ChemWiki - User contributions [en]

MRD:01512675

2020-05-08T22:57:51Z

Sa13018: /* Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from th */

= Exercise 1: H + H<math>_2</math> system =

== Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ==

The transition state is defined as the maximum point of the minimum energy path between the reactants and products. Therefore, on the potential energy surface diagram, it exists as a minimax; a saddle point.
This can therefore be described as the point where the partial differential of the potential energy with respect to all variables is zero.

In this case, due to a 3 dimensional surface plot, we have <math>
\frac{\partial V}{\partial r_1} = \frac{\partial V}{\partial r_2} = 0
</math>, where <math>r_1</math> and <math>r_2 </math> are the respective distances between AB and BC.

The local minimum point would see that a deviation in '''any''' direction would be an increase in potential energy. An example of this is the position of the reactants or products on the PES.

However, the transition state would see a decrease in potential energy along the reaction coordinate, and an increase in the coordinate orthogonal to this. This can be seen if we visualise the PES.

{|style="margin: 0 auto;"
| [[File:Transition state minimum.png|center|thumb|A 2D viewpoint of the PES, with the x-axis being the AB distance, and the y-axis being V(<math>r_1</math>). We see an established potential well. This runs across the entire PES as a valley of potential well. In this viewpoint, the transition state is seen as a minimum point. ]]
| [[File:Transition state maximum.png|center|thumb| A 3D visualisation of the PES, showing how the transition state is a maximum. The black curve is the reaction path, and maximum of this curve is the saddle point transition state.]]
|}

Due to the symmetry of this system, the transition state saddle point is observed at a point where <math>r_1</math> = <math>r_2</math>, but this is not necessarily always the case. In some asymmetric systems, the transition state will depend on the relative energy of the products and reactants; this can be further described by Hammond's postulate.

The saddle point of any-dimensional curve can be identified by computing a Hessian matrix.

<math>
H_{x,y} =
\begin{pmatrix}
\frac{\partial^{2} V}{\partial x^{2}} & \frac{\partial^{2} V}{\partial x\partial y } \\
\frac{\partial^{2} V}{\partial x \partial y } & \frac{\partial^{2} V}{\partial y^{2} } \\
\end{pmatrix}</math>

Upon finding the determinant (H). If H <0, it must be a saddle point, so to find the transition state, we just need to find the coordinates where this is the case.

== Question 2:Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ==

Due to the symmetry between the H and the \chem{H_2} molecule, and the fact that the interaction is happening at 180º, we know that <math>r_1</math> must equal <math>r_2</math> for any point on the potential for the transition state.

If we were to further set the momenta to zero for all 3 particles on the transition state, we know that the internuclear distances between the atoms will not change. This must hold true, as all the derivatives of the potential are zero; there are no forces acting on the particles.
As an equilibrium point, a plot of internuclear distances vs time will be constant for the transition state.

This was found for <math>r_{12}</math> = 90.775 pm. The internuclear distances vs time plot confirms this; it is shown below.

[[File:Internuclear dist v time 1.png]]
[[File:Contour plot trs1.png]]

== Question 3: Comment on how the MEP and the trajectory you just calculated differ.==

The Minimum Energy Path (MEP) plots the reaction path assuming that the particles have negligible intertial motion. It therefore ignores the vibration potential of the system, and hence only runs along the "valley" of the PES, rather than oscillating along the potential well as it does in the dynamical calculation. This makes sense; it would be the lowest energy path, but ignores the momenta of the particles at each point, and how it changes.

The differences are shown in the following two images; both of them tend towards a complete reaction from the transition state as predicted, but the dynamics plot shows oscillation along the reaction path in the y-axis as it moves along.

{|style="margin: 0 auto;"
| [[File:Dynamics plot sa13018.png|none|thumb| This shows the reaction path in a 'Dynamics' plot at an initial AB distance =91.775, and BC = 90.775. ]]
| [[File:Mep plot sa13018.png|none|thumb| This shows the reaction path in a 'MEP' plot at an initial AB distance =91.775, and BC = 90.775. ]]
|}

Furthermore, the MEP reaction path finishes at 193.1pm as the AB distance and 74 pm as the BC distance, whereas for the dynamical plot, the reaction path will continue after this. This makes intuitive physical sense; both have opposite velocities so will tend away from each other. However, for MEP, with no inertial motion, this is discounted, so the particles cease motion at this point, as there is negligible difference in energy at this point.

== Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ==

The following table assesses the plot for the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, with varying initial momenta.
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|The H-H molecule reacts with another coincidental H atom with the correct momentum, and a new H-H bond forms between atoms A and B.
| [[File:Table 1 sa13018.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The initial velocity of the atom is far too slow to collide with the molecule. The two systems approach each other initially, but the repulsion between them forces them to move in opposite directions, and with more velocity than before.
|[[File:Table 2 sa13018.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The two systems approach the transition state, and then the diatom splits, with the central atom pairing with the other terminal atom. The reaction is successful. They then proceed to move in opposite velocities.
|[[File:Table 3 sa13018.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|Initial velocity of particle C is so great that there is some repulsion between particle B and C, despite moving towards each other. As a result, particle B moves away from C after the initial attraction. It then oscillates around the transition state, moving from AB + C to A+ BC. It eventually settles with AB + C, the initial reactants. It can be seen by increasing the number of steps to 2000 that the final position of the two systems is greater separation. The particles AB and C are moving in opposite directions, so end up further apart than the initial conditions, and unreacted.
|[[File:Table 4 sa13018.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Oscillations around the transition state after an initial collision. This is very similar initially to the -10.1 p2 path. However, the rebound collision for A and B is even faster, resulting in another rebound for B. B then collides with particle C as they are both moving in the same directions, and the final state is that of the products, A + BC
|[[File:Table 5 sa13018.png]]
|}

== Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ==

Transition state theory can be used to determine the rate of a reaction, but it relies on some substantial assumptions. These contribute to deviations from the results observed in the GUI, resulting in overestimations.

The primary reason for overestimation is due to a lack of accounting for recrossing the energy threshold. TST allows for a 'quasi-equilibrium' - it allows molecules to reach the transition state from the side of both the reactants and products, but once the transition state has been crossed, it will be a successful reaction, regardless if the momentum would actually cause recrossing of the transition state. This is evidently not a realistic assumption, as we have seen in the previous section 1.4 ; e.g. the reaction pathway for the momenta p1 = -5.1 and p2 = -10.1, where despite the initial crossing of the barrier and consequent overcoming of the energy barrier, the system is not reactive.

This especially holds true for molecules with high momentum; so TST will deviate from experimental results at high temperatures. It should only be used for low temperature scenarios.

As a result, TST will incorrectly predict the number of successful reactions; it will give a higher prediction than reality, thus also '''overestimating the reaction rate'''.

Furthermore, it also assumes atoms obey the laws of '''classical''' mechanics using the Born-Oppenheimer approximation, and that each reaction only has 1 saddle point and 1 transition state. These assumptions will likely not hold true for other systems, where reactions will have more than one transition states. Quantum effects such as tunnelling of the atoms would also lead to different numbers than predicted from TST; more successful reactions would occur than TST would account for, as not all the molecules have to overcome the activation barrier to cross it.

= Exercise 2: F-H-H system =

== Question 1: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?==

Let's set the initial position as <math>r_1</math> = <math>r_2</math>, where <math>r_1</math> is F-H, and <math>r_2</math> is H-H.

If we run a MEP plot from this position, the particle will end up at the lower energy state, regardless of the initial reactants or products.

We can see in the image below, the minimum energy path is such that <math>r_1</math> is minimised, and <math>r_2</math> is maximised. With <math>r_1</math> as F-H, this implies that the lower energy state is H + H-F.

[[File:MEP F-H-H.png|center|thumb| An MEP plot showing the exothermic reaction is the formation of F-H, as this is the energy minimised product.]]

As a result, we can classify the F + H2 --> H + HF reaction as exothermic, as it is progressing from a higher energy state to a lower energy state.

H + HF ---> F + H2 would therefore be an endothermic process; it would require an input of energy to get this product.

This order of potential energies suggests that the bond strength between F-H is greater than for H2.

This is in agreement with experimental results. The strength of the H-F bond (<math> 565 kJmol^{-1}</math> ) is greater than the H-H ( <math> 432 kJmol^{-1}</math> ) bond; more energy is required to break the H-F bond than is gained forming an H-H bond.

== Question 2: Locate the approximate position of the transition state. ==

As before, the transition state will be the saddle point on the surface plot. However, as the system is no longer symmetric as before, the transition state is no longer at the point where <math>r_1</math> = <math>r_2</math>.

Hammond's postulate suggests that the transition state will be closer to the higher energy system in the reaction. We know that this is the case for F + H2, so the transition state must be such that the bond distance for F-H is greater than for H2.

Knowing this, we can surmise that if AB is F-H and BC is H2, the distances at the transition state are: AB =184 pm and BC = 74.4 pm, with momenta = 0. This complies with the logic that the reaction where FH is the product is exothermic, as this is supported by Hammond's postulate.
We once again know that the internuclear distance will not change at this position with respect to time, as the derivative of the potential V is zero. As force is the negative derivative of the potential, this implies that the transition state has no force components if initial momenta = 0, so no movement will occur at this point.

{|style="margin: 0 auto;"
| [[File:Transition state ex2 contour.png|thumb| The transition state on the contour plot. It is shown to have a much greater AB distance than BC.]]
| [[File:Transition state ex2 surface.png|thumb| The transition state on the PES.]]
|}

We can once again, further confirm that this is the saddle point by using the same point that the forces along AB and BC at this point are both zero.

== Question 3: Report the activation energy for both reactions. ==

To find the activation energies of each product F-H and H2 (2 mutually exclusive scenarios), we simply need to find the difference between the transition state energy and the energy of the products. This can be calculated for the reaction only (excluding vibration) by using MEP to identify the shortest route.

Due to the asymmetry of the atoms, we have 2 activation energies.

Starting from the <math>r_{ts}</math>where AB =184 pm and BC = 74.4 pm, with a perturbation of ± 2 pm in the AB axis, and a step count of 2500, we get these MEP results.

{|style="margin: 0 auto;"
| [[File:Transition state to minimum.png|thumb| The contour plot showing the MEP path from the transition state to the exothermic product. ]]
| [[File:Transition state to min en vs time.png|thumb| The energy vs time plot showing how the energy changes as it travels along the MEP path to the exothermic product.]]
|}

The plots above show the activation energy for the enothermic formation of the product F + H2 . The plots below show the activation energy for the exothermic formation of the products F-H + H.

{|style="margin: 0 auto;"
| [[File:Transition state to min 2.png|thumb| The contour plot showing the MEP path from the transition state to the endothermic product. ]]
| [[File:Screenshot 2020-05-08 at 22.47.21.png|thumb| The energy vs time plot showing how the energy changes as it travels along the MEP path to the endothermic product. There is very little change in energy as it reaches this, due to the early transition state in the exothermic reaction path. It agrees with Hammond's postulate.]]
|}

{| class="wikitable"
!Product
!Energy of Transition State (<math>kJmol^{-1}</math>)
!Energy of product (<math>kJmol^{-1}</math>)
!ΔE (<math>kJmol^{-1}</math>
|-
|F-H + H
|<nowiki>-433.991</nowiki>
|<nowiki>-435.271</nowiki>
|1.28
|-
|F + H2
|<nowiki>-433.979</nowiki>
|<nowiki>-556.895</nowiki>
|122.92
|}

This is because we know that the energy of the transition state is the maximum energy state on the reaction path, and doesn't consider oscillations along the potential well; so we can just use the MEP.

These results are consistent with our understanding of the system, and the contour plot; it is an extremely early TS for the exothermic formation of HF + H, so the activation energy for this reaction is extremely low compared to the reverse, endothermic process. This can be further shown by the contour plot; the difference in energy is shown to be almost negligible as it is along the same contour line.

== Question 4: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. ==

With the exothermic reaction <math>H_2 + F --> HF + H</math>, we will observe a significant release of energy. In this scenario, we can confirm that potential energy is converted to kinetic energy, as is seen in the plots below.

These plots use the initial conditions:
FH = 174 pm with p = -1 g.mol-1.pm.fs-1, HH = 74 pm with p = -4 g.mol-1.pm.fs-1, and 2000 steps.
This successfully reacts to give the exothermic reaction defined above.

{|style="margin: 0 auto;"
| [[File:Reactive path endo contour.png|thumb|A contour plot showing a reactive path on the Dynamics plot.]]
| [[File:Reactive path endo PES.png|thumb|A PES plot showing the reactive path. A highly vibrationally active molecule is shown as the end product.]]
| [[File:Momenta vs time plot endo.png|thumb| A momentum vs time plot for the reactive endothermic reaction. A-B is HF, and B-C is HH. The oscillation around zero is clearly vibration, as the momenta change direction consistently for the molecule, but not for the individual atom.]]
| [[File:Energy vs time endo.png|thumb|An energy vs time plot showing the successful exothermic reaction]]
|}

It is observed in the momenta vs time plot that a vibrating hydrogen molecule initially approaches the F atom. There are two bounces and recrossings, as observed in the animation, and then the molecule HF is formed. This has a lot of vibrational energy, as is seen in the PES surface plot, whereas the leaving H atom cannot vibrate as it is an individual atom.
Therefore, we see that the released energy from the exothermic reaction is realised in vibrations of the newly formed molecule.
The energy conversion from potential to kinetic is also shown in the energy vs time plot, where the dynamical relationship is shown.

This would correspond to an increase in temperature with heat being given off in the reaction, and can be observed by using IR spectroscopy, or with calorimetry methods (specically bomb calorimetry).
In the IR, overtones would be observed, corresponding to vibrational excitation at higher energy levels, and translational motion would also be observed, as is shown in the contour plot.

== Question 5: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. ==

For the reaction of HF + H -> H2 + F, a higher vibrational energy of the HF molecule is required, as stated by polyani's rules. This is due to the relatively late transition state of the endothermic reaction. Similarly for the reverse reaction, very little vibrational energy is required and only a slight amount of translational energy is required due to the low activation energy of the exothermic reaction.

Polanyi's rules empirically state that in an early transition state, translational energy is more useful than vibrational energy with regards to the generation of a successful reaction. The converse argument applies with a late transition state. These only hold true when there is enough energy to overcome the activation energy, and favourable conditions are present.

This was largely observed for the endothermic reaction of HF + H -> H2 + F (late transition state); a higher vibrational energy resulted in more readily complete reaction. This also held true for the opposite case.
These are shown in the plots below.

{|style="margin: 0 auto;"
| [[File:01513089 HF high vibration success.png|left|thumb|405x405px|Endothermic reaction (late transition state) with high vibrational energy and relatively low translational energy leading to a successful H2 reaction: HF distance 75.3 pm, -0.61 g.mol-1.pm.fs-1: HH distance 225.1 pm, -3.117 g.mol-1.pm.fs-1]]
| [[File:01513089 Plot.png|left|thumb|403x403px|Endothermic reaction (late transition state) with low vibrational energy and high translational energy not leading to successful H2 production: HF distance 100 pm, -20 g.mol-1.pm.fs-1: HH distance 225 pm, -20 g.mol-1.pm.fs-1]]
|}

{|style="margin: 0 auto;"
| [[File:01513089 HH low vibration success.png|thumb|411x411px|Exothermic reaction (early transition state) successfully producing HF with almost no vibrational energy and some translational energy: HF distance 210 pm, -2 g.mol-1.pm.fs-1: HH distance 74 pm, -1 g.mol-1.pm.fs-1]]
| [[File:01513089 HH high vibration usuccessful.png|thumb|405x405px|Exothermic reaction (early transition state) unsuccessful in producing HF with very high vibrational energy and some translational energy: HF distance 210 pm, -1 g.mol-1.pm.fs-1: HH distance 100 pm, -1 g.mol-1.pm.fs-1]]
|}

MRD:01512675

2020-05-08T22:56:39Z

= Exercise 1: H + H<math>_2</math> system =

== Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ==

The transition state is defined as the maximum point of the minimum energy path between the reactants and products. Therefore, on the potential energy surface diagram, it exists as a minimax; a saddle point.
This can therefore be described as the point where the partial differential of the potential energy with respect to all variables is zero.

In this case, due to a 3 dimensional surface plot, we have <math>
\frac{\partial V}{\partial r_1} = \frac{\partial V}{\partial r_2} = 0
</math>, where <math>r_1</math> and <math>r_2 </math> are the respective distances between AB and BC.

The local minimum point would see that a deviation in '''any''' direction would be an increase in potential energy. An example of this is the position of the reactants or products on the PES.

However, the transition state would see a decrease in potential energy along the reaction coordinate, and an increase in the coordinate orthogonal to this. This can be seen if we visualise the PES.

{|style="margin: 0 auto;"
| [[File:Transition state minimum.png|center|thumb|A 2D viewpoint of the PES, with the x-axis being the AB distance, and the y-axis being V(<math>r_1</math>). We see an established potential well. This runs across the entire PES as a valley of potential well. In this viewpoint, the transition state is seen as a minimum point. ]]
| [[File:Transition state maximum.png|center|thumb| A 3D visualisation of the PES, showing how the transition state is a maximum. The black curve is the reaction path, and maximum of this curve is the saddle point transition state.]]
|}

Due to the symmetry of this system, the transition state saddle point is observed at a point where <math>r_1</math> = <math>r_2</math>, but this is not necessarily always the case. In some asymmetric systems, the transition state will depend on the relative energy of the products and reactants; this can be further described by Hammond's postulate.

The saddle point of any-dimensional curve can be identified by computing a Hessian matrix.

<math>
H_{x,y} =
\begin{pmatrix}
\frac{\partial^{2} V}{\partial x^{2}} & \frac{\partial^{2} V}{\partial x\partial y } \\
\frac{\partial^{2} V}{\partial x \partial y } & \frac{\partial^{2} V}{\partial y^{2} } \\
\end{pmatrix}</math>

Upon finding the determinant (H). If H <0, it must be a saddle point, so to find the transition state, we just need to find the coordinates where this is the case.

== Question 2:Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ==

Due to the symmetry between the H and the \chem{H_2} molecule, and the fact that the interaction is happening at 180º, we know that <math>r_1</math> must equal <math>r_2</math> for any point on the potential for the transition state.

If we were to further set the momenta to zero for all 3 particles on the transition state, we know that the internuclear distances between the atoms will not change. This must hold true, as all the derivatives of the potential are zero; there are no forces acting on the particles.
As an equilibrium point, a plot of internuclear distances vs time will be constant for the transition state.

This was found for <math>r_{12}</math> = 90.775 pm. The internuclear distances vs time plot confirms this; it is shown below.

[[File:Internuclear dist v time 1.png]]
[[File:Contour plot trs1.png]]

== Question 3: Comment on how the MEP and the trajectory you just calculated differ.==

The Minimum Energy Path (MEP) plots the reaction path assuming that the particles have negligible intertial motion. It therefore ignores the vibration potential of the system, and hence only runs along the "valley" of the PES, rather than oscillating along the potential well as it does in the dynamical calculation. This makes sense; it would be the lowest energy path, but ignores the momenta of the particles at each point, and how it changes.

The differences are shown in the following two images; both of them tend towards a complete reaction from the transition state as predicted, but the dynamics plot shows oscillation along the reaction path in the y-axis as it moves along.

{|style="margin: 0 auto;"
| [[File:Dynamics plot sa13018.png|none|thumb| This shows the reaction path in a 'Dynamics' plot at an initial AB distance =91.775, and BC = 90.775. ]]
| [[File:Mep plot sa13018.png|none|thumb| This shows the reaction path in a 'MEP' plot at an initial AB distance =91.775, and BC = 90.775. ]]
|}

Furthermore, the MEP reaction path finishes at 193.1pm as the AB distance and 74 pm as the BC distance, whereas for the dynamical plot, the reaction path will continue after this. This makes intuitive physical sense; both have opposite velocities so will tend away from each other. However, for MEP, with no inertial motion, this is discounted, so the particles cease motion at this point, as there is negligible difference in energy at this point.

== Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ==

The following table assesses the plot for the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, with varying initial momenta.
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|The H-H molecule reacts with another coincidental H atom with the correct momentum, and a new H-H bond forms between atoms A and B.
| [[File:Table 1 sa13018.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The initial velocity of the atom is far too slow to collide with the molecule. The two systems approach each other initially, but the repulsion between them forces them to move in opposite directions, and with more velocity than before.
|[[File:Table 2 sa13018.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The two systems approach the transition state, and then the diatom splits, with the central atom pairing with the other terminal atom. They then proceed to move in opposite velocities.
|[[File:Table 3 sa13018.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|Initial velocity of particle C is so great that there is some repulsion between particle B and C, despite moving towards each other. As a result, particle B moves away from C after the initial attraction. It then oscillates around the transition state, moving from AB + C to A+ BC. It eventually settles with AB + C, the initial reactants. It can be seen by increasing the number of steps to 2000 that the final position of the two systems is greater separation. The particles AB and C are moving in opposite directions, so end up further apart than the initial conditions.
|[[File:Table 4 sa13018.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Oscillations around the transition state after an initial collision. This is very similar initially to the -10.1 p2 path. However, the rebound collision for A and B is even faster, resulting in another rebound for B. B then collides with particle C as they are both moving in the same directions, and the final state is that of the products, A + BC
|[[File:Table 5 sa13018.png]]
|}

== Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ==

Transition state theory can be used to determine the rate of a reaction, but it relies on some substantial assumptions. These contribute to deviations from the results observed in the GUI, resulting in overestimations.

The primary reason for overestimation is due to a lack of accounting for recrossing the energy threshold. TST allows for a 'quasi-equilibrium' - it allows molecules to reach the transition state from the side of both the reactants and products, but once the transition state has been crossed, it will be a successful reaction, regardless if the momentum would actually cause recrossing of the transition state. This is evidently not a realistic assumption, as we have seen in the previous section 1.4 ; e.g. the reaction pathway for the momenta p1 = -5.1 and p2 = -10.1, where despite the initial crossing of the barrier and consequent overcoming of the energy barrier, the system is not reactive.

This especially holds true for molecules with high momentum; so TST will deviate from experimental results at high temperatures. It should only be used for low temperature scenarios.

As a result, TST will incorrectly predict the number of successful reactions; it will give a higher prediction than reality, thus also '''overestimating the reaction rate'''.

Furthermore, it also assumes atoms obey the laws of '''classical''' mechanics using the Born-Oppenheimer approximation, and that each reaction only has 1 saddle point and 1 transition state. These assumptions will likely not hold true for other systems, where reactions will have more than one transition states. Quantum effects such as tunnelling of the atoms would also lead to different numbers than predicted from TST; more successful reactions would occur than TST would account for, as not all the molecules have to overcome the activation barrier to cross it.

= Exercise 2: F-H-H system =

== Question 1: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?==

Let's set the initial position as <math>r_1</math> = <math>r_2</math>, where <math>r_1</math> is F-H, and <math>r_2</math> is H-H.

If we run a MEP plot from this position, the particle will end up at the lower energy state, regardless of the initial reactants or products.

We can see in the image below, the minimum energy path is such that <math>r_1</math> is minimised, and <math>r_2</math> is maximised. With <math>r_1</math> as F-H, this implies that the lower energy state is H + H-F.

[[File:MEP F-H-H.png|center|thumb| An MEP plot showing the exothermic reaction is the formation of F-H, as this is the energy minimised product.]]

As a result, we can classify the F + H2 --> H + HF reaction as exothermic, as it is progressing from a higher energy state to a lower energy state.

H + HF ---> F + H2 would therefore be an endothermic process; it would require an input of energy to get this product.

This order of potential energies suggests that the bond strength between F-H is greater than for H2.

This is in agreement with experimental results. The strength of the H-F bond (<math> 565 kJmol^{-1}</math> ) is greater than the H-H ( <math> 432 kJmol^{-1}</math> ) bond; more energy is required to break the H-F bond than is gained forming an H-H bond.

== Question 2: Locate the approximate position of the transition state. ==

As before, the transition state will be the saddle point on the surface plot. However, as the system is no longer symmetric as before, the transition state is no longer at the point where <math>r_1</math> = <math>r_2</math>.

Hammond's postulate suggests that the transition state will be closer to the higher energy system in the reaction. We know that this is the case for F + H2, so the transition state must be such that the bond distance for F-H is greater than for H2.

Knowing this, we can surmise that if AB is F-H and BC is H2, the distances at the transition state are: AB =184 pm and BC = 74.4 pm, with momenta = 0. This complies with the logic that the reaction where FH is the product is exothermic, as this is supported by Hammond's postulate.
We once again know that the internuclear distance will not change at this position with respect to time, as the derivative of the potential V is zero. As force is the negative derivative of the potential, this implies that the transition state has no force components if initial momenta = 0, so no movement will occur at this point.

{|style="margin: 0 auto;"
| [[File:Transition state ex2 contour.png|thumb| The transition state on the contour plot. It is shown to have a much greater AB distance than BC.]]
| [[File:Transition state ex2 surface.png|thumb| The transition state on the PES.]]
|}

We can once again, further confirm that this is the saddle point by using the same point that the forces along AB and BC at this point are both zero.

== Question 3: Report the activation energy for both reactions. ==

To find the activation energies of each product F-H and H2 (2 mutually exclusive scenarios), we simply need to find the difference between the transition state energy and the energy of the products. This can be calculated for the reaction only (excluding vibration) by using MEP to identify the shortest route.

Due to the asymmetry of the atoms, we have 2 activation energies.

Starting from the <math>r_{ts}</math>where AB =184 pm and BC = 74.4 pm, with a perturbation of ± 2 pm in the AB axis, and a step count of 2500, we get these MEP results.

{|style="margin: 0 auto;"
| [[File:Transition state to minimum.png|thumb| The contour plot showing the MEP path from the transition state to the exothermic product. ]]
| [[File:Transition state to min en vs time.png|thumb| The energy vs time plot showing how the energy changes as it travels along the MEP path to the exothermic product.]]
|}

The plots above show the activation energy for the enothermic formation of the product F + H2 . The plots below show the activation energy for the exothermic formation of the products F-H + H.

{|style="margin: 0 auto;"
| [[File:Transition state to min 2.png|thumb| The contour plot showing the MEP path from the transition state to the endothermic product. ]]
| [[File:Screenshot 2020-05-08 at 22.47.21.png|thumb| The energy vs time plot showing how the energy changes as it travels along the MEP path to the endothermic product. There is very little change in energy as it reaches this, due to the early transition state in the exothermic reaction path. It agrees with Hammond's postulate.]]
|}

{| class="wikitable"
!Product
!Energy of Transition State (<math>kJmol^{-1}</math>)
!Energy of product (<math>kJmol^{-1}</math>)
!ΔE (<math>kJmol^{-1}</math>
|-
|F-H + H
|<nowiki>-433.991</nowiki>
|<nowiki>-435.271</nowiki>
|1.28
|-
|F + H2
|<nowiki>-433.979</nowiki>
|<nowiki>-556.895</nowiki>
|122.92
|}

This is because we know that the energy of the transition state is the maximum energy state on the reaction path, and doesn't consider oscillations along the potential well; so we can just use the MEP.

These results are consistent with our understanding of the system, and the contour plot; it is an extremely early TS for the exothermic formation of HF + H, so the activation energy for this reaction is extremely low compared to the reverse, endothermic process. This can be further shown by the contour plot; the difference in energy is shown to be almost negligible as it is along the same contour line.

== Question 4: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. ==

With the exothermic reaction <math>H_2 + F --> HF + H</math>, we will observe a significant release of energy. In this scenario, we can confirm that potential energy is converted to kinetic energy, as is seen in the plots below.

These plots use the initial conditions:
FH = 174 pm with p = -1 g.mol-1.pm.fs-1, HH = 74 pm with p = -4 g.mol-1.pm.fs-1, and 2000 steps.
This successfully reacts to give the exothermic reaction defined above.

