(2)'''===

Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.

'''Answer:'''

The approximate transition state position (r(ts)) was determined through analysis of the surface plot. This was confirmed by examining the internuclear distance vs time plot, to identify the point of intersection between the maxima and the minima - r(ts) is shown in the table below:

{| class="wikitable" border="1"
|+ Table 1
! r1 !! r2 !! r(ts)
|-
| 1.20 || 1.20 || 0.9075
|-
| 0.91 || 0.91 || 0.9075
|}

[[File:ab6114Figure1.PNG|680px|thumb|centre|Figure 1 - Internuclear distance vs Time plot of transition state position]]

A transition state is a point where the internuclear distance between all bonds remain constant and thus do not oscillate. Figure 1 shows that the internuclear distance does not oscillate at 0.9075 Å - hence this is the transition state position.

----

==='''Question(3)'''===

Comment on how the mep and the trajectory you just calculated differ.

'''Answer:'''

Dynamic calculations and mep are distinct for two main reasons. Calculations of mep span only the distance engaged by the transition state, while the dynamic calculation incorporates more of the reaction pathway - the dynamic calculation also takes mass into account, thus resulting in an oscillating reaction pathway line. No mass is included in the mep calculation, so no oscillation is observed.

----

==='''Question(4)'''===

Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.

'''Answer:'''

The effect of momentum variation on whether a trajectory is reactive or unreactive (for the initial positions r1 = 0.74 and r2 = 2.0):

{| class="wikitable" border="1"
|+ Table 2
! p1 !! p2 !! Trajectory !! Figure No.
|-
| -1.25 || -2.5 || Reactive || 2
|-
| -1.5 || -2.0 || Unreactive || 3
|-
| -1.5 || -2.5 || Reactive || 4
|-
| -2.5 || -5.0 || Unreactive || 5
|-
| -2.5 || -5.2 || Reactive || 6
|}

'''Figure 2'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 2. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc while at the same time the B-C bond distance remains relatively constant. This line continues through the transition state (r(A-B)=r(B-C)) and r(A-B) reaches a minimum and oscillates thus indicating that the Ha-Hb bond has formed. Simultaneously r(B-C) increases, confirming that the Hb-Hc bond has broken and Hc now exists as a separate hydrogen atom.

[[File:Ab6114_figure2.PNG|680px|thumb|centre|Figure 2 - The reactive trajectory associated with the specific parameters for Figure 2]]

'''Figure 3'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 3. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc. Since Ha has a momentum insufficient to overcome the activation energy barrier, the trajectory fails to reach the transition state - this is shown by the plot through r(A-B) returning to a larger distance without reaching transition state, and r(B-C) only oscillating due to vibrational fluctuations within the molecule.

[[File:Ab6114_figure3test2.PNG|680px|thumb|centre|Figure 3 - The unreactive trajectory associated with the specific parameters for Figure 3]]

'''Figure 4'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 4. The trajectory line illustrates a reaction where Ha had a momentum sufficient to overcome the activation energy barrier required to successfully interact with Hb-Hc, pass through the transition state, and form the molecule Ha-Hb.

[[File:Ab6114_figure4.PNG|680px|thumb|centre|Figure 4 - The reactive trajectory associated with the specific parameters for Figure 4]]

'''Figure 5'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 5. The trajectory line for these set parameters, initially show Ha to approach Hb-Hc with sufficient momentum to overcome the activation energy barrier and reach the transition state. However, instead of forming the product Ha-Hb, the trajectory line reverses and returns back to the reactants, with Hb-Hc exhibiting large vibrational energy fluctuations.

[[File:Ab6114_figure5.PNG|680px|thumb|centre|Figure 5 - The unreactive trajectory associated with the specific parameters for Figure 5]]

'''Figure 6'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 6. The trajectory line illustrates that Ha has sufficient momentum to overcome the activation energy barrier. However after reaching the transition state, the trajectory line goes back towards the reactants which is shown by r(B-C) decreasing and r(A-B) increasing in figure 6. The r(A-B) rose to a maxima before decreasing and oscillating, thus illustrating the formation of Ha-Hb. This reaction trajectory suggests that potentially an alternative reaction pathway was taken, one which does not involve a 'saddle point' on the PE plot corresponding with the transition states that normally produce the desired Ha and Hb-Hc.

[[File:Ab6114_figure6.PNG|680px|thumb|centre|Figure 6 - The reactive trajectory associated with the specific parameters for Figure 6]]

----

==='''Question(5)'''===

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

'''Answer:'''

The Transition State Theory's main assumption is that the reactants and activated transition states are at quasi-equilibrium. A quasi-equilibrium utilises thermodynamic molecular driving forces that modify the systems state being very small and slow, allowing for the existence of an internal equilibrium that undergoes only small deviations from this system. Classical Transition State Theory explains that an atom with a momentum satisfying the activation energy barrier requirements, will react upon collision with a given molecule and synthesize products. As this theory however ignores potential quantum mechanical interactions, reaction rates calculated will be slower than what is found in experimental data. Quantum mechanical contributing calculations allow for the tunneling behaviour of particles to be factored into an alternative reaction pathway. When an atom and a molecule lack the sufficient activation energy to overcome this barrier, the atom is able to 'tunnel' through this barrier, due to the particle-wave duality observed in quantum-scale objects.<ref name="thermodynamics" />

----

==Exercise 2: F - H - H system==

==='''Question(6)'''===

Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

'''Answer:'''

Analysis:

- H-F Bond Energy = 565 KJ/Mol<ref name="bond energies" />

- H-H Bond Energy = 432 KJ/Mol<ref name="bond energies" />

- H-F is a stronger bond than H-H

- ΔE = E(bonds cleaved)-E(bonds formed)

'''F + H2''' --> ΔE = 432 - 565 = -133 KJ/mol '''THUS EXOTHERMIC'''

'''H + H-F''' --> ΔE = 565 - 432 = 133 KJ/mol '''THUS ENDOTHERMIC'''

----

==='''Question(7)'''===

Locate the approximate position of the transition state.

'''Answer:'''

In order to locate the position of the transition state, the radii corresponding to a stationary total system energy (gradient=0) were identified. The positions of the radii have a zero net force acting upon them, so plotting the internuclear distance vs time results in a non-oscillating constant internuclear separation - see the internuclear separations below and the corresponding plot vs time in figure 7:

- r(H-F) = 1.815Å

- r(H-H) = 0.74Å

[[File:ab6114 figure7.PNG|680px|thumb|centre|Figure 7 - Internuclear distance vs Time plot for the transition state position]]

----

==='''Question(8)'''===

Report the activation energy for both reactions.

'''Answer:'''

Determining the activation energy for both reactions, required chemical intuition and trial and error in order to identify an adequate reaction trajectory. The activation energy is equal to the energy difference between the reactants and the transition state - activation energy values calculated in absence of mathematical analysis, thus all energy values involved would actually have +/- values.

'''Reaction 1: F + H2 --> HF + H'''

- Ea = -103.7 - (-103.9) = 0.2 Kcal/Mol

The energy values used for this calculation were obtained from Figure 8 (reactants) and Figure 9 (transition state) below:

[[File:ab6114 figure8.PNG|470px|thumb|centre|Figure 8 - Surface plot illustrating energy of the reactants]]

[[File:ab6114 figure9.PNG|470px|thumb|centre|Figure 9 - Surface plot illustrating the energy of the transition state]]

'''Reaction 2: H + HF --> F + H2'''

- Ea = -103.7 - (-133.9) = 30.2 Kcal/Mol

The energy values used for this calculation were obtained from Figure 10 (reactants) and Figure 9 (transition state) as shown below:

[[File:HZ4315 Figure9.PNG|500px|thumb|centre|Figure 9 - Surface plot showing the energy at the transition state]]

[[File:HZ4315 Figure10.PNG|500px|thumb|centre|Figure 10 - Surface plot showing the energy of the reactants in the reverse reaction]]

----

==='''Q(9)'''===

In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

'''A:'''

According to the Law of Conservation of Energy, the total energy of an isolated system must be conserved as energy is neither created nor destroyed, just converted from one form to another.<ref name="lawE" /> For this specific reaction system, the total energy is the sum of the kinetic and potential energies of each substrate involved in the reaction. For the exothermic reaction F + H2, the potential energy of the products is lower than the reactants hence, the kinetic energy of the products must be higher. This is illustrated by determining a successful reaction trajectory and plotting a kinetic energy vs time graph as shown below:

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 11 - A reaction pathway for an ultimately successful reaction (parameters determined using trial and error)]]

[[File:KEvsTIME.PNG|500px|thumb|centre|Figure 12 - Kinetic energy vs Time plot of the successful reaction - the small oscillations on the left correspond to the reactants which are clearly converted to products as the kinetic energy increases significantly]]

The increased kinetic energy allows the products to be detected by IR analysis. Since the H-F bond is oscillating, it will release energy in the form of vibrational energy, resulting in a peak on the spectrum. The H-H vibrational peak present on the reactant spectrum will now be absent. Another possible experimental technique is calorimetry. Since the reaction is exothermic, the amount of heat released can be measured and since, at constant pressure, the enthalpy change is equal to the net change in energy, the net free energy change of the reaction can be determined.

----

==='''Q(10)'''===

Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

'''A:'''

Using trial and error, a successful reaction trajectory was found for both reactions shown below. The parameters used can be seen within the figures.

'''Reaction 1: F + H2 --> HF + H'''

'''Lower momentum'''

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 13 - An ultimately successful reaction pathway at high momentum indicating the preference of an early transition state]]

'''Higher momentum'''

[[File:HZ4315 Figure12.PNG|500px|thumb|centre|Figure 14 - The reaction pathway at an even larger momentum to confirm the theory]]

Reaction 1, which is an exothermic reaction, exhibits an increased translational energy in response to increasing the momentum of the reactants. According to Polanyi's rules, higher translational energies are more efficient at activating an early transition state (ie. an early barrier) than vibrational energy.<ref name="polanyi" /> Since early transition states resemble the reactants (as stated by Hammond's Postulate)<ref name="hammond" />, the reaction trajectory may be expected to bounce back towards the reactants. Figure 12 confirms this as an increased momentum (and therefore increased translational energy) sees the trajectory bounce back several times before eventually proceeding to the form the products. Figures 11 and 12 are an example of a set of inefficient reaction trajectories as the barrier is recrossed several times which is not ideal in the formation of the product.

The exothermic nature of the reaction also alters the position of the transition state away from the lowest 'saddle point'. This goes against the assumptions of the Transition State Theory which states that reaction trajectories should pass through this. As temperature increases, a greater proportion of excited vibrational states become occupied, leading to transition states occuring at different energies. As a result, the associated reaction trajectory at higher temperature travels through different activated complexes, making the reaction harder to model and any associated results more difficult to compare.

'''Reaction 2: H + HF --> F + H2'''

Reaction 2, which is the reverse reaction of Reaction 1, is endothermic with the energy of the products being higher than the reactants. As a result, according the Hammond's Postulate, the position of the transition state is later in the reaction trajectory and more closely resembles the products. Contrary to the preferred conditions for exothermic reactions, a larger amount of vibrational energy is needed. To achieve this, lower momentum would be adopted. Providing this is achieved, the set of reaction trajectories would be an example of an efficient reaction as the reaction trajectories would pass through the transition state once and to the products.

----

==References==

<references>
<ref name="transition">P. Jonathan P. K. Doye, David J. Wales, Journal of mathematical physics, 1996, '''48''', 843 - 846.</ref>
<ref name="thermodynamics">Chem Libre texts - Wave-particle duality (2015) [online] Available at: https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Quantum_Mechanics/09._The_Hydrogen_Atom/Atomic_Theory/Electrons_in_Atoms/Wave-Particle_Duality [Assessed 28 Sep 2015].</ref>
<ref name="bond energies">The Department of Chemistry Illinois (2017). Average Bond Energies [online] Available at: http://butane.chem.uiuc.edu/cyerkes/Chem104ACSpring2009/Genchemref/bondenergies.html [Accessed 9 Feb 2017].</ref>
<ref name="lawE">Mach, E. (1872). History and Root of the Principles of the Conservation of Energy. Open Court Pub. Co., Illinois.</ref>
<ref name="polanyi">Yan, S., Y.-T. Wu, and K. Liu. "Tracking The Energy Flow Along The Reaction Path". Proceedings of the National Academy of Sciences 105.35 (2008): 12667-12672.</ref>
<ref name="hammond">Chemistry LibreTexts. (2017). Hammond’s Postulate. [online] Available at: https://chem.libretexts.org/Reference/Organic_Chemistry_Glossary/Hammond’s_Postulate [Accessed 11 May 2017].</ref>
<references/>

File:Ab6114 figure9test3.PNG

2017-05-26T14:21:09Z

Ab6114:

MRD:ab6114

2017-05-26T14:19:29Z

Ab6114:

=='''Molecular Reaction Dynamics'''==

==Exercise 1: H + H2 system==

==='''Question(1)'''===

What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

'''Answer:'''

Analysis of the PES curvature shows a maxima between two minimum points, all being stationary points with a gradient of zero. The first derivative of the curve, where the rate of change is zero, is found to confirm this. As Transition structure's inherently exist at the maximum point, it can be inferred that the transition structure lies exists at this point. In order to differentiate between the transition state and the minima, second derivative calculations can be performed, producing a negative value for a maxima and a positive value for a minima.<ref name="transition" />

----

==='''Question(2)'''===

Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.

'''Answer:'''

The approximate transition state position (r(ts)) was determined through analysis of the surface plot. This was confirmed by examining the internuclear distance vs time plot, to identify the point of intersection between the maxima and the minima - r(ts) is shown in the table below:

{| class="wikitable" border="1"
|+ Table 1
! r1 !! r2 !! r(ts)
|-
| 1.20 || 1.20 || 0.9075
|-
| 0.91 || 0.91 || 0.9075
|}

[[File:ab6114Figure1.PNG|680px|thumb|centre|Figure 1 - Internuclear distance vs Time plot of transition state position]]

A transition state is a point where the internuclear distance between all bonds remain constant and thus do not oscillate. Figure 1 shows that the internuclear distance does not oscillate at 0.9075 Å - hence this is the transition state position.

----

==='''Question(3)'''===

Comment on how the mep and the trajectory you just calculated differ.

'''Answer:'''

Dynamic calculations and mep are distinct for two main reasons. Calculations of mep span only the distance engaged by the transition state, while the dynamic calculation incorporates more of the reaction pathway - the dynamic calculation also takes mass into account, thus resulting in an oscillating reaction pathway line. No mass is included in the mep calculation, so no oscillation is observed.

----

==='''Question(4)'''===

Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.

'''Answer:'''

The effect of momentum variation on whether a trajectory is reactive or unreactive (for the initial positions r1 = 0.74 and r2 = 2.0):

{| class="wikitable" border="1"
|+ Table 2
! p1 !! p2 !! Trajectory !! Figure No.
|-
| -1.25 || -2.5 || Reactive || 2
|-
| -1.5 || -2.0 || Unreactive || 3
|-
| -1.5 || -2.5 || Reactive || 4
|-
| -2.5 || -5.0 || Unreactive || 5
|-
| -2.5 || -5.2 || Reactive || 6
|}

'''Figure 2'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 2. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc while at the same time the B-C bond distance remains relatively constant. This line continues through the transition state (r(A-B)=r(B-C)) and r(A-B) reaches a minimum and oscillates thus indicating that the Ha-Hb bond has formed. Simultaneously r(B-C) increases, confirming that the Hb-Hc bond has broken and Hc now exists as a separate hydrogen atom.

[[File:Ab6114_figure2.PNG|680px|thumb|centre|Figure 2 - The reactive trajectory associated with the specific parameters for Figure 2]]

'''Figure 3'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 3. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc. Since Ha has a momentum insufficient to overcome the activation energy barrier, the trajectory fails to reach the transition state - this is shown by the plot through r(A-B) returning to a larger distance without reaching transition state, and r(B-C) only oscillating due to vibrational fluctuations within the molecule.

[[File:Ab6114_figure3test2.PNG|680px|thumb|centre|Figure 3 - The unreactive trajectory associated with the specific parameters for Figure 3]]

'''Figure 4'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 4. The trajectory line illustrates a reaction where Ha had a momentum sufficient to overcome the activation energy barrier required to successfully interact with Hb-Hc, pass through the transition state, and form the molecule Ha-Hb.

[[File:Ab6114_figure4.PNG|680px|thumb|centre|Figure 4 - The reactive trajectory associated with the specific parameters for Figure 4]]

'''Figure 5'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 5. The trajectory line for these set parameters, initially show Ha to approach Hb-Hc with sufficient momentum to overcome the activation energy barrier and reach the transition state. However, instead of forming the product Ha-Hb, the trajectory line reverses and returns back to the reactants, with Hb-Hc exhibiting large vibrational energy fluctuations.

[[File:Ab6114_figure5.PNG|680px|thumb|centre|Figure 5 - The unreactive trajectory associated with the specific parameters for Figure 5]]

'''Figure 6'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 6. The trajectory line illustrates that Ha has sufficient momentum to overcome the activation energy barrier. However after reaching the transition state, the trajectory line goes back towards the reactants which is shown by r(B-C) decreasing and r(A-B) increasing in figure 6. The r(A-B) rose to a maxima before decreasing and oscillating, thus illustrating the formation of Ha-Hb. This reaction trajectory suggests that potentially an alternative reaction pathway was taken, one which does not involve a 'saddle point' on the PE plot corresponding with the transition states that normally produce the desired Ha and Hb-Hc.

[[File:Ab6114_figure6.PNG|680px|thumb|centre|Figure 6 - The reactive trajectory associated with the specific parameters for Figure 6]]

----

==='''Question(5)'''===

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

'''Answer:'''

The Transition State Theory's main assumption is that the reactants and activated transition states are at quasi-equilibrium. A quasi-equilibrium utilises thermodynamic molecular driving forces that modify the systems state being very small and slow, allowing for the existence of an internal equilibrium that undergoes only small deviations from this system. Classical Transition State Theory explains that an atom with a momentum satisfying the activation energy barrier requirements, will react upon collision with a given molecule and synthesize products. As this theory however ignores potential quantum mechanical interactions, reaction rates calculated will be slower than what is found in experimental data. Quantum mechanical contributing calculations allow for the tunneling behaviour of particles to be factored into an alternative reaction pathway. When an atom and a molecule lack the sufficient activation energy to overcome this barrier, the atom is able to 'tunnel' through this barrier, due to the particle-wave duality observed in quantum-scale objects.<ref name="thermodynamics" />

----

==Exercise 2: F - H - H system==

==='''Question(6)'''===

Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

'''Answer:'''

Analysis:

- H-F Bond Energy = 565 KJ/Mol<ref name="bond energies" />

- H-H Bond Energy = 432 KJ/Mol<ref name="bond energies" />

- H-F is a stronger bond than H-H

- ΔE = E(bonds cleaved)-E(bonds formed)

'''F + H2''' --> ΔE = 432 - 565 = -133 KJ/mol '''THUS EXOTHERMIC'''

'''H + H-F''' --> ΔE = 565 - 432 = 133 KJ/mol '''THUS ENDOTHERMIC'''

----

==='''Question(7)'''===

Locate the approximate position of the transition state.

'''Answer:'''

In order to locate the position of the transition state, the radii corresponding to a stationary total system energy (gradient=0) were identified. The positions of the radii have a zero net force acting upon them, so plotting the internuclear distance vs time results in a non-oscillating constant internuclear separation - see the internuclear separations below and the corresponding plot vs time in figure 7:

- r(H-F) = 1.815Å

- r(H-H) = 0.74Å

[[File:ab6114 figure7.PNG|680px|thumb|centre|Figure 7 - Internuclear distance vs Time plot for the transition state position]]

----

==='''Question(8)'''===

Report the activation energy for both reactions.

'''Answer:'''

Determining the activation energy for both reactions, required chemical intuition and trial and error in order to identify an adequate reaction trajectory. The activation energy is equal to the energy difference between the reactants and the transition state - activation energy values calculated in absence of mathematical analysis, thus all energy values involved would actually have +/- values.

'''Reaction 1: F + H2 --> HF + H'''

- Ea = -103.7 - (-103.9) = 0.2 Kcal/Mol

The energy values used for this calculation were obtained from Figure 8 (reactants) and Figure 9 (transition state) below:

[[File:ab6114 figure8.PNG|470px|thumb|centre|Figure 8 - Surface plot illustrating energy of the reactants]]

[[File:ab6114 figure9test3.PNG|470px|thumb|centre|Figure 9 - Surface plot illustrating the energy of the transition state]]

'''Reaction 2: H + HF --> F + H2'''

- Ea = -103.7 - (-133.9) = 30.2 Kcal/Mol

The energy values used for this calculation were obtained from Figure 10 (reactants) and Figure 9 (transition state) as shown below:

[[File:HZ4315 Figure9.PNG|500px|thumb|centre|Figure 9 - Surface plot showing the energy at the transition state]]

[[File:HZ4315 Figure10.PNG|500px|thumb|centre|Figure 10 - Surface plot showing the energy of the reactants in the reverse reaction]]

----

==='''Q(9)'''===

In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

'''A:'''

According to the Law of Conservation of Energy, the total energy of an isolated system must be conserved as energy is neither created nor destroyed, just converted from one form to another.<ref name="lawE" /> For this specific reaction system, the total energy is the sum of the kinetic and potential energies of each substrate involved in the reaction. For the exothermic reaction F + H2, the potential energy of the products is lower than the reactants hence, the kinetic energy of the products must be higher. This is illustrated by determining a successful reaction trajectory and plotting a kinetic energy vs time graph as shown below:

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 11 - A reaction pathway for an ultimately successful reaction (parameters determined using trial and error)]]

[[File:KEvsTIME.PNG|500px|thumb|centre|Figure 12 - Kinetic energy vs Time plot of the successful reaction - the small oscillations on the left correspond to the reactants which are clearly converted to products as the kinetic energy increases significantly]]

The increased kinetic energy allows the products to be detected by IR analysis. Since the H-F bond is oscillating, it will release energy in the form of vibrational energy, resulting in a peak on the spectrum. The H-H vibrational peak present on the reactant spectrum will now be absent. Another possible experimental technique is calorimetry. Since the reaction is exothermic, the amount of heat released can be measured and since, at constant pressure, the enthalpy change is equal to the net change in energy, the net free energy change of the reaction can be determined.

----

==='''Q(10)'''===

Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

'''A:'''

Using trial and error, a successful reaction trajectory was found for both reactions shown below. The parameters used can be seen within the figures.

'''Reaction 1: F + H2 --> HF + H'''

'''Lower momentum'''

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 13 - An ultimately successful reaction pathway at high momentum indicating the preference of an early transition state]]

'''Higher momentum'''

[[File:HZ4315 Figure12.PNG|500px|thumb|centre|Figure 14 - The reaction pathway at an even larger momentum to confirm the theory]]

Reaction 1, which is an exothermic reaction, exhibits an increased translational energy in response to increasing the momentum of the reactants. According to Polanyi's rules, higher translational energies are more efficient at activating an early transition state (ie. an early barrier) than vibrational energy.<ref name="polanyi" /> Since early transition states resemble the reactants (as stated by Hammond's Postulate)<ref name="hammond" />, the reaction trajectory may be expected to bounce back towards the reactants. Figure 12 confirms this as an increased momentum (and therefore increased translational energy) sees the trajectory bounce back several times before eventually proceeding to the form the products. Figures 11 and 12 are an example of a set of inefficient reaction trajectories as the barrier is recrossed several times which is not ideal in the formation of the product.

The exothermic nature of the reaction also alters the position of the transition state away from the lowest 'saddle point'. This goes against the assumptions of the Transition State Theory which states that reaction trajectories should pass through this. As temperature increases, a greater proportion of excited vibrational states become occupied, leading to transition states occuring at different energies. As a result, the associated reaction trajectory at higher temperature travels through different activated complexes, making the reaction harder to model and any associated results more difficult to compare.

'''Reaction 2: H + HF --> F + H2'''

Reaction 2, which is the reverse reaction of Reaction 1, is endothermic with the energy of the products being higher than the reactants. As a result, according the Hammond's Postulate, the position of the transition state is later in the reaction trajectory and more closely resembles the products. Contrary to the preferred conditions for exothermic reactions, a larger amount of vibrational energy is needed. To achieve this, lower momentum would be adopted. Providing this is achieved, the set of reaction trajectories would be an example of an efficient reaction as the reaction trajectories would pass through the transition state once and to the products.