{|style="margin: 0 auto;"
| [[File:Reactive path endo contour.png|thumb|A contour plot showing a reactive path on the Dynamics plot.]]
| [[File:Reactive path endo PES.png|thumb|A PES plot showing the reactive path. A highly vibrationally active molecule is shown as the end product.]]
| [[File:Momenta vs time plot endo.png|thumb| A momentum vs time plot for the reactive endothermic reaction. A-B is HF, and B-C is HH. The oscillation around zero is clearly vibration, as the momenta change direction consistently for the molecule, but not for the individual atom.]]
| [[File:Energy vs time endo.png|thumb|An energy vs time plot showing the successful exothermic reaction]]
|}

It is observed in the momenta vs time plot that a vibrating hydrogen molecule initially approaches the F atom. There are two bounces and recrossings, as observed in the animation, and then the molecule HF is formed. This has a lot of vibrational energy, as is seen in the PES surface plot, whereas the leaving H atom cannot vibrate as it is an individual atom.
Therefore, we see that the released energy from the exothermic reaction is realised in vibrations of the newly formed molecule.
The energy conversion from potential to kinetic is also shown in the energy vs time plot, where the dynamical relationship is shown.

This would correspond to an increase in temperature with heat being given off in the reaction, and can be observed by using IR spectroscopy, or with calorimetry methods (specically bomb calorimetry).
In the IR, overtones would be observed, corresponding to vibrational excitation at higher energy levels, and translational motion would also be observed, as is shown in the contour plot.

== Question 5: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. ==

For the reaction of HF + H -> H2 + F, a higher vibrational energy of the HF molecule is required, as stated by polyani's rules. This is due to the relatively late transition state of the endothermic reaction. Similarly for the reverse reaction, very little vibrational energy is required and only a slight amount of translational energy is required due to the low activation energy of the exothermic reaction.

Polanyi's rules empirically state that in an early transition state, translational energy is more useful than vibrational energy with regards to the generation of a successful reaction. The converse argument applies with a late transition state. These only hold true when there is enough energy to overcome the activation energy, and favourable conditions are present.

This was largely observed for the endothermic reaction of HF + H -> H2 + F (late transition state); a higher vibrational energy resulted in more readily complete reaction. This also held true for the opposite case.
These are shown in the plots below.

{|style="margin: 0 auto;"
| [[File:01513089 HF high vibration success.png|left|thumb|405x405px|Endothermic reaction (late transition state) with high vibrational energy and relatively low translational energy leading to a successful H2 reaction: HF distance 75.3 pm, -0.61 g.mol-1.pm.fs-1: HH distance 225.1 pm, -3.117 g.mol-1.pm.fs-1]]
| [[File:01513089 Plot.png|left|thumb|403x403px|Endothermic reaction (late transition state) with low vibrational energy and high translational energy not leading to successful H2 production: HF distance 100 pm, -20 g.mol-1.pm.fs-1: HH distance 225 pm, -20 g.mol-1.pm.fs-1]]
|}

{|style="margin: 0 auto;"
| [[File:01513089 HH low vibration success.png|thumb|411x411px|Exothermic reaction (early transition state) successfully producing HF with almost no vibrational energy and some translational energy: HF distance 210 pm, -2 g.mol-1.pm.fs-1: HH distance 74 pm, -1 g.mol-1.pm.fs-1]]
| [[File:01513089 HH high vibration usuccessful.png|thumb|405x405px|Exothermic reaction (early transition state) unsuccessful in producing HF with very high vibrational energy and some translational energy: HF distance 210 pm, -1 g.mol-1.pm.fs-1: HH distance 100 pm, -1 g.mol-1.pm.fs-1]]
|}

MRD:01512675

2020-05-08T22:55:02Z

= Exercise 1: H + H<math>_2</math> system =

== Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ==

The transition state is defined as the maximum point of the minimum energy path between the reactants and products. Therefore, on the potential energy surface diagram, it exists as a minimax; a saddle point.
This can therefore be described as the point where the partial differential of the potential energy with respect to all variables is zero.

In this case, due to a 3 dimensional surface plot, we have <math>
\frac{\partial V}{\partial r_1} = \frac{\partial V}{\partial r_2} = 0
</math>, where <math>r_1</math> and <math>r_2 </math> are the respective distances between AB and BC.

The local minimum point would see that a deviation in '''any''' direction would be an increase in potential energy. An example of this is the position of the reactants or products on the PES.

However, the transition state would see a decrease in potential energy along the reaction coordinate, and an increase in the coordinate orthogonal to this. This can be seen if we visualise the PES.

{|style="margin: 0 auto;"
| [[File:Transition state minimum.png|center|thumb|A 2D viewpoint of the PES, with the x-axis being the AB distance, and the y-axis being V(<math>r_1</math>). We see an established potential well. This runs across the entire PES as a valley of potential well. In this viewpoint, the transition state is seen as a minimum point. ]]
| [[File:Transition state maximum.png|center|thumb| A 3D visualisation of the PES, showing how the transition state is a maximum. The black curve is the reaction path, and maximum of this curve is the saddle point transition state.]]
|}

Due to the symmetry of this system, the transition state saddle point is observed at a point where <math>r_1</math> = <math>r_2</math>, but this is not necessarily always the case. In some asymmetric systems, the transition state will depend on the relative energy of the products and reactants; this can be further described by Hammond's postulate.

The saddle point of any-dimensional curve can be identified by computing a Hessian matrix.

<math>
H_{x,y} =
\begin{pmatrix}
\frac{\partial^{2} V}{\partial x^{2}} & \frac{\partial^{2} V}{\partial x\partial y } \\
\frac{\partial^{2} V}{\partial x \partial y } & \frac{\partial^{2} V}{\partial y^{2} } \\
\end{pmatrix}</math>

Upon finding the determinant (H). If H <0, it must be a saddle point, so to find the transition state, we just need to find the coordinates where this is the case.

== Question 2:Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ==

Due to the symmetry between the H and the \chem{H_2} molecule, and the fact that the interaction is happening at 180º, we know that <math>r_1</math> must equal <math>r_2</math> for any point on the potential for the transition state.

If we were to further set the momenta to zero for all 3 particles on the transition state, we know that the internuclear distances between the atoms will not change. This must hold true, as all the derivatives of the potential are zero; there are no forces acting on the particles.
As an equilibrium point, a plot of internuclear distances vs time will be constant for the transition state.

This was found for <math>r_{12}</math> = 90.775 pm. The internuclear distances vs time plot confirms this; it is shown below.

[[File:Internuclear dist v time 1.png]]
[[File:Contour plot trs1.png]]

== Question 3: Comment on how the MEP and the trajectory you just calculated differ.==

The Minimum Energy Path (MEP) plots the reaction path assuming that the particles have negligible intertial motion. It therefore ignores the vibration potential of the system, and hence only runs along the "valley" of the PES, rather than oscillating along the potential well as it does in the dynamical calculation. This makes sense; it would be the lowest energy path, but ignores the momenta of the particles at each point, and how it changes.

The differences are shown in the following two images; both of them tend towards a complete reaction from the transition state as predicted, but the dynamics plot shows oscillation along the reaction path in the y-axis as it moves along.

{|style="margin: 0 auto;"
| [[File:Dynamics plot sa13018.png|none|thumb| This shows the reaction path in a 'Dynamics' plot at an initial AB distance =91.775, and BC = 90.775. ]]
| [[File:Mep plot sa13018.png|none|thumb| This shows the reaction path in a 'MEP' plot at an initial AB distance =91.775, and BC = 90.775. ]]
|}

Furthermore, the MEP reaction path finishes at 193.1pm as the AB distance and 74 pm as the BC distance, whereas for the dynamical plot, the reaction path will continue after this. This makes intuitive physical sense; both have opposite velocities so will tend away from each other. However, for MEP, with no inertial motion, this is discounted, so the particles cease motion at this point, as there is negligible difference in energy at this point.

== Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ==

The following table assesses the plot for the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, with varying initial momenta.
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Proceeds as normal. No initial collision, but repulsion between the two systems is overcome, etc.
| [[File:Table 1 sa13018.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The initial velocity of the atom is far too slow to collide with the molecule. The two systems approach each other initially, but the repulsion between them forces them to move in opposite directions, and with more velocity than
|[[File:Table 2 sa13018.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The two systems approach the transition state, and then the diatom splits, with the central atom pairing with the other terminal atom. They then proceed to move in opposite velocities.
|[[File:Table 3 sa13018.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|Initial velocity of particle C is so great that there is some repulsion between particle B and C, despite moving towards each other. As a result, particle B moves away from C after the initial attraction. It then oscillates around the transition state, moving from AB + C to A+ BC. It eventually settles with AB + C, the initial reactants. It can be seen by increasing the number of steps to 2000 that the final position of the two systems is greater separation. The particles AB and C are moving in opposite directions, so end up further apart than the initial conditions.
|[[File:Table 4 sa13018.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Oscillations around the transition state after an initial collision. This is very similar initially to the -10.1 p2 path. However, the rebound collision for A and B is even faster, resulting in another rebound for B. B then collides with particle C as they are both moving in the same directions, and the final state is that of the products, A + BC
|[[File:Table 5 sa13018.png]]
|}

== Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ==

Transition state theory can be used to determine the rate of a reaction, but it relies on some substantial assumptions. These contribute to deviations from the results observed in the GUI, resulting in overestimations.

The primary reason for overestimation is due to a lack of accounting for recrossing the energy threshold. TST allows for a 'quasi-equilibrium' - it allows molecules to reach the transition state from the side of both the reactants and products, but once the transition state has been crossed, it will be a successful reaction, regardless if the momentum would actually cause recrossing of the transition state. This is evidently not a realistic assumption, as we have seen in the previous section 1.4 ; e.g. the reaction pathway for the momenta p1 = -5.1 and p2 = -10.1, where despite the initial crossing of the barrier and consequent overcoming of the energy barrier, the system is not reactive.

This especially holds true for molecules with high momentum; so TST will deviate from experimental results at high temperatures. It should only be used for low temperature scenarios.

As a result, TST will incorrectly predict the number of successful reactions; it will give a higher prediction than reality, thus also '''overestimating the reaction rate'''.

Furthermore, it also assumes atoms obey the laws of '''classical''' mechanics using the Born-Oppenheimer approximation, and that each reaction only has 1 saddle point and 1 transition state. These assumptions will likely not hold true for other systems, where reactions will have more than one transition states. Quantum effects such as tunnelling of the atoms would also lead to different numbers than predicted from TST; more successful reactions would occur than TST would account for, as not all the molecules have to overcome the activation barrier to cross it.

= Exercise 2: F-H-H system =

== Question 1: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?==

Let's set the initial position as <math>r_1</math> = <math>r_2</math>, where <math>r_1</math> is F-H, and <math>r_2</math> is H-H.

If we run a MEP plot from this position, the particle will end up at the lower energy state, regardless of the initial reactants or products.

We can see in the image below, the minimum energy path is such that <math>r_1</math> is minimised, and <math>r_2</math> is maximised. With <math>r_1</math> as F-H, this implies that the lower energy state is H + H-F.

[[File:MEP F-H-H.png|center|thumb| An MEP plot showing the exothermic reaction is the formation of F-H, as this is the energy minimised product.]]

As a result, we can classify the F + H2 --> H + HF reaction as exothermic, as it is progressing from a higher energy state to a lower energy state.

H + HF ---> F + H2 would therefore be an endothermic process; it would require an input of energy to get this product.

This order of potential energies suggests that the bond strength between F-H is greater than for H2.

This is in agreement with experimental results. The strength of the H-F bond (<math> 565 kJmol^{-1}</math> ) is greater than the H-H ( <math> 432 kJmol^{-1}</math> ) bond; more energy is required to break the H-F bond than is gained forming an H-H bond.

== Question 2: Locate the approximate position of the transition state. ==

As before, the transition state will be the saddle point on the surface plot. However, as the system is no longer symmetric as before, the transition state is no longer at the point where <math>r_1</math> = <math>r_2</math>.

Hammond's postulate suggests that the transition state will be closer to the higher energy system in the reaction. We know that this is the case for F + H2, so the transition state must be such that the bond distance for F-H is greater than for H2.

Knowing this, we can surmise that if AB is F-H and BC is H2, the distances at the transition state are: AB =184 pm and BC = 74.4 pm, with momenta = 0. This complies with the logic that the reaction where FH is the product is exothermic, as this is supported by Hammond's postulate.
We once again know that the internuclear distance will not change at this position with respect to time, as the derivative of the potential V is zero. As force is the negative derivative of the potential, this implies that the transition state has no force components if initial momenta = 0, so no movement will occur at this point.

{|style="margin: 0 auto;"
| [[File:Transition state ex2 contour.png|thumb| The transition state on the contour plot. It is shown to have a much greater AB distance than BC.]]
| [[File:Transition state ex2 surface.png|thumb| The transition state on the PES.]]
|}

We can once again, further confirm that this is the saddle point by using the same point that the forces along AB and BC at this point are both zero.

== Question 3: Report the activation energy for both reactions. ==

To find the activation energies of each product F-H and H2 (2 mutually exclusive scenarios), we simply need to find the difference between the transition state energy and the energy of the products. This can be calculated for the reaction only (excluding vibration) by using MEP to identify the shortest route.

Due to the asymmetry of the atoms, we have 2 activation energies.

Starting from the <math>r_{ts}</math>where AB =184 pm and BC = 74.4 pm, with a perturbation of ± 2 pm in the AB axis, and a step count of 2500, we get these MEP results.

{|style="margin: 0 auto;"
| [[File:Transition state to minimum.png|thumb| The contour plot showing the MEP path from the transition state to the exothermic product. ]]
| [[File:Transition state to min en vs time.png|thumb| The energy vs time plot showing how the energy changes as it travels along the MEP path to the exothermic product.]]
|}

The plots above show the activation energy for the enothermic formation of the product F + H2 . The plots below show the activation energy for the exothermic formation of the products F-H + H.

{|style="margin: 0 auto;"
| [[File:Transition state to min 2.png|thumb| The contour plot showing the MEP path from the transition state to the endothermic product. ]]
| [[File:Screenshot 2020-05-08 at 22.47.21.png|thumb| The energy vs time plot showing how the energy changes as it travels along the MEP path to the endothermic product. There is very little change in energy as it reaches this, due to the early transition state in the exothermic reaction path. It agrees with Hammond's postulate.]]
|}

{| class="wikitable"
!Product
!Energy of Transition State (<math>kJmol^{-1}</math>)
!Energy of product (<math>kJmol^{-1}</math>)
!ΔE (<math>kJmol^{-1}</math>
|-
|F-H + H
|<nowiki>-433.991</nowiki>
|<nowiki>-435.271</nowiki>
|1.28
|-
|F + H2
|<nowiki>-433.979</nowiki>
|<nowiki>-556.895</nowiki>
|122.92
|}

This is because we know that the energy of the transition state is the maximum energy state on the reaction path, and doesn't consider oscillations along the potential well; so we can just use the MEP.

These results are consistent with our understanding of the system, and the contour plot; it is an extremely early TS for the exothermic formation of HF + H, so the activation energy for this reaction is extremely low compared to the reverse, endothermic process. This can be further shown by the contour plot; the difference in energy is shown to be almost negligible as it is along the same contour line.

== Question 4: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. ==

With the exothermic reaction <math>H_2 + F --> HF + H</math>, we will observe a significant release of energy. In this scenario, we can confirm that potential energy is converted to kinetic energy, as is seen in the plots below.

These plots use the initial conditions:
FH = 174 pm with p = -1 g.mol-1.pm.fs-1, HH = 74 pm with p = -4 g.mol-1.pm.fs-1, and 2000 steps.
This successfully reacts to give the exothermic reaction defined above.

{|style="margin: 0 auto;"
| [[File:Reactive path endo contour.png|thumb|A contour plot showing a reactive path on the Dynamics plot.]]
| [[File:Reactive path endo PES.png|thumb|A PES plot showing the reactive path. A highly vibrationally active molecule is shown as the end product.]]
| [[File:Momenta vs time plot endo.png|thumb| A momentum vs time plot for the reactive endothermic reaction. A-B is HF, and B-C is HH. The oscillation around zero is clearly vibration, as the momenta change direction consistently for the molecule, but not for the individual atom.]]
| [[File:Energy vs time endo.png|thumb|An energy vs time plot showing the successful exothermic reaction]]
|}

It is observed in the momenta vs time plot that a vibrating hydrogen molecule initially approaches the F atom. There are two bounces and recrossings, as observed in the animation, and then the molecule HF is formed. This has a lot of vibrational energy, as is seen in the PES surface plot, whereas the leaving H atom cannot vibrate as it is an individual atom.
Therefore, we see that the released energy from the exothermic reaction is realised in vibrations of the newly formed molecule.
The energy conversion from potential to kinetic is also shown in the energy vs time plot, where the dynamical relationship is shown.

This would correspond to an increase in temperature with heat being given off in the reaction, and can be observed by using IR spectroscopy, or with calorimetry methods (specically bomb calorimetry).
In the IR, overtones would be observed, corresponding to vibrational excitation at higher energy levels, and translational motion would also be observed, as is shown in the contour plot.

== Question 5: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. ==

For the reaction of HF + H -> H2 + F, a higher vibrational energy of the HF molecule is required, as stated by polyani's rules. This is due to the relatively late transition state of the endothermic reaction. Similarly for the reverse reaction, very little vibrational energy is required and only a slight amount of translational energy is required due to the low activation energy of the exothermic reaction.

Polanyi's rules empirically state that in an early transition state, translational energy is more useful than vibrational energy with regards to the generation of a successful reaction. The converse argument applies with a late transition state. These only hold true when there is enough energy to overcome the activation energy, and favourable conditions are present.

This was largely observed for the endothermic reaction of HF + H -> H2 + F (late transition state); a higher vibrational energy resulted in more readily complete reaction. This also held true for the opposite case.
These are shown in the plots below.

{|style="margin: 0 auto;"
| [[File:01513089 HF high vibration success.png|left|thumb|405x405px|Endothermic reaction (late transition state) with high vibrational energy and relatively low translational energy leading to a successful H2 reaction: HF distance 75.3 pm, -0.61 g.mol-1.pm.fs-1: HH distance 225.1 pm, -3.117 g.mol-1.pm.fs-1]]
| [[File:01513089 Plot.png|left|thumb|403x403px|Endothermic reaction (late transition state) with low vibrational energy and high translational energy not leading to successful H2 production: HF distance 100 pm, -20 g.mol-1.pm.fs-1: HH distance 225 pm, -20 g.mol-1.pm.fs-1]]
|}

{|style="margin: 0 auto;"
| [[File:01513089 HH low vibration success.png|thumb|411x411px|Exothermic reaction (early transition state) successfully producing HF with almost no vibrational energy and some translational energy: HF distance 210 pm, -2 g.mol-1.pm.fs-1: HH distance 74 pm, -1 g.mol-1.pm.fs-1]]
| [[File:01513089 HH high vibration usuccessful.png|thumb|405x405px|Exothermic reaction (early transition state) unsuccessful in producing HF with very high vibrational energy and some translational energy: HF distance 210 pm, -1 g.mol-1.pm.fs-1: HH distance 100 pm, -1 g.mol-1.pm.fs-1]]
|}

File:Table 5 sa13018.png

2020-05-08T22:54:47Z

Sa13018:

File:Table 4 sa13018.png

2020-05-08T22:54:13Z

Sa13018:

File:Table 3 sa13018.png

2020-05-08T22:53:39Z

Sa13018:

File:Table 2 sa13018.png

2020-05-08T22:53:00Z

Sa13018:

File:Table 1 sa13018.png

2020-05-08T22:52:15Z

Sa13018:

MRD:01512675

2020-05-08T22:47:49Z

Sa13018: /* Question 5: Efficiency of reaction */

= Exercise 1: H + H<math>_2</math> system =

== Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ==

The transition state is defined as the maximum point of the minimum energy path between the reactants and products. Therefore, on the potential energy surface diagram, it exists as a minimax; a saddle point.
This can therefore be described as the point where the partial differential of the potential energy with respect to all variables is zero.

In this case, due to a 3 dimensional surface plot, we have <math>
\frac{\partial V}{\partial r_1} = \frac{\partial V}{\partial r_2} = 0
</math>, where <math>r_1</math> and <math>r_2 </math> are the respective distances between AB and BC.

The local minimum point would see that a deviation in '''any''' direction would be an increase in potential energy. An example of this is the position of the reactants or products on the PES.

However, the transition state would see a decrease in potential energy along the reaction coordinate, and an increase in the coordinate orthogonal to this. This can be seen if we visualise the PES.

{|style="margin: 0 auto;"
| [[File:Transition state minimum.png|center|thumb|A 2D viewpoint of the PES, with the x-axis being the AB distance, and the y-axis being V(<math>r_1</math>). We see an established potential well. This runs across the entire PES as a valley of potential well. In this viewpoint, the transition state is seen as a minimum point. ]]
| [[File:Transition state maximum.png|center|thumb| A 3D visualisation of the PES, showing how the transition state is a maximum. The black curve is the reaction path, and maximum of this curve is the saddle point transition state.]]
|}

Due to the symmetry of this system, the transition state saddle point is observed at a point where <math>r_1</math> = <math>r_2</math>, but this is not necessarily always the case. In some asymmetric systems, the transition state will depend on the relative energy of the products and reactants; this can be further described by Hammond's postulate.

The saddle point of any-dimensional curve can be identified by computing a Hessian matrix.

<math>
H_{x,y} =
\begin{pmatrix}
\frac{\partial^{2} V}{\partial x^{2}} & \frac{\partial^{2} V}{\partial x\partial y } \\
\frac{\partial^{2} V}{\partial x \partial y } & \frac{\partial^{2} V}{\partial y^{2} } \\
\end{pmatrix}</math>

Upon finding the determinant (H). If H <0, it must be a saddle point, so to find the transition state, we just need to find the coordinates where this is the case.

== Question 2:Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ==

Due to the symmetry between the H and the \chem{H_2} molecule, and the fact that the interaction is happening at 180º, we know that <math>r_1</math> must equal <math>r_2</math> for any point on the potential for the transition state.

If we were to further set the momenta to zero for all 3 particles on the transition state, we know that the internuclear distances between the atoms will not change. This must hold true, as all the derivatives of the potential are zero; there are no forces acting on the particles.
As an equilibrium point, a plot of internuclear distances vs time will be constant for the transition state.

This was found for <math>r_{12}</math> = 90.775 pm. The internuclear distances vs time plot confirms this; it is shown below.

[[File:Internuclear dist v time 1.png]]
[[File:Contour plot trs1.png]]

== Question 3: Comment on how the MEP and the trajectory you just calculated differ.==

The Minimum Energy Path (MEP) plots the reaction path assuming that the particles have negligible intertial motion. It therefore ignores the vibration potential of the system, and hence only runs along the "valley" of the PES, rather than oscillating along the potential well as it does in the dynamical calculation. This makes sense; it would be the lowest energy path, but ignores the momenta of the particles at each point, and how it changes.

The differences are shown in the following two images; both of them tend towards a complete reaction from the transition state as predicted, but the dynamics plot shows oscillation along the reaction path in the y-axis as it moves along.

{|style="margin: 0 auto;"
| [[File:Dynamics plot sa13018.png|none|thumb| This shows the reaction path in a 'Dynamics' plot at an initial AB distance =91.775, and BC = 90.775. ]]
| [[File:Mep plot sa13018.png|none|thumb| This shows the reaction path in a 'MEP' plot at an initial AB distance =91.775, and BC = 90.775. ]]
|}

Furthermore, the MEP reaction path finishes at 193.1pm as the AB distance and 74 pm as the BC distance, whereas for the dynamical plot, the reaction path will continue after this. This makes intuitive physical sense; both have opposite velocities so will tend away from each other. However, for MEP, with no inertial motion, this is discounted, so the particles cease motion at this point, as there is negligible difference in energy at this point.

== Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ==

The following table assesses the plot for the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, with varying initial momenta.
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Proceeds as normal. No initial collision, but repulsion between the two systems is overcome, etc.
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The initial velocity of the atom is far too slow to collide with the molecule. The two systems approach each other initially, but the repulsion between them forces them to move in opposite directions, and with more velocity than
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The two systems approach the transition state, and then the diatom splits, with the central atom pairing with the other terminal atom. They then proceed to move in opposite velocities.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|Initial velocity of particle C is so great that there is some repulsion between particle B and C, despite moving towards each other. As a result, particle B moves away from C after the initial attraction. It then oscillates around the transition state, moving from AB + C to A+ BC. It eventually settles with AB + C, the initial reactants. It can be seen by increasing the number of steps to 2000 that the final position of the two systems is greater separation. The particles AB and C are moving in opposite directions, so end up further apart than the initial conditions.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Oscillations around the transition state after an initial collision. This is very similar initially to the -10.1 p2 path. However, the rebound collision for A and B is even faster, resulting in another rebound for B. B then collides with particle C as they are both moving in the same directions, and the final state is that of the products, A + BC
|
|}

== Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ==

Transition state theory can be used to determine the rate of a reaction, but it relies on some substantial assumptions. These contribute to deviations from the results observed in the GUI, resulting in overestimations.

The primary reason for overestimation is due to a lack of accounting for recrossing the energy threshold. TST allows for a 'quasi-equilibrium' - it allows molecules to reach the transition state from the side of both the reactants and products, but once the transition state has been crossed, it will be a successful reaction, regardless if the momentum would actually cause recrossing of the transition state. This is evidently not a realistic assumption, as we have seen in the previous section 1.4 ; e.g. the reaction pathway for the momenta p1 = -5.1 and p2 = -10.1, where despite the initial crossing of the barrier and consequent overcoming of the energy barrier, the system is not reactive.

This especially holds true for molecules with high momentum; so TST will deviate from experimental results at high temperatures. It should only be used for low temperature scenarios.

As a result, TST will incorrectly predict the number of successful reactions; it will give a higher prediction than reality, thus also '''overestimating the reaction rate'''.

Furthermore, it also assumes atoms obey the laws of '''classical''' mechanics using the Born-Oppenheimer approximation, and that each reaction only has 1 saddle point and 1 transition state. These assumptions will likely not hold true for other systems, where reactions will have more than one transition states. Quantum effects such as tunnelling of the atoms would also lead to different numbers than predicted from TST; more successful reactions would occur than TST would account for, as not all the molecules have to overcome the activation barrier to cross it.

= Exercise 2: F-H-H system =

== Question 1: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?==

Let's set the initial position as <math>r_1</math> = <math>r_2</math>, where <math>r_1</math> is F-H, and <math>r_2</math> is H-H.

If we run a MEP plot from this position, the particle will end up at the lower energy state, regardless of the initial reactants or products.

We can see in the image below, the minimum energy path is such that <math>r_1</math> is minimised, and <math>r_2</math> is maximised. With <math>r_1</math> as F-H, this implies that the lower energy state is H + H-F.

[[File:MEP F-H-H.png|center|thumb| An MEP plot showing the exothermic reaction is the formation of F-H, as this is the energy minimised product.]]

As a result, we can classify the F + H2 --> H + HF reaction as exothermic, as it is progressing from a higher energy state to a lower energy state.

H + HF ---> F + H2 would therefore be an endothermic process; it would require an input of energy to get this product.

This order of potential energies suggests that the bond strength between F-H is greater than for H2.

This is in agreement with experimental results. The strength of the H-F bond (<math> 565 kJmol^{-1}</math> ) is greater than the H-H ( <math> 432 kJmol^{-1}</math> ) bond; more energy is required to break the H-F bond than is gained forming an H-H bond.

== Question 2: Locate the approximate position of the transition state. ==

As before, the transition state will be the saddle point on the surface plot. However, as the system is no longer symmetric as before, the transition state is no longer at the point where <math>r_1</math> = <math>r_2</math>.

Hammond's postulate suggests that the transition state will be closer to the higher energy system in the reaction. We know that this is the case for F + H2, so the transition state must be such that the bond distance for F-H is greater than for H2.

Knowing this, we can surmise that if AB is F-H and BC is H2, the distances at the transition state are: AB =184 pm and BC = 74.4 pm, with momenta = 0. This complies with the logic that the reaction where FH is the product is exothermic, as this is supported by Hammond's postulate.
We once again know that the internuclear distance will not change at this position with respect to time, as the derivative of the potential V is zero. As force is the negative derivative of the potential, this implies that the transition state has no force components if initial momenta = 0, so no movement will occur at this point.

{|style="margin: 0 auto;"
| [[File:Transition state ex2 contour.png|thumb| The transition state on the contour plot. It is shown to have a much greater AB distance than BC.]]
| [[File:Transition state ex2 surface.png|thumb| The transition state on the PES.]]
|}

We can once again, further confirm that this is the saddle point by using the same point that the forces along AB and BC at this point are both zero.

== Question 3: Report the activation energy for both reactions. ==

To find the activation energies of each product F-H and H2 (2 mutually exclusive scenarios), we simply need to find the difference between the transition state energy and the energy of the products. This can be calculated for the reaction only (excluding vibration) by using MEP to identify the shortest route.

Due to the asymmetry of the atoms, we have 2 activation energies.

Starting from the <math>r_{ts}</math>where AB =184 pm and BC = 74.4 pm, with a perturbation of ± 2 pm in the AB axis, and a step count of 2500, we get these MEP results.

{|style="margin: 0 auto;"
| [[File:Transition state to minimum.png|thumb| The contour plot showing the MEP path from the transition state to the exothermic product. ]]
| [[File:Transition state to min en vs time.png|thumb| The energy vs time plot showing how the energy changes as it travels along the MEP path to the exothermic product.]]
|}

The plots above show the activation energy for the enothermic formation of the product F + H2 . The plots below show the activation energy for the exothermic formation of the products F-H + H.

{|style="margin: 0 auto;"
| [[File:Transition state to min 2.png|thumb| The contour plot showing the MEP path from the transition state to the endothermic product. ]]
| [[File:Screenshot 2020-05-08 at 22.47.21.png|thumb| The energy vs time plot showing how the energy changes as it travels along the MEP path to the endothermic product. There is very little change in energy as it reaches this, due to the early transition state in the exothermic reaction path. It agrees with Hammond's postulate.]]
|}

{| class="wikitable"
!Product
!Energy of Transition State (<math>kJmol^{-1}</math>)
!Energy of product (<math>kJmol^{-1}</math>)
!ΔE (<math>kJmol^{-1}</math>
|-
|F-H + H
|<nowiki>-433.991</nowiki>
|<nowiki>-435.271</nowiki>
|1.28
|-
|F + H2
|<nowiki>-433.979</nowiki>
|<nowiki>-556.895</nowiki>
|122.92
|}

This is because we know that the energy of the transition state is the maximum energy state on the reaction path, and doesn't consider oscillations along the potential well; so we can just use the MEP.

These results are consistent with our understanding of the system, and the contour plot; it is an extremely early TS for the exothermic formation of HF + H, so the activation energy for this reaction is extremely low compared to the reverse, endothermic process. This can be further shown by the contour plot; the difference in energy is shown to be almost negligible as it is along the same contour line.

== Question 4: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. ==

With the exothermic reaction <math>H_2 + F --> HF + H</math>, we will observe a significant release of energy. In this scenario, we can confirm that potential energy is converted to kinetic energy, as is seen in the plots below.

These plots use the initial conditions:
FH = 174 pm with p = -1 g.mol-1.pm.fs-1, HH = 74 pm with p = -4 g.mol-1.pm.fs-1, and 2000 steps.
This successfully reacts to give the exothermic reaction defined above.

{|style="margin: 0 auto;"
| [[File:Reactive path endo contour.png|thumb|A contour plot showing a reactive path on the Dynamics plot.]]
| [[File:Reactive path endo PES.png|thumb|A PES plot showing the reactive path. A highly vibrationally active molecule is shown as the end product.]]
| [[File:Momenta vs time plot endo.png|thumb| A momentum vs time plot for the reactive endothermic reaction. A-B is HF, and B-C is HH. The oscillation around zero is clearly vibration, as the momenta change direction consistently for the molecule, but not for the individual atom.]]
| [[File:Energy vs time endo.png|thumb|An energy vs time plot showing the successful exothermic reaction]]
|}

It is observed in the momenta vs time plot that a vibrating hydrogen molecule initially approaches the F atom. There are two bounces and recrossings, as observed in the animation, and then the molecule HF is formed. This has a lot of vibrational energy, as is seen in the PES surface plot, whereas the leaving H atom cannot vibrate as it is an individual atom.
Therefore, we see that the released energy from the exothermic reaction is realised in vibrations of the newly formed molecule.
The energy conversion from potential to kinetic is also shown in the energy vs time plot, where the dynamical relationship is shown.

This would correspond to an increase in temperature with heat being given off in the reaction, and can be observed by using IR spectroscopy, or with calorimetry methods (specically bomb calorimetry).
In the IR, overtones would be observed, corresponding to vibrational excitation at higher energy levels, and translational motion would also be observed, as is shown in the contour plot.

== Question 5: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. ==

For the reaction of HF + H -> H2 + F, a higher vibrational energy of the HF molecule is required, as stated by polyani's rules. This is due to the relatively late transition state of the endothermic reaction. Similarly for the reverse reaction, very little vibrational energy is required and only a slight amount of translational energy is required due to the low activation energy of the exothermic reaction.

Polanyi's rules empirically state that in an early transition state, translational energy is more useful than vibrational energy with regards to the generation of a successful reaction. The converse argument applies with a late transition state. These only hold true when there is enough energy to overcome the activation energy, and favourable conditions are present.

This was largely observed for the endothermic reaction of HF + H -> H2 + F (late transition state); a higher vibrational energy resulted in more readily complete reaction. This also held true for the opposite case.
These are shown in the plots below.

{|style="margin: 0 auto;"
| [[File:01513089 HF high vibration success.png|left|thumb|405x405px|Endothermic reaction (late transition state) with high vibrational energy and relatively low translational energy leading to a successful H2 reaction: HF distance 75.3 pm, -0.61 g.mol-1.pm.fs-1: HH distance 225.1 pm, -3.117 g.mol-1.pm.fs-1]]
| [[File:01513089 Plot.png|left|thumb|403x403px|Endothermic reaction (late transition state) with low vibrational energy and high translational energy not leading to successful H2 production: HF distance 100 pm, -20 g.mol-1.pm.fs-1: HH distance 225 pm, -20 g.mol-1.pm.fs-1]]
|}

{|style="margin: 0 auto;"
| [[File:01513089 HH low vibration success.png|thumb|411x411px|Exothermic reaction (early transition state) successfully producing HF with almost no vibrational energy and some translational energy: HF distance 210 pm, -2 g.mol-1.pm.fs-1: HH distance 74 pm, -1 g.mol-1.pm.fs-1]]
| [[File:01513089 HH high vibration usuccessful.png|thumb|405x405px|Exothermic reaction (early transition state) unsuccessful in producing HF with very high vibrational energy and some translational energy: HF distance 210 pm, -1 g.mol-1.pm.fs-1: HH distance 100 pm, -1 g.mol-1.pm.fs-1]]
|}

MRD:01512675

2020-05-08T22:36:09Z

Sa13018: /* Question 4: The release of Reaction Energy */

= Exercise 1: H + H<math>_2</math> system =

== Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ==

The transition state is defined as the maximum point of the minimum energy path between the reactants and products. Therefore, on the potential energy surface diagram, it exists as a minimax; a saddle point.
This can therefore be described as the point where the partial differential of the potential energy with respect to all variables is zero.

In this case, due to a 3 dimensional surface plot, we have <math>
\frac{\partial V}{\partial r_1} = \frac{\partial V}{\partial r_2} = 0
</math>, where <math>r_1</math> and <math>r_2 </math> are the respective distances between AB and BC.

The local minimum point would see that a deviation in '''any''' direction would be an increase in potential energy. An example of this is the position of the reactants or products on the PES.

However, the transition state would see a decrease in potential energy along the reaction coordinate, and an increase in the coordinate orthogonal to this. This can be seen if we visualise the PES.

{|style="margin: 0 auto;"
| [[File:Transition state minimum.png|center|thumb|A 2D viewpoint of the PES, with the x-axis being the AB distance, and the y-axis being V(<math>r_1</math>). We see an established potential well. This runs across the entire PES as a valley of potential well. In this viewpoint, the transition state is seen as a minimum point. ]]
| [[File:Transition state maximum.png|center|thumb| A 3D visualisation of the PES, showing how the transition state is a maximum. The black curve is the reaction path, and maximum of this curve is the saddle point transition state.]]
|}

Due to the symmetry of this system, the transition state saddle point is observed at a point where <math>r_1</math> = <math>r_2</math>, but this is not necessarily always the case. In some asymmetric systems, the transition state will depend on the relative energy of the products and reactants; this can be further described by Hammond's postulate.

The saddle point of any-dimensional curve can be identified by computing a Hessian matrix.

<math>
H_{x,y} =
\begin{pmatrix}
\frac{\partial^{2} V}{\partial x^{2}} & \frac{\partial^{2} V}{\partial x\partial y } \\
\frac{\partial^{2} V}{\partial x \partial y } & \frac{\partial^{2} V}{\partial y^{2} } \\
\end{pmatrix}</math>

Upon finding the determinant (H). If H <0, it must be a saddle point, so to find the transition state, we just need to find the coordinates where this is the case.

== Question 2:Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ==

Due to the symmetry between the H and the \chem{H_2} molecule, and the fact that the interaction is happening at 180º, we know that <math>r_1</math> must equal <math>r_2</math> for any point on the potential for the transition state.

If we were to further set the momenta to zero for all 3 particles on the transition state, we know that the internuclear distances between the atoms will not change. This must hold true, as all the derivatives of the potential are zero; there are no forces acting on the particles.
As an equilibrium point, a plot of internuclear distances vs time will be constant for the transition state.

This was found for <math>r_{12}</math> = 90.775 pm. The internuclear distances vs time plot confirms this; it is shown below.

[[File:Internuclear dist v time 1.png]]
[[File:Contour plot trs1.png]]

== Question 3: Comment on how the MEP and the trajectory you just calculated differ.==

The Minimum Energy Path (MEP) plots the reaction path assuming that the particles have negligible intertial motion. It therefore ignores the vibration potential of the system, and hence only runs along the "valley" of the PES, rather than oscillating along the potential well as it does in the dynamical calculation. This makes sense; it would be the lowest energy path, but ignores the momenta of the particles at each point, and how it changes.

The differences are shown in the following two images; both of them tend towards a complete reaction from the transition state as predicted, but the dynamics plot shows oscillation along the reaction path in the y-axis as it moves along.

{|style="margin: 0 auto;"
| [[File:Dynamics plot sa13018.png|none|thumb| This shows the reaction path in a 'Dynamics' plot at an initial AB distance =91.775, and BC = 90.775. ]]
| [[File:Mep plot sa13018.png|none|thumb| This shows the reaction path in a 'MEP' plot at an initial AB distance =91.775, and BC = 90.775. ]]
|}

Furthermore, the MEP reaction path finishes at 193.1pm as the AB distance and 74 pm as the BC distance, whereas for the dynamical plot, the reaction path will continue after this. This makes intuitive physical sense; both have opposite velocities so will tend away from each other. However, for MEP, with no inertial motion, this is discounted, so the particles cease motion at this point, as there is negligible difference in energy at this point.

== Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ==

The following table assesses the plot for the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, with varying initial momenta.
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Proceeds as normal. No initial collision, but repulsion between the two systems is overcome, etc.
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The initial velocity of the atom is far too slow to collide with the molecule. The two systems approach each other initially, but the repulsion between them forces them to move in opposite directions, and with more velocity than
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The two systems approach the transition state, and then the diatom splits, with the central atom pairing with the other terminal atom. They then proceed to move in opposite velocities.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|Initial velocity of particle C is so great that there is some repulsion between particle B and C, despite moving towards each other. As a result, particle B moves away from C after the initial attraction. It then oscillates around the transition state, moving from AB + C to A+ BC. It eventually settles with AB + C, the initial reactants. It can be seen by increasing the number of steps to 2000 that the final position of the two systems is greater separation. The particles AB and C are moving in opposite directions, so end up further apart than the initial conditions.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Oscillations around the transition state after an initial collision. This is very similar initially to the -10.1 p2 path. However, the rebound collision for A and B is even faster, resulting in another rebound for B. B then collides with particle C as they are both moving in the same directions, and the final state is that of the products, A + BC
|
|}

== Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ==

Transition state theory can be used to determine the rate of a reaction, but it relies on some substantial assumptions. These contribute to deviations from the results observed in the GUI, resulting in overestimations.

The primary reason for overestimation is due to a lack of accounting for recrossing the energy threshold. TST allows for a 'quasi-equilibrium' - it allows molecules to reach the transition state from the side of both the reactants and products, but once the transition state has been crossed, it will be a successful reaction, regardless if the momentum would actually cause recrossing of the transition state. This is evidently not a realistic assumption, as we have seen in the previous section 1.4 ; e.g. the reaction pathway for the momenta p1 = -5.1 and p2 = -10.1, where despite the initial crossing of the barrier and consequent overcoming of the energy barrier, the system is not reactive.

This especially holds true for molecules with high momentum; so TST will deviate from experimental results at high temperatures. It should only be used for low temperature scenarios.

As a result, TST will incorrectly predict the number of successful reactions; it will give a higher prediction than reality, thus also '''overestimating the reaction rate'''.

Furthermore, it also assumes atoms obey the laws of '''classical''' mechanics using the Born-Oppenheimer approximation, and that each reaction only has 1 saddle point and 1 transition state. These assumptions will likely not hold true for other systems, where reactions will have more than one transition states. Quantum effects such as tunnelling of the atoms would also lead to different numbers than predicted from TST; more successful reactions would occur than TST would account for, as not all the molecules have to overcome the activation barrier to cross it.

= Exercise 2: F-H-H system =

== Question 1: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?==

Let's set the initial position as <math>r_1</math> = <math>r_2</math>, where <math>r_1</math> is F-H, and <math>r_2</math> is H-H.

If we run a MEP plot from this position, the particle will end up at the lower energy state, regardless of the initial reactants or products.

We can see in the image below, the minimum energy path is such that <math>r_1</math> is minimised, and <math>r_2</math> is maximised. With <math>r_1</math> as F-H, this implies that the lower energy state is H + H-F.

[[File:MEP F-H-H.png|center|thumb| An MEP plot showing the exothermic reaction is the formation of F-H, as this is the energy minimised product.]]

As a result, we can classify the F + H2 --> H + HF reaction as exothermic, as it is progressing from a higher energy state to a lower energy state.

H + HF ---> F + H2 would therefore be an endothermic process; it would require an input of energy to get this product.

This order of potential energies suggests that the bond strength between F-H is greater than for H2.

This is in agreement with experimental results. The strength of the H-F bond (<math> 565 kJmol^{-1}</math> ) is greater than the H-H ( <math> 432 kJmol^{-1}</math> ) bond; more energy is required to break the H-F bond than is gained forming an H-H bond.

== Question 2: Locate the approximate position of the transition state. ==

As before, the transition state will be the saddle point on the surface plot. However, as the system is no longer symmetric as before, the transition state is no longer at the point where <math>r_1</math> = <math>r_2</math>.

Hammond's postulate suggests that the transition state will be closer to the higher energy system in the reaction. We know that this is the case for F + H2, so the transition state must be such that the bond distance for F-H is greater than for H2.

Knowing this, we can surmise that if AB is F-H and BC is H2, the distances at the transition state are: AB =184 pm and BC = 74.4 pm, with momenta = 0. This complies with the logic that the reaction where FH is the product is exothermic, as this is supported by Hammond's postulate.
We once again know that the internuclear distance will not change at this position with respect to time, as the derivative of the potential V is zero. As force is the negative derivative of the potential, this implies that the transition state has no force components if initial momenta = 0, so no movement will occur at this point.

{|style="margin: 0 auto;"
| [[File:Transition state ex2 contour.png|thumb| The transition state on the contour plot. It is shown to have a much greater AB distance than BC.]]
| [[File:Transition state ex2 surface.png|thumb| The transition state on the PES.]]
|}

We can once again, further confirm that this is the saddle point by using the same point that the forces along AB and BC at this point are both zero.

== Question 3: Report the activation energy for both reactions. ==

To find the activation energies of each product F-H and H2 (2 mutually exclusive scenarios), we simply need to find the difference between the transition state energy and the energy of the products. This can be calculated for the reaction only (excluding vibration) by using MEP to identify the shortest route.

Due to the asymmetry of the atoms, we have 2 activation energies.

Starting from the <math>r_{ts}</math>where AB =184 pm and BC = 74.4 pm, with a perturbation of ± 2 pm in the AB axis, and a step count of 2500, we get these MEP results.

{|style="margin: 0 auto;"
| [[File:Transition state to minimum.png|thumb| The contour plot showing the MEP path from the transition state to the exothermic product. ]]
| [[File:Transition state to min en vs time.png|thumb| The energy vs time plot showing how the energy changes as it travels along the MEP path to the exothermic product.]]
|}

The plots above show the activation energy for the enothermic formation of the product F + H2 . The plots below show the activation energy for the exothermic formation of the products F-H + H.

{|style="margin: 0 auto;"
| [[File:Transition state to min 2.png|thumb| The contour plot showing the MEP path from the transition state to the endothermic product. ]]
| [[File:Screenshot 2020-05-08 at 22.47.21.png|thumb| The energy vs time plot showing how the energy changes as it travels along the MEP path to the endothermic product. There is very little change in energy as it reaches this, due to the early transition state in the exothermic reaction path. It agrees with Hammond's postulate.]]
|}

{| class="wikitable"
!Product
!Energy of Transition State (<math>kJmol^{-1}</math>)
!Energy of product (<math>kJmol^{-1}</math>)
!ΔE (<math>kJmol^{-1}</math>
|-
|F-H + H
|<nowiki>-433.991</nowiki>
|<nowiki>-435.271</nowiki>
|1.28
|-
|F + H2
|<nowiki>-433.979</nowiki>
|<nowiki>-556.895</nowiki>
|122.92
|}

This is because we know that the energy of the transition state is the maximum energy state on the reaction path, and doesn't consider oscillations along the potential well; so we can just use the MEP.

These results are consistent with our understanding of the system, and the contour plot; it is an extremely early TS for the exothermic formation of HF + H, so the activation energy for this reaction is extremely low compared to the reverse, endothermic process. This can be further shown by the contour plot; the difference in energy is shown to be almost negligible as it is along the same contour line.

== Question 4: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. ==

With the exothermic reaction <math>H_2 + F --> HF + H</math>, we will observe a significant release of energy. In this scenario, we can confirm that potential energy is converted to kinetic energy, as is seen in the plots below.

These plots use the initial conditions:
FH = 174 pm with p = -1 g.mol-1.pm.fs-1, HH = 74 pm with p = -4 g.mol-1.pm.fs-1, and 2000 steps.
This successfully reacts to give the exothermic reaction defined above.

{|style="margin: 0 auto;"
| [[File:Reactive path endo contour.png|thumb|A contour plot showing a reactive path on the Dynamics plot.]]
| [[File:Reactive path endo PES.png|thumb|A PES plot showing the reactive path. A highly vibrationally active molecule is shown as the end product.]]
| [[File:Momenta vs time plot endo.png|thumb| A momentum vs time plot for the reactive endothermic reaction. A-B is HF, and B-C is HH. The oscillation around zero is clearly vibration, as the momenta change direction consistently for the molecule, but not for the individual atom.]]
| [[File:Energy vs time endo.png|thumb|An energy vs time plot showing the successful exothermic reaction]]
|}

It is observed in the momenta vs time plot that a vibrating hydrogen molecule initially approaches the F atom. There are two bounces and recrossings, as observed in the animation, and then the molecule HF is formed. This has a lot of vibrational energy, as is seen in the PES surface plot, whereas the leaving H atom cannot vibrate as it is an individual atom.
Therefore, we see that the released energy from the exothermic reaction is realised in vibrations of the newly formed molecule.
The energy conversion from potential to kinetic is also shown in the energy vs time plot, where the dynamical relationship is shown.

This would correspond to an increase in temperature with heat being given off in the reaction, and can be observed by using IR spectroscopy, or with calorimetry methods (specically bomb calorimetry).
In the IR, overtones would be observed, corresponding to vibrational excitation at higher energy levels, and translational motion would also be observed, as is shown in the contour plot.

== Question 5: Efficiency of reaction ==

hello

File:Energy vs time endo.png

2020-05-08T22:31:50Z

Sa13018:

File:Momenta vs time plot endo.png

2020-05-08T22:23:22Z

Sa13018:

File:Reactive path endo PES.png

2020-05-08T22:21:42Z

Sa13018:

File:Reactive path endo contour.png

2020-05-08T22:19:16Z

Sa13018:

MRD:01512675

2020-05-08T21:58:12Z

Sa13018: /* Question 3: Report the activation energy for both reactions. */

= Exercise 1: H + H<math>_2</math> system =

== Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ==

The transition state is defined as the maximum point of the minimum energy path between the reactants and products. Therefore, on the potential energy surface diagram, it exists as a minimax; a saddle point.
This can therefore be described as the point where the partial differential of the potential energy with respect to all variables is zero.

In this case, due to a 3 dimensional surface plot, we have <math>
\frac{\partial V}{\partial r_1} = \frac{\partial V}{\partial r_2} = 0
</math>, where <math>r_1</math> and <math>r_2 </math> are the respective distances between AB and BC.

The local minimum point would see that a deviation in '''any''' direction would be an increase in potential energy. An example of this is the position of the reactants or products on the PES.

However, the transition state would see a decrease in potential energy along the reaction coordinate, and an increase in the coordinate orthogonal to this. This can be seen if we visualise the PES.

{|style="margin: 0 auto;"
| [[File:Transition state minimum.png|center|thumb|A 2D viewpoint of the PES, with the x-axis being the AB distance, and the y-axis being V(<math>r_1</math>). We see an established potential well. This runs across the entire PES as a valley of potential well. In this viewpoint, the transition state is seen as a minimum point. ]]
| [[File:Transition state maximum.png|center|thumb| A 3D visualisation of the PES, showing how the transition state is a maximum. The black curve is the reaction path, and maximum of this curve is the saddle point transition state.]]
|}

Due to the symmetry of this system, the transition state saddle point is observed at a point where <math>r_1</math> = <math>r_2</math>, but this is not necessarily always the case. In some asymmetric systems, the transition state will depend on the relative energy of the products and reactants; this can be further described by Hammond's postulate.

The saddle point of any-dimensional curve can be identified by computing a Hessian matrix.

<math>
H_{x,y} =
\begin{pmatrix}
\frac{\partial^{2} V}{\partial x^{2}} & \frac{\partial^{2} V}{\partial x\partial y } \\
\frac{\partial^{2} V}{\partial x \partial y } & \frac{\partial^{2} V}{\partial y^{2} } \\
\end{pmatrix}</math>

Upon finding the determinant (H). If H <0, it must be a saddle point, so to find the transition state, we just need to find the coordinates where this is the case.

== Question 2:Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ==

Due to the symmetry between the H and the \chem{H_2} molecule, and the fact that the interaction is happening at 180º, we know that <math>r_1</math> must equal <math>r_2</math> for any point on the potential for the transition state.

If we were to further set the momenta to zero for all 3 particles on the transition state, we know that the internuclear distances between the atoms will not change. This must hold true, as all the derivatives of the potential are zero; there are no forces acting on the particles.
As an equilibrium point, a plot of internuclear distances vs time will be constant for the transition state.

This was found for <math>r_{12}</math> = 90.775 pm. The internuclear distances vs time plot confirms this; it is shown below.

[[File:Internuclear dist v time 1.png]]
[[File:Contour plot trs1.png]]

== Question 3: Comment on how the MEP and the trajectory you just calculated differ.==

The Minimum Energy Path (MEP) plots the reaction path assuming that the particles have negligible intertial motion. It therefore ignores the vibration potential of the system, and hence only runs along the "valley" of the PES, rather than oscillating along the potential well as it does in the dynamical calculation. This makes sense; it would be the lowest energy path, but ignores the momenta of the particles at each point, and how it changes.

The differences are shown in the following two images; both of them tend towards a complete reaction from the transition state as predicted, but the dynamics plot shows oscillation along the reaction path in the y-axis as it moves along.

{|style="margin: 0 auto;"
| [[File:Dynamics plot sa13018.png|none|thumb| This shows the reaction path in a 'Dynamics' plot at an initial AB distance =91.775, and BC = 90.775. ]]
| [[File:Mep plot sa13018.png|none|thumb| This shows the reaction path in a 'MEP' plot at an initial AB distance =91.775, and BC = 90.775. ]]
|}

Furthermore, the MEP reaction path finishes at 193.1pm as the AB distance and 74 pm as the BC distance, whereas for the dynamical plot, the reaction path will continue after this. This makes intuitive physical sense; both have opposite velocities so will tend away from each other. However, for MEP, with no inertial motion, this is discounted, so the particles cease motion at this point, as there is negligible difference in energy at this point.

== Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ==

The following table assesses the plot for the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, with varying initial momenta.
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Proceeds as normal. No initial collision, but repulsion between the two systems is overcome, etc.
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The initial velocity of the atom is far too slow to collide with the molecule. The two systems approach each other initially, but the repulsion between them forces them to move in opposite directions, and with more velocity than
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The two systems approach the transition state, and then the diatom splits, with the central atom pairing with the other terminal atom. They then proceed to move in opposite velocities.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|Initial velocity of particle C is so great that there is some repulsion between particle B and C, despite moving towards each other. As a result, particle B moves away from C after the initial attraction. It then oscillates around the transition state, moving from AB + C to A+ BC. It eventually settles with AB + C, the initial reactants. It can be seen by increasing the number of steps to 2000 that the final position of the two systems is greater separation. The particles AB and C are moving in opposite directions, so end up further apart than the initial conditions.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Oscillations around the transition state after an initial collision. This is very similar initially to the -10.1 p2 path. However, the rebound collision for A and B is even faster, resulting in another rebound for B. B then collides with particle C as they are both moving in the same directions, and the final state is that of the products, A + BC
|
|}

== Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ==

Transition state theory can be used to determine the rate of a reaction, but it relies on some substantial assumptions. These contribute to deviations from the results observed in the GUI, resulting in overestimations.

The primary reason for overestimation is due to a lack of accounting for recrossing the energy threshold. TST allows for a 'quasi-equilibrium' - it allows molecules to reach the transition state from the side of both the reactants and products, but once the transition state has been crossed, it will be a successful reaction, regardless if the momentum would actually cause recrossing of the transition state. This is evidently not a realistic assumption, as we have seen in the previous section 1.4 ; e.g. the reaction pathway for the momenta p1 = -5.1 and p2 = -10.1, where despite the initial crossing of the barrier and consequent overcoming of the energy barrier, the system is not reactive.

This especially holds true for molecules with high momentum; so TST will deviate from experimental results at high temperatures. It should only be used for low temperature scenarios.

As a result, TST will incorrectly predict the number of successful reactions; it will give a higher prediction than reality, thus also '''overestimating the reaction rate'''.

Furthermore, it also assumes atoms obey the laws of '''classical''' mechanics using the Born-Oppenheimer approximation, and that each reaction only has 1 saddle point and 1 transition state. These assumptions will likely not hold true for other systems, where reactions will have more than one transition states. Quantum effects such as tunnelling of the atoms would also lead to different numbers than predicted from TST; more successful reactions would occur than TST would account for, as not all the molecules have to overcome the activation barrier to cross it.

= Exercise 2: F-H-H system =

== Question 1: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?==

Let's set the initial position as <math>r_1</math> = <math>r_2</math>, where <math>r_1</math> is F-H, and <math>r_2</math> is H-H.

If we run a MEP plot from this position, the particle will end up at the lower energy state, regardless of the initial reactants or products.

We can see in the image below, the minimum energy path is such that <math>r_1</math> is minimised, and <math>r_2</math> is maximised. With <math>r_1</math> as F-H, this implies that the lower energy state is H + H-F.

[[File:MEP F-H-H.png|center|thumb| An MEP plot showing the exothermic reaction is the formation of F-H, as this is the energy minimised product.]]

As a result, we can classify the F + H2 --> H + HF reaction as exothermic, as it is progressing from a higher energy state to a lower energy state.