----

==References==

<references>
<ref name="transition">P. Jonathan P. K. Doye, David J. Wales, Journal of mathematical physics, 1996, '''48''', 843 - 846.</ref>
<ref name="thermodynamics">Chem Libre texts - Wave-particle duality (2015) [online] Available at: https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Quantum_Mechanics/09._The_Hydrogen_Atom/Atomic_Theory/Electrons_in_Atoms/Wave-Particle_Duality [Assessed 28 Sep 2015].</ref>
<ref name="bond energies">The Department of Chemistry Illinois (2017). Average Bond Energies [online] Available at: http://butane.chem.uiuc.edu/cyerkes/Chem104ACSpring2009/Genchemref/bondenergies.html [Accessed 9 Feb 2017].</ref>
<ref name="lawE">Mach, E. (1872). History and Root of the Principles of the Conservation of Energy. Open Court Pub. Co., Illinois.</ref>
<ref name="polanyi">Yan, S., Y.-T. Wu, and K. Liu. "Tracking The Energy Flow Along The Reaction Path". Proceedings of the National Academy of Sciences 105.35 (2008): 12667-12672.</ref>
<ref name="hammond">Chemistry LibreTexts. (2017). Hammond’s Postulate. [online] Available at: https://chem.libretexts.org/Reference/Organic_Chemistry_Glossary/Hammond’s_Postulate [Accessed 11 May 2017].</ref>
<references/>

File:Ab6114 figure9test2.PNG

2017-05-26T14:17:16Z

Ab6114:

MRD:ab6114

2017-05-26T14:16:37Z

Ab6114:

=='''Molecular Reaction Dynamics'''==

==Exercise 1: H + H2 system==

==='''Question(1)'''===

What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

'''Answer:'''

Analysis of the PES curvature shows a maxima between two minimum points, all being stationary points with a gradient of zero. The first derivative of the curve, where the rate of change is zero, is found to confirm this. As Transition structure's inherently exist at the maximum point, it can be inferred that the transition structure lies exists at this point. In order to differentiate between the transition state and the minima, second derivative calculations can be performed, producing a negative value for a maxima and a positive value for a minima.<ref name="transition" />

----

==='''Question(2)'''===

Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.

'''Answer:'''

The approximate transition state position (r(ts)) was determined through analysis of the surface plot. This was confirmed by examining the internuclear distance vs time plot, to identify the point of intersection between the maxima and the minima - r(ts) is shown in the table below:

{| class="wikitable" border="1"
|+ Table 1
! r1 !! r2 !! r(ts)
|-
| 1.20 || 1.20 || 0.9075
|-
| 0.91 || 0.91 || 0.9075
|}

[[File:ab6114Figure1.PNG|680px|thumb|centre|Figure 1 - Internuclear distance vs Time plot of transition state position]]

A transition state is a point where the internuclear distance between all bonds remain constant and thus do not oscillate. Figure 1 shows that the internuclear distance does not oscillate at 0.9075 Å - hence this is the transition state position.

----

==='''Question(3)'''===

Comment on how the mep and the trajectory you just calculated differ.

'''Answer:'''

Dynamic calculations and mep are distinct for two main reasons. Calculations of mep span only the distance engaged by the transition state, while the dynamic calculation incorporates more of the reaction pathway - the dynamic calculation also takes mass into account, thus resulting in an oscillating reaction pathway line. No mass is included in the mep calculation, so no oscillation is observed.

----

==='''Question(4)'''===

Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.

'''Answer:'''

The effect of momentum variation on whether a trajectory is reactive or unreactive (for the initial positions r1 = 0.74 and r2 = 2.0):

{| class="wikitable" border="1"
|+ Table 2
! p1 !! p2 !! Trajectory !! Figure No.
|-
| -1.25 || -2.5 || Reactive || 2
|-
| -1.5 || -2.0 || Unreactive || 3
|-
| -1.5 || -2.5 || Reactive || 4
|-
| -2.5 || -5.0 || Unreactive || 5
|-
| -2.5 || -5.2 || Reactive || 6
|}

'''Figure 2'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 2. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc while at the same time the B-C bond distance remains relatively constant. This line continues through the transition state (r(A-B)=r(B-C)) and r(A-B) reaches a minimum and oscillates thus indicating that the Ha-Hb bond has formed. Simultaneously r(B-C) increases, confirming that the Hb-Hc bond has broken and Hc now exists as a separate hydrogen atom.

[[File:Ab6114_figure2.PNG|680px|thumb|centre|Figure 2 - The reactive trajectory associated with the specific parameters for Figure 2]]

'''Figure 3'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 3. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc. Since Ha has a momentum insufficient to overcome the activation energy barrier, the trajectory fails to reach the transition state - this is shown by the plot through r(A-B) returning to a larger distance without reaching transition state, and r(B-C) only oscillating due to vibrational fluctuations within the molecule.

[[File:Ab6114_figure3test2.PNG|680px|thumb|centre|Figure 3 - The unreactive trajectory associated with the specific parameters for Figure 3]]

'''Figure 4'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 4. The trajectory line illustrates a reaction where Ha had a momentum sufficient to overcome the activation energy barrier required to successfully interact with Hb-Hc, pass through the transition state, and form the molecule Ha-Hb.

[[File:Ab6114_figure4.PNG|680px|thumb|centre|Figure 4 - The reactive trajectory associated with the specific parameters for Figure 4]]

'''Figure 5'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 5. The trajectory line for these set parameters, initially show Ha to approach Hb-Hc with sufficient momentum to overcome the activation energy barrier and reach the transition state. However, instead of forming the product Ha-Hb, the trajectory line reverses and returns back to the reactants, with Hb-Hc exhibiting large vibrational energy fluctuations.

[[File:Ab6114_figure5.PNG|680px|thumb|centre|Figure 5 - The unreactive trajectory associated with the specific parameters for Figure 5]]

'''Figure 6'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 6. The trajectory line illustrates that Ha has sufficient momentum to overcome the activation energy barrier. However after reaching the transition state, the trajectory line goes back towards the reactants which is shown by r(B-C) decreasing and r(A-B) increasing in figure 6. The r(A-B) rose to a maxima before decreasing and oscillating, thus illustrating the formation of Ha-Hb. This reaction trajectory suggests that potentially an alternative reaction pathway was taken, one which does not involve a 'saddle point' on the PE plot corresponding with the transition states that normally produce the desired Ha and Hb-Hc.

[[File:Ab6114_figure6.PNG|680px|thumb|centre|Figure 6 - The reactive trajectory associated with the specific parameters for Figure 6]]

----

==='''Question(5)'''===

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

'''Answer:'''

The Transition State Theory's main assumption is that the reactants and activated transition states are at quasi-equilibrium. A quasi-equilibrium utilises thermodynamic molecular driving forces that modify the systems state being very small and slow, allowing for the existence of an internal equilibrium that undergoes only small deviations from this system. Classical Transition State Theory explains that an atom with a momentum satisfying the activation energy barrier requirements, will react upon collision with a given molecule and synthesize products. As this theory however ignores potential quantum mechanical interactions, reaction rates calculated will be slower than what is found in experimental data. Quantum mechanical contributing calculations allow for the tunneling behaviour of particles to be factored into an alternative reaction pathway. When an atom and a molecule lack the sufficient activation energy to overcome this barrier, the atom is able to 'tunnel' through this barrier, due to the particle-wave duality observed in quantum-scale objects.<ref name="thermodynamics" />

----

==Exercise 2: F - H - H system==

==='''Question(6)'''===

Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

'''Answer:'''

Analysis:

- H-F Bond Energy = 565 KJ/Mol<ref name="bond energies" />

- H-H Bond Energy = 432 KJ/Mol<ref name="bond energies" />

- H-F is a stronger bond than H-H

- ΔE = E(bonds cleaved)-E(bonds formed)

'''F + H2''' --> ΔE = 432 - 565 = -133 KJ/mol '''THUS EXOTHERMIC'''

'''H + H-F''' --> ΔE = 565 - 432 = 133 KJ/mol '''THUS ENDOTHERMIC'''

----

==='''Question(7)'''===

Locate the approximate position of the transition state.

'''Answer:'''

In order to locate the position of the transition state, the radii corresponding to a stationary total system energy (gradient=0) were identified. The positions of the radii have a zero net force acting upon them, so plotting the internuclear distance vs time results in a non-oscillating constant internuclear separation - see the internuclear separations below and the corresponding plot vs time in figure 7:

- r(H-F) = 1.815Å

- r(H-H) = 0.74Å

[[File:ab6114 figure7.PNG|680px|thumb|centre|Figure 7 - Internuclear distance vs Time plot for the transition state position]]

----

==='''Question(8)'''===

Report the activation energy for both reactions.

'''Answer:'''

Determining the activation energy for both reactions, required chemical intuition and trial and error in order to identify an adequate reaction trajectory. The activation energy is equal to the energy difference between the reactants and the transition state - activation energy values calculated in absence of mathematical analysis, thus all energy values involved would actually have +/- values.

'''Reaction 1: F + H2 --> HF + H'''

- Ea = -103.7 - (-103.9) = 0.2 Kcal/Mol

The energy values used for this calculation were obtained from Figure 8 (reactants) and Figure 9 (transition state) below:

[[File:ab6114 figure8.PNG|470px|thumb|centre|Figure 8 - Surface plot illustrating energy of the reactants]]

[[File:ab6114 figure9test2.PNG|470px|thumb|centre|Figure 9 - Surface plot illustrating the energy of the transition state]]

'''Reaction 2: H + HF --> F + H2'''

- Ea = -103.7 - (-133.9) = 30.2 Kcal/Mol

The energy values used for this calculation were obtained from Figure 10 (reactants) and Figure 9 (transition state) as shown below:

[[File:HZ4315 Figure9.PNG|500px|thumb|centre|Figure 9 - Surface plot showing the energy at the transition state]]

[[File:HZ4315 Figure10.PNG|500px|thumb|centre|Figure 10 - Surface plot showing the energy of the reactants in the reverse reaction]]

----

==='''Q(9)'''===

In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

'''A:'''

According to the Law of Conservation of Energy, the total energy of an isolated system must be conserved as energy is neither created nor destroyed, just converted from one form to another.<ref name="lawE" /> For this specific reaction system, the total energy is the sum of the kinetic and potential energies of each substrate involved in the reaction. For the exothermic reaction F + H2, the potential energy of the products is lower than the reactants hence, the kinetic energy of the products must be higher. This is illustrated by determining a successful reaction trajectory and plotting a kinetic energy vs time graph as shown below:

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 11 - A reaction pathway for an ultimately successful reaction (parameters determined using trial and error)]]

[[File:KEvsTIME.PNG|500px|thumb|centre|Figure 12 - Kinetic energy vs Time plot of the successful reaction - the small oscillations on the left correspond to the reactants which are clearly converted to products as the kinetic energy increases significantly]]

The increased kinetic energy allows the products to be detected by IR analysis. Since the H-F bond is oscillating, it will release energy in the form of vibrational energy, resulting in a peak on the spectrum. The H-H vibrational peak present on the reactant spectrum will now be absent. Another possible experimental technique is calorimetry. Since the reaction is exothermic, the amount of heat released can be measured and since, at constant pressure, the enthalpy change is equal to the net change in energy, the net free energy change of the reaction can be determined.

----

==='''Q(10)'''===

Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

'''A:'''

Using trial and error, a successful reaction trajectory was found for both reactions shown below. The parameters used can be seen within the figures.

'''Reaction 1: F + H2 --> HF + H'''

'''Lower momentum'''

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 13 - An ultimately successful reaction pathway at high momentum indicating the preference of an early transition state]]

'''Higher momentum'''

[[File:HZ4315 Figure12.PNG|500px|thumb|centre|Figure 14 - The reaction pathway at an even larger momentum to confirm the theory]]

Reaction 1, which is an exothermic reaction, exhibits an increased translational energy in response to increasing the momentum of the reactants. According to Polanyi's rules, higher translational energies are more efficient at activating an early transition state (ie. an early barrier) than vibrational energy.<ref name="polanyi" /> Since early transition states resemble the reactants (as stated by Hammond's Postulate)<ref name="hammond" />, the reaction trajectory may be expected to bounce back towards the reactants. Figure 12 confirms this as an increased momentum (and therefore increased translational energy) sees the trajectory bounce back several times before eventually proceeding to the form the products. Figures 11 and 12 are an example of a set of inefficient reaction trajectories as the barrier is recrossed several times which is not ideal in the formation of the product.

The exothermic nature of the reaction also alters the position of the transition state away from the lowest 'saddle point'. This goes against the assumptions of the Transition State Theory which states that reaction trajectories should pass through this. As temperature increases, a greater proportion of excited vibrational states become occupied, leading to transition states occuring at different energies. As a result, the associated reaction trajectory at higher temperature travels through different activated complexes, making the reaction harder to model and any associated results more difficult to compare.

'''Reaction 2: H + HF --> F + H2'''

Reaction 2, which is the reverse reaction of Reaction 1, is endothermic with the energy of the products being higher than the reactants. As a result, according the Hammond's Postulate, the position of the transition state is later in the reaction trajectory and more closely resembles the products. Contrary to the preferred conditions for exothermic reactions, a larger amount of vibrational energy is needed. To achieve this, lower momentum would be adopted. Providing this is achieved, the set of reaction trajectories would be an example of an efficient reaction as the reaction trajectories would pass through the transition state once and to the products.

----

==References==

<references>
<ref name="transition">P. Jonathan P. K. Doye, David J. Wales, Journal of mathematical physics, 1996, '''48''', 843 - 846.</ref>
<ref name="thermodynamics">Chem Libre texts - Wave-particle duality (2015) [online] Available at: https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Quantum_Mechanics/09._The_Hydrogen_Atom/Atomic_Theory/Electrons_in_Atoms/Wave-Particle_Duality [Assessed 28 Sep 2015].</ref>
<ref name="bond energies">The Department of Chemistry Illinois (2017). Average Bond Energies [online] Available at: http://butane.chem.uiuc.edu/cyerkes/Chem104ACSpring2009/Genchemref/bondenergies.html [Accessed 9 Feb 2017].</ref>
<ref name="lawE">Mach, E. (1872). History and Root of the Principles of the Conservation of Energy. Open Court Pub. Co., Illinois.</ref>
<ref name="polanyi">Yan, S., Y.-T. Wu, and K. Liu. "Tracking The Energy Flow Along The Reaction Path". Proceedings of the National Academy of Sciences 105.35 (2008): 12667-12672.</ref>
<ref name="hammond">Chemistry LibreTexts. (2017). Hammond’s Postulate. [online] Available at: https://chem.libretexts.org/Reference/Organic_Chemistry_Glossary/Hammond’s_Postulate [Accessed 11 May 2017].</ref>
<references/>

File:Ab6114 figure9.PNG

2017-05-26T14:15:24Z

Ab6114:

MRD:ab6114

2017-05-26T14:14:57Z

Ab6114:

=='''Molecular Reaction Dynamics'''==

==Exercise 1: H + H2 system==

==='''Question(1)'''===

What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

'''Answer:'''

Analysis of the PES curvature shows a maxima between two minimum points, all being stationary points with a gradient of zero. The first derivative of the curve, where the rate of change is zero, is found to confirm this. As Transition structure's inherently exist at the maximum point, it can be inferred that the transition structure lies exists at this point. In order to differentiate between the transition state and the minima, second derivative calculations can be performed, producing a negative value for a maxima and a positive value for a minima.<ref name="transition" />

----

==='''Question(2)'''===

Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.

'''Answer:'''

The approximate transition state position (r(ts)) was determined through analysis of the surface plot. This was confirmed by examining the internuclear distance vs time plot, to identify the point of intersection between the maxima and the minima - r(ts) is shown in the table below:

{| class="wikitable" border="1"
|+ Table 1
! r1 !! r2 !! r(ts)
|-
| 1.20 || 1.20 || 0.9075
|-
| 0.91 || 0.91 || 0.9075
|}

[[File:ab6114Figure1.PNG|680px|thumb|centre|Figure 1 - Internuclear distance vs Time plot of transition state position]]

A transition state is a point where the internuclear distance between all bonds remain constant and thus do not oscillate. Figure 1 shows that the internuclear distance does not oscillate at 0.9075 Å - hence this is the transition state position.

----

==='''Question(3)'''===

Comment on how the mep and the trajectory you just calculated differ.

'''Answer:'''

Dynamic calculations and mep are distinct for two main reasons. Calculations of mep span only the distance engaged by the transition state, while the dynamic calculation incorporates more of the reaction pathway - the dynamic calculation also takes mass into account, thus resulting in an oscillating reaction pathway line. No mass is included in the mep calculation, so no oscillation is observed.

----

==='''Question(4)'''===

Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.

'''Answer:'''

The effect of momentum variation on whether a trajectory is reactive or unreactive (for the initial positions r1 = 0.74 and r2 = 2.0):

{| class="wikitable" border="1"
|+ Table 2
! p1 !! p2 !! Trajectory !! Figure No.
|-
| -1.25 || -2.5 || Reactive || 2
|-
| -1.5 || -2.0 || Unreactive || 3
|-
| -1.5 || -2.5 || Reactive || 4
|-
| -2.5 || -5.0 || Unreactive || 5
|-
| -2.5 || -5.2 || Reactive || 6
|}

'''Figure 2'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 2. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc while at the same time the B-C bond distance remains relatively constant. This line continues through the transition state (r(A-B)=r(B-C)) and r(A-B) reaches a minimum and oscillates thus indicating that the Ha-Hb bond has formed. Simultaneously r(B-C) increases, confirming that the Hb-Hc bond has broken and Hc now exists as a separate hydrogen atom.

[[File:Ab6114_figure2.PNG|680px|thumb|centre|Figure 2 - The reactive trajectory associated with the specific parameters for Figure 2]]

'''Figure 3'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 3. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc. Since Ha has a momentum insufficient to overcome the activation energy barrier, the trajectory fails to reach the transition state - this is shown by the plot through r(A-B) returning to a larger distance without reaching transition state, and r(B-C) only oscillating due to vibrational fluctuations within the molecule.

[[File:Ab6114_figure3test2.PNG|680px|thumb|centre|Figure 3 - The unreactive trajectory associated with the specific parameters for Figure 3]]

'''Figure 4'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 4. The trajectory line illustrates a reaction where Ha had a momentum sufficient to overcome the activation energy barrier required to successfully interact with Hb-Hc, pass through the transition state, and form the molecule Ha-Hb.

[[File:Ab6114_figure4.PNG|680px|thumb|centre|Figure 4 - The reactive trajectory associated with the specific parameters for Figure 4]]

'''Figure 5'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 5. The trajectory line for these set parameters, initially show Ha to approach Hb-Hc with sufficient momentum to overcome the activation energy barrier and reach the transition state. However, instead of forming the product Ha-Hb, the trajectory line reverses and returns back to the reactants, with Hb-Hc exhibiting large vibrational energy fluctuations.

[[File:Ab6114_figure5.PNG|680px|thumb|centre|Figure 5 - The unreactive trajectory associated with the specific parameters for Figure 5]]

'''Figure 6'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 6. The trajectory line illustrates that Ha has sufficient momentum to overcome the activation energy barrier. However after reaching the transition state, the trajectory line goes back towards the reactants which is shown by r(B-C) decreasing and r(A-B) increasing in figure 6. The r(A-B) rose to a maxima before decreasing and oscillating, thus illustrating the formation of Ha-Hb. This reaction trajectory suggests that potentially an alternative reaction pathway was taken, one which does not involve a 'saddle point' on the PE plot corresponding with the transition states that normally produce the desired Ha and Hb-Hc.

[[File:Ab6114_figure6.PNG|680px|thumb|centre|Figure 6 - The reactive trajectory associated with the specific parameters for Figure 6]]

----

==='''Question(5)'''===

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

'''Answer:'''

The Transition State Theory's main assumption is that the reactants and activated transition states are at quasi-equilibrium. A quasi-equilibrium utilises thermodynamic molecular driving forces that modify the systems state being very small and slow, allowing for the existence of an internal equilibrium that undergoes only small deviations from this system. Classical Transition State Theory explains that an atom with a momentum satisfying the activation energy barrier requirements, will react upon collision with a given molecule and synthesize products. As this theory however ignores potential quantum mechanical interactions, reaction rates calculated will be slower than what is found in experimental data. Quantum mechanical contributing calculations allow for the tunneling behaviour of particles to be factored into an alternative reaction pathway. When an atom and a molecule lack the sufficient activation energy to overcome this barrier, the atom is able to 'tunnel' through this barrier, due to the particle-wave duality observed in quantum-scale objects.<ref name="thermodynamics" />

----

==Exercise 2: F - H - H system==

==='''Question(6)'''===

Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

'''Answer:'''

Analysis:

- H-F Bond Energy = 565 KJ/Mol<ref name="bond energies" />

- H-H Bond Energy = 432 KJ/Mol<ref name="bond energies" />

- H-F is a stronger bond than H-H

- ΔE = E(bonds cleaved)-E(bonds formed)

'''F + H2''' --> ΔE = 432 - 565 = -133 KJ/mol '''THUS EXOTHERMIC'''

'''H + H-F''' --> ΔE = 565 - 432 = 133 KJ/mol '''THUS ENDOTHERMIC'''

----

==='''Question(7)'''===

Locate the approximate position of the transition state.

'''Answer:'''

In order to locate the position of the transition state, the radii corresponding to a stationary total system energy (gradient=0) were identified. The positions of the radii have a zero net force acting upon them, so plotting the internuclear distance vs time results in a non-oscillating constant internuclear separation - see the internuclear separations below and the corresponding plot vs time in figure 7:

- r(H-F) = 1.815Å

- r(H-H) = 0.74Å

[[File:ab6114 figure7.PNG|680px|thumb|centre|Figure 7 - Internuclear distance vs Time plot for the transition state position]]

----

==='''Question(8)'''===

Report the activation energy for both reactions.

'''Answer:'''

Determining the activation energy for both reactions, required chemical intuition and trial and error in order to identify an adequate reaction trajectory. The activation energy is equal to the energy difference between the reactants and the transition state - activation energy values calculated in absence of mathematical analysis, thus all energy values involved would actually have +/- values.

'''Reaction 1: F + H2 --> HF + H'''

- Ea = -103.7 - (-103.9) = 0.2 Kcal/Mol

The energy values used for this calculation were obtained from Figure 8 (reactants) and Figure 9 (transition state) below:

[[File:ab6114 figure8.PNG|470px|thumb|centre|Figure 8 - Surface plot illustrating energy of the reactants]]

[[File:ab6114 figure9.PNG|470px|thumb|centre|Figure 9 - Surface plot illustrating the energy of the transition state]]

'''Reaction 2: H + HF --> F + H2'''

- Ea = -103.7 - (-133.9) = 30.2 Kcal/Mol

The energy values used for this calculation were obtained from Figure 10 (reactants) and Figure 9 (transition state) as shown below:

[[File:HZ4315 Figure9.PNG|500px|thumb|centre|Figure 9 - Surface plot showing the energy at the transition state]]

[[File:HZ4315 Figure10.PNG|500px|thumb|centre|Figure 10 - Surface plot showing the energy of the reactants in the reverse reaction]]

----

==='''Q(9)'''===

In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

'''A:'''

According to the Law of Conservation of Energy, the total energy of an isolated system must be conserved as energy is neither created nor destroyed, just converted from one form to another.<ref name="lawE" /> For this specific reaction system, the total energy is the sum of the kinetic and potential energies of each substrate involved in the reaction. For the exothermic reaction F + H2, the potential energy of the products is lower than the reactants hence, the kinetic energy of the products must be higher. This is illustrated by determining a successful reaction trajectory and plotting a kinetic energy vs time graph as shown below:

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 11 - A reaction pathway for an ultimately successful reaction (parameters determined using trial and error)]]

[[File:KEvsTIME.PNG|500px|thumb|centre|Figure 12 - Kinetic energy vs Time plot of the successful reaction - the small oscillations on the left correspond to the reactants which are clearly converted to products as the kinetic energy increases significantly]]

The increased kinetic energy allows the products to be detected by IR analysis. Since the H-F bond is oscillating, it will release energy in the form of vibrational energy, resulting in a peak on the spectrum. The H-H vibrational peak present on the reactant spectrum will now be absent. Another possible experimental technique is calorimetry. Since the reaction is exothermic, the amount of heat released can be measured and since, at constant pressure, the enthalpy change is equal to the net change in energy, the net free energy change of the reaction can be determined.

----

==='''Q(10)'''===

Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

'''A:'''

Using trial and error, a successful reaction trajectory was found for both reactions shown below. The parameters used can be seen within the figures.

'''Reaction 1: F + H2 --> HF + H'''

'''Lower momentum'''

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 13 - An ultimately successful reaction pathway at high momentum indicating the preference of an early transition state]]

'''Higher momentum'''

[[File:HZ4315 Figure12.PNG|500px|thumb|centre|Figure 14 - The reaction pathway at an even larger momentum to confirm the theory]]

Reaction 1, which is an exothermic reaction, exhibits an increased translational energy in response to increasing the momentum of the reactants. According to Polanyi's rules, higher translational energies are more efficient at activating an early transition state (ie. an early barrier) than vibrational energy.<ref name="polanyi" /> Since early transition states resemble the reactants (as stated by Hammond's Postulate)<ref name="hammond" />, the reaction trajectory may be expected to bounce back towards the reactants. Figure 12 confirms this as an increased momentum (and therefore increased translational energy) sees the trajectory bounce back several times before eventually proceeding to the form the products. Figures 11 and 12 are an example of a set of inefficient reaction trajectories as the barrier is recrossed several times which is not ideal in the formation of the product.