H + HF ---> F + H2 would therefore be an endothermic process; it would require an input of energy to get this product.

This order of potential energies suggests that the bond strength between F-H is greater than for H2.

This is in agreement with experimental results. The strength of the H-F bond (<math> 565 kJmol^{-1}</math> ) is greater than the H-H ( <math> 432 kJmol^{-1}</math> ) bond; more energy is required to break the H-F bond than is gained forming an H-H bond.

== Question 2: Locate the approximate position of the transition state. ==

As before, the transition state will be the saddle point on the surface plot. However, as the system is no longer symmetric as before, the transition state is no longer at the point where <math>r_1</math> = <math>r_2</math>.

Hammond's postulate suggests that the transition state will be closer to the higher energy system in the reaction. We know that this is the case for F + H2, so the transition state must be such that the bond distance for F-H is greater than for H2.

Knowing this, we can surmise that if AB is F-H and BC is H2, the distances at the transition state are: AB =184 pm and BC = 74.4 pm, with momenta = 0. This complies with the logic that the reaction where FH is the product is exothermic, as this is supported by Hammond's postulate.
We once again know that the internuclear distance will not change at this position with respect to time, as the derivative of the potential V is zero. As force is the negative derivative of the potential, this implies that the transition state has no force components if initial momenta = 0, so no movement will occur at this point.

{|style="margin: 0 auto;"
| [[File:Transition state ex2 contour.png|thumb| The transition state on the contour plot. It is shown to have a much greater AB distance than BC.]]
| [[File:Transition state ex2 surface.png|thumb| The transition state on the PES.]]
|}

We can once again, further confirm that this is the saddle point by using the same point that the forces along AB and BC at this point are both zero.

== Question 3: Report the activation energy for both reactions. ==

To find the activation energies of each product F-H and H2 (2 mutually exclusive scenarios), we simply need to find the difference between the transition state energy and the energy of the products. This can be calculated for the reaction only (excluding vibration) by using MEP to identify the shortest route.

Due to the asymmetry of the atoms, we have 2 activation energies.

Starting from the <math>r_{ts}</math>where AB =184 pm and BC = 74.4 pm, with a perturbation of ± 2 pm in the AB axis, and a step count of 2500, we get these MEP results.

{|style="margin: 0 auto;"
| [[File:Transition state to minimum.png|thumb| The contour plot showing the MEP path from the transition state to the exothermic product. ]]
| [[File:Transition state to min en vs time.png|thumb| The energy vs time plot showing how the energy changes as it travels along the MEP path to the exothermic product.]]
|}

The plots above show the activation energy for the enothermic formation of the product F + H2 . The plots below show the activation energy for the exothermic formation of the products F-H + H.

{|style="margin: 0 auto;"
| [[File:Transition state to min 2.png|thumb| The contour plot showing the MEP path from the transition state to the endothermic product. ]]
| [[File:Screenshot 2020-05-08 at 22.47.21.png|thumb| The energy vs time plot showing how the energy changes as it travels along the MEP path to the endothermic product. There is very little change in energy as it reaches this, due to the early transition state in the exothermic reaction path. It agrees with Hammond's postulate.]]
|}

{| class="wikitable"
!Product
!Energy of Transition State (<math>kJmol^{-1}</math>)
!Energy of product (<math>kJmol^{-1}</math>)
!ΔE (<math>kJmol^{-1}</math>
|-
|F-H + H
|<nowiki>-433.991</nowiki>
|<nowiki>-435.271</nowiki>
|1.28
|-
|F + H2
|<nowiki>-433.979</nowiki>
|<nowiki>-556.895</nowiki>
|122.92
|}

This is because we know that the energy of the transition state is the maximum energy state on the reaction path, and doesn't consider oscillations along the potential well; so we can just use the MEP.

These results are consistent with our understanding of the system, and the contour plot; it is an extremely early TS for the exothermic formation of HF + H, so the activation energy for this reaction is extremely low compared to the reverse, endothermic process. This can be further shown by the contour plot; the difference in energy is shown to be almost negligible as it is along the same contour line.

== Question 4: The release of Reaction Energy ==

hello

== Question 5: Efficiency of reaction ==

hello

MRD:01512675

2020-05-08T21:57:33Z

Sa13018: /* Question 3: Activation energy */

= Exercise 1: H + H<math>_2</math> system =

== Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ==

The transition state is defined as the maximum point of the minimum energy path between the reactants and products. Therefore, on the potential energy surface diagram, it exists as a minimax; a saddle point.
This can therefore be described as the point where the partial differential of the potential energy with respect to all variables is zero.

In this case, due to a 3 dimensional surface plot, we have <math>
\frac{\partial V}{\partial r_1} = \frac{\partial V}{\partial r_2} = 0
</math>, where <math>r_1</math> and <math>r_2 </math> are the respective distances between AB and BC.

The local minimum point would see that a deviation in '''any''' direction would be an increase in potential energy. An example of this is the position of the reactants or products on the PES.

However, the transition state would see a decrease in potential energy along the reaction coordinate, and an increase in the coordinate orthogonal to this. This can be seen if we visualise the PES.

{|style="margin: 0 auto;"
| [[File:Transition state minimum.png|center|thumb|A 2D viewpoint of the PES, with the x-axis being the AB distance, and the y-axis being V(<math>r_1</math>). We see an established potential well. This runs across the entire PES as a valley of potential well. In this viewpoint, the transition state is seen as a minimum point. ]]
| [[File:Transition state maximum.png|center|thumb| A 3D visualisation of the PES, showing how the transition state is a maximum. The black curve is the reaction path, and maximum of this curve is the saddle point transition state.]]
|}

Due to the symmetry of this system, the transition state saddle point is observed at a point where <math>r_1</math> = <math>r_2</math>, but this is not necessarily always the case. In some asymmetric systems, the transition state will depend on the relative energy of the products and reactants; this can be further described by Hammond's postulate.

The saddle point of any-dimensional curve can be identified by computing a Hessian matrix.

<math>
H_{x,y} =
\begin{pmatrix}
\frac{\partial^{2} V}{\partial x^{2}} & \frac{\partial^{2} V}{\partial x\partial y } \\
\frac{\partial^{2} V}{\partial x \partial y } & \frac{\partial^{2} V}{\partial y^{2} } \\
\end{pmatrix}</math>

Upon finding the determinant (H). If H <0, it must be a saddle point, so to find the transition state, we just need to find the coordinates where this is the case.

== Question 2:Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ==

Due to the symmetry between the H and the \chem{H_2} molecule, and the fact that the interaction is happening at 180º, we know that <math>r_1</math> must equal <math>r_2</math> for any point on the potential for the transition state.

If we were to further set the momenta to zero for all 3 particles on the transition state, we know that the internuclear distances between the atoms will not change. This must hold true, as all the derivatives of the potential are zero; there are no forces acting on the particles.
As an equilibrium point, a plot of internuclear distances vs time will be constant for the transition state.

This was found for <math>r_{12}</math> = 90.775 pm. The internuclear distances vs time plot confirms this; it is shown below.

[[File:Internuclear dist v time 1.png]]
[[File:Contour plot trs1.png]]

== Question 3: Comment on how the MEP and the trajectory you just calculated differ.==

The Minimum Energy Path (MEP) plots the reaction path assuming that the particles have negligible intertial motion. It therefore ignores the vibration potential of the system, and hence only runs along the "valley" of the PES, rather than oscillating along the potential well as it does in the dynamical calculation. This makes sense; it would be the lowest energy path, but ignores the momenta of the particles at each point, and how it changes.

The differences are shown in the following two images; both of them tend towards a complete reaction from the transition state as predicted, but the dynamics plot shows oscillation along the reaction path in the y-axis as it moves along.

{|style="margin: 0 auto;"
| [[File:Dynamics plot sa13018.png|none|thumb| This shows the reaction path in a 'Dynamics' plot at an initial AB distance =91.775, and BC = 90.775. ]]
| [[File:Mep plot sa13018.png|none|thumb| This shows the reaction path in a 'MEP' plot at an initial AB distance =91.775, and BC = 90.775. ]]
|}

Furthermore, the MEP reaction path finishes at 193.1pm as the AB distance and 74 pm as the BC distance, whereas for the dynamical plot, the reaction path will continue after this. This makes intuitive physical sense; both have opposite velocities so will tend away from each other. However, for MEP, with no inertial motion, this is discounted, so the particles cease motion at this point, as there is negligible difference in energy at this point.

== Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ==

The following table assesses the plot for the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, with varying initial momenta.
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Proceeds as normal. No initial collision, but repulsion between the two systems is overcome, etc.
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The initial velocity of the atom is far too slow to collide with the molecule. The two systems approach each other initially, but the repulsion between them forces them to move in opposite directions, and with more velocity than
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The two systems approach the transition state, and then the diatom splits, with the central atom pairing with the other terminal atom. They then proceed to move in opposite velocities.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|Initial velocity of particle C is so great that there is some repulsion between particle B and C, despite moving towards each other. As a result, particle B moves away from C after the initial attraction. It then oscillates around the transition state, moving from AB + C to A+ BC. It eventually settles with AB + C, the initial reactants. It can be seen by increasing the number of steps to 2000 that the final position of the two systems is greater separation. The particles AB and C are moving in opposite directions, so end up further apart than the initial conditions.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Oscillations around the transition state after an initial collision. This is very similar initially to the -10.1 p2 path. However, the rebound collision for A and B is even faster, resulting in another rebound for B. B then collides with particle C as they are both moving in the same directions, and the final state is that of the products, A + BC
|
|}

== Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ==

Transition state theory can be used to determine the rate of a reaction, but it relies on some substantial assumptions. These contribute to deviations from the results observed in the GUI, resulting in overestimations.

The primary reason for overestimation is due to a lack of accounting for recrossing the energy threshold. TST allows for a 'quasi-equilibrium' - it allows molecules to reach the transition state from the side of both the reactants and products, but once the transition state has been crossed, it will be a successful reaction, regardless if the momentum would actually cause recrossing of the transition state. This is evidently not a realistic assumption, as we have seen in the previous section 1.4 ; e.g. the reaction pathway for the momenta p1 = -5.1 and p2 = -10.1, where despite the initial crossing of the barrier and consequent overcoming of the energy barrier, the system is not reactive.

This especially holds true for molecules with high momentum; so TST will deviate from experimental results at high temperatures. It should only be used for low temperature scenarios.

As a result, TST will incorrectly predict the number of successful reactions; it will give a higher prediction than reality, thus also '''overestimating the reaction rate'''.

Furthermore, it also assumes atoms obey the laws of '''classical''' mechanics using the Born-Oppenheimer approximation, and that each reaction only has 1 saddle point and 1 transition state. These assumptions will likely not hold true for other systems, where reactions will have more than one transition states. Quantum effects such as tunnelling of the atoms would also lead to different numbers than predicted from TST; more successful reactions would occur than TST would account for, as not all the molecules have to overcome the activation barrier to cross it.

= Exercise 2: F-H-H system =

== Question 1: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?==

Let's set the initial position as <math>r_1</math> = <math>r_2</math>, where <math>r_1</math> is F-H, and <math>r_2</math> is H-H.

If we run a MEP plot from this position, the particle will end up at the lower energy state, regardless of the initial reactants or products.

We can see in the image below, the minimum energy path is such that <math>r_1</math> is minimised, and <math>r_2</math> is maximised. With <math>r_1</math> as F-H, this implies that the lower energy state is H + H-F.

[[File:MEP F-H-H.png|center|thumb| An MEP plot showing the exothermic reaction is the formation of F-H, as this is the energy minimised product.]]

As a result, we can classify the F + H2 --> H + HF reaction as exothermic, as it is progressing from a higher energy state to a lower energy state.

H + HF ---> F + H2 would therefore be an endothermic process; it would require an input of energy to get this product.

This order of potential energies suggests that the bond strength between F-H is greater than for H2.

This is in agreement with experimental results. The strength of the H-F bond (<math> 565 kJmol^{-1}</math> ) is greater than the H-H ( <math> 432 kJmol^{-1}</math> ) bond; more energy is required to break the H-F bond than is gained forming an H-H bond.

== Question 2: Locate the approximate position of the transition state. ==

As before, the transition state will be the saddle point on the surface plot. However, as the system is no longer symmetric as before, the transition state is no longer at the point where <math>r_1</math> = <math>r_2</math>.

Hammond's postulate suggests that the transition state will be closer to the higher energy system in the reaction. We know that this is the case for F + H2, so the transition state must be such that the bond distance for F-H is greater than for H2.

Knowing this, we can surmise that if AB is F-H and BC is H2, the distances at the transition state are: AB =184 pm and BC = 74.4 pm, with momenta = 0. This complies with the logic that the reaction where FH is the product is exothermic, as this is supported by Hammond's postulate.
We once again know that the internuclear distance will not change at this position with respect to time, as the derivative of the potential V is zero. As force is the negative derivative of the potential, this implies that the transition state has no force components if initial momenta = 0, so no movement will occur at this point.

{|style="margin: 0 auto;"
| [[File:Transition state ex2 contour.png|thumb| The transition state on the contour plot. It is shown to have a much greater AB distance than BC.]]
| [[File:Transition state ex2 surface.png|thumb| The transition state on the PES.]]
|}

We can once again, further confirm that this is the saddle point by using the same point that the forces along AB and BC at this point are both zero.

== Question 3: Report the activation energy for both reactions. ==

To find the activation energies of each product F-H and H2 (2 mutually exclusive scenarios), we simply need to find the difference between the transition state energy and the energy of the products. This can be calculated for the reaction only (excluding vibration) by using MEP to identify the shortest route.

Due to the asymmetry of the atoms, we have 2 activation energies.

Starting from the <math>r_{ts}</math>where AB =184 pm and BC = 74.4 pm, with a perturbation of ± 2 pm in the AB axis, and a step count of 2500, we get these MEP results.

{|style="margin: 0 auto;"
| [[File:Transition state to minimum.png|thumb| The contour plot showing the MEP path from the transition state to the exothermic product. ]]
| [[File:Transition state to min en vs time.png|thumb| The energy vs time plot showing how the energy changes as it travels along the MEP path to the exothermic product.]]
|}

The plots above show the activation energy for the enothermic formation of the product F + H2 . The plots below show the activation energy for the exothermic formation of the products F-H + H.

{|style="margin: 0 auto;"
| [[File:Transition state to min 2.png|thumb| The contour plot showing the MEP path from the transition state to the endothermic product. ]]
| [[File:Screenshot 2020-05-08 at 22.47.21.png|thumb| The energy vs time plot showing how the energy changes as it travels along the MEP path to the endothermic product. There is very little change in energy as it reaches this, due to the early transition state in the exothermic reaction path. It agrees with Hammond's postulate.]]
|}

{| class="wikitable"
!Product
!Energy of Transition State (<math>kJmol^{-1}</math>)
!Energy of product
!ΔE (<math>kJmol^{-1}</math>
|-
|F-H + H
|<nowiki>-433.991</nowiki>
|<nowiki>-435.271</nowiki>
|1.28
|-
|F + H2
|<nowiki>-433.979</nowiki>
|<nowiki>-556.895</nowiki>
|122.92
|}

This is because we know that the energy of the transition state is the maximum energy state on the reaction path, and doesn't consider oscillations along the potential well; so we can just use the MEP.

These results are consistent with our understanding of the system, and the contour plot; it is an extremely early TS for the exothermic formation of HF + H, so the activation energy for this reaction is extremely low compared to the reverse, endothermic process. This can be further shown by the contour plot; the difference in energy is shown to be almost negligible as it is along the same contour line.

== Question 4: The release of Reaction Energy ==

hello

== Question 5: Efficiency of reaction ==

hello

File:Screenshot 2020-05-08 at 22.47.21.png

2020-05-08T21:49:41Z

Sa13018:

MRD:01512675

2020-05-08T21:44:16Z

Sa13018: /* Question 2: Transition state distance */

= Exercise 1: H + H<math>_2</math> system =

== Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ==

The transition state is defined as the maximum point of the minimum energy path between the reactants and products. Therefore, on the potential energy surface diagram, it exists as a minimax; a saddle point.
This can therefore be described as the point where the partial differential of the potential energy with respect to all variables is zero.

In this case, due to a 3 dimensional surface plot, we have <math>
\frac{\partial V}{\partial r_1} = \frac{\partial V}{\partial r_2} = 0
</math>, where <math>r_1</math> and <math>r_2 </math> are the respective distances between AB and BC.

The local minimum point would see that a deviation in '''any''' direction would be an increase in potential energy. An example of this is the position of the reactants or products on the PES.

However, the transition state would see a decrease in potential energy along the reaction coordinate, and an increase in the coordinate orthogonal to this. This can be seen if we visualise the PES.

{|style="margin: 0 auto;"
| [[File:Transition state minimum.png|center|thumb|A 2D viewpoint of the PES, with the x-axis being the AB distance, and the y-axis being V(<math>r_1</math>). We see an established potential well. This runs across the entire PES as a valley of potential well. In this viewpoint, the transition state is seen as a minimum point. ]]
| [[File:Transition state maximum.png|center|thumb| A 3D visualisation of the PES, showing how the transition state is a maximum. The black curve is the reaction path, and maximum of this curve is the saddle point transition state.]]
|}

Due to the symmetry of this system, the transition state saddle point is observed at a point where <math>r_1</math> = <math>r_2</math>, but this is not necessarily always the case. In some asymmetric systems, the transition state will depend on the relative energy of the products and reactants; this can be further described by Hammond's postulate.

The saddle point of any-dimensional curve can be identified by computing a Hessian matrix.

<math>
H_{x,y} =
\begin{pmatrix}
\frac{\partial^{2} V}{\partial x^{2}} & \frac{\partial^{2} V}{\partial x\partial y } \\
\frac{\partial^{2} V}{\partial x \partial y } & \frac{\partial^{2} V}{\partial y^{2} } \\
\end{pmatrix}</math>

Upon finding the determinant (H). If H <0, it must be a saddle point, so to find the transition state, we just need to find the coordinates where this is the case.

== Question 2:Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ==

Due to the symmetry between the H and the \chem{H_2} molecule, and the fact that the interaction is happening at 180º, we know that <math>r_1</math> must equal <math>r_2</math> for any point on the potential for the transition state.

If we were to further set the momenta to zero for all 3 particles on the transition state, we know that the internuclear distances between the atoms will not change. This must hold true, as all the derivatives of the potential are zero; there are no forces acting on the particles.
As an equilibrium point, a plot of internuclear distances vs time will be constant for the transition state.

This was found for <math>r_{12}</math> = 90.775 pm. The internuclear distances vs time plot confirms this; it is shown below.

[[File:Internuclear dist v time 1.png]]
[[File:Contour plot trs1.png]]

== Question 3: Comment on how the MEP and the trajectory you just calculated differ.==

The Minimum Energy Path (MEP) plots the reaction path assuming that the particles have negligible intertial motion. It therefore ignores the vibration potential of the system, and hence only runs along the "valley" of the PES, rather than oscillating along the potential well as it does in the dynamical calculation. This makes sense; it would be the lowest energy path, but ignores the momenta of the particles at each point, and how it changes.

The differences are shown in the following two images; both of them tend towards a complete reaction from the transition state as predicted, but the dynamics plot shows oscillation along the reaction path in the y-axis as it moves along.

{|style="margin: 0 auto;"
| [[File:Dynamics plot sa13018.png|none|thumb| This shows the reaction path in a 'Dynamics' plot at an initial AB distance =91.775, and BC = 90.775. ]]
| [[File:Mep plot sa13018.png|none|thumb| This shows the reaction path in a 'MEP' plot at an initial AB distance =91.775, and BC = 90.775. ]]
|}

Furthermore, the MEP reaction path finishes at 193.1pm as the AB distance and 74 pm as the BC distance, whereas for the dynamical plot, the reaction path will continue after this. This makes intuitive physical sense; both have opposite velocities so will tend away from each other. However, for MEP, with no inertial motion, this is discounted, so the particles cease motion at this point, as there is negligible difference in energy at this point.

== Question 4: Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? ==

The following table assesses the plot for the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, with varying initial momenta.
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Proceeds as normal. No initial collision, but repulsion between the two systems is overcome, etc.
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The initial velocity of the atom is far too slow to collide with the molecule. The two systems approach each other initially, but the repulsion between them forces them to move in opposite directions, and with more velocity than
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The two systems approach the transition state, and then the diatom splits, with the central atom pairing with the other terminal atom. They then proceed to move in opposite velocities.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|Initial velocity of particle C is so great that there is some repulsion between particle B and C, despite moving towards each other. As a result, particle B moves away from C after the initial attraction. It then oscillates around the transition state, moving from AB + C to A+ BC. It eventually settles with AB + C, the initial reactants. It can be seen by increasing the number of steps to 2000 that the final position of the two systems is greater separation. The particles AB and C are moving in opposite directions, so end up further apart than the initial conditions.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Oscillations around the transition state after an initial collision. This is very similar initially to the -10.1 p2 path. However, the rebound collision for A and B is even faster, resulting in another rebound for B. B then collides with particle C as they are both moving in the same directions, and the final state is that of the products, A + BC
|
|}

== Question 5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ==

Transition state theory can be used to determine the rate of a reaction, but it relies on some substantial assumptions. These contribute to deviations from the results observed in the GUI, resulting in overestimations.

The primary reason for overestimation is due to a lack of accounting for recrossing the energy threshold. TST allows for a 'quasi-equilibrium' - it allows molecules to reach the transition state from the side of both the reactants and products, but once the transition state has been crossed, it will be a successful reaction, regardless if the momentum would actually cause recrossing of the transition state. This is evidently not a realistic assumption, as we have seen in the previous section 1.4 ; e.g. the reaction pathway for the momenta p1 = -5.1 and p2 = -10.1, where despite the initial crossing of the barrier and consequent overcoming of the energy barrier, the system is not reactive.

This especially holds true for molecules with high momentum; so TST will deviate from experimental results at high temperatures. It should only be used for low temperature scenarios.

As a result, TST will incorrectly predict the number of successful reactions; it will give a higher prediction than reality, thus also '''overestimating the reaction rate'''.

Furthermore, it also assumes atoms obey the laws of '''classical''' mechanics using the Born-Oppenheimer approximation, and that each reaction only has 1 saddle point and 1 transition state. These assumptions will likely not hold true for other systems, where reactions will have more than one transition states. Quantum effects such as tunnelling of the atoms would also lead to different numbers than predicted from TST; more successful reactions would occur than TST would account for, as not all the molecules have to overcome the activation barrier to cross it.

= Exercise 2: F-H-H system =

== Question 1: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?==

Let's set the initial position as <math>r_1</math> = <math>r_2</math>, where <math>r_1</math> is F-H, and <math>r_2</math> is H-H.

If we run a MEP plot from this position, the particle will end up at the lower energy state, regardless of the initial reactants or products.

We can see in the image below, the minimum energy path is such that <math>r_1</math> is minimised, and <math>r_2</math> is maximised. With <math>r_1</math> as F-H, this implies that the lower energy state is H + H-F.

[[File:MEP F-H-H.png|center|thumb| An MEP plot showing the exothermic reaction is the formation of F-H, as this is the energy minimised product.]]

As a result, we can classify the F + H2 --> H + HF reaction as exothermic, as it is progressing from a higher energy state to a lower energy state.

H + HF ---> F + H2 would therefore be an endothermic process; it would require an input of energy to get this product.

This order of potential energies suggests that the bond strength between F-H is greater than for H2.

This is in agreement with experimental results. The strength of the H-F bond (<math> 565 kJmol^{-1}</math> ) is greater than the H-H ( <math> 432 kJmol^{-1}</math> ) bond; more energy is required to break the H-F bond than is gained forming an H-H bond.

== Question 2: Locate the approximate position of the transition state. ==

As before, the transition state will be the saddle point on the surface plot. However, as the system is no longer symmetric as before, the transition state is no longer at the point where <math>r_1</math> = <math>r_2</math>.

Hammond's postulate suggests that the transition state will be closer to the higher energy system in the reaction. We know that this is the case for F + H2, so the transition state must be such that the bond distance for F-H is greater than for H2.

Knowing this, we can surmise that if AB is F-H and BC is H2, the distances at the transition state are: AB =184 pm and BC = 74.4 pm, with momenta = 0. This complies with the logic that the reaction where FH is the product is exothermic, as this is supported by Hammond's postulate.
We once again know that the internuclear distance will not change at this position with respect to time, as the derivative of the potential V is zero. As force is the negative derivative of the potential, this implies that the transition state has no force components if initial momenta = 0, so no movement will occur at this point.

{|style="margin: 0 auto;"
| [[File:Transition state ex2 contour.png|thumb| The transition state on the contour plot. It is shown to have a much greater AB distance than BC.]]
| [[File:Transition state ex2 surface.png|thumb| The transition state on the PES.]]
|}

We can once again, further confirm that this is the saddle point by using the same point that the forces along AB and BC at this point are both zero.

== Question 3: Activation energy ==

To find the activation energies of each product F-H and H2 (2 mutually exclusive scenarios), we simply need to find the difference between the transition state energy and the energy of the products. This can be calculated for the reaction only (excluding vibration) by using MEP to identify the shortest route.

Due to the asymmetry of the atoms, we have 2 activation energies.

Starting from the <math>r_{ts}</math>where AB =184 pm and BC = 74.4 pm, with a perturbation of ± 2 pm in the AB axis, and a step count of 2500, we get these MEP results.

[[File:Transition state to minimum.png]]
[[File:Transition state to min en vs time.png]]

The plots above show the activation energy for the enothermic formation of the product F + H2 . The plots below show the activation energy for the exothermic formation of the products F-H + H.

[[File:Transition state to min 2.png]]

{| class="wikitable"
!Product
!Energy of Transition State (<math>kJmol^{-1}</math>)
!Energy of product
!ΔE
|-
|F-H + H
|<nowiki>-433.991</nowiki>
|<nowiki>-434.155</nowiki>
|0.164
|-
|F + H2
|<nowiki>-433.979</nowiki>
|<nowiki>-556.895</nowiki>
|122.916
|}

This is because we know that the energy of the transition state is the maximum energy state on the reaction path, and doesn't consider oscillations along the potential well; so we can just use the MEP.

These results are consistent with our understanding of the system, and the contour plot; it is an extremely early TS for the exothermic formation of HF + H, so the activation energy for this reaction is extremely low compared to the reverse, endothermic process. This can be further shown by the contour plot; the difference in energy is shown to be almost negligible as it is along the same contour line.

== Question 4: The release of Reaction Energy ==

hello

== Question 5: Efficiency of reaction ==

hello

MRD:01512675

2020-05-08T21:38:30Z

Sa13018: /* Question 1: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? */

MRD:01512675

2020-05-08T21:36:51Z

Sa13018: /* Question 1: Energetics and bond strength */

MRD:01512675

2020-05-08T21:30:38Z

Sa13018: /* Question 5: Transition State Theory Predictions */

MRD:01512675

2020-05-08T21:30:13Z

MRD:01512675

2020-05-08T20:44:23Z

Sa13018: /* Question 4: Reactive and Unreactive Trajectories */

MRD:01512675

2020-05-08T20:41:42Z

Sa13018: /* Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? */

MRD:01512675

2020-05-08T20:40:29Z

Sa13018: /* Question 3: Comment on how the MEP and the trajectory you just calculated differ. */

MRD:01512675

2020-05-08T20:37:18Z

Sa13018: /* Question 3: Comment on how the MEP and the trajectory you just calculated differ. */

MRD:01512675

2020-05-08T20:35:20Z

Sa13018: /* Question 3: The Minimum Energy Path (MEP) */

File:Mep plot sa13018.png

2020-05-08T20:31:13Z

Sa13018:

File:Dynamics plot sa13018.png

2020-05-08T20:30:43Z

Sa13018:

MRD:01512675

2020-05-08T20:19:16Z

Sa13018: /* Question 2:Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. */

MRD:01512675

2020-05-08T20:18:56Z

Sa13018: /* Question 2: Locating the transition state */

File:Contour plot trs1.png

2020-05-08T20:18:21Z

Sa13018:

File:Internuclear dist v time 1.png

2020-05-08T20:17:18Z

Sa13018:

MRD:01512675

2020-05-08T20:06:13Z

MRD:01512675

2020-05-08T20:04:26Z

Sa13018: /* Question 1: The Transition State Region */

MRD:01512675

2020-05-08T14:19:29Z

Sa13018: /* Question 3: Activation energy */

= Exercise 1: H + H<math>_2</math> system =

== Question 1: The Transition State Region ==

The transition state is defined as the maximum point of the minimum energy path between the reactants and products. Therefore, on the potential energy surface diagram, it exists as a minimax; a saddle point.
This can therefore be described as the point where the partial differential of the potential energy with respect to all variables is zero.

In this case, due to a 3 dimensional surface plot, we have <math>
\frac{\partial V}{\partial r_1} = \frac{\partial V}{\partial r_2} = 0
</math>, where <math>r_1</math> and <math>r_2 </math> are the respective distances between AB and BC.

The local minimum point would see that a deviation in any direction would be an increase in potential energy.
However, the transition state is such that this would only happen in the axis where <math>r_1</math> = <math>r_2</math>, and would decrease in energy in any deviation of the AB or BC position.

This is shown in the images shown below.

[[File:Transition state minimum.png]]
[[File:Transition state maximum.png]]

The photo on the left shows the energy potential curve in a 2D viewpoint, as a function of the AB distance <math>r_1</math>. We can see that there is a steep increase of the V(<math>r_1</math>) function on either side of <math>r_1</math> = 74 pm, indicating the minimum in a potential well at this point. This continues as a valley between the reactants and products path.

The photo on the right shows a clear maximum point being mapped out by the black line. Despite being at a minimum point on the potential valley floor, this is a maximum point, indicating a minimax state. This is therefore a saddle point, and the transition state. Due to the symmetry of the system only consisting of H atoms, the transition state is observed at a point where <math>r_1</math> = <math>r_2</math>.

This can further be identified using the second partial derivative test; computing a Hessian matrix. If H <0, it must be a saddle.

== Question 2: Locating the transition state ==

Due to the symmetry between the H and the \chem{H_2} molecule, and the fact that the interaction is happening at 180º, we know that <math>r_1</math> must equal <math>r_2</math> for any point on the potential for the transition state. As a result, if both particles carry zero initial velocity and are on the transition state only, they will drop down the potential well to the transition state distance <math>r_{12}</math>.
Thus, in a scenario where <math>r_1</math> = <math>r_2</math>, and <math>v_2</math> = <math>v_1</math> = 0, the transition state will act only in a 2D potential plot; the dynamics will reach the minimum in the potential well.

This distance <math>r_12</math> is shown to be 90.775 pm, as there is no force along AB or BC at this distance when v = 0.
Furthermore, the contour plot shows the final position to be the same as the initial position; it is a point of equilibrium.
The internuclear distances vs time plot are also shown to constant here; this must therefore be the transition state.

[[File:Example.jpg]]
[[File:Example.jpg]]
[[File:Example.jpg]]

== Question 3: The Minimum Energy Path (MEP) ==

The MEP shows a clear difference in its plot and animation compared to the Dynamic calculation.
# There is no vibration/ oscillation in the MEP plot. This makes sense; it would be going against minimum energy route, as the oscillations would mean that the particle is not always at the minimum of the potential well.
# The MEP plot finishes once the minimum point of the energy potential surface is reached; after 193.1pm as the AB distance and 74 pm as the BC distance, the contour is consistent. However, we know that the particles must keep moving from this, as they carry opposite velocities. This is shown in the Dynamics plot, where the reaction path continues long after reaching the end of the x-label in the plot. This confirms that the MEP is relying only on the minimum of the potential energy surface; not the distances AB and BC.
This is shown in the comparison below.

It is also interesting to note the action of the MEP when we start from the reactants, as before, with enough energy to go to the products in the Dynamics. This is shown in the comparison below.

In this scenario, despite having enough initial velocity in one direction, the MEP doesn't follow that path. It reverses the direction of the initial velocity and instead ends up with no reaction taking place. This implies that it has followed the minimum energy path at every point in the plot, ignoring the vectors that the particle is following. This further exemplifies the difference between Dynamics and MEP, showing that MEP is only taking the mathematically logical route, whereas the Dynamics considers the physical process that is occuring.

== Question 4: Reactive and Unreactive Trajectories ==

The following table assesses the plot for the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, with varying initial momenta.
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Proceeds as normal. No initial collision, but repulsion between the two systems is overcome, etc.
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The initial velocity of the atom is far too slow to collide with the molecule. The two systems approach each other initially, but the repulsion between them forces them to move in opposite directions, and with more velocity than
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The two systems approach the transition state, and then the diatom splits, with the central atom pairing with the other terminal atom. They then proceed to move in opposite velocities.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|Initial velocity of particle C is so great that there is some repulsion between particle B and C, despite moving towards each other. As a result, particle B moves away from C after the initial attraction. It then oscillates around the transition state, moving from AB + C to A+ BC. It eventually settles with AB + C, the initial reactants. It can be seen by increasing the number of steps to 2000 that the final position of the two systems is greater separation. The particles AB and C are moving in opposite directions, so end up further apart than the initial conditions.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Oscillations around the transition state after an initial collision. This is very similar initially to the -10.1 p2 path. However, the rebound collision for A and B is even faster, resulting in another rebound for B. B then collides with particle C as they are both moving in the same directions, and the final state is that of the products, A + BC
|
|}

== Question 5: Transition State Theory Predictions ==

Transition state theory can be used to determine the rate of a reaction, but it relies on 5 assumptions:
# Once the kinetic energy has crossed the potential barrier, the reaction will proceed; there is no recrossing of the barrier.
# There are no quantum effects {specifically, no quantum tunnelling will take place}
# It is modelled on the Boltzmann distribution
# ask
# ask

Assumptions 1 and 2 are broken in these reaction dynamics, as we are considering classical collisions with some forces of attraction and repulsion between the atoms. As a result, despite the fact that crossing of the barrier can occur in a certain scenario, that activation barrier may be recrossed later. This is shown in the reaction pathway for the momenta p1 = -5.1 and p2 = -10.1, where despite the initial crossing of the barrier and consequent overcoming of the energy barrier, the system is not reactive. This therefore breaks from the mold of TST.

As TST suggests that some reactions that are not reactive are in fact reactive, it suggests that TST will overestimate how many reactions are successful. As a result, we can surmise that TST will overestimate the rate of reactions, compared to the GUI that we are using.

If the GUI agreed with assumption 1, TST would underestimate the reaction rate, as Tunnelling of the atoms would lead to a higher rate than the classical picture, as not all of the atoms have to reach the potential barrier to overcome it.
= Exercise 2: F-H-H system =

== Question 1: Energetics and bond strength ==

Let's set the initial position as <math>r_1</math> = <math>r_2</math>, where <math>r_1</math> is F-H, and <math>r_2</math> is H-H.

If we run a MEP plot from this position, the particle will end up at the lower energy state, regardless of the initial reactants or products.

We can see in the image below, the minimum energy path is such that <math>r_1</math> is minimised, and <math>r_2</math> is maximised. With <math>r_1</math> as F-H, this implies that the lower energy state is H + H-F.

[[File:MEP F-H-H.png]]

As a result, we can classify the F + H2 --> H + HF reaction as exothermic, as it is progressing from a higher energy state to a lower energy state.

H + HF ---> F + H2 would therefore be an endothermic process; it would require an input of energy to get this product.

This order of potential energies suggests that the bond strength between F-H is greater than for H2.

== Question 2: Transition state distance ==

As before, the transition state will be the saddle point on the surface plot. However, as the system is no longer symmetric as before, the transition state is no longer at the point where <math>r_1</math> = <math>r_2</math>.

Hammond's postulate suggests that the transition state will be closer to the higher energy system in the reaction. We know that this is the case for F + H2, so the transition state must be such that the bond distance for F-H is greater than for H2.

Knowing this, we can surmise that if AB is F-H and BC is H2, the distances at the transition state are: AB =184 pm and BC = 74.4 pm. This complies with the logic that the reaction where FH is the product is exothermic, as this is supported by Hammond's postulate.

[[File:Transition state ex2 contour.png]][[File:Transition state ex2 surface.png]]

We can once again, further confirm that this is the saddle point by using the same point that the forces along AB and BC at this point are both zero.

== Question 3: Activation energy ==

To find the activation energies of each product F-H and H2 (2 mutually exclusive scenarios), we simply need to find the difference between the transition state energy and the energy of the products. This can be calculated for the reaction only (excluding vibration) by using MEP to identify the shortest route.

Due to the asymmetry of the atoms, we have 2 activation energies.

Starting from the <math>r_{ts}</math>where AB =184 pm and BC = 74.4 pm, with a perturbation of ± 2 pm in the AB axis, and a step count of 2500, we get these MEP results.

[[File:Transition state to minimum.png]]
[[File:Transition state to min en vs time.png]]

The plots above show the activation energy for the enothermic formation of the product F + H2 . The plots below show the activation energy for the exothermic formation of the products F-H + H.

[[File:Transition state to min 2.png]]

{| class="wikitable"
!Product
!Energy of Transition State (<math>kJmol^{-1}</math>)
!Energy of product
!ΔE
|-
|F-H + H
|<nowiki>-433.991</nowiki>
|<nowiki>-434.155</nowiki>
|0.164
|-
|F + H2
|<nowiki>-433.979</nowiki>
|<nowiki>-556.895</nowiki>
|122.916
|}

This is because we know that the energy of the transition state is the maximum energy state on the reaction path, and doesn't consider oscillations along the potential well; so we can just use the MEP.

These results are consistent with our understanding of the system, and the contour plot; it is an extremely early TS for the exothermic formation of HF + H, so the activation energy for this reaction is extremely low compared to the reverse, endothermic process. This can be further shown by the contour plot; the difference in energy is shown to be almost negligible as it is along the same contour line.

== Question 4: The release of Reaction Energy ==

hello

== Question 5: Efficiency of reaction ==

hello

MRD:01512675

2020-05-08T14:18:18Z

Sa13018: /* Question 3: Activation energy */

= Exercise 1: H + H<math>_2</math> system =

== Question 1: The Transition State Region ==

The transition state is defined as the maximum point of the minimum energy path between the reactants and products. Therefore, on the potential energy surface diagram, it exists as a minimax; a saddle point.
This can therefore be described as the point where the partial differential of the potential energy with respect to all variables is zero.

In this case, due to a 3 dimensional surface plot, we have <math>
\frac{\partial V}{\partial r_1} = \frac{\partial V}{\partial r_2} = 0
</math>, where <math>r_1</math> and <math>r_2 </math> are the respective distances between AB and BC.

The local minimum point would see that a deviation in any direction would be an increase in potential energy.
However, the transition state is such that this would only happen in the axis where <math>r_1</math> = <math>r_2</math>, and would decrease in energy in any deviation of the AB or BC position.

This is shown in the images shown below.

[[File:Transition state minimum.png]]
[[File:Transition state maximum.png]]

The photo on the left shows the energy potential curve in a 2D viewpoint, as a function of the AB distance <math>r_1</math>. We can see that there is a steep increase of the V(<math>r_1</math>) function on either side of <math>r_1</math> = 74 pm, indicating the minimum in a potential well at this point. This continues as a valley between the reactants and products path.

The photo on the right shows a clear maximum point being mapped out by the black line. Despite being at a minimum point on the potential valley floor, this is a maximum point, indicating a minimax state. This is therefore a saddle point, and the transition state. Due to the symmetry of the system only consisting of H atoms, the transition state is observed at a point where <math>r_1</math> = <math>r_2</math>.

This can further be identified using the second partial derivative test; computing a Hessian matrix. If H <0, it must be a saddle.

== Question 2: Locating the transition state ==

Due to the symmetry between the H and the \chem{H_2} molecule, and the fact that the interaction is happening at 180º, we know that <math>r_1</math> must equal <math>r_2</math> for any point on the potential for the transition state. As a result, if both particles carry zero initial velocity and are on the transition state only, they will drop down the potential well to the transition state distance <math>r_{12}</math>.
Thus, in a scenario where <math>r_1</math> = <math>r_2</math>, and <math>v_2</math> = <math>v_1</math> = 0, the transition state will act only in a 2D potential plot; the dynamics will reach the minimum in the potential well.

This distance <math>r_12</math> is shown to be 90.775 pm, as there is no force along AB or BC at this distance when v = 0.
Furthermore, the contour plot shows the final position to be the same as the initial position; it is a point of equilibrium.
The internuclear distances vs time plot are also shown to constant here; this must therefore be the transition state.

[[File:Example.jpg]]
[[File:Example.jpg]]
[[File:Example.jpg]]

== Question 3: The Minimum Energy Path (MEP) ==

The MEP shows a clear difference in its plot and animation compared to the Dynamic calculation.
# There is no vibration/ oscillation in the MEP plot. This makes sense; it would be going against minimum energy route, as the oscillations would mean that the particle is not always at the minimum of the potential well.
# The MEP plot finishes once the minimum point of the energy potential surface is reached; after 193.1pm as the AB distance and 74 pm as the BC distance, the contour is consistent. However, we know that the particles must keep moving from this, as they carry opposite velocities. This is shown in the Dynamics plot, where the reaction path continues long after reaching the end of the x-label in the plot. This confirms that the MEP is relying only on the minimum of the potential energy surface; not the distances AB and BC.
This is shown in the comparison below.

It is also interesting to note the action of the MEP when we start from the reactants, as before, with enough energy to go to the products in the Dynamics. This is shown in the comparison below.

In this scenario, despite having enough initial velocity in one direction, the MEP doesn't follow that path. It reverses the direction of the initial velocity and instead ends up with no reaction taking place. This implies that it has followed the minimum energy path at every point in the plot, ignoring the vectors that the particle is following. This further exemplifies the difference between Dynamics and MEP, showing that MEP is only taking the mathematically logical route, whereas the Dynamics considers the physical process that is occuring.

== Question 4: Reactive and Unreactive Trajectories ==

The following table assesses the plot for the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, with varying initial momenta.
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Proceeds as normal. No initial collision, but repulsion between the two systems is overcome, etc.
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The initial velocity of the atom is far too slow to collide with the molecule. The two systems approach each other initially, but the repulsion between them forces them to move in opposite directions, and with more velocity than
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The two systems approach the transition state, and then the diatom splits, with the central atom pairing with the other terminal atom. They then proceed to move in opposite velocities.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|Initial velocity of particle C is so great that there is some repulsion between particle B and C, despite moving towards each other. As a result, particle B moves away from C after the initial attraction. It then oscillates around the transition state, moving from AB + C to A+ BC. It eventually settles with AB + C, the initial reactants. It can be seen by increasing the number of steps to 2000 that the final position of the two systems is greater separation. The particles AB and C are moving in opposite directions, so end up further apart than the initial conditions.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Oscillations around the transition state after an initial collision. This is very similar initially to the -10.1 p2 path. However, the rebound collision for A and B is even faster, resulting in another rebound for B. B then collides with particle C as they are both moving in the same directions, and the final state is that of the products, A + BC
|
|}

== Question 5: Transition State Theory Predictions ==

Transition state theory can be used to determine the rate of a reaction, but it relies on 5 assumptions:
# Once the kinetic energy has crossed the potential barrier, the reaction will proceed; there is no recrossing of the barrier.
# There are no quantum effects {specifically, no quantum tunnelling will take place}
# It is modelled on the Boltzmann distribution
# ask
# ask

Assumptions 1 and 2 are broken in these reaction dynamics, as we are considering classical collisions with some forces of attraction and repulsion between the atoms. As a result, despite the fact that crossing of the barrier can occur in a certain scenario, that activation barrier may be recrossed later. This is shown in the reaction pathway for the momenta p1 = -5.1 and p2 = -10.1, where despite the initial crossing of the barrier and consequent overcoming of the energy barrier, the system is not reactive. This therefore breaks from the mold of TST.

As TST suggests that some reactions that are not reactive are in fact reactive, it suggests that TST will overestimate how many reactions are successful. As a result, we can surmise that TST will overestimate the rate of reactions, compared to the GUI that we are using.

If the GUI agreed with assumption 1, TST would underestimate the reaction rate, as Tunnelling of the atoms would lead to a higher rate than the classical picture, as not all of the atoms have to reach the potential barrier to overcome it.
= Exercise 2: F-H-H system =

== Question 1: Energetics and bond strength ==

Let's set the initial position as <math>r_1</math> = <math>r_2</math>, where <math>r_1</math> is F-H, and <math>r_2</math> is H-H.

If we run a MEP plot from this position, the particle will end up at the lower energy state, regardless of the initial reactants or products.

We can see in the image below, the minimum energy path is such that <math>r_1</math> is minimised, and <math>r_2</math> is maximised. With <math>r_1</math> as F-H, this implies that the lower energy state is H + H-F.

[[File:MEP F-H-H.png]]

As a result, we can classify the F + H2 --> H + HF reaction as exothermic, as it is progressing from a higher energy state to a lower energy state.

H + HF ---> F + H2 would therefore be an endothermic process; it would require an input of energy to get this product.

This order of potential energies suggests that the bond strength between F-H is greater than for H2.

== Question 2: Transition state distance ==

As before, the transition state will be the saddle point on the surface plot. However, as the system is no longer symmetric as before, the transition state is no longer at the point where <math>r_1</math> = <math>r_2</math>.

Hammond's postulate suggests that the transition state will be closer to the higher energy system in the reaction. We know that this is the case for F + H2, so the transition state must be such that the bond distance for F-H is greater than for H2.

Knowing this, we can surmise that if AB is F-H and BC is H2, the distances at the transition state are: AB =184 pm and BC = 74.4 pm. This complies with the logic that the reaction where FH is the product is exothermic, as this is supported by Hammond's postulate.

[[File:Transition state ex2 contour.png]][[File:Transition state ex2 surface.png]]

We can once again, further confirm that this is the saddle point by using the same point that the forces along AB and BC at this point are both zero.

== Question 3: Activation energy ==

To find the activation energies of each product F-H and H2 (2 mutually exclusive scenarios), we simply need to find the difference between the transition state energy and the energy of the products. This can be calculated for the reaction only (excluding vibration) by using MEP to identify the shortest route.

Due to the asymmetry of the atoms, we have 2 activation energies.

Starting from the <math>r_{ts}</math>where AB =184 pm and BC = 74.4 pm, with a perturbation of ± 2 pm in the AB axis, and a step count of 2500, we get these results.

[[File:Transition state to minimum.png]]
[[File:Transition state to min en vs time.png]]

The plots above show the activation energy for the enothermic formation of the product F + H2 . The plots below show the activation energy for the exothermic formation of the products F-H + H.

[[File:Transition state to min 2.png]]

{| class="wikitable"
!Product
!Energy of Transition State (<math>kJmol^{-1}</math>)
!Energy of product
!ΔE
|-
|F-H + H
|<nowiki>-433.991</nowiki>
|<nowiki>-434.155</nowiki>
|0.164
|-
|F + H2
|<nowiki>-433.979</nowiki>
|<nowiki>-556.895</nowiki>
|122.916
|}

This is because we know that the energy of the transition state is the maximum energy state on the reaction path, and doesn't consider oscillations along the potential well; so we can just use the MEP.

These results are consistent with our understanding of the system, and the contour plot; it is an extremely early TS for the exothermic formation of HF + H, so the activation energy for this reaction is extremely low compared to the reverse, endothermic process. This can be further shown by the contour plot; the difference in energy is shown to be almost negligible as it is along the same contour line.

== Question 4: The release of Reaction Energy ==

hello

== Question 5: Efficiency of reaction ==

hello

MRD:01512675

2020-05-08T14:17:04Z

Sa13018: /* Question 2: Transition state distance */

= Exercise 1: H + H<math>_2</math> system =

== Question 1: The Transition State Region ==

The transition state is defined as the maximum point of the minimum energy path between the reactants and products. Therefore, on the potential energy surface diagram, it exists as a minimax; a saddle point.
This can therefore be described as the point where the partial differential of the potential energy with respect to all variables is zero.

In this case, due to a 3 dimensional surface plot, we have <math>
\frac{\partial V}{\partial r_1} = \frac{\partial V}{\partial r_2} = 0
</math>, where <math>r_1</math> and <math>r_2 </math> are the respective distances between AB and BC.

The local minimum point would see that a deviation in any direction would be an increase in potential energy.
However, the transition state is such that this would only happen in the axis where <math>r_1</math> = <math>r_2</math>, and would decrease in energy in any deviation of the AB or BC position.

This is shown in the images shown below.

[[File:Transition state minimum.png]]
[[File:Transition state maximum.png]]

The photo on the left shows the energy potential curve in a 2D viewpoint, as a function of the AB distance <math>r_1</math>. We can see that there is a steep increase of the V(<math>r_1</math>) function on either side of <math>r_1</math> = 74 pm, indicating the minimum in a potential well at this point. This continues as a valley between the reactants and products path.

The photo on the right shows a clear maximum point being mapped out by the black line. Despite being at a minimum point on the potential valley floor, this is a maximum point, indicating a minimax state. This is therefore a saddle point, and the transition state. Due to the symmetry of the system only consisting of H atoms, the transition state is observed at a point where <math>r_1</math> = <math>r_2</math>.

This can further be identified using the second partial derivative test; computing a Hessian matrix. If H <0, it must be a saddle.

== Question 2: Locating the transition state ==

Due to the symmetry between the H and the \chem{H_2} molecule, and the fact that the interaction is happening at 180º, we know that <math>r_1</math> must equal <math>r_2</math> for any point on the potential for the transition state. As a result, if both particles carry zero initial velocity and are on the transition state only, they will drop down the potential well to the transition state distance <math>r_{12}</math>.
Thus, in a scenario where <math>r_1</math> = <math>r_2</math>, and <math>v_2</math> = <math>v_1</math> = 0, the transition state will act only in a 2D potential plot; the dynamics will reach the minimum in the potential well.

This distance <math>r_12</math> is shown to be 90.775 pm, as there is no force along AB or BC at this distance when v = 0.
Furthermore, the contour plot shows the final position to be the same as the initial position; it is a point of equilibrium.
The internuclear distances vs time plot are also shown to constant here; this must therefore be the transition state.

[[File:Example.jpg]]
[[File:Example.jpg]]
[[File:Example.jpg]]

== Question 3: The Minimum Energy Path (MEP) ==

The MEP shows a clear difference in its plot and animation compared to the Dynamic calculation.
# There is no vibration/ oscillation in the MEP plot. This makes sense; it would be going against minimum energy route, as the oscillations would mean that the particle is not always at the minimum of the potential well.
# The MEP plot finishes once the minimum point of the energy potential surface is reached; after 193.1pm as the AB distance and 74 pm as the BC distance, the contour is consistent. However, we know that the particles must keep moving from this, as they carry opposite velocities. This is shown in the Dynamics plot, where the reaction path continues long after reaching the end of the x-label in the plot. This confirms that the MEP is relying only on the minimum of the potential energy surface; not the distances AB and BC.
This is shown in the comparison below.

It is also interesting to note the action of the MEP when we start from the reactants, as before, with enough energy to go to the products in the Dynamics. This is shown in the comparison below.

In this scenario, despite having enough initial velocity in one direction, the MEP doesn't follow that path. It reverses the direction of the initial velocity and instead ends up with no reaction taking place. This implies that it has followed the minimum energy path at every point in the plot, ignoring the vectors that the particle is following. This further exemplifies the difference between Dynamics and MEP, showing that MEP is only taking the mathematically logical route, whereas the Dynamics considers the physical process that is occuring.

== Question 4: Reactive and Unreactive Trajectories ==

The following table assesses the plot for the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, with varying initial momenta.
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Proceeds as normal. No initial collision, but repulsion between the two systems is overcome, etc.
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The initial velocity of the atom is far too slow to collide with the molecule. The two systems approach each other initially, but the repulsion between them forces them to move in opposite directions, and with more velocity than
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The two systems approach the transition state, and then the diatom splits, with the central atom pairing with the other terminal atom. They then proceed to move in opposite velocities.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|Initial velocity of particle C is so great that there is some repulsion between particle B and C, despite moving towards each other. As a result, particle B moves away from C after the initial attraction. It then oscillates around the transition state, moving from AB + C to A+ BC. It eventually settles with AB + C, the initial reactants. It can be seen by increasing the number of steps to 2000 that the final position of the two systems is greater separation. The particles AB and C are moving in opposite directions, so end up further apart than the initial conditions.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Oscillations around the transition state after an initial collision. This is very similar initially to the -10.1 p2 path. However, the rebound collision for A and B is even faster, resulting in another rebound for B. B then collides with particle C as they are both moving in the same directions, and the final state is that of the products, A + BC
|
|}

== Question 5: Transition State Theory Predictions ==

Transition state theory can be used to determine the rate of a reaction, but it relies on 5 assumptions:
# Once the kinetic energy has crossed the potential barrier, the reaction will proceed; there is no recrossing of the barrier.
# There are no quantum effects {specifically, no quantum tunnelling will take place}
# It is modelled on the Boltzmann distribution
# ask
# ask

Assumptions 1 and 2 are broken in these reaction dynamics, as we are considering classical collisions with some forces of attraction and repulsion between the atoms. As a result, despite the fact that crossing of the barrier can occur in a certain scenario, that activation barrier may be recrossed later. This is shown in the reaction pathway for the momenta p1 = -5.1 and p2 = -10.1, where despite the initial crossing of the barrier and consequent overcoming of the energy barrier, the system is not reactive. This therefore breaks from the mold of TST.

As TST suggests that some reactions that are not reactive are in fact reactive, it suggests that TST will overestimate how many reactions are successful. As a result, we can surmise that TST will overestimate the rate of reactions, compared to the GUI that we are using.

If the GUI agreed with assumption 1, TST would underestimate the reaction rate, as Tunnelling of the atoms would lead to a higher rate than the classical picture, as not all of the atoms have to reach the potential barrier to overcome it.
= Exercise 2: F-H-H system =

== Question 1: Energetics and bond strength ==

Let's set the initial position as <math>r_1</math> = <math>r_2</math>, where <math>r_1</math> is F-H, and <math>r_2</math> is H-H.

If we run a MEP plot from this position, the particle will end up at the lower energy state, regardless of the initial reactants or products.

We can see in the image below, the minimum energy path is such that <math>r_1</math> is minimised, and <math>r_2</math> is maximised. With <math>r_1</math> as F-H, this implies that the lower energy state is H + H-F.

[[File:MEP F-H-H.png]]

As a result, we can classify the F + H2 --> H + HF reaction as exothermic, as it is progressing from a higher energy state to a lower energy state.

H + HF ---> F + H2 would therefore be an endothermic process; it would require an input of energy to get this product.

This order of potential energies suggests that the bond strength between F-H is greater than for H2.

== Question 2: Transition state distance ==

As before, the transition state will be the saddle point on the surface plot. However, as the system is no longer symmetric as before, the transition state is no longer at the point where <math>r_1</math> = <math>r_2</math>.

Hammond's postulate suggests that the transition state will be closer to the higher energy system in the reaction. We know that this is the case for F + H2, so the transition state must be such that the bond distance for F-H is greater than for H2.

Knowing this, we can surmise that if AB is F-H and BC is H2, the distances at the transition state are: AB =184 pm and BC = 74.4 pm. This complies with the logic that the reaction where FH is the product is exothermic, as this is supported by Hammond's postulate.

[[File:Transition state ex2 contour.png]][[File:Transition state ex2 surface.png]]

We can once again, further confirm that this is the saddle point by using the same point that the forces along AB and BC at this point are both zero.

== Question 3: Activation energy ==

To find the activation energies of each product F-H and H2 (2 mutually exclusive scenarios), we simply need to find the difference between the transition state energy and the energy of the products. This can be calculated for the reaction only (excluding vibration) by using MEP to identify the shortest route.

Due to the asymmetry of the atoms, we have 2 activation energies.

Starting from the <math>r_{ts}</math>where AB =184 pm and BC = 74.4 pm, with a perturbation of ± 2 pm in the AB axis, and a step count of 2500, we get these results.