The exothermic nature of the reaction also alters the position of the transition state away from the lowest 'saddle point'. This goes against the assumptions of the Transition State Theory which states that reaction trajectories should pass through this. As temperature increases, a greater proportion of excited vibrational states become occupied, leading to transition states occuring at different energies. As a result, the associated reaction trajectory at higher temperature travels through different activated complexes, making the reaction harder to model and any associated results more difficult to compare.

'''Reaction 2: H + HF --> F + H2'''

Reaction 2, which is the reverse reaction of Reaction 1, is endothermic with the energy of the products being higher than the reactants. As a result, according the Hammond's Postulate, the position of the transition state is later in the reaction trajectory and more closely resembles the products. Contrary to the preferred conditions for exothermic reactions, a larger amount of vibrational energy is needed. To achieve this, lower momentum would be adopted. Providing this is achieved, the set of reaction trajectories would be an example of an efficient reaction as the reaction trajectories would pass through the transition state once and to the products.

----

==References==

<references>
<ref name="transition">P. Jonathan P. K. Doye, David J. Wales, Journal of mathematical physics, 1996, '''48''', 843 - 846.</ref>
<ref name="thermodynamics">Chem Libre texts - Wave-particle duality (2015) [online] Available at: https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Quantum_Mechanics/09._The_Hydrogen_Atom/Atomic_Theory/Electrons_in_Atoms/Wave-Particle_Duality [Assessed 28 Sep 2015].</ref>
<ref name="bond energies">The Department of Chemistry Illinois (2017). Average Bond Energies [online] Available at: http://butane.chem.uiuc.edu/cyerkes/Chem104ACSpring2009/Genchemref/bondenergies.html [Accessed 9 Feb 2017].</ref>
<ref name="lawE">Mach, E. (1872). History and Root of the Principles of the Conservation of Energy. Open Court Pub. Co., Illinois.</ref>
<ref name="polanyi">Yan, S., Y.-T. Wu, and K. Liu. "Tracking The Energy Flow Along The Reaction Path". Proceedings of the National Academy of Sciences 105.35 (2008): 12667-12672.</ref>
<ref name="hammond">Chemistry LibreTexts. (2017). Hammond’s Postulate. [online] Available at: https://chem.libretexts.org/Reference/Organic_Chemistry_Glossary/Hammond’s_Postulate [Accessed 11 May 2017].</ref>
<references/>

File:Ab6114 figure9.png

2017-05-26T14:14:12Z

Ab6114: Ab6114 uploaded a new version of File:Ab6114 figure9.png

File:Ab6114 figure9.png

2017-05-26T14:13:08Z

Ab6114:

File:Ab6114 figure8.PNG

2017-05-26T14:11:35Z

Ab6114:

MRD:ab6114

2017-05-26T14:10:54Z

Ab6114:

=='''Molecular Reaction Dynamics'''==

==Exercise 1: H + H2 system==

==='''Question(1)'''===

What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

'''Answer:'''

Analysis of the PES curvature shows a maxima between two minimum points, all being stationary points with a gradient of zero. The first derivative of the curve, where the rate of change is zero, is found to confirm this. As Transition structure's inherently exist at the maximum point, it can be inferred that the transition structure lies exists at this point. In order to differentiate between the transition state and the minima, second derivative calculations can be performed, producing a negative value for a maxima and a positive value for a minima.<ref name="transition" />

----

==='''Question(2)'''===

Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.

'''Answer:'''

The approximate transition state position (r(ts)) was determined through analysis of the surface plot. This was confirmed by examining the internuclear distance vs time plot, to identify the point of intersection between the maxima and the minima - r(ts) is shown in the table below:

{| class="wikitable" border="1"
|+ Table 1
! r1 !! r2 !! r(ts)
|-
| 1.20 || 1.20 || 0.9075
|-
| 0.91 || 0.91 || 0.9075
|}

[[File:ab6114Figure1.PNG|680px|thumb|centre|Figure 1 - Internuclear distance vs Time plot of transition state position]]

A transition state is a point where the internuclear distance between all bonds remain constant and thus do not oscillate. Figure 1 shows that the internuclear distance does not oscillate at 0.9075 Å - hence this is the transition state position.

----

==='''Question(3)'''===

Comment on how the mep and the trajectory you just calculated differ.

'''Answer:'''

Dynamic calculations and mep are distinct for two main reasons. Calculations of mep span only the distance engaged by the transition state, while the dynamic calculation incorporates more of the reaction pathway - the dynamic calculation also takes mass into account, thus resulting in an oscillating reaction pathway line. No mass is included in the mep calculation, so no oscillation is observed.

----

==='''Question(4)'''===

Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.

'''Answer:'''

The effect of momentum variation on whether a trajectory is reactive or unreactive (for the initial positions r1 = 0.74 and r2 = 2.0):

{| class="wikitable" border="1"
|+ Table 2
! p1 !! p2 !! Trajectory !! Figure No.
|-
| -1.25 || -2.5 || Reactive || 2
|-
| -1.5 || -2.0 || Unreactive || 3
|-
| -1.5 || -2.5 || Reactive || 4
|-
| -2.5 || -5.0 || Unreactive || 5
|-
| -2.5 || -5.2 || Reactive || 6
|}

'''Figure 2'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 2. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc while at the same time the B-C bond distance remains relatively constant. This line continues through the transition state (r(A-B)=r(B-C)) and r(A-B) reaches a minimum and oscillates thus indicating that the Ha-Hb bond has formed. Simultaneously r(B-C) increases, confirming that the Hb-Hc bond has broken and Hc now exists as a separate hydrogen atom.

[[File:Ab6114_figure2.PNG|680px|thumb|centre|Figure 2 - The reactive trajectory associated with the specific parameters for Figure 2]]

'''Figure 3'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 3. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc. Since Ha has a momentum insufficient to overcome the activation energy barrier, the trajectory fails to reach the transition state - this is shown by the plot through r(A-B) returning to a larger distance without reaching transition state, and r(B-C) only oscillating due to vibrational fluctuations within the molecule.

[[File:Ab6114_figure3test2.PNG|680px|thumb|centre|Figure 3 - The unreactive trajectory associated with the specific parameters for Figure 3]]

'''Figure 4'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 4. The trajectory line illustrates a reaction where Ha had a momentum sufficient to overcome the activation energy barrier required to successfully interact with Hb-Hc, pass through the transition state, and form the molecule Ha-Hb.

[[File:Ab6114_figure4.PNG|680px|thumb|centre|Figure 4 - The reactive trajectory associated with the specific parameters for Figure 4]]

'''Figure 5'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 5. The trajectory line for these set parameters, initially show Ha to approach Hb-Hc with sufficient momentum to overcome the activation energy barrier and reach the transition state. However, instead of forming the product Ha-Hb, the trajectory line reverses and returns back to the reactants, with Hb-Hc exhibiting large vibrational energy fluctuations.

[[File:Ab6114_figure5.PNG|680px|thumb|centre|Figure 5 - The unreactive trajectory associated with the specific parameters for Figure 5]]

'''Figure 6'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 6. The trajectory line illustrates that Ha has sufficient momentum to overcome the activation energy barrier. However after reaching the transition state, the trajectory line goes back towards the reactants which is shown by r(B-C) decreasing and r(A-B) increasing in figure 6. The r(A-B) rose to a maxima before decreasing and oscillating, thus illustrating the formation of Ha-Hb. This reaction trajectory suggests that potentially an alternative reaction pathway was taken, one which does not involve a 'saddle point' on the PE plot corresponding with the transition states that normally produce the desired Ha and Hb-Hc.

[[File:Ab6114_figure6.PNG|680px|thumb|centre|Figure 6 - The reactive trajectory associated with the specific parameters for Figure 6]]

----

==='''Question(5)'''===

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

'''Answer:'''

The Transition State Theory's main assumption is that the reactants and activated transition states are at quasi-equilibrium. A quasi-equilibrium utilises thermodynamic molecular driving forces that modify the systems state being very small and slow, allowing for the existence of an internal equilibrium that undergoes only small deviations from this system. Classical Transition State Theory explains that an atom with a momentum satisfying the activation energy barrier requirements, will react upon collision with a given molecule and synthesize products. As this theory however ignores potential quantum mechanical interactions, reaction rates calculated will be slower than what is found in experimental data. Quantum mechanical contributing calculations allow for the tunneling behaviour of particles to be factored into an alternative reaction pathway. When an atom and a molecule lack the sufficient activation energy to overcome this barrier, the atom is able to 'tunnel' through this barrier, due to the particle-wave duality observed in quantum-scale objects.<ref name="thermodynamics" />

----

==Exercise 2: F - H - H system==

==='''Question(6)'''===

Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

'''Answer:'''

Analysis:

- H-F Bond Energy = 565 KJ/Mol<ref name="bond energies" />

- H-H Bond Energy = 432 KJ/Mol<ref name="bond energies" />

- H-F is a stronger bond than H-H

- ΔE = E(bonds cleaved)-E(bonds formed)

'''F + H2''' --> ΔE = 432 - 565 = -133 KJ/mol '''THUS EXOTHERMIC'''

'''H + H-F''' --> ΔE = 565 - 432 = 133 KJ/mol '''THUS ENDOTHERMIC'''

----

==='''Question(7)'''===

Locate the approximate position of the transition state.

'''Answer:'''

In order to locate the position of the transition state, the radii corresponding to a stationary total system energy (gradient=0) were identified. The positions of the radii have a zero net force acting upon them, so plotting the internuclear distance vs time results in a non-oscillating constant internuclear separation - see the internuclear separations below and the corresponding plot vs time in figure 7:

- r(H-F) = 1.815Å

- r(H-H) = 0.74Å

[[File:ab6114 figure7.PNG|680px|thumb|centre|Figure 7 - Internuclear distance vs Time plot for the transition state position]]

----

==='''Question(8)'''===

Report the activation energy for both reactions.

'''Answer:'''

Determining the activation energy for both reactions, required chemical intuition and trial and error in order to identify an adequate reaction trajectory. The activation energy is equal to the energy difference between the reactants and the transition state - activation energy values calculated in absence of mathematical analysis, thus all energy values involved would actually have +/- values.

'''Reaction 1: F + H2 --> HF + H'''

- Ea = -103.7 - (-103.9) = 0.2 Kcal/Mol

The energy values used for this calculation were obtained from Figure 8 (reactants) and Figure 9 (transition state) below:

[[File:ab6114 figure8.PNG|470px|thumb|centre|Figure 8 - Surface plot illustrating energy of the reactants]]

[[File:ab6114 figure9|470px|thumb|centre|Figure 9 - Surface plot illustrating the energy of the transition state]]

'''Reaction 2: H + HF --> F + H2'''

- Ea = -103.7 - (-133.9) = 30.2 Kcal/Mol

The energy values used for this calculation were obtained from Figure 10 (reactants) and Figure 9 (transition state) as shown below:

[[File:HZ4315 Figure9.PNG|500px|thumb|centre|Figure 9 - Surface plot showing the energy at the transition state]]

[[File:HZ4315 Figure10.PNG|500px|thumb|centre|Figure 10 - Surface plot showing the energy of the reactants in the reverse reaction]]

----

==='''Q(9)'''===

In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

'''A:'''

According to the Law of Conservation of Energy, the total energy of an isolated system must be conserved as energy is neither created nor destroyed, just converted from one form to another.<ref name="lawE" /> For this specific reaction system, the total energy is the sum of the kinetic and potential energies of each substrate involved in the reaction. For the exothermic reaction F + H2, the potential energy of the products is lower than the reactants hence, the kinetic energy of the products must be higher. This is illustrated by determining a successful reaction trajectory and plotting a kinetic energy vs time graph as shown below:

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 11 - A reaction pathway for an ultimately successful reaction (parameters determined using trial and error)]]

[[File:KEvsTIME.PNG|500px|thumb|centre|Figure 12 - Kinetic energy vs Time plot of the successful reaction - the small oscillations on the left correspond to the reactants which are clearly converted to products as the kinetic energy increases significantly]]

The increased kinetic energy allows the products to be detected by IR analysis. Since the H-F bond is oscillating, it will release energy in the form of vibrational energy, resulting in a peak on the spectrum. The H-H vibrational peak present on the reactant spectrum will now be absent. Another possible experimental technique is calorimetry. Since the reaction is exothermic, the amount of heat released can be measured and since, at constant pressure, the enthalpy change is equal to the net change in energy, the net free energy change of the reaction can be determined.

----

==='''Q(10)'''===

Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

'''A:'''

Using trial and error, a successful reaction trajectory was found for both reactions shown below. The parameters used can be seen within the figures.

'''Reaction 1: F + H2 --> HF + H'''

'''Lower momentum'''

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 13 - An ultimately successful reaction pathway at high momentum indicating the preference of an early transition state]]

'''Higher momentum'''

[[File:HZ4315 Figure12.PNG|500px|thumb|centre|Figure 14 - The reaction pathway at an even larger momentum to confirm the theory]]

Reaction 1, which is an exothermic reaction, exhibits an increased translational energy in response to increasing the momentum of the reactants. According to Polanyi's rules, higher translational energies are more efficient at activating an early transition state (ie. an early barrier) than vibrational energy.<ref name="polanyi" /> Since early transition states resemble the reactants (as stated by Hammond's Postulate)<ref name="hammond" />, the reaction trajectory may be expected to bounce back towards the reactants. Figure 12 confirms this as an increased momentum (and therefore increased translational energy) sees the trajectory bounce back several times before eventually proceeding to the form the products. Figures 11 and 12 are an example of a set of inefficient reaction trajectories as the barrier is recrossed several times which is not ideal in the formation of the product.

The exothermic nature of the reaction also alters the position of the transition state away from the lowest 'saddle point'. This goes against the assumptions of the Transition State Theory which states that reaction trajectories should pass through this. As temperature increases, a greater proportion of excited vibrational states become occupied, leading to transition states occuring at different energies. As a result, the associated reaction trajectory at higher temperature travels through different activated complexes, making the reaction harder to model and any associated results more difficult to compare.

'''Reaction 2: H + HF --> F + H2'''

Reaction 2, which is the reverse reaction of Reaction 1, is endothermic with the energy of the products being higher than the reactants. As a result, according the Hammond's Postulate, the position of the transition state is later in the reaction trajectory and more closely resembles the products. Contrary to the preferred conditions for exothermic reactions, a larger amount of vibrational energy is needed. To achieve this, lower momentum would be adopted. Providing this is achieved, the set of reaction trajectories would be an example of an efficient reaction as the reaction trajectories would pass through the transition state once and to the products.

----

==References==

<references>
<ref name="transition">P. Jonathan P. K. Doye, David J. Wales, Journal of mathematical physics, 1996, '''48''', 843 - 846.</ref>
<ref name="thermodynamics">Chem Libre texts - Wave-particle duality (2015) [online] Available at: https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Quantum_Mechanics/09._The_Hydrogen_Atom/Atomic_Theory/Electrons_in_Atoms/Wave-Particle_Duality [Assessed 28 Sep 2015].</ref>
<ref name="bond energies">The Department of Chemistry Illinois (2017). Average Bond Energies [online] Available at: http://butane.chem.uiuc.edu/cyerkes/Chem104ACSpring2009/Genchemref/bondenergies.html [Accessed 9 Feb 2017].</ref>
<ref name="lawE">Mach, E. (1872). History and Root of the Principles of the Conservation of Energy. Open Court Pub. Co., Illinois.</ref>
<ref name="polanyi">Yan, S., Y.-T. Wu, and K. Liu. "Tracking The Energy Flow Along The Reaction Path". Proceedings of the National Academy of Sciences 105.35 (2008): 12667-12672.</ref>
<ref name="hammond">Chemistry LibreTexts. (2017). Hammond’s Postulate. [online] Available at: https://chem.libretexts.org/Reference/Organic_Chemistry_Glossary/Hammond’s_Postulate [Accessed 11 May 2017].</ref>
<references/>

File:Ab6114 figure7.PNG

2017-05-26T13:41:40Z

Ab6114:

MRD:ab6114

2017-05-26T13:41:07Z

Ab6114:

=='''Molecular Reaction Dynamics'''==

==Exercise 1: H + H2 system==

==='''Question(1)'''===

What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

'''Answer:'''

Analysis of the PES curvature shows a maxima between two minimum points, all being stationary points with a gradient of zero. The first derivative of the curve, where the rate of change is zero, is found to confirm this. As Transition structure's inherently exist at the maximum point, it can be inferred that the transition structure lies exists at this point. In order to differentiate between the transition state and the minima, second derivative calculations can be performed, producing a negative value for a maxima and a positive value for a minima.<ref name="transition" />

----

==='''Question(2)'''===

Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.

'''Answer:'''

The approximate transition state position (r(ts)) was determined through analysis of the surface plot. This was confirmed by examining the internuclear distance vs time plot, to identify the point of intersection between the maxima and the minima - r(ts) is shown in the table below:

{| class="wikitable" border="1"
|+ Table 1
! r1 !! r2 !! r(ts)
|-
| 1.20 || 1.20 || 0.9075
|-
| 0.91 || 0.91 || 0.9075
|}

[[File:ab6114Figure1.PNG|680px|thumb|centre|Figure 1 - Internuclear distance vs Time plot of transition state position]]

A transition state is a point where the internuclear distance between all bonds remain constant and thus do not oscillate. Figure 1 shows that the internuclear distance does not oscillate at 0.9075 Å - hence this is the transition state position.

----

==='''Question(3)'''===

Comment on how the mep and the trajectory you just calculated differ.

'''Answer:'''

Dynamic calculations and mep are distinct for two main reasons. Calculations of mep span only the distance engaged by the transition state, while the dynamic calculation incorporates more of the reaction pathway - the dynamic calculation also takes mass into account, thus resulting in an oscillating reaction pathway line. No mass is included in the mep calculation, so no oscillation is observed.

----

==='''Question(4)'''===

Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.

'''Answer:'''

The effect of momentum variation on whether a trajectory is reactive or unreactive (for the initial positions r1 = 0.74 and r2 = 2.0):

{| class="wikitable" border="1"
|+ Table 2
! p1 !! p2 !! Trajectory !! Figure No.
|-
| -1.25 || -2.5 || Reactive || 2
|-
| -1.5 || -2.0 || Unreactive || 3
|-
| -1.5 || -2.5 || Reactive || 4
|-
| -2.5 || -5.0 || Unreactive || 5
|-
| -2.5 || -5.2 || Reactive || 6
|}

'''Figure 2'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 2. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc while at the same time the B-C bond distance remains relatively constant. This line continues through the transition state (r(A-B)=r(B-C)) and r(A-B) reaches a minimum and oscillates thus indicating that the Ha-Hb bond has formed. Simultaneously r(B-C) increases, confirming that the Hb-Hc bond has broken and Hc now exists as a separate hydrogen atom.

[[File:Ab6114_figure2.PNG|680px|thumb|centre|Figure 2 - The reactive trajectory associated with the specific parameters for Figure 2]]

'''Figure 3'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 3. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc. Since Ha has a momentum insufficient to overcome the activation energy barrier, the trajectory fails to reach the transition state - this is shown by the plot through r(A-B) returning to a larger distance without reaching transition state, and r(B-C) only oscillating due to vibrational fluctuations within the molecule.

[[File:Ab6114_figure3test2.PNG|680px|thumb|centre|Figure 3 - The unreactive trajectory associated with the specific parameters for Figure 3]]

'''Figure 4'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 4. The trajectory line illustrates a reaction where Ha had a momentum sufficient to overcome the activation energy barrier required to successfully interact with Hb-Hc, pass through the transition state, and form the molecule Ha-Hb.

[[File:Ab6114_figure4.PNG|680px|thumb|centre|Figure 4 - The reactive trajectory associated with the specific parameters for Figure 4]]

'''Figure 5'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 5. The trajectory line for these set parameters, initially show Ha to approach Hb-Hc with sufficient momentum to overcome the activation energy barrier and reach the transition state. However, instead of forming the product Ha-Hb, the trajectory line reverses and returns back to the reactants, with Hb-Hc exhibiting large vibrational energy fluctuations.

[[File:Ab6114_figure5.PNG|680px|thumb|centre|Figure 5 - The unreactive trajectory associated with the specific parameters for Figure 5]]

'''Figure 6'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 6. The trajectory line illustrates that Ha has sufficient momentum to overcome the activation energy barrier. However after reaching the transition state, the trajectory line goes back towards the reactants which is shown by r(B-C) decreasing and r(A-B) increasing in figure 6. The r(A-B) rose to a maxima before decreasing and oscillating, thus illustrating the formation of Ha-Hb. This reaction trajectory suggests that potentially an alternative reaction pathway was taken, one which does not involve a 'saddle point' on the PE plot corresponding with the transition states that normally produce the desired Ha and Hb-Hc.

[[File:Ab6114_figure6.PNG|680px|thumb|centre|Figure 6 - The reactive trajectory associated with the specific parameters for Figure 6]]

----

==='''Question(5)'''===

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

'''Answer:'''

The Transition State Theory's main assumption is that the reactants and activated transition states are at quasi-equilibrium. A quasi-equilibrium utilises thermodynamic molecular driving forces that modify the systems state being very small and slow, allowing for the existence of an internal equilibrium that undergoes only small deviations from this system. Classical Transition State Theory explains that an atom with a momentum satisfying the activation energy barrier requirements, will react upon collision with a given molecule and synthesize products. As this theory however ignores potential quantum mechanical interactions, reaction rates calculated will be slower than what is found in experimental data. Quantum mechanical contributing calculations allow for the tunneling behaviour of particles to be factored into an alternative reaction pathway. When an atom and a molecule lack the sufficient activation energy to overcome this barrier, the atom is able to 'tunnel' through this barrier, due to the particle-wave duality observed in quantum-scale objects.<ref name="thermodynamics" />

----

==Exercise 2: F - H - H system==

==='''Question(6)'''===

Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

'''Answer:'''

Analysis:

- H-F Bond Energy = 565 KJ/Mol<ref name="bond energies" />

- H-H Bond Energy = 432 KJ/Mol<ref name="bond energies" />

- H-F is a stronger bond than H-H

- ΔE = E(bonds cleaved)-E(bonds formed)

'''F + H2''' --> ΔE = 432 - 565 = -133 KJ/mol '''THUS EXOTHERMIC'''

'''H + H-F''' --> ΔE = 565 - 432 = 133 KJ/mol '''THUS ENDOTHERMIC'''

----

==='''Question(7)'''===

Locate the approximate position of the transition state.

'''Answer:'''

In order to locate the position of the transition state, the radii corresponding to a stationary total system energy (gradient=0) were identified. The positions of the radii have a zero net force acting upon them, so plotting the internuclear distance vs time results in a non-oscillating constant internuclear separation - see the internuclear separations below and the corresponding plot vs time in figure 7:

- r(H-F) = 1.815Å

- r(H-H) = 0.74Å

[[File:ab6114 figure7.PNG|680px|thumb|centre|Figure 7 - Internuclear distance vs Time plot for the transition state position]]

----

==='''Q(8)'''===

Report the activation energy for both reactions.

'''A:'''

In order to determine the activation energy for both reactions, chemical intuition and trial and error were used to find a suitable reaction trajectory. The activation energy was then equal to the difference in energy between that of the reactants and the transition state. The activation energy values obtained were done without mathematical analysis therefore, all energy values involved would actually have a +/- value.

'''Reaction 1: F + H2 --> HF + H'''

- Ea = -103.7 - (-103.9) = 0.2 KCal/Mol

The energy values used for this calculation were obtained from Figure 8 (reactants) and Figure 9 (transition state) as shown below:

[[File:HZ4315 Figure8.PNG|500px|thumb|centre|Figure 8 - Surface plot showing the energy of the reactants]]

[[File:HZ4315 Figure9.PNG|500px|thumb|centre|Figure 9 - Surface plot showing the energy at the transition state]]

'''Reaction 2: H + HF --> F + H2'''

- Ea = -103.7 - (-133.9) = 30.2 KCal/Mol

The energy values used for this calculation were obtained from Figure 10 (reactants) and Figure 9 (transition state) as shown below:

[[File:HZ4315 Figure9.PNG|500px|thumb|centre|Figure 9 - Surface plot showing the energy at the transition state]]

[[File:HZ4315 Figure10.PNG|500px|thumb|centre|Figure 10 - Surface plot showing the energy of the reactants in the reverse reaction]]

----

==='''Q(9)'''===

In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

'''A:'''

According to the Law of Conservation of Energy, the total energy of an isolated system must be conserved as energy is neither created nor destroyed, just converted from one form to another.<ref name="lawE" /> For this specific reaction system, the total energy is the sum of the kinetic and potential energies of each substrate involved in the reaction. For the exothermic reaction F + H2, the potential energy of the products is lower than the reactants hence, the kinetic energy of the products must be higher. This is illustrated by determining a successful reaction trajectory and plotting a kinetic energy vs time graph as shown below:

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 11 - A reaction pathway for an ultimately successful reaction (parameters determined using trial and error)]]

[[File:KEvsTIME.PNG|500px|thumb|centre|Figure 12 - Kinetic energy vs Time plot of the successful reaction - the small oscillations on the left correspond to the reactants which are clearly converted to products as the kinetic energy increases significantly]]

The increased kinetic energy allows the products to be detected by IR analysis. Since the H-F bond is oscillating, it will release energy in the form of vibrational energy, resulting in a peak on the spectrum. The H-H vibrational peak present on the reactant spectrum will now be absent. Another possible experimental technique is calorimetry. Since the reaction is exothermic, the amount of heat released can be measured and since, at constant pressure, the enthalpy change is equal to the net change in energy, the net free energy change of the reaction can be determined.