[[File:Transition state to minimum.png]]
[[File:Transition state to min en vs time.png]]

The plots above show the activation energy for the enothermic formation of the product F + H2 . The plots below show the activation energy for the exothermic formation of the products F-H + H.

[[File:Transition state to min 2.png]]

{| class="wikitable"
!Product
!Energy of Transition State (kJmol^-1)
!Energy of product
!ΔE
|-
|F-H + H
|<nowiki>-433.991</nowiki>
|<nowiki>-434.155</nowiki>
|0.164
|-
|F + H2
|<nowiki>-433.979</nowiki>
|<nowiki>-556.895</nowiki>
|122.916
|}

This is because we know that the energy of the transition state is the maximum energy state on the reaction path, and doesn't consider oscillations along the potential well; so we can just use the MEP.

These results are consistent with our understanding of the system, and the contour plot; it is an extremely early TS for the exothermic formation of HF + H, so the activation energy for this reaction is extremely low compared to the reverse, endothermic process. This can be further shown by the contour plot; the difference in energy is shown to be almost negligible as it is along the same contour line.

== Question 4: The release of Reaction Energy ==

hello

== Question 5: Efficiency of reaction ==

hello

MRD:01512675

2020-05-08T14:16:30Z

Sa13018: /* Question 3: Activation energy */

= Exercise 1: H + H<math>_2</math> system =

== Question 1: The Transition State Region ==

The transition state is defined as the maximum point of the minimum energy path between the reactants and products. Therefore, on the potential energy surface diagram, it exists as a minimax; a saddle point.
This can therefore be described as the point where the partial differential of the potential energy with respect to all variables is zero.

In this case, due to a 3 dimensional surface plot, we have <math>
\frac{\partial V}{\partial r_1} = \frac{\partial V}{\partial r_2} = 0
</math>, where <math>r_1</math> and <math>r_2 </math> are the respective distances between AB and BC.

The local minimum point would see that a deviation in any direction would be an increase in potential energy.
However, the transition state is such that this would only happen in the axis where <math>r_1</math> = <math>r_2</math>, and would decrease in energy in any deviation of the AB or BC position.

This is shown in the images shown below.

[[File:Transition state minimum.png]]
[[File:Transition state maximum.png]]

The photo on the left shows the energy potential curve in a 2D viewpoint, as a function of the AB distance <math>r_1</math>. We can see that there is a steep increase of the V(<math>r_1</math>) function on either side of <math>r_1</math> = 74 pm, indicating the minimum in a potential well at this point. This continues as a valley between the reactants and products path.

The photo on the right shows a clear maximum point being mapped out by the black line. Despite being at a minimum point on the potential valley floor, this is a maximum point, indicating a minimax state. This is therefore a saddle point, and the transition state. Due to the symmetry of the system only consisting of H atoms, the transition state is observed at a point where <math>r_1</math> = <math>r_2</math>.

This can further be identified using the second partial derivative test; computing a Hessian matrix. If H <0, it must be a saddle.

== Question 2: Locating the transition state ==

Due to the symmetry between the H and the \chem{H_2} molecule, and the fact that the interaction is happening at 180º, we know that <math>r_1</math> must equal <math>r_2</math> for any point on the potential for the transition state. As a result, if both particles carry zero initial velocity and are on the transition state only, they will drop down the potential well to the transition state distance <math>r_{12}</math>.
Thus, in a scenario where <math>r_1</math> = <math>r_2</math>, and <math>v_2</math> = <math>v_1</math> = 0, the transition state will act only in a 2D potential plot; the dynamics will reach the minimum in the potential well.

This distance <math>r_12</math> is shown to be 90.775 pm, as there is no force along AB or BC at this distance when v = 0.
Furthermore, the contour plot shows the final position to be the same as the initial position; it is a point of equilibrium.
The internuclear distances vs time plot are also shown to constant here; this must therefore be the transition state.

[[File:Example.jpg]]
[[File:Example.jpg]]
[[File:Example.jpg]]

== Question 3: The Minimum Energy Path (MEP) ==

The MEP shows a clear difference in its plot and animation compared to the Dynamic calculation.
# There is no vibration/ oscillation in the MEP plot. This makes sense; it would be going against minimum energy route, as the oscillations would mean that the particle is not always at the minimum of the potential well.
# The MEP plot finishes once the minimum point of the energy potential surface is reached; after 193.1pm as the AB distance and 74 pm as the BC distance, the contour is consistent. However, we know that the particles must keep moving from this, as they carry opposite velocities. This is shown in the Dynamics plot, where the reaction path continues long after reaching the end of the x-label in the plot. This confirms that the MEP is relying only on the minimum of the potential energy surface; not the distances AB and BC.
This is shown in the comparison below.

It is also interesting to note the action of the MEP when we start from the reactants, as before, with enough energy to go to the products in the Dynamics. This is shown in the comparison below.

In this scenario, despite having enough initial velocity in one direction, the MEP doesn't follow that path. It reverses the direction of the initial velocity and instead ends up with no reaction taking place. This implies that it has followed the minimum energy path at every point in the plot, ignoring the vectors that the particle is following. This further exemplifies the difference between Dynamics and MEP, showing that MEP is only taking the mathematically logical route, whereas the Dynamics considers the physical process that is occuring.

== Question 4: Reactive and Unreactive Trajectories ==

The following table assesses the plot for the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, with varying initial momenta.
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Proceeds as normal. No initial collision, but repulsion between the two systems is overcome, etc.
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The initial velocity of the atom is far too slow to collide with the molecule. The two systems approach each other initially, but the repulsion between them forces them to move in opposite directions, and with more velocity than
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The two systems approach the transition state, and then the diatom splits, with the central atom pairing with the other terminal atom. They then proceed to move in opposite velocities.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|Initial velocity of particle C is so great that there is some repulsion between particle B and C, despite moving towards each other. As a result, particle B moves away from C after the initial attraction. It then oscillates around the transition state, moving from AB + C to A+ BC. It eventually settles with AB + C, the initial reactants. It can be seen by increasing the number of steps to 2000 that the final position of the two systems is greater separation. The particles AB and C are moving in opposite directions, so end up further apart than the initial conditions.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Oscillations around the transition state after an initial collision. This is very similar initially to the -10.1 p2 path. However, the rebound collision for A and B is even faster, resulting in another rebound for B. B then collides with particle C as they are both moving in the same directions, and the final state is that of the products, A + BC
|
|}

== Question 5: Transition State Theory Predictions ==

Transition state theory can be used to determine the rate of a reaction, but it relies on 5 assumptions:
# Once the kinetic energy has crossed the potential barrier, the reaction will proceed; there is no recrossing of the barrier.
# There are no quantum effects {specifically, no quantum tunnelling will take place}
# It is modelled on the Boltzmann distribution
# ask
# ask

Assumptions 1 and 2 are broken in these reaction dynamics, as we are considering classical collisions with some forces of attraction and repulsion between the atoms. As a result, despite the fact that crossing of the barrier can occur in a certain scenario, that activation barrier may be recrossed later. This is shown in the reaction pathway for the momenta p1 = -5.1 and p2 = -10.1, where despite the initial crossing of the barrier and consequent overcoming of the energy barrier, the system is not reactive. This therefore breaks from the mold of TST.

As TST suggests that some reactions that are not reactive are in fact reactive, it suggests that TST will overestimate how many reactions are successful. As a result, we can surmise that TST will overestimate the rate of reactions, compared to the GUI that we are using.

If the GUI agreed with assumption 1, TST would underestimate the reaction rate, as Tunnelling of the atoms would lead to a higher rate than the classical picture, as not all of the atoms have to reach the potential barrier to overcome it.
= Exercise 2: F-H-H system =

== Question 1: Energetics and bond strength ==

Let's set the initial position as <math>r_1</math> = <math>r_2</math>, where <math>r_1</math> is F-H, and <math>r_2</math> is H-H.

If we run a MEP plot from this position, the particle will end up at the lower energy state, regardless of the initial reactants or products.

We can see in the image below, the minimum energy path is such that <math>r_1</math> is minimised, and <math>r_2</math> is maximised. With <math>r_1</math> as F-H, this implies that the lower energy state is H + H-F.

[[File:MEP F-H-H.png]]

As a result, we can classify the F + H2 --> H + HF reaction as exothermic, as it is progressing from a higher energy state to a lower energy state.

H + HF ---> F + H2 would therefore be an endothermic process; it would require an input of energy to get this product.

This order of potential energies suggests that the bond strength between F-H is greater than for H2.

== Question 2: Transition state distance ==

As before, the transition state will be the saddle point on the surface plot. However, as the system is no longer symmetric as before, the transition state is no longer at the point where <math>r_1</math> = <math>r_2</math>.

Hammond's postulate suggests that the transition state will be closer to the higher energy system in the reaction. We know that this is the case for F + H2, so the transition state must be such that the bond distance for F-H is greater than for H2.

Knowing this, we can surmise that if AB is F-H and BC is H2, the distances at the transition state are: AB =184 pm and BC = 74.4 pm. This complies with the logic that the reaction where FH is the product is exothermic, as this is supported by Hammond's postulate.

[[File:Transition state ex2 contour.png]][[File:Transition state ex2 surface.png]]

We can once again, further confirm that this is the saddle point by using the same point that the forces along AB and BC at this point are both zero.

== Question 3: Activation energy ==

To find the activation energies of each product F-H and H2 (2 mutually exclusive scenarios), we simply need to find the difference between the transition state energy and the energy of the products. This can be calculated for the reaction only (excluding vibration) by using MEP to identify the shortest route.

Due to the asymmetry of the atoms, we have 2 activation energies.

Starting from the <math>r_{ts}</math>where AB =184 pm and BC = 74.4 pm, with a perturbation of ± 2 pm in the AB axis, and a step count of 2500, we get these results.

[[File:Transition state to minimum.png]]
[[File:Transition state to min en vs time.png]]

The plots above show the activation energy for the enothermic formation of the product F + H2 . The plots below show the activation energy for the exothermic formation of the products F-H + H.

[[File:Transition state to min 2.png]]

{| class="wikitable"
!Product
!Energy of Transition State (kJmol^-1)
!Energy of product
!ΔE
|-
|F-H + H
|<nowiki>-433.991</nowiki>
|<nowiki>-434.155</nowiki>
|0.164
|-
|F + H2
|<nowiki>-433.979</nowiki>
|<nowiki>-556.895</nowiki>
|122.916
|}

This is because we know that the energy of the transition state is the maximum energy state on the reaction path, and doesn't consider oscillations along the potential well; so we can just use the MEP.

These results are consistent with our understanding of the system, and the contour plot; it is an extremely early TS for the exothermic formation of HF + H, so the activation energy for this reaction is extremely low compared to the reverse, endothermic process. This can be further shown by the contour plot; the difference in energy is shown to be almost negligible as it is along the same contour line.

== Question 4: The release of Reaction Energy ==

hello

== Question 5: Efficiency of reaction ==

hello

File:Transition state to min 2.png

2020-05-08T14:14:55Z

Sa13018:

MRD:01512675

2020-05-08T14:14:23Z

Sa13018: /* Question 3: Activation energy */

= Exercise 1: H + H<math>_2</math> system =

== Question 1: The Transition State Region ==

The transition state is defined as the maximum point of the minimum energy path between the reactants and products. Therefore, on the potential energy surface diagram, it exists as a minimax; a saddle point.
This can therefore be described as the point where the partial differential of the potential energy with respect to all variables is zero.

In this case, due to a 3 dimensional surface plot, we have <math>
\frac{\partial V}{\partial r_1} = \frac{\partial V}{\partial r_2} = 0
</math>, where <math>r_1</math> and <math>r_2 </math> are the respective distances between AB and BC.

The local minimum point would see that a deviation in any direction would be an increase in potential energy.
However, the transition state is such that this would only happen in the axis where <math>r_1</math> = <math>r_2</math>, and would decrease in energy in any deviation of the AB or BC position.

This is shown in the images shown below.

[[File:Transition state minimum.png]]
[[File:Transition state maximum.png]]

The photo on the left shows the energy potential curve in a 2D viewpoint, as a function of the AB distance <math>r_1</math>. We can see that there is a steep increase of the V(<math>r_1</math>) function on either side of <math>r_1</math> = 74 pm, indicating the minimum in a potential well at this point. This continues as a valley between the reactants and products path.

The photo on the right shows a clear maximum point being mapped out by the black line. Despite being at a minimum point on the potential valley floor, this is a maximum point, indicating a minimax state. This is therefore a saddle point, and the transition state. Due to the symmetry of the system only consisting of H atoms, the transition state is observed at a point where <math>r_1</math> = <math>r_2</math>.

This can further be identified using the second partial derivative test; computing a Hessian matrix. If H <0, it must be a saddle.

== Question 2: Locating the transition state ==

Due to the symmetry between the H and the \chem{H_2} molecule, and the fact that the interaction is happening at 180º, we know that <math>r_1</math> must equal <math>r_2</math> for any point on the potential for the transition state. As a result, if both particles carry zero initial velocity and are on the transition state only, they will drop down the potential well to the transition state distance <math>r_{12}</math>.
Thus, in a scenario where <math>r_1</math> = <math>r_2</math>, and <math>v_2</math> = <math>v_1</math> = 0, the transition state will act only in a 2D potential plot; the dynamics will reach the minimum in the potential well.

This distance <math>r_12</math> is shown to be 90.775 pm, as there is no force along AB or BC at this distance when v = 0.
Furthermore, the contour plot shows the final position to be the same as the initial position; it is a point of equilibrium.
The internuclear distances vs time plot are also shown to constant here; this must therefore be the transition state.

[[File:Example.jpg]]
[[File:Example.jpg]]
[[File:Example.jpg]]

== Question 3: The Minimum Energy Path (MEP) ==

The MEP shows a clear difference in its plot and animation compared to the Dynamic calculation.
# There is no vibration/ oscillation in the MEP plot. This makes sense; it would be going against minimum energy route, as the oscillations would mean that the particle is not always at the minimum of the potential well.
# The MEP plot finishes once the minimum point of the energy potential surface is reached; after 193.1pm as the AB distance and 74 pm as the BC distance, the contour is consistent. However, we know that the particles must keep moving from this, as they carry opposite velocities. This is shown in the Dynamics plot, where the reaction path continues long after reaching the end of the x-label in the plot. This confirms that the MEP is relying only on the minimum of the potential energy surface; not the distances AB and BC.
This is shown in the comparison below.

It is also interesting to note the action of the MEP when we start from the reactants, as before, with enough energy to go to the products in the Dynamics. This is shown in the comparison below.

In this scenario, despite having enough initial velocity in one direction, the MEP doesn't follow that path. It reverses the direction of the initial velocity and instead ends up with no reaction taking place. This implies that it has followed the minimum energy path at every point in the plot, ignoring the vectors that the particle is following. This further exemplifies the difference between Dynamics and MEP, showing that MEP is only taking the mathematically logical route, whereas the Dynamics considers the physical process that is occuring.

== Question 4: Reactive and Unreactive Trajectories ==

The following table assesses the plot for the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, with varying initial momenta.
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Proceeds as normal. No initial collision, but repulsion between the two systems is overcome, etc.
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The initial velocity of the atom is far too slow to collide with the molecule. The two systems approach each other initially, but the repulsion between them forces them to move in opposite directions, and with more velocity than
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The two systems approach the transition state, and then the diatom splits, with the central atom pairing with the other terminal atom. They then proceed to move in opposite velocities.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|Initial velocity of particle C is so great that there is some repulsion between particle B and C, despite moving towards each other. As a result, particle B moves away from C after the initial attraction. It then oscillates around the transition state, moving from AB + C to A+ BC. It eventually settles with AB + C, the initial reactants. It can be seen by increasing the number of steps to 2000 that the final position of the two systems is greater separation. The particles AB and C are moving in opposite directions, so end up further apart than the initial conditions.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Oscillations around the transition state after an initial collision. This is very similar initially to the -10.1 p2 path. However, the rebound collision for A and B is even faster, resulting in another rebound for B. B then collides with particle C as they are both moving in the same directions, and the final state is that of the products, A + BC
|
|}

== Question 5: Transition State Theory Predictions ==

Transition state theory can be used to determine the rate of a reaction, but it relies on 5 assumptions:
# Once the kinetic energy has crossed the potential barrier, the reaction will proceed; there is no recrossing of the barrier.
# There are no quantum effects {specifically, no quantum tunnelling will take place}
# It is modelled on the Boltzmann distribution
# ask
# ask

Assumptions 1 and 2 are broken in these reaction dynamics, as we are considering classical collisions with some forces of attraction and repulsion between the atoms. As a result, despite the fact that crossing of the barrier can occur in a certain scenario, that activation barrier may be recrossed later. This is shown in the reaction pathway for the momenta p1 = -5.1 and p2 = -10.1, where despite the initial crossing of the barrier and consequent overcoming of the energy barrier, the system is not reactive. This therefore breaks from the mold of TST.

As TST suggests that some reactions that are not reactive are in fact reactive, it suggests that TST will overestimate how many reactions are successful. As a result, we can surmise that TST will overestimate the rate of reactions, compared to the GUI that we are using.

If the GUI agreed with assumption 1, TST would underestimate the reaction rate, as Tunnelling of the atoms would lead to a higher rate than the classical picture, as not all of the atoms have to reach the potential barrier to overcome it.
= Exercise 2: F-H-H system =

== Question 1: Energetics and bond strength ==

Let's set the initial position as <math>r_1</math> = <math>r_2</math>, where <math>r_1</math> is F-H, and <math>r_2</math> is H-H.

If we run a MEP plot from this position, the particle will end up at the lower energy state, regardless of the initial reactants or products.

We can see in the image below, the minimum energy path is such that <math>r_1</math> is minimised, and <math>r_2</math> is maximised. With <math>r_1</math> as F-H, this implies that the lower energy state is H + H-F.

[[File:MEP F-H-H.png]]

As a result, we can classify the F + H2 --> H + HF reaction as exothermic, as it is progressing from a higher energy state to a lower energy state.

H + HF ---> F + H2 would therefore be an endothermic process; it would require an input of energy to get this product.

This order of potential energies suggests that the bond strength between F-H is greater than for H2.

== Question 2: Transition state distance ==

As before, the transition state will be the saddle point on the surface plot. However, as the system is no longer symmetric as before, the transition state is no longer at the point where <math>r_1</math> = <math>r_2</math>.

Hammond's postulate suggests that the transition state will be closer to the higher energy system in the reaction. We know that this is the case for F + H2, so the transition state must be such that the bond distance for F-H is greater than for H2.

Knowing this, we can surmise that if AB is F-H and BC is H2, the distances at the transition state are: AB =184 pm and BC = 74.4 pm. This complies with the logic that the reaction where FH is the product is exothermic, as this is supported by Hammond's postulate.

[[File:Transition state ex2 contour.png]][[File:Transition state ex2 surface.png]]

We can once again, further confirm that this is the saddle point by using the same point that the forces along AB and BC at this point are both zero.

== Question 3: Activation energy ==

To find the activation energies of each product F-H and H2 (2 mutually exclusive scenarios), we simply need to find the difference between the transition state energy and the energy of the products. This can be calculated for the reaction only (excluding vibration) by using MEP to identify the shortest route.

Due to the asymmetry of the atoms, we have 2 activation energies.

Starting from the <math>r_{ts}</math>where AB =184 pm and BC = 74.4 pm, with a perturbation of ± 2 pm in the AB axis, and a step count of 2500, we get these results.

[[File:Transition state to minimum.png]]
[[File:Transition state to min en vs time.png]]

The plots above show the activation energy for the enothermic formation of the product F + H2 . The plots below show the activation energy for the exothermic formation of the products F-H + H.

[[File:Example.jpg]][[File:Example.jpg]]

{| class="wikitable"
!Product
!Energy of Transition State (kJmol^-1)
!Energy of product
!ΔE
|-
|F-H + H
|<nowiki>-433.991</nowiki>
|<nowiki>-434.155</nowiki>
|0.164
|-
|F + H2
|<nowiki>-433.979</nowiki>
|<nowiki>-556.895</nowiki>
|122.916
|}

This is because we know that the energy of the transition state is the maximum energy state on the reaction path, and doesn't consider oscillations along the potential well; so we can just use the MEP.

These results are consistent with our understanding of the system, and the contour plot; it is an extremely early TS for the exothermic formation of HF + H, so the activation energy for this reaction is extremely low compared to the reverse, endothermic process. This can be further shown by the contour plot; the difference in energy is shown to be almost negligible as it is along the same contour line.

== Question 4: The release of Reaction Energy ==

hello

== Question 5: Efficiency of reaction ==

hello

MRD:01512675

2020-05-08T14:03:15Z

Sa13018: /* Question 3: Activation energy */

= Exercise 1: H + H<math>_2</math> system =

== Question 1: The Transition State Region ==

The transition state is defined as the maximum point of the minimum energy path between the reactants and products. Therefore, on the potential energy surface diagram, it exists as a minimax; a saddle point.
This can therefore be described as the point where the partial differential of the potential energy with respect to all variables is zero.

In this case, due to a 3 dimensional surface plot, we have <math>
\frac{\partial V}{\partial r_1} = \frac{\partial V}{\partial r_2} = 0
</math>, where <math>r_1</math> and <math>r_2 </math> are the respective distances between AB and BC.

The local minimum point would see that a deviation in any direction would be an increase in potential energy.
However, the transition state is such that this would only happen in the axis where <math>r_1</math> = <math>r_2</math>, and would decrease in energy in any deviation of the AB or BC position.

This is shown in the images shown below.

[[File:Transition state minimum.png]]
[[File:Transition state maximum.png]]

The photo on the left shows the energy potential curve in a 2D viewpoint, as a function of the AB distance <math>r_1</math>. We can see that there is a steep increase of the V(<math>r_1</math>) function on either side of <math>r_1</math> = 74 pm, indicating the minimum in a potential well at this point. This continues as a valley between the reactants and products path.

The photo on the right shows a clear maximum point being mapped out by the black line. Despite being at a minimum point on the potential valley floor, this is a maximum point, indicating a minimax state. This is therefore a saddle point, and the transition state. Due to the symmetry of the system only consisting of H atoms, the transition state is observed at a point where <math>r_1</math> = <math>r_2</math>.

This can further be identified using the second partial derivative test; computing a Hessian matrix. If H <0, it must be a saddle.

== Question 2: Locating the transition state ==

Due to the symmetry between the H and the \chem{H_2} molecule, and the fact that the interaction is happening at 180º, we know that <math>r_1</math> must equal <math>r_2</math> for any point on the potential for the transition state. As a result, if both particles carry zero initial velocity and are on the transition state only, they will drop down the potential well to the transition state distance <math>r_{12}</math>.
Thus, in a scenario where <math>r_1</math> = <math>r_2</math>, and <math>v_2</math> = <math>v_1</math> = 0, the transition state will act only in a 2D potential plot; the dynamics will reach the minimum in the potential well.

This distance <math>r_12</math> is shown to be 90.775 pm, as there is no force along AB or BC at this distance when v = 0.
Furthermore, the contour plot shows the final position to be the same as the initial position; it is a point of equilibrium.
The internuclear distances vs time plot are also shown to constant here; this must therefore be the transition state.

[[File:Example.jpg]]
[[File:Example.jpg]]
[[File:Example.jpg]]

== Question 3: The Minimum Energy Path (MEP) ==

The MEP shows a clear difference in its plot and animation compared to the Dynamic calculation.
# There is no vibration/ oscillation in the MEP plot. This makes sense; it would be going against minimum energy route, as the oscillations would mean that the particle is not always at the minimum of the potential well.
# The MEP plot finishes once the minimum point of the energy potential surface is reached; after 193.1pm as the AB distance and 74 pm as the BC distance, the contour is consistent. However, we know that the particles must keep moving from this, as they carry opposite velocities. This is shown in the Dynamics plot, where the reaction path continues long after reaching the end of the x-label in the plot. This confirms that the MEP is relying only on the minimum of the potential energy surface; not the distances AB and BC.
This is shown in the comparison below.

It is also interesting to note the action of the MEP when we start from the reactants, as before, with enough energy to go to the products in the Dynamics. This is shown in the comparison below.

In this scenario, despite having enough initial velocity in one direction, the MEP doesn't follow that path. It reverses the direction of the initial velocity and instead ends up with no reaction taking place. This implies that it has followed the minimum energy path at every point in the plot, ignoring the vectors that the particle is following. This further exemplifies the difference between Dynamics and MEP, showing that MEP is only taking the mathematically logical route, whereas the Dynamics considers the physical process that is occuring.

== Question 4: Reactive and Unreactive Trajectories ==

The following table assesses the plot for the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, with varying initial momenta.
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Proceeds as normal. No initial collision, but repulsion between the two systems is overcome, etc.
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The initial velocity of the atom is far too slow to collide with the molecule. The two systems approach each other initially, but the repulsion between them forces them to move in opposite directions, and with more velocity than
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The two systems approach the transition state, and then the diatom splits, with the central atom pairing with the other terminal atom. They then proceed to move in opposite velocities.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|Initial velocity of particle C is so great that there is some repulsion between particle B and C, despite moving towards each other. As a result, particle B moves away from C after the initial attraction. It then oscillates around the transition state, moving from AB + C to A+ BC. It eventually settles with AB + C, the initial reactants. It can be seen by increasing the number of steps to 2000 that the final position of the two systems is greater separation. The particles AB and C are moving in opposite directions, so end up further apart than the initial conditions.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Oscillations around the transition state after an initial collision. This is very similar initially to the -10.1 p2 path. However, the rebound collision for A and B is even faster, resulting in another rebound for B. B then collides with particle C as they are both moving in the same directions, and the final state is that of the products, A + BC
|
|}

== Question 5: Transition State Theory Predictions ==

Transition state theory can be used to determine the rate of a reaction, but it relies on 5 assumptions:
# Once the kinetic energy has crossed the potential barrier, the reaction will proceed; there is no recrossing of the barrier.
# There are no quantum effects {specifically, no quantum tunnelling will take place}
# It is modelled on the Boltzmann distribution
# ask
# ask

Assumptions 1 and 2 are broken in these reaction dynamics, as we are considering classical collisions with some forces of attraction and repulsion between the atoms. As a result, despite the fact that crossing of the barrier can occur in a certain scenario, that activation barrier may be recrossed later. This is shown in the reaction pathway for the momenta p1 = -5.1 and p2 = -10.1, where despite the initial crossing of the barrier and consequent overcoming of the energy barrier, the system is not reactive. This therefore breaks from the mold of TST.

As TST suggests that some reactions that are not reactive are in fact reactive, it suggests that TST will overestimate how many reactions are successful. As a result, we can surmise that TST will overestimate the rate of reactions, compared to the GUI that we are using.

If the GUI agreed with assumption 1, TST would underestimate the reaction rate, as Tunnelling of the atoms would lead to a higher rate than the classical picture, as not all of the atoms have to reach the potential barrier to overcome it.
= Exercise 2: F-H-H system =

== Question 1: Energetics and bond strength ==

Let's set the initial position as <math>r_1</math> = <math>r_2</math>, where <math>r_1</math> is F-H, and <math>r_2</math> is H-H.

If we run a MEP plot from this position, the particle will end up at the lower energy state, regardless of the initial reactants or products.

We can see in the image below, the minimum energy path is such that <math>r_1</math> is minimised, and <math>r_2</math> is maximised. With <math>r_1</math> as F-H, this implies that the lower energy state is H + H-F.

[[File:MEP F-H-H.png]]

As a result, we can classify the F + H2 --> H + HF reaction as exothermic, as it is progressing from a higher energy state to a lower energy state.

H + HF ---> F + H2 would therefore be an endothermic process; it would require an input of energy to get this product.

This order of potential energies suggests that the bond strength between F-H is greater than for H2.

== Question 2: Transition state distance ==

As before, the transition state will be the saddle point on the surface plot. However, as the system is no longer symmetric as before, the transition state is no longer at the point where <math>r_1</math> = <math>r_2</math>.

Hammond's postulate suggests that the transition state will be closer to the higher energy system in the reaction. We know that this is the case for F + H2, so the transition state must be such that the bond distance for F-H is greater than for H2.

Knowing this, we can surmise that if AB is F-H and BC is H2, the distances at the transition state are: AB =184 pm and BC = 74.4 pm. This complies with the logic that the reaction where FH is the product is exothermic, as this is supported by Hammond's postulate.

[[File:Transition state ex2 contour.png]][[File:Transition state ex2 surface.png]]

We can once again, further confirm that this is the saddle point by using the same point that the forces along AB and BC at this point are both zero.

== Question 3: Activation energy ==

To find the activation energies of each product F-H and H2 (2 mutually exclusive scenarios), we simply need to find the difference between the transition state energy and the energy of the products. This can be calculated for the reaction only (excluding vibration) by using MEP to identify the shortest route.

Due to the asymmetry of the atoms, we have 2 activation energies.

Starting from the <math>r_{ts}</math>where AB =184 pm and BC = 74.4 pm, with a perturbation of ± 2 pm in the AB axis, and a step count of 2500, we get these results.

[[File:Transition state to minimum.png]]
[[File:Transition state to min en vs time.png]]

The plots above show the activation energy for the enothermic formation of the product F + H2 . The plots below show the activation energy for the exothermic formation of the products F-H + H.

[[File:Example.jpg]][[File:Example.jpg]]

{| class="wikitable"
!Product
!Energy of Transition State
!Energy of product
!ΔE
|-
|F-H + H
|
|
|
|-
|F + H2
|
|
|
|}

This is because we know that the energy of the transition state is the maximum energy state on the reaction path, and doesn't consider oscillations along the potential well.

== Question 4: The release of Reaction Energy ==

hello

== Question 5: Efficiency of reaction ==

hello

MRD:01512675

2020-05-08T13:03:46Z

Sa13018: /* Question 3: Activation energy */

= Exercise 1: H + H<math>_2</math> system =

== Question 1: The Transition State Region ==

The transition state is defined as the maximum point of the minimum energy path between the reactants and products. Therefore, on the potential energy surface diagram, it exists as a minimax; a saddle point.
This can therefore be described as the point where the partial differential of the potential energy with respect to all variables is zero.

In this case, due to a 3 dimensional surface plot, we have <math>
\frac{\partial V}{\partial r_1} = \frac{\partial V}{\partial r_2} = 0
</math>, where <math>r_1</math> and <math>r_2 </math> are the respective distances between AB and BC.

The local minimum point would see that a deviation in any direction would be an increase in potential energy.
However, the transition state is such that this would only happen in the axis where <math>r_1</math> = <math>r_2</math>, and would decrease in energy in any deviation of the AB or BC position.

This is shown in the images shown below.

[[File:Transition state minimum.png]]
[[File:Transition state maximum.png]]

The photo on the left shows the energy potential curve in a 2D viewpoint, as a function of the AB distance <math>r_1</math>. We can see that there is a steep increase of the V(<math>r_1</math>) function on either side of <math>r_1</math> = 74 pm, indicating the minimum in a potential well at this point. This continues as a valley between the reactants and products path.

The photo on the right shows a clear maximum point being mapped out by the black line. Despite being at a minimum point on the potential valley floor, this is a maximum point, indicating a minimax state. This is therefore a saddle point, and the transition state. Due to the symmetry of the system only consisting of H atoms, the transition state is observed at a point where <math>r_1</math> = <math>r_2</math>.

This can further be identified using the second partial derivative test; computing a Hessian matrix. If H <0, it must be a saddle.

== Question 2: Locating the transition state ==

Due to the symmetry between the H and the \chem{H_2} molecule, and the fact that the interaction is happening at 180º, we know that <math>r_1</math> must equal <math>r_2</math> for any point on the potential for the transition state. As a result, if both particles carry zero initial velocity and are on the transition state only, they will drop down the potential well to the transition state distance <math>r_{12}</math>.
Thus, in a scenario where <math>r_1</math> = <math>r_2</math>, and <math>v_2</math> = <math>v_1</math> = 0, the transition state will act only in a 2D potential plot; the dynamics will reach the minimum in the potential well.

This distance <math>r_12</math> is shown to be 90.775 pm, as there is no force along AB or BC at this distance when v = 0.
Furthermore, the contour plot shows the final position to be the same as the initial position; it is a point of equilibrium.
The internuclear distances vs time plot are also shown to constant here; this must therefore be the transition state.

[[File:Example.jpg]]
[[File:Example.jpg]]
[[File:Example.jpg]]

== Question 3: The Minimum Energy Path (MEP) ==

The MEP shows a clear difference in its plot and animation compared to the Dynamic calculation.
# There is no vibration/ oscillation in the MEP plot. This makes sense; it would be going against minimum energy route, as the oscillations would mean that the particle is not always at the minimum of the potential well.
# The MEP plot finishes once the minimum point of the energy potential surface is reached; after 193.1pm as the AB distance and 74 pm as the BC distance, the contour is consistent. However, we know that the particles must keep moving from this, as they carry opposite velocities. This is shown in the Dynamics plot, where the reaction path continues long after reaching the end of the x-label in the plot. This confirms that the MEP is relying only on the minimum of the potential energy surface; not the distances AB and BC.
This is shown in the comparison below.

It is also interesting to note the action of the MEP when we start from the reactants, as before, with enough energy to go to the products in the Dynamics. This is shown in the comparison below.

In this scenario, despite having enough initial velocity in one direction, the MEP doesn't follow that path. It reverses the direction of the initial velocity and instead ends up with no reaction taking place. This implies that it has followed the minimum energy path at every point in the plot, ignoring the vectors that the particle is following. This further exemplifies the difference between Dynamics and MEP, showing that MEP is only taking the mathematically logical route, whereas the Dynamics considers the physical process that is occuring.

== Question 4: Reactive and Unreactive Trajectories ==

The following table assesses the plot for the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, with varying initial momenta.
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Proceeds as normal. No initial collision, but repulsion between the two systems is overcome, etc.
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The initial velocity of the atom is far too slow to collide with the molecule. The two systems approach each other initially, but the repulsion between them forces them to move in opposite directions, and with more velocity than
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The two systems approach the transition state, and then the diatom splits, with the central atom pairing with the other terminal atom. They then proceed to move in opposite velocities.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|Initial velocity of particle C is so great that there is some repulsion between particle B and C, despite moving towards each other. As a result, particle B moves away from C after the initial attraction. It then oscillates around the transition state, moving from AB + C to A+ BC. It eventually settles with AB + C, the initial reactants. It can be seen by increasing the number of steps to 2000 that the final position of the two systems is greater separation. The particles AB and C are moving in opposite directions, so end up further apart than the initial conditions.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Oscillations around the transition state after an initial collision. This is very similar initially to the -10.1 p2 path. However, the rebound collision for A and B is even faster, resulting in another rebound for B. B then collides with particle C as they are both moving in the same directions, and the final state is that of the products, A + BC
|
|}

== Question 5: Transition State Theory Predictions ==

Transition state theory can be used to determine the rate of a reaction, but it relies on 5 assumptions:
# Once the kinetic energy has crossed the potential barrier, the reaction will proceed; there is no recrossing of the barrier.
# There are no quantum effects {specifically, no quantum tunnelling will take place}
# It is modelled on the Boltzmann distribution
# ask
# ask

Assumptions 1 and 2 are broken in these reaction dynamics, as we are considering classical collisions with some forces of attraction and repulsion between the atoms. As a result, despite the fact that crossing of the barrier can occur in a certain scenario, that activation barrier may be recrossed later. This is shown in the reaction pathway for the momenta p1 = -5.1 and p2 = -10.1, where despite the initial crossing of the barrier and consequent overcoming of the energy barrier, the system is not reactive. This therefore breaks from the mold of TST.

As TST suggests that some reactions that are not reactive are in fact reactive, it suggests that TST will overestimate how many reactions are successful. As a result, we can surmise that TST will overestimate the rate of reactions, compared to the GUI that we are using.

If the GUI agreed with assumption 1, TST would underestimate the reaction rate, as Tunnelling of the atoms would lead to a higher rate than the classical picture, as not all of the atoms have to reach the potential barrier to overcome it.
= Exercise 2: F-H-H system =

== Question 1: Energetics and bond strength ==

Let's set the initial position as <math>r_1</math> = <math>r_2</math>, where <math>r_1</math> is F-H, and <math>r_2</math> is H-H.

If we run a MEP plot from this position, the particle will end up at the lower energy state, regardless of the initial reactants or products.

We can see in the image below, the minimum energy path is such that <math>r_1</math> is minimised, and <math>r_2</math> is maximised. With <math>r_1</math> as F-H, this implies that the lower energy state is H + H-F.

[[File:MEP F-H-H.png]]

As a result, we can classify the F + H2 --> H + HF reaction as exothermic, as it is progressing from a higher energy state to a lower energy state.

H + HF ---> F + H2 would therefore be an endothermic process; it would require an input of energy to get this product.

This order of potential energies suggests that the bond strength between F-H is greater than for H2.

== Question 2: Transition state distance ==

As before, the transition state will be the saddle point on the surface plot. However, as the system is no longer symmetric as before, the transition state is no longer at the point where <math>r_1</math> = <math>r_2</math>.

Hammond's postulate suggests that the transition state will be closer to the higher energy system in the reaction. We know that this is the case for F + H2, so the transition state must be such that the bond distance for F-H is greater than for H2.

Knowing this, we can surmise that if AB is F-H and BC is H2, the distances at the transition state are: AB =184 pm and BC = 74.4 pm. This complies with the logic that the reaction where FH is the product is exothermic, as this is supported by Hammond's postulate.

[[File:Transition state ex2 contour.png]][[File:Transition state ex2 surface.png]]

We can once again, further confirm that this is the saddle point by using the same point that the forces along AB and BC at this point are both zero.

== Question 3: Activation energy ==

To find the activation energies of each product F-H and H2 (2 mutually exclusive scenarios), we simply need to find the difference between the transition state energy and the energy of the products. This can be calculated for the reaction only (excluding vibration) by using MEP to identify the shortest route.

Due to the asymmetry of the atoms, we have 2 activation energies.

Starting from the <math>r_{ts}</math>where AB =184 pm and BC = 74.4 pm, with a perturbation of ± 2 pm in the AB axis, and a step count of 2500, we get these results.

[[File:Transition state to minimum.png]]
[[File:Transition state to min en vs time.png]]

The plots above show the activation energy for the enothermic formation of the product F + H2 . The plots below show the activation energy for the exothermic formation of the products F-H + H.

[[File:Example.jpg]][[File:Example.jpg]]

{| class="wikitable"
!Product
!Energy initial
!Energy final
!ΔE
|-
|F-H + H
|
|
|
|-
|F + H2
|
|
|
|}

This is because we know that the energy of the transition state is the maximum energy state on the reaction path, and doesn't consider oscillations along the potential well.

== Question 4: The release of Reaction Energy ==

hello

== Question 5: Efficiency of reaction ==

hello

MRD:01512675

2020-05-08T12:58:22Z

Sa13018: /* Question 3: Activation energy */

= Exercise 1: H + H<math>_2</math> system =

== Question 1: The Transition State Region ==

The transition state is defined as the maximum point of the minimum energy path between the reactants and products. Therefore, on the potential energy surface diagram, it exists as a minimax; a saddle point.
This can therefore be described as the point where the partial differential of the potential energy with respect to all variables is zero.

In this case, due to a 3 dimensional surface plot, we have <math>
\frac{\partial V}{\partial r_1} = \frac{\partial V}{\partial r_2} = 0
</math>, where <math>r_1</math> and <math>r_2 </math> are the respective distances between AB and BC.

The local minimum point would see that a deviation in any direction would be an increase in potential energy.
However, the transition state is such that this would only happen in the axis where <math>r_1</math> = <math>r_2</math>, and would decrease in energy in any deviation of the AB or BC position.

This is shown in the images shown below.

[[File:Transition state minimum.png]]
[[File:Transition state maximum.png]]

The photo on the left shows the energy potential curve in a 2D viewpoint, as a function of the AB distance <math>r_1</math>. We can see that there is a steep increase of the V(<math>r_1</math>) function on either side of <math>r_1</math> = 74 pm, indicating the minimum in a potential well at this point. This continues as a valley between the reactants and products path.

The photo on the right shows a clear maximum point being mapped out by the black line. Despite being at a minimum point on the potential valley floor, this is a maximum point, indicating a minimax state. This is therefore a saddle point, and the transition state. Due to the symmetry of the system only consisting of H atoms, the transition state is observed at a point where <math>r_1</math> = <math>r_2</math>.

This can further be identified using the second partial derivative test; computing a Hessian matrix. If H <0, it must be a saddle.

== Question 2: Locating the transition state ==

Due to the symmetry between the H and the \chem{H_2} molecule, and the fact that the interaction is happening at 180º, we know that <math>r_1</math> must equal <math>r_2</math> for any point on the potential for the transition state. As a result, if both particles carry zero initial velocity and are on the transition state only, they will drop down the potential well to the transition state distance <math>r_{12}</math>.
Thus, in a scenario where <math>r_1</math> = <math>r_2</math>, and <math>v_2</math> = <math>v_1</math> = 0, the transition state will act only in a 2D potential plot; the dynamics will reach the minimum in the potential well.

This distance <math>r_12</math> is shown to be 90.775 pm, as there is no force along AB or BC at this distance when v = 0.
Furthermore, the contour plot shows the final position to be the same as the initial position; it is a point of equilibrium.
The internuclear distances vs time plot are also shown to constant here; this must therefore be the transition state.

[[File:Example.jpg]]
[[File:Example.jpg]]
[[File:Example.jpg]]

== Question 3: The Minimum Energy Path (MEP) ==

The MEP shows a clear difference in its plot and animation compared to the Dynamic calculation.
# There is no vibration/ oscillation in the MEP plot. This makes sense; it would be going against minimum energy route, as the oscillations would mean that the particle is not always at the minimum of the potential well.
# The MEP plot finishes once the minimum point of the energy potential surface is reached; after 193.1pm as the AB distance and 74 pm as the BC distance, the contour is consistent. However, we know that the particles must keep moving from this, as they carry opposite velocities. This is shown in the Dynamics plot, where the reaction path continues long after reaching the end of the x-label in the plot. This confirms that the MEP is relying only on the minimum of the potential energy surface; not the distances AB and BC.
This is shown in the comparison below.

It is also interesting to note the action of the MEP when we start from the reactants, as before, with enough energy to go to the products in the Dynamics. This is shown in the comparison below.

In this scenario, despite having enough initial velocity in one direction, the MEP doesn't follow that path. It reverses the direction of the initial velocity and instead ends up with no reaction taking place. This implies that it has followed the minimum energy path at every point in the plot, ignoring the vectors that the particle is following. This further exemplifies the difference between Dynamics and MEP, showing that MEP is only taking the mathematically logical route, whereas the Dynamics considers the physical process that is occuring.

== Question 4: Reactive and Unreactive Trajectories ==

The following table assesses the plot for the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, with varying initial momenta.
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Proceeds as normal. No initial collision, but repulsion between the two systems is overcome, etc.
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The initial velocity of the atom is far too slow to collide with the molecule. The two systems approach each other initially, but the repulsion between them forces them to move in opposite directions, and with more velocity than
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The two systems approach the transition state, and then the diatom splits, with the central atom pairing with the other terminal atom. They then proceed to move in opposite velocities.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|Initial velocity of particle C is so great that there is some repulsion between particle B and C, despite moving towards each other. As a result, particle B moves away from C after the initial attraction. It then oscillates around the transition state, moving from AB + C to A+ BC. It eventually settles with AB + C, the initial reactants. It can be seen by increasing the number of steps to 2000 that the final position of the two systems is greater separation. The particles AB and C are moving in opposite directions, so end up further apart than the initial conditions.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Oscillations around the transition state after an initial collision. This is very similar initially to the -10.1 p2 path. However, the rebound collision for A and B is even faster, resulting in another rebound for B. B then collides with particle C as they are both moving in the same directions, and the final state is that of the products, A + BC
|
|}

== Question 5: Transition State Theory Predictions ==

Transition state theory can be used to determine the rate of a reaction, but it relies on 5 assumptions:
# Once the kinetic energy has crossed the potential barrier, the reaction will proceed; there is no recrossing of the barrier.
# There are no quantum effects {specifically, no quantum tunnelling will take place}
# It is modelled on the Boltzmann distribution
# ask
# ask

Assumptions 1 and 2 are broken in these reaction dynamics, as we are considering classical collisions with some forces of attraction and repulsion between the atoms. As a result, despite the fact that crossing of the barrier can occur in a certain scenario, that activation barrier may be recrossed later. This is shown in the reaction pathway for the momenta p1 = -5.1 and p2 = -10.1, where despite the initial crossing of the barrier and consequent overcoming of the energy barrier, the system is not reactive. This therefore breaks from the mold of TST.

As TST suggests that some reactions that are not reactive are in fact reactive, it suggests that TST will overestimate how many reactions are successful. As a result, we can surmise that TST will overestimate the rate of reactions, compared to the GUI that we are using.

If the GUI agreed with assumption 1, TST would underestimate the reaction rate, as Tunnelling of the atoms would lead to a higher rate than the classical picture, as not all of the atoms have to reach the potential barrier to overcome it.
= Exercise 2: F-H-H system =

== Question 1: Energetics and bond strength ==

Let's set the initial position as <math>r_1</math> = <math>r_2</math>, where <math>r_1</math> is F-H, and <math>r_2</math> is H-H.

If we run a MEP plot from this position, the particle will end up at the lower energy state, regardless of the initial reactants or products.

We can see in the image below, the minimum energy path is such that <math>r_1</math> is minimised, and <math>r_2</math> is maximised. With <math>r_1</math> as F-H, this implies that the lower energy state is H + H-F.

[[File:MEP F-H-H.png]]

As a result, we can classify the F + H2 --> H + HF reaction as exothermic, as it is progressing from a higher energy state to a lower energy state.

H + HF ---> F + H2 would therefore be an endothermic process; it would require an input of energy to get this product.

This order of potential energies suggests that the bond strength between F-H is greater than for H2.

== Question 2: Transition state distance ==

As before, the transition state will be the saddle point on the surface plot. However, as the system is no longer symmetric as before, the transition state is no longer at the point where <math>r_1</math> = <math>r_2</math>.

Hammond's postulate suggests that the transition state will be closer to the higher energy system in the reaction. We know that this is the case for F + H2, so the transition state must be such that the bond distance for F-H is greater than for H2.

Knowing this, we can surmise that if AB is F-H and BC is H2, the distances at the transition state are: AB =184 pm and BC = 74.4 pm. This complies with the logic that the reaction where FH is the product is exothermic, as this is supported by Hammond's postulate.

[[File:Transition state ex2 contour.png]][[File:Transition state ex2 surface.png]]

We can once again, further confirm that this is the saddle point by using the same point that the forces along AB and BC at this point are both zero.

== Question 3: Activation energy ==

To find the activation energies of each product F-H and H2 (2 mutually exclusive scenarios), we simply need to find the difference between the transition state energy and the energy of the products. This can be calculated for the reaction only (excluding vibration) by using MEP to identify the shortest route.

Due to the asymmetry of the atoms, we have 2 activation energies.

[[File:Transition state to minimum.png]]
[[File:Transition state to min en vs time.png]]

[[File:Example.jpg]][[File:Example.jpg]]

Starting from the <math>r_{ts}</math> = AB =184 pm and BC = 74.4 pm, with a perturbation of ± 5 pm in the AB axis.
{| class="wikitable"
!Product
!Energy initial
!Energy final
!ΔE
|-
|F-H + H
|
|
|
|-
|F + H2
|
|
|
|}

This is because we know that the energy of the transition state is the maximum energy state on the reaction path, and doesn't consider oscillations along the potential well.

== Question 4: Mechanism of the Release of reaction energy ==

hello

== Question 5: Efficiency of reaction ==

hello

MRD:01512675

2020-05-08T12:58:03Z

Sa13018: /* Question 3: Activation energy */

= Exercise 1: H + H<math>_2</math> system =

== Question 1: The Transition State Region ==

The transition state is defined as the maximum point of the minimum energy path between the reactants and products. Therefore, on the potential energy surface diagram, it exists as a minimax; a saddle point.
This can therefore be described as the point where the partial differential of the potential energy with respect to all variables is zero.

In this case, due to a 3 dimensional surface plot, we have <math>
\frac{\partial V}{\partial r_1} = \frac{\partial V}{\partial r_2} = 0
</math>, where <math>r_1</math> and <math>r_2 </math> are the respective distances between AB and BC.

The local minimum point would see that a deviation in any direction would be an increase in potential energy.
However, the transition state is such that this would only happen in the axis where <math>r_1</math> = <math>r_2</math>, and would decrease in energy in any deviation of the AB or BC position.

This is shown in the images shown below.

[[File:Transition state minimum.png]]
[[File:Transition state maximum.png]]

The photo on the left shows the energy potential curve in a 2D viewpoint, as a function of the AB distance <math>r_1</math>. We can see that there is a steep increase of the V(<math>r_1</math>) function on either side of <math>r_1</math> = 74 pm, indicating the minimum in a potential well at this point. This continues as a valley between the reactants and products path.

The photo on the right shows a clear maximum point being mapped out by the black line. Despite being at a minimum point on the potential valley floor, this is a maximum point, indicating a minimax state. This is therefore a saddle point, and the transition state. Due to the symmetry of the system only consisting of H atoms, the transition state is observed at a point where <math>r_1</math> = <math>r_2</math>.

This can further be identified using the second partial derivative test; computing a Hessian matrix. If H <0, it must be a saddle.

== Question 2: Locating the transition state ==

Due to the symmetry between the H and the \chem{H_2} molecule, and the fact that the interaction is happening at 180º, we know that <math>r_1</math> must equal <math>r_2</math> for any point on the potential for the transition state. As a result, if both particles carry zero initial velocity and are on the transition state only, they will drop down the potential well to the transition state distance <math>r_{12}</math>.
Thus, in a scenario where <math>r_1</math> = <math>r_2</math>, and <math>v_2</math> = <math>v_1</math> = 0, the transition state will act only in a 2D potential plot; the dynamics will reach the minimum in the potential well.

This distance <math>r_12</math> is shown to be 90.775 pm, as there is no force along AB or BC at this distance when v = 0.
Furthermore, the contour plot shows the final position to be the same as the initial position; it is a point of equilibrium.
The internuclear distances vs time plot are also shown to constant here; this must therefore be the transition state.

[[File:Example.jpg]]
[[File:Example.jpg]]
[[File:Example.jpg]]

== Question 3: The Minimum Energy Path (MEP) ==

The MEP shows a clear difference in its plot and animation compared to the Dynamic calculation.
# There is no vibration/ oscillation in the MEP plot. This makes sense; it would be going against minimum energy route, as the oscillations would mean that the particle is not always at the minimum of the potential well.
# The MEP plot finishes once the minimum point of the energy potential surface is reached; after 193.1pm as the AB distance and 74 pm as the BC distance, the contour is consistent. However, we know that the particles must keep moving from this, as they carry opposite velocities. This is shown in the Dynamics plot, where the reaction path continues long after reaching the end of the x-label in the plot. This confirms that the MEP is relying only on the minimum of the potential energy surface; not the distances AB and BC.
This is shown in the comparison below.

It is also interesting to note the action of the MEP when we start from the reactants, as before, with enough energy to go to the products in the Dynamics. This is shown in the comparison below.

In this scenario, despite having enough initial velocity in one direction, the MEP doesn't follow that path. It reverses the direction of the initial velocity and instead ends up with no reaction taking place. This implies that it has followed the minimum energy path at every point in the plot, ignoring the vectors that the particle is following. This further exemplifies the difference between Dynamics and MEP, showing that MEP is only taking the mathematically logical route, whereas the Dynamics considers the physical process that is occuring.

== Question 4: Reactive and Unreactive Trajectories ==

The following table assesses the plot for the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, with varying initial momenta.
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Proceeds as normal. No initial collision, but repulsion between the two systems is overcome, etc.
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The initial velocity of the atom is far too slow to collide with the molecule. The two systems approach each other initially, but the repulsion between them forces them to move in opposite directions, and with more velocity than
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The two systems approach the transition state, and then the diatom splits, with the central atom pairing with the other terminal atom. They then proceed to move in opposite velocities.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|Initial velocity of particle C is so great that there is some repulsion between particle B and C, despite moving towards each other. As a result, particle B moves away from C after the initial attraction. It then oscillates around the transition state, moving from AB + C to A+ BC. It eventually settles with AB + C, the initial reactants. It can be seen by increasing the number of steps to 2000 that the final position of the two systems is greater separation. The particles AB and C are moving in opposite directions, so end up further apart than the initial conditions.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Oscillations around the transition state after an initial collision. This is very similar initially to the -10.1 p2 path. However, the rebound collision for A and B is even faster, resulting in another rebound for B. B then collides with particle C as they are both moving in the same directions, and the final state is that of the products, A + BC
|
|}

== Question 5: Transition State Theory Predictions ==

Transition state theory can be used to determine the rate of a reaction, but it relies on 5 assumptions:
# Once the kinetic energy has crossed the potential barrier, the reaction will proceed; there is no recrossing of the barrier.
# There are no quantum effects {specifically, no quantum tunnelling will take place}
# It is modelled on the Boltzmann distribution
# ask
# ask

Assumptions 1 and 2 are broken in these reaction dynamics, as we are considering classical collisions with some forces of attraction and repulsion between the atoms. As a result, despite the fact that crossing of the barrier can occur in a certain scenario, that activation barrier may be recrossed later. This is shown in the reaction pathway for the momenta p1 = -5.1 and p2 = -10.1, where despite the initial crossing of the barrier and consequent overcoming of the energy barrier, the system is not reactive. This therefore breaks from the mold of TST.

As TST suggests that some reactions that are not reactive are in fact reactive, it suggests that TST will overestimate how many reactions are successful. As a result, we can surmise that TST will overestimate the rate of reactions, compared to the GUI that we are using.

If the GUI agreed with assumption 1, TST would underestimate the reaction rate, as Tunnelling of the atoms would lead to a higher rate than the classical picture, as not all of the atoms have to reach the potential barrier to overcome it.
= Exercise 2: F-H-H system =

== Question 1: Energetics and bond strength ==

Let's set the initial position as <math>r_1</math> = <math>r_2</math>, where <math>r_1</math> is F-H, and <math>r_2</math> is H-H.

If we run a MEP plot from this position, the particle will end up at the lower energy state, regardless of the initial reactants or products.

We can see in the image below, the minimum energy path is such that <math>r_1</math> is minimised, and <math>r_2</math> is maximised. With <math>r_1</math> as F-H, this implies that the lower energy state is H + H-F.

[[File:MEP F-H-H.png]]

As a result, we can classify the F + H2 --> H + HF reaction as exothermic, as it is progressing from a higher energy state to a lower energy state.

H + HF ---> F + H2 would therefore be an endothermic process; it would require an input of energy to get this product.

This order of potential energies suggests that the bond strength between F-H is greater than for H2.

== Question 2: Transition state distance ==

As before, the transition state will be the saddle point on the surface plot. However, as the system is no longer symmetric as before, the transition state is no longer at the point where <math>r_1</math> = <math>r_2</math>.

Hammond's postulate suggests that the transition state will be closer to the higher energy system in the reaction. We know that this is the case for F + H2, so the transition state must be such that the bond distance for F-H is greater than for H2.

Knowing this, we can surmise that if AB is F-H and BC is H2, the distances at the transition state are: AB =184 pm and BC = 74.4 pm. This complies with the logic that the reaction where FH is the product is exothermic, as this is supported by Hammond's postulate.

[[File:Transition state ex2 contour.png]][[File:Transition state ex2 surface.png]]

We can once again, further confirm that this is the saddle point by using the same point that the forces along AB and BC at this point are both zero.