----

==='''Q(10)'''===

Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

'''A:'''

Using trial and error, a successful reaction trajectory was found for both reactions shown below. The parameters used can be seen within the figures.

'''Reaction 1: F + H2 --> HF + H'''

'''Lower momentum'''

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 13 - An ultimately successful reaction pathway at high momentum indicating the preference of an early transition state]]

'''Higher momentum'''

[[File:HZ4315 Figure12.PNG|500px|thumb|centre|Figure 14 - The reaction pathway at an even larger momentum to confirm the theory]]

Reaction 1, which is an exothermic reaction, exhibits an increased translational energy in response to increasing the momentum of the reactants. According to Polanyi's rules, higher translational energies are more efficient at activating an early transition state (ie. an early barrier) than vibrational energy.<ref name="polanyi" /> Since early transition states resemble the reactants (as stated by Hammond's Postulate)<ref name="hammond" />, the reaction trajectory may be expected to bounce back towards the reactants. Figure 12 confirms this as an increased momentum (and therefore increased translational energy) sees the trajectory bounce back several times before eventually proceeding to the form the products. Figures 11 and 12 are an example of a set of inefficient reaction trajectories as the barrier is recrossed several times which is not ideal in the formation of the product.

The exothermic nature of the reaction also alters the position of the transition state away from the lowest 'saddle point'. This goes against the assumptions of the Transition State Theory which states that reaction trajectories should pass through this. As temperature increases, a greater proportion of excited vibrational states become occupied, leading to transition states occuring at different energies. As a result, the associated reaction trajectory at higher temperature travels through different activated complexes, making the reaction harder to model and any associated results more difficult to compare.

'''Reaction 2: H + HF --> F + H2'''

Reaction 2, which is the reverse reaction of Reaction 1, is endothermic with the energy of the products being higher than the reactants. As a result, according the Hammond's Postulate, the position of the transition state is later in the reaction trajectory and more closely resembles the products. Contrary to the preferred conditions for exothermic reactions, a larger amount of vibrational energy is needed. To achieve this, lower momentum would be adopted. Providing this is achieved, the set of reaction trajectories would be an example of an efficient reaction as the reaction trajectories would pass through the transition state once and to the products.

----

==References==

<references>
<ref name="transition">P. Jonathan P. K. Doye, David J. Wales, Journal of mathematical physics, 1996, '''48''', 843 - 846.</ref>
<ref name="thermodynamics">Chem Libre texts - Wave-particle duality (2015) [online] Available at: https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Quantum_Mechanics/09._The_Hydrogen_Atom/Atomic_Theory/Electrons_in_Atoms/Wave-Particle_Duality [Assessed 28 Sep 2015].</ref>
<ref name="bond energies">The Department of Chemistry Illinois (2017). Average Bond Energies [online] Available at: http://butane.chem.uiuc.edu/cyerkes/Chem104ACSpring2009/Genchemref/bondenergies.html [Accessed 9 Feb 2017].</ref>
<ref name="lawE">Mach, E. (1872). History and Root of the Principles of the Conservation of Energy. Open Court Pub. Co., Illinois.</ref>
<ref name="polanyi">Yan, S., Y.-T. Wu, and K. Liu. "Tracking The Energy Flow Along The Reaction Path". Proceedings of the National Academy of Sciences 105.35 (2008): 12667-12672.</ref>
<ref name="hammond">Chemistry LibreTexts. (2017). Hammond’s Postulate. [online] Available at: https://chem.libretexts.org/Reference/Organic_Chemistry_Glossary/Hammond’s_Postulate [Accessed 11 May 2017].</ref>
<references/>

MRD:ab6114

2017-05-26T13:21:52Z

Ab6114:

=='''Molecular Reaction Dynamics'''==

==Exercise 1: H + H2 system==

==='''Question(1)'''===

What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

'''Answer:'''

Analysis of the PES curvature shows a maxima between two minimum points, all being stationary points with a gradient of zero. The first derivative of the curve, where the rate of change is zero, is found to confirm this. As Transition structure's inherently exist at the maximum point, it can be inferred that the transition structure lies exists at this point. In order to differentiate between the transition state and the minima, second derivative calculations can be performed, producing a negative value for a maxima and a positive value for a minima.<ref name="transition" />

----

==='''Question(2)'''===

Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.

'''Answer:'''

The approximate transition state position (r(ts)) was determined through analysis of the surface plot. This was confirmed by examining the internuclear distance vs time plot, to identify the point of intersection between the maxima and the minima - r(ts) is shown in the table below:

{| class="wikitable" border="1"
|+ Table 1
! r1 !! r2 !! r(ts)
|-
| 1.20 || 1.20 || 0.9075
|-
| 0.91 || 0.91 || 0.9075
|}

[[File:ab6114Figure1.PNG|680px|thumb|centre|Figure 1 - Internuclear distance vs Time plot of transition state position]]

A transition state is a point where the internuclear distance between all bonds remain constant and thus do not oscillate. Figure 1 shows that the internuclear distance does not oscillate at 0.9075 Å - hence this is the transition state position.

----

==='''Question(3)'''===

Comment on how the mep and the trajectory you just calculated differ.

'''Answer:'''

Dynamic calculations and mep are distinct for two main reasons. Calculations of mep span only the distance engaged by the transition state, while the dynamic calculation incorporates more of the reaction pathway - the dynamic calculation also takes mass into account, thus resulting in an oscillating reaction pathway line. No mass is included in the mep calculation, so no oscillation is observed.

----

==='''Question(4)'''===

Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.

'''Answer:'''

The effect of momentum variation on whether a trajectory is reactive or unreactive (for the initial positions r1 = 0.74 and r2 = 2.0):

{| class="wikitable" border="1"
|+ Table 2
! p1 !! p2 !! Trajectory !! Figure No.
|-
| -1.25 || -2.5 || Reactive || 2
|-
| -1.5 || -2.0 || Unreactive || 3
|-
| -1.5 || -2.5 || Reactive || 4
|-
| -2.5 || -5.0 || Unreactive || 5
|-
| -2.5 || -5.2 || Reactive || 6
|}

'''Figure 2'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 2. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc while at the same time the B-C bond distance remains relatively constant. This line continues through the transition state (r(A-B)=r(B-C)) and r(A-B) reaches a minimum and oscillates thus indicating that the Ha-Hb bond has formed. Simultaneously r(B-C) increases, confirming that the Hb-Hc bond has broken and Hc now exists as a separate hydrogen atom.

[[File:Ab6114_figure2.PNG|680px|thumb|centre|Figure 2 - The reactive trajectory associated with the specific parameters for Figure 2]]

'''Figure 3'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 3. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc. Since Ha has a momentum insufficient to overcome the activation energy barrier, the trajectory fails to reach the transition state - this is shown by the plot through r(A-B) returning to a larger distance without reaching transition state, and r(B-C) only oscillating due to vibrational fluctuations within the molecule.

[[File:Ab6114_figure3test2.PNG|680px|thumb|centre|Figure 3 - The unreactive trajectory associated with the specific parameters for Figure 3]]

'''Figure 4'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 4. The trajectory line illustrates a reaction where Ha had a momentum sufficient to overcome the activation energy barrier required to successfully interact with Hb-Hc, pass through the transition state, and form the molecule Ha-Hb.

[[File:Ab6114_figure4.PNG|680px|thumb|centre|Figure 4 - The reactive trajectory associated with the specific parameters for Figure 4]]

'''Figure 5'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 5. The trajectory line for these set parameters, initially show Ha to approach Hb-Hc with sufficient momentum to overcome the activation energy barrier and reach the transition state. However, instead of forming the product Ha-Hb, the trajectory line reverses and returns back to the reactants, with Hb-Hc exhibiting large vibrational energy fluctuations.

[[File:Ab6114_figure5.PNG|680px|thumb|centre|Figure 5 - The unreactive trajectory associated with the specific parameters for Figure 5]]

'''Figure 6'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 6. The trajectory line illustrates that Ha has sufficient momentum to overcome the activation energy barrier. However after reaching the transition state, the trajectory line goes back towards the reactants which is shown by r(B-C) decreasing and r(A-B) increasing in figure 6. The r(A-B) rose to a maxima before decreasing and oscillating, thus illustrating the formation of Ha-Hb. This reaction trajectory suggests that potentially an alternative reaction pathway was taken, one which does not involve a 'saddle point' on the PE plot corresponding with the transition states that normally produce the desired Ha and Hb-Hc.

[[File:Ab6114_figure6.PNG|680px|thumb|centre|Figure 6 - The reactive trajectory associated with the specific parameters for Figure 6]]

----

==='''Question(5)'''===

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

'''Answer:'''

The Transition State Theory's main assumption is that the reactants and activated transition states are at quasi-equilibrium. A quasi-equilibrium utilises thermodynamic molecular driving forces that modify the systems state being very small and slow, allowing for the existence of an internal equilibrium that undergoes only small deviations from this system. Classical Transition State Theory explains that an atom with a momentum satisfying the activation energy barrier requirements, will react upon collision with a given molecule and synthesize products. As this theory however ignores potential quantum mechanical interactions, reaction rates calculated will be slower than what is found in experimental data. Quantum mechanical contributing calculations allow for the tunneling behaviour of particles to be factored into an alternative reaction pathway. When an atom and a molecule lack the sufficient activation energy to overcome this barrier, the atom is able to 'tunnel' through this barrier, due to the particle-wave duality observed in quantum-scale objects.<ref name="thermodynamics" />

----

==Exercise 2: F - H - H system==

==='''Question(6)'''===

Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

'''Answer:'''

Analysis:

- H-F Bond Energy = 565 KJ/Mol<ref name="bond energies" />

- H-H Bond Energy = 432 KJ/Mol<ref name="bond energies" />

- H-F is a stronger bond than H-H

- ΔE = E(bonds cleaved)-E(bonds formed)

'''F + H2''' --> ΔE = 432 - 565 = -133 KJ/mol '''THUS EXOTHERMIC'''

'''H + H-F''' --> ΔE = 565 - 432 = 133 KJ/mol '''THUS ENDOTHERMIC'''

----

==='''Q(7)'''===

Locate the approximate position of the transition state.

'''A:'''

To determine the position of the transition state, the radii at which the total energy of the system is stationary (ie. the gradient is zero) were found. The location of these radii is such that the net force acting upon them is zero and plotting the internuclear distance against time results in a constant internuclear separation which does not oscillate. The internuclear separations are shown below and the associated plot against time can be seen in figure 7:

- r(H-H) = 0.74Å

- r(H-F) = 1.815Å

[[File:HZ4315 Figure 7.PNG|700px|thumb|centre|Figure 7 - The Internuclear distance vs Time plot for the transition state position]]

----

==='''Q(8)'''===

Report the activation energy for both reactions.

'''A:'''

In order to determine the activation energy for both reactions, chemical intuition and trial and error were used to find a suitable reaction trajectory. The activation energy was then equal to the difference in energy between that of the reactants and the transition state. The activation energy values obtained were done without mathematical analysis therefore, all energy values involved would actually have a +/- value.

'''Reaction 1: F + H2 --> HF + H'''

- Ea = -103.7 - (-103.9) = 0.2 KCal/Mol

The energy values used for this calculation were obtained from Figure 8 (reactants) and Figure 9 (transition state) as shown below:

[[File:HZ4315 Figure8.PNG|500px|thumb|centre|Figure 8 - Surface plot showing the energy of the reactants]]

[[File:HZ4315 Figure9.PNG|500px|thumb|centre|Figure 9 - Surface plot showing the energy at the transition state]]

'''Reaction 2: H + HF --> F + H2'''

- Ea = -103.7 - (-133.9) = 30.2 KCal/Mol

The energy values used for this calculation were obtained from Figure 10 (reactants) and Figure 9 (transition state) as shown below:

[[File:HZ4315 Figure9.PNG|500px|thumb|centre|Figure 9 - Surface plot showing the energy at the transition state]]

[[File:HZ4315 Figure10.PNG|500px|thumb|centre|Figure 10 - Surface plot showing the energy of the reactants in the reverse reaction]]

----

==='''Q(9)'''===

In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

'''A:'''

According to the Law of Conservation of Energy, the total energy of an isolated system must be conserved as energy is neither created nor destroyed, just converted from one form to another.<ref name="lawE" /> For this specific reaction system, the total energy is the sum of the kinetic and potential energies of each substrate involved in the reaction. For the exothermic reaction F + H2, the potential energy of the products is lower than the reactants hence, the kinetic energy of the products must be higher. This is illustrated by determining a successful reaction trajectory and plotting a kinetic energy vs time graph as shown below:

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 11 - A reaction pathway for an ultimately successful reaction (parameters determined using trial and error)]]

[[File:KEvsTIME.PNG|500px|thumb|centre|Figure 12 - Kinetic energy vs Time plot of the successful reaction - the small oscillations on the left correspond to the reactants which are clearly converted to products as the kinetic energy increases significantly]]

The increased kinetic energy allows the products to be detected by IR analysis. Since the H-F bond is oscillating, it will release energy in the form of vibrational energy, resulting in a peak on the spectrum. The H-H vibrational peak present on the reactant spectrum will now be absent. Another possible experimental technique is calorimetry. Since the reaction is exothermic, the amount of heat released can be measured and since, at constant pressure, the enthalpy change is equal to the net change in energy, the net free energy change of the reaction can be determined.

----

==='''Q(10)'''===

Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

'''A:'''

Using trial and error, a successful reaction trajectory was found for both reactions shown below. The parameters used can be seen within the figures.

'''Reaction 1: F + H2 --> HF + H'''

'''Lower momentum'''

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 13 - An ultimately successful reaction pathway at high momentum indicating the preference of an early transition state]]

'''Higher momentum'''

[[File:HZ4315 Figure12.PNG|500px|thumb|centre|Figure 14 - The reaction pathway at an even larger momentum to confirm the theory]]

Reaction 1, which is an exothermic reaction, exhibits an increased translational energy in response to increasing the momentum of the reactants. According to Polanyi's rules, higher translational energies are more efficient at activating an early transition state (ie. an early barrier) than vibrational energy.<ref name="polanyi" /> Since early transition states resemble the reactants (as stated by Hammond's Postulate)<ref name="hammond" />, the reaction trajectory may be expected to bounce back towards the reactants. Figure 12 confirms this as an increased momentum (and therefore increased translational energy) sees the trajectory bounce back several times before eventually proceeding to the form the products. Figures 11 and 12 are an example of a set of inefficient reaction trajectories as the barrier is recrossed several times which is not ideal in the formation of the product.

The exothermic nature of the reaction also alters the position of the transition state away from the lowest 'saddle point'. This goes against the assumptions of the Transition State Theory which states that reaction trajectories should pass through this. As temperature increases, a greater proportion of excited vibrational states become occupied, leading to transition states occuring at different energies. As a result, the associated reaction trajectory at higher temperature travels through different activated complexes, making the reaction harder to model and any associated results more difficult to compare.

'''Reaction 2: H + HF --> F + H2'''

Reaction 2, which is the reverse reaction of Reaction 1, is endothermic with the energy of the products being higher than the reactants. As a result, according the Hammond's Postulate, the position of the transition state is later in the reaction trajectory and more closely resembles the products. Contrary to the preferred conditions for exothermic reactions, a larger amount of vibrational energy is needed. To achieve this, lower momentum would be adopted. Providing this is achieved, the set of reaction trajectories would be an example of an efficient reaction as the reaction trajectories would pass through the transition state once and to the products.

----

==References==

<references>
<ref name="transition">P. Jonathan P. K. Doye, David J. Wales, Journal of mathematical physics, 1996, '''48''', 843 - 846.</ref>
<ref name="thermodynamics">Chem Libre texts - Wave-particle duality (2015) [online] Available at: https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Quantum_Mechanics/09._The_Hydrogen_Atom/Atomic_Theory/Electrons_in_Atoms/Wave-Particle_Duality [Assessed 28 Sep 2015].</ref>
<ref name="bond energies">The Department of Chemistry Illinois (2017). Average Bond Energies [online] Available at: http://butane.chem.uiuc.edu/cyerkes/Chem104ACSpring2009/Genchemref/bondenergies.html [Accessed 9 Feb 2017].</ref>
<ref name="lawE">Mach, E. (1872). History and Root of the Principles of the Conservation of Energy. Open Court Pub. Co., Illinois.</ref>
<ref name="polanyi">Yan, S., Y.-T. Wu, and K. Liu. "Tracking The Energy Flow Along The Reaction Path". Proceedings of the National Academy of Sciences 105.35 (2008): 12667-12672.</ref>
<ref name="hammond">Chemistry LibreTexts. (2017). Hammond’s Postulate. [online] Available at: https://chem.libretexts.org/Reference/Organic_Chemistry_Glossary/Hammond’s_Postulate [Accessed 11 May 2017].</ref>
<references/>

MRD:ab6114

2017-05-26T12:56:57Z

Ab6114:

=='''Molecular Reaction Dynamics'''==

==Exercise 1: H + H2 system==

==='''Question(1)'''===

What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

'''Answer:'''

Analysis of the PES curvature shows a maxima between two minimum points, all being stationary points with a gradient of zero. The first derivative of the curve, where the rate of change is zero, is found to confirm this. As Transition structure's inherently exist at the maximum point, it can be inferred that the transition structure lies exists at this point. In order to differentiate between the transition state and the minima, second derivative calculations can be performed, producing a negative value for a maxima and a positive value for a minima.<ref name="transition" />

----

==='''Question(2)'''===

Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.

'''Answer:'''

The approximate transition state position (r(ts)) was determined through analysis of the surface plot. This was confirmed by examining the internuclear distance vs time plot, to identify the point of intersection between the maxima and the minima - r(ts) is shown in the table below:

{| class="wikitable" border="1"
|+ Table 1
! r1 !! r2 !! r(ts)
|-
| 1.20 || 1.20 || 0.9075
|-
| 0.91 || 0.91 || 0.9075
|}

[[File:ab6114Figure1.PNG|680px|thumb|centre|Figure 1 - Internuclear distance vs Time plot of transition state position]]

A transition state is a point where the internuclear distance between all bonds remain constant and thus do not oscillate. Figure 1 shows that the internuclear distance does not oscillate at 0.9075 Å - hence this is the transition state position.

----

==='''Question(3)'''===

Comment on how the mep and the trajectory you just calculated differ.

'''Answer:'''

Dynamic calculations and mep are distinct for two main reasons. Calculations of mep span only the distance engaged by the transition state, while the dynamic calculation incorporates more of the reaction pathway - the dynamic calculation also takes mass into account, thus resulting in an oscillating reaction pathway line. No mass is included in the mep calculation, so no oscillation is observed.

----

==='''Question(4)'''===

Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.

'''Answer:'''

The effect of momentum variation on whether a trajectory is reactive or unreactive (for the initial positions r1 = 0.74 and r2 = 2.0):

{| class="wikitable" border="1"
|+ Table 2
! p1 !! p2 !! Trajectory !! Figure No.
|-
| -1.25 || -2.5 || Reactive || 2
|-
| -1.5 || -2.0 || Unreactive || 3
|-
| -1.5 || -2.5 || Reactive || 4
|-
| -2.5 || -5.0 || Unreactive || 5
|-
| -2.5 || -5.2 || Reactive || 6
|}

'''Figure 2'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 2. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc while at the same time the B-C bond distance remains relatively constant. This line continues through the transition state (r(A-B)=r(B-C)) and r(A-B) reaches a minimum and oscillates thus indicating that the Ha-Hb bond has formed. Simultaneously r(B-C) increases, confirming that the Hb-Hc bond has broken and Hc now exists as a separate hydrogen atom.

[[File:Ab6114_figure2.PNG|680px|thumb|centre|Figure 2 - The reactive trajectory associated with the specific parameters for Figure 2]]

'''Figure 3'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 3. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc. Since Ha has a momentum insufficient to overcome the activation energy barrier, the trajectory fails to reach the transition state - this is shown by the plot through r(A-B) returning to a larger distance without reaching transition state, and r(B-C) only oscillating due to vibrational fluctuations within the molecule.

[[File:Ab6114_figure3test2.PNG|680px|thumb|centre|Figure 3 - The unreactive trajectory associated with the specific parameters for Figure 3]]

'''Figure 4'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 4. The trajectory line illustrates a reaction where Ha had a momentum sufficient to overcome the activation energy barrier required to successfully interact with Hb-Hc, pass through the transition state, and form the molecule Ha-Hb.

[[File:Ab6114_figure4.PNG|680px|thumb|centre|Figure 4 - The reactive trajectory associated with the specific parameters for Figure 4]]

'''Figure 5'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 5. The trajectory line for these set parameters, initially show Ha to approach Hb-Hc with sufficient momentum to overcome the activation energy barrier and reach the transition state. However, instead of forming the product Ha-Hb, the trajectory line reverses and returns back to the reactants, with Hb-Hc exhibiting large vibrational energy fluctuations.

[[File:Ab6114_figure5.PNG|680px|thumb|centre|Figure 5 - The unreactive trajectory associated with the specific parameters for Figure 5]]

'''Figure 6'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 6. The trajectory line illustrates that Ha has sufficient momentum to overcome the activation energy barrier. However after reaching the transition state, the trajectory line goes back towards the reactants which is shown by r(B-C) decreasing and r(A-B) increasing in figure 6. The r(A-B) rose to a maxima before decreasing and oscillating, thus illustrating the formation of Ha-Hb. This reaction trajectory suggests that potentially an alternative reaction pathway was taken, one which does not involve a 'saddle point' on the PE plot corresponding with the transition states that normally produce the desired Ha and Hb-Hc.

[[File:Ab6114_figure6.PNG|680px|thumb|centre|Figure 6 - The reactive trajectory associated with the specific parameters for Figure 6]]

----

==='''Question(5)'''===

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

'''Answer:'''

The Transition State Theory's main assumption is that the reactants and activated transition states are at quasi-equilibrium. A quasi-equilibrium utilises thermodynamic molecular driving forces that modify the systems state being very small and slow, allowing for the existence of an internal equilibrium that undergoes only small deviations from this system. Classical Transition State Theory explains that an atom with a momentum satisfying the activation energy barrier requirements, will react upon collision with a given molecule and synthesize products. As this theory however ignores potential quantum mechanical interactions, reaction rates calculated will be slower than what is found in experimental data. Quantum mechanical contributing calculations allow for the tunneling behaviour of particles to be factored into an alternative reaction pathway. When an atom and a molecule lack the sufficient activation energy to overcome this barrier, the atom is able to 'tunnel' through this barrier, due to the particle-wave duality observed in quantum-scale objects.<ref name="thermodynamics" />

----

==Exercise 2: F - H - H system==

==='''Q(6)'''===

Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

'''A:'''

Key points:

- H-F Bond Energy = 565 KJ/Mol<ref name="energies" />

- H-H Bond Energy = 432 KJ/Mol<ref name="energies" />

- H-F is a stronger bond than H-H

- ΔE = E(bonds broken)-E(bonds formed)

'''F + H2''' --> ΔE = 432 - 565 = -133 KJ/Mol '''THEREFORE EXOTHERMIC'''

'''H + H-F''' --> ΔE = 565 - 432 = 133 KJ/Mol '''THEREFORE ENDOTHERMIC'''

----

==='''Q(7)'''===

Locate the approximate position of the transition state.

'''A:'''

To determine the position of the transition state, the radii at which the total energy of the system is stationary (ie. the gradient is zero) were found. The location of these radii is such that the net force acting upon them is zero and plotting the internuclear distance against time results in a constant internuclear separation which does not oscillate. The internuclear separations are shown below and the associated plot against time can be seen in figure 7:

- r(H-H) = 0.74Å

- r(H-F) = 1.815Å

[[File:HZ4315 Figure 7.PNG|700px|thumb|centre|Figure 7 - The Internuclear distance vs Time plot for the transition state position]]

----

==='''Q(8)'''===

Report the activation energy for both reactions.

'''A:'''

In order to determine the activation energy for both reactions, chemical intuition and trial and error were used to find a suitable reaction trajectory. The activation energy was then equal to the difference in energy between that of the reactants and the transition state. The activation energy values obtained were done without mathematical analysis therefore, all energy values involved would actually have a +/- value.

'''Reaction 1: F + H2 --> HF + H'''

- Ea = -103.7 - (-103.9) = 0.2 KCal/Mol

The energy values used for this calculation were obtained from Figure 8 (reactants) and Figure 9 (transition state) as shown below:

[[File:HZ4315 Figure8.PNG|500px|thumb|centre|Figure 8 - Surface plot showing the energy of the reactants]]

[[File:HZ4315 Figure9.PNG|500px|thumb|centre|Figure 9 - Surface plot showing the energy at the transition state]]

'''Reaction 2: H + HF --> F + H2'''

- Ea = -103.7 - (-133.9) = 30.2 KCal/Mol

The energy values used for this calculation were obtained from Figure 10 (reactants) and Figure 9 (transition state) as shown below:

[[File:HZ4315 Figure9.PNG|500px|thumb|centre|Figure 9 - Surface plot showing the energy at the transition state]]

[[File:HZ4315 Figure10.PNG|500px|thumb|centre|Figure 10 - Surface plot showing the energy of the reactants in the reverse reaction]]

----

==='''Q(9)'''===

In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

'''A:'''

According to the Law of Conservation of Energy, the total energy of an isolated system must be conserved as energy is neither created nor destroyed, just converted from one form to another.<ref name="lawE" /> For this specific reaction system, the total energy is the sum of the kinetic and potential energies of each substrate involved in the reaction. For the exothermic reaction F + H2, the potential energy of the products is lower than the reactants hence, the kinetic energy of the products must be higher. This is illustrated by determining a successful reaction trajectory and plotting a kinetic energy vs time graph as shown below:

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 11 - A reaction pathway for an ultimately successful reaction (parameters determined using trial and error)]]

[[File:KEvsTIME.PNG|500px|thumb|centre|Figure 12 - Kinetic energy vs Time plot of the successful reaction - the small oscillations on the left correspond to the reactants which are clearly converted to products as the kinetic energy increases significantly]]

The increased kinetic energy allows the products to be detected by IR analysis. Since the H-F bond is oscillating, it will release energy in the form of vibrational energy, resulting in a peak on the spectrum. The H-H vibrational peak present on the reactant spectrum will now be absent. Another possible experimental technique is calorimetry. Since the reaction is exothermic, the amount of heat released can be measured and since, at constant pressure, the enthalpy change is equal to the net change in energy, the net free energy change of the reaction can be determined.

----

==='''Q(10)'''===

Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

'''A:'''

Using trial and error, a successful reaction trajectory was found for both reactions shown below. The parameters used can be seen within the figures.

'''Reaction 1: F + H2 --> HF + H'''

'''Lower momentum'''

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 13 - An ultimately successful reaction pathway at high momentum indicating the preference of an early transition state]]

'''Higher momentum'''

[[File:HZ4315 Figure12.PNG|500px|thumb|centre|Figure 14 - The reaction pathway at an even larger momentum to confirm the theory]]

Reaction 1, which is an exothermic reaction, exhibits an increased translational energy in response to increasing the momentum of the reactants. According to Polanyi's rules, higher translational energies are more efficient at activating an early transition state (ie. an early barrier) than vibrational energy.<ref name="polanyi" /> Since early transition states resemble the reactants (as stated by Hammond's Postulate)<ref name="hammond" />, the reaction trajectory may be expected to bounce back towards the reactants. Figure 12 confirms this as an increased momentum (and therefore increased translational energy) sees the trajectory bounce back several times before eventually proceeding to the form the products. Figures 11 and 12 are an example of a set of inefficient reaction trajectories as the barrier is recrossed several times which is not ideal in the formation of the product.

The exothermic nature of the reaction also alters the position of the transition state away from the lowest 'saddle point'. This goes against the assumptions of the Transition State Theory which states that reaction trajectories should pass through this. As temperature increases, a greater proportion of excited vibrational states become occupied, leading to transition states occuring at different energies. As a result, the associated reaction trajectory at higher temperature travels through different activated complexes, making the reaction harder to model and any associated results more difficult to compare.

'''Reaction 2: H + HF --> F + H2'''

Reaction 2, which is the reverse reaction of Reaction 1, is endothermic with the energy of the products being higher than the reactants. As a result, according the Hammond's Postulate, the position of the transition state is later in the reaction trajectory and more closely resembles the products. Contrary to the preferred conditions for exothermic reactions, a larger amount of vibrational energy is needed. To achieve this, lower momentum would be adopted. Providing this is achieved, the set of reaction trajectories would be an example of an efficient reaction as the reaction trajectories would pass through the transition state once and to the products.

----

==References==

<references>
<ref name="transition">P. Jonathan P. K. Doye, David J. Wales, Journal of mathematical physics, 1996, '''48''', 843 - 846.</ref>
<ref name="thermodynamics">Chem Libre texts - Wave-particle duality (2015) [online] Available at: https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Quantum_Mechanics/09._The_Hydrogen_Atom/Atomic_Theory/Electrons_in_Atoms/Wave-Particle_Duality [Assessed 28 Sep 2015].</ref>
<ref name="energies">Wiredchemist.com. (2017). Common Bond Energies (D. [online] Available at: http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html [Accessed 11 May 2017].</ref>
<ref name="lawE">Mach, E. (1872). History and Root of the Principles of the Conservation of Energy. Open Court Pub. Co., Illinois.</ref>
<ref name="polanyi">Yan, S., Y.-T. Wu, and K. Liu. "Tracking The Energy Flow Along The Reaction Path". Proceedings of the National Academy of Sciences 105.35 (2008): 12667-12672.</ref>
<ref name="hammond">Chemistry LibreTexts. (2017). Hammond’s Postulate. [online] Available at: https://chem.libretexts.org/Reference/Organic_Chemistry_Glossary/Hammond’s_Postulate [Accessed 11 May 2017].</ref>
<references/>

File:Ab6114 figure6.PNG

2017-05-26T11:49:18Z

Ab6114:

MRD:ab6114

2017-05-26T11:48:36Z

Ab6114:

=='''Molecular Reaction Dynamics'''==

==Exercise 1: H + H2 system==

==='''Question(1)'''===

What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

'''Answer:'''

Analysis of the PES curvature shows a maxima between two minimum points, all being stationary points with a gradient of zero. The first derivative of the curve, where the rate of change is zero, is found to confirm this. As Transition structure's inherently exist at the maximum point, it can be inferred that the transition structure lies exists at this point. In order to differentiate between the transition state and the minima, second derivative calculations can be performed, producing a negative value for a maxima and a positive value for a minima.<ref name="transition" />

----

==='''Question(2)'''===

Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.

'''Answer:'''

The approximate transition state position (r(ts)) was determined through analysis of the surface plot. This was confirmed by examining the internuclear distance vs time plot, to identify the point of intersection between the maxima and the minima - r(ts) is shown in the table below:

{| class="wikitable" border="1"
|+ Table 1
! r1 !! r2 !! r(ts)
|-
| 1.20 || 1.20 || 0.9075
|-
| 0.91 || 0.91 || 0.9075
|}

[[File:ab6114Figure1.PNG|680px|thumb|centre|Figure 1 - Internuclear distance vs Time plot of transition state position]]

A transition state is a point where the internuclear distance between all bonds remain constant and thus do not oscillate. Figure 1 shows that the internuclear distance does not oscillate at 0.9075 Å - hence this is the transition state position.

----

==='''Question(3)'''===

Comment on how the mep and the trajectory you just calculated differ.

'''Answer:'''

Dynamic calculations and mep are distinct for two main reasons. Calculations of mep span only the distance engaged by the transition state, while the dynamic calculation incorporates more of the reaction pathway - the dynamic calculation also takes mass into account, thus resulting in an oscillating reaction pathway line. No mass is included in the mep calculation, so no oscillation is observed.

----

==='''Question(4)'''===

Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.

'''Answer:'''

The effect of momentum variation on whether a trajectory is reactive or unreactive (for the initial positions r1 = 0.74 and r2 = 2.0):

{| class="wikitable" border="1"
|+ Table 2
! p1 !! p2 !! Trajectory !! Figure No.
|-
| -1.25 || -2.5 || Reactive || 2
|-
| -1.5 || -2.0 || Unreactive || 3
|-
| -1.5 || -2.5 || Reactive || 4
|-
| -2.5 || -5.0 || Unreactive || 5
|-
| -2.5 || -5.2 || Reactive || 6
|}

'''Figure 2'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 2. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc while at the same time the B-C bond distance remains relatively constant. This line continues through the transition state (r(A-B)=r(B-C)) and r(A-B) reaches a minimum and oscillates thus indicating that the Ha-Hb bond has formed. Simultaneously r(B-C) increases, confirming that the Hb-Hc bond has broken and Hc now exists as a separate hydrogen atom.

[[File:Ab6114_figure2.PNG|680px|thumb|centre|Figure 2 - The reactive trajectory associated with the specific parameters for Figure 2]]

'''Figure 3'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 3. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc. Since Ha has a momentum insufficient to overcome the activation energy barrier, the trajectory fails to reach the transition state - this is shown by the plot through r(A-B) returning to a larger distance without reaching transition state, and r(B-C) only oscillating due to vibrational fluctuations within the molecule.

[[File:Ab6114_figure3test2.PNG|680px|thumb|centre|Figure 3 - The unreactive trajectory associated with the specific parameters for Figure 3]]

'''Figure 4'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 4. The trajectory line illustrates a reaction where Ha had a momentum sufficient to overcome the activation energy barrier required to successfully interact with Hb-Hc, pass through the transition state, and form the molecule Ha-Hb.

[[File:Ab6114_figure4.PNG|680px|thumb|centre|Figure 4 - The reactive trajectory associated with the specific parameters for Figure 4]]

'''Figure 5'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 5. The trajectory line for these set parameters, initially show Ha to approach Hb-Hc with sufficient momentum to overcome the activation energy barrier and reach the transition state. However, instead of forming the product Ha-Hb, the trajectory line reverses and returns back to the reactants, with Hb-Hc exhibiting large vibrational energy fluctuations.

[[File:Ab6114_figure5.PNG|680px|thumb|centre|Figure 5 - The unreactive trajectory associated with the specific parameters for Figure 5]]

'''Figure 6'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 6. The trajectory line illustrates that Ha has sufficient momentum to overcome the activation energy barrier. However after reaching the transition state, the trajectory line goes back towards the reactants which is shown by r(B-C) decreasing and r(A-B) increasing in figure 6. The r(A-B) rose to a maxima before decreasing and oscillating, thus illustrating the formation of Ha-Hb. This reaction trajectory suggests that potentially an alternative reaction pathway was taken, one which does not involve a 'saddle point' on the PE plot corresponding with the transition states that normally produce the desired Ha and Hb-Hc.

[[File:ab6114figure6.PNG|680px|thumb|centre|Figure 6 - The reactive trajectory associated with the specific parameters for Figure 6]]

----

==='''Q(5)'''===

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

'''A:'''

One assumption of the Transition State Theory is that the reactants and activated transition states are in quasi-equilibrium. Quasi-equilibrium involves the molecular driving forces which would alter the state of the system being very small, resulting in an internal equilibrium being reached. Deviations from this system are very small. In addition to this, according to the Classical Transition State Theory, providing an atom possesses sufficient momentum to overcome the activation barrier, upon collision with another molecule it will react, yielding products. However, since the theory does not account for potential quantum mechanical effects, the resulting reaction rates will be lower than expected. Quantum mechanical contribution can provide alternative reaction pathways through the introduction of tunneling. This can occur when the atom and molecule do not have the required energy to overcome the activation barrier however, due to the wave like properties of the particles, they can 'tunnel' through the barrier instead. As a result, quantum mechanics allows certain reactions to avoid the classical 'saddle point' on the potential energy surface plots associated with the transition state and instead take up an alternative reaction pathway through the barrier.<ref name="quantum" />

----

==Exercise 2: F - H - H system==

==='''Q(6)'''===

Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

'''A:'''

Key points:

- H-F Bond Energy = 565 KJ/Mol<ref name="energies" />

- H-H Bond Energy = 432 KJ/Mol<ref name="energies" />

- H-F is a stronger bond than H-H

- ΔE = E(bonds broken)-E(bonds formed)

'''F + H2''' --> ΔE = 432 - 565 = -133 KJ/Mol '''THEREFORE EXOTHERMIC'''

'''H + H-F''' --> ΔE = 565 - 432 = 133 KJ/Mol '''THEREFORE ENDOTHERMIC'''

----

==='''Q(7)'''===

Locate the approximate position of the transition state.

'''A:'''

To determine the position of the transition state, the radii at which the total energy of the system is stationary (ie. the gradient is zero) were found. The location of these radii is such that the net force acting upon them is zero and plotting the internuclear distance against time results in a constant internuclear separation which does not oscillate. The internuclear separations are shown below and the associated plot against time can be seen in figure 7:

- r(H-H) = 0.74Å

- r(H-F) = 1.815Å

[[File:HZ4315 Figure 7.PNG|700px|thumb|centre|Figure 7 - The Internuclear distance vs Time plot for the transition state position]]

----

==='''Q(8)'''===

Report the activation energy for both reactions.

'''A:'''

In order to determine the activation energy for both reactions, chemical intuition and trial and error were used to find a suitable reaction trajectory. The activation energy was then equal to the difference in energy between that of the reactants and the transition state. The activation energy values obtained were done without mathematical analysis therefore, all energy values involved would actually have a +/- value.

'''Reaction 1: F + H2 --> HF + H'''

- Ea = -103.7 - (-103.9) = 0.2 KCal/Mol

The energy values used for this calculation were obtained from Figure 8 (reactants) and Figure 9 (transition state) as shown below:

[[File:HZ4315 Figure8.PNG|500px|thumb|centre|Figure 8 - Surface plot showing the energy of the reactants]]

[[File:HZ4315 Figure9.PNG|500px|thumb|centre|Figure 9 - Surface plot showing the energy at the transition state]]

'''Reaction 2: H + HF --> F + H2'''

- Ea = -103.7 - (-133.9) = 30.2 KCal/Mol

The energy values used for this calculation were obtained from Figure 10 (reactants) and Figure 9 (transition state) as shown below:

[[File:HZ4315 Figure9.PNG|500px|thumb|centre|Figure 9 - Surface plot showing the energy at the transition state]]

[[File:HZ4315 Figure10.PNG|500px|thumb|centre|Figure 10 - Surface plot showing the energy of the reactants in the reverse reaction]]

----

==='''Q(9)'''===

In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

'''A:'''

According to the Law of Conservation of Energy, the total energy of an isolated system must be conserved as energy is neither created nor destroyed, just converted from one form to another.<ref name="lawE" /> For this specific reaction system, the total energy is the sum of the kinetic and potential energies of each substrate involved in the reaction. For the exothermic reaction F + H2, the potential energy of the products is lower than the reactants hence, the kinetic energy of the products must be higher. This is illustrated by determining a successful reaction trajectory and plotting a kinetic energy vs time graph as shown below:

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 11 - A reaction pathway for an ultimately successful reaction (parameters determined using trial and error)]]

[[File:KEvsTIME.PNG|500px|thumb|centre|Figure 12 - Kinetic energy vs Time plot of the successful reaction - the small oscillations on the left correspond to the reactants which are clearly converted to products as the kinetic energy increases significantly]]

The increased kinetic energy allows the products to be detected by IR analysis. Since the H-F bond is oscillating, it will release energy in the form of vibrational energy, resulting in a peak on the spectrum. The H-H vibrational peak present on the reactant spectrum will now be absent. Another possible experimental technique is calorimetry. Since the reaction is exothermic, the amount of heat released can be measured and since, at constant pressure, the enthalpy change is equal to the net change in energy, the net free energy change of the reaction can be determined.

----

==='''Q(10)'''===

Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

'''A:'''

Using trial and error, a successful reaction trajectory was found for both reactions shown below. The parameters used can be seen within the figures.

'''Reaction 1: F + H2 --> HF + H'''

'''Lower momentum'''

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 13 - An ultimately successful reaction pathway at high momentum indicating the preference of an early transition state]]

'''Higher momentum'''

[[File:HZ4315 Figure12.PNG|500px|thumb|centre|Figure 14 - The reaction pathway at an even larger momentum to confirm the theory]]

Reaction 1, which is an exothermic reaction, exhibits an increased translational energy in response to increasing the momentum of the reactants. According to Polanyi's rules, higher translational energies are more efficient at activating an early transition state (ie. an early barrier) than vibrational energy.<ref name="polanyi" /> Since early transition states resemble the reactants (as stated by Hammond's Postulate)<ref name="hammond" />, the reaction trajectory may be expected to bounce back towards the reactants. Figure 12 confirms this as an increased momentum (and therefore increased translational energy) sees the trajectory bounce back several times before eventually proceeding to the form the products. Figures 11 and 12 are an example of a set of inefficient reaction trajectories as the barrier is recrossed several times which is not ideal in the formation of the product.

The exothermic nature of the reaction also alters the position of the transition state away from the lowest 'saddle point'. This goes against the assumptions of the Transition State Theory which states that reaction trajectories should pass through this. As temperature increases, a greater proportion of excited vibrational states become occupied, leading to transition states occuring at different energies. As a result, the associated reaction trajectory at higher temperature travels through different activated complexes, making the reaction harder to model and any associated results more difficult to compare.

'''Reaction 2: H + HF --> F + H2'''

Reaction 2, which is the reverse reaction of Reaction 1, is endothermic with the energy of the products being higher than the reactants. As a result, according the Hammond's Postulate, the position of the transition state is later in the reaction trajectory and more closely resembles the products. Contrary to the preferred conditions for exothermic reactions, a larger amount of vibrational energy is needed. To achieve this, lower momentum would be adopted. Providing this is achieved, the set of reaction trajectories would be an example of an efficient reaction as the reaction trajectories would pass through the transition state once and to the products.

----

==References==

<references>
<ref name="transition">P. Jonathan P. K. Doye, David J. Wales, Journal of mathematical physics, 1996, '''48''', 843 - 846.</ref>
<ref name="quantum">Physicsoftheuniverse.com. (2017). Quantum Tunnelling and the Uncertainty Principle - Quantum Theory and the Uncertainty Principle - The Physics of the Universe. [online] Available at: http://www.physicsoftheuniverse.com/topics_quantum_uncertainty.html [Accessed 11 May 2017].</ref>
<ref name="energies">Wiredchemist.com. (2017). Common Bond Energies (D. [online] Available at: http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html [Accessed 11 May 2017].</ref>
<ref name="lawE">Mach, E. (1872). History and Root of the Principles of the Conservation of Energy. Open Court Pub. Co., Illinois.</ref>
<ref name="polanyi">Yan, S., Y.-T. Wu, and K. Liu. "Tracking The Energy Flow Along The Reaction Path". Proceedings of the National Academy of Sciences 105.35 (2008): 12667-12672.</ref>
<ref name="hammond">Chemistry LibreTexts. (2017). Hammond’s Postulate. [online] Available at: https://chem.libretexts.org/Reference/Organic_Chemistry_Glossary/Hammond’s_Postulate [Accessed 11 May 2017].</ref>
<references/>

File:Ab6114 figure5.PNG

2017-05-26T11:34:55Z

Ab6114:

MRD:ab6114

2017-05-26T11:34:12Z

Ab6114:

=='''Molecular Reaction Dynamics'''==

==Exercise 1: H + H2 system==

==='''Question(1)'''===

What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

'''Answer:'''

Analysis of the PES curvature shows a maxima between two minimum points, all being stationary points with a gradient of zero. The first derivative of the curve, where the rate of change is zero, is found to confirm this. As Transition structure's inherently exist at the maximum point, it can be inferred that the transition structure lies exists at this point. In order to differentiate between the transition state and the minima, second derivative calculations can be performed, producing a negative value for a maxima and a positive value for a minima.<ref name="transition" />

----

==='''Question(2)'''===

Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.

'''Answer:'''

The approximate transition state position (r(ts)) was determined through analysis of the surface plot. This was confirmed by examining the internuclear distance vs time plot, to identify the point of intersection between the maxima and the minima - r(ts) is shown in the table below:

{| class="wikitable" border="1"
|+ Table 1
! r1 !! r2 !! r(ts)
|-
| 1.20 || 1.20 || 0.9075
|-
| 0.91 || 0.91 || 0.9075
|}

[[File:ab6114Figure1.PNG|680px|thumb|centre|Figure 1 - Internuclear distance vs Time plot of transition state position]]

A transition state is a point where the internuclear distance between all bonds remain constant and thus do not oscillate. Figure 1 shows that the internuclear distance does not oscillate at 0.9075 Å - hence this is the transition state position.

----

==='''Question(3)'''===

Comment on how the mep and the trajectory you just calculated differ.

'''Answer:'''

Dynamic calculations and mep are distinct for two main reasons. Calculations of mep span only the distance engaged by the transition state, while the dynamic calculation incorporates more of the reaction pathway - the dynamic calculation also takes mass into account, thus resulting in an oscillating reaction pathway line. No mass is included in the mep calculation, so no oscillation is observed.

----

==='''Question(4)'''===

Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.

'''Answer:'''

The effect of momentum variation on whether a trajectory is reactive or unreactive (for the initial positions r1 = 0.74 and r2 = 2.0):

{| class="wikitable" border="1"
|+ Table 2
! p1 !! p2 !! Trajectory !! Figure No.
|-
| -1.25 || -2.5 || Reactive || 2
|-
| -1.5 || -2.0 || Unreactive || 3
|-
| -1.5 || -2.5 || Reactive || 4
|-
| -2.5 || -5.0 || Unreactive || 5
|-
| -2.5 || -5.2 || Reactive || 6
|}

'''Figure 2'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 2. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc while at the same time the B-C bond distance remains relatively constant. This line continues through the transition state (r(A-B)=r(B-C)) and r(A-B) reaches a minimum and oscillates thus indicating that the Ha-Hb bond has formed. Simultaneously r(B-C) increases, confirming that the Hb-Hc bond has broken and Hc now exists as a separate hydrogen atom.

[[File:Ab6114_figure2.PNG|680px|thumb|centre|Figure 2 - The reactive trajectory associated with the specific parameters for Figure 2]]

'''Figure 3'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 3. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc. Since Ha has a momentum insufficient to overcome the activation energy barrier, the trajectory fails to reach the transition state - this is shown by the plot through r(A-B) returning to a larger distance without reaching transition state, and r(B-C) only oscillating due to vibrational fluctuations within the molecule.

[[File:Ab6114_figure3test2.PNG|680px|thumb|centre|Figure 3 - The unreactive trajectory associated with the specific parameters for Figure 3]]

'''Figure 4'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 4. The trajectory line illustrates a reaction where Ha had a momentum sufficient to overcome the activation energy barrier required to successfully interact with Hb-Hc, pass through the transition state, and form the molecule Ha-Hb.

[[File:Ab6114_figure4.PNG|680px|thumb|centre|Figure 4 - The reactive trajectory associated with the specific parameters for Figure 4]]

'''Figure 5'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 5. The trajectory line for these set parameters, initially show Ha to approach Hb-Hc with sufficient momentum to overcome the activation energy barrier and reach the transition state. However, instead of forming the product Ha-Hb, the trajectory line reverses and returns back to the reactants, with Hb-Hc exhibiting large vibrational energy fluctuations.

[[File:Ab6114figure5.PNG|680px|thumb|centre|Figure 5 - The unreactive trajectory associated with the set parameters for Figure 5]]

'''Figure 6'''

The plot below shows the reactive trajectory associated with the set parameters for Figure 6. The reaction trajectory indicates that Ha has sufficient momentum to overcome the activation barrier however, upon reaching the transition state, the trajectory is driven back towards the reactants which is illustrated by r(B-C) decreasing and r(A-B) increasing. r(A-B) reaches a maximum value before decreasing and oscillating, indicating the formation of Ha-Hb. This trajectory suggests that the reaction has taken an alternative reaction pathway which avoids the 'saddle point' on the potential energy plot associated with the transition states to yield the desired products.

[[File:HZ4315 Figure6.PNG|700px|thumb|centre|Figure 6 - The reactive trajectory associated with the set parameters for Figure 6]]

----

==='''Q(5)'''===

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

'''A:'''

One assumption of the Transition State Theory is that the reactants and activated transition states are in quasi-equilibrium. Quasi-equilibrium involves the molecular driving forces which would alter the state of the system being very small, resulting in an internal equilibrium being reached. Deviations from this system are very small. In addition to this, according to the Classical Transition State Theory, providing an atom possesses sufficient momentum to overcome the activation barrier, upon collision with another molecule it will react, yielding products. However, since the theory does not account for potential quantum mechanical effects, the resulting reaction rates will be lower than expected. Quantum mechanical contribution can provide alternative reaction pathways through the introduction of tunneling. This can occur when the atom and molecule do not have the required energy to overcome the activation barrier however, due to the wave like properties of the particles, they can 'tunnel' through the barrier instead. As a result, quantum mechanics allows certain reactions to avoid the classical 'saddle point' on the potential energy surface plots associated with the transition state and instead take up an alternative reaction pathway through the barrier.<ref name="quantum" />

----

==Exercise 2: F - H - H system==

==='''Q(6)'''===

Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

'''A:'''

Key points:

- H-F Bond Energy = 565 KJ/Mol<ref name="energies" />

- H-H Bond Energy = 432 KJ/Mol<ref name="energies" />

- H-F is a stronger bond than H-H

- ΔE = E(bonds broken)-E(bonds formed)

'''F + H2''' --> ΔE = 432 - 565 = -133 KJ/Mol '''THEREFORE EXOTHERMIC'''

'''H + H-F''' --> ΔE = 565 - 432 = 133 KJ/Mol '''THEREFORE ENDOTHERMIC'''

----

==='''Q(7)'''===

Locate the approximate position of the transition state.

'''A:'''

To determine the position of the transition state, the radii at which the total energy of the system is stationary (ie. the gradient is zero) were found. The location of these radii is such that the net force acting upon them is zero and plotting the internuclear distance against time results in a constant internuclear separation which does not oscillate. The internuclear separations are shown below and the associated plot against time can be seen in figure 7:

- r(H-H) = 0.74Å

- r(H-F) = 1.815Å

[[File:HZ4315 Figure 7.PNG|700px|thumb|centre|Figure 7 - The Internuclear distance vs Time plot for the transition state position]]

----

==='''Q(8)'''===

Report the activation energy for both reactions.

'''A:'''

In order to determine the activation energy for both reactions, chemical intuition and trial and error were used to find a suitable reaction trajectory. The activation energy was then equal to the difference in energy between that of the reactants and the transition state. The activation energy values obtained were done without mathematical analysis therefore, all energy values involved would actually have a +/- value.

'''Reaction 1: F + H2 --> HF + H'''

- Ea = -103.7 - (-103.9) = 0.2 KCal/Mol

The energy values used for this calculation were obtained from Figure 8 (reactants) and Figure 9 (transition state) as shown below:

[[File:HZ4315 Figure8.PNG|500px|thumb|centre|Figure 8 - Surface plot showing the energy of the reactants]]

[[File:HZ4315 Figure9.PNG|500px|thumb|centre|Figure 9 - Surface plot showing the energy at the transition state]]

'''Reaction 2: H + HF --> F + H2'''

- Ea = -103.7 - (-133.9) = 30.2 KCal/Mol

The energy values used for this calculation were obtained from Figure 10 (reactants) and Figure 9 (transition state) as shown below:

[[File:HZ4315 Figure9.PNG|500px|thumb|centre|Figure 9 - Surface plot showing the energy at the transition state]]

[[File:HZ4315 Figure10.PNG|500px|thumb|centre|Figure 10 - Surface plot showing the energy of the reactants in the reverse reaction]]

----

==='''Q(9)'''===

In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

'''A:'''

According to the Law of Conservation of Energy, the total energy of an isolated system must be conserved as energy is neither created nor destroyed, just converted from one form to another.<ref name="lawE" /> For this specific reaction system, the total energy is the sum of the kinetic and potential energies of each substrate involved in the reaction. For the exothermic reaction F + H2, the potential energy of the products is lower than the reactants hence, the kinetic energy of the products must be higher. This is illustrated by determining a successful reaction trajectory and plotting a kinetic energy vs time graph as shown below:

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 11 - A reaction pathway for an ultimately successful reaction (parameters determined using trial and error)]]

[[File:KEvsTIME.PNG|500px|thumb|centre|Figure 12 - Kinetic energy vs Time plot of the successful reaction - the small oscillations on the left correspond to the reactants which are clearly converted to products as the kinetic energy increases significantly]]

The increased kinetic energy allows the products to be detected by IR analysis. Since the H-F bond is oscillating, it will release energy in the form of vibrational energy, resulting in a peak on the spectrum. The H-H vibrational peak present on the reactant spectrum will now be absent. Another possible experimental technique is calorimetry. Since the reaction is exothermic, the amount of heat released can be measured and since, at constant pressure, the enthalpy change is equal to the net change in energy, the net free energy change of the reaction can be determined.

----

==='''Q(10)'''===

Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

'''A:'''

Using trial and error, a successful reaction trajectory was found for both reactions shown below. The parameters used can be seen within the figures.

'''Reaction 1: F + H2 --> HF + H'''

'''Lower momentum'''

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 13 - An ultimately successful reaction pathway at high momentum indicating the preference of an early transition state]]

'''Higher momentum'''

[[File:HZ4315 Figure12.PNG|500px|thumb|centre|Figure 14 - The reaction pathway at an even larger momentum to confirm the theory]]

Reaction 1, which is an exothermic reaction, exhibits an increased translational energy in response to increasing the momentum of the reactants. According to Polanyi's rules, higher translational energies are more efficient at activating an early transition state (ie. an early barrier) than vibrational energy.<ref name="polanyi" /> Since early transition states resemble the reactants (as stated by Hammond's Postulate)<ref name="hammond" />, the reaction trajectory may be expected to bounce back towards the reactants. Figure 12 confirms this as an increased momentum (and therefore increased translational energy) sees the trajectory bounce back several times before eventually proceeding to the form the products. Figures 11 and 12 are an example of a set of inefficient reaction trajectories as the barrier is recrossed several times which is not ideal in the formation of the product.

The exothermic nature of the reaction also alters the position of the transition state away from the lowest 'saddle point'. This goes against the assumptions of the Transition State Theory which states that reaction trajectories should pass through this. As temperature increases, a greater proportion of excited vibrational states become occupied, leading to transition states occuring at different energies. As a result, the associated reaction trajectory at higher temperature travels through different activated complexes, making the reaction harder to model and any associated results more difficult to compare.

'''Reaction 2: H + HF --> F + H2'''

Reaction 2, which is the reverse reaction of Reaction 1, is endothermic with the energy of the products being higher than the reactants. As a result, according the Hammond's Postulate, the position of the transition state is later in the reaction trajectory and more closely resembles the products. Contrary to the preferred conditions for exothermic reactions, a larger amount of vibrational energy is needed. To achieve this, lower momentum would be adopted. Providing this is achieved, the set of reaction trajectories would be an example of an efficient reaction as the reaction trajectories would pass through the transition state once and to the products.

----

==References==

<references>
<ref name="transition">P. Jonathan P. K. Doye, David J. Wales, Journal of mathematical physics, 1996, '''48''', 843 - 846.</ref>
<ref name="quantum">Physicsoftheuniverse.com. (2017). Quantum Tunnelling and the Uncertainty Principle - Quantum Theory and the Uncertainty Principle - The Physics of the Universe. [online] Available at: http://www.physicsoftheuniverse.com/topics_quantum_uncertainty.html [Accessed 11 May 2017].</ref>
<ref name="energies">Wiredchemist.com. (2017). Common Bond Energies (D. [online] Available at: http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html [Accessed 11 May 2017].</ref>
<ref name="lawE">Mach, E. (1872). History and Root of the Principles of the Conservation of Energy. Open Court Pub. Co., Illinois.</ref>
<ref name="polanyi">Yan, S., Y.-T. Wu, and K. Liu. "Tracking The Energy Flow Along The Reaction Path". Proceedings of the National Academy of Sciences 105.35 (2008): 12667-12672.</ref>
<ref name="hammond">Chemistry LibreTexts. (2017). Hammond’s Postulate. [online] Available at: https://chem.libretexts.org/Reference/Organic_Chemistry_Glossary/Hammond’s_Postulate [Accessed 11 May 2017].</ref>
<references/>

File:Ab6114 figure4.PNG

2017-05-26T11:25:49Z

Ab6114:

MRD:ab6114

2017-05-26T11:24:39Z

Ab6114:

=='''Molecular Reaction Dynamics'''==

==Exercise 1: H + H2 system==

==='''Question(1)'''===

What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

'''Answer:'''

Analysis of the PES curvature shows a maxima between two minimum points, all being stationary points with a gradient of zero. The first derivative of the curve, where the rate of change is zero, is found to confirm this. As Transition structure's inherently exist at the maximum point, it can be inferred that the transition structure lies exists at this point. In order to differentiate between the transition state and the minima, second derivative calculations can be performed, producing a negative value for a maxima and a positive value for a minima.<ref name="transition" />

----

==='''Question(2)'''===

Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.

'''Answer:'''

The approximate transition state position (r(ts)) was determined through analysis of the surface plot. This was confirmed by examining the internuclear distance vs time plot, to identify the point of intersection between the maxima and the minima - r(ts) is shown in the table below:

{| class="wikitable" border="1"
|+ Table 1
! r1 !! r2 !! r(ts)
|-
| 1.20 || 1.20 || 0.9075
|-
| 0.91 || 0.91 || 0.9075
|}

[[File:ab6114Figure1.PNG|680px|thumb|centre|Figure 1 - Internuclear distance vs Time plot of transition state position]]

A transition state is a point where the internuclear distance between all bonds remain constant and thus do not oscillate. Figure 1 shows that the internuclear distance does not oscillate at 0.9075 Å - hence this is the transition state position.

----

==='''Question(3)'''===

Comment on how the mep and the trajectory you just calculated differ.

'''Answer:'''

Dynamic calculations and mep are distinct for two main reasons. Calculations of mep span only the distance engaged by the transition state, while the dynamic calculation incorporates more of the reaction pathway - the dynamic calculation also takes mass into account, thus resulting in an oscillating reaction pathway line. No mass is included in the mep calculation, so no oscillation is observed.

----

==='''Question(4)'''===

Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.

'''Answer:'''

The effect of momentum variation on whether a trajectory is reactive or unreactive (for the initial positions r1 = 0.74 and r2 = 2.0):

{| class="wikitable" border="1"
|+ Table 2
! p1 !! p2 !! Trajectory !! Figure No.
|-
| -1.25 || -2.5 || Reactive || 2
|-
| -1.5 || -2.0 || Unreactive || 3
|-
| -1.5 || -2.5 || Reactive || 4
|-
| -2.5 || -5.0 || Unreactive || 5
|-
| -2.5 || -5.2 || Reactive || 6
|}

'''Figure 2'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 2. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc while at the same time the B-C bond distance remains relatively constant. This line continues through the transition state (r(A-B)=r(B-C)) and r(A-B) reaches a minimum and oscillates thus indicating that the Ha-Hb bond has formed. Simultaneously r(B-C) increases, confirming that the Hb-Hc bond has broken and Hc now exists as a separate hydrogen atom.

[[File:Ab6114_figure2.PNG|680px|thumb|centre|Figure 2 - The reactive trajectory associated with the specific parameters for Figure 2]]

'''Figure 3'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 3. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc. Since Ha has a momentum insufficient to overcome the activation energy barrier, the trajectory fails to reach the transition state - this is shown by the plot through r(A-B) returning to a larger distance without reaching transition state, and r(B-C) only oscillating due to vibrational fluctuations within the molecule.

[[File:Ab6114_figure3test2.PNG|680px|thumb|centre|Figure 3 - The unreactive trajectory associated with the specific parameters for Figure 3]]

'''Figure 4'''

The surface plot below shows the reactive trajectory corresponding with the set parameters for figure 4. The trajectory line illustrates a reaction where Ha had a momentum sufficient to overcome the activation energy barrier required to successfully interact with Hb-Hc, pass through the transition state, and form the molecule Ha-Hb.

[[File:HZ4315 Figure4.PNG|700px|thumb|centre|Figure 4 - The reactive trajectory associated with the set parameters for Figure 4]]

'''Figure 5'''

The plot below shows the unreactive trajectory associated with the set parameters for Figure 5. The reaction trajectory for these parameters initially shows that Ha approaches Hb-Hc with sufficient momentum to overcome the activation barrier and reach the transition state. However, instead of proceeding to the products, the reaction trajectory returns back to the reactants with molecule Hb-Hc possessing a large amount of vibrational energy.

[[File:HZ4315 Figure5.PNG|700px|thumb|centre|Figure 5 - The unreactive trajectory associated with the set parameters for Figure 5]]

'''Figure 6'''

The plot below shows the reactive trajectory associated with the set parameters for Figure 6. The reaction trajectory indicates that Ha has sufficient momentum to overcome the activation barrier however, upon reaching the transition state, the trajectory is driven back towards the reactants which is illustrated by r(B-C) decreasing and r(A-B) increasing. r(A-B) reaches a maximum value before decreasing and oscillating, indicating the formation of Ha-Hb. This trajectory suggests that the reaction has taken an alternative reaction pathway which avoids the 'saddle point' on the potential energy plot associated with the transition states to yield the desired products.

[[File:HZ4315 Figure6.PNG|700px|thumb|centre|Figure 6 - The reactive trajectory associated with the set parameters for Figure 6]]

----

==='''Q(5)'''===

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

'''A:'''

One assumption of the Transition State Theory is that the reactants and activated transition states are in quasi-equilibrium. Quasi-equilibrium involves the molecular driving forces which would alter the state of the system being very small, resulting in an internal equilibrium being reached. Deviations from this system are very small. In addition to this, according to the Classical Transition State Theory, providing an atom possesses sufficient momentum to overcome the activation barrier, upon collision with another molecule it will react, yielding products. However, since the theory does not account for potential quantum mechanical effects, the resulting reaction rates will be lower than expected. Quantum mechanical contribution can provide alternative reaction pathways through the introduction of tunneling. This can occur when the atom and molecule do not have the required energy to overcome the activation barrier however, due to the wave like properties of the particles, they can 'tunnel' through the barrier instead. As a result, quantum mechanics allows certain reactions to avoid the classical 'saddle point' on the potential energy surface plots associated with the transition state and instead take up an alternative reaction pathway through the barrier.<ref name="quantum" />

----

==Exercise 2: F - H - H system==

==='''Q(6)'''===

Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

'''A:'''

Key points:

- H-F Bond Energy = 565 KJ/Mol<ref name="energies" />

- H-H Bond Energy = 432 KJ/Mol<ref name="energies" />

- H-F is a stronger bond than H-H

- ΔE = E(bonds broken)-E(bonds formed)

'''F + H2''' --> ΔE = 432 - 565 = -133 KJ/Mol '''THEREFORE EXOTHERMIC'''

'''H + H-F''' --> ΔE = 565 - 432 = 133 KJ/Mol '''THEREFORE ENDOTHERMIC'''

----

==='''Q(7)'''===

Locate the approximate position of the transition state.

'''A:'''

To determine the position of the transition state, the radii at which the total energy of the system is stationary (ie. the gradient is zero) were found. The location of these radii is such that the net force acting upon them is zero and plotting the internuclear distance against time results in a constant internuclear separation which does not oscillate. The internuclear separations are shown below and the associated plot against time can be seen in figure 7:

- r(H-H) = 0.74Å

- r(H-F) = 1.815Å

[[File:HZ4315 Figure 7.PNG|700px|thumb|centre|Figure 7 - The Internuclear distance vs Time plot for the transition state position]]

----

==='''Q(8)'''===

Report the activation energy for both reactions.

'''A:'''

In order to determine the activation energy for both reactions, chemical intuition and trial and error were used to find a suitable reaction trajectory. The activation energy was then equal to the difference in energy between that of the reactants and the transition state. The activation energy values obtained were done without mathematical analysis therefore, all energy values involved would actually have a +/- value.

'''Reaction 1: F + H2 --> HF + H'''

- Ea = -103.7 - (-103.9) = 0.2 KCal/Mol

The energy values used for this calculation were obtained from Figure 8 (reactants) and Figure 9 (transition state) as shown below:

[[File:HZ4315 Figure8.PNG|500px|thumb|centre|Figure 8 - Surface plot showing the energy of the reactants]]

[[File:HZ4315 Figure9.PNG|500px|thumb|centre|Figure 9 - Surface plot showing the energy at the transition state]]

'''Reaction 2: H + HF --> F + H2'''

- Ea = -103.7 - (-133.9) = 30.2 KCal/Mol

The energy values used for this calculation were obtained from Figure 10 (reactants) and Figure 9 (transition state) as shown below:

[[File:HZ4315 Figure9.PNG|500px|thumb|centre|Figure 9 - Surface plot showing the energy at the transition state]]

[[File:HZ4315 Figure10.PNG|500px|thumb|centre|Figure 10 - Surface plot showing the energy of the reactants in the reverse reaction]]

----

==='''Q(9)'''===

In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

'''A:'''

According to the Law of Conservation of Energy, the total energy of an isolated system must be conserved as energy is neither created nor destroyed, just converted from one form to another.<ref name="lawE" /> For this specific reaction system, the total energy is the sum of the kinetic and potential energies of each substrate involved in the reaction. For the exothermic reaction F + H2, the potential energy of the products is lower than the reactants hence, the kinetic energy of the products must be higher. This is illustrated by determining a successful reaction trajectory and plotting a kinetic energy vs time graph as shown below:

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 11 - A reaction pathway for an ultimately successful reaction (parameters determined using trial and error)]]

[[File:KEvsTIME.PNG|500px|thumb|centre|Figure 12 - Kinetic energy vs Time plot of the successful reaction - the small oscillations on the left correspond to the reactants which are clearly converted to products as the kinetic energy increases significantly]]

The increased kinetic energy allows the products to be detected by IR analysis. Since the H-F bond is oscillating, it will release energy in the form of vibrational energy, resulting in a peak on the spectrum. The H-H vibrational peak present on the reactant spectrum will now be absent. Another possible experimental technique is calorimetry. Since the reaction is exothermic, the amount of heat released can be measured and since, at constant pressure, the enthalpy change is equal to the net change in energy, the net free energy change of the reaction can be determined.

----

==='''Q(10)'''===

Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

'''A:'''

Using trial and error, a successful reaction trajectory was found for both reactions shown below. The parameters used can be seen within the figures.

'''Reaction 1: F + H2 --> HF + H'''

'''Lower momentum'''

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 13 - An ultimately successful reaction pathway at high momentum indicating the preference of an early transition state]]

'''Higher momentum'''

[[File:HZ4315 Figure12.PNG|500px|thumb|centre|Figure 14 - The reaction pathway at an even larger momentum to confirm the theory]]

Reaction 1, which is an exothermic reaction, exhibits an increased translational energy in response to increasing the momentum of the reactants. According to Polanyi's rules, higher translational energies are more efficient at activating an early transition state (ie. an early barrier) than vibrational energy.<ref name="polanyi" /> Since early transition states resemble the reactants (as stated by Hammond's Postulate)<ref name="hammond" />, the reaction trajectory may be expected to bounce back towards the reactants. Figure 12 confirms this as an increased momentum (and therefore increased translational energy) sees the trajectory bounce back several times before eventually proceeding to the form the products. Figures 11 and 12 are an example of a set of inefficient reaction trajectories as the barrier is recrossed several times which is not ideal in the formation of the product.

The exothermic nature of the reaction also alters the position of the transition state away from the lowest 'saddle point'. This goes against the assumptions of the Transition State Theory which states that reaction trajectories should pass through this. As temperature increases, a greater proportion of excited vibrational states become occupied, leading to transition states occuring at different energies. As a result, the associated reaction trajectory at higher temperature travels through different activated complexes, making the reaction harder to model and any associated results more difficult to compare.

'''Reaction 2: H + HF --> F + H2'''

Reaction 2, which is the reverse reaction of Reaction 1, is endothermic with the energy of the products being higher than the reactants. As a result, according the Hammond's Postulate, the position of the transition state is later in the reaction trajectory and more closely resembles the products. Contrary to the preferred conditions for exothermic reactions, a larger amount of vibrational energy is needed. To achieve this, lower momentum would be adopted. Providing this is achieved, the set of reaction trajectories would be an example of an efficient reaction as the reaction trajectories would pass through the transition state once and to the products.

----

==References==

<references>
<ref name="transition">P. Jonathan P. K. Doye, David J. Wales, Journal of mathematical physics, 1996, '''48''', 843 - 846.</ref>
<ref name="quantum">Physicsoftheuniverse.com. (2017). Quantum Tunnelling and the Uncertainty Principle - Quantum Theory and the Uncertainty Principle - The Physics of the Universe. [online] Available at: http://www.physicsoftheuniverse.com/topics_quantum_uncertainty.html [Accessed 11 May 2017].</ref>
<ref name="energies">Wiredchemist.com. (2017). Common Bond Energies (D. [online] Available at: http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html [Accessed 11 May 2017].</ref>
<ref name="lawE">Mach, E. (1872). History and Root of the Principles of the Conservation of Energy. Open Court Pub. Co., Illinois.</ref>
<ref name="polanyi">Yan, S., Y.-T. Wu, and K. Liu. "Tracking The Energy Flow Along The Reaction Path". Proceedings of the National Academy of Sciences 105.35 (2008): 12667-12672.</ref>
<ref name="hammond">Chemistry LibreTexts. (2017). Hammond’s Postulate. [online] Available at: https://chem.libretexts.org/Reference/Organic_Chemistry_Glossary/Hammond’s_Postulate [Accessed 11 May 2017].</ref>
<references/>

File:Ab6114 figure3test2.PNG

2017-05-26T11:18:37Z

Ab6114:

File:Ab6114 figure3.png

2017-05-26T11:14:26Z

Ab6114:

MRD:ab6114

2017-05-26T11:13:58Z

Ab6114:

=='''Molecular Reaction Dynamics'''==

==Exercise 1: H + H2 system==

==='''Question(1)'''===

What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

'''Answer:'''

Analysis of the PES curvature shows a maxima between two minimum points, all being stationary points with a gradient of zero. The first derivative of the curve, where the rate of change is zero, is found to confirm this. As Transition structure's inherently exist at the maximum point, it can be inferred that the transition structure lies exists at this point. In order to differentiate between the transition state and the minima, second derivative calculations can be performed, producing a negative value for a maxima and a positive value for a minima.<ref name="transition" />

----

==='''Question(2)'''===

Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.

'''Answer:'''

The approximate transition state position (r(ts)) was determined through analysis of the surface plot. This was confirmed by examining the internuclear distance vs time plot, to identify the point of intersection between the maxima and the minima - r(ts) is shown in the table below:

{| class="wikitable" border="1"
|+ Table 1
! r1 !! r2 !! r(ts)
|-
| 1.20 || 1.20 || 0.9075
|-
| 0.91 || 0.91 || 0.9075
|}

[[File:ab6114Figure1.PNG|680px|thumb|centre|Figure 1 - Internuclear distance vs Time plot of transition state position]]

A transition state is a point where the internuclear distance between all bonds remain constant and thus do not oscillate. Figure 1 shows that the internuclear distance does not oscillate at 0.9075 Å - hence this is the transition state position.

----

==='''Question(3)'''===

Comment on how the mep and the trajectory you just calculated differ.

'''Answer:'''

Dynamic calculations and mep are distinct for two main reasons. Calculations of mep span only the distance engaged by the transition state, while the dynamic calculation incorporates more of the reaction pathway - the dynamic calculation also takes mass into account, thus resulting in an oscillating reaction pathway line. No mass is included in the mep calculation, so no oscillation is observed.

----

==='''Question(4)'''===

Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.

'''Answer:'''

The effect of momentum variation on whether a trajectory is reactive or unreactive (for the initial positions r1 = 0.74 and r2 = 2.0):

{| class="wikitable" border="1"
|+ Table 2
! p1 !! p2 !! Trajectory !! Figure No.
|-
| -1.25 || -2.5 || Reactive || 2
|-
| -1.5 || -2.0 || Unreactive || 3
|-
| -1.5 || -2.5 || Reactive || 4
|-
| -2.5 || -5.0 || Unreactive || 5
|-
| -2.5 || -5.2 || Reactive || 6
|}

'''Figure 2'''

The surface plot shows the reactive trajectory corresponding with the set parameters for figure 2. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc while at the same time the B-C bond distance remains relatively constant. This line continues through the transition state (r(A-B)=r(B-C)) and r(A-B) reaches a minimum and oscillates thus indicating that the Ha-Hb bond has formed. Simultaneously r(B-C) increases, confirming that the Hb-Hc bond has broken and Hc now exists as a separate hydrogen atom.

[[File:Ab6114_figure2.PNG|680px|thumb|centre|Figure 2 - The reactive trajectory associated with the specific parameters for Figure 2]]

'''Figure 3'''

The surface plot below shows the unreactive trajectory corresponding with the set parameters for figure 3. The trajectory line shows that the r(A-B) bond distance decreases as atom Ha approaches molecule Hb-Hc. Since Ha has a momentum insufficient to overcome the activation energy barrier, the trajectory fails to reach the transition state - this is shown by the plot through r(A-B) returning to a larger distance without reaching transition state, and r(B-C) only oscillating due to vibrational fluctuations within the molecule.

[[File:HZ4315 Figure3.PNG|700px|thumb|centre|Figure 3 - The unreactive trajectory associated with the set parameters for Figure 3]]

'''Figure 4'''

The plot below shows the reactive trajectory associated with the set parameters for Figure 4. The reaction trajectory shows a successful reaction where Ha possesses enough momentum to overcome the activation barrier when approaching Hb-Hc hence, it passes through the transition state. The reaction then proceeds to form the molecule Ha-Hb.

[[File:HZ4315 Figure4.PNG|700px|thumb|centre|Figure 4 - The reactive trajectory associated with the set parameters for Figure 4]]

'''Figure 5'''

The plot below shows the unreactive trajectory associated with the set parameters for Figure 5. The reaction trajectory for these parameters initially shows that Ha approaches Hb-Hc with sufficient momentum to overcome the activation barrier and reach the transition state. However, instead of proceeding to the products, the reaction trajectory returns back to the reactants with molecule Hb-Hc possessing a large amount of vibrational energy.

[[File:HZ4315 Figure5.PNG|700px|thumb|centre|Figure 5 - The unreactive trajectory associated with the set parameters for Figure 5]]

'''Figure 6'''

The plot below shows the reactive trajectory associated with the set parameters for Figure 6. The reaction trajectory indicates that Ha has sufficient momentum to overcome the activation barrier however, upon reaching the transition state, the trajectory is driven back towards the reactants which is illustrated by r(B-C) decreasing and r(A-B) increasing. r(A-B) reaches a maximum value before decreasing and oscillating, indicating the formation of Ha-Hb. This trajectory suggests that the reaction has taken an alternative reaction pathway which avoids the 'saddle point' on the potential energy plot associated with the transition states to yield the desired products.

[[File:HZ4315 Figure6.PNG|700px|thumb|centre|Figure 6 - The reactive trajectory associated with the set parameters for Figure 6]]

----

==='''Q(5)'''===

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

'''A:'''

One assumption of the Transition State Theory is that the reactants and activated transition states are in quasi-equilibrium. Quasi-equilibrium involves the molecular driving forces which would alter the state of the system being very small, resulting in an internal equilibrium being reached. Deviations from this system are very small. In addition to this, according to the Classical Transition State Theory, providing an atom possesses sufficient momentum to overcome the activation barrier, upon collision with another molecule it will react, yielding products. However, since the theory does not account for potential quantum mechanical effects, the resulting reaction rates will be lower than expected. Quantum mechanical contribution can provide alternative reaction pathways through the introduction of tunneling. This can occur when the atom and molecule do not have the required energy to overcome the activation barrier however, due to the wave like properties of the particles, they can 'tunnel' through the barrier instead. As a result, quantum mechanics allows certain reactions to avoid the classical 'saddle point' on the potential energy surface plots associated with the transition state and instead take up an alternative reaction pathway through the barrier.<ref name="quantum" />

----

==Exercise 2: F - H - H system==

==='''Q(6)'''===

Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

'''A:'''

Key points:

- H-F Bond Energy = 565 KJ/Mol<ref name="energies" />

- H-H Bond Energy = 432 KJ/Mol<ref name="energies" />

- H-F is a stronger bond than H-H

- ΔE = E(bonds broken)-E(bonds formed)

'''F + H2''' --> ΔE = 432 - 565 = -133 KJ/Mol '''THEREFORE EXOTHERMIC'''

'''H + H-F''' --> ΔE = 565 - 432 = 133 KJ/Mol '''THEREFORE ENDOTHERMIC'''

----

==='''Q(7)'''===

Locate the approximate position of the transition state.

'''A:'''

To determine the position of the transition state, the radii at which the total energy of the system is stationary (ie. the gradient is zero) were found. The location of these radii is such that the net force acting upon them is zero and plotting the internuclear distance against time results in a constant internuclear separation which does not oscillate. The internuclear separations are shown below and the associated plot against time can be seen in figure 7:

- r(H-H) = 0.74Å

- r(H-F) = 1.815Å

[[File:HZ4315 Figure 7.PNG|700px|thumb|centre|Figure 7 - The Internuclear distance vs Time plot for the transition state position]]

----

==='''Q(8)'''===

Report the activation energy for both reactions.

'''A:'''

In order to determine the activation energy for both reactions, chemical intuition and trial and error were used to find a suitable reaction trajectory. The activation energy was then equal to the difference in energy between that of the reactants and the transition state. The activation energy values obtained were done without mathematical analysis therefore, all energy values involved would actually have a +/- value.

'''Reaction 1: F + H2 --> HF + H'''

- Ea = -103.7 - (-103.9) = 0.2 KCal/Mol

The energy values used for this calculation were obtained from Figure 8 (reactants) and Figure 9 (transition state) as shown below:

[[File:HZ4315 Figure8.PNG|500px|thumb|centre|Figure 8 - Surface plot showing the energy of the reactants]]

[[File:HZ4315 Figure9.PNG|500px|thumb|centre|Figure 9 - Surface plot showing the energy at the transition state]]

'''Reaction 2: H + HF --> F + H2'''

- Ea = -103.7 - (-133.9) = 30.2 KCal/Mol

The energy values used for this calculation were obtained from Figure 10 (reactants) and Figure 9 (transition state) as shown below:

[[File:HZ4315 Figure9.PNG|500px|thumb|centre|Figure 9 - Surface plot showing the energy at the transition state]]

[[File:HZ4315 Figure10.PNG|500px|thumb|centre|Figure 10 - Surface plot showing the energy of the reactants in the reverse reaction]]

----

==='''Q(9)'''===

In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

'''A:'''

According to the Law of Conservation of Energy, the total energy of an isolated system must be conserved as energy is neither created nor destroyed, just converted from one form to another.<ref name="lawE" /> For this specific reaction system, the total energy is the sum of the kinetic and potential energies of each substrate involved in the reaction. For the exothermic reaction F + H2, the potential energy of the products is lower than the reactants hence, the kinetic energy of the products must be higher. This is illustrated by determining a successful reaction trajectory and plotting a kinetic energy vs time graph as shown below:

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 11 - A reaction pathway for an ultimately successful reaction (parameters determined using trial and error)]]

[[File:KEvsTIME.PNG|500px|thumb|centre|Figure 12 - Kinetic energy vs Time plot of the successful reaction - the small oscillations on the left correspond to the reactants which are clearly converted to products as the kinetic energy increases significantly]]

The increased kinetic energy allows the products to be detected by IR analysis. Since the H-F bond is oscillating, it will release energy in the form of vibrational energy, resulting in a peak on the spectrum. The H-H vibrational peak present on the reactant spectrum will now be absent. Another possible experimental technique is calorimetry. Since the reaction is exothermic, the amount of heat released can be measured and since, at constant pressure, the enthalpy change is equal to the net change in energy, the net free energy change of the reaction can be determined.

----

==='''Q(10)'''===

Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

'''A:'''

Using trial and error, a successful reaction trajectory was found for both reactions shown below. The parameters used can be seen within the figures.

'''Reaction 1: F + H2 --> HF + H'''

'''Lower momentum'''

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 13 - An ultimately successful reaction pathway at high momentum indicating the preference of an early transition state]]

'''Higher momentum'''

[[File:HZ4315 Figure12.PNG|500px|thumb|centre|Figure 14 - The reaction pathway at an even larger momentum to confirm the theory]]

Reaction 1, which is an exothermic reaction, exhibits an increased translational energy in response to increasing the momentum of the reactants. According to Polanyi's rules, higher translational energies are more efficient at activating an early transition state (ie. an early barrier) than vibrational energy.<ref name="polanyi" /> Since early transition states resemble the reactants (as stated by Hammond's Postulate)<ref name="hammond" />, the reaction trajectory may be expected to bounce back towards the reactants. Figure 12 confirms this as an increased momentum (and therefore increased translational energy) sees the trajectory bounce back several times before eventually proceeding to the form the products. Figures 11 and 12 are an example of a set of inefficient reaction trajectories as the barrier is recrossed several times which is not ideal in the formation of the product.

The exothermic nature of the reaction also alters the position of the transition state away from the lowest 'saddle point'. This goes against the assumptions of the Transition State Theory which states that reaction trajectories should pass through this. As temperature increases, a greater proportion of excited vibrational states become occupied, leading to transition states occuring at different energies. As a result, the associated reaction trajectory at higher temperature travels through different activated complexes, making the reaction harder to model and any associated results more difficult to compare.

'''Reaction 2: H + HF --> F + H2'''

Reaction 2, which is the reverse reaction of Reaction 1, is endothermic with the energy of the products being higher than the reactants. As a result, according the Hammond's Postulate, the position of the transition state is later in the reaction trajectory and more closely resembles the products. Contrary to the preferred conditions for exothermic reactions, a larger amount of vibrational energy is needed. To achieve this, lower momentum would be adopted. Providing this is achieved, the set of reaction trajectories would be an example of an efficient reaction as the reaction trajectories would pass through the transition state once and to the products.

----

==References==

<references>
<ref name="transition">P. Jonathan P. K. Doye, David J. Wales, Journal of mathematical physics, 1996, '''48''', 843 - 846.</ref>
<ref name="quantum">Physicsoftheuniverse.com. (2017). Quantum Tunnelling and the Uncertainty Principle - Quantum Theory and the Uncertainty Principle - The Physics of the Universe. [online] Available at: http://www.physicsoftheuniverse.com/topics_quantum_uncertainty.html [Accessed 11 May 2017].</ref>
<ref name="energies">Wiredchemist.com. (2017). Common Bond Energies (D. [online] Available at: http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html [Accessed 11 May 2017].</ref>
<ref name="lawE">Mach, E. (1872). History and Root of the Principles of the Conservation of Energy. Open Court Pub. Co., Illinois.</ref>
<ref name="polanyi">Yan, S., Y.-T. Wu, and K. Liu. "Tracking The Energy Flow Along The Reaction Path". Proceedings of the National Academy of Sciences 105.35 (2008): 12667-12672.</ref>
<ref name="hammond">Chemistry LibreTexts. (2017). Hammond’s Postulate. [online] Available at: https://chem.libretexts.org/Reference/Organic_Chemistry_Glossary/Hammond’s_Postulate [Accessed 11 May 2017].</ref>
<references/>

File:Ab6114 figure2.PNG

2017-05-26T11:02:02Z

Ab6114:

MRD:ab6114

2017-05-26T10:57:55Z

Ab6114:

=='''Molecular Reaction Dynamics'''==

==Exercise 1: H + H2 system==

==='''Question(1)'''===

What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

'''Answer:'''

Analysis of the PES curvature shows a maxima between two minimum points, all being stationary points with a gradient of zero. The first derivative of the curve, where the rate of change is zero, is found to confirm this. As Transition structure's inherently exist at the maximum point, it can be inferred that the transition structure lies exists at this point. In order to differentiate between the transition state and the minima, second derivative calculations can be performed, producing a negative value for a maxima and a positive value for a minima.<ref name="transition" />

----

==='''Question(2)'''===

Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.

'''Answer:'''

The approximate transition state position (r(ts)) was determined through analysis of the surface plot. This was confirmed by examining the internuclear distance vs time plot, to identify the point of intersection between the maxima and the minima - r(ts) is shown in the table below:

{| class="wikitable" border="1"
|+ Table 1
! r1 !! r2 !! r(ts)
|-
| 1.20 || 1.20 || 0.9075
|-
| 0.91 || 0.91 || 0.9075
|}

[[File:ab6114Figure1.PNG|680px|thumb|centre|Figure 1 - Internuclear distance vs Time plot of transition state position]]

A transition state is a point where the internuclear distance between all bonds remain constant and thus do not oscillate. Figure 1 shows that the internuclear distance does not oscillate at 0.9075 Å - hence this is the transition state position.

----

==='''Question(3)'''===

Comment on how the mep and the trajectory you just calculated differ.

'''Answer:'''

Dynamic calculations and mep are distinct for two main reasons. Calculations of mep span only the distance engaged by the transition state, while the dynamic calculation incorporates more of the reaction pathway - the dynamic calculation also takes mass into account, thus resulting in an oscillating reaction pathway line. No mass is included in the mep calculation, so no oscillation is observed.

----

==='''Question(4)'''===

Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.

'''Answer:'''

The effect of momentum variation on whether a trajectory is reactive or unreactive (for the initial positions r1 = 0.74 and r2 = 2.0):

{| class="wikitable" border="1"
|+ Table 2
! p1 !! p2 !! Trajectory !! Figure No.
|-
| -1.25 || -2.5 || Reactive || 2
|-
| -1.5 || -2.0 || Unreactive || 3
|-
| -1.5 || -2.5 || Reactive || 4
|-
| -2.5 || -5.0 || Unreactive || 5
|-
| -2.5 || -5.2 || Reactive || 6
|}

'''Figure 2'''

The surface plot shows the reactive trajectory corresponding with the set parameters for figure 2. The trajectory line shows that the A-B bond distance decreases as atom Ha approaches molecule Hb-Hc while at the same time the B-C bond distance remains relatively constant. This line continues through the transition state (r(A-B)=r(B-C)) and r(A-B) reaches a minimum and oscillates thus indicating that the Ha-Hb bond has formed. Simultaneously r(B-C) increases, confirming that the Hb-Hc bond has broken and Hc now exists as a separate hydrogen atom.

[[File:ab6114figure2.PNG|680px|thumb|centre|Figure 2 - The reactive trajectory associated with the specific parameters for Figure 2]]

'''Figure 3'''

The surface plot below shows the unreactive trajectory associated with the set parameters for Figure 3. The trajectory line shows r(A-B) decreasing as Ha approaches Hb-Hc. However, since Ha does not possess sufficient momentum to overcome the activation barrier, the trajectory never reaches the transition state. The plot reflects this as it shows r(A-B) returning to a larger distance before the transition state is reached. r(B-C) oscillates due to vibrational energy within the molecule.

[[File:HZ4315 Figure3.PNG|700px|thumb|centre|Figure 3 - The unreactive trajectory associated with the set parameters for Figure 3]]

'''Figure 4'''

The plot below shows the reactive trajectory associated with the set parameters for Figure 4. The reaction trajectory shows a successful reaction where Ha possesses enough momentum to overcome the activation barrier when approaching Hb-Hc hence, it passes through the transition state. The reaction then proceeds to form the molecule Ha-Hb.

[[File:HZ4315 Figure4.PNG|700px|thumb|centre|Figure 4 - The reactive trajectory associated with the set parameters for Figure 4]]

'''Figure 5'''

The plot below shows the unreactive trajectory associated with the set parameters for Figure 5. The reaction trajectory for these parameters initially shows that Ha approaches Hb-Hc with sufficient momentum to overcome the activation barrier and reach the transition state. However, instead of proceeding to the products, the reaction trajectory returns back to the reactants with molecule Hb-Hc possessing a large amount of vibrational energy.

[[File:HZ4315 Figure5.PNG|700px|thumb|centre|Figure 5 - The unreactive trajectory associated with the set parameters for Figure 5]]

'''Figure 6'''

The plot below shows the reactive trajectory associated with the set parameters for Figure 6. The reaction trajectory indicates that Ha has sufficient momentum to overcome the activation barrier however, upon reaching the transition state, the trajectory is driven back towards the reactants which is illustrated by r(B-C) decreasing and r(A-B) increasing. r(A-B) reaches a maximum value before decreasing and oscillating, indicating the formation of Ha-Hb. This trajectory suggests that the reaction has taken an alternative reaction pathway which avoids the 'saddle point' on the potential energy plot associated with the transition states to yield the desired products.

[[File:HZ4315 Figure6.PNG|700px|thumb|centre|Figure 6 - The reactive trajectory associated with the set parameters for Figure 6]]

----

==='''Q(5)'''===

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

'''A:'''

One assumption of the Transition State Theory is that the reactants and activated transition states are in quasi-equilibrium. Quasi-equilibrium involves the molecular driving forces which would alter the state of the system being very small, resulting in an internal equilibrium being reached. Deviations from this system are very small. In addition to this, according to the Classical Transition State Theory, providing an atom possesses sufficient momentum to overcome the activation barrier, upon collision with another molecule it will react, yielding products. However, since the theory does not account for potential quantum mechanical effects, the resulting reaction rates will be lower than expected. Quantum mechanical contribution can provide alternative reaction pathways through the introduction of tunneling. This can occur when the atom and molecule do not have the required energy to overcome the activation barrier however, due to the wave like properties of the particles, they can 'tunnel' through the barrier instead. As a result, quantum mechanics allows certain reactions to avoid the classical 'saddle point' on the potential energy surface plots associated with the transition state and instead take up an alternative reaction pathway through the barrier.<ref name="quantum" />

----

==Exercise 2: F - H - H system==

==='''Q(6)'''===

Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

'''A:'''

Key points:

- H-F Bond Energy = 565 KJ/Mol<ref name="energies" />

- H-H Bond Energy = 432 KJ/Mol<ref name="energies" />

- H-F is a stronger bond than H-H

- ΔE = E(bonds broken)-E(bonds formed)

'''F + H2''' --> ΔE = 432 - 565 = -133 KJ/Mol '''THEREFORE EXOTHERMIC'''

'''H + H-F''' --> ΔE = 565 - 432 = 133 KJ/Mol '''THEREFORE ENDOTHERMIC'''

----

==='''Q(7)'''===

Locate the approximate position of the transition state.

'''A:'''

To determine the position of the transition state, the radii at which the total energy of the system is stationary (ie. the gradient is zero) were found. The location of these radii is such that the net force acting upon them is zero and plotting the internuclear distance against time results in a constant internuclear separation which does not oscillate. The internuclear separations are shown below and the associated plot against time can be seen in figure 7:

- r(H-H) = 0.74Å

- r(H-F) = 1.815Å

[[File:HZ4315 Figure 7.PNG|700px|thumb|centre|Figure 7 - The Internuclear distance vs Time plot for the transition state position]]

----

==='''Q(8)'''===

Report the activation energy for both reactions.

'''A:'''

In order to determine the activation energy for both reactions, chemical intuition and trial and error were used to find a suitable reaction trajectory. The activation energy was then equal to the difference in energy between that of the reactants and the transition state. The activation energy values obtained were done without mathematical analysis therefore, all energy values involved would actually have a +/- value.

'''Reaction 1: F + H2 --> HF + H'''

- Ea = -103.7 - (-103.9) = 0.2 KCal/Mol

The energy values used for this calculation were obtained from Figure 8 (reactants) and Figure 9 (transition state) as shown below:

[[File:HZ4315 Figure8.PNG|500px|thumb|centre|Figure 8 - Surface plot showing the energy of the reactants]]

[[File:HZ4315 Figure9.PNG|500px|thumb|centre|Figure 9 - Surface plot showing the energy at the transition state]]

'''Reaction 2: H + HF --> F + H2'''

- Ea = -103.7 - (-133.9) = 30.2 KCal/Mol

The energy values used for this calculation were obtained from Figure 10 (reactants) and Figure 9 (transition state) as shown below:

[[File:HZ4315 Figure9.PNG|500px|thumb|centre|Figure 9 - Surface plot showing the energy at the transition state]]

[[File:HZ4315 Figure10.PNG|500px|thumb|centre|Figure 10 - Surface plot showing the energy of the reactants in the reverse reaction]]

----

==='''Q(9)'''===

In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

'''A:'''

According to the Law of Conservation of Energy, the total energy of an isolated system must be conserved as energy is neither created nor destroyed, just converted from one form to another.<ref name="lawE" /> For this specific reaction system, the total energy is the sum of the kinetic and potential energies of each substrate involved in the reaction. For the exothermic reaction F + H2, the potential energy of the products is lower than the reactants hence, the kinetic energy of the products must be higher. This is illustrated by determining a successful reaction trajectory and plotting a kinetic energy vs time graph as shown below:

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 11 - A reaction pathway for an ultimately successful reaction (parameters determined using trial and error)]]

[[File:KEvsTIME.PNG|500px|thumb|centre|Figure 12 - Kinetic energy vs Time plot of the successful reaction - the small oscillations on the left correspond to the reactants which are clearly converted to products as the kinetic energy increases significantly]]

The increased kinetic energy allows the products to be detected by IR analysis. Since the H-F bond is oscillating, it will release energy in the form of vibrational energy, resulting in a peak on the spectrum. The H-H vibrational peak present on the reactant spectrum will now be absent. Another possible experimental technique is calorimetry. Since the reaction is exothermic, the amount of heat released can be measured and since, at constant pressure, the enthalpy change is equal to the net change in energy, the net free energy change of the reaction can be determined.

----

==='''Q(10)'''===

Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

'''A:'''

Using trial and error, a successful reaction trajectory was found for both reactions shown below. The parameters used can be seen within the figures.

'''Reaction 1: F + H2 --> HF + H'''

'''Lower momentum'''

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 13 - An ultimately successful reaction pathway at high momentum indicating the preference of an early transition state]]

'''Higher momentum'''

[[File:HZ4315 Figure12.PNG|500px|thumb|centre|Figure 14 - The reaction pathway at an even larger momentum to confirm the theory]]

Reaction 1, which is an exothermic reaction, exhibits an increased translational energy in response to increasing the momentum of the reactants. According to Polanyi's rules, higher translational energies are more efficient at activating an early transition state (ie. an early barrier) than vibrational energy.<ref name="polanyi" /> Since early transition states resemble the reactants (as stated by Hammond's Postulate)<ref name="hammond" />, the reaction trajectory may be expected to bounce back towards the reactants. Figure 12 confirms this as an increased momentum (and therefore increased translational energy) sees the trajectory bounce back several times before eventually proceeding to the form the products. Figures 11 and 12 are an example of a set of inefficient reaction trajectories as the barrier is recrossed several times which is not ideal in the formation of the product.

The exothermic nature of the reaction also alters the position of the transition state away from the lowest 'saddle point'. This goes against the assumptions of the Transition State Theory which states that reaction trajectories should pass through this. As temperature increases, a greater proportion of excited vibrational states become occupied, leading to transition states occuring at different energies. As a result, the associated reaction trajectory at higher temperature travels through different activated complexes, making the reaction harder to model and any associated results more difficult to compare.

'''Reaction 2: H + HF --> F + H2'''

Reaction 2, which is the reverse reaction of Reaction 1, is endothermic with the energy of the products being higher than the reactants. As a result, according the Hammond's Postulate, the position of the transition state is later in the reaction trajectory and more closely resembles the products. Contrary to the preferred conditions for exothermic reactions, a larger amount of vibrational energy is needed. To achieve this, lower momentum would be adopted. Providing this is achieved, the set of reaction trajectories would be an example of an efficient reaction as the reaction trajectories would pass through the transition state once and to the products.

----

==References==

<references>
<ref name="transition">P. Jonathan P. K. Doye, David J. Wales, Journal of mathematical physics, 1996, '''48''', 843 - 846.</ref>
<ref name="quantum">Physicsoftheuniverse.com. (2017). Quantum Tunnelling and the Uncertainty Principle - Quantum Theory and the Uncertainty Principle - The Physics of the Universe. [online] Available at: http://www.physicsoftheuniverse.com/topics_quantum_uncertainty.html [Accessed 11 May 2017].</ref>
<ref name="energies">Wiredchemist.com. (2017). Common Bond Energies (D. [online] Available at: http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html [Accessed 11 May 2017].</ref>
<ref name="lawE">Mach, E. (1872). History and Root of the Principles of the Conservation of Energy. Open Court Pub. Co., Illinois.</ref>
<ref name="polanyi">Yan, S., Y.-T. Wu, and K. Liu. "Tracking The Energy Flow Along The Reaction Path". Proceedings of the National Academy of Sciences 105.35 (2008): 12667-12672.</ref>
<ref name="hammond">Chemistry LibreTexts. (2017). Hammond’s Postulate. [online] Available at: https://chem.libretexts.org/Reference/Organic_Chemistry_Glossary/Hammond’s_Postulate [Accessed 11 May 2017].</ref>
<references/>

File:Ab6114figure6.png

2017-05-26T10:44:02Z

Ab6114:

File:Ab6114figure5.png

2017-05-26T10:43:27Z

Ab6114:

File:Ab6114figure4.png

2017-05-26T10:43:07Z

Ab6114:

File:Ab6114figure3.png

2017-05-26T10:42:51Z

Ab6114:

File:Ab6114figure2.png

2017-05-26T10:42:29Z

Ab6114:

MRD:ab6114

2017-05-26T10:41:30Z

Ab6114:

=='''Molecular Reaction Dynamics'''==

==Exercise 1: H + H2 system==

==='''Question(1)'''===

What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

'''Answer:'''

Analysis of the PES curvature shows a maxima between two minimum points, all being stationary points with a gradient of zero. The first derivative of the curve, where the rate of change is zero, is found to confirm this. As Transition structure's inherently exist at the maximum point, it can be inferred that the transition structure lies exists at this point. In order to differentiate between the transition state and the minima, second derivative calculations can be performed, producing a negative value for a maxima and a positive value for a minima.<ref name="transition" />

----

==='''Question(2)'''===

Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.

'''Answer:'''

The approximate transition state position (r(ts)) was determined through analysis of the surface plot. This was confirmed by examining the internuclear distance vs time plot, to identify the point of intersection between the maxima and the minima - r(ts) is shown in the table below:

{| class="wikitable" border="1"
|+ Table 1
! r1 !! r2 !! r(ts)
|-
| 1.20 || 1.20 || 0.9075
|-
| 0.91 || 0.91 || 0.9075
|}

[[File:ab6114Figure1.PNG|680px|thumb|centre|Figure 1 - Internuclear distance vs Time plot of transition state position]]

A transition state is a point where the internuclear distance between all bonds remain constant and thus do not oscillate. Figure 1 shows that the internuclear distance does not oscillate at 0.9075 Å - hence this is the transition state position.

----

==='''Question(3)'''===

Comment on how the mep and the trajectory you just calculated differ.

'''Answer:'''

Dynamic calculations and mep are distinct for two main reasons. Calculations of mep span only the distance engaged by the transition state, while the dynamic calculation incorporates more of the reaction pathway - the dynamic calculation also takes mass into account, thus resulting in an oscillating reaction pathway line. No mass is included in the mep calculation, so no oscillation is observed.

----

==='''Question(4)'''===

Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.

'''Answer:'''

The effect of momentum variation on whether a trajectory is reactive or unreactive (for the initial positions r1 = 0.74 and r2 = 2.0):

{| class="wikitable" border="1"
|+ Table 2
! p1 !! p2 !! Trajectory !! Figure No.
|-
| -1.25 || -2.5 || Reactive || 2
|-
| -1.5 || -2.0 || Unreactive || 3
|-
| -1.5 || -2.5 || Reactive || 4
|-
| -2.5 || -5.0 || Unreactive || 5
|-
| -2.5 || -5.2 || Reactive || 6
|}

'''Figure 2'''

The plot below shows the reactive trajectory associated with the set parameters for Figure 2. The trajectory line indicates that the A-B bond distance decreases as atom Ha approaches molecule Hb-Hc whilst the B-C bond distance stays relatively constant. The line continues through the transition state where r(A-B)=r(B-C) after which r(A-B) reaches a minimum and oscillates which indicates that the Ha-Hb bond has formed whilst r(B-C) simultaneously increases, showing that the Hb-Hc bond has broken and Hc now exists as a separate atom.

[[File:HZ4315 FIgure2.PNG|700px|thumb|centre|Figure 2 - The reactive trajectory associated with the set parameters for Figure 2]]

'''Figure 3'''

The plot below shows the unreactive trajectory associated with the set parameters for Figure 3. The trajectory line shows r(A-B) decreasing as Ha approaches Hb-Hc. However, since Ha does not possess sufficient momentum to overcome the activation barrier, the trajectory never reaches the transition state. The plot reflects this as it shows r(A-B) returning to a larger distance before the transition state is reached. r(B-C) oscillates due to vibrational energy within the molecule.

[[File:HZ4315 Figure3.PNG|700px|thumb|centre|Figure 3 - The unreactive trajectory associated with the set parameters for Figure 3]]

'''Figure 4'''

The plot below shows the reactive trajectory associated with the set parameters for Figure 4. The reaction trajectory shows a successful reaction where Ha possesses enough momentum to overcome the activation barrier when approaching Hb-Hc hence, it passes through the transition state. The reaction then proceeds to form the molecule Ha-Hb.

[[File:HZ4315 Figure4.PNG|700px|thumb|centre|Figure 4 - The reactive trajectory associated with the set parameters for Figure 4]]

'''Figure 5'''

The plot below shows the unreactive trajectory associated with the set parameters for Figure 5. The reaction trajectory for these parameters initially shows that Ha approaches Hb-Hc with sufficient momentum to overcome the activation barrier and reach the transition state. However, instead of proceeding to the products, the reaction trajectory returns back to the reactants with molecule Hb-Hc possessing a large amount of vibrational energy.

[[File:HZ4315 Figure5.PNG|700px|thumb|centre|Figure 5 - The unreactive trajectory associated with the set parameters for Figure 5]]

'''Figure 6'''

The plot below shows the reactive trajectory associated with the set parameters for Figure 6. The reaction trajectory indicates that Ha has sufficient momentum to overcome the activation barrier however, upon reaching the transition state, the trajectory is driven back towards the reactants which is illustrated by r(B-C) decreasing and r(A-B) increasing. r(A-B) reaches a maximum value before decreasing and oscillating, indicating the formation of Ha-Hb. This trajectory suggests that the reaction has taken an alternative reaction pathway which avoids the 'saddle point' on the potential energy plot associated with the transition states to yield the desired products.

[[File:HZ4315 Figure6.PNG|700px|thumb|centre|Figure 6 - The reactive trajectory associated with the set parameters for Figure 6]]

----

==='''Q(5)'''===

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

'''A:'''

One assumption of the Transition State Theory is that the reactants and activated transition states are in quasi-equilibrium. Quasi-equilibrium involves the molecular driving forces which would alter the state of the system being very small, resulting in an internal equilibrium being reached. Deviations from this system are very small. In addition to this, according to the Classical Transition State Theory, providing an atom possesses sufficient momentum to overcome the activation barrier, upon collision with another molecule it will react, yielding products. However, since the theory does not account for potential quantum mechanical effects, the resulting reaction rates will be lower than expected. Quantum mechanical contribution can provide alternative reaction pathways through the introduction of tunneling. This can occur when the atom and molecule do not have the required energy to overcome the activation barrier however, due to the wave like properties of the particles, they can 'tunnel' through the barrier instead. As a result, quantum mechanics allows certain reactions to avoid the classical 'saddle point' on the potential energy surface plots associated with the transition state and instead take up an alternative reaction pathway through the barrier.<ref name="quantum" />

----

==Exercise 2: F - H - H system==

==='''Q(6)'''===

Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

'''A:'''

Key points:

- H-F Bond Energy = 565 KJ/Mol<ref name="energies" />

- H-H Bond Energy = 432 KJ/Mol<ref name="energies" />

- H-F is a stronger bond than H-H

- ΔE = E(bonds broken)-E(bonds formed)

'''F + H2''' --> ΔE = 432 - 565 = -133 KJ/Mol '''THEREFORE EXOTHERMIC'''

'''H + H-F''' --> ΔE = 565 - 432 = 133 KJ/Mol '''THEREFORE ENDOTHERMIC'''

----

==='''Q(7)'''===

Locate the approximate position of the transition state.

'''A:'''

To determine the position of the transition state, the radii at which the total energy of the system is stationary (ie. the gradient is zero) were found. The location of these radii is such that the net force acting upon them is zero and plotting the internuclear distance against time results in a constant internuclear separation which does not oscillate. The internuclear separations are shown below and the associated plot against time can be seen in figure 7:

- r(H-H) = 0.74Å

- r(H-F) = 1.815Å

[[File:HZ4315 Figure 7.PNG|700px|thumb|centre|Figure 7 - The Internuclear distance vs Time plot for the transition state position]]

----

==='''Q(8)'''===

Report the activation energy for both reactions.

'''A:'''

In order to determine the activation energy for both reactions, chemical intuition and trial and error were used to find a suitable reaction trajectory. The activation energy was then equal to the difference in energy between that of the reactants and the transition state. The activation energy values obtained were done without mathematical analysis therefore, all energy values involved would actually have a +/- value.

'''Reaction 1: F + H2 --> HF + H'''

- Ea = -103.7 - (-103.9) = 0.2 KCal/Mol

The energy values used for this calculation were obtained from Figure 8 (reactants) and Figure 9 (transition state) as shown below:

[[File:HZ4315 Figure8.PNG|500px|thumb|centre|Figure 8 - Surface plot showing the energy of the reactants]]

[[File:HZ4315 Figure9.PNG|500px|thumb|centre|Figure 9 - Surface plot showing the energy at the transition state]]

'''Reaction 2: H + HF --> F + H2'''

- Ea = -103.7 - (-133.9) = 30.2 KCal/Mol

The energy values used for this calculation were obtained from Figure 10 (reactants) and Figure 9 (transition state) as shown below:

[[File:HZ4315 Figure9.PNG|500px|thumb|centre|Figure 9 - Surface plot showing the energy at the transition state]]

[[File:HZ4315 Figure10.PNG|500px|thumb|centre|Figure 10 - Surface plot showing the energy of the reactants in the reverse reaction]]

----

==='''Q(9)'''===

In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

'''A:'''

According to the Law of Conservation of Energy, the total energy of an isolated system must be conserved as energy is neither created nor destroyed, just converted from one form to another.<ref name="lawE" /> For this specific reaction system, the total energy is the sum of the kinetic and potential energies of each substrate involved in the reaction. For the exothermic reaction F + H2, the potential energy of the products is lower than the reactants hence, the kinetic energy of the products must be higher. This is illustrated by determining a successful reaction trajectory and plotting a kinetic energy vs time graph as shown below:

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 11 - A reaction pathway for an ultimately successful reaction (parameters determined using trial and error)]]

[[File:KEvsTIME.PNG|500px|thumb|centre|Figure 12 - Kinetic energy vs Time plot of the successful reaction - the small oscillations on the left correspond to the reactants which are clearly converted to products as the kinetic energy increases significantly]]

The increased kinetic energy allows the products to be detected by IR analysis. Since the H-F bond is oscillating, it will release energy in the form of vibrational energy, resulting in a peak on the spectrum. The H-H vibrational peak present on the reactant spectrum will now be absent. Another possible experimental technique is calorimetry. Since the reaction is exothermic, the amount of heat released can be measured and since, at constant pressure, the enthalpy change is equal to the net change in energy, the net free energy change of the reaction can be determined.

----

==='''Q(10)'''===

Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

'''A:'''

Using trial and error, a successful reaction trajectory was found for both reactions shown below. The parameters used can be seen within the figures.

'''Reaction 1: F + H2 --> HF + H'''

'''Lower momentum'''

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 13 - An ultimately successful reaction pathway at high momentum indicating the preference of an early transition state]]

'''Higher momentum'''

[[File:HZ4315 Figure12.PNG|500px|thumb|centre|Figure 14 - The reaction pathway at an even larger momentum to confirm the theory]]

Reaction 1, which is an exothermic reaction, exhibits an increased translational energy in response to increasing the momentum of the reactants. According to Polanyi's rules, higher translational energies are more efficient at activating an early transition state (ie. an early barrier) than vibrational energy.<ref name="polanyi" /> Since early transition states resemble the reactants (as stated by Hammond's Postulate)<ref name="hammond" />, the reaction trajectory may be expected to bounce back towards the reactants. Figure 12 confirms this as an increased momentum (and therefore increased translational energy) sees the trajectory bounce back several times before eventually proceeding to the form the products. Figures 11 and 12 are an example of a set of inefficient reaction trajectories as the barrier is recrossed several times which is not ideal in the formation of the product.

The exothermic nature of the reaction also alters the position of the transition state away from the lowest 'saddle point'. This goes against the assumptions of the Transition State Theory which states that reaction trajectories should pass through this. As temperature increases, a greater proportion of excited vibrational states become occupied, leading to transition states occuring at different energies. As a result, the associated reaction trajectory at higher temperature travels through different activated complexes, making the reaction harder to model and any associated results more difficult to compare.

'''Reaction 2: H + HF --> F + H2'''

Reaction 2, which is the reverse reaction of Reaction 1, is endothermic with the energy of the products being higher than the reactants. As a result, according the Hammond's Postulate, the position of the transition state is later in the reaction trajectory and more closely resembles the products. Contrary to the preferred conditions for exothermic reactions, a larger amount of vibrational energy is needed. To achieve this, lower momentum would be adopted. Providing this is achieved, the set of reaction trajectories would be an example of an efficient reaction as the reaction trajectories would pass through the transition state once and to the products.

----

==References==

<references>
<ref name="transition">P. Jonathan P. K. Doye, David J. Wales, Journal of mathematical physics, 1996, '''48''', 843 - 846.</ref>
<ref name="quantum">Physicsoftheuniverse.com. (2017). Quantum Tunnelling and the Uncertainty Principle - Quantum Theory and the Uncertainty Principle - The Physics of the Universe. [online] Available at: http://www.physicsoftheuniverse.com/topics_quantum_uncertainty.html [Accessed 11 May 2017].</ref>
<ref name="energies">Wiredchemist.com. (2017). Common Bond Energies (D. [online] Available at: http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html [Accessed 11 May 2017].</ref>
<ref name="lawE">Mach, E. (1872). History and Root of the Principles of the Conservation of Energy. Open Court Pub. Co., Illinois.</ref>
<ref name="polanyi">Yan, S., Y.-T. Wu, and K. Liu. "Tracking The Energy Flow Along The Reaction Path". Proceedings of the National Academy of Sciences 105.35 (2008): 12667-12672.</ref>
<ref name="hammond">Chemistry LibreTexts. (2017). Hammond’s Postulate. [online] Available at: https://chem.libretexts.org/Reference/Organic_Chemistry_Glossary/Hammond’s_Postulate [Accessed 11 May 2017].</ref>
<references/>

File:Ab6114Figure1.PNG

2017-05-26T09:59:58Z

Ab6114:

MRD:ab6114

2017-05-26T09:58:44Z

=='''Molecular Reaction Dynamics'''==

==Exercise 1: H + H2 system==

==='''Question(1)'''===

What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

'''Answer:'''

Analysis of the PES curvature shows a maxima between two minimum points, all being stationary points with a gradient of zero. The first derivative of the curve, where the rate of change is zero, is found to confirm this. As Transition structure's inherently exist at the maximum point, it can be inferred that the transition structure lies exists at this point. In order to differentiate between the transition state and the minima, second derivative calculations can be performed, producing a negative value for a maxima and a positive value for a minima.<ref name="transition" />

----

==='''Question(2)'''===

Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.

'''Answer:'''

The approximate transition state position (r(ts)) was determined through analysis of the surface plot. This was confirmed by examining the internuclear distance vs time plot, to identify the point of intersection between the maxima and the minima - r(ts) is shown in the table below:

{| class="wikitable" border="1"
|+ Table 1
! r1 !! r2 !! r(ts)
|-
| 1.20 || 1.20 || 0.9075
|-
| 0.91 || 0.91 || 0.9075
|}

[[File:Figure1.PNG|680px|thumb|centre|Figure 1 - Internuclear distance vs Time plot of transition state position]]

By definition, a transition state is a point where the internuclear distance between all bonds remains constant and therefore does not oscillate. As shown by figure 1, the internuclear distance does not oscillate at 0.908Å hence, this is the transition state position.

----

==='''Q(3)'''===

Comment on how the mep and the trajectory you just calculated differ.

'''A:'''

There are two distinct differences between the mep and dynamic calculations. The mep calculation only spans the distance occupied by the transition state whereas the dynamic calculation spans more of the reaction pathway. Furthermore, the dynamic calculation incorporates mass hence, the resulting reaction pathway line exhibits oscillation. In contrast, mep calculations do not use mass and therefore no oscillation is witnessed.

----

==='''Q(4)'''===

Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.

'''A:'''

For the initial positions r1 = 0.74 and r2 = 2.0, the following results were obtained:

The effect of changing momentum on whether a trajectory is reactive or unreactive:

{| class="wikitable" border="1"
|+ Table 2
! p1 !! p2 !! Trajectory !! Figure No.
|-
| -1.25 || -2.5 || Reactive || 2
|-
| -1.5 || -2.0 || Unreactive || 3
|-
| -1.5 || -2.5 || Reactive || 4
|-
| -2.5 || -5.0 || Unreactive || 5
|-
| -2.5 || -5.2 || Reactive || 6
|}

'''Figure 2'''

The plot below shows the reactive trajectory associated with the set parameters for Figure 2. The trajectory line indicates that the A-B bond distance decreases as atom Ha approaches molecule Hb-Hc whilst the B-C bond distance stays relatively constant. The line continues through the transition state where r(A-B)=r(B-C) after which r(A-B) reaches a minimum and oscillates which indicates that the Ha-Hb bond has formed whilst r(B-C) simultaneously increases, showing that the Hb-Hc bond has broken and Hc now exists as a separate atom.

[[File:HZ4315 FIgure2.PNG|700px|thumb|centre|Figure 2 - The reactive trajectory associated with the set parameters for Figure 2]]

'''Figure 3'''

The plot below shows the unreactive trajectory associated with the set parameters for Figure 3. The trajectory line shows r(A-B) decreasing as Ha approaches Hb-Hc. However, since Ha does not possess sufficient momentum to overcome the activation barrier, the trajectory never reaches the transition state. The plot reflects this as it shows r(A-B) returning to a larger distance before the transition state is reached. r(B-C) oscillates due to vibrational energy within the molecule.

[[File:HZ4315 Figure3.PNG|700px|thumb|centre|Figure 3 - The unreactive trajectory associated with the set parameters for Figure 3]]

'''Figure 4'''

The plot below shows the reactive trajectory associated with the set parameters for Figure 4. The reaction trajectory shows a successful reaction where Ha possesses enough momentum to overcome the activation barrier when approaching Hb-Hc hence, it passes through the transition state. The reaction then proceeds to form the molecule Ha-Hb.

[[File:HZ4315 Figure4.PNG|700px|thumb|centre|Figure 4 - The reactive trajectory associated with the set parameters for Figure 4]]

'''Figure 5'''

The plot below shows the unreactive trajectory associated with the set parameters for Figure 5. The reaction trajectory for these parameters initially shows that Ha approaches Hb-Hc with sufficient momentum to overcome the activation barrier and reach the transition state. However, instead of proceeding to the products, the reaction trajectory returns back to the reactants with molecule Hb-Hc possessing a large amount of vibrational energy.

[[File:HZ4315 Figure5.PNG|700px|thumb|centre|Figure 5 - The unreactive trajectory associated with the set parameters for Figure 5]]

'''Figure 6'''

The plot below shows the reactive trajectory associated with the set parameters for Figure 6. The reaction trajectory indicates that Ha has sufficient momentum to overcome the activation barrier however, upon reaching the transition state, the trajectory is driven back towards the reactants which is illustrated by r(B-C) decreasing and r(A-B) increasing. r(A-B) reaches a maximum value before decreasing and oscillating, indicating the formation of Ha-Hb. This trajectory suggests that the reaction has taken an alternative reaction pathway which avoids the 'saddle point' on the potential energy plot associated with the transition states to yield the desired products.

[[File:HZ4315 Figure6.PNG|700px|thumb|centre|Figure 6 - The reactive trajectory associated with the set parameters for Figure 6]]

----

==='''Q(5)'''===

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

'''A:'''

One assumption of the Transition State Theory is that the reactants and activated transition states are in quasi-equilibrium. Quasi-equilibrium involves the molecular driving forces which would alter the state of the system being very small, resulting in an internal equilibrium being reached. Deviations from this system are very small. In addition to this, according to the Classical Transition State Theory, providing an atom possesses sufficient momentum to overcome the activation barrier, upon collision with another molecule it will react, yielding products. However, since the theory does not account for potential quantum mechanical effects, the resulting reaction rates will be lower than expected. Quantum mechanical contribution can provide alternative reaction pathways through the introduction of tunneling. This can occur when the atom and molecule do not have the required energy to overcome the activation barrier however, due to the wave like properties of the particles, they can 'tunnel' through the barrier instead. As a result, quantum mechanics allows certain reactions to avoid the classical 'saddle point' on the potential energy surface plots associated with the transition state and instead take up an alternative reaction pathway through the barrier.<ref name="quantum" />

----

==Exercise 2: F - H - H system==

==='''Q(6)'''===

Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

'''A:'''

Key points:

- H-F Bond Energy = 565 KJ/Mol<ref name="energies" />

- H-H Bond Energy = 432 KJ/Mol<ref name="energies" />

- H-F is a stronger bond than H-H

- ΔE = E(bonds broken)-E(bonds formed)

'''F + H2''' --> ΔE = 432 - 565 = -133 KJ/Mol '''THEREFORE EXOTHERMIC'''

'''H + H-F''' --> ΔE = 565 - 432 = 133 KJ/Mol '''THEREFORE ENDOTHERMIC'''

----

==='''Q(7)'''===

Locate the approximate position of the transition state.

'''A:'''

To determine the position of the transition state, the radii at which the total energy of the system is stationary (ie. the gradient is zero) were found. The location of these radii is such that the net force acting upon them is zero and plotting the internuclear distance against time results in a constant internuclear separation which does not oscillate. The internuclear separations are shown below and the associated plot against time can be seen in figure 7:

- r(H-H) = 0.74Å

- r(H-F) = 1.815Å

[[File:HZ4315 Figure 7.PNG|700px|thumb|centre|Figure 7 - The Internuclear distance vs Time plot for the transition state position]]

----

==='''Q(8)'''===

Report the activation energy for both reactions.

'''A:'''

In order to determine the activation energy for both reactions, chemical intuition and trial and error were used to find a suitable reaction trajectory. The activation energy was then equal to the difference in energy between that of the reactants and the transition state. The activation energy values obtained were done without mathematical analysis therefore, all energy values involved would actually have a +/- value.

'''Reaction 1: F + H2 --> HF + H'''

- Ea = -103.7 - (-103.9) = 0.2 KCal/Mol

The energy values used for this calculation were obtained from Figure 8 (reactants) and Figure 9 (transition state) as shown below:

[[File:HZ4315 Figure8.PNG|500px|thumb|centre|Figure 8 - Surface plot showing the energy of the reactants]]

[[File:HZ4315 Figure9.PNG|500px|thumb|centre|Figure 9 - Surface plot showing the energy at the transition state]]

'''Reaction 2: H + HF --> F + H2'''

- Ea = -103.7 - (-133.9) = 30.2 KCal/Mol

The energy values used for this calculation were obtained from Figure 10 (reactants) and Figure 9 (transition state) as shown below:

[[File:HZ4315 Figure9.PNG|500px|thumb|centre|Figure 9 - Surface plot showing the energy at the transition state]]

[[File:HZ4315 Figure10.PNG|500px|thumb|centre|Figure 10 - Surface plot showing the energy of the reactants in the reverse reaction]]

----

==='''Q(9)'''===

In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

'''A:'''

According to the Law of Conservation of Energy, the total energy of an isolated system must be conserved as energy is neither created nor destroyed, just converted from one form to another.<ref name="lawE" /> For this specific reaction system, the total energy is the sum of the kinetic and potential energies of each substrate involved in the reaction. For the exothermic reaction F + H2, the potential energy of the products is lower than the reactants hence, the kinetic energy of the products must be higher. This is illustrated by determining a successful reaction trajectory and plotting a kinetic energy vs time graph as shown below:

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 11 - A reaction pathway for an ultimately successful reaction (parameters determined using trial and error)]]

[[File:KEvsTIME.PNG|500px|thumb|centre|Figure 12 - Kinetic energy vs Time plot of the successful reaction - the small oscillations on the left correspond to the reactants which are clearly converted to products as the kinetic energy increases significantly]]

The increased kinetic energy allows the products to be detected by IR analysis. Since the H-F bond is oscillating, it will release energy in the form of vibrational energy, resulting in a peak on the spectrum. The H-H vibrational peak present on the reactant spectrum will now be absent. Another possible experimental technique is calorimetry. Since the reaction is exothermic, the amount of heat released can be measured and since, at constant pressure, the enthalpy change is equal to the net change in energy, the net free energy change of the reaction can be determined.

----

==='''Q(10)'''===

Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.

'''A:'''

Using trial and error, a successful reaction trajectory was found for both reactions shown below. The parameters used can be seen within the figures.

'''Reaction 1: F + H2 --> HF + H'''

'''Lower momentum'''

[[File:HZ4315 Figure11.PNG|500px|thumb|centre|Figure 13 - An ultimately successful reaction pathway at high momentum indicating the preference of an early transition state]]

'''Higher momentum'''

[[File:HZ4315 Figure12.PNG|500px|thumb|centre|Figure 14 - The reaction pathway at an even larger momentum to confirm the theory]]

Reaction 1, which is an exothermic reaction, exhibits an increased translational energy in response to increasing the momentum of the reactants. According to Polanyi's rules, higher translational energies are more efficient at activating an early transition state (ie. an early barrier) than vibrational energy.<ref name="polanyi" /> Since early transition states resemble the reactants (as stated by Hammond's Postulate)<ref name="hammond" />, the reaction trajectory may be expected to bounce back towards the reactants. Figure 12 confirms this as an increased momentum (and therefore increased translational energy) sees the trajectory bounce back several times before eventually proceeding to the form the products. Figures 11 and 12 are an example of a set of inefficient reaction trajectories as the barrier is recrossed several times which is not ideal in the formation of the product.

The exothermic nature of the reaction also alters the position of the transition state away from the lowest 'saddle point'. This goes against the assumptions of the Transition State Theory which states that reaction trajectories should pass through this. As temperature increases, a greater proportion of excited vibrational states become occupied, leading to transition states occuring at different energies. As a result, the associated reaction trajectory at higher temperature travels through different activated complexes, making the reaction harder to model and any associated results more difficult to compare.

'''Reaction 2: H + HF --> F + H2'''

Reaction 2, which is the reverse reaction of Reaction 1, is endothermic with the energy of the products being higher than the reactants. As a result, according the Hammond's Postulate, the position of the transition state is later in the reaction trajectory and more closely resembles the products. Contrary to the preferred conditions for exothermic reactions, a larger amount of vibrational energy is needed. To achieve this, lower momentum would be adopted. Providing this is achieved, the set of reaction trajectories would be an example of an efficient reaction as the reaction trajectories would pass through the transition state once and to the products.

----

==References==

<references>
<ref name="transition">P. Jonathan P. K. Doye, David J. Wales, Journal of mathematical physics, 1996, '''48''', 843 - 846.</ref>
<ref name="quantum">Physicsoftheuniverse.com. (2017). Quantum Tunnelling and the Uncertainty Principle - Quantum Theory and the Uncertainty Principle - The Physics of the Universe. [online] Available at: http://www.physicsoftheuniverse.com/topics_quantum_uncertainty.html [Accessed 11 May 2017].</ref>
<ref name="energies">Wiredchemist.com. (2017). Common Bond Energies (D. [online] Available at: http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html [Accessed 11 May 2017].</ref>
<ref name="lawE">Mach, E. (1872). History and Root of the Principles of the Conservation of Energy. Open Court Pub. Co., Illinois.</ref>
<ref name="polanyi">Yan, S., Y.-T. Wu, and K. Liu. "Tracking The Energy Flow Along The Reaction Path". Proceedings of the National Academy of Sciences 105.35 (2008): 12667-12672.</ref>
<ref name="hammond">Chemistry LibreTexts. (2017). Hammond’s Postulate. [online] Available at: https://chem.libretexts.org/Reference/Organic_Chemistry_Glossary/Hammond’s_Postulate [Accessed 11 May 2017].</ref>
<references/>