== Question 3: Activation energy ==

To find the activation energies of each product F-H and H2 (2 mutually exclusive scenarios), we simply need to find the difference between the transition state energy and the energy of the products. This can be calculated for the reaction only (excluding vibration) by using MEP to identify the shortest route.

Due to the asymmetry of the atoms, we have 2 activation energies.

[[File:Transition state to minimum.png]]
[[File:Transition state to min en vs time.png]]

[[File:Example.jpg]][[File:Example.jpg]]

Starting from the r <math>_{ts}</math> = AB =184 pm and BC = 74.4 pm, with a perturbation of ± 5 pm in the AB axis.
{| class="wikitable"
!Product
!Energy initial
!Energy final
!ΔE
|-
|F-H + H
|
|
|
|-
|F + H2
|
|
|
|}

This is because we know that the energy of the transition state is the maximum energy state on the reaction path, and doesn't consider oscillations along the potential well.

== Question 4: Mechanism of the Release of reaction energy ==

hello

== Question 5: Efficiency of reaction ==

hello

MRD:01512675

2020-05-08T12:56:49Z

Sa13018: /* Question 3: Activation energy */

= Exercise 1: H + H<math>_2</math> system =

== Question 1: The Transition State Region ==

The transition state is defined as the maximum point of the minimum energy path between the reactants and products. Therefore, on the potential energy surface diagram, it exists as a minimax; a saddle point.
This can therefore be described as the point where the partial differential of the potential energy with respect to all variables is zero.

In this case, due to a 3 dimensional surface plot, we have <math>
\frac{\partial V}{\partial r_1} = \frac{\partial V}{\partial r_2} = 0
</math>, where <math>r_1</math> and <math>r_2 </math> are the respective distances between AB and BC.

The local minimum point would see that a deviation in any direction would be an increase in potential energy.
However, the transition state is such that this would only happen in the axis where <math>r_1</math> = <math>r_2</math>, and would decrease in energy in any deviation of the AB or BC position.

This is shown in the images shown below.

[[File:Transition state minimum.png]]
[[File:Transition state maximum.png]]

The photo on the left shows the energy potential curve in a 2D viewpoint, as a function of the AB distance <math>r_1</math>. We can see that there is a steep increase of the V(<math>r_1</math>) function on either side of <math>r_1</math> = 74 pm, indicating the minimum in a potential well at this point. This continues as a valley between the reactants and products path.

The photo on the right shows a clear maximum point being mapped out by the black line. Despite being at a minimum point on the potential valley floor, this is a maximum point, indicating a minimax state. This is therefore a saddle point, and the transition state. Due to the symmetry of the system only consisting of H atoms, the transition state is observed at a point where <math>r_1</math> = <math>r_2</math>.

This can further be identified using the second partial derivative test; computing a Hessian matrix. If H <0, it must be a saddle.

== Question 2: Locating the transition state ==

Due to the symmetry between the H and the \chem{H_2} molecule, and the fact that the interaction is happening at 180º, we know that <math>r_1</math> must equal <math>r_2</math> for any point on the potential for the transition state. As a result, if both particles carry zero initial velocity and are on the transition state only, they will drop down the potential well to the transition state distance <math>r_{12}</math>.
Thus, in a scenario where <math>r_1</math> = <math>r_2</math>, and <math>v_2</math> = <math>v_1</math> = 0, the transition state will act only in a 2D potential plot; the dynamics will reach the minimum in the potential well.

This distance <math>r_12</math> is shown to be 90.775 pm, as there is no force along AB or BC at this distance when v = 0.
Furthermore, the contour plot shows the final position to be the same as the initial position; it is a point of equilibrium.
The internuclear distances vs time plot are also shown to constant here; this must therefore be the transition state.

[[File:Example.jpg]]
[[File:Example.jpg]]
[[File:Example.jpg]]

== Question 3: The Minimum Energy Path (MEP) ==

The MEP shows a clear difference in its plot and animation compared to the Dynamic calculation.
# There is no vibration/ oscillation in the MEP plot. This makes sense; it would be going against minimum energy route, as the oscillations would mean that the particle is not always at the minimum of the potential well.
# The MEP plot finishes once the minimum point of the energy potential surface is reached; after 193.1pm as the AB distance and 74 pm as the BC distance, the contour is consistent. However, we know that the particles must keep moving from this, as they carry opposite velocities. This is shown in the Dynamics plot, where the reaction path continues long after reaching the end of the x-label in the plot. This confirms that the MEP is relying only on the minimum of the potential energy surface; not the distances AB and BC.
This is shown in the comparison below.

It is also interesting to note the action of the MEP when we start from the reactants, as before, with enough energy to go to the products in the Dynamics. This is shown in the comparison below.

In this scenario, despite having enough initial velocity in one direction, the MEP doesn't follow that path. It reverses the direction of the initial velocity and instead ends up with no reaction taking place. This implies that it has followed the minimum energy path at every point in the plot, ignoring the vectors that the particle is following. This further exemplifies the difference between Dynamics and MEP, showing that MEP is only taking the mathematically logical route, whereas the Dynamics considers the physical process that is occuring.

== Question 4: Reactive and Unreactive Trajectories ==

The following table assesses the plot for the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, with varying initial momenta.
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Proceeds as normal. No initial collision, but repulsion between the two systems is overcome, etc.
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The initial velocity of the atom is far too slow to collide with the molecule. The two systems approach each other initially, but the repulsion between them forces them to move in opposite directions, and with more velocity than
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The two systems approach the transition state, and then the diatom splits, with the central atom pairing with the other terminal atom. They then proceed to move in opposite velocities.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|Initial velocity of particle C is so great that there is some repulsion between particle B and C, despite moving towards each other. As a result, particle B moves away from C after the initial attraction. It then oscillates around the transition state, moving from AB + C to A+ BC. It eventually settles with AB + C, the initial reactants. It can be seen by increasing the number of steps to 2000 that the final position of the two systems is greater separation. The particles AB and C are moving in opposite directions, so end up further apart than the initial conditions.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Oscillations around the transition state after an initial collision. This is very similar initially to the -10.1 p2 path. However, the rebound collision for A and B is even faster, resulting in another rebound for B. B then collides with particle C as they are both moving in the same directions, and the final state is that of the products, A + BC
|
|}

== Question 5: Transition State Theory Predictions ==

Transition state theory can be used to determine the rate of a reaction, but it relies on 5 assumptions:
# Once the kinetic energy has crossed the potential barrier, the reaction will proceed; there is no recrossing of the barrier.
# There are no quantum effects {specifically, no quantum tunnelling will take place}
# It is modelled on the Boltzmann distribution
# ask
# ask

Assumptions 1 and 2 are broken in these reaction dynamics, as we are considering classical collisions with some forces of attraction and repulsion between the atoms. As a result, despite the fact that crossing of the barrier can occur in a certain scenario, that activation barrier may be recrossed later. This is shown in the reaction pathway for the momenta p1 = -5.1 and p2 = -10.1, where despite the initial crossing of the barrier and consequent overcoming of the energy barrier, the system is not reactive. This therefore breaks from the mold of TST.

As TST suggests that some reactions that are not reactive are in fact reactive, it suggests that TST will overestimate how many reactions are successful. As a result, we can surmise that TST will overestimate the rate of reactions, compared to the GUI that we are using.

If the GUI agreed with assumption 1, TST would underestimate the reaction rate, as Tunnelling of the atoms would lead to a higher rate than the classical picture, as not all of the atoms have to reach the potential barrier to overcome it.
= Exercise 2: F-H-H system =

== Question 1: Energetics and bond strength ==

Let's set the initial position as <math>r_1</math> = <math>r_2</math>, where <math>r_1</math> is F-H, and <math>r_2</math> is H-H.

If we run a MEP plot from this position, the particle will end up at the lower energy state, regardless of the initial reactants or products.

We can see in the image below, the minimum energy path is such that <math>r_1</math> is minimised, and <math>r_2</math> is maximised. With <math>r_1</math> as F-H, this implies that the lower energy state is H + H-F.

[[File:MEP F-H-H.png]]

As a result, we can classify the F + H2 --> H + HF reaction as exothermic, as it is progressing from a higher energy state to a lower energy state.

H + HF ---> F + H2 would therefore be an endothermic process; it would require an input of energy to get this product.

This order of potential energies suggests that the bond strength between F-H is greater than for H2.

== Question 2: Transition state distance ==

As before, the transition state will be the saddle point on the surface plot. However, as the system is no longer symmetric as before, the transition state is no longer at the point where <math>r_1</math> = <math>r_2</math>.

Hammond's postulate suggests that the transition state will be closer to the higher energy system in the reaction. We know that this is the case for F + H2, so the transition state must be such that the bond distance for F-H is greater than for H2.

Knowing this, we can surmise that if AB is F-H and BC is H2, the distances at the transition state are: AB =184 pm and BC = 74.4 pm. This complies with the logic that the reaction where FH is the product is exothermic, as this is supported by Hammond's postulate.

[[File:Transition state ex2 contour.png]][[File:Transition state ex2 surface.png]]

We can once again, further confirm that this is the saddle point by using the same point that the forces along AB and BC at this point are both zero.

== Question 3: Activation energy ==

To find the activation energies of each product F-H and H2 (2 mutually exclusive scenarios), we simply need to find the difference between the transition state energy and the energy of the products. This can be calculated for the reaction only (excluding vibration) by using MEP to identify the shortest route.

Due to the asymmetry of the atoms, we have 2 activation energies.

[[File:Transition state to minimum.png]]
[[File:Transition state to min en vs time.png]]

[[File:Example.jpg]][[File:Example.jpg]]

Starting from the r {ts} = AB =184 pm and BC = 74.4 pm, with a perturbation of ± 5 pm in the AB axis.
{| class="wikitable"
!Product
!Energy initial
!Energy final
!ΔE
|-
|F-H + H
|
|
|
|-
|F + H2
|
|
|
|}

This is because we know that the energy of the transition state is the maximum energy state on the reaction path, and doesn't consider oscillations along the potential well.

== Question 4: Mechanism of the Release of reaction energy ==

hello

== Question 5: Efficiency of reaction ==

hello

File:Transition state to min en vs time.png

2020-05-08T12:56:31Z

Sa13018:

MRD:01512675

2020-05-08T12:55:00Z

Sa13018: /* Question 3: Activation energy */

= Exercise 1: H + H<math>_2</math> system =

== Question 1: The Transition State Region ==

The transition state is defined as the maximum point of the minimum energy path between the reactants and products. Therefore, on the potential energy surface diagram, it exists as a minimax; a saddle point.
This can therefore be described as the point where the partial differential of the potential energy with respect to all variables is zero.

In this case, due to a 3 dimensional surface plot, we have <math>
\frac{\partial V}{\partial r_1} = \frac{\partial V}{\partial r_2} = 0
</math>, where <math>r_1</math> and <math>r_2 </math> are the respective distances between AB and BC.

The local minimum point would see that a deviation in any direction would be an increase in potential energy.
However, the transition state is such that this would only happen in the axis where <math>r_1</math> = <math>r_2</math>, and would decrease in energy in any deviation of the AB or BC position.

This is shown in the images shown below.

[[File:Transition state minimum.png]]
[[File:Transition state maximum.png]]

The photo on the left shows the energy potential curve in a 2D viewpoint, as a function of the AB distance <math>r_1</math>. We can see that there is a steep increase of the V(<math>r_1</math>) function on either side of <math>r_1</math> = 74 pm, indicating the minimum in a potential well at this point. This continues as a valley between the reactants and products path.

The photo on the right shows a clear maximum point being mapped out by the black line. Despite being at a minimum point on the potential valley floor, this is a maximum point, indicating a minimax state. This is therefore a saddle point, and the transition state. Due to the symmetry of the system only consisting of H atoms, the transition state is observed at a point where <math>r_1</math> = <math>r_2</math>.

This can further be identified using the second partial derivative test; computing a Hessian matrix. If H <0, it must be a saddle.

== Question 2: Locating the transition state ==

Due to the symmetry between the H and the \chem{H_2} molecule, and the fact that the interaction is happening at 180º, we know that <math>r_1</math> must equal <math>r_2</math> for any point on the potential for the transition state. As a result, if both particles carry zero initial velocity and are on the transition state only, they will drop down the potential well to the transition state distance <math>r_{12}</math>.
Thus, in a scenario where <math>r_1</math> = <math>r_2</math>, and <math>v_2</math> = <math>v_1</math> = 0, the transition state will act only in a 2D potential plot; the dynamics will reach the minimum in the potential well.

This distance <math>r_12</math> is shown to be 90.775 pm, as there is no force along AB or BC at this distance when v = 0.
Furthermore, the contour plot shows the final position to be the same as the initial position; it is a point of equilibrium.
The internuclear distances vs time plot are also shown to constant here; this must therefore be the transition state.

[[File:Example.jpg]]
[[File:Example.jpg]]
[[File:Example.jpg]]

== Question 3: The Minimum Energy Path (MEP) ==

The MEP shows a clear difference in its plot and animation compared to the Dynamic calculation.
# There is no vibration/ oscillation in the MEP plot. This makes sense; it would be going against minimum energy route, as the oscillations would mean that the particle is not always at the minimum of the potential well.
# The MEP plot finishes once the minimum point of the energy potential surface is reached; after 193.1pm as the AB distance and 74 pm as the BC distance, the contour is consistent. However, we know that the particles must keep moving from this, as they carry opposite velocities. This is shown in the Dynamics plot, where the reaction path continues long after reaching the end of the x-label in the plot. This confirms that the MEP is relying only on the minimum of the potential energy surface; not the distances AB and BC.
This is shown in the comparison below.

It is also interesting to note the action of the MEP when we start from the reactants, as before, with enough energy to go to the products in the Dynamics. This is shown in the comparison below.

In this scenario, despite having enough initial velocity in one direction, the MEP doesn't follow that path. It reverses the direction of the initial velocity and instead ends up with no reaction taking place. This implies that it has followed the minimum energy path at every point in the plot, ignoring the vectors that the particle is following. This further exemplifies the difference between Dynamics and MEP, showing that MEP is only taking the mathematically logical route, whereas the Dynamics considers the physical process that is occuring.

== Question 4: Reactive and Unreactive Trajectories ==

The following table assesses the plot for the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, with varying initial momenta.
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Proceeds as normal. No initial collision, but repulsion between the two systems is overcome, etc.
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The initial velocity of the atom is far too slow to collide with the molecule. The two systems approach each other initially, but the repulsion between them forces them to move in opposite directions, and with more velocity than
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The two systems approach the transition state, and then the diatom splits, with the central atom pairing with the other terminal atom. They then proceed to move in opposite velocities.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|Initial velocity of particle C is so great that there is some repulsion between particle B and C, despite moving towards each other. As a result, particle B moves away from C after the initial attraction. It then oscillates around the transition state, moving from AB + C to A+ BC. It eventually settles with AB + C, the initial reactants. It can be seen by increasing the number of steps to 2000 that the final position of the two systems is greater separation. The particles AB and C are moving in opposite directions, so end up further apart than the initial conditions.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Oscillations around the transition state after an initial collision. This is very similar initially to the -10.1 p2 path. However, the rebound collision for A and B is even faster, resulting in another rebound for B. B then collides with particle C as they are both moving in the same directions, and the final state is that of the products, A + BC
|
|}

== Question 5: Transition State Theory Predictions ==

Transition state theory can be used to determine the rate of a reaction, but it relies on 5 assumptions:
# Once the kinetic energy has crossed the potential barrier, the reaction will proceed; there is no recrossing of the barrier.
# There are no quantum effects {specifically, no quantum tunnelling will take place}
# It is modelled on the Boltzmann distribution
# ask
# ask

Assumptions 1 and 2 are broken in these reaction dynamics, as we are considering classical collisions with some forces of attraction and repulsion between the atoms. As a result, despite the fact that crossing of the barrier can occur in a certain scenario, that activation barrier may be recrossed later. This is shown in the reaction pathway for the momenta p1 = -5.1 and p2 = -10.1, where despite the initial crossing of the barrier and consequent overcoming of the energy barrier, the system is not reactive. This therefore breaks from the mold of TST.

As TST suggests that some reactions that are not reactive are in fact reactive, it suggests that TST will overestimate how many reactions are successful. As a result, we can surmise that TST will overestimate the rate of reactions, compared to the GUI that we are using.

If the GUI agreed with assumption 1, TST would underestimate the reaction rate, as Tunnelling of the atoms would lead to a higher rate than the classical picture, as not all of the atoms have to reach the potential barrier to overcome it.
= Exercise 2: F-H-H system =

== Question 1: Energetics and bond strength ==

Let's set the initial position as <math>r_1</math> = <math>r_2</math>, where <math>r_1</math> is F-H, and <math>r_2</math> is H-H.

If we run a MEP plot from this position, the particle will end up at the lower energy state, regardless of the initial reactants or products.

We can see in the image below, the minimum energy path is such that <math>r_1</math> is minimised, and <math>r_2</math> is maximised. With <math>r_1</math> as F-H, this implies that the lower energy state is H + H-F.

[[File:MEP F-H-H.png]]

As a result, we can classify the F + H2 --> H + HF reaction as exothermic, as it is progressing from a higher energy state to a lower energy state.

H + HF ---> F + H2 would therefore be an endothermic process; it would require an input of energy to get this product.

This order of potential energies suggests that the bond strength between F-H is greater than for H2.

== Question 2: Transition state distance ==

As before, the transition state will be the saddle point on the surface plot. However, as the system is no longer symmetric as before, the transition state is no longer at the point where <math>r_1</math> = <math>r_2</math>.

Hammond's postulate suggests that the transition state will be closer to the higher energy system in the reaction. We know that this is the case for F + H2, so the transition state must be such that the bond distance for F-H is greater than for H2.

Knowing this, we can surmise that if AB is F-H and BC is H2, the distances at the transition state are: AB =184 pm and BC = 74.4 pm. This complies with the logic that the reaction where FH is the product is exothermic, as this is supported by Hammond's postulate.

[[File:Transition state ex2 contour.png]][[File:Transition state ex2 surface.png]]

We can once again, further confirm that this is the saddle point by using the same point that the forces along AB and BC at this point are both zero.

== Question 3: Activation energy ==

To find the activation energies of each product F-H and H2 (2 mutually exclusive scenarios), we simply need to find the difference between the transition state energy and the energy of the products. This can be calculated for the reaction only (excluding vibration) by using MEP to identify the shortest route.

Due to the asymmetry of the atoms, we have 2 activation energies.

[[File:Transition state to minimum.png.]][[File:Transition_state_to_min_en_vs_time.png]]

[[File:Example.jpg]][[File:Example.jpg]]

Starting from the r {ts} = AB =184 pm and BC = 74.4 pm, with a perturbation of ± 5 pm in the AB axis.
{| class="wikitable"
!Product
!Energy initial
!Energy final
!ΔE
|-
|F-H + H
|
|
|
|-
|F + H2
|
|
|
|}

This is because we know that the energy of the transition state is the maximum energy state on the reaction path, and doesn't consider oscillations along the potential well.

== Question 4: Mechanism of the Release of reaction energy ==

hello

== Question 5: Efficiency of reaction ==

hello

MRD:01512675

2020-05-08T12:54:28Z

Sa13018: /* Question 3: Activation energy */

= Exercise 1: H + H<math>_2</math> system =

== Question 1: The Transition State Region ==

The transition state is defined as the maximum point of the minimum energy path between the reactants and products. Therefore, on the potential energy surface diagram, it exists as a minimax; a saddle point.
This can therefore be described as the point where the partial differential of the potential energy with respect to all variables is zero.

In this case, due to a 3 dimensional surface plot, we have <math>
\frac{\partial V}{\partial r_1} = \frac{\partial V}{\partial r_2} = 0
</math>, where <math>r_1</math> and <math>r_2 </math> are the respective distances between AB and BC.

The local minimum point would see that a deviation in any direction would be an increase in potential energy.
However, the transition state is such that this would only happen in the axis where <math>r_1</math> = <math>r_2</math>, and would decrease in energy in any deviation of the AB or BC position.

This is shown in the images shown below.

[[File:Transition state minimum.png]]
[[File:Transition state maximum.png]]

The photo on the left shows the energy potential curve in a 2D viewpoint, as a function of the AB distance <math>r_1</math>. We can see that there is a steep increase of the V(<math>r_1</math>) function on either side of <math>r_1</math> = 74 pm, indicating the minimum in a potential well at this point. This continues as a valley between the reactants and products path.

The photo on the right shows a clear maximum point being mapped out by the black line. Despite being at a minimum point on the potential valley floor, this is a maximum point, indicating a minimax state. This is therefore a saddle point, and the transition state. Due to the symmetry of the system only consisting of H atoms, the transition state is observed at a point where <math>r_1</math> = <math>r_2</math>.

This can further be identified using the second partial derivative test; computing a Hessian matrix. If H <0, it must be a saddle.

== Question 2: Locating the transition state ==

Due to the symmetry between the H and the \chem{H_2} molecule, and the fact that the interaction is happening at 180º, we know that <math>r_1</math> must equal <math>r_2</math> for any point on the potential for the transition state. As a result, if both particles carry zero initial velocity and are on the transition state only, they will drop down the potential well to the transition state distance <math>r_{12}</math>.
Thus, in a scenario where <math>r_1</math> = <math>r_2</math>, and <math>v_2</math> = <math>v_1</math> = 0, the transition state will act only in a 2D potential plot; the dynamics will reach the minimum in the potential well.

This distance <math>r_12</math> is shown to be 90.775 pm, as there is no force along AB or BC at this distance when v = 0.
Furthermore, the contour plot shows the final position to be the same as the initial position; it is a point of equilibrium.
The internuclear distances vs time plot are also shown to constant here; this must therefore be the transition state.

[[File:Example.jpg]]
[[File:Example.jpg]]
[[File:Example.jpg]]

== Question 3: The Minimum Energy Path (MEP) ==

The MEP shows a clear difference in its plot and animation compared to the Dynamic calculation.
# There is no vibration/ oscillation in the MEP plot. This makes sense; it would be going against minimum energy route, as the oscillations would mean that the particle is not always at the minimum of the potential well.
# The MEP plot finishes once the minimum point of the energy potential surface is reached; after 193.1pm as the AB distance and 74 pm as the BC distance, the contour is consistent. However, we know that the particles must keep moving from this, as they carry opposite velocities. This is shown in the Dynamics plot, where the reaction path continues long after reaching the end of the x-label in the plot. This confirms that the MEP is relying only on the minimum of the potential energy surface; not the distances AB and BC.
This is shown in the comparison below.

It is also interesting to note the action of the MEP when we start from the reactants, as before, with enough energy to go to the products in the Dynamics. This is shown in the comparison below.

In this scenario, despite having enough initial velocity in one direction, the MEP doesn't follow that path. It reverses the direction of the initial velocity and instead ends up with no reaction taking place. This implies that it has followed the minimum energy path at every point in the plot, ignoring the vectors that the particle is following. This further exemplifies the difference between Dynamics and MEP, showing that MEP is only taking the mathematically logical route, whereas the Dynamics considers the physical process that is occuring.

== Question 4: Reactive and Unreactive Trajectories ==

The following table assesses the plot for the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, with varying initial momenta.
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Proceeds as normal. No initial collision, but repulsion between the two systems is overcome, etc.
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The initial velocity of the atom is far too slow to collide with the molecule. The two systems approach each other initially, but the repulsion between them forces them to move in opposite directions, and with more velocity than
|
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The two systems approach the transition state, and then the diatom splits, with the central atom pairing with the other terminal atom. They then proceed to move in opposite velocities.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|Initial velocity of particle C is so great that there is some repulsion between particle B and C, despite moving towards each other. As a result, particle B moves away from C after the initial attraction. It then oscillates around the transition state, moving from AB + C to A+ BC. It eventually settles with AB + C, the initial reactants. It can be seen by increasing the number of steps to 2000 that the final position of the two systems is greater separation. The particles AB and C are moving in opposite directions, so end up further apart than the initial conditions.
|
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|Oscillations around the transition state after an initial collision. This is very similar initially to the -10.1 p2 path. However, the rebound collision for A and B is even faster, resulting in another rebound for B. B then collides with particle C as they are both moving in the same directions, and the final state is that of the products, A + BC
|
|}

== Question 5: Transition State Theory Predictions ==

Transition state theory can be used to determine the rate of a reaction, but it relies on 5 assumptions:
# Once the kinetic energy has crossed the potential barrier, the reaction will proceed; there is no recrossing of the barrier.
# There are no quantum effects {specifically, no quantum tunnelling will take place}
# It is modelled on the Boltzmann distribution
# ask
# ask

Assumptions 1 and 2 are broken in these reaction dynamics, as we are considering classical collisions with some forces of attraction and repulsion between the atoms. As a result, despite the fact that crossing of the barrier can occur in a certain scenario, that activation barrier may be recrossed later. This is shown in the reaction pathway for the momenta p1 = -5.1 and p2 = -10.1, where despite the initial crossing of the barrier and consequent overcoming of the energy barrier, the system is not reactive. This therefore breaks from the mold of TST.

As TST suggests that some reactions that are not reactive are in fact reactive, it suggests that TST will overestimate how many reactions are successful. As a result, we can surmise that TST will overestimate the rate of reactions, compared to the GUI that we are using.

If the GUI agreed with assumption 1, TST would underestimate the reaction rate, as Tunnelling of the atoms would lead to a higher rate than the classical picture, as not all of the atoms have to reach the potential barrier to overcome it.
= Exercise 2: F-H-H system =

== Question 1: Energetics and bond strength ==

Let's set the initial position as <math>r_1</math> = <math>r_2</math>, where <math>r_1</math> is F-H, and <math>r_2</math> is H-H.

If we run a MEP plot from this position, the particle will end up at the lower energy state, regardless of the initial reactants or products.

We can see in the image below, the minimum energy path is such that <math>r_1</math> is minimised, and <math>r_2</math> is maximised. With <math>r_1</math> as F-H, this implies that the lower energy state is H + H-F.

[[File:MEP F-H-H.png]]

As a result, we can classify the F + H2 --> H + HF reaction as exothermic, as it is progressing from a higher energy state to a lower energy state.

H + HF ---> F + H2 would therefore be an endothermic process; it would require an input of energy to get this product.

This order of potential energies suggests that the bond strength between F-H is greater than for H2.

== Question 2: Transition state distance ==

As before, the transition state will be the saddle point on the surface plot. However, as the system is no longer symmetric as before, the transition state is no longer at the point where <math>r_1</math> = <math>r_2</math>.

Hammond's postulate suggests that the transition state will be closer to the higher energy system in the reaction. We know that this is the case for F + H2, so the transition state must be such that the bond distance for F-H is greater than for H2.

Knowing this, we can surmise that if AB is F-H and BC is H2, the distances at the transition state are: AB =184 pm and BC = 74.4 pm. This complies with the logic that the reaction where FH is the product is exothermic, as this is supported by Hammond's postulate.

[[File:Transition state ex2 contour.png]][[File:Transition state ex2 surface.png]]

We can once again, further confirm that this is the saddle point by using the same point that the forces along AB and BC at this point are both zero.

== Question 3: Activation energy ==

To find the activation energies of each product F-H and H2 (2 mutually exclusive scenarios), we simply need to find the difference between the transition state energy and the energy of the products. This can be calculated for the reaction only (excluding vibration) by using MEP to identify the shortest route.

Due to the asymmetry of the atoms, we have 2 activation energies.

[[Transition state to minimum.png.]][[File:Transition_state_to_min_en_vs_time.png]]

[[File:Example.jpg]][[File:Example.jpg]]

Starting from the r {ts} = AB =184 pm and BC = 74.4 pm, with a perturbation of ± 5 pm in the AB axis.
{| class="wikitable"
!Product
!Energy initial
!Energy final
!ΔE
|-
|F-H + H
|
|
|
|-
|F + H2
|
|
|
|}

This is because we know that the energy of the transition state is the maximum energy state on the reaction path, and doesn't consider oscillations along the potential well.

== Question 4: Mechanism of the Release of reaction energy ==

hello

== Question 5: Efficiency of reaction ==

hello

File:Transition state to minimum.png

2020-05-08T12:53:02Z

Sa13018: