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MRD:00940575

2020-07-07T09:21:38Z

Mak214: /* Molecular Reaction Dynamics Write-up */

== '''Molecular Reaction Dynamics Write-up''' ==

=== Exercise 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 energy as a function of the positons of atoms needs to be worked out first. The first order derivative of this gives stationary points. These points correspond to the physically stable reactant and product, and transition state. The transition state is defined by the saddle point on a potential energy surface diagram. This point has the highest energy on the reaction pathway line, so it can be distinguished easily from other stationary points.

''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.''<gallery>
File:00940575_1.png
</gallery><gallery>
File:00940575_2.png
</gallery>At rab and rbc = 90. both bonds has an oscillation that superimpose on each other as they have the same length. Many values of r was put in until the value converged to a flat line with no oscillation that showed that rts= 90.76 pm {{fontcolor1|red|Good. well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 10:21, 7 July 2020 (BST)}} <gallery>
[[File:00940575_1]]
</gallery>

''Comment on how the mep and the trajectory you just calculated differ.''<gallery>
File:00940575_mep_1.png
</gallery><gallery>
File:00940575_mep_2.png
</gallery><gallery>
File:00940575_p.png
</gallery>In the previous trajectory, rab and rbc was 90.76pm, but when rbc was chaged to 91.76 pm, it increases and rab decreases and asymptotes to a rab=74.04743. Momentum for ab asymptotes too. {{fontcolor1|red|MEP and Dynamics are different calculations which treat the momentum of the system in completely different ways. In MEP calculations the momentum is reset to 0 at every timestep, meaning the system only responds to the gradient of the potential energy surface at each point - so it will find the nearest energy minimum. Dynamics trajectories are more realistic simulations where the system will hold on to momentum and so we see vibration etc with dynamics calculations. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 10:21, 7 July 2020 (BST)}}

''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?''

rbc =74 pm , rab = 200 pm
{| 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>-414.28</nowiki>
|reactive
|rab and rbc has crossed once at the transition state. r ab is oscillating and rbc is increasing so it has reacted.
|<gallery>
File:00940575 31.png
</gallery>
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.08</nowiki>
|unreactive
|rab and rbc does not cross, as the momentum was not enough to overcome the barrier it does not react
|<gallery>
File:00940575 32.png
</gallery>
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.97</nowiki>
|reactive
|rab and rbc has crossed once at the transition state. r ab is oscillating and rbc is increasing so it has reacted.
|<gallery>
File:00940575 33.png
</gallery>
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.26</nowiki>
|unreactive
|rab and rbc has crossed twice . rab is smaller than rbc for a short while, this means p2 was too large so even the energy was enough to overcome the barrier, the transition state was not formed.
|<gallery>
File:00940575 34.png
</gallery>
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.47</nowiki>
|reactive
|rab and rbc has crossed three times. The saddle point where the transition state is formed is when they meet for the third time.
|<gallery>
File:00940575 35.png
</gallery>
|}

''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''
{{fontcolor1|red| The key postulate of TST which is proven to be an overestimate of the rate here is the assumption that once the transition state has been reached then the reaction will always go to the products. Your fourth entry in the table above shows that this is not always the case, and that an equilibrium may exist between the TS and the products as well as between Reactants and TS... There is also an oversight in TST which does not account for quantum tunnelling, although we could not model for this in this lab, and the overall effect would be small. I would advise you to read up on TST, maybe you could read a classmates wiki page to learn more!! [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 10:21, 7 July 2020 (BST)}}
The transition state theory uses the surface that intersects with the saddle point. This overestimates the rate of reaction.

{{fontcolor1|red| It is such a shame that you didnt have time to finish this lab, there are some important concepts that were introduced, please take some time to read through the literature linked to the lab so you don't miss out!! Overall there is not a lot of time that has gone into this report and you have not engaged with the theory to fully answer the questions. See my comments above. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 10:21, 7 July 2020 (BST)}}

MRD:yw14815

2020-07-07T09:07:35Z

Mak214: /* The Polanyi Rules */

=Introduction=

This Wiki page outlines the study of the molecular reaction dynamics of two simple triatomic systems: H + H2 and F + H2. The atoms are 180° to each other, with a 1D space symplification. The atoms are labelled A, B and C, with vectors r1 and r2 representing vectors BA and BC in pm respectively. Momentum are also measured in p1 and p2 in g mol-1 pm fs-1. Below is the systematic diagram of the triatomic system model, which will be referred to throughout the whole wiki. <ref name="CP3MD"/>

[[File:Fig1_Yanda.png|400px|thumb|center|Figure 1. A triatomic system model. Diagram from Imperial College London, https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD (accessed May 2020.)]]

=EXERCISE 1: H + H2 system=
== Potential Energy Surfaces ==

A potential energy surface is a visual representation that describes the potential energy of a system, such as a collective of atoms, in terms of parameters, which normally is the relative positions of the reactants and the products. A reaction path is shown to be the gradient line on a PES diagram, and represents both potential energy and internuclear distances between atoms in the system as a function of the reaction coordinate, thereby modelling their changes through the progress of the reaction.

The PES diagram for Figure 1 above was calculated for the reaction system H + H2 using the parameters given:
{| class="wikitable"
!Calculations
!Steps
!r1 / pm
!r2 / pm
!p1 / gmol-1pmfs-1
!p2 / gmol-1pmfs-1
|-
|Dynamics
|500
|74
|230
|0.0
|<nowiki>-5.2</nowiki>
|}

[[File:Fig2_Yanda.png|400px|thumb|center|Figure 2.Potential Energy Surface for the H + H2 reaction path.]]

The reaction path, which passes through each potential energy minimum, including the transition state, which is a first order saddle point at which the potential energy is at its maximum (highest energy minimum) and where its gradient is 0, hence ∂V('''r''')/∂'''r '''=0 and the second differential at the TS is less than 0.Furthermore, the determinant of the Hessian Matrix is smaller than 0, with the eigenvalues of the Hessian Matrix being one positive and one negative

Local minimum, on the other hand,while also have the first derivative ∂V('''r''')/∂'''r '''=0, its second differential at the minimum is greater than 0, with the eigenvalues of the Hessian Matrix both being positive and the determinant of the Hessian Matrix greater than 0.

Literally speaking, the main difference between transition state and a local minimum is that the transition state is in fact a local energy maximum. Initially, with 0 initial momentum, both transition state and local minimum will remain there forever. However, with a very small momentum added towards the reactants direction, the transition state will be passed and will roll over towards the reactants, and vice versa for products. Testing the trajectories near the transition state is a method used to identify the presence of the transition state. On the other hand, a slightly perturbation from the local minimum will only result in rolling back towards the local minimum, because only a sufficient energy is needed to overcome the energy barrier to escape the "well" of the local minimum. In the PES diagram, a local minimum represents an intermediate, which can be represented in a 2D diagram below.<ref name="Plot"/>

[[File:Fig3_Yanda.PNG|400px|thumb|center|Figure 3. Potential Energy Surface Diagram showing the Transition State. Diagram from http://vergil.chemistry.gatech.edu/courses/chem6485/pdf/pes-lecture.pdf (accessed May 2020).]]

Another interesting observation is that the gradient along the reaction pathway is least steep close to both sides of the transition state, suggesting the point at which bonds are in the process of breaking and forming, hence lower degree of association and less vibrational oscillations. {{fontcolor1|red| Great. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:50, 7 July 2020 (BST)}}

== Locating the Transition State ==

The best estimate of the transition state position (rts) is 90.77 pm. The transition state should be symmetric as it represents the state of the reaction where old bonds and new bonds are in the middle of being broken and formed. As such, given the initial conditions of both momentum p1=0 and p2=0, the two internuclear distances r1 and r2 must be equal at the transitions state. The constant internuclear distances can be understood and found with a "Internuclear Distances vs Time plot" shown in Figure 4 below, where the straight horizontal line suggests constant potential energy with the absence of any vibrational oscillations. {{fontcolor1|red| Yes. Good job. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:53, 7 July 2020 (BST)}}

[[File:Fig4_Yanda.png|400px|thumb|center|Figure 4. Internuclear Distance vs Time plot for finding transition state position.]]

== Calculation of Reaction Path ==

{| class="wikitable" border="1"
! Calculations !! Steps !! r1 / pm !! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| MEP || 50000 || 91.77 || 90.77 || 0 || 0
|}

The Potential Energy Surface for MEP Calculation is shown in Figure 5 below.

[[File:Fig5_Yanda.png|400px|thumb|center|Figure 5.Potential Energy Surface Diagram with MEP.]]

{| class="wikitable" border="1"
! Calculations !! Steps !! r1 / pm !! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| Dynamics || 500 || 91.77 || 90.77 || 0 || 0
|}

Similarly, the Potential Energy Surface for Dynamics Calculation is shown in Figure 6 below.

[[File:Fig6_Yanda.png|400px|thumb|center|Figure 6.Potential Energy Surface Diagram with Dynamics.]]

Comparing the Dynamics (Figure 6) and the MEP (Figure 5), it can be seen that the MEP does not take into account of vibrational oscillations within the atoms, but just passing through all the potential minimum. {{fontcolor1|red| yes. the MEP seeks the nearest local minimum. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:54, 7 July 2020 (BST)}} On the other hand, the trajectory just calculated takes into account of the momentum changes during the reaction by oscillating about the minimum potential energy. In addition, during the calculations, 50000 steps were required to complete the minimum energy path in MEP, while 500 steps were required in Dynamics.

The "Internuclear Distances vs Time" and "Momenta vs Time" graphs for Dynamics calculations are respectively plotted in Figure 7 and 8 below.

[[File:Fig7_Yanda.png|400px|thumb|center|Figure 7. Internuclear Distances vs Time using Dynamics.]]

[[File:Fig8_Yanda.png|400px|thumb|center|Figure 8. Momenta vs Time using Dynamics.]]

Figure 7 and 8 can help to describe the reaction. Distance r1 (B-C) and r2 (A-C) are seen seen to increase as A moves further away from B-C after the transition state. After an extended period of time, it can be seen that A-C and B-C distances increases linearly, suggesting that the distance A-B is constant and that B-C is moving further away from A. Also,initially, A-B decreases slightly before plateauing/oscillating, suggesting that the bond between A-B has formed and that they are vibrating up and down from equilibrium. The vibration is further suggested by the oscillation of A-B in the Momenta vs. Time graph in Figure 8.

The "Internuclear Distances vs Time" and "Momenta vs Time" graphs for MEP calculations are respectively plotted in Figure 9 and 10 below.

[[File:Fig9_Yanda.png|400px|thumb|center|Figure 9. Internuclear Distances vs Time using MEP.]]

[[File:Fig10_Yanda.png|400px|thumb|center|Figure 10. Momenta vs Time using MEP.]]

With reference to Figure 9 and 10, the internuclear distance r2 (A-B) decreases rapidly before converging to a limit and plateauing, while r1 (B-C), hence A-C increases rapidly. The whole system has 0 momentum because the momenta/velocities are always reset to zero in each time step.

Also, the final values of the positions r1(t), r2(t), p1(t) and p2(t) for the trajectory for large enough t, at t=100fs are calculated in the table below.

{| class="wikitable" border="1"
! r1 / pm!! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| 735.414|| 75.4847|| 5.08082 || 2.02769
|}

If the initial conditions were r1 = rts and r2 = rts + 1 pm instead, the reaction path will move in the opposite direction, exiting from the opposite direction in the potential energy surface (Figure 11) whereby distance r2 (A-B) increases while the distance r1 (B-C) stays constant with only vibrations of the bond similar to the previous conditions (Figure 12 and 13). Hence, in the first case the transition state moves towards AB + C, while the reverse will be in this case favoring the formation of products A + BC, as A start off further to B than that of C to B.

[[File:Fig11_Yanda.png|400px|thumb|center|Figure 11. Potential Energy Surface with Dynamics with initial conditions reversed.]]
[[File:Fig12_Yanda.png|400px|thumb|center|Figure 12. Internuclear Distances vs Time with Dynamics with initial conditions reversed.]]
[[File:Fig13_Yanda.png|400px|thumb|center|Figure 13. Momenta vs Time with Dynamics with initial conditions reversed.]]

The distorted transition state, where the values of r1=rts and r2=rts + 1 pm would be inverted for as the previous calculations, with the graphs for internuclear distance and momenta also inverted, such that graph of B-C appears as the original A-B.

Finally,a calculation was setup where the initial positions correspond to the final positions of the trajectory calculative, above, the same final momenta values but with their signs reversed.

{| class="wikitable" border="1"
! r1 / pm!! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| 735.414|| 75.4847|| -5.08082 || -2.02769
|}

[[File:Fig14_Yanda.png|400px|thumb|center|Figure 14. Potential Energy Surface with Dynamics moving from products to estimated Transition State.]]
[[File:Fig15_Yanda.png|400px|thumb|center|Figure 15. Internuclear Distances vs Time with Dynamics moving from products to estimated Transition State.]]
[[File:Fig16_Yanda.png|400px|thumb|center|Figure 16. Momenta vs Time with Dynamics moving from products to estimated Transition State.]]

This situation is the reverse of the original situation in which now, the species start off as reactants with just the right starting positions and momenta to reach the slightly distorted transition state. Technically, the three graphs in this situation should be the mirror image of the three graphs in the original situation. However, a perfect symmetrical relationship is not observed due to random errors in estimating through rounding off or sampling errors of the final positions and momenta from the Internuclear Distances vs Time graph and Momenta vs Time graph respectively.For example, in Figure 15, the distance between B-C should be constant after the transition state, but was not the case due to a errors in the initial input data.

{{fontcolor1|red| Fantastic. Really good use of figures here, clear discussion. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:57, 7 July 2020 (BST)}}

== Reactive and Unreactive Trajectories ==

For the initial positions r1 = 74 pm and r2 = 200 pm, various trajectories were ran with different momentum values and combinations as tabulated below.

{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280|| Reactive || Reactive as reaction path can get over maximum at TS and form products. Straight trajectory centered around Dynamics suggest molecule B-C does not vibrate until A collides into it, with sufficient momentum that subsequently resulted in molecule A-B with vibrational oscillation after TS || [[File:Fig17_Yanda.png|300px]] [[File:Fig18_Yanda.png|300px]]
|-
| -3.1 || -4.1 || -420.077 || Unreactive || Not reactive as reactive path unable to get over maximum at TS and form products. The PES curve suggests that the trajectory is vibrational/oscillating throughout and that the reaction path does not move past the TS. It turns back before the TS once the potential energy exceeds the kinetic energy of the species of the momenta given || [[File:Fig19_Yanda.png|300px]] [[File:Fig20_Yanda.png|300px]]
|-
| -3.1 || -5.1|| -413.977 || Reactive|| Reactive as similar to the first scenario. Note that higher initial momentum of molecule B-C results in larger vibrations as visible in the PES curve || [[File:Fig21_Yanda.png|300px]] [[File:Fig22_Yanda.png|300px]]
|-
| -5.1|| -10.1|| -357.277 || Unreactive || Initial momenta large enough for reaction path to attain TS, but the activation barrier was crossed twice such that eventually the reactants were reformed again. At TS, A-B-C, the initial momentum was not large enough to form stable products. As such, reaction path recrosses the TS, as visible from the contour map, back into the reactions, reforming A + BC. Much larger vibrational oscillations that the previous scenarios due to large initial momentum as visible from the PES curve || [[File:Fig23_Yanda.png|300px]] [[File:Fig24_Yanda.png|300px]]
|-
| -5.1|| -10.6|| -349.477 || Reactive || Similar to the previous scenario, but that now the initial momenta are larger enough to allow three crossing of the activation barrier, hence passing through the TS thrice, eventually forming the products A + BC|| [[File:Fig25_Yanda.png|300px]] [[File:Fig26_Yanda.png|300px]]
|}

From the table, it can be seen that while a minimum initial momentum of the species are needed to cross the activation barrier and over the TS to form the products, a much initial momentum does not guarantee product formation as can be seen in the second last scenario. As such, the conditions required for enough momentum given the initial preconditions as listed are: -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1. {{fontcolor1|red| Great. Did you check the bounds for P2 momentum also? Your last trajectory falls outside of these bounds... }}

== Transition State Theory ==

The deviations from the experimental values and the values describes in the Transition State Theory (TST) arises because of the assumptions within the theory itself. First of all, the TST adheres to the Born-Oppenheimer approximation, where the nuclear and electronic motions are decoupled. The Transition State Theory assumes that only the reactants are in constant equilibrium with the transition state structure, and that the energy of the particles follow a Boltzmann distribution. In addition, once the reactants have trajectories with a kinetic energy along the reaction coordinate greater than the activation energy, it will be reactive, and that the products will definitely form, ignoring the possibility that the transition state structure will collapse back to the reactants. However, experimentally, it has been seen in the few scenarios above where the initial momentum (p1=-5.1, p2=-10.1), hence kinetic energy is high enough to result in multiple crossing of the activation energy barrier through multiple forward and backward directions. Theoretically, the theory suggests only a single crossing of the activation energy barrier in the forward direction is possible. Also, note that in the transition state, the motion along the reaction coordinate can be treated classically. <ref name="MRD"/>

Finally, the theory fails for some reactions at elevated temperatures due to more complex molecule motions or low temperatures due to quantum tunneling, whereby the transition state need not be reached to form products.<ref name="MDACK"/> {{fontcolor1|red| Yes. Great response. Thank you for including references [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 10:02, 7 July 2020 (BST)}}

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

The energetics of both F + H2 and the reverse H + HF were investigated and the findings are summarized in the table below. The method to extract these data are explained further in this section.

{| class="wikitable" border="1"
! Reaction !! Transition State Energy / kJ/mol !! Reactants Energy / kJ/mol !! Activation Energy / kJ/mol !! Type of Reaction
|-
| F + H2 || -433.980 || -435.100|| +1.120 || Exothermic
|-
| H + HF || -433.980 || -560.700 || +126.720 || Endothermic
|}

The reaction between F and H2 is an exothermic reaction owing to the stronger H-F bond relative to the H-H bond. Since enthalpy of reaction is calculated from energy of bond breaking - energy of bond making, the stronger negative value arising from breaking the stronger H-F bond resulted in an exothermic reaction. The stronger H-F bond can be explained due to the larger electronegativity difference between H and F, for which F= 4.0 and H = 2.1. <ref name="Physical Chemistry"/> As a result of the electronegativity difference, the F atom pulls electron density closer to itself, resulting in a highly polarized H-F bond, adding on an extra contribution to the bond strength. H-H bond, on the other hand, is a simple covalent bond. Literature value suggests that the bond dissociation energies of H-H and H-F are 432 kJ/mol and 565 kJ/mol respectively, further suggesting that the enthalpy released from breaking H-F bond is much higher than that of H-H bonds. Therefore, the reaction between F and H2 to form HF and H is energetically favorable, with an overall exothermic reaction.

The reverse reaction, H + HF is therefore endothermic, thereby requiring external energy in order to break the H-F bond, whereby the formation of the H-H bond does not compensate for.

The Hammond's postulate states that the transition state of a reaction resembles either the reactants or the products depending on which it is closer in energy. Hence, in an exothermic reaction, the transition state is closer to the reactants in energy. In an endothermic reaction, the transition state is closer to the products in energy.<ref name="OC"/>

Referring to the table above, for the forward reaction between F and H2, since the reactants energy is much closer to the transition state energy as compared to the reactions energy and the transition state energy in H + HF backward reaction, the forward reaction is the exothermic reaction, and the backward reaction is the endothermic reaction. This can also be determined graphically. With reference to the PES diagram and Energy vs Time plot below for the reaction between F and H2, it shows that the transition state has been attained, with no net change in potential energy. Furthermore, the PES suggests that the reactants are higher in energy than the products, indicating an exothermic reaction.

[[File:Fig27_Yanda.png|center|300px|thumb|Figure 27. Energy vs. Time Plot for F + H2. ]]
[[File:Fig28_Yanda.png|center|300px|thumb|Figure 28. Potential Energy Surface Diagram for F + H2. ]]

Similarly, Figure 29 below shows the transition state from the reverse reaction H + HF. Comparatively, it can be seen that the transition state closely resembles the products, which are now higher in energy than the reactants, thereby deeming it an endothermic reaction.

[[File:Fig29_Yanda.png|center|300px|thumb|Figure 29. Potential Energy Surface Diagram for H + HF. ]]

The approximate position of the transition state, is found to be r1 = BC = 181.1 pm and r2 = AB = 74.5 pm .The potential energy at the TS is then found to be -433.980 kJ/mol.

The activation energy of the reaction was found by slightly perturbing it from the TS and finding the energy difference of the products to the transition state energy as shown in the tables and Figures 30-33 below.

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| MEP || 2000 || 0.1 || 181.1 || 74.5 || 0 || 0
|}

[[File:Fig30_Yanda.png|center|300px|thumb|Figure 30. Contour Map for the reaction from H2 + F to HF + H. ]]
[[File:Fig31_Yanda.png|center|300px|thumb|Figure 31. Energy vs Time diagram for the reaction from H2 + F to HF + H. ]]

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| MEP || 2000 || 0.1 || 182.1 || 74.5 || 0 || 0
|}

[[File:Fig32_Yanda.png|center|300px|thumb|Figure 32. Contour Map for the reaction from HF + H to H2 + F. ]] [[File:Fig33_Yanda.png|center|300px|thumb|Figure 33. Energy vs Time diagram for the reaction from HF + H to H2 + F. ]]

By finding the trajectory down the MEP with the initial conditions, then inserting the last geometry as the initial condition. After which, replace one of the bond distance of the reactants to an extremely huge number, while keeping the bond distance of the product the same, the potential energy of the reactants and the products can be found. Though, it must be noted that this is just an approximation, as the exact potential energy is when the potential energy curve stabilizes. Ultimately, the potential energy of the reactants are found, can be used to find the activation energies, as summarized in the first table above.

{{fontcolor1|red| Brilliant. Well documented method with references and figures to show where your arguments can be drawn from. Well done.
[[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 10:05, 7 July 2020 (BST)}}

==Reaction Dynamics==

The following reaction condition resulted in the reaction from H2 + F -> HF + H

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| Dynamics || 1400 || 0.5 || 220 || 71 || -1.5 || -0.3
|}

The conservation of energy suggests that total energy is constant through the course of reaction. Hence, in the exothermic reaction between H2 and F to form HF + H, the loss in potential energy from the breaking of bonds is replaced by the increase in kinetic energy as seen in the Energy Vs Time graph below. The kinetic energy can be of three forms: translational, rotational or vibrational. Rotational motion is ignored here as according to the animation, it is currently working on a 1D system.

A bomb calorimeter can be used to measure the overall gain in kinetic energy, both translational and kinetic by measuring the transfer of these energy from the products to the walls of the calorimeter in the form of heat. The heat transferred can then be measured in the form of change in temperature with a thermal sensor. <ref name="Physical Chemistry"/>

However, in order to differentiate the respective contributions of translational and vibrational energy, further experimental is required. To determine the vibrational energy, IR spectroscopy can be utilized. Before the reactants collide, most of the bonds occupy lowest energy vibrational levels. However, during a collision, higher level vibrational modes are excited as potential energy is convereted into vibrational kinetic energy. In IR spectroscopy, the higher level modes are characterized as overtones signals where energy levels move from initially 0 to 1, to 0 to 2 or 0 to 3, or even higher. These overtones can be detected and seen with higher wavenumbers as compared to the normal 0 to 1 transition. <ref name="Physical Chemistry"/> {{fontcolor1|red| Yes. Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 10:06, 7 July 2020 (BST)}}

[[File:Fig34_Yanda.png|center|300px|thumb|Figure 34. Contour Map of reaction from H2 + F to HF + H ]]
[[File:Fig35_Yanda.png|center|300px|thumb|Figure 35. Internuclear Distance vs Time Diagram of reaction from H2 + F to HF + H ]]
[[File:Fig36_Yanda.png|center|300px|thumb|Figure 36. Energy vs Time Diagram of reaction from H2 + F to HF + H ]]

==The Polanyi Rules==

The Polanyi rules state that for reactants with a given momentum along the direction of the reaction path, the translational energy, as opposed to the vibrational energy, is preferred in overcoming the early transition state in an exothermic reaction. The opposite will be true for endothermic reactions, where vibrational energy takes priority over translational energy to successfully form stable products.<ref name="J. Phys. Chem."/>

Reaction trajectories were calculated for the exothermic reaction of F + H2 with 1000 steps, but by varying the momentum of r1, the momentum of H-H bond, with all other conditions kept constant such that only the vibration energy of the H2 is varied. This is to investigate an early TS, since it is exothermic.

{| class="wikitable" border=1
! rH-F / pm !! rH-H / pm !! pHF/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1 !! Reactive Trajectory? || Contour Plot
|-
| 230 || 74 || -1.0 || -3.0 || No || [[File:Fig37_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || -2.0 || No || [[File:Fig38_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || -1.0 || No || [[File:Fig39_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 0 || No || [[File:Fig40_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 1.0 || No || [[File:Fig41_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 2.0 || No || [[File:Fig42_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 3.0 || No || [[File:Fig43_Yanda.png|200px]]
|}

However, keeping the same initial positions and everything else constant, except that the momentum pHF is increased to -1.5 while the momentum of pHH is reduced to 0 and is shown in the figure below. It resulted in a reactive trajectory, which suggested that the increase in translational energy of the system has greater effect in allowing the transition state to be overcome to form stable products.

[[File:Fig44_Yanda.png|300px|thumb|center|Figure 44. Reactive trajectory of F + H2 to HF + H reaction.]]

Next, the endothermic reaction of HF + H to F + H2, which is a late TS is investigated. Similarly, everything is kept constant, apart from pHF before keeping it constant and increasing pHH. The bond distance of HF is kept at 92 nm according to literature.

{| class="wikitable" border=1
! rH-F / pm !! rH-H / pm !! pHF/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1 !! Reactive Trajectory? || Contour Plot
|-
| 230 || 92 || 0 || -1.0 || No || [[File:Fig45_Yanda.png|200px]]
|-
| 230 || 92 || -5.0 || -1.0 || No || [[File:Fig46_Yanda.png|200px]]
|-
| 230 || 92 || -10.0 || -1.0 || No || [[File:Fig47_Yanda.png|200px]]
|-
| 230 || 92 || -15.0 || -1.0 || No || [[File:Fig48_Yanda.png|200px]]
|-
| 230 || 92 || -20.0 || -1.0 || No || [[File:Fig50_Yanda.png|200px]]
|-
| 230 || 92 || -25.0 || -1.0 || No || [[File:Fig51_Yanda.png|200px]]
|-
| 230 || 92 || -30.0 || -1.0 || Yes || [[File:Fig52_Yanda.png|200px]]
|}

Hence, it has shown that by increasing the vibrational energy of the H-F bond for the endothermic reaction while keep translational energy fixed at a low vaue resulted in the formation of products.

As such, Polanyi's empirical rules have been shown for both exothermic reactions (early TS) and endothermic reactions (late TS). However, it should be noted that such rules might not work for all cases due to imbalance distribution of translational and vibrational energies or errors in calculating energies, be it bond energies or trajectory calculations.
{{fontcolor1|red| Great report, really thorough and well presented with references throughout. 5/5 [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 10:07, 7 July 2020 (BST)}}

=References=

<references>
<ref name="CP3MD">Imperial College London, https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD, (accessed May 2018).</ref>
<ref name="Plot"> C.D., Sherrill, http://vergil.chemistry.gatech.edu/courses/chem6485/pdf/pes-lecture.pdf, (accessed May 2018). </ref>
<ref name="MRD">R. D. Levine, Molecular Reaction Dynamics, Cambridge University Press, Cambridge, 2005, pp. 202-203.</ref>
<ref name="MDACK">G. D. Billing and K. V. Mikkelsen, Introduction to Molecular Dynamics and Chemical Kinetics, John Wiley & Sons, New York, 1996, p. 31.</ref>
<ref name="Physical Chemistry">P. Atkins and J. de Paula, Atkins' Physical Chemistry, Oxford University Press, Oxford, 10th edn., 2014, pp. 65, 71 and 986.</ref>
<ref name="OC">J. Clayden, N. Greeves and S. Warren, Organic Chemistry, Oxford University Press, Oxford, 2nd edn., 2012, pp. 989.</ref>
<ref name="J. Phys. Chem.">C. Daniel et al., J. Phys. Chem., 1993, 97, 12485-12490.</ref>
</references>

MRD:yw14815

2020-07-07T09:06:13Z

Mak214: /* Reaction Dynamics */

=Introduction=

This Wiki page outlines the study of the molecular reaction dynamics of two simple triatomic systems: H + H2 and F + H2. The atoms are 180° to each other, with a 1D space symplification. The atoms are labelled A, B and C, with vectors r1 and r2 representing vectors BA and BC in pm respectively. Momentum are also measured in p1 and p2 in g mol-1 pm fs-1. Below is the systematic diagram of the triatomic system model, which will be referred to throughout the whole wiki. <ref name="CP3MD"/>

[[File:Fig1_Yanda.png|400px|thumb|center|Figure 1. A triatomic system model. Diagram from Imperial College London, https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD (accessed May 2020.)]]

=EXERCISE 1: H + H2 system=
== Potential Energy Surfaces ==

A potential energy surface is a visual representation that describes the potential energy of a system, such as a collective of atoms, in terms of parameters, which normally is the relative positions of the reactants and the products. A reaction path is shown to be the gradient line on a PES diagram, and represents both potential energy and internuclear distances between atoms in the system as a function of the reaction coordinate, thereby modelling their changes through the progress of the reaction.

The PES diagram for Figure 1 above was calculated for the reaction system H + H2 using the parameters given:
{| class="wikitable"
!Calculations
!Steps
!r1 / pm
!r2 / pm
!p1 / gmol-1pmfs-1
!p2 / gmol-1pmfs-1
|-
|Dynamics
|500
|74
|230
|0.0
|<nowiki>-5.2</nowiki>
|}

[[File:Fig2_Yanda.png|400px|thumb|center|Figure 2.Potential Energy Surface for the H + H2 reaction path.]]

The reaction path, which passes through each potential energy minimum, including the transition state, which is a first order saddle point at which the potential energy is at its maximum (highest energy minimum) and where its gradient is 0, hence ∂V('''r''')/∂'''r '''=0 and the second differential at the TS is less than 0.Furthermore, the determinant of the Hessian Matrix is smaller than 0, with the eigenvalues of the Hessian Matrix being one positive and one negative

Local minimum, on the other hand,while also have the first derivative ∂V('''r''')/∂'''r '''=0, its second differential at the minimum is greater than 0, with the eigenvalues of the Hessian Matrix both being positive and the determinant of the Hessian Matrix greater than 0.

Literally speaking, the main difference between transition state and a local minimum is that the transition state is in fact a local energy maximum. Initially, with 0 initial momentum, both transition state and local minimum will remain there forever. However, with a very small momentum added towards the reactants direction, the transition state will be passed and will roll over towards the reactants, and vice versa for products. Testing the trajectories near the transition state is a method used to identify the presence of the transition state. On the other hand, a slightly perturbation from the local minimum will only result in rolling back towards the local minimum, because only a sufficient energy is needed to overcome the energy barrier to escape the "well" of the local minimum. In the PES diagram, a local minimum represents an intermediate, which can be represented in a 2D diagram below.<ref name="Plot"/>

[[File:Fig3_Yanda.PNG|400px|thumb|center|Figure 3. Potential Energy Surface Diagram showing the Transition State. Diagram from http://vergil.chemistry.gatech.edu/courses/chem6485/pdf/pes-lecture.pdf (accessed May 2020).]]

Another interesting observation is that the gradient along the reaction pathway is least steep close to both sides of the transition state, suggesting the point at which bonds are in the process of breaking and forming, hence lower degree of association and less vibrational oscillations. {{fontcolor1|red| Great. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:50, 7 July 2020 (BST)}}

== Locating the Transition State ==

The best estimate of the transition state position (rts) is 90.77 pm. The transition state should be symmetric as it represents the state of the reaction where old bonds and new bonds are in the middle of being broken and formed. As such, given the initial conditions of both momentum p1=0 and p2=0, the two internuclear distances r1 and r2 must be equal at the transitions state. The constant internuclear distances can be understood and found with a "Internuclear Distances vs Time plot" shown in Figure 4 below, where the straight horizontal line suggests constant potential energy with the absence of any vibrational oscillations. {{fontcolor1|red| Yes. Good job. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:53, 7 July 2020 (BST)}}

[[File:Fig4_Yanda.png|400px|thumb|center|Figure 4. Internuclear Distance vs Time plot for finding transition state position.]]

== Calculation of Reaction Path ==

{| class="wikitable" border="1"
! Calculations !! Steps !! r1 / pm !! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| MEP || 50000 || 91.77 || 90.77 || 0 || 0
|}

The Potential Energy Surface for MEP Calculation is shown in Figure 5 below.

[[File:Fig5_Yanda.png|400px|thumb|center|Figure 5.Potential Energy Surface Diagram with MEP.]]

{| class="wikitable" border="1"
! Calculations !! Steps !! r1 / pm !! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| Dynamics || 500 || 91.77 || 90.77 || 0 || 0
|}

Similarly, the Potential Energy Surface for Dynamics Calculation is shown in Figure 6 below.

[[File:Fig6_Yanda.png|400px|thumb|center|Figure 6.Potential Energy Surface Diagram with Dynamics.]]

Comparing the Dynamics (Figure 6) and the MEP (Figure 5), it can be seen that the MEP does not take into account of vibrational oscillations within the atoms, but just passing through all the potential minimum. {{fontcolor1|red| yes. the MEP seeks the nearest local minimum. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:54, 7 July 2020 (BST)}} On the other hand, the trajectory just calculated takes into account of the momentum changes during the reaction by oscillating about the minimum potential energy. In addition, during the calculations, 50000 steps were required to complete the minimum energy path in MEP, while 500 steps were required in Dynamics.

The "Internuclear Distances vs Time" and "Momenta vs Time" graphs for Dynamics calculations are respectively plotted in Figure 7 and 8 below.

[[File:Fig7_Yanda.png|400px|thumb|center|Figure 7. Internuclear Distances vs Time using Dynamics.]]

[[File:Fig8_Yanda.png|400px|thumb|center|Figure 8. Momenta vs Time using Dynamics.]]

Figure 7 and 8 can help to describe the reaction. Distance r1 (B-C) and r2 (A-C) are seen seen to increase as A moves further away from B-C after the transition state. After an extended period of time, it can be seen that A-C and B-C distances increases linearly, suggesting that the distance A-B is constant and that B-C is moving further away from A. Also,initially, A-B decreases slightly before plateauing/oscillating, suggesting that the bond between A-B has formed and that they are vibrating up and down from equilibrium. The vibration is further suggested by the oscillation of A-B in the Momenta vs. Time graph in Figure 8.

The "Internuclear Distances vs Time" and "Momenta vs Time" graphs for MEP calculations are respectively plotted in Figure 9 and 10 below.

[[File:Fig9_Yanda.png|400px|thumb|center|Figure 9. Internuclear Distances vs Time using MEP.]]

[[File:Fig10_Yanda.png|400px|thumb|center|Figure 10. Momenta vs Time using MEP.]]

With reference to Figure 9 and 10, the internuclear distance r2 (A-B) decreases rapidly before converging to a limit and plateauing, while r1 (B-C), hence A-C increases rapidly. The whole system has 0 momentum because the momenta/velocities are always reset to zero in each time step.

Also, the final values of the positions r1(t), r2(t), p1(t) and p2(t) for the trajectory for large enough t, at t=100fs are calculated in the table below.

{| class="wikitable" border="1"
! r1 / pm!! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| 735.414|| 75.4847|| 5.08082 || 2.02769
|}

If the initial conditions were r1 = rts and r2 = rts + 1 pm instead, the reaction path will move in the opposite direction, exiting from the opposite direction in the potential energy surface (Figure 11) whereby distance r2 (A-B) increases while the distance r1 (B-C) stays constant with only vibrations of the bond similar to the previous conditions (Figure 12 and 13). Hence, in the first case the transition state moves towards AB + C, while the reverse will be in this case favoring the formation of products A + BC, as A start off further to B than that of C to B.

[[File:Fig11_Yanda.png|400px|thumb|center|Figure 11. Potential Energy Surface with Dynamics with initial conditions reversed.]]
[[File:Fig12_Yanda.png|400px|thumb|center|Figure 12. Internuclear Distances vs Time with Dynamics with initial conditions reversed.]]
[[File:Fig13_Yanda.png|400px|thumb|center|Figure 13. Momenta vs Time with Dynamics with initial conditions reversed.]]

The distorted transition state, where the values of r1=rts and r2=rts + 1 pm would be inverted for as the previous calculations, with the graphs for internuclear distance and momenta also inverted, such that graph of B-C appears as the original A-B.

Finally,a calculation was setup where the initial positions correspond to the final positions of the trajectory calculative, above, the same final momenta values but with their signs reversed.

{| class="wikitable" border="1"
! r1 / pm!! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| 735.414|| 75.4847|| -5.08082 || -2.02769
|}

[[File:Fig14_Yanda.png|400px|thumb|center|Figure 14. Potential Energy Surface with Dynamics moving from products to estimated Transition State.]]
[[File:Fig15_Yanda.png|400px|thumb|center|Figure 15. Internuclear Distances vs Time with Dynamics moving from products to estimated Transition State.]]
[[File:Fig16_Yanda.png|400px|thumb|center|Figure 16. Momenta vs Time with Dynamics moving from products to estimated Transition State.]]

This situation is the reverse of the original situation in which now, the species start off as reactants with just the right starting positions and momenta to reach the slightly distorted transition state. Technically, the three graphs in this situation should be the mirror image of the three graphs in the original situation. However, a perfect symmetrical relationship is not observed due to random errors in estimating through rounding off or sampling errors of the final positions and momenta from the Internuclear Distances vs Time graph and Momenta vs Time graph respectively.For example, in Figure 15, the distance between B-C should be constant after the transition state, but was not the case due to a errors in the initial input data.

{{fontcolor1|red| Fantastic. Really good use of figures here, clear discussion. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:57, 7 July 2020 (BST)}}

== Reactive and Unreactive Trajectories ==

For the initial positions r1 = 74 pm and r2 = 200 pm, various trajectories were ran with different momentum values and combinations as tabulated below.

{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280|| Reactive || Reactive as reaction path can get over maximum at TS and form products. Straight trajectory centered around Dynamics suggest molecule B-C does not vibrate until A collides into it, with sufficient momentum that subsequently resulted in molecule A-B with vibrational oscillation after TS || [[File:Fig17_Yanda.png|300px]] [[File:Fig18_Yanda.png|300px]]
|-
| -3.1 || -4.1 || -420.077 || Unreactive || Not reactive as reactive path unable to get over maximum at TS and form products. The PES curve suggests that the trajectory is vibrational/oscillating throughout and that the reaction path does not move past the TS. It turns back before the TS once the potential energy exceeds the kinetic energy of the species of the momenta given || [[File:Fig19_Yanda.png|300px]] [[File:Fig20_Yanda.png|300px]]
|-
| -3.1 || -5.1|| -413.977 || Reactive|| Reactive as similar to the first scenario. Note that higher initial momentum of molecule B-C results in larger vibrations as visible in the PES curve || [[File:Fig21_Yanda.png|300px]] [[File:Fig22_Yanda.png|300px]]
|-
| -5.1|| -10.1|| -357.277 || Unreactive || Initial momenta large enough for reaction path to attain TS, but the activation barrier was crossed twice such that eventually the reactants were reformed again. At TS, A-B-C, the initial momentum was not large enough to form stable products. As such, reaction path recrosses the TS, as visible from the contour map, back into the reactions, reforming A + BC. Much larger vibrational oscillations that the previous scenarios due to large initial momentum as visible from the PES curve || [[File:Fig23_Yanda.png|300px]] [[File:Fig24_Yanda.png|300px]]
|-
| -5.1|| -10.6|| -349.477 || Reactive || Similar to the previous scenario, but that now the initial momenta are larger enough to allow three crossing of the activation barrier, hence passing through the TS thrice, eventually forming the products A + BC|| [[File:Fig25_Yanda.png|300px]] [[File:Fig26_Yanda.png|300px]]
|}

From the table, it can be seen that while a minimum initial momentum of the species are needed to cross the activation barrier and over the TS to form the products, a much initial momentum does not guarantee product formation as can be seen in the second last scenario. As such, the conditions required for enough momentum given the initial preconditions as listed are: -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1. {{fontcolor1|red| Great. Did you check the bounds for P2 momentum also? Your last trajectory falls outside of these bounds... }}

== Transition State Theory ==

The deviations from the experimental values and the values describes in the Transition State Theory (TST) arises because of the assumptions within the theory itself. First of all, the TST adheres to the Born-Oppenheimer approximation, where the nuclear and electronic motions are decoupled. The Transition State Theory assumes that only the reactants are in constant equilibrium with the transition state structure, and that the energy of the particles follow a Boltzmann distribution. In addition, once the reactants have trajectories with a kinetic energy along the reaction coordinate greater than the activation energy, it will be reactive, and that the products will definitely form, ignoring the possibility that the transition state structure will collapse back to the reactants. However, experimentally, it has been seen in the few scenarios above where the initial momentum (p1=-5.1, p2=-10.1), hence kinetic energy is high enough to result in multiple crossing of the activation energy barrier through multiple forward and backward directions. Theoretically, the theory suggests only a single crossing of the activation energy barrier in the forward direction is possible. Also, note that in the transition state, the motion along the reaction coordinate can be treated classically. <ref name="MRD"/>

Finally, the theory fails for some reactions at elevated temperatures due to more complex molecule motions or low temperatures due to quantum tunneling, whereby the transition state need not be reached to form products.<ref name="MDACK"/> {{fontcolor1|red| Yes. Great response. Thank you for including references [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 10:02, 7 July 2020 (BST)}}

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

The energetics of both F + H2 and the reverse H + HF were investigated and the findings are summarized in the table below. The method to extract these data are explained further in this section.

{| class="wikitable" border="1"
! Reaction !! Transition State Energy / kJ/mol !! Reactants Energy / kJ/mol !! Activation Energy / kJ/mol !! Type of Reaction
|-
| F + H2 || -433.980 || -435.100|| +1.120 || Exothermic
|-
| H + HF || -433.980 || -560.700 || +126.720 || Endothermic
|}

The reaction between F and H2 is an exothermic reaction owing to the stronger H-F bond relative to the H-H bond. Since enthalpy of reaction is calculated from energy of bond breaking - energy of bond making, the stronger negative value arising from breaking the stronger H-F bond resulted in an exothermic reaction. The stronger H-F bond can be explained due to the larger electronegativity difference between H and F, for which F= 4.0 and H = 2.1. <ref name="Physical Chemistry"/> As a result of the electronegativity difference, the F atom pulls electron density closer to itself, resulting in a highly polarized H-F bond, adding on an extra contribution to the bond strength. H-H bond, on the other hand, is a simple covalent bond. Literature value suggests that the bond dissociation energies of H-H and H-F are 432 kJ/mol and 565 kJ/mol respectively, further suggesting that the enthalpy released from breaking H-F bond is much higher than that of H-H bonds. Therefore, the reaction between F and H2 to form HF and H is energetically favorable, with an overall exothermic reaction.

The reverse reaction, H + HF is therefore endothermic, thereby requiring external energy in order to break the H-F bond, whereby the formation of the H-H bond does not compensate for.

The Hammond's postulate states that the transition state of a reaction resembles either the reactants or the products depending on which it is closer in energy. Hence, in an exothermic reaction, the transition state is closer to the reactants in energy. In an endothermic reaction, the transition state is closer to the products in energy.<ref name="OC"/>

Referring to the table above, for the forward reaction between F and H2, since the reactants energy is much closer to the transition state energy as compared to the reactions energy and the transition state energy in H + HF backward reaction, the forward reaction is the exothermic reaction, and the backward reaction is the endothermic reaction. This can also be determined graphically. With reference to the PES diagram and Energy vs Time plot below for the reaction between F and H2, it shows that the transition state has been attained, with no net change in potential energy. Furthermore, the PES suggests that the reactants are higher in energy than the products, indicating an exothermic reaction.

[[File:Fig27_Yanda.png|center|300px|thumb|Figure 27. Energy vs. Time Plot for F + H2. ]]
[[File:Fig28_Yanda.png|center|300px|thumb|Figure 28. Potential Energy Surface Diagram for F + H2. ]]

Similarly, Figure 29 below shows the transition state from the reverse reaction H + HF. Comparatively, it can be seen that the transition state closely resembles the products, which are now higher in energy than the reactants, thereby deeming it an endothermic reaction.

[[File:Fig29_Yanda.png|center|300px|thumb|Figure 29. Potential Energy Surface Diagram for H + HF. ]]

The approximate position of the transition state, is found to be r1 = BC = 181.1 pm and r2 = AB = 74.5 pm .The potential energy at the TS is then found to be -433.980 kJ/mol.

The activation energy of the reaction was found by slightly perturbing it from the TS and finding the energy difference of the products to the transition state energy as shown in the tables and Figures 30-33 below.

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| MEP || 2000 || 0.1 || 181.1 || 74.5 || 0 || 0
|}

[[File:Fig30_Yanda.png|center|300px|thumb|Figure 30. Contour Map for the reaction from H2 + F to HF + H. ]]
[[File:Fig31_Yanda.png|center|300px|thumb|Figure 31. Energy vs Time diagram for the reaction from H2 + F to HF + H. ]]

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| MEP || 2000 || 0.1 || 182.1 || 74.5 || 0 || 0
|}

[[File:Fig32_Yanda.png|center|300px|thumb|Figure 32. Contour Map for the reaction from HF + H to H2 + F. ]] [[File:Fig33_Yanda.png|center|300px|thumb|Figure 33. Energy vs Time diagram for the reaction from HF + H to H2 + F. ]]

By finding the trajectory down the MEP with the initial conditions, then inserting the last geometry as the initial condition. After which, replace one of the bond distance of the reactants to an extremely huge number, while keeping the bond distance of the product the same, the potential energy of the reactants and the products can be found. Though, it must be noted that this is just an approximation, as the exact potential energy is when the potential energy curve stabilizes. Ultimately, the potential energy of the reactants are found, can be used to find the activation energies, as summarized in the first table above.

{{fontcolor1|red| Brilliant. Well documented method with references and figures to show where your arguments can be drawn from. Well done.
[[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 10:05, 7 July 2020 (BST)}}

==Reaction Dynamics==

The following reaction condition resulted in the reaction from H2 + F -> HF + H

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| Dynamics || 1400 || 0.5 || 220 || 71 || -1.5 || -0.3
|}

The conservation of energy suggests that total energy is constant through the course of reaction. Hence, in the exothermic reaction between H2 and F to form HF + H, the loss in potential energy from the breaking of bonds is replaced by the increase in kinetic energy as seen in the Energy Vs Time graph below. The kinetic energy can be of three forms: translational, rotational or vibrational. Rotational motion is ignored here as according to the animation, it is currently working on a 1D system.

A bomb calorimeter can be used to measure the overall gain in kinetic energy, both translational and kinetic by measuring the transfer of these energy from the products to the walls of the calorimeter in the form of heat. The heat transferred can then be measured in the form of change in temperature with a thermal sensor. <ref name="Physical Chemistry"/>

However, in order to differentiate the respective contributions of translational and vibrational energy, further experimental is required. To determine the vibrational energy, IR spectroscopy can be utilized. Before the reactants collide, most of the bonds occupy lowest energy vibrational levels. However, during a collision, higher level vibrational modes are excited as potential energy is convereted into vibrational kinetic energy. In IR spectroscopy, the higher level modes are characterized as overtones signals where energy levels move from initially 0 to 1, to 0 to 2 or 0 to 3, or even higher. These overtones can be detected and seen with higher wavenumbers as compared to the normal 0 to 1 transition. <ref name="Physical Chemistry"/> {{fontcolor1|red| Yes. Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 10:06, 7 July 2020 (BST)}}

[[File:Fig34_Yanda.png|center|300px|thumb|Figure 34. Contour Map of reaction from H2 + F to HF + H ]]
[[File:Fig35_Yanda.png|center|300px|thumb|Figure 35. Internuclear Distance vs Time Diagram of reaction from H2 + F to HF + H ]]
[[File:Fig36_Yanda.png|center|300px|thumb|Figure 36. Energy vs Time Diagram of reaction from H2 + F to HF + H ]]

==The Polanyi Rules==

The Polanyi rules state that for reactants with a given momentum along the direction of the reaction path, the translational energy, as opposed to the vibrational energy, is preferred in overcoming the early transition state in an exothermic reaction. The opposite will be true for endothermic reactions, where vibrational energy takes priority over translational energy to successfully form stable products.<ref name="J. Phys. Chem."/>

Reaction trajectories were calculated for the exothermic reaction of F + H2 with 1000 steps, but by varying the momentum of r1, the momentum of H-H bond, with all other conditions kept constant such that only the vibration energy of the H2 is varied. This is to investigate an early TS, since it is exothermic.

{| class="wikitable" border=1
! rH-F / pm !! rH-H / pm !! pHF/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1 !! Reactive Trajectory? || Contour Plot
|-
| 230 || 74 || -1.0 || -3.0 || No || [[File:Fig37_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || -2.0 || No || [[File:Fig38_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || -1.0 || No || [[File:Fig39_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 0 || No || [[File:Fig40_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 1.0 || No || [[File:Fig41_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 2.0 || No || [[File:Fig42_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 3.0 || No || [[File:Fig43_Yanda.png|200px]]
|}

However, keeping the same initial positions and everything else constant, except that the momentum pHF is increased to -1.5 while the momentum of pHH is reduced to 0 and is shown in the figure below. It resulted in a reactive trajectory, which suggested that the increase in translational energy of the system has greater effect in allowing the transition state to be overcome to form stable products.

[[File:Fig44_Yanda.png|300px|thumb|center|Figure 44. Reactive trajectory of F + H2 to HF + H reaction.]]

Next, the endothermic reaction of HF + H to F + H2, which is a late TS is investigated. Similarly, everything is kept constant, apart from pHF before keeping it constant and increasing pHH. The bond distance of HF is kept at 92 nm according to literature.

{| class="wikitable" border=1
! rH-F / pm !! rH-H / pm !! pHF/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1 !! Reactive Trajectory? || Contour Plot
|-
| 230 || 92 || 0 || -1.0 || No || [[File:Fig45_Yanda.png|200px]]
|-
| 230 || 92 || -5.0 || -1.0 || No || [[File:Fig46_Yanda.png|200px]]
|-
| 230 || 92 || -10.0 || -1.0 || No || [[File:Fig47_Yanda.png|200px]]
|-
| 230 || 92 || -15.0 || -1.0 || No || [[File:Fig48_Yanda.png|200px]]
|-
| 230 || 92 || -20.0 || -1.0 || No || [[File:Fig50_Yanda.png|200px]]
|-
| 230 || 92 || -25.0 || -1.0 || No || [[File:Fig51_Yanda.png|200px]]
|-
| 230 || 92 || -30.0 || -1.0 || Yes || [[File:Fig52_Yanda.png|200px]]
|}

Hence, it has shown that by increasing the vibrational energy of the H-F bond for the endothermic reaction while keep translational energy fixed at a low vaue resulted in the formation of products.

As such, Polanyi's empirical rules have been shown for both exothermic reactions (early TS) and endothermic reactions (late TS). However, it should be noted that such rules might not work for all cases due to imbalance distribution of translational and vibrational energies or errors in calculating energies, be it bond energies or trajectory calculations.

=References=

<references>
<ref name="CP3MD">Imperial College London, https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD, (accessed May 2018).</ref>
<ref name="Plot"> C.D., Sherrill, http://vergil.chemistry.gatech.edu/courses/chem6485/pdf/pes-lecture.pdf, (accessed May 2018). </ref>
<ref name="MRD">R. D. Levine, Molecular Reaction Dynamics, Cambridge University Press, Cambridge, 2005, pp. 202-203.</ref>
<ref name="MDACK">G. D. Billing and K. V. Mikkelsen, Introduction to Molecular Dynamics and Chemical Kinetics, John Wiley & Sons, New York, 1996, p. 31.</ref>
<ref name="Physical Chemistry">P. Atkins and J. de Paula, Atkins' Physical Chemistry, Oxford University Press, Oxford, 10th edn., 2014, pp. 65, 71 and 986.</ref>
<ref name="OC">J. Clayden, N. Greeves and S. Warren, Organic Chemistry, Oxford University Press, Oxford, 2nd edn., 2012, pp. 989.</ref>
<ref name="J. Phys. Chem.">C. Daniel et al., J. Phys. Chem., 1993, 97, 12485-12490.</ref>
</references>

MRD:yw14815

2020-07-07T09:05:07Z

Mak214: /* PES Inspection */

=Introduction=

This Wiki page outlines the study of the molecular reaction dynamics of two simple triatomic systems: H + H2 and F + H2. The atoms are 180° to each other, with a 1D space symplification. The atoms are labelled A, B and C, with vectors r1 and r2 representing vectors BA and BC in pm respectively. Momentum are also measured in p1 and p2 in g mol-1 pm fs-1. Below is the systematic diagram of the triatomic system model, which will be referred to throughout the whole wiki. <ref name="CP3MD"/>

[[File:Fig1_Yanda.png|400px|thumb|center|Figure 1. A triatomic system model. Diagram from Imperial College London, https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD (accessed May 2020.)]]

=EXERCISE 1: H + H2 system=
== Potential Energy Surfaces ==

A potential energy surface is a visual representation that describes the potential energy of a system, such as a collective of atoms, in terms of parameters, which normally is the relative positions of the reactants and the products. A reaction path is shown to be the gradient line on a PES diagram, and represents both potential energy and internuclear distances between atoms in the system as a function of the reaction coordinate, thereby modelling their changes through the progress of the reaction.

The PES diagram for Figure 1 above was calculated for the reaction system H + H2 using the parameters given:
{| class="wikitable"
!Calculations
!Steps
!r1 / pm
!r2 / pm
!p1 / gmol-1pmfs-1
!p2 / gmol-1pmfs-1
|-
|Dynamics
|500
|74
|230
|0.0
|<nowiki>-5.2</nowiki>
|}

[[File:Fig2_Yanda.png|400px|thumb|center|Figure 2.Potential Energy Surface for the H + H2 reaction path.]]

The reaction path, which passes through each potential energy minimum, including the transition state, which is a first order saddle point at which the potential energy is at its maximum (highest energy minimum) and where its gradient is 0, hence ∂V('''r''')/∂'''r '''=0 and the second differential at the TS is less than 0.Furthermore, the determinant of the Hessian Matrix is smaller than 0, with the eigenvalues of the Hessian Matrix being one positive and one negative

Local minimum, on the other hand,while also have the first derivative ∂V('''r''')/∂'''r '''=0, its second differential at the minimum is greater than 0, with the eigenvalues of the Hessian Matrix both being positive and the determinant of the Hessian Matrix greater than 0.

Literally speaking, the main difference between transition state and a local minimum is that the transition state is in fact a local energy maximum. Initially, with 0 initial momentum, both transition state and local minimum will remain there forever. However, with a very small momentum added towards the reactants direction, the transition state will be passed and will roll over towards the reactants, and vice versa for products. Testing the trajectories near the transition state is a method used to identify the presence of the transition state. On the other hand, a slightly perturbation from the local minimum will only result in rolling back towards the local minimum, because only a sufficient energy is needed to overcome the energy barrier to escape the "well" of the local minimum. In the PES diagram, a local minimum represents an intermediate, which can be represented in a 2D diagram below.<ref name="Plot"/>

[[File:Fig3_Yanda.PNG|400px|thumb|center|Figure 3. Potential Energy Surface Diagram showing the Transition State. Diagram from http://vergil.chemistry.gatech.edu/courses/chem6485/pdf/pes-lecture.pdf (accessed May 2020).]]

Another interesting observation is that the gradient along the reaction pathway is least steep close to both sides of the transition state, suggesting the point at which bonds are in the process of breaking and forming, hence lower degree of association and less vibrational oscillations. {{fontcolor1|red| Great. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:50, 7 July 2020 (BST)}}

== Locating the Transition State ==

The best estimate of the transition state position (rts) is 90.77 pm. The transition state should be symmetric as it represents the state of the reaction where old bonds and new bonds are in the middle of being broken and formed. As such, given the initial conditions of both momentum p1=0 and p2=0, the two internuclear distances r1 and r2 must be equal at the transitions state. The constant internuclear distances can be understood and found with a "Internuclear Distances vs Time plot" shown in Figure 4 below, where the straight horizontal line suggests constant potential energy with the absence of any vibrational oscillations. {{fontcolor1|red| Yes. Good job. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:53, 7 July 2020 (BST)}}

[[File:Fig4_Yanda.png|400px|thumb|center|Figure 4. Internuclear Distance vs Time plot for finding transition state position.]]

== Calculation of Reaction Path ==

{| class="wikitable" border="1"
! Calculations !! Steps !! r1 / pm !! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| MEP || 50000 || 91.77 || 90.77 || 0 || 0
|}

The Potential Energy Surface for MEP Calculation is shown in Figure 5 below.

[[File:Fig5_Yanda.png|400px|thumb|center|Figure 5.Potential Energy Surface Diagram with MEP.]]

{| class="wikitable" border="1"
! Calculations !! Steps !! r1 / pm !! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| Dynamics || 500 || 91.77 || 90.77 || 0 || 0
|}

Similarly, the Potential Energy Surface for Dynamics Calculation is shown in Figure 6 below.

[[File:Fig6_Yanda.png|400px|thumb|center|Figure 6.Potential Energy Surface Diagram with Dynamics.]]

Comparing the Dynamics (Figure 6) and the MEP (Figure 5), it can be seen that the MEP does not take into account of vibrational oscillations within the atoms, but just passing through all the potential minimum. {{fontcolor1|red| yes. the MEP seeks the nearest local minimum. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:54, 7 July 2020 (BST)}} On the other hand, the trajectory just calculated takes into account of the momentum changes during the reaction by oscillating about the minimum potential energy. In addition, during the calculations, 50000 steps were required to complete the minimum energy path in MEP, while 500 steps were required in Dynamics.

The "Internuclear Distances vs Time" and "Momenta vs Time" graphs for Dynamics calculations are respectively plotted in Figure 7 and 8 below.

[[File:Fig7_Yanda.png|400px|thumb|center|Figure 7. Internuclear Distances vs Time using Dynamics.]]

[[File:Fig8_Yanda.png|400px|thumb|center|Figure 8. Momenta vs Time using Dynamics.]]

Figure 7 and 8 can help to describe the reaction. Distance r1 (B-C) and r2 (A-C) are seen seen to increase as A moves further away from B-C after the transition state. After an extended period of time, it can be seen that A-C and B-C distances increases linearly, suggesting that the distance A-B is constant and that B-C is moving further away from A. Also,initially, A-B decreases slightly before plateauing/oscillating, suggesting that the bond between A-B has formed and that they are vibrating up and down from equilibrium. The vibration is further suggested by the oscillation of A-B in the Momenta vs. Time graph in Figure 8.

The "Internuclear Distances vs Time" and "Momenta vs Time" graphs for MEP calculations are respectively plotted in Figure 9 and 10 below.

[[File:Fig9_Yanda.png|400px|thumb|center|Figure 9. Internuclear Distances vs Time using MEP.]]

[[File:Fig10_Yanda.png|400px|thumb|center|Figure 10. Momenta vs Time using MEP.]]

With reference to Figure 9 and 10, the internuclear distance r2 (A-B) decreases rapidly before converging to a limit and plateauing, while r1 (B-C), hence A-C increases rapidly. The whole system has 0 momentum because the momenta/velocities are always reset to zero in each time step.

Also, the final values of the positions r1(t), r2(t), p1(t) and p2(t) for the trajectory for large enough t, at t=100fs are calculated in the table below.

{| class="wikitable" border="1"
! r1 / pm!! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| 735.414|| 75.4847|| 5.08082 || 2.02769
|}

If the initial conditions were r1 = rts and r2 = rts + 1 pm instead, the reaction path will move in the opposite direction, exiting from the opposite direction in the potential energy surface (Figure 11) whereby distance r2 (A-B) increases while the distance r1 (B-C) stays constant with only vibrations of the bond similar to the previous conditions (Figure 12 and 13). Hence, in the first case the transition state moves towards AB + C, while the reverse will be in this case favoring the formation of products A + BC, as A start off further to B than that of C to B.

[[File:Fig11_Yanda.png|400px|thumb|center|Figure 11. Potential Energy Surface with Dynamics with initial conditions reversed.]]
[[File:Fig12_Yanda.png|400px|thumb|center|Figure 12. Internuclear Distances vs Time with Dynamics with initial conditions reversed.]]
[[File:Fig13_Yanda.png|400px|thumb|center|Figure 13. Momenta vs Time with Dynamics with initial conditions reversed.]]

The distorted transition state, where the values of r1=rts and r2=rts + 1 pm would be inverted for as the previous calculations, with the graphs for internuclear distance and momenta also inverted, such that graph of B-C appears as the original A-B.

Finally,a calculation was setup where the initial positions correspond to the final positions of the trajectory calculative, above, the same final momenta values but with their signs reversed.

{| class="wikitable" border="1"
! r1 / pm!! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| 735.414|| 75.4847|| -5.08082 || -2.02769
|}

[[File:Fig14_Yanda.png|400px|thumb|center|Figure 14. Potential Energy Surface with Dynamics moving from products to estimated Transition State.]]
[[File:Fig15_Yanda.png|400px|thumb|center|Figure 15. Internuclear Distances vs Time with Dynamics moving from products to estimated Transition State.]]
[[File:Fig16_Yanda.png|400px|thumb|center|Figure 16. Momenta vs Time with Dynamics moving from products to estimated Transition State.]]

This situation is the reverse of the original situation in which now, the species start off as reactants with just the right starting positions and momenta to reach the slightly distorted transition state. Technically, the three graphs in this situation should be the mirror image of the three graphs in the original situation. However, a perfect symmetrical relationship is not observed due to random errors in estimating through rounding off or sampling errors of the final positions and momenta from the Internuclear Distances vs Time graph and Momenta vs Time graph respectively.For example, in Figure 15, the distance between B-C should be constant after the transition state, but was not the case due to a errors in the initial input data.

{{fontcolor1|red| Fantastic. Really good use of figures here, clear discussion. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:57, 7 July 2020 (BST)}}

== Reactive and Unreactive Trajectories ==

For the initial positions r1 = 74 pm and r2 = 200 pm, various trajectories were ran with different momentum values and combinations as tabulated below.

{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280|| Reactive || Reactive as reaction path can get over maximum at TS and form products. Straight trajectory centered around Dynamics suggest molecule B-C does not vibrate until A collides into it, with sufficient momentum that subsequently resulted in molecule A-B with vibrational oscillation after TS || [[File:Fig17_Yanda.png|300px]] [[File:Fig18_Yanda.png|300px]]
|-
| -3.1 || -4.1 || -420.077 || Unreactive || Not reactive as reactive path unable to get over maximum at TS and form products. The PES curve suggests that the trajectory is vibrational/oscillating throughout and that the reaction path does not move past the TS. It turns back before the TS once the potential energy exceeds the kinetic energy of the species of the momenta given || [[File:Fig19_Yanda.png|300px]] [[File:Fig20_Yanda.png|300px]]
|-
| -3.1 || -5.1|| -413.977 || Reactive|| Reactive as similar to the first scenario. Note that higher initial momentum of molecule B-C results in larger vibrations as visible in the PES curve || [[File:Fig21_Yanda.png|300px]] [[File:Fig22_Yanda.png|300px]]
|-
| -5.1|| -10.1|| -357.277 || Unreactive || Initial momenta large enough for reaction path to attain TS, but the activation barrier was crossed twice such that eventually the reactants were reformed again. At TS, A-B-C, the initial momentum was not large enough to form stable products. As such, reaction path recrosses the TS, as visible from the contour map, back into the reactions, reforming A + BC. Much larger vibrational oscillations that the previous scenarios due to large initial momentum as visible from the PES curve || [[File:Fig23_Yanda.png|300px]] [[File:Fig24_Yanda.png|300px]]
|-
| -5.1|| -10.6|| -349.477 || Reactive || Similar to the previous scenario, but that now the initial momenta are larger enough to allow three crossing of the activation barrier, hence passing through the TS thrice, eventually forming the products A + BC|| [[File:Fig25_Yanda.png|300px]] [[File:Fig26_Yanda.png|300px]]
|}

From the table, it can be seen that while a minimum initial momentum of the species are needed to cross the activation barrier and over the TS to form the products, a much initial momentum does not guarantee product formation as can be seen in the second last scenario. As such, the conditions required for enough momentum given the initial preconditions as listed are: -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1. {{fontcolor1|red| Great. Did you check the bounds for P2 momentum also? Your last trajectory falls outside of these bounds... }}

== Transition State Theory ==

The deviations from the experimental values and the values describes in the Transition State Theory (TST) arises because of the assumptions within the theory itself. First of all, the TST adheres to the Born-Oppenheimer approximation, where the nuclear and electronic motions are decoupled. The Transition State Theory assumes that only the reactants are in constant equilibrium with the transition state structure, and that the energy of the particles follow a Boltzmann distribution. In addition, once the reactants have trajectories with a kinetic energy along the reaction coordinate greater than the activation energy, it will be reactive, and that the products will definitely form, ignoring the possibility that the transition state structure will collapse back to the reactants. However, experimentally, it has been seen in the few scenarios above where the initial momentum (p1=-5.1, p2=-10.1), hence kinetic energy is high enough to result in multiple crossing of the activation energy barrier through multiple forward and backward directions. Theoretically, the theory suggests only a single crossing of the activation energy barrier in the forward direction is possible. Also, note that in the transition state, the motion along the reaction coordinate can be treated classically. <ref name="MRD"/>

Finally, the theory fails for some reactions at elevated temperatures due to more complex molecule motions or low temperatures due to quantum tunneling, whereby the transition state need not be reached to form products.<ref name="MDACK"/> {{fontcolor1|red| Yes. Great response. Thank you for including references [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 10:02, 7 July 2020 (BST)}}

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

The energetics of both F + H2 and the reverse H + HF were investigated and the findings are summarized in the table below. The method to extract these data are explained further in this section.

{| class="wikitable" border="1"
! Reaction !! Transition State Energy / kJ/mol !! Reactants Energy / kJ/mol !! Activation Energy / kJ/mol !! Type of Reaction
|-
| F + H2 || -433.980 || -435.100|| +1.120 || Exothermic
|-
| H + HF || -433.980 || -560.700 || +126.720 || Endothermic
|}

The reaction between F and H2 is an exothermic reaction owing to the stronger H-F bond relative to the H-H bond. Since enthalpy of reaction is calculated from energy of bond breaking - energy of bond making, the stronger negative value arising from breaking the stronger H-F bond resulted in an exothermic reaction. The stronger H-F bond can be explained due to the larger electronegativity difference between H and F, for which F= 4.0 and H = 2.1. <ref name="Physical Chemistry"/> As a result of the electronegativity difference, the F atom pulls electron density closer to itself, resulting in a highly polarized H-F bond, adding on an extra contribution to the bond strength. H-H bond, on the other hand, is a simple covalent bond. Literature value suggests that the bond dissociation energies of H-H and H-F are 432 kJ/mol and 565 kJ/mol respectively, further suggesting that the enthalpy released from breaking H-F bond is much higher than that of H-H bonds. Therefore, the reaction between F and H2 to form HF and H is energetically favorable, with an overall exothermic reaction.

The reverse reaction, H + HF is therefore endothermic, thereby requiring external energy in order to break the H-F bond, whereby the formation of the H-H bond does not compensate for.

The Hammond's postulate states that the transition state of a reaction resembles either the reactants or the products depending on which it is closer in energy. Hence, in an exothermic reaction, the transition state is closer to the reactants in energy. In an endothermic reaction, the transition state is closer to the products in energy.<ref name="OC"/>

Referring to the table above, for the forward reaction between F and H2, since the reactants energy is much closer to the transition state energy as compared to the reactions energy and the transition state energy in H + HF backward reaction, the forward reaction is the exothermic reaction, and the backward reaction is the endothermic reaction. This can also be determined graphically. With reference to the PES diagram and Energy vs Time plot below for the reaction between F and H2, it shows that the transition state has been attained, with no net change in potential energy. Furthermore, the PES suggests that the reactants are higher in energy than the products, indicating an exothermic reaction.

[[File:Fig27_Yanda.png|center|300px|thumb|Figure 27. Energy vs. Time Plot for F + H2. ]]
[[File:Fig28_Yanda.png|center|300px|thumb|Figure 28. Potential Energy Surface Diagram for F + H2. ]]

Similarly, Figure 29 below shows the transition state from the reverse reaction H + HF. Comparatively, it can be seen that the transition state closely resembles the products, which are now higher in energy than the reactants, thereby deeming it an endothermic reaction.

[[File:Fig29_Yanda.png|center|300px|thumb|Figure 29. Potential Energy Surface Diagram for H + HF. ]]

The approximate position of the transition state, is found to be r1 = BC = 181.1 pm and r2 = AB = 74.5 pm .The potential energy at the TS is then found to be -433.980 kJ/mol.

The activation energy of the reaction was found by slightly perturbing it from the TS and finding the energy difference of the products to the transition state energy as shown in the tables and Figures 30-33 below.

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| MEP || 2000 || 0.1 || 181.1 || 74.5 || 0 || 0
|}

[[File:Fig30_Yanda.png|center|300px|thumb|Figure 30. Contour Map for the reaction from H2 + F to HF + H. ]]
[[File:Fig31_Yanda.png|center|300px|thumb|Figure 31. Energy vs Time diagram for the reaction from H2 + F to HF + H. ]]

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| MEP || 2000 || 0.1 || 182.1 || 74.5 || 0 || 0
|}

[[File:Fig32_Yanda.png|center|300px|thumb|Figure 32. Contour Map for the reaction from HF + H to H2 + F. ]] [[File:Fig33_Yanda.png|center|300px|thumb|Figure 33. Energy vs Time diagram for the reaction from HF + H to H2 + F. ]]

By finding the trajectory down the MEP with the initial conditions, then inserting the last geometry as the initial condition. After which, replace one of the bond distance of the reactants to an extremely huge number, while keeping the bond distance of the product the same, the potential energy of the reactants and the products can be found. Though, it must be noted that this is just an approximation, as the exact potential energy is when the potential energy curve stabilizes. Ultimately, the potential energy of the reactants are found, can be used to find the activation energies, as summarized in the first table above.

{{fontcolor1|red| Brilliant. Well documented method with references and figures to show where your arguments can be drawn from. Well done.
[[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 10:05, 7 July 2020 (BST)}}

==Reaction Dynamics==

The following reaction condition resulted in the reaction from H2 + F -> HF + H

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| Dynamics || 1400 || 0.5 || 220 || 71 || -1.5 || -0.3
|}

The conservation of energy suggests that total energy is constant through the course of reaction. Hence, in the exothermic reaction between H2 and F to form HF + H, the loss in potential energy from the breaking of bonds is replaced by the increase in kinetic energy as seen in the Energy Vs Time graph below. The kinetic energy can be of three forms: translational, rotational or vibrational. Rotational motion is ignored here as according to the animation, it is currently working on a 1D system.

A bomb calorimeter can be used to measure the overall gain in kinetic energy, both translational and kinetic by measuring the transfer of these energy from the products to the walls of the calorimeter in the form of heat. The heat transferred can then be measured in the form of change in temperature with a thermal sensor. <ref name="Physical Chemistry"/>

However, in order to differentiate the respective contributions of translational and vibrational energy, further experimental is required. To determine the vibrational energy, IR spectroscopy can be utilized. Before the reactants collide, most of the bonds occupy lowest energy vibrational levels. However, during a collision, higher level vibrational modes are excited as potential energy is convereted into vibrational kinetic energy. In IR spectroscopy, the higher level modes are characterized as overtones signals where energy levels move from initially 0 to 1, to 0 to 2 or 0 to 3, or even higher. These overtones can be detected and seen with higher wavenumbers as compared to the normal 0 to 1 transition. <ref name="Physical Chemistry"/>

[[File:Fig34_Yanda.png|center|300px|thumb|Figure 34. Contour Map of reaction from H2 + F to HF + H ]]
[[File:Fig35_Yanda.png|center|300px|thumb|Figure 35. Internuclear Distance vs Time Diagram of reaction from H2 + F to HF + H ]]
[[File:Fig36_Yanda.png|center|300px|thumb|Figure 36. Energy vs Time Diagram of reaction from H2 + F to HF + H ]]

==The Polanyi Rules==

The Polanyi rules state that for reactants with a given momentum along the direction of the reaction path, the translational energy, as opposed to the vibrational energy, is preferred in overcoming the early transition state in an exothermic reaction. The opposite will be true for endothermic reactions, where vibrational energy takes priority over translational energy to successfully form stable products.<ref name="J. Phys. Chem."/>

Reaction trajectories were calculated for the exothermic reaction of F + H2 with 1000 steps, but by varying the momentum of r1, the momentum of H-H bond, with all other conditions kept constant such that only the vibration energy of the H2 is varied. This is to investigate an early TS, since it is exothermic.

{| class="wikitable" border=1
! rH-F / pm !! rH-H / pm !! pHF/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1 !! Reactive Trajectory? || Contour Plot
|-
| 230 || 74 || -1.0 || -3.0 || No || [[File:Fig37_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || -2.0 || No || [[File:Fig38_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || -1.0 || No || [[File:Fig39_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 0 || No || [[File:Fig40_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 1.0 || No || [[File:Fig41_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 2.0 || No || [[File:Fig42_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 3.0 || No || [[File:Fig43_Yanda.png|200px]]
|}

However, keeping the same initial positions and everything else constant, except that the momentum pHF is increased to -1.5 while the momentum of pHH is reduced to 0 and is shown in the figure below. It resulted in a reactive trajectory, which suggested that the increase in translational energy of the system has greater effect in allowing the transition state to be overcome to form stable products.

[[File:Fig44_Yanda.png|300px|thumb|center|Figure 44. Reactive trajectory of F + H2 to HF + H reaction.]]

Next, the endothermic reaction of HF + H to F + H2, which is a late TS is investigated. Similarly, everything is kept constant, apart from pHF before keeping it constant and increasing pHH. The bond distance of HF is kept at 92 nm according to literature.

{| class="wikitable" border=1
! rH-F / pm !! rH-H / pm !! pHF/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1 !! Reactive Trajectory? || Contour Plot
|-
| 230 || 92 || 0 || -1.0 || No || [[File:Fig45_Yanda.png|200px]]
|-
| 230 || 92 || -5.0 || -1.0 || No || [[File:Fig46_Yanda.png|200px]]
|-
| 230 || 92 || -10.0 || -1.0 || No || [[File:Fig47_Yanda.png|200px]]
|-
| 230 || 92 || -15.0 || -1.0 || No || [[File:Fig48_Yanda.png|200px]]
|-
| 230 || 92 || -20.0 || -1.0 || No || [[File:Fig50_Yanda.png|200px]]
|-
| 230 || 92 || -25.0 || -1.0 || No || [[File:Fig51_Yanda.png|200px]]
|-
| 230 || 92 || -30.0 || -1.0 || Yes || [[File:Fig52_Yanda.png|200px]]
|}

Hence, it has shown that by increasing the vibrational energy of the H-F bond for the endothermic reaction while keep translational energy fixed at a low vaue resulted in the formation of products.

As such, Polanyi's empirical rules have been shown for both exothermic reactions (early TS) and endothermic reactions (late TS). However, it should be noted that such rules might not work for all cases due to imbalance distribution of translational and vibrational energies or errors in calculating energies, be it bond energies or trajectory calculations.

=References=

<references>
<ref name="CP3MD">Imperial College London, https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD, (accessed May 2018).</ref>
<ref name="Plot"> C.D., Sherrill, http://vergil.chemistry.gatech.edu/courses/chem6485/pdf/pes-lecture.pdf, (accessed May 2018). </ref>
<ref name="MRD">R. D. Levine, Molecular Reaction Dynamics, Cambridge University Press, Cambridge, 2005, pp. 202-203.</ref>
<ref name="MDACK">G. D. Billing and K. V. Mikkelsen, Introduction to Molecular Dynamics and Chemical Kinetics, John Wiley & Sons, New York, 1996, p. 31.</ref>
<ref name="Physical Chemistry">P. Atkins and J. de Paula, Atkins' Physical Chemistry, Oxford University Press, Oxford, 10th edn., 2014, pp. 65, 71 and 986.</ref>
<ref name="OC">J. Clayden, N. Greeves and S. Warren, Organic Chemistry, Oxford University Press, Oxford, 2nd edn., 2012, pp. 989.</ref>
<ref name="J. Phys. Chem.">C. Daniel et al., J. Phys. Chem., 1993, 97, 12485-12490.</ref>
</references>

MRD:yw14815

2020-07-07T09:02:43Z

Mak214: /* Transition State Theory */

=Introduction=

This Wiki page outlines the study of the molecular reaction dynamics of two simple triatomic systems: H + H2 and F + H2. The atoms are 180° to each other, with a 1D space symplification. The atoms are labelled A, B and C, with vectors r1 and r2 representing vectors BA and BC in pm respectively. Momentum are also measured in p1 and p2 in g mol-1 pm fs-1. Below is the systematic diagram of the triatomic system model, which will be referred to throughout the whole wiki. <ref name="CP3MD"/>

[[File:Fig1_Yanda.png|400px|thumb|center|Figure 1. A triatomic system model. Diagram from Imperial College London, https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD (accessed May 2020.)]]

=EXERCISE 1: H + H2 system=
== Potential Energy Surfaces ==

A potential energy surface is a visual representation that describes the potential energy of a system, such as a collective of atoms, in terms of parameters, which normally is the relative positions of the reactants and the products. A reaction path is shown to be the gradient line on a PES diagram, and represents both potential energy and internuclear distances between atoms in the system as a function of the reaction coordinate, thereby modelling their changes through the progress of the reaction.

The PES diagram for Figure 1 above was calculated for the reaction system H + H2 using the parameters given:
{| class="wikitable"
!Calculations
!Steps
!r1 / pm
!r2 / pm
!p1 / gmol-1pmfs-1
!p2 / gmol-1pmfs-1
|-
|Dynamics
|500
|74
|230
|0.0
|<nowiki>-5.2</nowiki>
|}

[[File:Fig2_Yanda.png|400px|thumb|center|Figure 2.Potential Energy Surface for the H + H2 reaction path.]]

The reaction path, which passes through each potential energy minimum, including the transition state, which is a first order saddle point at which the potential energy is at its maximum (highest energy minimum) and where its gradient is 0, hence ∂V('''r''')/∂'''r '''=0 and the second differential at the TS is less than 0.Furthermore, the determinant of the Hessian Matrix is smaller than 0, with the eigenvalues of the Hessian Matrix being one positive and one negative

Local minimum, on the other hand,while also have the first derivative ∂V('''r''')/∂'''r '''=0, its second differential at the minimum is greater than 0, with the eigenvalues of the Hessian Matrix both being positive and the determinant of the Hessian Matrix greater than 0.

Literally speaking, the main difference between transition state and a local minimum is that the transition state is in fact a local energy maximum. Initially, with 0 initial momentum, both transition state and local minimum will remain there forever. However, with a very small momentum added towards the reactants direction, the transition state will be passed and will roll over towards the reactants, and vice versa for products. Testing the trajectories near the transition state is a method used to identify the presence of the transition state. On the other hand, a slightly perturbation from the local minimum will only result in rolling back towards the local minimum, because only a sufficient energy is needed to overcome the energy barrier to escape the "well" of the local minimum. In the PES diagram, a local minimum represents an intermediate, which can be represented in a 2D diagram below.<ref name="Plot"/>

[[File:Fig3_Yanda.PNG|400px|thumb|center|Figure 3. Potential Energy Surface Diagram showing the Transition State. Diagram from http://vergil.chemistry.gatech.edu/courses/chem6485/pdf/pes-lecture.pdf (accessed May 2020).]]

Another interesting observation is that the gradient along the reaction pathway is least steep close to both sides of the transition state, suggesting the point at which bonds are in the process of breaking and forming, hence lower degree of association and less vibrational oscillations. {{fontcolor1|red| Great. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:50, 7 July 2020 (BST)}}

== Locating the Transition State ==

The best estimate of the transition state position (rts) is 90.77 pm. The transition state should be symmetric as it represents the state of the reaction where old bonds and new bonds are in the middle of being broken and formed. As such, given the initial conditions of both momentum p1=0 and p2=0, the two internuclear distances r1 and r2 must be equal at the transitions state. The constant internuclear distances can be understood and found with a "Internuclear Distances vs Time plot" shown in Figure 4 below, where the straight horizontal line suggests constant potential energy with the absence of any vibrational oscillations. {{fontcolor1|red| Yes. Good job. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:53, 7 July 2020 (BST)}}

[[File:Fig4_Yanda.png|400px|thumb|center|Figure 4. Internuclear Distance vs Time plot for finding transition state position.]]

== Calculation of Reaction Path ==

{| class="wikitable" border="1"
! Calculations !! Steps !! r1 / pm !! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| MEP || 50000 || 91.77 || 90.77 || 0 || 0
|}

The Potential Energy Surface for MEP Calculation is shown in Figure 5 below.

[[File:Fig5_Yanda.png|400px|thumb|center|Figure 5.Potential Energy Surface Diagram with MEP.]]

{| class="wikitable" border="1"
! Calculations !! Steps !! r1 / pm !! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| Dynamics || 500 || 91.77 || 90.77 || 0 || 0
|}

Similarly, the Potential Energy Surface for Dynamics Calculation is shown in Figure 6 below.

[[File:Fig6_Yanda.png|400px|thumb|center|Figure 6.Potential Energy Surface Diagram with Dynamics.]]

Comparing the Dynamics (Figure 6) and the MEP (Figure 5), it can be seen that the MEP does not take into account of vibrational oscillations within the atoms, but just passing through all the potential minimum. {{fontcolor1|red| yes. the MEP seeks the nearest local minimum. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:54, 7 July 2020 (BST)}} On the other hand, the trajectory just calculated takes into account of the momentum changes during the reaction by oscillating about the minimum potential energy. In addition, during the calculations, 50000 steps were required to complete the minimum energy path in MEP, while 500 steps were required in Dynamics.

The "Internuclear Distances vs Time" and "Momenta vs Time" graphs for Dynamics calculations are respectively plotted in Figure 7 and 8 below.

[[File:Fig7_Yanda.png|400px|thumb|center|Figure 7. Internuclear Distances vs Time using Dynamics.]]

[[File:Fig8_Yanda.png|400px|thumb|center|Figure 8. Momenta vs Time using Dynamics.]]

Figure 7 and 8 can help to describe the reaction. Distance r1 (B-C) and r2 (A-C) are seen seen to increase as A moves further away from B-C after the transition state. After an extended period of time, it can be seen that A-C and B-C distances increases linearly, suggesting that the distance A-B is constant and that B-C is moving further away from A. Also,initially, A-B decreases slightly before plateauing/oscillating, suggesting that the bond between A-B has formed and that they are vibrating up and down from equilibrium. The vibration is further suggested by the oscillation of A-B in the Momenta vs. Time graph in Figure 8.

The "Internuclear Distances vs Time" and "Momenta vs Time" graphs for MEP calculations are respectively plotted in Figure 9 and 10 below.

[[File:Fig9_Yanda.png|400px|thumb|center|Figure 9. Internuclear Distances vs Time using MEP.]]

[[File:Fig10_Yanda.png|400px|thumb|center|Figure 10. Momenta vs Time using MEP.]]

With reference to Figure 9 and 10, the internuclear distance r2 (A-B) decreases rapidly before converging to a limit and plateauing, while r1 (B-C), hence A-C increases rapidly. The whole system has 0 momentum because the momenta/velocities are always reset to zero in each time step.

Also, the final values of the positions r1(t), r2(t), p1(t) and p2(t) for the trajectory for large enough t, at t=100fs are calculated in the table below.

{| class="wikitable" border="1"
! r1 / pm!! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| 735.414|| 75.4847|| 5.08082 || 2.02769
|}

If the initial conditions were r1 = rts and r2 = rts + 1 pm instead, the reaction path will move in the opposite direction, exiting from the opposite direction in the potential energy surface (Figure 11) whereby distance r2 (A-B) increases while the distance r1 (B-C) stays constant with only vibrations of the bond similar to the previous conditions (Figure 12 and 13). Hence, in the first case the transition state moves towards AB + C, while the reverse will be in this case favoring the formation of products A + BC, as A start off further to B than that of C to B.

[[File:Fig11_Yanda.png|400px|thumb|center|Figure 11. Potential Energy Surface with Dynamics with initial conditions reversed.]]
[[File:Fig12_Yanda.png|400px|thumb|center|Figure 12. Internuclear Distances vs Time with Dynamics with initial conditions reversed.]]
[[File:Fig13_Yanda.png|400px|thumb|center|Figure 13. Momenta vs Time with Dynamics with initial conditions reversed.]]

The distorted transition state, where the values of r1=rts and r2=rts + 1 pm would be inverted for as the previous calculations, with the graphs for internuclear distance and momenta also inverted, such that graph of B-C appears as the original A-B.

Finally,a calculation was setup where the initial positions correspond to the final positions of the trajectory calculative, above, the same final momenta values but with their signs reversed.

{| class="wikitable" border="1"
! r1 / pm!! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| 735.414|| 75.4847|| -5.08082 || -2.02769
|}

[[File:Fig14_Yanda.png|400px|thumb|center|Figure 14. Potential Energy Surface with Dynamics moving from products to estimated Transition State.]]
[[File:Fig15_Yanda.png|400px|thumb|center|Figure 15. Internuclear Distances vs Time with Dynamics moving from products to estimated Transition State.]]
[[File:Fig16_Yanda.png|400px|thumb|center|Figure 16. Momenta vs Time with Dynamics moving from products to estimated Transition State.]]

This situation is the reverse of the original situation in which now, the species start off as reactants with just the right starting positions and momenta to reach the slightly distorted transition state. Technically, the three graphs in this situation should be the mirror image of the three graphs in the original situation. However, a perfect symmetrical relationship is not observed due to random errors in estimating through rounding off or sampling errors of the final positions and momenta from the Internuclear Distances vs Time graph and Momenta vs Time graph respectively.For example, in Figure 15, the distance between B-C should be constant after the transition state, but was not the case due to a errors in the initial input data.

{{fontcolor1|red| Fantastic. Really good use of figures here, clear discussion. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:57, 7 July 2020 (BST)}}

== Reactive and Unreactive Trajectories ==

For the initial positions r1 = 74 pm and r2 = 200 pm, various trajectories were ran with different momentum values and combinations as tabulated below.

{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280|| Reactive || Reactive as reaction path can get over maximum at TS and form products. Straight trajectory centered around Dynamics suggest molecule B-C does not vibrate until A collides into it, with sufficient momentum that subsequently resulted in molecule A-B with vibrational oscillation after TS || [[File:Fig17_Yanda.png|300px]] [[File:Fig18_Yanda.png|300px]]
|-
| -3.1 || -4.1 || -420.077 || Unreactive || Not reactive as reactive path unable to get over maximum at TS and form products. The PES curve suggests that the trajectory is vibrational/oscillating throughout and that the reaction path does not move past the TS. It turns back before the TS once the potential energy exceeds the kinetic energy of the species of the momenta given || [[File:Fig19_Yanda.png|300px]] [[File:Fig20_Yanda.png|300px]]
|-
| -3.1 || -5.1|| -413.977 || Reactive|| Reactive as similar to the first scenario. Note that higher initial momentum of molecule B-C results in larger vibrations as visible in the PES curve || [[File:Fig21_Yanda.png|300px]] [[File:Fig22_Yanda.png|300px]]
|-
| -5.1|| -10.1|| -357.277 || Unreactive || Initial momenta large enough for reaction path to attain TS, but the activation barrier was crossed twice such that eventually the reactants were reformed again. At TS, A-B-C, the initial momentum was not large enough to form stable products. As such, reaction path recrosses the TS, as visible from the contour map, back into the reactions, reforming A + BC. Much larger vibrational oscillations that the previous scenarios due to large initial momentum as visible from the PES curve || [[File:Fig23_Yanda.png|300px]] [[File:Fig24_Yanda.png|300px]]
|-
| -5.1|| -10.6|| -349.477 || Reactive || Similar to the previous scenario, but that now the initial momenta are larger enough to allow three crossing of the activation barrier, hence passing through the TS thrice, eventually forming the products A + BC|| [[File:Fig25_Yanda.png|300px]] [[File:Fig26_Yanda.png|300px]]
|}

From the table, it can be seen that while a minimum initial momentum of the species are needed to cross the activation barrier and over the TS to form the products, a much initial momentum does not guarantee product formation as can be seen in the second last scenario. As such, the conditions required for enough momentum given the initial preconditions as listed are: -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1. {{fontcolor1|red| Great. Did you check the bounds for P2 momentum also? Your last trajectory falls outside of these bounds... }}

== Transition State Theory ==

The deviations from the experimental values and the values describes in the Transition State Theory (TST) arises because of the assumptions within the theory itself. First of all, the TST adheres to the Born-Oppenheimer approximation, where the nuclear and electronic motions are decoupled. The Transition State Theory assumes that only the reactants are in constant equilibrium with the transition state structure, and that the energy of the particles follow a Boltzmann distribution. In addition, once the reactants have trajectories with a kinetic energy along the reaction coordinate greater than the activation energy, it will be reactive, and that the products will definitely form, ignoring the possibility that the transition state structure will collapse back to the reactants. However, experimentally, it has been seen in the few scenarios above where the initial momentum (p1=-5.1, p2=-10.1), hence kinetic energy is high enough to result in multiple crossing of the activation energy barrier through multiple forward and backward directions. Theoretically, the theory suggests only a single crossing of the activation energy barrier in the forward direction is possible. Also, note that in the transition state, the motion along the reaction coordinate can be treated classically. <ref name="MRD"/>

Finally, the theory fails for some reactions at elevated temperatures due to more complex molecule motions or low temperatures due to quantum tunneling, whereby the transition state need not be reached to form products.<ref name="MDACK"/> {{fontcolor1|red| Yes. Great response. Thank you for including references [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 10:02, 7 July 2020 (BST)}}

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

The energetics of both F + H2 and the reverse H + HF were investigated and the findings are summarized in the table below. The method to extract these data are explained further in this section.

{| class="wikitable" border="1"
! Reaction !! Transition State Energy / kJ/mol !! Reactants Energy / kJ/mol !! Activation Energy / kJ/mol !! Type of Reaction
|-
| F + H2 || -433.980 || -435.100|| +1.120 || Exothermic
|-
| H + HF || -433.980 || -560.700 || +126.720 || Endothermic
|}

The reaction between F and H2 is an exothermic reaction owing to the stronger H-F bond relative to the H-H bond. Since enthalpy of reaction is calculated from energy of bond breaking - energy of bond making, the stronger negative value arising from breaking the stronger H-F bond resulted in an exothermic reaction. The stronger H-F bond can be explained due to the larger electronegativity difference between H and F, for which F= 4.0 and H = 2.1. <ref name="Physical Chemistry"/> As a result of the electronegativity difference, the F atom pulls electron density closer to itself, resulting in a highly polarized H-F bond, adding on an extra contribution to the bond strength. H-H bond, on the other hand, is a simple covalent bond. Literature value suggests that the bond dissociation energies of H-H and H-F are 432 kJ/mol and 565 kJ/mol respectively, further suggesting that the enthalpy released from breaking H-F bond is much higher than that of H-H bonds. Therefore, the reaction between F and H2 to form HF and H is energetically favorable, with an overall exothermic reaction.

The reverse reaction, H + HF is therefore endothermic, thereby requiring external energy in order to break the H-F bond, whereby the formation of the H-H bond does not compensate for.

The Hammond's postulate states that the transition state of a reaction resembles either the reactants or the products depending on which it is closer in energy. Hence, in an exothermic reaction, the transition state is closer to the reactants in energy. In an endothermic reaction, the transition state is closer to the products in energy.<ref name="OC"/>

Referring to the table above, for the forward reaction between F and H2, since the reactants energy is much closer to the transition state energy as compared to the reactions energy and the transition state energy in H + HF backward reaction, the forward reaction is the exothermic reaction, and the backward reaction is the endothermic reaction. This can also be determined graphically. With reference to the PES diagram and Energy vs Time plot below for the reaction between F and H2, it shows that the transition state has been attained, with no net change in potential energy. Furthermore, the PES suggests that the reactants are higher in energy than the products, indicating an exothermic reaction.

[[File:Fig27_Yanda.png|center|300px|thumb|Figure 27. Energy vs. Time Plot for F + H2. ]]
[[File:Fig28_Yanda.png|center|300px|thumb|Figure 28. Potential Energy Surface Diagram for F + H2. ]]

Similarly, Figure 29 below shows the transition state from the reverse reaction H + HF. Comparatively, it can be seen that the transition state closely resembles the products, which are now higher in energy than the reactants, thereby deeming it an endothermic reaction.

[[File:Fig29_Yanda.png|center|300px|thumb|Figure 29. Potential Energy Surface Diagram for H + HF. ]]

The approximate position of the transition state, is found to be r1 = BC = 181.1 pm and r2 = AB = 74.5 pm .The potential energy at the TS is then found to be -433.980 kJ/mol.

The activation energy of the reaction was found by slightly perturbing it from the TS and finding the energy difference of the products to the transition state energy as shown in the tables and Figures 30-33 below.

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| MEP || 2000 || 0.1 || 181.1 || 74.5 || 0 || 0
|}

[[File:Fig30_Yanda.png|center|300px|thumb|Figure 30. Contour Map for the reaction from H2 + F to HF + H. ]]
[[File:Fig31_Yanda.png|center|300px|thumb|Figure 31. Energy vs Time diagram for the reaction from H2 + F to HF + H. ]]

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| MEP || 2000 || 0.1 || 182.1 || 74.5 || 0 || 0
|}

[[File:Fig32_Yanda.png|center|300px|thumb|Figure 32. Contour Map for the reaction from HF + H to H2 + F. ]] [[File:Fig33_Yanda.png|center|300px|thumb|Figure 33. Energy vs Time diagram for the reaction from HF + H to H2 + F. ]]

By finding the trajectory down the MEP with the initial conditions, then inserting the last geometry as the initial condition. After which, replace one of the bond distance of the reactants to an extremely huge number, while keeping the bond distance of the product the same, the potential energy of the reactants and the products can be found. Though, it must be noted that this is just an approximation, as the exact potential energy is when the potential energy curve stabilizes. Ultimately, the potential energy of the reactants are found, can be used to find the activation energies, as summarized in the first table above.

==Reaction Dynamics==

The following reaction condition resulted in the reaction from H2 + F -> HF + H

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| Dynamics || 1400 || 0.5 || 220 || 71 || -1.5 || -0.3
|}

The conservation of energy suggests that total energy is constant through the course of reaction. Hence, in the exothermic reaction between H2 and F to form HF + H, the loss in potential energy from the breaking of bonds is replaced by the increase in kinetic energy as seen in the Energy Vs Time graph below. The kinetic energy can be of three forms: translational, rotational or vibrational. Rotational motion is ignored here as according to the animation, it is currently working on a 1D system.

A bomb calorimeter can be used to measure the overall gain in kinetic energy, both translational and kinetic by measuring the transfer of these energy from the products to the walls of the calorimeter in the form of heat. The heat transferred can then be measured in the form of change in temperature with a thermal sensor. <ref name="Physical Chemistry"/>

However, in order to differentiate the respective contributions of translational and vibrational energy, further experimental is required. To determine the vibrational energy, IR spectroscopy can be utilized. Before the reactants collide, most of the bonds occupy lowest energy vibrational levels. However, during a collision, higher level vibrational modes are excited as potential energy is convereted into vibrational kinetic energy. In IR spectroscopy, the higher level modes are characterized as overtones signals where energy levels move from initially 0 to 1, to 0 to 2 or 0 to 3, or even higher. These overtones can be detected and seen with higher wavenumbers as compared to the normal 0 to 1 transition. <ref name="Physical Chemistry"/>

[[File:Fig34_Yanda.png|center|300px|thumb|Figure 34. Contour Map of reaction from H2 + F to HF + H ]]
[[File:Fig35_Yanda.png|center|300px|thumb|Figure 35. Internuclear Distance vs Time Diagram of reaction from H2 + F to HF + H ]]
[[File:Fig36_Yanda.png|center|300px|thumb|Figure 36. Energy vs Time Diagram of reaction from H2 + F to HF + H ]]

==The Polanyi Rules==

The Polanyi rules state that for reactants with a given momentum along the direction of the reaction path, the translational energy, as opposed to the vibrational energy, is preferred in overcoming the early transition state in an exothermic reaction. The opposite will be true for endothermic reactions, where vibrational energy takes priority over translational energy to successfully form stable products.<ref name="J. Phys. Chem."/>

Reaction trajectories were calculated for the exothermic reaction of F + H2 with 1000 steps, but by varying the momentum of r1, the momentum of H-H bond, with all other conditions kept constant such that only the vibration energy of the H2 is varied. This is to investigate an early TS, since it is exothermic.

{| class="wikitable" border=1
! rH-F / pm !! rH-H / pm !! pHF/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1 !! Reactive Trajectory? || Contour Plot
|-
| 230 || 74 || -1.0 || -3.0 || No || [[File:Fig37_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || -2.0 || No || [[File:Fig38_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || -1.0 || No || [[File:Fig39_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 0 || No || [[File:Fig40_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 1.0 || No || [[File:Fig41_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 2.0 || No || [[File:Fig42_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 3.0 || No || [[File:Fig43_Yanda.png|200px]]
|}

However, keeping the same initial positions and everything else constant, except that the momentum pHF is increased to -1.5 while the momentum of pHH is reduced to 0 and is shown in the figure below. It resulted in a reactive trajectory, which suggested that the increase in translational energy of the system has greater effect in allowing the transition state to be overcome to form stable products.

[[File:Fig44_Yanda.png|300px|thumb|center|Figure 44. Reactive trajectory of F + H2 to HF + H reaction.]]

Next, the endothermic reaction of HF + H to F + H2, which is a late TS is investigated. Similarly, everything is kept constant, apart from pHF before keeping it constant and increasing pHH. The bond distance of HF is kept at 92 nm according to literature.

{| class="wikitable" border=1
! rH-F / pm !! rH-H / pm !! pHF/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1 !! Reactive Trajectory? || Contour Plot
|-
| 230 || 92 || 0 || -1.0 || No || [[File:Fig45_Yanda.png|200px]]
|-
| 230 || 92 || -5.0 || -1.0 || No || [[File:Fig46_Yanda.png|200px]]
|-
| 230 || 92 || -10.0 || -1.0 || No || [[File:Fig47_Yanda.png|200px]]
|-
| 230 || 92 || -15.0 || -1.0 || No || [[File:Fig48_Yanda.png|200px]]
|-
| 230 || 92 || -20.0 || -1.0 || No || [[File:Fig50_Yanda.png|200px]]
|-
| 230 || 92 || -25.0 || -1.0 || No || [[File:Fig51_Yanda.png|200px]]
|-
| 230 || 92 || -30.0 || -1.0 || Yes || [[File:Fig52_Yanda.png|200px]]
|}

Hence, it has shown that by increasing the vibrational energy of the H-F bond for the endothermic reaction while keep translational energy fixed at a low vaue resulted in the formation of products.

As such, Polanyi's empirical rules have been shown for both exothermic reactions (early TS) and endothermic reactions (late TS). However, it should be noted that such rules might not work for all cases due to imbalance distribution of translational and vibrational energies or errors in calculating energies, be it bond energies or trajectory calculations.

=References=

<references>
<ref name="CP3MD">Imperial College London, https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD, (accessed May 2018).</ref>
<ref name="Plot"> C.D., Sherrill, http://vergil.chemistry.gatech.edu/courses/chem6485/pdf/pes-lecture.pdf, (accessed May 2018). </ref>
<ref name="MRD">R. D. Levine, Molecular Reaction Dynamics, Cambridge University Press, Cambridge, 2005, pp. 202-203.</ref>
<ref name="MDACK">G. D. Billing and K. V. Mikkelsen, Introduction to Molecular Dynamics and Chemical Kinetics, John Wiley & Sons, New York, 1996, p. 31.</ref>
<ref name="Physical Chemistry">P. Atkins and J. de Paula, Atkins' Physical Chemistry, Oxford University Press, Oxford, 10th edn., 2014, pp. 65, 71 and 986.</ref>
<ref name="OC">J. Clayden, N. Greeves and S. Warren, Organic Chemistry, Oxford University Press, Oxford, 2nd edn., 2012, pp. 989.</ref>
<ref name="J. Phys. Chem.">C. Daniel et al., J. Phys. Chem., 1993, 97, 12485-12490.</ref>
</references>

MRD:yw14815

2020-07-07T09:01:11Z

Mak214: /* Reactive and Unreactive Trajectories */

=Introduction=

This Wiki page outlines the study of the molecular reaction dynamics of two simple triatomic systems: H + H2 and F + H2. The atoms are 180° to each other, with a 1D space symplification. The atoms are labelled A, B and C, with vectors r1 and r2 representing vectors BA and BC in pm respectively. Momentum are also measured in p1 and p2 in g mol-1 pm fs-1. Below is the systematic diagram of the triatomic system model, which will be referred to throughout the whole wiki. <ref name="CP3MD"/>

[[File:Fig1_Yanda.png|400px|thumb|center|Figure 1. A triatomic system model. Diagram from Imperial College London, https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD (accessed May 2020.)]]

=EXERCISE 1: H + H2 system=
== Potential Energy Surfaces ==

A potential energy surface is a visual representation that describes the potential energy of a system, such as a collective of atoms, in terms of parameters, which normally is the relative positions of the reactants and the products. A reaction path is shown to be the gradient line on a PES diagram, and represents both potential energy and internuclear distances between atoms in the system as a function of the reaction coordinate, thereby modelling their changes through the progress of the reaction.

The PES diagram for Figure 1 above was calculated for the reaction system H + H2 using the parameters given:
{| class="wikitable"
!Calculations
!Steps
!r1 / pm
!r2 / pm
!p1 / gmol-1pmfs-1
!p2 / gmol-1pmfs-1
|-
|Dynamics
|500
|74
|230
|0.0
|<nowiki>-5.2</nowiki>
|}

[[File:Fig2_Yanda.png|400px|thumb|center|Figure 2.Potential Energy Surface for the H + H2 reaction path.]]

The reaction path, which passes through each potential energy minimum, including the transition state, which is a first order saddle point at which the potential energy is at its maximum (highest energy minimum) and where its gradient is 0, hence ∂V('''r''')/∂'''r '''=0 and the second differential at the TS is less than 0.Furthermore, the determinant of the Hessian Matrix is smaller than 0, with the eigenvalues of the Hessian Matrix being one positive and one negative

Local minimum, on the other hand,while also have the first derivative ∂V('''r''')/∂'''r '''=0, its second differential at the minimum is greater than 0, with the eigenvalues of the Hessian Matrix both being positive and the determinant of the Hessian Matrix greater than 0.

Literally speaking, the main difference between transition state and a local minimum is that the transition state is in fact a local energy maximum. Initially, with 0 initial momentum, both transition state and local minimum will remain there forever. However, with a very small momentum added towards the reactants direction, the transition state will be passed and will roll over towards the reactants, and vice versa for products. Testing the trajectories near the transition state is a method used to identify the presence of the transition state. On the other hand, a slightly perturbation from the local minimum will only result in rolling back towards the local minimum, because only a sufficient energy is needed to overcome the energy barrier to escape the "well" of the local minimum. In the PES diagram, a local minimum represents an intermediate, which can be represented in a 2D diagram below.<ref name="Plot"/>

[[File:Fig3_Yanda.PNG|400px|thumb|center|Figure 3. Potential Energy Surface Diagram showing the Transition State. Diagram from http://vergil.chemistry.gatech.edu/courses/chem6485/pdf/pes-lecture.pdf (accessed May 2020).]]

Another interesting observation is that the gradient along the reaction pathway is least steep close to both sides of the transition state, suggesting the point at which bonds are in the process of breaking and forming, hence lower degree of association and less vibrational oscillations. {{fontcolor1|red| Great. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:50, 7 July 2020 (BST)}}

== Locating the Transition State ==

The best estimate of the transition state position (rts) is 90.77 pm. The transition state should be symmetric as it represents the state of the reaction where old bonds and new bonds are in the middle of being broken and formed. As such, given the initial conditions of both momentum p1=0 and p2=0, the two internuclear distances r1 and r2 must be equal at the transitions state. The constant internuclear distances can be understood and found with a "Internuclear Distances vs Time plot" shown in Figure 4 below, where the straight horizontal line suggests constant potential energy with the absence of any vibrational oscillations. {{fontcolor1|red| Yes. Good job. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:53, 7 July 2020 (BST)}}

[[File:Fig4_Yanda.png|400px|thumb|center|Figure 4. Internuclear Distance vs Time plot for finding transition state position.]]

== Calculation of Reaction Path ==

{| class="wikitable" border="1"
! Calculations !! Steps !! r1 / pm !! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| MEP || 50000 || 91.77 || 90.77 || 0 || 0
|}

The Potential Energy Surface for MEP Calculation is shown in Figure 5 below.

[[File:Fig5_Yanda.png|400px|thumb|center|Figure 5.Potential Energy Surface Diagram with MEP.]]

{| class="wikitable" border="1"
! Calculations !! Steps !! r1 / pm !! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| Dynamics || 500 || 91.77 || 90.77 || 0 || 0
|}

Similarly, the Potential Energy Surface for Dynamics Calculation is shown in Figure 6 below.

[[File:Fig6_Yanda.png|400px|thumb|center|Figure 6.Potential Energy Surface Diagram with Dynamics.]]

Comparing the Dynamics (Figure 6) and the MEP (Figure 5), it can be seen that the MEP does not take into account of vibrational oscillations within the atoms, but just passing through all the potential minimum. {{fontcolor1|red| yes. the MEP seeks the nearest local minimum. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:54, 7 July 2020 (BST)}} On the other hand, the trajectory just calculated takes into account of the momentum changes during the reaction by oscillating about the minimum potential energy. In addition, during the calculations, 50000 steps were required to complete the minimum energy path in MEP, while 500 steps were required in Dynamics.

The "Internuclear Distances vs Time" and "Momenta vs Time" graphs for Dynamics calculations are respectively plotted in Figure 7 and 8 below.

[[File:Fig7_Yanda.png|400px|thumb|center|Figure 7. Internuclear Distances vs Time using Dynamics.]]

[[File:Fig8_Yanda.png|400px|thumb|center|Figure 8. Momenta vs Time using Dynamics.]]

Figure 7 and 8 can help to describe the reaction. Distance r1 (B-C) and r2 (A-C) are seen seen to increase as A moves further away from B-C after the transition state. After an extended period of time, it can be seen that A-C and B-C distances increases linearly, suggesting that the distance A-B is constant and that B-C is moving further away from A. Also,initially, A-B decreases slightly before plateauing/oscillating, suggesting that the bond between A-B has formed and that they are vibrating up and down from equilibrium. The vibration is further suggested by the oscillation of A-B in the Momenta vs. Time graph in Figure 8.

The "Internuclear Distances vs Time" and "Momenta vs Time" graphs for MEP calculations are respectively plotted in Figure 9 and 10 below.

[[File:Fig9_Yanda.png|400px|thumb|center|Figure 9. Internuclear Distances vs Time using MEP.]]

[[File:Fig10_Yanda.png|400px|thumb|center|Figure 10. Momenta vs Time using MEP.]]

With reference to Figure 9 and 10, the internuclear distance r2 (A-B) decreases rapidly before converging to a limit and plateauing, while r1 (B-C), hence A-C increases rapidly. The whole system has 0 momentum because the momenta/velocities are always reset to zero in each time step.

Also, the final values of the positions r1(t), r2(t), p1(t) and p2(t) for the trajectory for large enough t, at t=100fs are calculated in the table below.

{| class="wikitable" border="1"
! r1 / pm!! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| 735.414|| 75.4847|| 5.08082 || 2.02769
|}

If the initial conditions were r1 = rts and r2 = rts + 1 pm instead, the reaction path will move in the opposite direction, exiting from the opposite direction in the potential energy surface (Figure 11) whereby distance r2 (A-B) increases while the distance r1 (B-C) stays constant with only vibrations of the bond similar to the previous conditions (Figure 12 and 13). Hence, in the first case the transition state moves towards AB + C, while the reverse will be in this case favoring the formation of products A + BC, as A start off further to B than that of C to B.

[[File:Fig11_Yanda.png|400px|thumb|center|Figure 11. Potential Energy Surface with Dynamics with initial conditions reversed.]]
[[File:Fig12_Yanda.png|400px|thumb|center|Figure 12. Internuclear Distances vs Time with Dynamics with initial conditions reversed.]]
[[File:Fig13_Yanda.png|400px|thumb|center|Figure 13. Momenta vs Time with Dynamics with initial conditions reversed.]]

The distorted transition state, where the values of r1=rts and r2=rts + 1 pm would be inverted for as the previous calculations, with the graphs for internuclear distance and momenta also inverted, such that graph of B-C appears as the original A-B.

Finally,a calculation was setup where the initial positions correspond to the final positions of the trajectory calculative, above, the same final momenta values but with their signs reversed.

{| class="wikitable" border="1"
! r1 / pm!! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| 735.414|| 75.4847|| -5.08082 || -2.02769
|}

[[File:Fig14_Yanda.png|400px|thumb|center|Figure 14. Potential Energy Surface with Dynamics moving from products to estimated Transition State.]]
[[File:Fig15_Yanda.png|400px|thumb|center|Figure 15. Internuclear Distances vs Time with Dynamics moving from products to estimated Transition State.]]
[[File:Fig16_Yanda.png|400px|thumb|center|Figure 16. Momenta vs Time with Dynamics moving from products to estimated Transition State.]]

This situation is the reverse of the original situation in which now, the species start off as reactants with just the right starting positions and momenta to reach the slightly distorted transition state. Technically, the three graphs in this situation should be the mirror image of the three graphs in the original situation. However, a perfect symmetrical relationship is not observed due to random errors in estimating through rounding off or sampling errors of the final positions and momenta from the Internuclear Distances vs Time graph and Momenta vs Time graph respectively.For example, in Figure 15, the distance between B-C should be constant after the transition state, but was not the case due to a errors in the initial input data.

{{fontcolor1|red| Fantastic. Really good use of figures here, clear discussion. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:57, 7 July 2020 (BST)}}

== Reactive and Unreactive Trajectories ==

For the initial positions r1 = 74 pm and r2 = 200 pm, various trajectories were ran with different momentum values and combinations as tabulated below.

{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280|| Reactive || Reactive as reaction path can get over maximum at TS and form products. Straight trajectory centered around Dynamics suggest molecule B-C does not vibrate until A collides into it, with sufficient momentum that subsequently resulted in molecule A-B with vibrational oscillation after TS || [[File:Fig17_Yanda.png|300px]] [[File:Fig18_Yanda.png|300px]]
|-
| -3.1 || -4.1 || -420.077 || Unreactive || Not reactive as reactive path unable to get over maximum at TS and form products. The PES curve suggests that the trajectory is vibrational/oscillating throughout and that the reaction path does not move past the TS. It turns back before the TS once the potential energy exceeds the kinetic energy of the species of the momenta given || [[File:Fig19_Yanda.png|300px]] [[File:Fig20_Yanda.png|300px]]
|-
| -3.1 || -5.1|| -413.977 || Reactive|| Reactive as similar to the first scenario. Note that higher initial momentum of molecule B-C results in larger vibrations as visible in the PES curve || [[File:Fig21_Yanda.png|300px]] [[File:Fig22_Yanda.png|300px]]
|-
| -5.1|| -10.1|| -357.277 || Unreactive || Initial momenta large enough for reaction path to attain TS, but the activation barrier was crossed twice such that eventually the reactants were reformed again. At TS, A-B-C, the initial momentum was not large enough to form stable products. As such, reaction path recrosses the TS, as visible from the contour map, back into the reactions, reforming A + BC. Much larger vibrational oscillations that the previous scenarios due to large initial momentum as visible from the PES curve || [[File:Fig23_Yanda.png|300px]] [[File:Fig24_Yanda.png|300px]]
|-
| -5.1|| -10.6|| -349.477 || Reactive || Similar to the previous scenario, but that now the initial momenta are larger enough to allow three crossing of the activation barrier, hence passing through the TS thrice, eventually forming the products A + BC|| [[File:Fig25_Yanda.png|300px]] [[File:Fig26_Yanda.png|300px]]
|}

From the table, it can be seen that while a minimum initial momentum of the species are needed to cross the activation barrier and over the TS to form the products, a much initial momentum does not guarantee product formation as can be seen in the second last scenario. As such, the conditions required for enough momentum given the initial preconditions as listed are: -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1. {{fontcolor1|red| Great. Did you check the bounds for P2 momentum also? Your last trajectory falls outside of these bounds... }}

== Transition State Theory ==

The deviations from the experimental values and the values describes in the Transition State Theory (TST) arises because of the assumptions within the theory itself. First of all, the TST adheres to the Born-Oppenheimer approximation, where the nuclear and electronic motions are decoupled. The Transition State Theory assumes that only the reactants are in constant equilibrium with the transition state structure, and that the energy of the particles follow a Boltzmann distribution. In addition, once the reactants have trajectories with a kinetic energy along the reaction coordinate greater than the activation energy, it will be reactive, and that the products will definitely form, ignoring the possibility that the transition state structure will collapse back to the reactants. However, experimentally, it has been seen in the few scenarios above where the initial momentum (p1=-5.1, p2=-10.1), hence kinetic energy is high enough to result in multiple crossing of the activation energy barrier through multiple forward and backward directions. Theoretically, the theory suggests only a single crossing of the activation energy barrier in the forward direction is possible. Also, note that in the transition state, the motion along the reaction coordinate can be treated classically. <ref name="MRD"/>

Finally, the theory fails for some reactions at elevated temperatures due to more complex molecule motions or low temperatures due to quantum tunneling, whereby the transition state need not be reached to form products.<ref name="MDACK"/>

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

The energetics of both F + H2 and the reverse H + HF were investigated and the findings are summarized in the table below. The method to extract these data are explained further in this section.

{| class="wikitable" border="1"
! Reaction !! Transition State Energy / kJ/mol !! Reactants Energy / kJ/mol !! Activation Energy / kJ/mol !! Type of Reaction
|-
| F + H2 || -433.980 || -435.100|| +1.120 || Exothermic
|-
| H + HF || -433.980 || -560.700 || +126.720 || Endothermic
|}

The reaction between F and H2 is an exothermic reaction owing to the stronger H-F bond relative to the H-H bond. Since enthalpy of reaction is calculated from energy of bond breaking - energy of bond making, the stronger negative value arising from breaking the stronger H-F bond resulted in an exothermic reaction. The stronger H-F bond can be explained due to the larger electronegativity difference between H and F, for which F= 4.0 and H = 2.1. <ref name="Physical Chemistry"/> As a result of the electronegativity difference, the F atom pulls electron density closer to itself, resulting in a highly polarized H-F bond, adding on an extra contribution to the bond strength. H-H bond, on the other hand, is a simple covalent bond. Literature value suggests that the bond dissociation energies of H-H and H-F are 432 kJ/mol and 565 kJ/mol respectively, further suggesting that the enthalpy released from breaking H-F bond is much higher than that of H-H bonds. Therefore, the reaction between F and H2 to form HF and H is energetically favorable, with an overall exothermic reaction.

The reverse reaction, H + HF is therefore endothermic, thereby requiring external energy in order to break the H-F bond, whereby the formation of the H-H bond does not compensate for.

The Hammond's postulate states that the transition state of a reaction resembles either the reactants or the products depending on which it is closer in energy. Hence, in an exothermic reaction, the transition state is closer to the reactants in energy. In an endothermic reaction, the transition state is closer to the products in energy.<ref name="OC"/>

Referring to the table above, for the forward reaction between F and H2, since the reactants energy is much closer to the transition state energy as compared to the reactions energy and the transition state energy in H + HF backward reaction, the forward reaction is the exothermic reaction, and the backward reaction is the endothermic reaction. This can also be determined graphically. With reference to the PES diagram and Energy vs Time plot below for the reaction between F and H2, it shows that the transition state has been attained, with no net change in potential energy. Furthermore, the PES suggests that the reactants are higher in energy than the products, indicating an exothermic reaction.

[[File:Fig27_Yanda.png|center|300px|thumb|Figure 27. Energy vs. Time Plot for F + H2. ]]
[[File:Fig28_Yanda.png|center|300px|thumb|Figure 28. Potential Energy Surface Diagram for F + H2. ]]

Similarly, Figure 29 below shows the transition state from the reverse reaction H + HF. Comparatively, it can be seen that the transition state closely resembles the products, which are now higher in energy than the reactants, thereby deeming it an endothermic reaction.

[[File:Fig29_Yanda.png|center|300px|thumb|Figure 29. Potential Energy Surface Diagram for H + HF. ]]

The approximate position of the transition state, is found to be r1 = BC = 181.1 pm and r2 = AB = 74.5 pm .The potential energy at the TS is then found to be -433.980 kJ/mol.

The activation energy of the reaction was found by slightly perturbing it from the TS and finding the energy difference of the products to the transition state energy as shown in the tables and Figures 30-33 below.

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| MEP || 2000 || 0.1 || 181.1 || 74.5 || 0 || 0
|}

[[File:Fig30_Yanda.png|center|300px|thumb|Figure 30. Contour Map for the reaction from H2 + F to HF + H. ]]
[[File:Fig31_Yanda.png|center|300px|thumb|Figure 31. Energy vs Time diagram for the reaction from H2 + F to HF + H. ]]

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| MEP || 2000 || 0.1 || 182.1 || 74.5 || 0 || 0
|}

[[File:Fig32_Yanda.png|center|300px|thumb|Figure 32. Contour Map for the reaction from HF + H to H2 + F. ]] [[File:Fig33_Yanda.png|center|300px|thumb|Figure 33. Energy vs Time diagram for the reaction from HF + H to H2 + F. ]]

By finding the trajectory down the MEP with the initial conditions, then inserting the last geometry as the initial condition. After which, replace one of the bond distance of the reactants to an extremely huge number, while keeping the bond distance of the product the same, the potential energy of the reactants and the products can be found. Though, it must be noted that this is just an approximation, as the exact potential energy is when the potential energy curve stabilizes. Ultimately, the potential energy of the reactants are found, can be used to find the activation energies, as summarized in the first table above.

==Reaction Dynamics==

The following reaction condition resulted in the reaction from H2 + F -> HF + H

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| Dynamics || 1400 || 0.5 || 220 || 71 || -1.5 || -0.3
|}

The conservation of energy suggests that total energy is constant through the course of reaction. Hence, in the exothermic reaction between H2 and F to form HF + H, the loss in potential energy from the breaking of bonds is replaced by the increase in kinetic energy as seen in the Energy Vs Time graph below. The kinetic energy can be of three forms: translational, rotational or vibrational. Rotational motion is ignored here as according to the animation, it is currently working on a 1D system.

A bomb calorimeter can be used to measure the overall gain in kinetic energy, both translational and kinetic by measuring the transfer of these energy from the products to the walls of the calorimeter in the form of heat. The heat transferred can then be measured in the form of change in temperature with a thermal sensor. <ref name="Physical Chemistry"/>

However, in order to differentiate the respective contributions of translational and vibrational energy, further experimental is required. To determine the vibrational energy, IR spectroscopy can be utilized. Before the reactants collide, most of the bonds occupy lowest energy vibrational levels. However, during a collision, higher level vibrational modes are excited as potential energy is convereted into vibrational kinetic energy. In IR spectroscopy, the higher level modes are characterized as overtones signals where energy levels move from initially 0 to 1, to 0 to 2 or 0 to 3, or even higher. These overtones can be detected and seen with higher wavenumbers as compared to the normal 0 to 1 transition. <ref name="Physical Chemistry"/>

[[File:Fig34_Yanda.png|center|300px|thumb|Figure 34. Contour Map of reaction from H2 + F to HF + H ]]
[[File:Fig35_Yanda.png|center|300px|thumb|Figure 35. Internuclear Distance vs Time Diagram of reaction from H2 + F to HF + H ]]
[[File:Fig36_Yanda.png|center|300px|thumb|Figure 36. Energy vs Time Diagram of reaction from H2 + F to HF + H ]]

==The Polanyi Rules==

The Polanyi rules state that for reactants with a given momentum along the direction of the reaction path, the translational energy, as opposed to the vibrational energy, is preferred in overcoming the early transition state in an exothermic reaction. The opposite will be true for endothermic reactions, where vibrational energy takes priority over translational energy to successfully form stable products.<ref name="J. Phys. Chem."/>

Reaction trajectories were calculated for the exothermic reaction of F + H2 with 1000 steps, but by varying the momentum of r1, the momentum of H-H bond, with all other conditions kept constant such that only the vibration energy of the H2 is varied. This is to investigate an early TS, since it is exothermic.

{| class="wikitable" border=1
! rH-F / pm !! rH-H / pm !! pHF/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1 !! Reactive Trajectory? || Contour Plot
|-
| 230 || 74 || -1.0 || -3.0 || No || [[File:Fig37_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || -2.0 || No || [[File:Fig38_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || -1.0 || No || [[File:Fig39_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 0 || No || [[File:Fig40_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 1.0 || No || [[File:Fig41_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 2.0 || No || [[File:Fig42_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 3.0 || No || [[File:Fig43_Yanda.png|200px]]
|}

However, keeping the same initial positions and everything else constant, except that the momentum pHF is increased to -1.5 while the momentum of pHH is reduced to 0 and is shown in the figure below. It resulted in a reactive trajectory, which suggested that the increase in translational energy of the system has greater effect in allowing the transition state to be overcome to form stable products.

[[File:Fig44_Yanda.png|300px|thumb|center|Figure 44. Reactive trajectory of F + H2 to HF + H reaction.]]

Next, the endothermic reaction of HF + H to F + H2, which is a late TS is investigated. Similarly, everything is kept constant, apart from pHF before keeping it constant and increasing pHH. The bond distance of HF is kept at 92 nm according to literature.

{| class="wikitable" border=1
! rH-F / pm !! rH-H / pm !! pHF/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1 !! Reactive Trajectory? || Contour Plot
|-
| 230 || 92 || 0 || -1.0 || No || [[File:Fig45_Yanda.png|200px]]
|-
| 230 || 92 || -5.0 || -1.0 || No || [[File:Fig46_Yanda.png|200px]]
|-
| 230 || 92 || -10.0 || -1.0 || No || [[File:Fig47_Yanda.png|200px]]
|-
| 230 || 92 || -15.0 || -1.0 || No || [[File:Fig48_Yanda.png|200px]]
|-
| 230 || 92 || -20.0 || -1.0 || No || [[File:Fig50_Yanda.png|200px]]
|-
| 230 || 92 || -25.0 || -1.0 || No || [[File:Fig51_Yanda.png|200px]]
|-
| 230 || 92 || -30.0 || -1.0 || Yes || [[File:Fig52_Yanda.png|200px]]
|}

Hence, it has shown that by increasing the vibrational energy of the H-F bond for the endothermic reaction while keep translational energy fixed at a low vaue resulted in the formation of products.

As such, Polanyi's empirical rules have been shown for both exothermic reactions (early TS) and endothermic reactions (late TS). However, it should be noted that such rules might not work for all cases due to imbalance distribution of translational and vibrational energies or errors in calculating energies, be it bond energies or trajectory calculations.

=References=

<references>
<ref name="CP3MD">Imperial College London, https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD, (accessed May 2018).</ref>
<ref name="Plot"> C.D., Sherrill, http://vergil.chemistry.gatech.edu/courses/chem6485/pdf/pes-lecture.pdf, (accessed May 2018). </ref>
<ref name="MRD">R. D. Levine, Molecular Reaction Dynamics, Cambridge University Press, Cambridge, 2005, pp. 202-203.</ref>
<ref name="MDACK">G. D. Billing and K. V. Mikkelsen, Introduction to Molecular Dynamics and Chemical Kinetics, John Wiley & Sons, New York, 1996, p. 31.</ref>
<ref name="Physical Chemistry">P. Atkins and J. de Paula, Atkins' Physical Chemistry, Oxford University Press, Oxford, 10th edn., 2014, pp. 65, 71 and 986.</ref>
<ref name="OC">J. Clayden, N. Greeves and S. Warren, Organic Chemistry, Oxford University Press, Oxford, 2nd edn., 2012, pp. 989.</ref>
<ref name="J. Phys. Chem.">C. Daniel et al., J. Phys. Chem., 1993, 97, 12485-12490.</ref>
</references>

MRD:yw14815

2020-07-07T08:57:22Z

Mak214: /* Calculation of Reaction Path */

=Introduction=

This Wiki page outlines the study of the molecular reaction dynamics of two simple triatomic systems: H + H2 and F + H2. The atoms are 180° to each other, with a 1D space symplification. The atoms are labelled A, B and C, with vectors r1 and r2 representing vectors BA and BC in pm respectively. Momentum are also measured in p1 and p2 in g mol-1 pm fs-1. Below is the systematic diagram of the triatomic system model, which will be referred to throughout the whole wiki. <ref name="CP3MD"/>

[[File:Fig1_Yanda.png|400px|thumb|center|Figure 1. A triatomic system model. Diagram from Imperial College London, https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD (accessed May 2020.)]]

=EXERCISE 1: H + H2 system=
== Potential Energy Surfaces ==

A potential energy surface is a visual representation that describes the potential energy of a system, such as a collective of atoms, in terms of parameters, which normally is the relative positions of the reactants and the products. A reaction path is shown to be the gradient line on a PES diagram, and represents both potential energy and internuclear distances between atoms in the system as a function of the reaction coordinate, thereby modelling their changes through the progress of the reaction.

The PES diagram for Figure 1 above was calculated for the reaction system H + H2 using the parameters given:
{| class="wikitable"
!Calculations
!Steps
!r1 / pm
!r2 / pm
!p1 / gmol-1pmfs-1
!p2 / gmol-1pmfs-1
|-
|Dynamics
|500
|74
|230
|0.0
|<nowiki>-5.2</nowiki>
|}

[[File:Fig2_Yanda.png|400px|thumb|center|Figure 2.Potential Energy Surface for the H + H2 reaction path.]]

The reaction path, which passes through each potential energy minimum, including the transition state, which is a first order saddle point at which the potential energy is at its maximum (highest energy minimum) and where its gradient is 0, hence ∂V('''r''')/∂'''r '''=0 and the second differential at the TS is less than 0.Furthermore, the determinant of the Hessian Matrix is smaller than 0, with the eigenvalues of the Hessian Matrix being one positive and one negative

Local minimum, on the other hand,while also have the first derivative ∂V('''r''')/∂'''r '''=0, its second differential at the minimum is greater than 0, with the eigenvalues of the Hessian Matrix both being positive and the determinant of the Hessian Matrix greater than 0.

Literally speaking, the main difference between transition state and a local minimum is that the transition state is in fact a local energy maximum. Initially, with 0 initial momentum, both transition state and local minimum will remain there forever. However, with a very small momentum added towards the reactants direction, the transition state will be passed and will roll over towards the reactants, and vice versa for products. Testing the trajectories near the transition state is a method used to identify the presence of the transition state. On the other hand, a slightly perturbation from the local minimum will only result in rolling back towards the local minimum, because only a sufficient energy is needed to overcome the energy barrier to escape the "well" of the local minimum. In the PES diagram, a local minimum represents an intermediate, which can be represented in a 2D diagram below.<ref name="Plot"/>

[[File:Fig3_Yanda.PNG|400px|thumb|center|Figure 3. Potential Energy Surface Diagram showing the Transition State. Diagram from http://vergil.chemistry.gatech.edu/courses/chem6485/pdf/pes-lecture.pdf (accessed May 2020).]]

Another interesting observation is that the gradient along the reaction pathway is least steep close to both sides of the transition state, suggesting the point at which bonds are in the process of breaking and forming, hence lower degree of association and less vibrational oscillations. {{fontcolor1|red| Great. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:50, 7 July 2020 (BST)}}

== Locating the Transition State ==

The best estimate of the transition state position (rts) is 90.77 pm. The transition state should be symmetric as it represents the state of the reaction where old bonds and new bonds are in the middle of being broken and formed. As such, given the initial conditions of both momentum p1=0 and p2=0, the two internuclear distances r1 and r2 must be equal at the transitions state. The constant internuclear distances can be understood and found with a "Internuclear Distances vs Time plot" shown in Figure 4 below, where the straight horizontal line suggests constant potential energy with the absence of any vibrational oscillations. {{fontcolor1|red| Yes. Good job. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:53, 7 July 2020 (BST)}}

[[File:Fig4_Yanda.png|400px|thumb|center|Figure 4. Internuclear Distance vs Time plot for finding transition state position.]]

== Calculation of Reaction Path ==

{| class="wikitable" border="1"
! Calculations !! Steps !! r1 / pm !! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| MEP || 50000 || 91.77 || 90.77 || 0 || 0
|}

The Potential Energy Surface for MEP Calculation is shown in Figure 5 below.

[[File:Fig5_Yanda.png|400px|thumb|center|Figure 5.Potential Energy Surface Diagram with MEP.]]

{| class="wikitable" border="1"
! Calculations !! Steps !! r1 / pm !! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| Dynamics || 500 || 91.77 || 90.77 || 0 || 0
|}

Similarly, the Potential Energy Surface for Dynamics Calculation is shown in Figure 6 below.

[[File:Fig6_Yanda.png|400px|thumb|center|Figure 6.Potential Energy Surface Diagram with Dynamics.]]

Comparing the Dynamics (Figure 6) and the MEP (Figure 5), it can be seen that the MEP does not take into account of vibrational oscillations within the atoms, but just passing through all the potential minimum. {{fontcolor1|red| yes. the MEP seeks the nearest local minimum. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:54, 7 July 2020 (BST)}} On the other hand, the trajectory just calculated takes into account of the momentum changes during the reaction by oscillating about the minimum potential energy. In addition, during the calculations, 50000 steps were required to complete the minimum energy path in MEP, while 500 steps were required in Dynamics.

The "Internuclear Distances vs Time" and "Momenta vs Time" graphs for Dynamics calculations are respectively plotted in Figure 7 and 8 below.

[[File:Fig7_Yanda.png|400px|thumb|center|Figure 7. Internuclear Distances vs Time using Dynamics.]]

[[File:Fig8_Yanda.png|400px|thumb|center|Figure 8. Momenta vs Time using Dynamics.]]

Figure 7 and 8 can help to describe the reaction. Distance r1 (B-C) and r2 (A-C) are seen seen to increase as A moves further away from B-C after the transition state. After an extended period of time, it can be seen that A-C and B-C distances increases linearly, suggesting that the distance A-B is constant and that B-C is moving further away from A. Also,initially, A-B decreases slightly before plateauing/oscillating, suggesting that the bond between A-B has formed and that they are vibrating up and down from equilibrium. The vibration is further suggested by the oscillation of A-B in the Momenta vs. Time graph in Figure 8.

The "Internuclear Distances vs Time" and "Momenta vs Time" graphs for MEP calculations are respectively plotted in Figure 9 and 10 below.

[[File:Fig9_Yanda.png|400px|thumb|center|Figure 9. Internuclear Distances vs Time using MEP.]]

[[File:Fig10_Yanda.png|400px|thumb|center|Figure 10. Momenta vs Time using MEP.]]

With reference to Figure 9 and 10, the internuclear distance r2 (A-B) decreases rapidly before converging to a limit and plateauing, while r1 (B-C), hence A-C increases rapidly. The whole system has 0 momentum because the momenta/velocities are always reset to zero in each time step.

Also, the final values of the positions r1(t), r2(t), p1(t) and p2(t) for the trajectory for large enough t, at t=100fs are calculated in the table below.

{| class="wikitable" border="1"
! r1 / pm!! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| 735.414|| 75.4847|| 5.08082 || 2.02769
|}

If the initial conditions were r1 = rts and r2 = rts + 1 pm instead, the reaction path will move in the opposite direction, exiting from the opposite direction in the potential energy surface (Figure 11) whereby distance r2 (A-B) increases while the distance r1 (B-C) stays constant with only vibrations of the bond similar to the previous conditions (Figure 12 and 13). Hence, in the first case the transition state moves towards AB + C, while the reverse will be in this case favoring the formation of products A + BC, as A start off further to B than that of C to B.

[[File:Fig11_Yanda.png|400px|thumb|center|Figure 11. Potential Energy Surface with Dynamics with initial conditions reversed.]]
[[File:Fig12_Yanda.png|400px|thumb|center|Figure 12. Internuclear Distances vs Time with Dynamics with initial conditions reversed.]]
[[File:Fig13_Yanda.png|400px|thumb|center|Figure 13. Momenta vs Time with Dynamics with initial conditions reversed.]]

The distorted transition state, where the values of r1=rts and r2=rts + 1 pm would be inverted for as the previous calculations, with the graphs for internuclear distance and momenta also inverted, such that graph of B-C appears as the original A-B.

Finally,a calculation was setup where the initial positions correspond to the final positions of the trajectory calculative, above, the same final momenta values but with their signs reversed.

{| class="wikitable" border="1"
! r1 / pm!! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| 735.414|| 75.4847|| -5.08082 || -2.02769
|}

[[File:Fig14_Yanda.png|400px|thumb|center|Figure 14. Potential Energy Surface with Dynamics moving from products to estimated Transition State.]]
[[File:Fig15_Yanda.png|400px|thumb|center|Figure 15. Internuclear Distances vs Time with Dynamics moving from products to estimated Transition State.]]
[[File:Fig16_Yanda.png|400px|thumb|center|Figure 16. Momenta vs Time with Dynamics moving from products to estimated Transition State.]]

This situation is the reverse of the original situation in which now, the species start off as reactants with just the right starting positions and momenta to reach the slightly distorted transition state. Technically, the three graphs in this situation should be the mirror image of the three graphs in the original situation. However, a perfect symmetrical relationship is not observed due to random errors in estimating through rounding off or sampling errors of the final positions and momenta from the Internuclear Distances vs Time graph and Momenta vs Time graph respectively.For example, in Figure 15, the distance between B-C should be constant after the transition state, but was not the case due to a errors in the initial input data.

{{fontcolor1|red| Fantastic. Really good use of figures here, clear discussion. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:57, 7 July 2020 (BST)}}

== Reactive and Unreactive Trajectories ==

For the initial positions r1 = 74 pm and r2 = 200 pm, various trajectories were ran with different momentum values and combinations as tabulated below.

{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280|| Reactive || Reactive as reaction path can get over maximum at TS and form products. Straight trajectory centered around Dynamics suggest molecule B-C does not vibrate until A collides into it, with sufficient momentum that subsequently resulted in molecule A-B with vibrational oscillation after TS || [[File:Fig17_Yanda.png|300px]] [[File:Fig18_Yanda.png|300px]]
|-
| -3.1 || -4.1 || -420.077 || Unreactive || Not reactive as reactive path unable to get over maximum at TS and form products. The PES curve suggests that the trajectory is vibrational/oscillating throughout and that the reaction path does not move past the TS. It turns back before the TS once the potential energy exceeds the kinetic energy of the species of the momenta given || [[File:Fig19_Yanda.png|300px]] [[File:Fig20_Yanda.png|300px]]
|-
| -3.1 || -5.1|| -413.977 || Reactive|| Reactive as similar to the first scenario. Note that higher initial momentum of molecule B-C results in larger vibrations as visible in the PES curve || [[File:Fig21_Yanda.png|300px]] [[File:Fig22_Yanda.png|300px]]
|-
| -5.1|| -10.1|| -357.277 || Unreactive || Initial momenta large enough for reaction path to attain TS, but the activation barrier was crossed twice such that eventually the reactants were reformed again. At TS, A-B-C, the initial momentum was not large enough to form stable products. As such, reaction path recrosses the TS, as visible from the contour map, back into the reactions, reforming A + BC. Much larger vibrational oscillations that the previous scenarios due to large initial momentum as visible from the PES curve || [[File:Fig23_Yanda.png|300px]] [[File:Fig24_Yanda.png|300px]]
|-
| -5.1|| -10.6|| -349.477 || Reactive || Similar to the previous scenario, but that now the initial momenta are larger enough to allow three crossing of the activation barrier, hence passing through the TS thrice, eventually forming the products A + BC|| [[File:Fig25_Yanda.png|300px]] [[File:Fig26_Yanda.png|300px]]
|}

From the table, it can be seen that while a minimum initial momentum of the species are needed to cross the activation barrier and over the TS to form the products, a much initial momentum does not guarantee product formation as can be seen in the second last scenario. As such, the conditions required for enough momentum given the initial preconditions as listed are: -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1.

== Transition State Theory ==

The deviations from the experimental values and the values describes in the Transition State Theory (TST) arises because of the assumptions within the theory itself. First of all, the TST adheres to the Born-Oppenheimer approximation, where the nuclear and electronic motions are decoupled. The Transition State Theory assumes that only the reactants are in constant equilibrium with the transition state structure, and that the energy of the particles follow a Boltzmann distribution. In addition, once the reactants have trajectories with a kinetic energy along the reaction coordinate greater than the activation energy, it will be reactive, and that the products will definitely form, ignoring the possibility that the transition state structure will collapse back to the reactants. However, experimentally, it has been seen in the few scenarios above where the initial momentum (p1=-5.1, p2=-10.1), hence kinetic energy is high enough to result in multiple crossing of the activation energy barrier through multiple forward and backward directions. Theoretically, the theory suggests only a single crossing of the activation energy barrier in the forward direction is possible. Also, note that in the transition state, the motion along the reaction coordinate can be treated classically. <ref name="MRD"/>

Finally, the theory fails for some reactions at elevated temperatures due to more complex molecule motions or low temperatures due to quantum tunneling, whereby the transition state need not be reached to form products.<ref name="MDACK"/>

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

The energetics of both F + H2 and the reverse H + HF were investigated and the findings are summarized in the table below. The method to extract these data are explained further in this section.

{| class="wikitable" border="1"
! Reaction !! Transition State Energy / kJ/mol !! Reactants Energy / kJ/mol !! Activation Energy / kJ/mol !! Type of Reaction
|-
| F + H2 || -433.980 || -435.100|| +1.120 || Exothermic
|-
| H + HF || -433.980 || -560.700 || +126.720 || Endothermic
|}

The reaction between F and H2 is an exothermic reaction owing to the stronger H-F bond relative to the H-H bond. Since enthalpy of reaction is calculated from energy of bond breaking - energy of bond making, the stronger negative value arising from breaking the stronger H-F bond resulted in an exothermic reaction. The stronger H-F bond can be explained due to the larger electronegativity difference between H and F, for which F= 4.0 and H = 2.1. <ref name="Physical Chemistry"/> As a result of the electronegativity difference, the F atom pulls electron density closer to itself, resulting in a highly polarized H-F bond, adding on an extra contribution to the bond strength. H-H bond, on the other hand, is a simple covalent bond. Literature value suggests that the bond dissociation energies of H-H and H-F are 432 kJ/mol and 565 kJ/mol respectively, further suggesting that the enthalpy released from breaking H-F bond is much higher than that of H-H bonds. Therefore, the reaction between F and H2 to form HF and H is energetically favorable, with an overall exothermic reaction.

The reverse reaction, H + HF is therefore endothermic, thereby requiring external energy in order to break the H-F bond, whereby the formation of the H-H bond does not compensate for.

The Hammond's postulate states that the transition state of a reaction resembles either the reactants or the products depending on which it is closer in energy. Hence, in an exothermic reaction, the transition state is closer to the reactants in energy. In an endothermic reaction, the transition state is closer to the products in energy.<ref name="OC"/>

Referring to the table above, for the forward reaction between F and H2, since the reactants energy is much closer to the transition state energy as compared to the reactions energy and the transition state energy in H + HF backward reaction, the forward reaction is the exothermic reaction, and the backward reaction is the endothermic reaction. This can also be determined graphically. With reference to the PES diagram and Energy vs Time plot below for the reaction between F and H2, it shows that the transition state has been attained, with no net change in potential energy. Furthermore, the PES suggests that the reactants are higher in energy than the products, indicating an exothermic reaction.

[[File:Fig27_Yanda.png|center|300px|thumb|Figure 27. Energy vs. Time Plot for F + H2. ]]
[[File:Fig28_Yanda.png|center|300px|thumb|Figure 28. Potential Energy Surface Diagram for F + H2. ]]

Similarly, Figure 29 below shows the transition state from the reverse reaction H + HF. Comparatively, it can be seen that the transition state closely resembles the products, which are now higher in energy than the reactants, thereby deeming it an endothermic reaction.

[[File:Fig29_Yanda.png|center|300px|thumb|Figure 29. Potential Energy Surface Diagram for H + HF. ]]

The approximate position of the transition state, is found to be r1 = BC = 181.1 pm and r2 = AB = 74.5 pm .The potential energy at the TS is then found to be -433.980 kJ/mol.

The activation energy of the reaction was found by slightly perturbing it from the TS and finding the energy difference of the products to the transition state energy as shown in the tables and Figures 30-33 below.

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| MEP || 2000 || 0.1 || 181.1 || 74.5 || 0 || 0
|}

[[File:Fig30_Yanda.png|center|300px|thumb|Figure 30. Contour Map for the reaction from H2 + F to HF + H. ]]
[[File:Fig31_Yanda.png|center|300px|thumb|Figure 31. Energy vs Time diagram for the reaction from H2 + F to HF + H. ]]

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| MEP || 2000 || 0.1 || 182.1 || 74.5 || 0 || 0
|}

[[File:Fig32_Yanda.png|center|300px|thumb|Figure 32. Contour Map for the reaction from HF + H to H2 + F. ]] [[File:Fig33_Yanda.png|center|300px|thumb|Figure 33. Energy vs Time diagram for the reaction from HF + H to H2 + F. ]]

By finding the trajectory down the MEP with the initial conditions, then inserting the last geometry as the initial condition. After which, replace one of the bond distance of the reactants to an extremely huge number, while keeping the bond distance of the product the same, the potential energy of the reactants and the products can be found. Though, it must be noted that this is just an approximation, as the exact potential energy is when the potential energy curve stabilizes. Ultimately, the potential energy of the reactants are found, can be used to find the activation energies, as summarized in the first table above.

==Reaction Dynamics==

The following reaction condition resulted in the reaction from H2 + F -> HF + H

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| Dynamics || 1400 || 0.5 || 220 || 71 || -1.5 || -0.3
|}

The conservation of energy suggests that total energy is constant through the course of reaction. Hence, in the exothermic reaction between H2 and F to form HF + H, the loss in potential energy from the breaking of bonds is replaced by the increase in kinetic energy as seen in the Energy Vs Time graph below. The kinetic energy can be of three forms: translational, rotational or vibrational. Rotational motion is ignored here as according to the animation, it is currently working on a 1D system.

A bomb calorimeter can be used to measure the overall gain in kinetic energy, both translational and kinetic by measuring the transfer of these energy from the products to the walls of the calorimeter in the form of heat. The heat transferred can then be measured in the form of change in temperature with a thermal sensor. <ref name="Physical Chemistry"/>

However, in order to differentiate the respective contributions of translational and vibrational energy, further experimental is required. To determine the vibrational energy, IR spectroscopy can be utilized. Before the reactants collide, most of the bonds occupy lowest energy vibrational levels. However, during a collision, higher level vibrational modes are excited as potential energy is convereted into vibrational kinetic energy. In IR spectroscopy, the higher level modes are characterized as overtones signals where energy levels move from initially 0 to 1, to 0 to 2 or 0 to 3, or even higher. These overtones can be detected and seen with higher wavenumbers as compared to the normal 0 to 1 transition. <ref name="Physical Chemistry"/>

[[File:Fig34_Yanda.png|center|300px|thumb|Figure 34. Contour Map of reaction from H2 + F to HF + H ]]
[[File:Fig35_Yanda.png|center|300px|thumb|Figure 35. Internuclear Distance vs Time Diagram of reaction from H2 + F to HF + H ]]
[[File:Fig36_Yanda.png|center|300px|thumb|Figure 36. Energy vs Time Diagram of reaction from H2 + F to HF + H ]]

==The Polanyi Rules==

The Polanyi rules state that for reactants with a given momentum along the direction of the reaction path, the translational energy, as opposed to the vibrational energy, is preferred in overcoming the early transition state in an exothermic reaction. The opposite will be true for endothermic reactions, where vibrational energy takes priority over translational energy to successfully form stable products.<ref name="J. Phys. Chem."/>

Reaction trajectories were calculated for the exothermic reaction of F + H2 with 1000 steps, but by varying the momentum of r1, the momentum of H-H bond, with all other conditions kept constant such that only the vibration energy of the H2 is varied. This is to investigate an early TS, since it is exothermic.

{| class="wikitable" border=1
! rH-F / pm !! rH-H / pm !! pHF/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1 !! Reactive Trajectory? || Contour Plot
|-
| 230 || 74 || -1.0 || -3.0 || No || [[File:Fig37_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || -2.0 || No || [[File:Fig38_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || -1.0 || No || [[File:Fig39_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 0 || No || [[File:Fig40_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 1.0 || No || [[File:Fig41_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 2.0 || No || [[File:Fig42_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 3.0 || No || [[File:Fig43_Yanda.png|200px]]
|}

However, keeping the same initial positions and everything else constant, except that the momentum pHF is increased to -1.5 while the momentum of pHH is reduced to 0 and is shown in the figure below. It resulted in a reactive trajectory, which suggested that the increase in translational energy of the system has greater effect in allowing the transition state to be overcome to form stable products.

[[File:Fig44_Yanda.png|300px|thumb|center|Figure 44. Reactive trajectory of F + H2 to HF + H reaction.]]

Next, the endothermic reaction of HF + H to F + H2, which is a late TS is investigated. Similarly, everything is kept constant, apart from pHF before keeping it constant and increasing pHH. The bond distance of HF is kept at 92 nm according to literature.

{| class="wikitable" border=1
! rH-F / pm !! rH-H / pm !! pHF/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1 !! Reactive Trajectory? || Contour Plot
|-
| 230 || 92 || 0 || -1.0 || No || [[File:Fig45_Yanda.png|200px]]
|-
| 230 || 92 || -5.0 || -1.0 || No || [[File:Fig46_Yanda.png|200px]]
|-
| 230 || 92 || -10.0 || -1.0 || No || [[File:Fig47_Yanda.png|200px]]
|-
| 230 || 92 || -15.0 || -1.0 || No || [[File:Fig48_Yanda.png|200px]]
|-
| 230 || 92 || -20.0 || -1.0 || No || [[File:Fig50_Yanda.png|200px]]
|-
| 230 || 92 || -25.0 || -1.0 || No || [[File:Fig51_Yanda.png|200px]]
|-
| 230 || 92 || -30.0 || -1.0 || Yes || [[File:Fig52_Yanda.png|200px]]
|}

Hence, it has shown that by increasing the vibrational energy of the H-F bond for the endothermic reaction while keep translational energy fixed at a low vaue resulted in the formation of products.

As such, Polanyi's empirical rules have been shown for both exothermic reactions (early TS) and endothermic reactions (late TS). However, it should be noted that such rules might not work for all cases due to imbalance distribution of translational and vibrational energies or errors in calculating energies, be it bond energies or trajectory calculations.

=References=

<references>
<ref name="CP3MD">Imperial College London, https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD, (accessed May 2018).</ref>
<ref name="Plot"> C.D., Sherrill, http://vergil.chemistry.gatech.edu/courses/chem6485/pdf/pes-lecture.pdf, (accessed May 2018). </ref>
<ref name="MRD">R. D. Levine, Molecular Reaction Dynamics, Cambridge University Press, Cambridge, 2005, pp. 202-203.</ref>
<ref name="MDACK">G. D. Billing and K. V. Mikkelsen, Introduction to Molecular Dynamics and Chemical Kinetics, John Wiley & Sons, New York, 1996, p. 31.</ref>
<ref name="Physical Chemistry">P. Atkins and J. de Paula, Atkins' Physical Chemistry, Oxford University Press, Oxford, 10th edn., 2014, pp. 65, 71 and 986.</ref>
<ref name="OC">J. Clayden, N. Greeves and S. Warren, Organic Chemistry, Oxford University Press, Oxford, 2nd edn., 2012, pp. 989.</ref>
<ref name="J. Phys. Chem.">C. Daniel et al., J. Phys. Chem., 1993, 97, 12485-12490.</ref>
</references>

MRD:yw14815

2020-07-07T08:54:23Z

Mak214: /* Calculation of Reaction Path */

=Introduction=

This Wiki page outlines the study of the molecular reaction dynamics of two simple triatomic systems: H + H2 and F + H2. The atoms are 180° to each other, with a 1D space symplification. The atoms are labelled A, B and C, with vectors r1 and r2 representing vectors BA and BC in pm respectively. Momentum are also measured in p1 and p2 in g mol-1 pm fs-1. Below is the systematic diagram of the triatomic system model, which will be referred to throughout the whole wiki. <ref name="CP3MD"/>

[[File:Fig1_Yanda.png|400px|thumb|center|Figure 1. A triatomic system model. Diagram from Imperial College London, https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD (accessed May 2020.)]]

=EXERCISE 1: H + H2 system=
== Potential Energy Surfaces ==

A potential energy surface is a visual representation that describes the potential energy of a system, such as a collective of atoms, in terms of parameters, which normally is the relative positions of the reactants and the products. A reaction path is shown to be the gradient line on a PES diagram, and represents both potential energy and internuclear distances between atoms in the system as a function of the reaction coordinate, thereby modelling their changes through the progress of the reaction.

The PES diagram for Figure 1 above was calculated for the reaction system H + H2 using the parameters given:
{| class="wikitable"
!Calculations
!Steps
!r1 / pm
!r2 / pm
!p1 / gmol-1pmfs-1
!p2 / gmol-1pmfs-1
|-
|Dynamics
|500
|74
|230
|0.0
|<nowiki>-5.2</nowiki>
|}

[[File:Fig2_Yanda.png|400px|thumb|center|Figure 2.Potential Energy Surface for the H + H2 reaction path.]]

The reaction path, which passes through each potential energy minimum, including the transition state, which is a first order saddle point at which the potential energy is at its maximum (highest energy minimum) and where its gradient is 0, hence ∂V('''r''')/∂'''r '''=0 and the second differential at the TS is less than 0.Furthermore, the determinant of the Hessian Matrix is smaller than 0, with the eigenvalues of the Hessian Matrix being one positive and one negative

Local minimum, on the other hand,while also have the first derivative ∂V('''r''')/∂'''r '''=0, its second differential at the minimum is greater than 0, with the eigenvalues of the Hessian Matrix both being positive and the determinant of the Hessian Matrix greater than 0.

Literally speaking, the main difference between transition state and a local minimum is that the transition state is in fact a local energy maximum. Initially, with 0 initial momentum, both transition state and local minimum will remain there forever. However, with a very small momentum added towards the reactants direction, the transition state will be passed and will roll over towards the reactants, and vice versa for products. Testing the trajectories near the transition state is a method used to identify the presence of the transition state. On the other hand, a slightly perturbation from the local minimum will only result in rolling back towards the local minimum, because only a sufficient energy is needed to overcome the energy barrier to escape the "well" of the local minimum. In the PES diagram, a local minimum represents an intermediate, which can be represented in a 2D diagram below.<ref name="Plot"/>

[[File:Fig3_Yanda.PNG|400px|thumb|center|Figure 3. Potential Energy Surface Diagram showing the Transition State. Diagram from http://vergil.chemistry.gatech.edu/courses/chem6485/pdf/pes-lecture.pdf (accessed May 2020).]]

Another interesting observation is that the gradient along the reaction pathway is least steep close to both sides of the transition state, suggesting the point at which bonds are in the process of breaking and forming, hence lower degree of association and less vibrational oscillations. {{fontcolor1|red| Great. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:50, 7 July 2020 (BST)}}

== Locating the Transition State ==

The best estimate of the transition state position (rts) is 90.77 pm. The transition state should be symmetric as it represents the state of the reaction where old bonds and new bonds are in the middle of being broken and formed. As such, given the initial conditions of both momentum p1=0 and p2=0, the two internuclear distances r1 and r2 must be equal at the transitions state. The constant internuclear distances can be understood and found with a "Internuclear Distances vs Time plot" shown in Figure 4 below, where the straight horizontal line suggests constant potential energy with the absence of any vibrational oscillations. {{fontcolor1|red| Yes. Good job. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:53, 7 July 2020 (BST)}}

[[File:Fig4_Yanda.png|400px|thumb|center|Figure 4. Internuclear Distance vs Time plot for finding transition state position.]]

== Calculation of Reaction Path ==

{| class="wikitable" border="1"
! Calculations !! Steps !! r1 / pm !! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| MEP || 50000 || 91.77 || 90.77 || 0 || 0
|}

The Potential Energy Surface for MEP Calculation is shown in Figure 5 below.

[[File:Fig5_Yanda.png|400px|thumb|center|Figure 5.Potential Energy Surface Diagram with MEP.]]

{| class="wikitable" border="1"
! Calculations !! Steps !! r1 / pm !! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| Dynamics || 500 || 91.77 || 90.77 || 0 || 0
|}

Similarly, the Potential Energy Surface for Dynamics Calculation is shown in Figure 6 below.

[[File:Fig6_Yanda.png|400px|thumb|center|Figure 6.Potential Energy Surface Diagram with Dynamics.]]

Comparing the Dynamics (Figure 6) and the MEP (Figure 5), it can be seen that the MEP does not take into account of vibrational oscillations within the atoms, but just passing through all the potential minimum. {{fontcolor1|red| yes. the MEP seeks the nearest local minimum. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:54, 7 July 2020 (BST)}} On the other hand, the trajectory just calculated takes into account of the momentum changes during the reaction by oscillating about the minimum potential energy. In addition, during the calculations, 50000 steps were required to complete the minimum energy path in MEP, while 500 steps were required in Dynamics.

The "Internuclear Distances vs Time" and "Momenta vs Time" graphs for Dynamics calculations are respectively plotted in Figure 7 and 8 below.

[[File:Fig7_Yanda.png|400px|thumb|center|Figure 7. Internuclear Distances vs Time using Dynamics.]]

[[File:Fig8_Yanda.png|400px|thumb|center|Figure 8. Momenta vs Time using Dynamics.]]

Figure 7 and 8 can help to describe the reaction. Distance r1 (B-C) and r2 (A-C) are seen seen to increase as A moves further away from B-C after the transition state. After an extended period of time, it can be seen that A-C and B-C distances increases linearly, suggesting that the distance A-B is constant and that B-C is moving further away from A. Also,initially, A-B decreases slightly before plateauing/oscillating, suggesting that the bond between A-B has formed and that they are vibrating up and down from equilibrium. The vibration is further suggested by the oscillation of A-B in the Momenta vs. Time graph in Figure 8.

The "Internuclear Distances vs Time" and "Momenta vs Time" graphs for MEP calculations are respectively plotted in Figure 9 and 10 below.

[[File:Fig9_Yanda.png|400px|thumb|center|Figure 9. Internuclear Distances vs Time using MEP.]]

[[File:Fig10_Yanda.png|400px|thumb|center|Figure 10. Momenta vs Time using MEP.]]

With reference to Figure 9 and 10, the internuclear distance r2 (A-B) decreases rapidly before converging to a limit and plateauing, while r1 (B-C), hence A-C increases rapidly. The whole system has 0 momentum because the momenta/velocities are always reset to zero in each time step.

Also, the final values of the positions r1(t), r2(t), p1(t) and p2(t) for the trajectory for large enough t, at t=100fs are calculated in the table below.

{| class="wikitable" border="1"
! r1 / pm!! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| 735.414|| 75.4847|| 5.08082 || 2.02769
|}

If the initial conditions were r1 = rts and r2 = rts + 1 pm instead, the reaction path will move in the opposite direction, exiting from the opposite direction in the potential energy surface (Figure 11) whereby distance r2 (A-B) increases while the distance r1 (B-C) stays constant with only vibrations of the bond similar to the previous conditions (Figure 12 and 13). Hence, in the first case the transition state moves towards AB + C, while the reverse will be in this case favoring the formation of products A + BC, as A start off further to B than that of C to B.

[[File:Fig11_Yanda.png|400px|thumb|center|Figure 11. Potential Energy Surface with Dynamics with initial conditions reversed.]]
[[File:Fig12_Yanda.png|400px|thumb|center|Figure 12. Internuclear Distances vs Time with Dynamics with initial conditions reversed.]]
[[File:Fig13_Yanda.png|400px|thumb|center|Figure 13. Momenta vs Time with Dynamics with initial conditions reversed.]]

The distorted transition state, where the values of r1=rts and r2=rts + 1 pm would be inverted for as the previous calculations, with the graphs for internuclear distance and momenta also inverted, such that graph of B-C appears as the original A-B.

Finally,a calculation was setup where the initial positions correspond to the final positions of the trajectory calculative, above, the same final momenta values but with their signs reversed.

{| class="wikitable" border="1"
! r1 / pm!! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| 735.414|| 75.4847|| -5.08082 || -2.02769
|}

[[File:Fig14_Yanda.png|400px|thumb|center|Figure 14. Potential Energy Surface with Dynamics moving from products to estimated Transition State.]]
[[File:Fig15_Yanda.png|400px|thumb|center|Figure 15. Internuclear Distances vs Time with Dynamics moving from products to estimated Transition State.]]
[[File:Fig16_Yanda.png|400px|thumb|center|Figure 16. Momenta vs Time with Dynamics moving from products to estimated Transition State.]]

This situation is the reverse of the original situation in which now, the species start off as reactants with just the right starting positions and momenta to reach the slightly distorted transition state. Technically, the three graphs in this situation should be the mirror image of the three graphs in the original situation. However, a perfect symmetrical relationship is not observed due to random errors in estimating through rounding off or sampling errors of the final positions and momenta from the Internuclear Distances vs Time graph and Momenta vs Time graph respectively.For example, in Figure 15, the distance between B-C should be constant after the transition state, but was not the case due to a errors in the initial input data.

== Reactive and Unreactive Trajectories ==

For the initial positions r1 = 74 pm and r2 = 200 pm, various trajectories were ran with different momentum values and combinations as tabulated below.

{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280|| Reactive || Reactive as reaction path can get over maximum at TS and form products. Straight trajectory centered around Dynamics suggest molecule B-C does not vibrate until A collides into it, with sufficient momentum that subsequently resulted in molecule A-B with vibrational oscillation after TS || [[File:Fig17_Yanda.png|300px]] [[File:Fig18_Yanda.png|300px]]
|-
| -3.1 || -4.1 || -420.077 || Unreactive || Not reactive as reactive path unable to get over maximum at TS and form products. The PES curve suggests that the trajectory is vibrational/oscillating throughout and that the reaction path does not move past the TS. It turns back before the TS once the potential energy exceeds the kinetic energy of the species of the momenta given || [[File:Fig19_Yanda.png|300px]] [[File:Fig20_Yanda.png|300px]]
|-
| -3.1 || -5.1|| -413.977 || Reactive|| Reactive as similar to the first scenario. Note that higher initial momentum of molecule B-C results in larger vibrations as visible in the PES curve || [[File:Fig21_Yanda.png|300px]] [[File:Fig22_Yanda.png|300px]]
|-
| -5.1|| -10.1|| -357.277 || Unreactive || Initial momenta large enough for reaction path to attain TS, but the activation barrier was crossed twice such that eventually the reactants were reformed again. At TS, A-B-C, the initial momentum was not large enough to form stable products. As such, reaction path recrosses the TS, as visible from the contour map, back into the reactions, reforming A + BC. Much larger vibrational oscillations that the previous scenarios due to large initial momentum as visible from the PES curve || [[File:Fig23_Yanda.png|300px]] [[File:Fig24_Yanda.png|300px]]
|-
| -5.1|| -10.6|| -349.477 || Reactive || Similar to the previous scenario, but that now the initial momenta are larger enough to allow three crossing of the activation barrier, hence passing through the TS thrice, eventually forming the products A + BC|| [[File:Fig25_Yanda.png|300px]] [[File:Fig26_Yanda.png|300px]]
|}

From the table, it can be seen that while a minimum initial momentum of the species are needed to cross the activation barrier and over the TS to form the products, a much initial momentum does not guarantee product formation as can be seen in the second last scenario. As such, the conditions required for enough momentum given the initial preconditions as listed are: -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1.

== Transition State Theory ==

The deviations from the experimental values and the values describes in the Transition State Theory (TST) arises because of the assumptions within the theory itself. First of all, the TST adheres to the Born-Oppenheimer approximation, where the nuclear and electronic motions are decoupled. The Transition State Theory assumes that only the reactants are in constant equilibrium with the transition state structure, and that the energy of the particles follow a Boltzmann distribution. In addition, once the reactants have trajectories with a kinetic energy along the reaction coordinate greater than the activation energy, it will be reactive, and that the products will definitely form, ignoring the possibility that the transition state structure will collapse back to the reactants. However, experimentally, it has been seen in the few scenarios above where the initial momentum (p1=-5.1, p2=-10.1), hence kinetic energy is high enough to result in multiple crossing of the activation energy barrier through multiple forward and backward directions. Theoretically, the theory suggests only a single crossing of the activation energy barrier in the forward direction is possible. Also, note that in the transition state, the motion along the reaction coordinate can be treated classically. <ref name="MRD"/>

Finally, the theory fails for some reactions at elevated temperatures due to more complex molecule motions or low temperatures due to quantum tunneling, whereby the transition state need not be reached to form products.<ref name="MDACK"/>

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

The energetics of both F + H2 and the reverse H + HF were investigated and the findings are summarized in the table below. The method to extract these data are explained further in this section.

{| class="wikitable" border="1"
! Reaction !! Transition State Energy / kJ/mol !! Reactants Energy / kJ/mol !! Activation Energy / kJ/mol !! Type of Reaction
|-
| F + H2 || -433.980 || -435.100|| +1.120 || Exothermic
|-
| H + HF || -433.980 || -560.700 || +126.720 || Endothermic
|}

The reaction between F and H2 is an exothermic reaction owing to the stronger H-F bond relative to the H-H bond. Since enthalpy of reaction is calculated from energy of bond breaking - energy of bond making, the stronger negative value arising from breaking the stronger H-F bond resulted in an exothermic reaction. The stronger H-F bond can be explained due to the larger electronegativity difference between H and F, for which F= 4.0 and H = 2.1. <ref name="Physical Chemistry"/> As a result of the electronegativity difference, the F atom pulls electron density closer to itself, resulting in a highly polarized H-F bond, adding on an extra contribution to the bond strength. H-H bond, on the other hand, is a simple covalent bond. Literature value suggests that the bond dissociation energies of H-H and H-F are 432 kJ/mol and 565 kJ/mol respectively, further suggesting that the enthalpy released from breaking H-F bond is much higher than that of H-H bonds. Therefore, the reaction between F and H2 to form HF and H is energetically favorable, with an overall exothermic reaction.

The reverse reaction, H + HF is therefore endothermic, thereby requiring external energy in order to break the H-F bond, whereby the formation of the H-H bond does not compensate for.

The Hammond's postulate states that the transition state of a reaction resembles either the reactants or the products depending on which it is closer in energy. Hence, in an exothermic reaction, the transition state is closer to the reactants in energy. In an endothermic reaction, the transition state is closer to the products in energy.<ref name="OC"/>

Referring to the table above, for the forward reaction between F and H2, since the reactants energy is much closer to the transition state energy as compared to the reactions energy and the transition state energy in H + HF backward reaction, the forward reaction is the exothermic reaction, and the backward reaction is the endothermic reaction. This can also be determined graphically. With reference to the PES diagram and Energy vs Time plot below for the reaction between F and H2, it shows that the transition state has been attained, with no net change in potential energy. Furthermore, the PES suggests that the reactants are higher in energy than the products, indicating an exothermic reaction.

[[File:Fig27_Yanda.png|center|300px|thumb|Figure 27. Energy vs. Time Plot for F + H2. ]]
[[File:Fig28_Yanda.png|center|300px|thumb|Figure 28. Potential Energy Surface Diagram for F + H2. ]]

Similarly, Figure 29 below shows the transition state from the reverse reaction H + HF. Comparatively, it can be seen that the transition state closely resembles the products, which are now higher in energy than the reactants, thereby deeming it an endothermic reaction.

[[File:Fig29_Yanda.png|center|300px|thumb|Figure 29. Potential Energy Surface Diagram for H + HF. ]]

The approximate position of the transition state, is found to be r1 = BC = 181.1 pm and r2 = AB = 74.5 pm .The potential energy at the TS is then found to be -433.980 kJ/mol.

The activation energy of the reaction was found by slightly perturbing it from the TS and finding the energy difference of the products to the transition state energy as shown in the tables and Figures 30-33 below.

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| MEP || 2000 || 0.1 || 181.1 || 74.5 || 0 || 0
|}

[[File:Fig30_Yanda.png|center|300px|thumb|Figure 30. Contour Map for the reaction from H2 + F to HF + H. ]]
[[File:Fig31_Yanda.png|center|300px|thumb|Figure 31. Energy vs Time diagram for the reaction from H2 + F to HF + H. ]]

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| MEP || 2000 || 0.1 || 182.1 || 74.5 || 0 || 0
|}

[[File:Fig32_Yanda.png|center|300px|thumb|Figure 32. Contour Map for the reaction from HF + H to H2 + F. ]] [[File:Fig33_Yanda.png|center|300px|thumb|Figure 33. Energy vs Time diagram for the reaction from HF + H to H2 + F. ]]

By finding the trajectory down the MEP with the initial conditions, then inserting the last geometry as the initial condition. After which, replace one of the bond distance of the reactants to an extremely huge number, while keeping the bond distance of the product the same, the potential energy of the reactants and the products can be found. Though, it must be noted that this is just an approximation, as the exact potential energy is when the potential energy curve stabilizes. Ultimately, the potential energy of the reactants are found, can be used to find the activation energies, as summarized in the first table above.

==Reaction Dynamics==

The following reaction condition resulted in the reaction from H2 + F -> HF + H

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| Dynamics || 1400 || 0.5 || 220 || 71 || -1.5 || -0.3
|}

The conservation of energy suggests that total energy is constant through the course of reaction. Hence, in the exothermic reaction between H2 and F to form HF + H, the loss in potential energy from the breaking of bonds is replaced by the increase in kinetic energy as seen in the Energy Vs Time graph below. The kinetic energy can be of three forms: translational, rotational or vibrational. Rotational motion is ignored here as according to the animation, it is currently working on a 1D system.

A bomb calorimeter can be used to measure the overall gain in kinetic energy, both translational and kinetic by measuring the transfer of these energy from the products to the walls of the calorimeter in the form of heat. The heat transferred can then be measured in the form of change in temperature with a thermal sensor. <ref name="Physical Chemistry"/>

However, in order to differentiate the respective contributions of translational and vibrational energy, further experimental is required. To determine the vibrational energy, IR spectroscopy can be utilized. Before the reactants collide, most of the bonds occupy lowest energy vibrational levels. However, during a collision, higher level vibrational modes are excited as potential energy is convereted into vibrational kinetic energy. In IR spectroscopy, the higher level modes are characterized as overtones signals where energy levels move from initially 0 to 1, to 0 to 2 or 0 to 3, or even higher. These overtones can be detected and seen with higher wavenumbers as compared to the normal 0 to 1 transition. <ref name="Physical Chemistry"/>

[[File:Fig34_Yanda.png|center|300px|thumb|Figure 34. Contour Map of reaction from H2 + F to HF + H ]]
[[File:Fig35_Yanda.png|center|300px|thumb|Figure 35. Internuclear Distance vs Time Diagram of reaction from H2 + F to HF + H ]]
[[File:Fig36_Yanda.png|center|300px|thumb|Figure 36. Energy vs Time Diagram of reaction from H2 + F to HF + H ]]

==The Polanyi Rules==

The Polanyi rules state that for reactants with a given momentum along the direction of the reaction path, the translational energy, as opposed to the vibrational energy, is preferred in overcoming the early transition state in an exothermic reaction. The opposite will be true for endothermic reactions, where vibrational energy takes priority over translational energy to successfully form stable products.<ref name="J. Phys. Chem."/>

Reaction trajectories were calculated for the exothermic reaction of F + H2 with 1000 steps, but by varying the momentum of r1, the momentum of H-H bond, with all other conditions kept constant such that only the vibration energy of the H2 is varied. This is to investigate an early TS, since it is exothermic.

{| class="wikitable" border=1
! rH-F / pm !! rH-H / pm !! pHF/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1 !! Reactive Trajectory? || Contour Plot
|-
| 230 || 74 || -1.0 || -3.0 || No || [[File:Fig37_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || -2.0 || No || [[File:Fig38_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || -1.0 || No || [[File:Fig39_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 0 || No || [[File:Fig40_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 1.0 || No || [[File:Fig41_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 2.0 || No || [[File:Fig42_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 3.0 || No || [[File:Fig43_Yanda.png|200px]]
|}

However, keeping the same initial positions and everything else constant, except that the momentum pHF is increased to -1.5 while the momentum of pHH is reduced to 0 and is shown in the figure below. It resulted in a reactive trajectory, which suggested that the increase in translational energy of the system has greater effect in allowing the transition state to be overcome to form stable products.

[[File:Fig44_Yanda.png|300px|thumb|center|Figure 44. Reactive trajectory of F + H2 to HF + H reaction.]]

Next, the endothermic reaction of HF + H to F + H2, which is a late TS is investigated. Similarly, everything is kept constant, apart from pHF before keeping it constant and increasing pHH. The bond distance of HF is kept at 92 nm according to literature.

{| class="wikitable" border=1
! rH-F / pm !! rH-H / pm !! pHF/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1 !! Reactive Trajectory? || Contour Plot
|-
| 230 || 92 || 0 || -1.0 || No || [[File:Fig45_Yanda.png|200px]]
|-
| 230 || 92 || -5.0 || -1.0 || No || [[File:Fig46_Yanda.png|200px]]
|-
| 230 || 92 || -10.0 || -1.0 || No || [[File:Fig47_Yanda.png|200px]]
|-
| 230 || 92 || -15.0 || -1.0 || No || [[File:Fig48_Yanda.png|200px]]
|-
| 230 || 92 || -20.0 || -1.0 || No || [[File:Fig50_Yanda.png|200px]]
|-
| 230 || 92 || -25.0 || -1.0 || No || [[File:Fig51_Yanda.png|200px]]
|-
| 230 || 92 || -30.0 || -1.0 || Yes || [[File:Fig52_Yanda.png|200px]]
|}

Hence, it has shown that by increasing the vibrational energy of the H-F bond for the endothermic reaction while keep translational energy fixed at a low vaue resulted in the formation of products.

As such, Polanyi's empirical rules have been shown for both exothermic reactions (early TS) and endothermic reactions (late TS). However, it should be noted that such rules might not work for all cases due to imbalance distribution of translational and vibrational energies or errors in calculating energies, be it bond energies or trajectory calculations.

=References=

<references>
<ref name="CP3MD">Imperial College London, https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD, (accessed May 2018).</ref>
<ref name="Plot"> C.D., Sherrill, http://vergil.chemistry.gatech.edu/courses/chem6485/pdf/pes-lecture.pdf, (accessed May 2018). </ref>
<ref name="MRD">R. D. Levine, Molecular Reaction Dynamics, Cambridge University Press, Cambridge, 2005, pp. 202-203.</ref>
<ref name="MDACK">G. D. Billing and K. V. Mikkelsen, Introduction to Molecular Dynamics and Chemical Kinetics, John Wiley & Sons, New York, 1996, p. 31.</ref>
<ref name="Physical Chemistry">P. Atkins and J. de Paula, Atkins' Physical Chemistry, Oxford University Press, Oxford, 10th edn., 2014, pp. 65, 71 and 986.</ref>
<ref name="OC">J. Clayden, N. Greeves and S. Warren, Organic Chemistry, Oxford University Press, Oxford, 2nd edn., 2012, pp. 989.</ref>
<ref name="J. Phys. Chem.">C. Daniel et al., J. Phys. Chem., 1993, 97, 12485-12490.</ref>
</references>

MRD:yw14815

2020-07-07T08:53:19Z

Mak214: /* Locating the Transition State */

=Introduction=

This Wiki page outlines the study of the molecular reaction dynamics of two simple triatomic systems: H + H2 and F + H2. The atoms are 180° to each other, with a 1D space symplification. The atoms are labelled A, B and C, with vectors r1 and r2 representing vectors BA and BC in pm respectively. Momentum are also measured in p1 and p2 in g mol-1 pm fs-1. Below is the systematic diagram of the triatomic system model, which will be referred to throughout the whole wiki. <ref name="CP3MD"/>

[[File:Fig1_Yanda.png|400px|thumb|center|Figure 1. A triatomic system model. Diagram from Imperial College London, https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD (accessed May 2020.)]]

=EXERCISE 1: H + H2 system=
== Potential Energy Surfaces ==

A potential energy surface is a visual representation that describes the potential energy of a system, such as a collective of atoms, in terms of parameters, which normally is the relative positions of the reactants and the products. A reaction path is shown to be the gradient line on a PES diagram, and represents both potential energy and internuclear distances between atoms in the system as a function of the reaction coordinate, thereby modelling their changes through the progress of the reaction.

The PES diagram for Figure 1 above was calculated for the reaction system H + H2 using the parameters given:
{| class="wikitable"
!Calculations
!Steps
!r1 / pm
!r2 / pm
!p1 / gmol-1pmfs-1
!p2 / gmol-1pmfs-1
|-
|Dynamics
|500
|74
|230
|0.0
|<nowiki>-5.2</nowiki>
|}

[[File:Fig2_Yanda.png|400px|thumb|center|Figure 2.Potential Energy Surface for the H + H2 reaction path.]]

The reaction path, which passes through each potential energy minimum, including the transition state, which is a first order saddle point at which the potential energy is at its maximum (highest energy minimum) and where its gradient is 0, hence ∂V('''r''')/∂'''r '''=0 and the second differential at the TS is less than 0.Furthermore, the determinant of the Hessian Matrix is smaller than 0, with the eigenvalues of the Hessian Matrix being one positive and one negative

Local minimum, on the other hand,while also have the first derivative ∂V('''r''')/∂'''r '''=0, its second differential at the minimum is greater than 0, with the eigenvalues of the Hessian Matrix both being positive and the determinant of the Hessian Matrix greater than 0.

Literally speaking, the main difference between transition state and a local minimum is that the transition state is in fact a local energy maximum. Initially, with 0 initial momentum, both transition state and local minimum will remain there forever. However, with a very small momentum added towards the reactants direction, the transition state will be passed and will roll over towards the reactants, and vice versa for products. Testing the trajectories near the transition state is a method used to identify the presence of the transition state. On the other hand, a slightly perturbation from the local minimum will only result in rolling back towards the local minimum, because only a sufficient energy is needed to overcome the energy barrier to escape the "well" of the local minimum. In the PES diagram, a local minimum represents an intermediate, which can be represented in a 2D diagram below.<ref name="Plot"/>

[[File:Fig3_Yanda.PNG|400px|thumb|center|Figure 3. Potential Energy Surface Diagram showing the Transition State. Diagram from http://vergil.chemistry.gatech.edu/courses/chem6485/pdf/pes-lecture.pdf (accessed May 2020).]]

Another interesting observation is that the gradient along the reaction pathway is least steep close to both sides of the transition state, suggesting the point at which bonds are in the process of breaking and forming, hence lower degree of association and less vibrational oscillations. {{fontcolor1|red| Great. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:50, 7 July 2020 (BST)}}

== Locating the Transition State ==

The best estimate of the transition state position (rts) is 90.77 pm. The transition state should be symmetric as it represents the state of the reaction where old bonds and new bonds are in the middle of being broken and formed. As such, given the initial conditions of both momentum p1=0 and p2=0, the two internuclear distances r1 and r2 must be equal at the transitions state. The constant internuclear distances can be understood and found with a "Internuclear Distances vs Time plot" shown in Figure 4 below, where the straight horizontal line suggests constant potential energy with the absence of any vibrational oscillations. {{fontcolor1|red| Yes. Good job. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:53, 7 July 2020 (BST)}}

[[File:Fig4_Yanda.png|400px|thumb|center|Figure 4. Internuclear Distance vs Time plot for finding transition state position.]]

== Calculation of Reaction Path ==

{| class="wikitable" border="1"
! Calculations !! Steps !! r1 / pm !! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| MEP || 50000 || 91.77 || 90.77 || 0 || 0
|}

The Potential Energy Surface for MEP Calculation is shown in Figure 5 below.

[[File:Fig5_Yanda.png|400px|thumb|center|Figure 5.Potential Energy Surface Diagram with MEP.]]

{| class="wikitable" border="1"
! Calculations !! Steps !! r1 / pm !! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| Dynamics || 500 || 91.77 || 90.77 || 0 || 0
|}

Similarly, the Potential Energy Surface for Dynamics Calculation is shown in Figure 6 below.

[[File:Fig6_Yanda.png|400px|thumb|center|Figure 6.Potential Energy Surface Diagram with Dynamics.]]

Comparing the Dynamics (Figure 6) and the MEP (Figure 5), it can be seen that the MEP does not take into account of vibrational oscillations within the atoms, but just passing through all the potential minimum. On the other hand, the trajectory just calculated takes into account of the momentum changes during the reaction by oscillating about the minimum potential energy. In addition, during the calculations, 50000 steps were required to complete the minimum energy path in MEP, while 500 steps were required in Dynamics.

The "Internuclear Distances vs Time" and "Momenta vs Time" graphs for Dynamics calculations are respectively plotted in Figure 7 and 8 below.

[[File:Fig7_Yanda.png|400px|thumb|center|Figure 7. Internuclear Distances vs Time using Dynamics.]]

[[File:Fig8_Yanda.png|400px|thumb|center|Figure 8. Momenta vs Time using Dynamics.]]

Figure 7 and 8 can help to describe the reaction. Distance r1 (B-C) and r2 (A-C) are seen seen to increase as A moves further away from B-C after the transition state. After an extended period of time, it can be seen that A-C and B-C distances increases linearly, suggesting that the distance A-B is constant and that B-C is moving further away from A. Also,initially, A-B decreases slightly before plateauing/oscillating, suggesting that the bond between A-B has formed and that they are vibrating up and down from equilibrium. The vibration is further suggested by the oscillation of A-B in the Momenta vs. Time graph in Figure 8.

The "Internuclear Distances vs Time" and "Momenta vs Time" graphs for MEP calculations are respectively plotted in Figure 9 and 10 below.

[[File:Fig9_Yanda.png|400px|thumb|center|Figure 9. Internuclear Distances vs Time using MEP.]]

[[File:Fig10_Yanda.png|400px|thumb|center|Figure 10. Momenta vs Time using MEP.]]

With reference to Figure 9 and 10, the internuclear distance r2 (A-B) decreases rapidly before converging to a limit and plateauing, while r1 (B-C), hence A-C increases rapidly. The whole system has 0 momentum because the momenta/velocities are always reset to zero in each time step.

Also, the final values of the positions r1(t), r2(t), p1(t) and p2(t) for the trajectory for large enough t, at t=100fs are calculated in the table below.

{| class="wikitable" border="1"
! r1 / pm!! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| 735.414|| 75.4847|| 5.08082 || 2.02769
|}

If the initial conditions were r1 = rts and r2 = rts + 1 pm instead, the reaction path will move in the opposite direction, exiting from the opposite direction in the potential energy surface (Figure 11) whereby distance r2 (A-B) increases while the distance r1 (B-C) stays constant with only vibrations of the bond similar to the previous conditions (Figure 12 and 13). Hence, in the first case the transition state moves towards AB + C, while the reverse will be in this case favoring the formation of products A + BC, as A start off further to B than that of C to B.

[[File:Fig11_Yanda.png|400px|thumb|center|Figure 11. Potential Energy Surface with Dynamics with initial conditions reversed.]]
[[File:Fig12_Yanda.png|400px|thumb|center|Figure 12. Internuclear Distances vs Time with Dynamics with initial conditions reversed.]]
[[File:Fig13_Yanda.png|400px|thumb|center|Figure 13. Momenta vs Time with Dynamics with initial conditions reversed.]]

The distorted transition state, where the values of r1=rts and r2=rts + 1 pm would be inverted for as the previous calculations, with the graphs for internuclear distance and momenta also inverted, such that graph of B-C appears as the original A-B.

Finally,a calculation was setup where the initial positions correspond to the final positions of the trajectory calculative, above, the same final momenta values but with their signs reversed.

{| class="wikitable" border="1"
! r1 / pm!! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| 735.414|| 75.4847|| -5.08082 || -2.02769
|}

[[File:Fig14_Yanda.png|400px|thumb|center|Figure 14. Potential Energy Surface with Dynamics moving from products to estimated Transition State.]]
[[File:Fig15_Yanda.png|400px|thumb|center|Figure 15. Internuclear Distances vs Time with Dynamics moving from products to estimated Transition State.]]
[[File:Fig16_Yanda.png|400px|thumb|center|Figure 16. Momenta vs Time with Dynamics moving from products to estimated Transition State.]]

This situation is the reverse of the original situation in which now, the species start off as reactants with just the right starting positions and momenta to reach the slightly distorted transition state. Technically, the three graphs in this situation should be the mirror image of the three graphs in the original situation. However, a perfect symmetrical relationship is not observed due to random errors in estimating through rounding off or sampling errors of the final positions and momenta from the Internuclear Distances vs Time graph and Momenta vs Time graph respectively.For example, in Figure 15, the distance between B-C should be constant after the transition state, but was not the case due to a errors in the initial input data.

== Reactive and Unreactive Trajectories ==

For the initial positions r1 = 74 pm and r2 = 200 pm, various trajectories were ran with different momentum values and combinations as tabulated below.

{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280|| Reactive || Reactive as reaction path can get over maximum at TS and form products. Straight trajectory centered around Dynamics suggest molecule B-C does not vibrate until A collides into it, with sufficient momentum that subsequently resulted in molecule A-B with vibrational oscillation after TS || [[File:Fig17_Yanda.png|300px]] [[File:Fig18_Yanda.png|300px]]
|-
| -3.1 || -4.1 || -420.077 || Unreactive || Not reactive as reactive path unable to get over maximum at TS and form products. The PES curve suggests that the trajectory is vibrational/oscillating throughout and that the reaction path does not move past the TS. It turns back before the TS once the potential energy exceeds the kinetic energy of the species of the momenta given || [[File:Fig19_Yanda.png|300px]] [[File:Fig20_Yanda.png|300px]]
|-
| -3.1 || -5.1|| -413.977 || Reactive|| Reactive as similar to the first scenario. Note that higher initial momentum of molecule B-C results in larger vibrations as visible in the PES curve || [[File:Fig21_Yanda.png|300px]] [[File:Fig22_Yanda.png|300px]]
|-
| -5.1|| -10.1|| -357.277 || Unreactive || Initial momenta large enough for reaction path to attain TS, but the activation barrier was crossed twice such that eventually the reactants were reformed again. At TS, A-B-C, the initial momentum was not large enough to form stable products. As such, reaction path recrosses the TS, as visible from the contour map, back into the reactions, reforming A + BC. Much larger vibrational oscillations that the previous scenarios due to large initial momentum as visible from the PES curve || [[File:Fig23_Yanda.png|300px]] [[File:Fig24_Yanda.png|300px]]
|-
| -5.1|| -10.6|| -349.477 || Reactive || Similar to the previous scenario, but that now the initial momenta are larger enough to allow three crossing of the activation barrier, hence passing through the TS thrice, eventually forming the products A + BC|| [[File:Fig25_Yanda.png|300px]] [[File:Fig26_Yanda.png|300px]]
|}

From the table, it can be seen that while a minimum initial momentum of the species are needed to cross the activation barrier and over the TS to form the products, a much initial momentum does not guarantee product formation as can be seen in the second last scenario. As such, the conditions required for enough momentum given the initial preconditions as listed are: -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1.

== Transition State Theory ==

The deviations from the experimental values and the values describes in the Transition State Theory (TST) arises because of the assumptions within the theory itself. First of all, the TST adheres to the Born-Oppenheimer approximation, where the nuclear and electronic motions are decoupled. The Transition State Theory assumes that only the reactants are in constant equilibrium with the transition state structure, and that the energy of the particles follow a Boltzmann distribution. In addition, once the reactants have trajectories with a kinetic energy along the reaction coordinate greater than the activation energy, it will be reactive, and that the products will definitely form, ignoring the possibility that the transition state structure will collapse back to the reactants. However, experimentally, it has been seen in the few scenarios above where the initial momentum (p1=-5.1, p2=-10.1), hence kinetic energy is high enough to result in multiple crossing of the activation energy barrier through multiple forward and backward directions. Theoretically, the theory suggests only a single crossing of the activation energy barrier in the forward direction is possible. Also, note that in the transition state, the motion along the reaction coordinate can be treated classically. <ref name="MRD"/>

Finally, the theory fails for some reactions at elevated temperatures due to more complex molecule motions or low temperatures due to quantum tunneling, whereby the transition state need not be reached to form products.<ref name="MDACK"/>

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

The energetics of both F + H2 and the reverse H + HF were investigated and the findings are summarized in the table below. The method to extract these data are explained further in this section.

{| class="wikitable" border="1"
! Reaction !! Transition State Energy / kJ/mol !! Reactants Energy / kJ/mol !! Activation Energy / kJ/mol !! Type of Reaction
|-
| F + H2 || -433.980 || -435.100|| +1.120 || Exothermic
|-
| H + HF || -433.980 || -560.700 || +126.720 || Endothermic
|}

The reaction between F and H2 is an exothermic reaction owing to the stronger H-F bond relative to the H-H bond. Since enthalpy of reaction is calculated from energy of bond breaking - energy of bond making, the stronger negative value arising from breaking the stronger H-F bond resulted in an exothermic reaction. The stronger H-F bond can be explained due to the larger electronegativity difference between H and F, for which F= 4.0 and H = 2.1. <ref name="Physical Chemistry"/> As a result of the electronegativity difference, the F atom pulls electron density closer to itself, resulting in a highly polarized H-F bond, adding on an extra contribution to the bond strength. H-H bond, on the other hand, is a simple covalent bond. Literature value suggests that the bond dissociation energies of H-H and H-F are 432 kJ/mol and 565 kJ/mol respectively, further suggesting that the enthalpy released from breaking H-F bond is much higher than that of H-H bonds. Therefore, the reaction between F and H2 to form HF and H is energetically favorable, with an overall exothermic reaction.

The reverse reaction, H + HF is therefore endothermic, thereby requiring external energy in order to break the H-F bond, whereby the formation of the H-H bond does not compensate for.

The Hammond's postulate states that the transition state of a reaction resembles either the reactants or the products depending on which it is closer in energy. Hence, in an exothermic reaction, the transition state is closer to the reactants in energy. In an endothermic reaction, the transition state is closer to the products in energy.<ref name="OC"/>

Referring to the table above, for the forward reaction between F and H2, since the reactants energy is much closer to the transition state energy as compared to the reactions energy and the transition state energy in H + HF backward reaction, the forward reaction is the exothermic reaction, and the backward reaction is the endothermic reaction. This can also be determined graphically. With reference to the PES diagram and Energy vs Time plot below for the reaction between F and H2, it shows that the transition state has been attained, with no net change in potential energy. Furthermore, the PES suggests that the reactants are higher in energy than the products, indicating an exothermic reaction.

[[File:Fig27_Yanda.png|center|300px|thumb|Figure 27. Energy vs. Time Plot for F + H2. ]]
[[File:Fig28_Yanda.png|center|300px|thumb|Figure 28. Potential Energy Surface Diagram for F + H2. ]]

Similarly, Figure 29 below shows the transition state from the reverse reaction H + HF. Comparatively, it can be seen that the transition state closely resembles the products, which are now higher in energy than the reactants, thereby deeming it an endothermic reaction.

[[File:Fig29_Yanda.png|center|300px|thumb|Figure 29. Potential Energy Surface Diagram for H + HF. ]]

The approximate position of the transition state, is found to be r1 = BC = 181.1 pm and r2 = AB = 74.5 pm .The potential energy at the TS is then found to be -433.980 kJ/mol.

The activation energy of the reaction was found by slightly perturbing it from the TS and finding the energy difference of the products to the transition state energy as shown in the tables and Figures 30-33 below.

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| MEP || 2000 || 0.1 || 181.1 || 74.5 || 0 || 0
|}

[[File:Fig30_Yanda.png|center|300px|thumb|Figure 30. Contour Map for the reaction from H2 + F to HF + H. ]]
[[File:Fig31_Yanda.png|center|300px|thumb|Figure 31. Energy vs Time diagram for the reaction from H2 + F to HF + H. ]]

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| MEP || 2000 || 0.1 || 182.1 || 74.5 || 0 || 0
|}

[[File:Fig32_Yanda.png|center|300px|thumb|Figure 32. Contour Map for the reaction from HF + H to H2 + F. ]] [[File:Fig33_Yanda.png|center|300px|thumb|Figure 33. Energy vs Time diagram for the reaction from HF + H to H2 + F. ]]

By finding the trajectory down the MEP with the initial conditions, then inserting the last geometry as the initial condition. After which, replace one of the bond distance of the reactants to an extremely huge number, while keeping the bond distance of the product the same, the potential energy of the reactants and the products can be found. Though, it must be noted that this is just an approximation, as the exact potential energy is when the potential energy curve stabilizes. Ultimately, the potential energy of the reactants are found, can be used to find the activation energies, as summarized in the first table above.

==Reaction Dynamics==

The following reaction condition resulted in the reaction from H2 + F -> HF + H

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| Dynamics || 1400 || 0.5 || 220 || 71 || -1.5 || -0.3
|}

The conservation of energy suggests that total energy is constant through the course of reaction. Hence, in the exothermic reaction between H2 and F to form HF + H, the loss in potential energy from the breaking of bonds is replaced by the increase in kinetic energy as seen in the Energy Vs Time graph below. The kinetic energy can be of three forms: translational, rotational or vibrational. Rotational motion is ignored here as according to the animation, it is currently working on a 1D system.

A bomb calorimeter can be used to measure the overall gain in kinetic energy, both translational and kinetic by measuring the transfer of these energy from the products to the walls of the calorimeter in the form of heat. The heat transferred can then be measured in the form of change in temperature with a thermal sensor. <ref name="Physical Chemistry"/>

However, in order to differentiate the respective contributions of translational and vibrational energy, further experimental is required. To determine the vibrational energy, IR spectroscopy can be utilized. Before the reactants collide, most of the bonds occupy lowest energy vibrational levels. However, during a collision, higher level vibrational modes are excited as potential energy is convereted into vibrational kinetic energy. In IR spectroscopy, the higher level modes are characterized as overtones signals where energy levels move from initially 0 to 1, to 0 to 2 or 0 to 3, or even higher. These overtones can be detected and seen with higher wavenumbers as compared to the normal 0 to 1 transition. <ref name="Physical Chemistry"/>

[[File:Fig34_Yanda.png|center|300px|thumb|Figure 34. Contour Map of reaction from H2 + F to HF + H ]]
[[File:Fig35_Yanda.png|center|300px|thumb|Figure 35. Internuclear Distance vs Time Diagram of reaction from H2 + F to HF + H ]]
[[File:Fig36_Yanda.png|center|300px|thumb|Figure 36. Energy vs Time Diagram of reaction from H2 + F to HF + H ]]

==The Polanyi Rules==

The Polanyi rules state that for reactants with a given momentum along the direction of the reaction path, the translational energy, as opposed to the vibrational energy, is preferred in overcoming the early transition state in an exothermic reaction. The opposite will be true for endothermic reactions, where vibrational energy takes priority over translational energy to successfully form stable products.<ref name="J. Phys. Chem."/>

Reaction trajectories were calculated for the exothermic reaction of F + H2 with 1000 steps, but by varying the momentum of r1, the momentum of H-H bond, with all other conditions kept constant such that only the vibration energy of the H2 is varied. This is to investigate an early TS, since it is exothermic.

{| class="wikitable" border=1
! rH-F / pm !! rH-H / pm !! pHF/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1 !! Reactive Trajectory? || Contour Plot
|-
| 230 || 74 || -1.0 || -3.0 || No || [[File:Fig37_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || -2.0 || No || [[File:Fig38_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || -1.0 || No || [[File:Fig39_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 0 || No || [[File:Fig40_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 1.0 || No || [[File:Fig41_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 2.0 || No || [[File:Fig42_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 3.0 || No || [[File:Fig43_Yanda.png|200px]]
|}

However, keeping the same initial positions and everything else constant, except that the momentum pHF is increased to -1.5 while the momentum of pHH is reduced to 0 and is shown in the figure below. It resulted in a reactive trajectory, which suggested that the increase in translational energy of the system has greater effect in allowing the transition state to be overcome to form stable products.

[[File:Fig44_Yanda.png|300px|thumb|center|Figure 44. Reactive trajectory of F + H2 to HF + H reaction.]]

Next, the endothermic reaction of HF + H to F + H2, which is a late TS is investigated. Similarly, everything is kept constant, apart from pHF before keeping it constant and increasing pHH. The bond distance of HF is kept at 92 nm according to literature.

{| class="wikitable" border=1
! rH-F / pm !! rH-H / pm !! pHF/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1 !! Reactive Trajectory? || Contour Plot
|-
| 230 || 92 || 0 || -1.0 || No || [[File:Fig45_Yanda.png|200px]]
|-
| 230 || 92 || -5.0 || -1.0 || No || [[File:Fig46_Yanda.png|200px]]
|-
| 230 || 92 || -10.0 || -1.0 || No || [[File:Fig47_Yanda.png|200px]]
|-
| 230 || 92 || -15.0 || -1.0 || No || [[File:Fig48_Yanda.png|200px]]
|-
| 230 || 92 || -20.0 || -1.0 || No || [[File:Fig50_Yanda.png|200px]]
|-
| 230 || 92 || -25.0 || -1.0 || No || [[File:Fig51_Yanda.png|200px]]
|-
| 230 || 92 || -30.0 || -1.0 || Yes || [[File:Fig52_Yanda.png|200px]]
|}

Hence, it has shown that by increasing the vibrational energy of the H-F bond for the endothermic reaction while keep translational energy fixed at a low vaue resulted in the formation of products.

As such, Polanyi's empirical rules have been shown for both exothermic reactions (early TS) and endothermic reactions (late TS). However, it should be noted that such rules might not work for all cases due to imbalance distribution of translational and vibrational energies or errors in calculating energies, be it bond energies or trajectory calculations.

=References=

<references>
<ref name="CP3MD">Imperial College London, https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD, (accessed May 2018).</ref>
<ref name="Plot"> C.D., Sherrill, http://vergil.chemistry.gatech.edu/courses/chem6485/pdf/pes-lecture.pdf, (accessed May 2018). </ref>
<ref name="MRD">R. D. Levine, Molecular Reaction Dynamics, Cambridge University Press, Cambridge, 2005, pp. 202-203.</ref>
<ref name="MDACK">G. D. Billing and K. V. Mikkelsen, Introduction to Molecular Dynamics and Chemical Kinetics, John Wiley & Sons, New York, 1996, p. 31.</ref>
<ref name="Physical Chemistry">P. Atkins and J. de Paula, Atkins' Physical Chemistry, Oxford University Press, Oxford, 10th edn., 2014, pp. 65, 71 and 986.</ref>
<ref name="OC">J. Clayden, N. Greeves and S. Warren, Organic Chemistry, Oxford University Press, Oxford, 2nd edn., 2012, pp. 989.</ref>
<ref name="J. Phys. Chem.">C. Daniel et al., J. Phys. Chem., 1993, 97, 12485-12490.</ref>
</references>

MRD:yw14815

2020-07-07T08:52:08Z

Mak214:

=Introduction=

This Wiki page outlines the study of the molecular reaction dynamics of two simple triatomic systems: H + H2 and F + H2. The atoms are 180° to each other, with a 1D space symplification. The atoms are labelled A, B and C, with vectors r1 and r2 representing vectors BA and BC in pm respectively. Momentum are also measured in p1 and p2 in g mol-1 pm fs-1. Below is the systematic diagram of the triatomic system model, which will be referred to throughout the whole wiki. <ref name="CP3MD"/>

[[File:Fig1_Yanda.png|400px|thumb|center|Figure 1. A triatomic system model. Diagram from Imperial College London, https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD (accessed May 2020.)]]

=EXERCISE 1: H + H2 system=
== Potential Energy Surfaces ==

A potential energy surface is a visual representation that describes the potential energy of a system, such as a collective of atoms, in terms of parameters, which normally is the relative positions of the reactants and the products. A reaction path is shown to be the gradient line on a PES diagram, and represents both potential energy and internuclear distances between atoms in the system as a function of the reaction coordinate, thereby modelling their changes through the progress of the reaction.

The PES diagram for Figure 1 above was calculated for the reaction system H + H2 using the parameters given:
{| class="wikitable"
!Calculations
!Steps
!r1 / pm
!r2 / pm
!p1 / gmol-1pmfs-1
!p2 / gmol-1pmfs-1
|-
|Dynamics
|500
|74
|230
|0.0
|<nowiki>-5.2</nowiki>
|}

[[File:Fig2_Yanda.png|400px|thumb|center|Figure 2.Potential Energy Surface for the H + H2 reaction path.]]

The reaction path, which passes through each potential energy minimum, including the transition state, which is a first order saddle point at which the potential energy is at its maximum (highest energy minimum) and where its gradient is 0, hence ∂V('''r''')/∂'''r '''=0 and the second differential at the TS is less than 0.Furthermore, the determinant of the Hessian Matrix is smaller than 0, with the eigenvalues of the Hessian Matrix being one positive and one negative

Local minimum, on the other hand,while also have the first derivative ∂V('''r''')/∂'''r '''=0, its second differential at the minimum is greater than 0, with the eigenvalues of the Hessian Matrix both being positive and the determinant of the Hessian Matrix greater than 0.

Literally speaking, the main difference between transition state and a local minimum is that the transition state is in fact a local energy maximum. Initially, with 0 initial momentum, both transition state and local minimum will remain there forever. However, with a very small momentum added towards the reactants direction, the transition state will be passed and will roll over towards the reactants, and vice versa for products. Testing the trajectories near the transition state is a method used to identify the presence of the transition state. On the other hand, a slightly perturbation from the local minimum will only result in rolling back towards the local minimum, because only a sufficient energy is needed to overcome the energy barrier to escape the "well" of the local minimum. In the PES diagram, a local minimum represents an intermediate, which can be represented in a 2D diagram below.<ref name="Plot"/>

[[File:Fig3_Yanda.PNG|400px|thumb|center|Figure 3. Potential Energy Surface Diagram showing the Transition State. Diagram from http://vergil.chemistry.gatech.edu/courses/chem6485/pdf/pes-lecture.pdf (accessed May 2020).]]

Another interesting observation is that the gradient along the reaction pathway is least steep close to both sides of the transition state, suggesting the point at which bonds are in the process of breaking and forming, hence lower degree of association and less vibrational oscillations. {{fontcolor1|red| Great. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:50, 7 July 2020 (BST)}}

== Locating the Transition State ==

The best estimate of the transition state position (rts) is 90.77 pm. The transition state should be symmetric as it represents the state of the reaction where old bonds and new bonds are in the middle of being broken and formed. As such, given the initial conditions of both momentum p1=0 and p2=0, the two internuclear distances r1 and r2 must be equal at the transitions state. The constant internuclear distances can be understood and found with a "Internuclear Distances vs Time plot" shown in Figure 4 below, where the straight horizontal line suggests constant potential energy with the absence of any vibrational oscillations.

[[File:Fig4_Yanda.png|400px|thumb|center|Figure 4. Internuclear Distance vs Time plot for finding transition state position.]]

== Calculation of Reaction Path ==

{| class="wikitable" border="1"
! Calculations !! Steps !! r1 / pm !! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| MEP || 50000 || 91.77 || 90.77 || 0 || 0
|}

The Potential Energy Surface for MEP Calculation is shown in Figure 5 below.

[[File:Fig5_Yanda.png|400px|thumb|center|Figure 5.Potential Energy Surface Diagram with MEP.]]

{| class="wikitable" border="1"
! Calculations !! Steps !! r1 / pm !! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| Dynamics || 500 || 91.77 || 90.77 || 0 || 0
|}

Similarly, the Potential Energy Surface for Dynamics Calculation is shown in Figure 6 below.

[[File:Fig6_Yanda.png|400px|thumb|center|Figure 6.Potential Energy Surface Diagram with Dynamics.]]

Comparing the Dynamics (Figure 6) and the MEP (Figure 5), it can be seen that the MEP does not take into account of vibrational oscillations within the atoms, but just passing through all the potential minimum. On the other hand, the trajectory just calculated takes into account of the momentum changes during the reaction by oscillating about the minimum potential energy. In addition, during the calculations, 50000 steps were required to complete the minimum energy path in MEP, while 500 steps were required in Dynamics.

The "Internuclear Distances vs Time" and "Momenta vs Time" graphs for Dynamics calculations are respectively plotted in Figure 7 and 8 below.

[[File:Fig7_Yanda.png|400px|thumb|center|Figure 7. Internuclear Distances vs Time using Dynamics.]]

[[File:Fig8_Yanda.png|400px|thumb|center|Figure 8. Momenta vs Time using Dynamics.]]

Figure 7 and 8 can help to describe the reaction. Distance r1 (B-C) and r2 (A-C) are seen seen to increase as A moves further away from B-C after the transition state. After an extended period of time, it can be seen that A-C and B-C distances increases linearly, suggesting that the distance A-B is constant and that B-C is moving further away from A. Also,initially, A-B decreases slightly before plateauing/oscillating, suggesting that the bond between A-B has formed and that they are vibrating up and down from equilibrium. The vibration is further suggested by the oscillation of A-B in the Momenta vs. Time graph in Figure 8.

The "Internuclear Distances vs Time" and "Momenta vs Time" graphs for MEP calculations are respectively plotted in Figure 9 and 10 below.

[[File:Fig9_Yanda.png|400px|thumb|center|Figure 9. Internuclear Distances vs Time using MEP.]]

[[File:Fig10_Yanda.png|400px|thumb|center|Figure 10. Momenta vs Time using MEP.]]

With reference to Figure 9 and 10, the internuclear distance r2 (A-B) decreases rapidly before converging to a limit and plateauing, while r1 (B-C), hence A-C increases rapidly. The whole system has 0 momentum because the momenta/velocities are always reset to zero in each time step.

Also, the final values of the positions r1(t), r2(t), p1(t) and p2(t) for the trajectory for large enough t, at t=100fs are calculated in the table below.

{| class="wikitable" border="1"
! r1 / pm!! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| 735.414|| 75.4847|| 5.08082 || 2.02769
|}

If the initial conditions were r1 = rts and r2 = rts + 1 pm instead, the reaction path will move in the opposite direction, exiting from the opposite direction in the potential energy surface (Figure 11) whereby distance r2 (A-B) increases while the distance r1 (B-C) stays constant with only vibrations of the bond similar to the previous conditions (Figure 12 and 13). Hence, in the first case the transition state moves towards AB + C, while the reverse will be in this case favoring the formation of products A + BC, as A start off further to B than that of C to B.

[[File:Fig11_Yanda.png|400px|thumb|center|Figure 11. Potential Energy Surface with Dynamics with initial conditions reversed.]]
[[File:Fig12_Yanda.png|400px|thumb|center|Figure 12. Internuclear Distances vs Time with Dynamics with initial conditions reversed.]]
[[File:Fig13_Yanda.png|400px|thumb|center|Figure 13. Momenta vs Time with Dynamics with initial conditions reversed.]]

The distorted transition state, where the values of r1=rts and r2=rts + 1 pm would be inverted for as the previous calculations, with the graphs for internuclear distance and momenta also inverted, such that graph of B-C appears as the original A-B.

Finally,a calculation was setup where the initial positions correspond to the final positions of the trajectory calculative, above, the same final momenta values but with their signs reversed.

{| class="wikitable" border="1"
! r1 / pm!! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| 735.414|| 75.4847|| -5.08082 || -2.02769
|}

[[File:Fig14_Yanda.png|400px|thumb|center|Figure 14. Potential Energy Surface with Dynamics moving from products to estimated Transition State.]]
[[File:Fig15_Yanda.png|400px|thumb|center|Figure 15. Internuclear Distances vs Time with Dynamics moving from products to estimated Transition State.]]
[[File:Fig16_Yanda.png|400px|thumb|center|Figure 16. Momenta vs Time with Dynamics moving from products to estimated Transition State.]]

This situation is the reverse of the original situation in which now, the species start off as reactants with just the right starting positions and momenta to reach the slightly distorted transition state. Technically, the three graphs in this situation should be the mirror image of the three graphs in the original situation. However, a perfect symmetrical relationship is not observed due to random errors in estimating through rounding off or sampling errors of the final positions and momenta from the Internuclear Distances vs Time graph and Momenta vs Time graph respectively.For example, in Figure 15, the distance between B-C should be constant after the transition state, but was not the case due to a errors in the initial input data.

== Reactive and Unreactive Trajectories ==

For the initial positions r1 = 74 pm and r2 = 200 pm, various trajectories were ran with different momentum values and combinations as tabulated below.

{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280|| Reactive || Reactive as reaction path can get over maximum at TS and form products. Straight trajectory centered around Dynamics suggest molecule B-C does not vibrate until A collides into it, with sufficient momentum that subsequently resulted in molecule A-B with vibrational oscillation after TS || [[File:Fig17_Yanda.png|300px]] [[File:Fig18_Yanda.png|300px]]
|-
| -3.1 || -4.1 || -420.077 || Unreactive || Not reactive as reactive path unable to get over maximum at TS and form products. The PES curve suggests that the trajectory is vibrational/oscillating throughout and that the reaction path does not move past the TS. It turns back before the TS once the potential energy exceeds the kinetic energy of the species of the momenta given || [[File:Fig19_Yanda.png|300px]] [[File:Fig20_Yanda.png|300px]]
|-
| -3.1 || -5.1|| -413.977 || Reactive|| Reactive as similar to the first scenario. Note that higher initial momentum of molecule B-C results in larger vibrations as visible in the PES curve || [[File:Fig21_Yanda.png|300px]] [[File:Fig22_Yanda.png|300px]]
|-
| -5.1|| -10.1|| -357.277 || Unreactive || Initial momenta large enough for reaction path to attain TS, but the activation barrier was crossed twice such that eventually the reactants were reformed again. At TS, A-B-C, the initial momentum was not large enough to form stable products. As such, reaction path recrosses the TS, as visible from the contour map, back into the reactions, reforming A + BC. Much larger vibrational oscillations that the previous scenarios due to large initial momentum as visible from the PES curve || [[File:Fig23_Yanda.png|300px]] [[File:Fig24_Yanda.png|300px]]
|-
| -5.1|| -10.6|| -349.477 || Reactive || Similar to the previous scenario, but that now the initial momenta are larger enough to allow three crossing of the activation barrier, hence passing through the TS thrice, eventually forming the products A + BC|| [[File:Fig25_Yanda.png|300px]] [[File:Fig26_Yanda.png|300px]]
|}

From the table, it can be seen that while a minimum initial momentum of the species are needed to cross the activation barrier and over the TS to form the products, a much initial momentum does not guarantee product formation as can be seen in the second last scenario. As such, the conditions required for enough momentum given the initial preconditions as listed are: -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1.

== Transition State Theory ==

The deviations from the experimental values and the values describes in the Transition State Theory (TST) arises because of the assumptions within the theory itself. First of all, the TST adheres to the Born-Oppenheimer approximation, where the nuclear and electronic motions are decoupled. The Transition State Theory assumes that only the reactants are in constant equilibrium with the transition state structure, and that the energy of the particles follow a Boltzmann distribution. In addition, once the reactants have trajectories with a kinetic energy along the reaction coordinate greater than the activation energy, it will be reactive, and that the products will definitely form, ignoring the possibility that the transition state structure will collapse back to the reactants. However, experimentally, it has been seen in the few scenarios above where the initial momentum (p1=-5.1, p2=-10.1), hence kinetic energy is high enough to result in multiple crossing of the activation energy barrier through multiple forward and backward directions. Theoretically, the theory suggests only a single crossing of the activation energy barrier in the forward direction is possible. Also, note that in the transition state, the motion along the reaction coordinate can be treated classically. <ref name="MRD"/>

Finally, the theory fails for some reactions at elevated temperatures due to more complex molecule motions or low temperatures due to quantum tunneling, whereby the transition state need not be reached to form products.<ref name="MDACK"/>

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

The energetics of both F + H2 and the reverse H + HF were investigated and the findings are summarized in the table below. The method to extract these data are explained further in this section.

{| class="wikitable" border="1"
! Reaction !! Transition State Energy / kJ/mol !! Reactants Energy / kJ/mol !! Activation Energy / kJ/mol !! Type of Reaction
|-
| F + H2 || -433.980 || -435.100|| +1.120 || Exothermic
|-
| H + HF || -433.980 || -560.700 || +126.720 || Endothermic
|}

The reaction between F and H2 is an exothermic reaction owing to the stronger H-F bond relative to the H-H bond. Since enthalpy of reaction is calculated from energy of bond breaking - energy of bond making, the stronger negative value arising from breaking the stronger H-F bond resulted in an exothermic reaction. The stronger H-F bond can be explained due to the larger electronegativity difference between H and F, for which F= 4.0 and H = 2.1. <ref name="Physical Chemistry"/> As a result of the electronegativity difference, the F atom pulls electron density closer to itself, resulting in a highly polarized H-F bond, adding on an extra contribution to the bond strength. H-H bond, on the other hand, is a simple covalent bond. Literature value suggests that the bond dissociation energies of H-H and H-F are 432 kJ/mol and 565 kJ/mol respectively, further suggesting that the enthalpy released from breaking H-F bond is much higher than that of H-H bonds. Therefore, the reaction between F and H2 to form HF and H is energetically favorable, with an overall exothermic reaction.

The reverse reaction, H + HF is therefore endothermic, thereby requiring external energy in order to break the H-F bond, whereby the formation of the H-H bond does not compensate for.

The Hammond's postulate states that the transition state of a reaction resembles either the reactants or the products depending on which it is closer in energy. Hence, in an exothermic reaction, the transition state is closer to the reactants in energy. In an endothermic reaction, the transition state is closer to the products in energy.<ref name="OC"/>

Referring to the table above, for the forward reaction between F and H2, since the reactants energy is much closer to the transition state energy as compared to the reactions energy and the transition state energy in H + HF backward reaction, the forward reaction is the exothermic reaction, and the backward reaction is the endothermic reaction. This can also be determined graphically. With reference to the PES diagram and Energy vs Time plot below for the reaction between F and H2, it shows that the transition state has been attained, with no net change in potential energy. Furthermore, the PES suggests that the reactants are higher in energy than the products, indicating an exothermic reaction.

[[File:Fig27_Yanda.png|center|300px|thumb|Figure 27. Energy vs. Time Plot for F + H2. ]]
[[File:Fig28_Yanda.png|center|300px|thumb|Figure 28. Potential Energy Surface Diagram for F + H2. ]]

Similarly, Figure 29 below shows the transition state from the reverse reaction H + HF. Comparatively, it can be seen that the transition state closely resembles the products, which are now higher in energy than the reactants, thereby deeming it an endothermic reaction.

[[File:Fig29_Yanda.png|center|300px|thumb|Figure 29. Potential Energy Surface Diagram for H + HF. ]]

The approximate position of the transition state, is found to be r1 = BC = 181.1 pm and r2 = AB = 74.5 pm .The potential energy at the TS is then found to be -433.980 kJ/mol.

The activation energy of the reaction was found by slightly perturbing it from the TS and finding the energy difference of the products to the transition state energy as shown in the tables and Figures 30-33 below.

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| MEP || 2000 || 0.1 || 181.1 || 74.5 || 0 || 0
|}

[[File:Fig30_Yanda.png|center|300px|thumb|Figure 30. Contour Map for the reaction from H2 + F to HF + H. ]]
[[File:Fig31_Yanda.png|center|300px|thumb|Figure 31. Energy vs Time diagram for the reaction from H2 + F to HF + H. ]]

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| MEP || 2000 || 0.1 || 182.1 || 74.5 || 0 || 0
|}

[[File:Fig32_Yanda.png|center|300px|thumb|Figure 32. Contour Map for the reaction from HF + H to H2 + F. ]] [[File:Fig33_Yanda.png|center|300px|thumb|Figure 33. Energy vs Time diagram for the reaction from HF + H to H2 + F. ]]

By finding the trajectory down the MEP with the initial conditions, then inserting the last geometry as the initial condition. After which, replace one of the bond distance of the reactants to an extremely huge number, while keeping the bond distance of the product the same, the potential energy of the reactants and the products can be found. Though, it must be noted that this is just an approximation, as the exact potential energy is when the potential energy curve stabilizes. Ultimately, the potential energy of the reactants are found, can be used to find the activation energies, as summarized in the first table above.

==Reaction Dynamics==

The following reaction condition resulted in the reaction from H2 + F -> HF + H

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| Dynamics || 1400 || 0.5 || 220 || 71 || -1.5 || -0.3
|}

The conservation of energy suggests that total energy is constant through the course of reaction. Hence, in the exothermic reaction between H2 and F to form HF + H, the loss in potential energy from the breaking of bonds is replaced by the increase in kinetic energy as seen in the Energy Vs Time graph below. The kinetic energy can be of three forms: translational, rotational or vibrational. Rotational motion is ignored here as according to the animation, it is currently working on a 1D system.

A bomb calorimeter can be used to measure the overall gain in kinetic energy, both translational and kinetic by measuring the transfer of these energy from the products to the walls of the calorimeter in the form of heat. The heat transferred can then be measured in the form of change in temperature with a thermal sensor. <ref name="Physical Chemistry"/>

However, in order to differentiate the respective contributions of translational and vibrational energy, further experimental is required. To determine the vibrational energy, IR spectroscopy can be utilized. Before the reactants collide, most of the bonds occupy lowest energy vibrational levels. However, during a collision, higher level vibrational modes are excited as potential energy is convereted into vibrational kinetic energy. In IR spectroscopy, the higher level modes are characterized as overtones signals where energy levels move from initially 0 to 1, to 0 to 2 or 0 to 3, or even higher. These overtones can be detected and seen with higher wavenumbers as compared to the normal 0 to 1 transition. <ref name="Physical Chemistry"/>

[[File:Fig34_Yanda.png|center|300px|thumb|Figure 34. Contour Map of reaction from H2 + F to HF + H ]]
[[File:Fig35_Yanda.png|center|300px|thumb|Figure 35. Internuclear Distance vs Time Diagram of reaction from H2 + F to HF + H ]]
[[File:Fig36_Yanda.png|center|300px|thumb|Figure 36. Energy vs Time Diagram of reaction from H2 + F to HF + H ]]

==The Polanyi Rules==

The Polanyi rules state that for reactants with a given momentum along the direction of the reaction path, the translational energy, as opposed to the vibrational energy, is preferred in overcoming the early transition state in an exothermic reaction. The opposite will be true for endothermic reactions, where vibrational energy takes priority over translational energy to successfully form stable products.<ref name="J. Phys. Chem."/>

Reaction trajectories were calculated for the exothermic reaction of F + H2 with 1000 steps, but by varying the momentum of r1, the momentum of H-H bond, with all other conditions kept constant such that only the vibration energy of the H2 is varied. This is to investigate an early TS, since it is exothermic.

{| class="wikitable" border=1
! rH-F / pm !! rH-H / pm !! pHF/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1 !! Reactive Trajectory? || Contour Plot
|-
| 230 || 74 || -1.0 || -3.0 || No || [[File:Fig37_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || -2.0 || No || [[File:Fig38_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || -1.0 || No || [[File:Fig39_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 0 || No || [[File:Fig40_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 1.0 || No || [[File:Fig41_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 2.0 || No || [[File:Fig42_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 3.0 || No || [[File:Fig43_Yanda.png|200px]]
|}

However, keeping the same initial positions and everything else constant, except that the momentum pHF is increased to -1.5 while the momentum of pHH is reduced to 0 and is shown in the figure below. It resulted in a reactive trajectory, which suggested that the increase in translational energy of the system has greater effect in allowing the transition state to be overcome to form stable products.

[[File:Fig44_Yanda.png|300px|thumb|center|Figure 44. Reactive trajectory of F + H2 to HF + H reaction.]]

Next, the endothermic reaction of HF + H to F + H2, which is a late TS is investigated. Similarly, everything is kept constant, apart from pHF before keeping it constant and increasing pHH. The bond distance of HF is kept at 92 nm according to literature.

{| class="wikitable" border=1
! rH-F / pm !! rH-H / pm !! pHF/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1 !! Reactive Trajectory? || Contour Plot
|-
| 230 || 92 || 0 || -1.0 || No || [[File:Fig45_Yanda.png|200px]]
|-
| 230 || 92 || -5.0 || -1.0 || No || [[File:Fig46_Yanda.png|200px]]
|-
| 230 || 92 || -10.0 || -1.0 || No || [[File:Fig47_Yanda.png|200px]]
|-
| 230 || 92 || -15.0 || -1.0 || No || [[File:Fig48_Yanda.png|200px]]
|-
| 230 || 92 || -20.0 || -1.0 || No || [[File:Fig50_Yanda.png|200px]]
|-
| 230 || 92 || -25.0 || -1.0 || No || [[File:Fig51_Yanda.png|200px]]
|-
| 230 || 92 || -30.0 || -1.0 || Yes || [[File:Fig52_Yanda.png|200px]]
|}

Hence, it has shown that by increasing the vibrational energy of the H-F bond for the endothermic reaction while keep translational energy fixed at a low vaue resulted in the formation of products.

As such, Polanyi's empirical rules have been shown for both exothermic reactions (early TS) and endothermic reactions (late TS). However, it should be noted that such rules might not work for all cases due to imbalance distribution of translational and vibrational energies or errors in calculating energies, be it bond energies or trajectory calculations.

=References=

<references>
<ref name="CP3MD">Imperial College London, https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD, (accessed May 2018).</ref>
<ref name="Plot"> C.D., Sherrill, http://vergil.chemistry.gatech.edu/courses/chem6485/pdf/pes-lecture.pdf, (accessed May 2018). </ref>
<ref name="MRD">R. D. Levine, Molecular Reaction Dynamics, Cambridge University Press, Cambridge, 2005, pp. 202-203.</ref>
<ref name="MDACK">G. D. Billing and K. V. Mikkelsen, Introduction to Molecular Dynamics and Chemical Kinetics, John Wiley & Sons, New York, 1996, p. 31.</ref>
<ref name="Physical Chemistry">P. Atkins and J. de Paula, Atkins' Physical Chemistry, Oxford University Press, Oxford, 10th edn., 2014, pp. 65, 71 and 986.</ref>
<ref name="OC">J. Clayden, N. Greeves and S. Warren, Organic Chemistry, Oxford University Press, Oxford, 2nd edn., 2012, pp. 989.</ref>
<ref name="J. Phys. Chem.">C. Daniel et al., J. Phys. Chem., 1993, 97, 12485-12490.</ref>
</references>

MRD:yw14815

2020-07-07T08:50:54Z

Mak214: /* Potential Energy Surfaces */

=Introduction=

This Wiki page outlines the study of the molecular reaction dynamics of two simple triatomic systems: H + H2 and F + H2. The atoms are 180° to each other, with a 1D space symplification. The atoms are labelled A, B and C, with vectors r1 and r2 representing vectors BA and BC in pm respectively. Momentum are also measured in p1 and p2 in g mol-1 pm fs-1. Below is the systematic diagram of the triatomic system model, which will be referred to throughout the whole wiki. <ref name="CP3MD"/>

[[File:Fig1_Yanda.png|400px|thumb|center|Figure 1. A triatomic system model. Diagram from Imperial College London, https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD (accessed May 2020.)]]

=EXERCISE 1: H + H2 system=
== Potential Energy Surfaces ==

A potential energy surface is a visual representation that describes the potential energy of a system, such as a collective of atoms, in terms of parameters, which normally is the relative positions of the reactants and the products. A reaction path is shown to be the gradient line on a PES diagram, and represents both potential energy and internuclear distances between atoms in the system as a function of the reaction coordinate, thereby modelling their changes through the progress of the reaction.

The PES diagram for Figure 1 above was calculated for the reaction system H + H2 using the parameters given:
{| class="wikitable"
!Calculations
!Steps
!r1 / pm
!r2 / pm
!p1 / gmol-1pmfs-1
!p2 / gmol-1pmfs-1
|-
|Dynamics
|500
|74
|230
|0.0
|<nowiki>-5.2</nowiki>
|}

[[File:Fig2_Yanda.png|400px|thumb|center|Figure 2.Potential Energy Surface for the H + H2 reaction path.]]

The reaction path, which passes through each potential energy minimum, including the transition state, which is a first order saddle point at which the potential energy is at its maximum (highest energy minimum) and where its gradient is 0, hence ∂V('''r''')/∂'''r '''=0 and the second differential at the TS is less than 0.Furthermore, the determinant of the Hessian Matrix is smaller than 0, with the eigenvalues of the Hessian Matrix being one positive and one negative

Local minimum, on the other hand,while also have the first derivative ∂V('''r''')/∂'''r '''=0, its second differential at the minimum is greater than 0, with the eigenvalues of the Hessian Matrix both being positive and the determinant of the Hessian Matrix greater than 0.

Literally speaking, the main difference between transition state and a local minimum is that the transition state is in fact a local energy maximum. Initially, with 0 initial momentum, both transition state and local minimum will remain there forever. However, with a very small momentum added towards the reactants direction, the transition state will be passed and will roll over towards the reactants, and vice versa for products. Testing the trajectories near the transition state is a method used to identify the presence of the transition state. On the other hand, a slightly perturbation from the local minimum will only result in rolling back towards the local minimum, because only a sufficient energy is needed to overcome the energy barrier to escape the "well" of the local minimum. In the PES diagram, a local minimum represents an intermediate, which can be represented in a 2D diagram below.<ref name="Plot"/>

[[File:Fig3_Yanda.PNG|400px|thumb|center|Figure 3. Potential Energy Surface Diagram showing the Transition State. Diagram from http://vergil.chemistry.gatech.edu/courses/chem6485/pdf/pes-lecture.pdf (accessed May 2020).]]

Another interesting observation is that the gradient along the reaction pathway is least steep close to both sides of the transition state, suggesting the point at which bonds are in the process of breaking and forming, hence lower degree of association and less vibrational oscillations. {{fontcolor1|red Great. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 09:50, 7 July 2020 (BST)}}

== Locating the Transition State ==

The best estimate of the transition state position (rts) is 90.77 pm. The transition state should be symmetric as it represents the state of the reaction where old bonds and new bonds are in the middle of being broken and formed. As such, given the initial conditions of both momentum p1=0 and p2=0, the two internuclear distances r1 and r2 must be equal at the transitions state. The constant internuclear distances can be understood and found with a "Internuclear Distances vs Time plot" shown in Figure 4 below, where the straight horizontal line suggests constant potential energy with the absence of any vibrational oscillations.

[[File:Fig4_Yanda.png|400px|thumb|center|Figure 4. Internuclear Distance vs Time plot for finding transition state position.]]

== Calculation of Reaction Path ==

{| class="wikitable" border="1"
! Calculations !! Steps !! r1 / pm !! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| MEP || 50000 || 91.77 || 90.77 || 0 || 0
|}

The Potential Energy Surface for MEP Calculation is shown in Figure 5 below.

[[File:Fig5_Yanda.png|400px|thumb|center|Figure 5.Potential Energy Surface Diagram with MEP.]]

{| class="wikitable" border="1"
! Calculations !! Steps !! r1 / pm !! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| Dynamics || 500 || 91.77 || 90.77 || 0 || 0
|}

Similarly, the Potential Energy Surface for Dynamics Calculation is shown in Figure 6 below.

[[File:Fig6_Yanda.png|400px|thumb|center|Figure 6.Potential Energy Surface Diagram with Dynamics.]]

Comparing the Dynamics (Figure 6) and the MEP (Figure 5), it can be seen that the MEP does not take into account of vibrational oscillations within the atoms, but just passing through all the potential minimum. On the other hand, the trajectory just calculated takes into account of the momentum changes during the reaction by oscillating about the minimum potential energy. In addition, during the calculations, 50000 steps were required to complete the minimum energy path in MEP, while 500 steps were required in Dynamics.

The "Internuclear Distances vs Time" and "Momenta vs Time" graphs for Dynamics calculations are respectively plotted in Figure 7 and 8 below.

[[File:Fig7_Yanda.png|400px|thumb|center|Figure 7. Internuclear Distances vs Time using Dynamics.]]

[[File:Fig8_Yanda.png|400px|thumb|center|Figure 8. Momenta vs Time using Dynamics.]]

Figure 7 and 8 can help to describe the reaction. Distance r1 (B-C) and r2 (A-C) are seen seen to increase as A moves further away from B-C after the transition state. After an extended period of time, it can be seen that A-C and B-C distances increases linearly, suggesting that the distance A-B is constant and that B-C is moving further away from A. Also,initially, A-B decreases slightly before plateauing/oscillating, suggesting that the bond between A-B has formed and that they are vibrating up and down from equilibrium. The vibration is further suggested by the oscillation of A-B in the Momenta vs. Time graph in Figure 8.

The "Internuclear Distances vs Time" and "Momenta vs Time" graphs for MEP calculations are respectively plotted in Figure 9 and 10 below.

[[File:Fig9_Yanda.png|400px|thumb|center|Figure 9. Internuclear Distances vs Time using MEP.]]

[[File:Fig10_Yanda.png|400px|thumb|center|Figure 10. Momenta vs Time using MEP.]]

With reference to Figure 9 and 10, the internuclear distance r2 (A-B) decreases rapidly before converging to a limit and plateauing, while r1 (B-C), hence A-C increases rapidly. The whole system has 0 momentum because the momenta/velocities are always reset to zero in each time step.

Also, the final values of the positions r1(t), r2(t), p1(t) and p2(t) for the trajectory for large enough t, at t=100fs are calculated in the table below.

{| class="wikitable" border="1"
! r1 / pm!! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| 735.414|| 75.4847|| 5.08082 || 2.02769
|}

If the initial conditions were r1 = rts and r2 = rts + 1 pm instead, the reaction path will move in the opposite direction, exiting from the opposite direction in the potential energy surface (Figure 11) whereby distance r2 (A-B) increases while the distance r1 (B-C) stays constant with only vibrations of the bond similar to the previous conditions (Figure 12 and 13). Hence, in the first case the transition state moves towards AB + C, while the reverse will be in this case favoring the formation of products A + BC, as A start off further to B than that of C to B.

[[File:Fig11_Yanda.png|400px|thumb|center|Figure 11. Potential Energy Surface with Dynamics with initial conditions reversed.]]
[[File:Fig12_Yanda.png|400px|thumb|center|Figure 12. Internuclear Distances vs Time with Dynamics with initial conditions reversed.]]
[[File:Fig13_Yanda.png|400px|thumb|center|Figure 13. Momenta vs Time with Dynamics with initial conditions reversed.]]

The distorted transition state, where the values of r1=rts and r2=rts + 1 pm would be inverted for as the previous calculations, with the graphs for internuclear distance and momenta also inverted, such that graph of B-C appears as the original A-B.

Finally,a calculation was setup where the initial positions correspond to the final positions of the trajectory calculative, above, the same final momenta values but with their signs reversed.

{| class="wikitable" border="1"
! r1 / pm!! r2 / pm !! p1 / gmol-1pmfs-1 !! p2 / gmol-1pmfs-1
|-
| 735.414|| 75.4847|| -5.08082 || -2.02769
|}

[[File:Fig14_Yanda.png|400px|thumb|center|Figure 14. Potential Energy Surface with Dynamics moving from products to estimated Transition State.]]
[[File:Fig15_Yanda.png|400px|thumb|center|Figure 15. Internuclear Distances vs Time with Dynamics moving from products to estimated Transition State.]]
[[File:Fig16_Yanda.png|400px|thumb|center|Figure 16. Momenta vs Time with Dynamics moving from products to estimated Transition State.]]

This situation is the reverse of the original situation in which now, the species start off as reactants with just the right starting positions and momenta to reach the slightly distorted transition state. Technically, the three graphs in this situation should be the mirror image of the three graphs in the original situation. However, a perfect symmetrical relationship is not observed due to random errors in estimating through rounding off or sampling errors of the final positions and momenta from the Internuclear Distances vs Time graph and Momenta vs Time graph respectively.For example, in Figure 15, the distance between B-C should be constant after the transition state, but was not the case due to a errors in the initial input data.

== Reactive and Unreactive Trajectories ==

For the initial positions r1 = 74 pm and r2 = 200 pm, various trajectories were ran with different momentum values and combinations as tabulated below.

{| class="wikitable" border="1"
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280|| Reactive || Reactive as reaction path can get over maximum at TS and form products. Straight trajectory centered around Dynamics suggest molecule B-C does not vibrate until A collides into it, with sufficient momentum that subsequently resulted in molecule A-B with vibrational oscillation after TS || [[File:Fig17_Yanda.png|300px]] [[File:Fig18_Yanda.png|300px]]
|-
| -3.1 || -4.1 || -420.077 || Unreactive || Not reactive as reactive path unable to get over maximum at TS and form products. The PES curve suggests that the trajectory is vibrational/oscillating throughout and that the reaction path does not move past the TS. It turns back before the TS once the potential energy exceeds the kinetic energy of the species of the momenta given || [[File:Fig19_Yanda.png|300px]] [[File:Fig20_Yanda.png|300px]]
|-
| -3.1 || -5.1|| -413.977 || Reactive|| Reactive as similar to the first scenario. Note that higher initial momentum of molecule B-C results in larger vibrations as visible in the PES curve || [[File:Fig21_Yanda.png|300px]] [[File:Fig22_Yanda.png|300px]]
|-
| -5.1|| -10.1|| -357.277 || Unreactive || Initial momenta large enough for reaction path to attain TS, but the activation barrier was crossed twice such that eventually the reactants were reformed again. At TS, A-B-C, the initial momentum was not large enough to form stable products. As such, reaction path recrosses the TS, as visible from the contour map, back into the reactions, reforming A + BC. Much larger vibrational oscillations that the previous scenarios due to large initial momentum as visible from the PES curve || [[File:Fig23_Yanda.png|300px]] [[File:Fig24_Yanda.png|300px]]
|-
| -5.1|| -10.6|| -349.477 || Reactive || Similar to the previous scenario, but that now the initial momenta are larger enough to allow three crossing of the activation barrier, hence passing through the TS thrice, eventually forming the products A + BC|| [[File:Fig25_Yanda.png|300px]] [[File:Fig26_Yanda.png|300px]]
|}

From the table, it can be seen that while a minimum initial momentum of the species are needed to cross the activation barrier and over the TS to form the products, a much initial momentum does not guarantee product formation as can be seen in the second last scenario. As such, the conditions required for enough momentum given the initial preconditions as listed are: -3.1 < p1/ g.mol-1.pm.fs-1 < -1.6 and p2 = -5.1 g.mol-1.pm.fs-1.

== Transition State Theory ==

The deviations from the experimental values and the values describes in the Transition State Theory (TST) arises because of the assumptions within the theory itself. First of all, the TST adheres to the Born-Oppenheimer approximation, where the nuclear and electronic motions are decoupled. The Transition State Theory assumes that only the reactants are in constant equilibrium with the transition state structure, and that the energy of the particles follow a Boltzmann distribution. In addition, once the reactants have trajectories with a kinetic energy along the reaction coordinate greater than the activation energy, it will be reactive, and that the products will definitely form, ignoring the possibility that the transition state structure will collapse back to the reactants. However, experimentally, it has been seen in the few scenarios above where the initial momentum (p1=-5.1, p2=-10.1), hence kinetic energy is high enough to result in multiple crossing of the activation energy barrier through multiple forward and backward directions. Theoretically, the theory suggests only a single crossing of the activation energy barrier in the forward direction is possible. Also, note that in the transition state, the motion along the reaction coordinate can be treated classically. <ref name="MRD"/>

Finally, the theory fails for some reactions at elevated temperatures due to more complex molecule motions or low temperatures due to quantum tunneling, whereby the transition state need not be reached to form products.<ref name="MDACK"/>

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

The energetics of both F + H2 and the reverse H + HF were investigated and the findings are summarized in the table below. The method to extract these data are explained further in this section.

{| class="wikitable" border="1"
! Reaction !! Transition State Energy / kJ/mol !! Reactants Energy / kJ/mol !! Activation Energy / kJ/mol !! Type of Reaction
|-
| F + H2 || -433.980 || -435.100|| +1.120 || Exothermic
|-
| H + HF || -433.980 || -560.700 || +126.720 || Endothermic
|}

The reaction between F and H2 is an exothermic reaction owing to the stronger H-F bond relative to the H-H bond. Since enthalpy of reaction is calculated from energy of bond breaking - energy of bond making, the stronger negative value arising from breaking the stronger H-F bond resulted in an exothermic reaction. The stronger H-F bond can be explained due to the larger electronegativity difference between H and F, for which F= 4.0 and H = 2.1. <ref name="Physical Chemistry"/> As a result of the electronegativity difference, the F atom pulls electron density closer to itself, resulting in a highly polarized H-F bond, adding on an extra contribution to the bond strength. H-H bond, on the other hand, is a simple covalent bond. Literature value suggests that the bond dissociation energies of H-H and H-F are 432 kJ/mol and 565 kJ/mol respectively, further suggesting that the enthalpy released from breaking H-F bond is much higher than that of H-H bonds. Therefore, the reaction between F and H2 to form HF and H is energetically favorable, with an overall exothermic reaction.

The reverse reaction, H + HF is therefore endothermic, thereby requiring external energy in order to break the H-F bond, whereby the formation of the H-H bond does not compensate for.

The Hammond's postulate states that the transition state of a reaction resembles either the reactants or the products depending on which it is closer in energy. Hence, in an exothermic reaction, the transition state is closer to the reactants in energy. In an endothermic reaction, the transition state is closer to the products in energy.<ref name="OC"/>

Referring to the table above, for the forward reaction between F and H2, since the reactants energy is much closer to the transition state energy as compared to the reactions energy and the transition state energy in H + HF backward reaction, the forward reaction is the exothermic reaction, and the backward reaction is the endothermic reaction. This can also be determined graphically. With reference to the PES diagram and Energy vs Time plot below for the reaction between F and H2, it shows that the transition state has been attained, with no net change in potential energy. Furthermore, the PES suggests that the reactants are higher in energy than the products, indicating an exothermic reaction.

[[File:Fig27_Yanda.png|center|300px|thumb|Figure 27. Energy vs. Time Plot for F + H2. ]]
[[File:Fig28_Yanda.png|center|300px|thumb|Figure 28. Potential Energy Surface Diagram for F + H2. ]]

Similarly, Figure 29 below shows the transition state from the reverse reaction H + HF. Comparatively, it can be seen that the transition state closely resembles the products, which are now higher in energy than the reactants, thereby deeming it an endothermic reaction.

[[File:Fig29_Yanda.png|center|300px|thumb|Figure 29. Potential Energy Surface Diagram for H + HF. ]]

The approximate position of the transition state, is found to be r1 = BC = 181.1 pm and r2 = AB = 74.5 pm .The potential energy at the TS is then found to be -433.980 kJ/mol.

The activation energy of the reaction was found by slightly perturbing it from the TS and finding the energy difference of the products to the transition state energy as shown in the tables and Figures 30-33 below.

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| MEP || 2000 || 0.1 || 181.1 || 74.5 || 0 || 0
|}

[[File:Fig30_Yanda.png|center|300px|thumb|Figure 30. Contour Map for the reaction from H2 + F to HF + H. ]]
[[File:Fig31_Yanda.png|center|300px|thumb|Figure 31. Energy vs Time diagram for the reaction from H2 + F to HF + H. ]]

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| MEP || 2000 || 0.1 || 182.1 || 74.5 || 0 || 0
|}

[[File:Fig32_Yanda.png|center|300px|thumb|Figure 32. Contour Map for the reaction from HF + H to H2 + F. ]] [[File:Fig33_Yanda.png|center|300px|thumb|Figure 33. Energy vs Time diagram for the reaction from HF + H to H2 + F. ]]

By finding the trajectory down the MEP with the initial conditions, then inserting the last geometry as the initial condition. After which, replace one of the bond distance of the reactants to an extremely huge number, while keeping the bond distance of the product the same, the potential energy of the reactants and the products can be found. Though, it must be noted that this is just an approximation, as the exact potential energy is when the potential energy curve stabilizes. Ultimately, the potential energy of the reactants are found, can be used to find the activation energies, as summarized in the first table above.

==Reaction Dynamics==

The following reaction condition resulted in the reaction from H2 + F -> HF + H

{| class="wikitable" border="1"
! Calculation !! Steps !! Time/fs !! rFH / pm !! rHH / pm !! pFH/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1
|-
| Dynamics || 1400 || 0.5 || 220 || 71 || -1.5 || -0.3
|}

The conservation of energy suggests that total energy is constant through the course of reaction. Hence, in the exothermic reaction between H2 and F to form HF + H, the loss in potential energy from the breaking of bonds is replaced by the increase in kinetic energy as seen in the Energy Vs Time graph below. The kinetic energy can be of three forms: translational, rotational or vibrational. Rotational motion is ignored here as according to the animation, it is currently working on a 1D system.

A bomb calorimeter can be used to measure the overall gain in kinetic energy, both translational and kinetic by measuring the transfer of these energy from the products to the walls of the calorimeter in the form of heat. The heat transferred can then be measured in the form of change in temperature with a thermal sensor. <ref name="Physical Chemistry"/>

However, in order to differentiate the respective contributions of translational and vibrational energy, further experimental is required. To determine the vibrational energy, IR spectroscopy can be utilized. Before the reactants collide, most of the bonds occupy lowest energy vibrational levels. However, during a collision, higher level vibrational modes are excited as potential energy is convereted into vibrational kinetic energy. In IR spectroscopy, the higher level modes are characterized as overtones signals where energy levels move from initially 0 to 1, to 0 to 2 or 0 to 3, or even higher. These overtones can be detected and seen with higher wavenumbers as compared to the normal 0 to 1 transition. <ref name="Physical Chemistry"/>

[[File:Fig34_Yanda.png|center|300px|thumb|Figure 34. Contour Map of reaction from H2 + F to HF + H ]]
[[File:Fig35_Yanda.png|center|300px|thumb|Figure 35. Internuclear Distance vs Time Diagram of reaction from H2 + F to HF + H ]]
[[File:Fig36_Yanda.png|center|300px|thumb|Figure 36. Energy vs Time Diagram of reaction from H2 + F to HF + H ]]

==The Polanyi Rules==

The Polanyi rules state that for reactants with a given momentum along the direction of the reaction path, the translational energy, as opposed to the vibrational energy, is preferred in overcoming the early transition state in an exothermic reaction. The opposite will be true for endothermic reactions, where vibrational energy takes priority over translational energy to successfully form stable products.<ref name="J. Phys. Chem."/>

Reaction trajectories were calculated for the exothermic reaction of F + H2 with 1000 steps, but by varying the momentum of r1, the momentum of H-H bond, with all other conditions kept constant such that only the vibration energy of the H2 is varied. This is to investigate an early TS, since it is exothermic.

{| class="wikitable" border=1
! rH-F / pm !! rH-H / pm !! pHF/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1 !! Reactive Trajectory? || Contour Plot
|-
| 230 || 74 || -1.0 || -3.0 || No || [[File:Fig37_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || -2.0 || No || [[File:Fig38_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || -1.0 || No || [[File:Fig39_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 0 || No || [[File:Fig40_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 1.0 || No || [[File:Fig41_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 2.0 || No || [[File:Fig42_Yanda.png|200px]]
|-
| 230 || 74 || -1.0 || 3.0 || No || [[File:Fig43_Yanda.png|200px]]
|}

However, keeping the same initial positions and everything else constant, except that the momentum pHF is increased to -1.5 while the momentum of pHH is reduced to 0 and is shown in the figure below. It resulted in a reactive trajectory, which suggested that the increase in translational energy of the system has greater effect in allowing the transition state to be overcome to form stable products.

[[File:Fig44_Yanda.png|300px|thumb|center|Figure 44. Reactive trajectory of F + H2 to HF + H reaction.]]

Next, the endothermic reaction of HF + H to F + H2, which is a late TS is investigated. Similarly, everything is kept constant, apart from pHF before keeping it constant and increasing pHH. The bond distance of HF is kept at 92 nm according to literature.

{| class="wikitable" border=1
! rH-F / pm !! rH-H / pm !! pHF/gmol-1pmfs-1 !! pHH/gmol-1pmfs-1 !! Reactive Trajectory? || Contour Plot
|-
| 230 || 92 || 0 || -1.0 || No || [[File:Fig45_Yanda.png|200px]]
|-
| 230 || 92 || -5.0 || -1.0 || No || [[File:Fig46_Yanda.png|200px]]
|-
| 230 || 92 || -10.0 || -1.0 || No || [[File:Fig47_Yanda.png|200px]]
|-
| 230 || 92 || -15.0 || -1.0 || No || [[File:Fig48_Yanda.png|200px]]
|-
| 230 || 92 || -20.0 || -1.0 || No || [[File:Fig50_Yanda.png|200px]]
|-
| 230 || 92 || -25.0 || -1.0 || No || [[File:Fig51_Yanda.png|200px]]
|-
| 230 || 92 || -30.0 || -1.0 || Yes || [[File:Fig52_Yanda.png|200px]]
|}

Hence, it has shown that by increasing the vibrational energy of the H-F bond for the endothermic reaction while keep translational energy fixed at a low vaue resulted in the formation of products.

As such, Polanyi's empirical rules have been shown for both exothermic reactions (early TS) and endothermic reactions (late TS). However, it should be noted that such rules might not work for all cases due to imbalance distribution of translational and vibrational energies or errors in calculating energies, be it bond energies or trajectory calculations.

=References=

<references>
<ref name="CP3MD">Imperial College London, https://wiki.ch.ic.ac.uk/wiki/index.php?title=CP3MD, (accessed May 2018).</ref>
<ref name="Plot"> C.D., Sherrill, http://vergil.chemistry.gatech.edu/courses/chem6485/pdf/pes-lecture.pdf, (accessed May 2018). </ref>
<ref name="MRD">R. D. Levine, Molecular Reaction Dynamics, Cambridge University Press, Cambridge, 2005, pp. 202-203.</ref>
<ref name="MDACK">G. D. Billing and K. V. Mikkelsen, Introduction to Molecular Dynamics and Chemical Kinetics, John Wiley & Sons, New York, 1996, p. 31.</ref>
<ref name="Physical Chemistry">P. Atkins and J. de Paula, Atkins' Physical Chemistry, Oxford University Press, Oxford, 10th edn., 2014, pp. 65, 71 and 986.</ref>
<ref name="OC">J. Clayden, N. Greeves and S. Warren, Organic Chemistry, Oxford University Press, Oxford, 2nd edn., 2012, pp. 989.</ref>
<ref name="J. Phys. Chem.">C. Daniel et al., J. Phys. Chem., 1993, 97, 12485-12490.</ref>
</references>

MRD:mb7418

2020-05-29T20:40:04Z

Mak214: /* Reaction dynamics */

== EXERCISE 1: H + H2 system ==

===Dynamics from the transition state region===
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? {{fontcolor1|red| This text is repeated below which makes this section difficult to read. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:03, 29 May 2020 (BST)}} Potential energy surface diagrams show the potential energy the system has according to the coordinates of the atoms in the system.<ref>Chemistry LibreTexts. 2020. ''2.6: Potential Energy Surfaces''. [online] Available at: <<nowiki>https://chem.libretexts.org/Courses/University_of_California_Davis/UCD_Chem_107B%3A_Physical_Chemistry_for_Life_Scientists/Chapters/2%3A_Chemical_Kinetics/2.06%3A_Potential_Energy_Surfaces</nowiki>> [Accessed 19 May 2020].</ref> Figure 1 shows a potential energy surface diagram representing the reaction, equation 1: A + BC → AB + C.
{| class="wikitable"
|+Table 1. Molecules distances (r) and momenta (p)
!
!r / ppm
!p/ g.mol-1.pm.fs-1
|-
|r2=r(AB)
|230
|<nowiki>-5.2</nowiki>
|-
|r1=r(BC)
|74
|0
|}
In this case r(1)=r(BC) and r(2)=r(AB). BC is a hydrogen molecule consisting of two hydrogen atoms, B and C. Atom A is a hydrogen atom which has energy equal to or more than the activation energy, thus reaching the transition state. The transition state structure consists of atom A forming a bond with molecule BC while the BC bond is slightly dissociating. After the transition state, the bond between AB would form making a new molecule and the BC bond would be fully dissociated.
* Q1. 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?
[[File:PE with TS.png|left|thumb|[[Figure]] 1. Potential energy surface diagram with transition state labelled ]]
In a potential energy surface diagram, the transition state is mathematically defined as the first-order saddle point<ref>Chemistry LibreTexts. 2020. ''Hammond’S Postulate''. [online] Available at: <<nowiki>https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hammond%E2%80%99s_Postulate</nowiki>> [Accessed 22 May 2020].</ref>:

∂V(r1)/∂(r1)=0 and ∂V(r2)/∂r2=0

∂2V(r)/∂(r1)2 > 0 and ∂2V(r)/∂(r2)2 < 0

[∂2V(r)/∂(r1)2]*[ ∂2V(r)/∂(r2)2] - [ ∂2V/∂(r1)(r2)]2 < 0

Hessian matrix is shown in Table 2, resulting in DET = -0.999698 < 0
{| class="wikitable"
|+Table 1. Hessian matrix for the transition state
!
!f(x)
!f(y)
|-
|f(x)
|<nowiki>+0.707</nowiki>
|<nowiki>-0.707</nowiki>
|-
|f(y)
|<nowiki>-0.707</nowiki>
|<nowiki>-0.707</nowiki>
|}
In Figure 1, the black wavy lines shows that the molecule is vibrating. This is the minimum energy path required for the reactants to react and turn into the products via the transition state.<ref>Atkins, de Paula, Keeler: Physical Chemistry, 11th Edition. Section 18D.3 Potential Energy Surfaces (PESs)</ref> This suggests that the transition state is the peak in the minimum energy path. The transition state is the minimum energy required for the reactants to convert into the products. Hence once the transition state is reached, if the reaction was successful the products would be formed. Figure 1 shows the location of the transition state in the potential energy diagram.

The transition state can be distinguished from a local minimum of the potential energy surface because it is on the diagonal axis. {{fontcolor1|red| Because it is also a maximum (minimum AND maximum in orthogonal directions.. i.e a saddle point) therefore distinguishable from a local minimum. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:03, 29 May 2020 (BST)}} It is the saddle point.This is when everything in the system is stationary when there is no momentum.

===Locating the transition state r1=r2===
* Q2. 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.

<nowiki> </nowiki>When the transition state is formed, due to all the atoms being hydrogen, this suggest that the transition state is symmetrical. This means that the bond distance between A and B is the same as the bond distance of B and C. This can be used to locate the transition state when the momentum is 0. Table 3 shows an estimate of the transition state position (rts).
{| class="wikitable"
|+Table 3. Distances and momenta to locate the transition state
!
!r / ppm
!p/ g.mol-1.pm.fs-1
|-
|r2=r(AB)
|90.75
|0
|-
|r1=r(BC)
|90.75
|0
|}
[[File:Mb7418 dis vs time ts.png|thumb|[[Figure 2]]. Distance vs time plot of the transition state ]]
Figure 2 shows an Intermolecular distance vs time graph resulted from the initial conditions from Table 2. The line is flat, this suggest that everything in the system is stationary. Whereas, if it wasn't the transition state, in the intermolecular distance vs time graph you'd see oscillations representing the vibrational movements. The more oscillations would indicate the more vibrational movements. Hence the transition state is a flat line as there isn't any. {{fontcolor1|red| OK. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:33, 29 May 2020 (BST)}}
[[File:Mb7418 ts contour.png|thumb|[[Figure 3]]. Potential energy diagram as a contour plot of the transition state]]
The contour plot shown in Figure 3 shows the transition state as a red cross. This is the saddle point of the potential surface. Whenever r1 and r2 are the same, the point would be anywhere across the diagonal axis. This contour plot helped estimate the transition state position as the trajectory shows that the AB and BC bond lengths are the same. This is the point at which reactants have to overcome to react into the products.

The minimum energy path can differ depending on the relative momentum of the atoms/molecules and their vibrational excitation energy. The transition state of the reaction won't change its position. The reactants require sufficient translational kinetic energy and vibrational energy to overcome the transition state (saddle point) in order to react and turn into the products.

===Trajectories from r1 = rts+δ, r2 = rts ===
* Q3. Comment on how the ''mep'' and the trajectory you just calculated differ. '''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''' and the momenta '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1
The trajectory of the minimum energy path can be visualised slightly near the transition state, this is when atom A is approaching molecule BC to form a bond with it, A + BC.
{| class="wikitable"
|+Table 4. Initial conditions slightly displaced from transition state
!
!r /ppm
!p/ g.mol-1.pm.fs-1
|-
|r2=r(AB)
|90.75
|0
|-
|r1=r(BC)
|91.75
|0
|}
The initial conditions shown in Table 4 gives the potential energy diagram in Figure 4 and 5 when the calculation type is MEP. The minimum energy path shown by the black line isn't wavy. [[File:Mb7418 mep.png|left|thumb|[[Figure 4]]. MEP calculation diagram]]
[[File:Mb7418 mep 1.png|thumb|234x234px|[[Figure 5]]. MEP calculation diagram bottom view]]However, when the same initial conditions are applied when the calculation type is Dynamic, the minimum energy path is wavy shown in Figure 6. This shows that the MEP calculation gives a different result of the reaction trajectory. The molecule is constantly moving at a velocity since there is no external force acting upon the reactants. This is called inertia motion.<ref>Evans, M.G. and Polanyi, M., 1938. Inertia and driving force of chemical reactions. ''Transactions of the Faraday Society'', ''34'', pp.11-24.</ref> Whereas the potential energy diagram produced from the Dynamics calculation takes into account the reactants inertia by representing the pathway as non-linear. Additionally, the MEP calculation diagram has a smaller trajectory than the Dynamics calculation.The reason for this is beacause the MEP calculation sets the momentum to zero at every step whereas the dynamic calculation the momenta is developed classically<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., 8.3.1, Prentice-Hall, 1998.</ref>. The MEP calculation only takes the gradient into account. The amount of steps there are corresponds to the number of times it is calculating what the gradient is at that point. The only thing relative to the MEP calculation is the gradient. Whereas with dynamics, it takes into account how long it takes for the gradient to change. The momentum (velocity) is changing each time instead of setting it to zero every step. Hence it's longer as after the products are formed, they are still have motion such as vibrational and kinetic. {{fontcolor1|red| Yes. This does show an understanding of MEP vs Dynamics. Just to clarify - the amount of steps is the number of times the system moves during the simulation - this was a bit confused in your response. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:36, 29 May 2020 (BST)}} [[File:Mb7418 dyn 1.png|thumb|[[Figure 6]]. Dynamics calculation bottom view]]

== Reactive and unreactive trajectories ==

===Interpreting trajectories===

* Q4. For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, run trajectories with the following momenta combination and complete the table.

{| class="wikitable"
|+Table 5. Reactive and unreactive trajectories
!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>
|Reactive
|The starting point (represented by the red cross) starts at 200 pm between atoms A and B. Then the distance AB decreases. Atoms A and B are approaching each other and once it reaches the transition state, the AB distance and BC distance are the same and all the atoms are stationary. Then the AB bond is formed and atom C repels from atom B. The molecule AB is constantly vibrating. This could suggest that some translational kinetic energy got converted into vibrational energy after the transition state. Hence after the trajectory has oscillations after it passes the transition state.
|[[File:Mb7418 6.png|centre|thumb|200x200px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The trajectory has some oscillations, showing that the reactants are constantly vibrating. Hence it has vibrational energy. However, it doesn't go over the transition state so it doesn't react into the products. This suggests that the reactants approach each other but then they repel. While they do that they oscillate as they get closer. It ends up with a high AB distance (essentially the starting point of the reaction), this shows that the reactants haven't reacted due to the AB molecule not forming. This could be due to not having enough translational kinetic energy.
|[[File:Mb7418 2.png|centre|thumb|200x200px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The starting point starts at 200 pm between atoms A and B. Then the the atoms got closer and reaches the transition state. This is when the distance AB and BC are equal. Then the AB distance remains continuous while the BC distance begins to increase. The BC bond breaks and the AB bond forms. It only crosses the transition state once. It ends up in the products as the distance AB is short (molecule containing atoms A and B bonding) and BC is distance is long (reactant molecule BC bond has been broken).
|[[File:Mb7418 3.png|centre|thumb|200x200px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Unreactive
|The whole trajectory shows that it's unreactive because the starting point and ending point are similar. It started at a high AB distance value and also ended at a high AB distance value. This suggests this is end unreactive.
At the starting point, AB are far apart. Then they begin approaching each other and reach the transition state. However, they repel from each other and then start approaching each other again. This happens again and then the distance AB begins to increase again. The BC bond is formed again, resulting in the reactant molecule BC at the end.
|[[File:Mb7418 4.png|centre|thumb|200x200px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Reactive
|This crosses the transition point more than once. It's reactive as it's end point is at a high BC distance and low AB distance, suggesting that the BC bond broke and the AB bond has formed resulting in AB + B.

From the starting point, the AB distance starts to decrease. This suggests atom A and B approach each other, but then the AB distance begins to increase again so the A and B atoms repel. There is a moment when the distance between atoms A and B is the same as the distance between B and C (transition state) Then the B and C atoms approach each other and then repel B continuously. While atoms A and B get closer forming a bond. The molecule AB vibrates while atom C continuously repels away from it.
|[[File:Mb7418 5.png|centre|thumb|200x200px]]
|}

{{fontcolor1|red| OK. What do you conclude from the table? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:40, 29 May 2020 (BST)}}

=== Transition state theory ===
* Q5. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?
The three assumptions of the transition state theory<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Chapter 10 pg. 20, Prentice-Hall, 1998</ref> are:

1) The transition state will only be crossed over once. Once the transition state it reached, the products will be formed.

2) During the transition state, any motion along the reaction coordinate can be separate from the other motions and treated classically as a translation. The concentration of the transition state is in equilibrium with the reactants. (concentration constant relating them both, there's a Maxwell-Boltzmann distribution between the reactants and the transition state at any given time) This is defined as the Quasi equilibrium. It's assumed that the kinetic energy is predicted by the Boltzmann distribution. The transition states that are turning into products are distributed among their states according to the Maxwell-Boltzmann Laws<ref>Rowlinson*, J.S., 2005. The Maxwell–Boltzmann distribution. ''Molecular Physics'', ''103''(21-23), pp.2821-2828.</ref> even in absence of an equilibrium between the reactant and product molecules.

'''3)''' There's minimum effects of quantum tunneling.

The quantum tunneling effect<ref>Abyss.uoregon.edu. 2020. ''Quantum Tunneling''. [online] Available at: <<nowiki>http://abyss.uoregon.edu/~js/glossary/quantum_tunneling.html</nowiki>> [Accessed 22 May 2020].</ref> is when the reactants can overcome the barrier by moving across it to turn into the products. The transition state theory doesn't include quantum tunneling. If quantum tunneling was included, it would increase the rate of reaction as more reactants will turn into products. There are only classical motions in the transition state theory. Hence it is suggested that quantum tunneling isn't included in the model used in Table 5 because it's assumed there's a faster alternative route used.

Additionally, in the Table 5 model the transition state theory wasn't true because the transition state was crossed over more than once in the last two reactions in Table 5, this could decrease the rate of reaction due to the products being able to cross the transition state again and go back into the reactants. Therefore, the transition state theory is an over-estimate in comparison to the model used. {{fontcolor1|red| Yes. Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:41, 29 May 2020 (BST)}}

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

=== PES inspection ===
* Q6. 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?
{| class="wikitable"
|+Table 6. Initial conditions for Figure 7 (F + H2 → HF + H)
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|500
|0
|-
|r(BC)
|70
|0
|}

[[File:Mb7418 F + H2 exo edit CORRECT.png|left|thumb|[[Figure 7]]. F + H2 -> HF + H exothermic potential surface diagram]]
In the following equations, A = atom F, B = atom H, C = atom H.

The initial conditions in Table 6 gives the potential energy diagram for Figure 7. The reason for this is because the F and H bond is very far apart, suggesting that it's not bonded whereas the hydrogen atoms are closer together, suggesting a F atom and H2 molecule are in the initial conditions.

Equation 2: F + H2 → HF + H is an exothermic reaction<ref name=":0">Duff, J.W. and Truhlar, D.G., 1975. Effect of curvature of the reaction path on dynamic effects in endothermic chemical reactions and product energies in exothermic reactions. ''The Journal of Chemical Physics'', ''62''(6), pp.2477-2491.</ref> shown in Figure 7. The energy of the reactants is higher than the energy of the products. This shows that this is an exothermic reaction due to the process releasing energy. This suggests that the bond strength of H2 is weaker than the bond strength of HF as no additional energy was required to break its bond. This reaction has an early transition state<ref>Chemistry LibreTexts. 2020. ''Hammond’S Postulate''. [online] Available at: <<nowiki>https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hammond%E2%80%99s_Postulate</nowiki>> [Accessed 22 May 2020].</ref> due to the the fluorine atom being highly electro-negative so it attacks the hydrogen molecule very quickly.
{| class="wikitable"
|+Table 7. Initial conditions for Figure 8 (HF + H → F + H2)
!
!r / ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|70
|0
|-
|r(BC)
|500
|0
|}
[[File:Mb7418 NEW CORRECT ENDO.png|left|thumb|[[Figure 8]]. HF + H -> F + H2 endothermic potential surface diagram]]
The initial conditions in Table 7 gives the potential energy diagram for Figure 8. The reason for this is because the F and H atoms are very close together compared to both the hydrogens. This suggests that a HF molecule and one H atom are in the initial conditions.

Equation 3: HF + H → F + H2 is an endothermic reaction<ref name=":0" /> shown in Figure 8. The energy of the reactants is lower than the energy of the products. This suggests that this reaction is endothermic as the process used additional energy to break the H-F bond to form the H-H bond. Therefore, in the potential energy surface diagram there will be a lot of vibrational energy before the transition state compared to after. This also suggest that the HF bond is stronger than the HH bond as it requires additional energy to break its bond. This will give a late transition state.The translational energy could get converted into vibrational energy in order to break the bond between the strong H-F bond<ref name=":1">Polanyi, J.C. and Tardy, D., 1969. Energy Distribution in the Exothermic Reaction F+ H2 and the Endothermic Reaction HF+ H. ''The Journal of Chemical Physics'', ''51''(12), pp.5717-5719.</ref>. This favours the endothermic reaction because the reactants position will be aligned with the direction of the minimum energy path.''''' '''''

* Q7. Locate the approximate position of the transition state.
{| class="wikitable"
|+Table 8. Transition state position
!
!r /ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|180
|0
|-
|r(BC)
|74.5
|0
|}
The transition state position shown in Table 8 was estimated by looking at the initial conditions in Table 6 for the exothermic F + H2 → HF + H reaction. Due to exothermic reactions having early transition states, this suggested that the bond distance BC (the hydrogen atoms) would be small as it's the reactant. The initial bond distance of AB was reduced until the hessian matrix, shown in Table 9, gave a value of Det= -1.0 <0 which proved it was a saddle point. Additionally, there was no force. This also suggested it was the transition state.
{| class="wikitable"
|+Table 9. Hessian matrix for Transition state
|AB direction
|<nowiki>+1.000</nowiki>
|<nowiki>-0.024</nowiki>
|-
|BC direction
|<nowiki>-0.024</nowiki>
|<nowiki>-1.000</nowiki>
|}
[[File:Mb7418 dis vs time ts ex2.png|thumb|[[Figure 9]]. Intermolecular distance vs time graph for transition state for equations two and three]]
Additionally, Figure 9 shows the intermolecular distance vs time graph. This is a flat line, this could be due to everything in the system being stationary. There are no oscillations showing that there are any vibrational motions. This suggest that it's at the initial conditions in Table 8 represent the transition state position.

* Q8. Report the activation energy for both reactions.
Equation 2: F + H2 → HF + H the activation energy value is 1.0 kJ/mol.

Equation 3: HF + H → F + H2 the activation energy value is 126.7 kJ/mol.

The activation energy for equation 2 was calculated by getting the potential energy for transition state by using the initial conditions in Table 8. Then this was subtracted by the potential energy obtained by using the initial conditions in Table 10. Activation energy = TS(potential energy) - Reactants(potential energy) = (-433.981)-(435.056) = 1.075 kJ/mol
{| class="wikitable"
|+Table 10. Initial conditions for equation 2
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|500
|0
|-
|r(BC)
|74.5
|0
|}
The activation energy for equation 3 was calculated by getting the potential energy for for transition state by using the initial conditions in Table 8. Then this was subtracted by the potential energy obtained by using the initial conditions in Table 11. r(AB) which is the HF bond distance = 92 ppm<ref>Cccbdb.nist.gov. 2020. ''CCCBDB Listing Of Experimental Data Page 2''. [online] Available at: <<nowiki>https://cccbdb.nist.gov/exp2x.asp?casno=7664393&charge=0</nowiki>> [Accessed 22 May 2020].</ref>'''. '''The r(BC) bond which is the HH bond distance was set very high because the hydrogen atoms aren't bonded in the reactants.

Activation energy = TS(potential energy) - Reactants(potential energy) = (-433.981)-(560.698) = 126.7 kJ/mol
{| class="wikitable"
|+Table 11. Initial conditions for equation 3
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|92
|0
|-
|r(BC)
|500
|0
|}

{{fontcolor1|red| Good work. well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:28, 29 May 2020 (BST)}}

=== Reaction dynamics ===
* Q9. 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.
{| class="wikitable"
|+Table 12. Initial conditions for reactive trajectory
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|175
|<nowiki>-1.6</nowiki>
|-
|r(BC)
|74
|0.3
|}
[[File:Mb7418 THIS reactive.png|left|thumb|[[Figure 10]]. Reactive trajectory]]
The step number was increased to 4000 and the initial conditions in Table 12 gives the potential energy surface diagram in Figure 10. This shows a reactive trajectory for the exothermic reaction F + H2 → HF + H as it passes the transition state as well as having many oscillations. The end point is also far away from the starting point.

The starting point (reactants) has a high energy level compared to the products due to being exothermic<ref name=":0" />. Hence the potential energy converts into translational kinetic energy. Therefore, there are many oscillations in Figure 10 due to energy being conserved.

This can be explained by the difference in distribution of the energy levels between an excited and relaxed state. The excited state will have an occupied ground state (v=0) as well as an occupied first state (v=1) where v=vibrational number. Whereas in the relaxed state only the ground state will be occupied. This approach uses the harmonic oscillator<ref>Cccbdb.nist.gov. 2020. ''CCCBDB Listing Of N2 Experimental Data Page 2''. [online] Available at: <<nowiki>https://cccbdb.nist.gov/exp2x.asp?casno=7727379&charge=0</nowiki>> [Accessed 18 May 2020].</ref>. When energy is released, in the excited state there will be two transitions, v(0)→v(1) and v(1)→v(2), the fundamental peak and overtone. In the ground state it will only have transitions from v(0)→v(1).

Therefore, in an IR spectrum<ref>Chemistry LibreTexts. 2020. ''5.5: The Harmonic Oscillator And Infrared Spectra''. [online] Available at: <<nowiki>https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.05%3A_The_Harmonic_Oscillator_and_Infrared_Spectra</nowiki>> [Accessed 22 May 2020].</ref>, for the excited state there will be two peaks corresponding to the overtone and fundamental peak. The overtone peak intensity will be smaller than the fundamental's peak intensity. Hence there are many oscillations due to the overtone decreasing and then increasing in intensity while the fundamental increases and then decreases in intensity.

Therefore, this could be confirmed experimentally by Infrared spectroscopy (IR).
* Q10. 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.
[[File:Mb7418 last Q.png|left|thumb|[[Figure 11]]. Reactive potential energy surface contour plot]]
The position of the transition state affects whether the reaction is either exothermic or endothermic. An endothermic reaction will have a late transition state, suggesting that the vibrational energy efficiency will be higher<ref>Guo, H. and Liu, K., 2020. ''Control Of Chemical Reactivity By Transition-State And Beyond''.</ref>. The reason for this is because if it had a higher translational kinetic energy, the reactants wouldn't follow the minimum energy path and will just collide with each other but then immediately repel resulting in being back at the starting point. Hence, the translational kinetic energy converts into vibrational energy<ref name=":1" /> instead so the reactants collide and are aligned with the minimum energy pathway. The direction of the path taken is shown in Figure 11. The initial conditions in Table 12 was used to give the reactive trajectory. This suggests that endothermic reactions have higher vibrational energy to make their reactions more efficient so it's able to overcome the transition state. Hence there are a lot of oscillations before the transition state is reached because it requires more efficient vibrational energy in order to reach the transition state.

Whereas if the reaction has an early transition state, it will be an exothermic reaction. This suggest that there will be a higher translational kinetic energy compared to vibrational energy. This means that there will be less oscillations before it reaches the transition state as shown in Figure 11.

Overall, an efficient endothermic reaction will have a higher vibrational energy whereas an efficient exothermic reaction will have a higher translational kinetic energy. {{fontcolor1|red| Great. Please see a couple of comments of mine earlier on in the report. Otherwise I think this is a good job. Thank you especially for the extensive bibliography and for referencing throughout the text, good practice. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 21:40, 29 May 2020 (BST)}}

MRD:mb7418

2020-05-29T20:38:31Z

Mak214: /* Dynamics from the transition state region */

MRD:mb7418

2020-05-29T20:37:50Z

Mak214: /* EXERCISE 1: H + H2 system */

== EXERCISE 1: H + H2 system ==

===Dynamics from the transition state region===
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? {{fontcolor1|red| This text is repeated below which makes this section difficult to read. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:03, 29 May 2020 (BST)}} Potential energy surface diagrams show the potential energy the system has according to the coordinates of the atoms in the system.<ref>Chemistry LibreTexts. 2020. ''2.6: Potential Energy Surfaces''. [online] Available at: <<nowiki>https://chem.libretexts.org/Courses/University_of_California_Davis/UCD_Chem_107B%3A_Physical_Chemistry_for_Life_Scientists/Chapters/2%3A_Chemical_Kinetics/2.06%3A_Potential_Energy_Surfaces</nowiki>> [Accessed 19 May 2020].</ref> Figure 1 shows a potential energy surface diagram representing the reaction, equation 1: A + BC → AB + C.
{| class="wikitable"
|+Table 1. Molecules distances (r) and momenta (p)
!
!r / ppm
!p/ g.mol-1.pm.fs-1
|-
|r2=r(AB)
|230
|<nowiki>-5.2</nowiki>
|-
|r1=r(BC)
|74
|0
|}
In this case r(1)=r(BC) and r(2)=r(AB). BC is a hydrogen molecule consisting of two hydrogen atoms, B and C. Atom A is a hydrogen atom which has energy equal to or more than the activation energy, thus reaching the transition state. The transition state structure consists of atom A forming a bond with molecule BC while the BC bond is slightly dissociating. After the transition state, the bond between AB would form making a new molecule and the BC bond would be fully dissociated.
* Q1. 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?
[[File:PE with TS.png|left|thumb|[[Figure]] 1. Potential energy surface diagram with transition state labelled ]]
In a potential energy surface diagram, the transition state is mathematically defined as the first-order saddle point<ref>Chemistry LibreTexts. 2020. ''Hammond’S Postulate''. [online] Available at: <<nowiki>https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hammond%E2%80%99s_Postulate</nowiki>> [Accessed 22 May 2020].</ref>:

∂V(r1)/∂(r1)=0 and ∂V(r2)/∂r2=0

∂2V(r)/∂(r1)2 > 0 and ∂2V(r)/∂(r2)2 < 0

[∂2V(r)/∂(r1)2]*[ ∂2V(r)/∂(r2)2] - [ ∂2V/∂(r1)(r2)]2 < 0

Hessian matrix is shown in Table 2, resulting in DET = -0.999698 < 0
{| class="wikitable"
|+Table 1. Hessian matrix for the transition state
!
!f(x)
!f(y)
|-
|f(x)
|<nowiki>+0.707</nowiki>
|<nowiki>-0.707</nowiki>
|-
|f(y)
|<nowiki>-0.707</nowiki>
|<nowiki>-0.707</nowiki>
|}
In Figure 1, the black wavy lines shows that the molecule is vibrating. This is the minimum energy path required for the reactants to react and turn into the products via the transition state.<ref>Atkins, de Paula, Keeler: Physical Chemistry, 11th Edition. Section 18D.3 Potential Energy Surfaces (PESs)</ref> This suggests that the transition state is the peak in the minimum energy path. The transition state is the minimum energy required for the reactants to convert into the products. Hence once the transition state is reached, if the reaction was successful the products would be formed. Figure 1 shows the location of the transition state in the potential energy diagram.

The transition state can be distinguished from a local minimum of the potential energy surface because it is on the diagonal axis. {{fontcolor1|red| Because it is also a maximum (minimum AND maximum = saddle point) therefore distinguishable from a local minimum. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:03, 29 May 2020 (BST)}} It is the saddle point.This is when everything in the system is stationary when there is no momentum.

===Locating the transition state r1=r2===
* Q2. 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.

<nowiki> </nowiki>When the transition state is formed, due to all the atoms being hydrogen, this suggest that the transition state is symmetrical. This means that the bond distance between A and B is the same as the bond distance of B and C. This can be used to locate the transition state when the momentum is 0. Table 3 shows an estimate of the transition state position (rts).
{| class="wikitable"
|+Table 3. Distances and momenta to locate the transition state
!
!r / ppm
!p/ g.mol-1.pm.fs-1
|-
|r2=r(AB)
|90.75
|0
|-
|r1=r(BC)
|90.75
|0
|}
[[File:Mb7418 dis vs time ts.png|thumb|[[Figure 2]]. Distance vs time plot of the transition state ]]
Figure 2 shows an Intermolecular distance vs time graph resulted from the initial conditions from Table 2. The line is flat, this suggest that everything in the system is stationary. Whereas, if it wasn't the transition state, in the intermolecular distance vs time graph you'd see oscillations representing the vibrational movements. The more oscillations would indicate the more vibrational movements. Hence the transition state is a flat line as there isn't any. {{fontcolor1|red| OK. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:33, 29 May 2020 (BST)}}
[[File:Mb7418 ts contour.png|thumb|[[Figure 3]]. Potential energy diagram as a contour plot of the transition state]]
The contour plot shown in Figure 3 shows the transition state as a red cross. This is the saddle point of the potential surface. Whenever r1 and r2 are the same, the point would be anywhere across the diagonal axis. This contour plot helped estimate the transition state position as the trajectory shows that the AB and BC bond lengths are the same. This is the point at which reactants have to overcome to react into the products.

The minimum energy path can differ depending on the relative momentum of the atoms/molecules and their vibrational excitation energy. The transition state of the reaction won't change its position. The reactants require sufficient translational kinetic energy and vibrational energy to overcome the transition state (saddle point) in order to react and turn into the products.

===Trajectories from r1 = rts+δ, r2 = rts ===
* Q3. Comment on how the ''mep'' and the trajectory you just calculated differ. '''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''' and the momenta '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1
The trajectory of the minimum energy path can be visualised slightly near the transition state, this is when atom A is approaching molecule BC to form a bond with it, A + BC.
{| class="wikitable"
|+Table 4. Initial conditions slightly displaced from transition state
!
!r /ppm
!p/ g.mol-1.pm.fs-1
|-
|r2=r(AB)
|90.75
|0
|-
|r1=r(BC)
|91.75
|0
|}
The initial conditions shown in Table 4 gives the potential energy diagram in Figure 4 and 5 when the calculation type is MEP. The minimum energy path shown by the black line isn't wavy. [[File:Mb7418 mep.png|left|thumb|[[Figure 4]]. MEP calculation diagram]]
[[File:Mb7418 mep 1.png|thumb|234x234px|[[Figure 5]]. MEP calculation diagram bottom view]]However, when the same initial conditions are applied when the calculation type is Dynamic, the minimum energy path is wavy shown in Figure 6. This shows that the MEP calculation gives a different result of the reaction trajectory. The molecule is constantly moving at a velocity since there is no external force acting upon the reactants. This is called inertia motion.<ref>Evans, M.G. and Polanyi, M., 1938. Inertia and driving force of chemical reactions. ''Transactions of the Faraday Society'', ''34'', pp.11-24.</ref> Whereas the potential energy diagram produced from the Dynamics calculation takes into account the reactants inertia by representing the pathway as non-linear. Additionally, the MEP calculation diagram has a smaller trajectory than the Dynamics calculation.The reason for this is beacause the MEP calculation sets the momentum to zero at every step whereas the dynamic calculation the momenta is developed classically<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., 8.3.1, Prentice-Hall, 1998.</ref>. The MEP calculation only takes the gradient into account. The amount of steps there are corresponds to the number of times it is calculating what the gradient is at that point. The only thing relative to the MEP calculation is the gradient. Whereas with dynamics, it takes into account how long it takes for the gradient to change. The momentum (velocity) is changing each time instead of setting it to zero every step. Hence it's longer as after the products are formed, they are still have motion such as vibrational and kinetic. {{fontcolor1|red| Yes. This does show an understanding of MEP vs Dynamics. Just to clarify - the amount of steps is the number of times the system moves during the simulation - this was a bit confused in your response. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:36, 29 May 2020 (BST)}} [[File:Mb7418 dyn 1.png|thumb|[[Figure 6]]. Dynamics calculation bottom view]]

== Reactive and unreactive trajectories ==

===Interpreting trajectories===

* Q4. For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, run trajectories with the following momenta combination and complete the table.

{| class="wikitable"
|+Table 5. Reactive and unreactive trajectories
!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>
|Reactive
|The starting point (represented by the red cross) starts at 200 pm between atoms A and B. Then the distance AB decreases. Atoms A and B are approaching each other and once it reaches the transition state, the AB distance and BC distance are the same and all the atoms are stationary. Then the AB bond is formed and atom C repels from atom B. The molecule AB is constantly vibrating. This could suggest that some translational kinetic energy got converted into vibrational energy after the transition state. Hence after the trajectory has oscillations after it passes the transition state.
|[[File:Mb7418 6.png|centre|thumb|200x200px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The trajectory has some oscillations, showing that the reactants are constantly vibrating. Hence it has vibrational energy. However, it doesn't go over the transition state so it doesn't react into the products. This suggests that the reactants approach each other but then they repel. While they do that they oscillate as they get closer. It ends up with a high AB distance (essentially the starting point of the reaction), this shows that the reactants haven't reacted due to the AB molecule not forming. This could be due to not having enough translational kinetic energy.
|[[File:Mb7418 2.png|centre|thumb|200x200px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The starting point starts at 200 pm between atoms A and B. Then the the atoms got closer and reaches the transition state. This is when the distance AB and BC are equal. Then the AB distance remains continuous while the BC distance begins to increase. The BC bond breaks and the AB bond forms. It only crosses the transition state once. It ends up in the products as the distance AB is short (molecule containing atoms A and B bonding) and BC is distance is long (reactant molecule BC bond has been broken).
|[[File:Mb7418 3.png|centre|thumb|200x200px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Unreactive
|The whole trajectory shows that it's unreactive because the starting point and ending point are similar. It started at a high AB distance value and also ended at a high AB distance value. This suggests this is end unreactive.
At the starting point, AB are far apart. Then they begin approaching each other and reach the transition state. However, they repel from each other and then start approaching each other again. This happens again and then the distance AB begins to increase again. The BC bond is formed again, resulting in the reactant molecule BC at the end.
|[[File:Mb7418 4.png|centre|thumb|200x200px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Reactive
|This crosses the transition point more than once. It's reactive as it's end point is at a high BC distance and low AB distance, suggesting that the BC bond broke and the AB bond has formed resulting in AB + B.

From the starting point, the AB distance starts to decrease. This suggests atom A and B approach each other, but then the AB distance begins to increase again so the A and B atoms repel. There is a moment when the distance between atoms A and B is the same as the distance between B and C (transition state) Then the B and C atoms approach each other and then repel B continuously. While atoms A and B get closer forming a bond. The molecule AB vibrates while atom C continuously repels away from it.
|[[File:Mb7418 5.png|centre|thumb|200x200px]]
|}

{{fontcolor1|red| OK. What do you conclude from the table? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:40, 29 May 2020 (BST)}}

=== Transition state theory ===
* Q5. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?
The three assumptions of the transition state theory<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Chapter 10 pg. 20, Prentice-Hall, 1998</ref> are:

1) The transition state will only be crossed over once. Once the transition state it reached, the products will be formed.

2) During the transition state, any motion along the reaction coordinate can be separate from the other motions and treated classically as a translation. The concentration of the transition state is in equilibrium with the reactants. (concentration constant relating them both, there's a Maxwell-Boltzmann distribution between the reactants and the transition state at any given time) This is defined as the Quasi equilibrium. It's assumed that the kinetic energy is predicted by the Boltzmann distribution. The transition states that are turning into products are distributed among their states according to the Maxwell-Boltzmann Laws<ref>Rowlinson*, J.S., 2005. The Maxwell–Boltzmann distribution. ''Molecular Physics'', ''103''(21-23), pp.2821-2828.</ref> even in absence of an equilibrium between the reactant and product molecules.

'''3)''' There's minimum effects of quantum tunneling.

The quantum tunneling effect<ref>Abyss.uoregon.edu. 2020. ''Quantum Tunneling''. [online] Available at: <<nowiki>http://abyss.uoregon.edu/~js/glossary/quantum_tunneling.html</nowiki>> [Accessed 22 May 2020].</ref> is when the reactants can overcome the barrier by moving across it to turn into the products. The transition state theory doesn't include quantum tunneling. If quantum tunneling was included, it would increase the rate of reaction as more reactants will turn into products. There are only classical motions in the transition state theory. Hence it is suggested that quantum tunneling isn't included in the model used in Table 5 because it's assumed there's a faster alternative route used.

Additionally, in the Table 5 model the transition state theory wasn't true because the transition state was crossed over more than once in the last two reactions in Table 5, this could decrease the rate of reaction due to the products being able to cross the transition state again and go back into the reactants. Therefore, the transition state theory is an over-estimate in comparison to the model used. {{fontcolor1|red| Yes. Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:41, 29 May 2020 (BST)}}

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

=== PES inspection ===
* Q6. 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?
{| class="wikitable"
|+Table 6. Initial conditions for Figure 7 (F + H2 → HF + H)
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|500
|0
|-
|r(BC)
|70
|0
|}

[[File:Mb7418 F + H2 exo edit CORRECT.png|left|thumb|[[Figure 7]]. F + H2 -> HF + H exothermic potential surface diagram]]
In the following equations, A = atom F, B = atom H, C = atom H.

The initial conditions in Table 6 gives the potential energy diagram for Figure 7. The reason for this is because the F and H bond is very far apart, suggesting that it's not bonded whereas the hydrogen atoms are closer together, suggesting a F atom and H2 molecule are in the initial conditions.

Equation 2: F + H2 → HF + H is an exothermic reaction<ref name=":0">Duff, J.W. and Truhlar, D.G., 1975. Effect of curvature of the reaction path on dynamic effects in endothermic chemical reactions and product energies in exothermic reactions. ''The Journal of Chemical Physics'', ''62''(6), pp.2477-2491.</ref> shown in Figure 7. The energy of the reactants is higher than the energy of the products. This shows that this is an exothermic reaction due to the process releasing energy. This suggests that the bond strength of H2 is weaker than the bond strength of HF as no additional energy was required to break its bond. This reaction has an early transition state<ref>Chemistry LibreTexts. 2020. ''Hammond’S Postulate''. [online] Available at: <<nowiki>https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hammond%E2%80%99s_Postulate</nowiki>> [Accessed 22 May 2020].</ref> due to the the fluorine atom being highly electro-negative so it attacks the hydrogen molecule very quickly.
{| class="wikitable"
|+Table 7. Initial conditions for Figure 8 (HF + H → F + H2)
!
!r / ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|70
|0
|-
|r(BC)
|500
|0
|}
[[File:Mb7418 NEW CORRECT ENDO.png|left|thumb|[[Figure 8]]. HF + H -> F + H2 endothermic potential surface diagram]]
The initial conditions in Table 7 gives the potential energy diagram for Figure 8. The reason for this is because the F and H atoms are very close together compared to both the hydrogens. This suggests that a HF molecule and one H atom are in the initial conditions.

Equation 3: HF + H → F + H2 is an endothermic reaction<ref name=":0" /> shown in Figure 8. The energy of the reactants is lower than the energy of the products. This suggests that this reaction is endothermic as the process used additional energy to break the H-F bond to form the H-H bond. Therefore, in the potential energy surface diagram there will be a lot of vibrational energy before the transition state compared to after. This also suggest that the HF bond is stronger than the HH bond as it requires additional energy to break its bond. This will give a late transition state.The translational energy could get converted into vibrational energy in order to break the bond between the strong H-F bond<ref name=":1">Polanyi, J.C. and Tardy, D., 1969. Energy Distribution in the Exothermic Reaction F+ H2 and the Endothermic Reaction HF+ H. ''The Journal of Chemical Physics'', ''51''(12), pp.5717-5719.</ref>. This favours the endothermic reaction because the reactants position will be aligned with the direction of the minimum energy path.''''' '''''

* Q7. Locate the approximate position of the transition state.
{| class="wikitable"
|+Table 8. Transition state position
!
!r /ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|180
|0
|-
|r(BC)
|74.5
|0
|}
The transition state position shown in Table 8 was estimated by looking at the initial conditions in Table 6 for the exothermic F + H2 → HF + H reaction. Due to exothermic reactions having early transition states, this suggested that the bond distance BC (the hydrogen atoms) would be small as it's the reactant. The initial bond distance of AB was reduced until the hessian matrix, shown in Table 9, gave a value of Det= -1.0 <0 which proved it was a saddle point. Additionally, there was no force. This also suggested it was the transition state.
{| class="wikitable"
|+Table 9. Hessian matrix for Transition state
|AB direction
|<nowiki>+1.000</nowiki>
|<nowiki>-0.024</nowiki>
|-
|BC direction
|<nowiki>-0.024</nowiki>
|<nowiki>-1.000</nowiki>
|}
[[File:Mb7418 dis vs time ts ex2.png|thumb|[[Figure 9]]. Intermolecular distance vs time graph for transition state for equations two and three]]
Additionally, Figure 9 shows the intermolecular distance vs time graph. This is a flat line, this could be due to everything in the system being stationary. There are no oscillations showing that there are any vibrational motions. This suggest that it's at the initial conditions in Table 8 represent the transition state position.

* Q8. Report the activation energy for both reactions.
Equation 2: F + H2 → HF + H the activation energy value is 1.0 kJ/mol.

Equation 3: HF + H → F + H2 the activation energy value is 126.7 kJ/mol.

The activation energy for equation 2 was calculated by getting the potential energy for transition state by using the initial conditions in Table 8. Then this was subtracted by the potential energy obtained by using the initial conditions in Table 10. Activation energy = TS(potential energy) - Reactants(potential energy) = (-433.981)-(435.056) = 1.075 kJ/mol
{| class="wikitable"
|+Table 10. Initial conditions for equation 2
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|500
|0
|-
|r(BC)
|74.5
|0
|}
The activation energy for equation 3 was calculated by getting the potential energy for for transition state by using the initial conditions in Table 8. Then this was subtracted by the potential energy obtained by using the initial conditions in Table 11. r(AB) which is the HF bond distance = 92 ppm<ref>Cccbdb.nist.gov. 2020. ''CCCBDB Listing Of Experimental Data Page 2''. [online] Available at: <<nowiki>https://cccbdb.nist.gov/exp2x.asp?casno=7664393&charge=0</nowiki>> [Accessed 22 May 2020].</ref>'''. '''The r(BC) bond which is the HH bond distance was set very high because the hydrogen atoms aren't bonded in the reactants.

Activation energy = TS(potential energy) - Reactants(potential energy) = (-433.981)-(560.698) = 126.7 kJ/mol
{| class="wikitable"
|+Table 11. Initial conditions for equation 3
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|92
|0
|-
|r(BC)
|500
|0
|}

{{fontcolor1|red| Good work. well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:28, 29 May 2020 (BST)}}

=== Reaction dynamics ===
* Q9. 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.
{| class="wikitable"
|+Table 12. Initial conditions for reactive trajectory
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|175
|<nowiki>-1.6</nowiki>
|-
|r(BC)
|74
|0.3
|}
[[File:Mb7418 THIS reactive.png|left|thumb|[[Figure 10]]. Reactive trajectory]]
The step number was increased to 4000 and the initial conditions in Table 12 gives the potential energy surface diagram in Figure 10. This shows a reactive trajectory for the exothermic reaction F + H2 → HF + H as it passes the transition state as well as having many oscillations. The end point is also far away from the starting point.

The starting point (reactants) has a high energy level compared to the products due to being exothermic<ref name=":0" />. Hence the potential energy converts into translational kinetic energy. Therefore, there are many oscillations in Figure 10 due to energy being conserved.

This can be explained by the difference in distribution of the energy levels between an excited and relaxed state. The excited state will have an occupied ground state (v=0) as well as an occupied first state (v=1) where v=vibrational number. Whereas in the relaxed state only the ground state will be occupied. This approach uses the harmonic oscillator<ref>Cccbdb.nist.gov. 2020. ''CCCBDB Listing Of N2 Experimental Data Page 2''. [online] Available at: <<nowiki>https://cccbdb.nist.gov/exp2x.asp?casno=7727379&charge=0</nowiki>> [Accessed 18 May 2020].</ref>. When energy is released, in the excited state there will be two transitions, v(0)→v(1) and v(1)→v(2), the fundamental peak and overtone. In the ground state it will only have transitions from v(0)→v(1).

Therefore, in an IR spectrum<ref>Chemistry LibreTexts. 2020. ''5.5: The Harmonic Oscillator And Infrared Spectra''. [online] Available at: <<nowiki>https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.05%3A_The_Harmonic_Oscillator_and_Infrared_Spectra</nowiki>> [Accessed 22 May 2020].</ref>, for the excited state there will be two peaks corresponding to the overtone and fundamental peak. The overtone peak intensity will be smaller than the fundamental's peak intensity. Hence there are many oscillations due to the overtone decreasing and then increasing in intensity while the fundamental increases and then decreases in intensity.

Therefore, this could be confirmed experimentally by Infrared spectroscopy (IR).
* Q10. 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.
[[File:Mb7418 last Q.png|left|thumb|[[Figure 11]]. Reactive potential energy surface contour plot]]
The position of the transition state affects whether the reaction is either exothermic or endothermic. An endothermic reaction will have a late transition state, suggesting that the vibrational energy efficiency will be higher<ref>Guo, H. and Liu, K., 2020. ''Control Of Chemical Reactivity By Transition-State And Beyond''.</ref>. The reason for this is because if it had a higher translational kinetic energy, the reactants wouldn't follow the minimum energy path and will just collide with each other but then immediately repel resulting in being back at the starting point. Hence, the translational kinetic energy converts into vibrational energy<ref name=":1" /> instead so the reactants collide and are aligned with the minimum energy pathway. The direction of the path taken is shown in Figure 11. The initial conditions in Table 12 was used to give the reactive trajectory. This suggests that endothermic reactions have higher vibrational energy to make their reactions more efficient so it's able to overcome the transition state. Hence there are a lot of oscillations before the transition state is reached because it requires more efficient vibrational energy in order to reach the transition state.

Whereas if the reaction has an early transition state, it will be an exothermic reaction. This suggest that there will be a higher translational kinetic energy compared to vibrational energy. This means that there will be less oscillations before it reaches the transition state as shown in Figure 11.

Overall, an efficient endothermic reaction will have a higher vibrational energy whereas an efficient exothermic reaction will have a higher translational kinetic energy.

MRD:mb7418

2020-05-29T18:28:25Z

Mak214: /* PES inspection */

== EXERCISE 1: H + H2 system ==

===Dynamics from the transition state region===
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? {{fontcolor1|red| This text is repeated below which makes this section difficult to read. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:03, 29 May 2020 (BST)}} Potential energy surface diagrams show the potential energy the system has according to the coordinates of the atoms in the system.<ref>Chemistry LibreTexts. 2020. ''2.6: Potential Energy Surfaces''. [online] Available at: <<nowiki>https://chem.libretexts.org/Courses/University_of_California_Davis/UCD_Chem_107B%3A_Physical_Chemistry_for_Life_Scientists/Chapters/2%3A_Chemical_Kinetics/2.06%3A_Potential_Energy_Surfaces</nowiki>> [Accessed 19 May 2020].</ref> Figure 1 shows a potential energy surface diagram representing the reaction, equation 1: A + BC → AB + C.
{| class="wikitable"
|+Table 1. Molecules distances (r) and momenta (p)
!
!r / ppm
!p/ g.mol-1.pm.fs-1
|-
|r2=r(AB)
|230
|<nowiki>-5.2</nowiki>
|-
|r1=r(BC)
|74
|0
|}
In this case r(1)=r(BC) and r(2)=r(AB). BC is a hydrogen molecule consisting of two hydrogen atoms, B and C. Atom A is a hydrogen atom which has energy equal to or more than the activation energy, thus reaching the transition state. The transition state structure consists of atom A forming a bond with molecule BC while the BC bond is slightly dissociating. After the transition state, the bond between AB would form making a new molecule and the BC bond would be fully dissociated.
* Q1. 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?
[[File:PE with TS.png|left|thumb|[[Figure]] 1. Potential energy surface diagram with transition state labelled ]]
In a potential energy surface diagram, the transition state is mathematically defined as the first-order saddle point<ref>Chemistry LibreTexts. 2020. ''Hammond’S Postulate''. [online] Available at: <<nowiki>https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hammond%E2%80%99s_Postulate</nowiki>> [Accessed 22 May 2020].</ref>:

∂V(r1)/∂(r1)=0 and ∂V(r2)/∂r2=0

∂2V(r)/∂(r1)2 > 0 and ∂2V(r)/∂(r2)2 < 0

[∂2V(r)/∂(r1)2]*[ ∂2V(r)/∂(r2)2] - [ ∂2V/∂(r1)(r2)]2 < 0

Hessian matrix is shown in Table 2, resulting in DET = -0.999698 < 0
{| class="wikitable"
|+Table 1. Hessian matrix for the transition state
!
!f(x)
!f(y)
|-
|f(x)
|<nowiki>+0.707</nowiki>
|<nowiki>-0.707</nowiki>
|-
|f(y)
|<nowiki>-0.707</nowiki>
|<nowiki>-0.707</nowiki>
|}
In Figure 1, the black wavy lines shows that the molecule is vibrating. This is the minimum energy path required for the reactants to react and turn into the products via the transition state.<ref>Atkins, de Paula, Keeler: Physical Chemistry, 11th Edition. Section 18D.3 Potential Energy Surfaces (PESs)</ref> This suggests that the transition state is the peak in the minimum energy path. The transition state is the minimum energy required for the reactants to convert into the products. Hence once the transition state is reached, if the reaction was successful the products would be formed. Figure 1 shows the location of the transition state in the potential energy diagram.

The transition state can be distinguished from a local minimum of the potential energy surface because it is on the diagonal axis. {{fontcolor1|red|Because it is also a maximum (minimum AND maximum = saddle point) therefore distinguishable from a local minimum. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:03, 29 May 2020 (BST)}} It is the saddle point.This is when everything in the system is stationary when there is no momentum.

===Locating the transition state r1=r2===
* Q2. 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.

<nowiki> </nowiki>When the transition state is formed, due to all the atoms being hydrogen, this suggest that the transition state is symmetrical. This means that the bond distance between A and B is the same as the bond distance of B and C. This can be used to locate the transition state when the momentum is 0. Table 3 shows an estimate of the transition state position (rts).
{| class="wikitable"
|+Table 3. Distances and momenta to locate the transition state
!
!r / ppm
!p/ g.mol-1.pm.fs-1
|-
|r2=r(AB)
|90.75
|0
|-
|r1=r(BC)
|90.75
|0
|}
[[File:Mb7418 dis vs time ts.png|thumb|[[Figure 2]]. Distance vs time plot of the transition state ]]
Figure 2 shows an Intermolecular distance vs time graph resulted from the initial conditions from Table 2. The line is flat, this suggest that everything in the system is stationary. Whereas, if it wasn't the transition state, in the intermolecular distance vs time graph you'd see oscillations representing the vibrational movements. The more oscillations would indicate the more vibrational movements. Hence the transition state is a flat line as there isn't any. {{fontcolor1|red| OK. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:33, 29 May 2020 (BST)}}
[[File:Mb7418 ts contour.png|thumb|[[Figure 3]]. Potential energy diagram as a contour plot of the transition state]]
The contour plot shown in Figure 3 shows the transition state as a red cross. This is the saddle point of the potential surface. Whenever r1 and r2 are the same, the point would be anywhere across the diagonal axis. This contour plot helped estimate the transition state position as the trajectory shows that the AB and BC bond lengths are the same. This is the point at which reactants have to overcome to react into the products.

The minimum energy path can differ depending on the relative momentum of the atoms/molecules and their vibrational excitation energy. The transition state of the reaction won't change its position. The reactants require sufficient translational kinetic energy and vibrational energy to overcome the transition state (saddle point) in order to react and turn into the products.

===Trajectories from r1 = rts+δ, r2 = rts ===
* Q3. Comment on how the ''mep'' and the trajectory you just calculated differ. '''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''' and the momenta '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1
The trajectory of the minimum energy path can be visualised slightly near the transition state, this is when atom A is approaching molecule BC to form a bond with it, A + BC.
{| class="wikitable"
|+Table 4. Initial conditions slightly displaced from transition state
!
!r /ppm
!p/ g.mol-1.pm.fs-1
|-
|r2=r(AB)
|90.75
|0
|-
|r1=r(BC)
|91.75
|0
|}
The initial conditions shown in Table 4 gives the potential energy diagram in Figure 4 and 5 when the calculation type is MEP. The minimum energy path shown by the black line isn't wavy. [[File:Mb7418 mep.png|left|thumb|[[Figure 4]]. MEP calculation diagram]]
[[File:Mb7418 mep 1.png|thumb|234x234px|[[Figure 5]]. MEP calculation diagram bottom view]]However, when the same initial conditions are applied when the calculation type is Dynamic, the minimum energy path is wavy shown in Figure 6. This shows that the MEP calculation gives a different result of the reaction trajectory. The molecule is constantly moving at a velocity since there is no external force acting upon the reactants. This is called inertia motion.<ref>Evans, M.G. and Polanyi, M., 1938. Inertia and driving force of chemical reactions. ''Transactions of the Faraday Society'', ''34'', pp.11-24.</ref> Whereas the potential energy diagram produced from the Dynamics calculation takes into account the reactants inertia by representing the pathway as non-linear. Additionally, the MEP calculation diagram has a smaller trajectory than the Dynamics calculation.The reason for this is beacause the MEP calculation sets the momentum to zero at every step whereas the dynamic calculation the momenta is developed classically<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., 8.3.1, Prentice-Hall, 1998.</ref>. The MEP calculation only takes the gradient into account. The amount of steps there are corresponds to the number of times it is calculating what the gradient is at that point. The only thing relative to the MEP calculation is the gradient. Whereas with dynamics, it takes into account how long it takes for the gradient to change. The momentum (velocity) is changing each time instead of setting it to zero every step. Hence it's longer as after the products are formed, they are still have motion such as vibrational and kinetic. {{fontcolor1|red| Yes. This does show an understanding of MEP vs Dynamics. Just to clarify - the amount of steps is the number of times the system moves during the simulation - this was a bit confused in your response. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:36, 29 May 2020 (BST)}} [[File:Mb7418 dyn 1.png|thumb|[[Figure 6]]. Dynamics calculation bottom view]]

== Reactive and unreactive trajectories ==

===Interpreting trajectories===

* Q4. For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, run trajectories with the following momenta combination and complete the table.

{| class="wikitable"
|+Table 5. Reactive and unreactive trajectories
!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>
|Reactive
|The starting point (represented by the red cross) starts at 200 pm between atoms A and B. Then the distance AB decreases. Atoms A and B are approaching each other and once it reaches the transition state, the AB distance and BC distance are the same and all the atoms are stationary. Then the AB bond is formed and atom C repels from atom B. The molecule AB is constantly vibrating. This could suggest that some translational kinetic energy got converted into vibrational energy after the transition state. Hence after the trajectory has oscillations after it passes the transition state.
|[[File:Mb7418 6.png|centre|thumb|200x200px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The trajectory has some oscillations, showing that the reactants are constantly vibrating. Hence it has vibrational energy. However, it doesn't go over the transition state so it doesn't react into the products. This suggests that the reactants approach each other but then they repel. While they do that they oscillate as they get closer. It ends up with a high AB distance (essentially the starting point of the reaction), this shows that the reactants haven't reacted due to the AB molecule not forming. This could be due to not having enough translational kinetic energy.
|[[File:Mb7418 2.png|centre|thumb|200x200px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The starting point starts at 200 pm between atoms A and B. Then the the atoms got closer and reaches the transition state. This is when the distance AB and BC are equal. Then the AB distance remains continuous while the BC distance begins to increase. The BC bond breaks and the AB bond forms. It only crosses the transition state once. It ends up in the products as the distance AB is short (molecule containing atoms A and B bonding) and BC is distance is long (reactant molecule BC bond has been broken).
|[[File:Mb7418 3.png|centre|thumb|200x200px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Unreactive
|The whole trajectory shows that it's unreactive because the starting point and ending point are similar. It started at a high AB distance value and also ended at a high AB distance value. This suggests this is end unreactive.
At the starting point, AB are far apart. Then they begin approaching each other and reach the transition state. However, they repel from each other and then start approaching each other again. This happens again and then the distance AB begins to increase again. The BC bond is formed again, resulting in the reactant molecule BC at the end.
|[[File:Mb7418 4.png|centre|thumb|200x200px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Reactive
|This crosses the transition point more than once. It's reactive as it's end point is at a high BC distance and low AB distance, suggesting that the BC bond broke and the AB bond has formed resulting in AB + B.

From the starting point, the AB distance starts to decrease. This suggests atom A and B approach each other, but then the AB distance begins to increase again so the A and B atoms repel. There is a moment when the distance between atoms A and B is the same as the distance between B and C (transition state) Then the B and C atoms approach each other and then repel B continuously. While atoms A and B get closer forming a bond. The molecule AB vibrates while atom C continuously repels away from it.
|[[File:Mb7418 5.png|centre|thumb|200x200px]]
|}

{{fontcolor1|red| OK. What do you conclude from the table? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:40, 29 May 2020 (BST)}}

=== Transition state theory ===
* Q5. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?
The three assumptions of the transition state theory<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Chapter 10 pg. 20, Prentice-Hall, 1998</ref> are:

1) The transition state will only be crossed over once. Once the transition state it reached, the products will be formed.

2) During the transition state, any motion along the reaction coordinate can be separate from the other motions and treated classically as a translation. The concentration of the transition state is in equilibrium with the reactants. (concentration constant relating them both, there's a Maxwell-Boltzmann distribution between the reactants and the transition state at any given time) This is defined as the Quasi equilibrium. It's assumed that the kinetic energy is predicted by the Boltzmann distribution. The transition states that are turning into products are distributed among their states according to the Maxwell-Boltzmann Laws<ref>Rowlinson*, J.S., 2005. The Maxwell–Boltzmann distribution. ''Molecular Physics'', ''103''(21-23), pp.2821-2828.</ref> even in absence of an equilibrium between the reactant and product molecules.

'''3)''' There's minimum effects of quantum tunneling.

The quantum tunneling effect<ref>Abyss.uoregon.edu. 2020. ''Quantum Tunneling''. [online] Available at: <<nowiki>http://abyss.uoregon.edu/~js/glossary/quantum_tunneling.html</nowiki>> [Accessed 22 May 2020].</ref> is when the reactants can overcome the barrier by moving across it to turn into the products. The transition state theory doesn't include quantum tunneling. If quantum tunneling was included, it would increase the rate of reaction as more reactants will turn into products. There are only classical motions in the transition state theory. Hence it is suggested that quantum tunneling isn't included in the model used in Table 5 because it's assumed there's a faster alternative route used.

Additionally, in the Table 5 model the transition state theory wasn't true because the transition state was crossed over more than once in the last two reactions in Table 5, this could decrease the rate of reaction due to the products being able to cross the transition state again and go back into the reactants. Therefore, the transition state theory is an over-estimate in comparison to the model used. {{fontcolor1|red| Yes. Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:41, 29 May 2020 (BST)}}

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

=== PES inspection ===
* Q6. 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?
{| class="wikitable"
|+Table 6. Initial conditions for Figure 7 (F + H2 → HF + H)
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|500
|0
|-
|r(BC)
|70
|0
|}

[[File:Mb7418 F + H2 exo edit CORRECT.png|left|thumb|[[Figure 7]]. F + H2 -> HF + H exothermic potential surface diagram]]
In the following equations, A = atom F, B = atom H, C = atom H.

The initial conditions in Table 6 gives the potential energy diagram for Figure 7. The reason for this is because the F and H bond is very far apart, suggesting that it's not bonded whereas the hydrogen atoms are closer together, suggesting a F atom and H2 molecule are in the initial conditions.

Equation 2: F + H2 → HF + H is an exothermic reaction<ref name=":0">Duff, J.W. and Truhlar, D.G., 1975. Effect of curvature of the reaction path on dynamic effects in endothermic chemical reactions and product energies in exothermic reactions. ''The Journal of Chemical Physics'', ''62''(6), pp.2477-2491.</ref> shown in Figure 7. The energy of the reactants is higher than the energy of the products. This shows that this is an exothermic reaction due to the process releasing energy. This suggests that the bond strength of H2 is weaker than the bond strength of HF as no additional energy was required to break its bond. This reaction has an early transition state<ref>Chemistry LibreTexts. 2020. ''Hammond’S Postulate''. [online] Available at: <<nowiki>https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hammond%E2%80%99s_Postulate</nowiki>> [Accessed 22 May 2020].</ref> due to the the fluorine atom being highly electro-negative so it attacks the hydrogen molecule very quickly.
{| class="wikitable"
|+Table 7. Initial conditions for Figure 8 (HF + H → F + H2)
!
!r / ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|70
|0
|-
|r(BC)
|500
|0
|}
[[File:Mb7418 NEW CORRECT ENDO.png|left|thumb|[[Figure 8]]. HF + H -> F + H2 endothermic potential surface diagram]]
The initial conditions in Table 7 gives the potential energy diagram for Figure 8. The reason for this is because the F and H atoms are very close together compared to both the hydrogens. This suggests that a HF molecule and one H atom are in the initial conditions.

Equation 3: HF + H → F + H2 is an endothermic reaction<ref name=":0" /> shown in Figure 8. The energy of the reactants is lower than the energy of the products. This suggests that this reaction is endothermic as the process used additional energy to break the H-F bond to form the H-H bond. Therefore, in the potential energy surface diagram there will be a lot of vibrational energy before the transition state compared to after. This also suggest that the HF bond is stronger than the HH bond as it requires additional energy to break its bond. This will give a late transition state.The translational energy could get converted into vibrational energy in order to break the bond between the strong H-F bond<ref name=":1">Polanyi, J.C. and Tardy, D., 1969. Energy Distribution in the Exothermic Reaction F+ H2 and the Endothermic Reaction HF+ H. ''The Journal of Chemical Physics'', ''51''(12), pp.5717-5719.</ref>. This favours the endothermic reaction because the reactants position will be aligned with the direction of the minimum energy path.''''' '''''

* Q7. Locate the approximate position of the transition state.
{| class="wikitable"
|+Table 8. Transition state position
!
!r /ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|180
|0
|-
|r(BC)
|74.5
|0
|}
The transition state position shown in Table 8 was estimated by looking at the initial conditions in Table 6 for the exothermic F + H2 → HF + H reaction. Due to exothermic reactions having early transition states, this suggested that the bond distance BC (the hydrogen atoms) would be small as it's the reactant. The initial bond distance of AB was reduced until the hessian matrix, shown in Table 9, gave a value of Det= -1.0 <0 which proved it was a saddle point. Additionally, there was no force. This also suggested it was the transition state.
{| class="wikitable"
|+Table 9. Hessian matrix for Transition state
|AB direction
|<nowiki>+1.000</nowiki>
|<nowiki>-0.024</nowiki>
|-
|BC direction
|<nowiki>-0.024</nowiki>
|<nowiki>-1.000</nowiki>
|}
[[File:Mb7418 dis vs time ts ex2.png|thumb|[[Figure 9]]. Intermolecular distance vs time graph for transition state for equations two and three]]
Additionally, Figure 9 shows the intermolecular distance vs time graph. This is a flat line, this could be due to everything in the system being stationary. There are no oscillations showing that there are any vibrational motions. This suggest that it's at the initial conditions in Table 8 represent the transition state position.

* Q8. Report the activation energy for both reactions.
Equation 2: F + H2 → HF + H the activation energy value is 1.0 kJ/mol.

Equation 3: HF + H → F + H2 the activation energy value is 126.7 kJ/mol.

The activation energy for equation 2 was calculated by getting the potential energy for transition state by using the initial conditions in Table 8. Then this was subtracted by the potential energy obtained by using the initial conditions in Table 10. Activation energy = TS(potential energy) - Reactants(potential energy) = (-433.981)-(435.056) = 1.075 kJ/mol
{| class="wikitable"
|+Table 10. Initial conditions for equation 2
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|500
|0
|-
|r(BC)
|74.5
|0
|}
The activation energy for equation 3 was calculated by getting the potential energy for for transition state by using the initial conditions in Table 8. Then this was subtracted by the potential energy obtained by using the initial conditions in Table 11. r(AB) which is the HF bond distance = 92 ppm<ref>Cccbdb.nist.gov. 2020. ''CCCBDB Listing Of Experimental Data Page 2''. [online] Available at: <<nowiki>https://cccbdb.nist.gov/exp2x.asp?casno=7664393&charge=0</nowiki>> [Accessed 22 May 2020].</ref>'''. '''The r(BC) bond which is the HH bond distance was set very high because the hydrogen atoms aren't bonded in the reactants.

Activation energy = TS(potential energy) - Reactants(potential energy) = (-433.981)-(560.698) = 126.7 kJ/mol
{| class="wikitable"
|+Table 11. Initial conditions for equation 3
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|92
|0
|-
|r(BC)
|500
|0
|}

{{fontcolor1|red| Good work. well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:28, 29 May 2020 (BST)}}

=== Reaction dynamics ===
* Q9. 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.
{| class="wikitable"
|+Table 12. Initial conditions for reactive trajectory
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|175
|<nowiki>-1.6</nowiki>
|-
|r(BC)
|74
|0.3
|}
[[File:Mb7418 THIS reactive.png|left|thumb|[[Figure 10]]. Reactive trajectory]]
The step number was increased to 4000 and the initial conditions in Table 12 gives the potential energy surface diagram in Figure 10. This shows a reactive trajectory for the exothermic reaction F + H2 → HF + H as it passes the transition state as well as having many oscillations. The end point is also far away from the starting point.

The starting point (reactants) has a high energy level compared to the products due to being exothermic<ref name=":0" />. Hence the potential energy converts into translational kinetic energy. Therefore, there are many oscillations in Figure 10 due to energy being conserved.

This can be explained by the difference in distribution of the energy levels between an excited and relaxed state. The excited state will have an occupied ground state (v=0) as well as an occupied first state (v=1) where v=vibrational number. Whereas in the relaxed state only the ground state will be occupied. This approach uses the harmonic oscillator<ref>Cccbdb.nist.gov. 2020. ''CCCBDB Listing Of N2 Experimental Data Page 2''. [online] Available at: <<nowiki>https://cccbdb.nist.gov/exp2x.asp?casno=7727379&charge=0</nowiki>> [Accessed 18 May 2020].</ref>. When energy is released, in the excited state there will be two transitions, v(0)→v(1) and v(1)→v(2), the fundamental peak and overtone. In the ground state it will only have transitions from v(0)→v(1).

Therefore, in an IR spectrum<ref>Chemistry LibreTexts. 2020. ''5.5: The Harmonic Oscillator And Infrared Spectra''. [online] Available at: <<nowiki>https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.05%3A_The_Harmonic_Oscillator_and_Infrared_Spectra</nowiki>> [Accessed 22 May 2020].</ref>, for the excited state there will be two peaks corresponding to the overtone and fundamental peak. The overtone peak intensity will be smaller than the fundamental's peak intensity. Hence there are many oscillations due to the overtone decreasing and then increasing in intensity while the fundamental increases and then decreases in intensity.

Therefore, this could be confirmed experimentally by Infrared spectroscopy (IR).
* Q10. 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.
[[File:Mb7418 last Q.png|left|thumb|[[Figure 11]]. Reactive potential energy surface contour plot]]
The position of the transition state affects whether the reaction is either exothermic or endothermic. An endothermic reaction will have a late transition state, suggesting that the vibrational energy efficiency will be higher<ref>Guo, H. and Liu, K., 2020. ''Control Of Chemical Reactivity By Transition-State And Beyond''.</ref>. The reason for this is because if it had a higher translational kinetic energy, the reactants wouldn't follow the minimum energy path and will just collide with each other but then immediately repel resulting in being back at the starting point. Hence, the translational kinetic energy converts into vibrational energy<ref name=":1" /> instead so the reactants collide and are aligned with the minimum energy pathway. The direction of the path taken is shown in Figure 11. The initial conditions in Table 12 was used to give the reactive trajectory. This suggests that endothermic reactions have higher vibrational energy to make their reactions more efficient so it's able to overcome the transition state. Hence there are a lot of oscillations before the transition state is reached because it requires more efficient vibrational energy in order to reach the transition state.

Whereas if the reaction has an early transition state, it will be an exothermic reaction. This suggest that there will be a higher translational kinetic energy compared to vibrational energy. This means that there will be less oscillations before it reaches the transition state as shown in Figure 11.

Overall, an efficient endothermic reaction will have a higher vibrational energy whereas an efficient exothermic reaction will have a higher translational kinetic energy.

MRD:M4H1M4

2020-05-29T18:25:29Z

Mak214: /* Reaction dynamics */

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

=== Dynamics from the transition state region ===
[[File:Hdiagram.png|200px|thumb|left|H + H2]]
''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 first order saddle point on the potential energy surface (PES) diagram. The derivative of potential energy is equal to the force acting on the atoms/molecules. At this point the gradient is equal to 0 and therefore no force is acting on the particles. This allows us to use a constrained one dimensional optimisation to calculate transition state. In order to verify this stationary point as the saddle transition state the Hessian eigenvectors can be used. Only one of the second derivatives must be negative and the other positive so that (fxxfyy - f2xy) < 0. Therefore the determinent of the Hessian matrix less than 0.

This verifies the point as a saddle rather than a local minimum of the potential energy surface. {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:06, 29 May 2020 (BST)}}

==== Trajectories from r1 = r2: locating the transition state ====
''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.''

The the H + H2 ---> H2 + H surface is symmetrical, neither H2 or H atom are favoured.Therefore r1 is equal to r2 in the transition state. This restriction allows us to calculate the transition state position through a process of trial and error.

[[File:R1=R2.png|600px|thumb|left|Figure 1 - PES of transition state of reaction H + H2]]

[[File:Hessian.png|600px|thumb|Figure 2 - Hessian matrix and determinant for H + H2 reaction]]

[[File:Dvst01499529.png|600px|thumb|left|Figure 3 - Displacement vs Time for reaction at transition state]]Transition state position r1= r2 = 90.8 pm

As shown in figure 1, at these positions the forces action along AB and BC are approximately 0 suggesting a stationary point. Figure 3 shows two horizontal lines with gradient 0 which therefore means 0 acceleration and 0 forces acting on the atoms.

Looking at the Hessian matrix we can verify this as a saddle point. To further confirm this, trajectories were calculated with either r1 or r2 being slightly displaced from equilibrium. Figure 4 shows that increasing r2 by 1 pm pushing the reaction to the reactant side figure 5 shows increasing r2 by 1 pm favours the formation of the products. {{fontcolor1|red| Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:06, 29 May 2020 (BST)}}

[[File:Reactantchannel.01499529.png|200px|thumb|Figure 4 - Trajectory in reactant channel at '''r2''' = '''rts'''+1, '''r1''' = '''rts p1''' = '''p2''' = 0 g.mol-1.pm.fs-1]]

[[File:Productchannel.01499529.png|200px|thumb|Figure 5 - Trajectory in product channel at '''r1''' = '''rts'''+1 pm, '''r2''' = '''rts, p1''' = '''p2''' = 0 g.mol-1.pm.fs-1]]

==== Trajectories from r1 = rts+δ, r2 = rts ====
''Comment on how the mep and the trajectory you just calculated differ.''

Figure 6 shows a straight line suggesting the velocity and momentum do not change with time. Figure 7, the dynamic trajectory, accounts for the oscillation of the bonds. {{fontcolor1|red| Why? What are the conditions for the MEP calculation that lead you to see this. More discussion is needed here. MEP calculations involve resetting the momenta to zero at every time step so we end up finding the nearest local minimum.. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:06, 29 May 2020 (BST)}}

Final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t):

r1=74.0 pm r2= 352.6 pm p1= 3.2 g.mol-1.pm.fs-1 p2= 5.1 g.mol-1.pm.fs-1
* ''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.''
''What do you observe?''

The reaction does not occur atom C reaches a constant velocity and atom B and A vibrate with constant acceleration.
[[File:MEPMT.png|200px|thumb|middle|Figure 6 - Momentum vs Time MEP]]
[[File:DYNMT.png|200px|thumb|middle|Figure 7 - Momentum vs Time dynamic]]

=== Reactive and unreactive trajectories ===
{| 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
|-
|1
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|Yes
|Initially r1 < r2, and therefore atoms A and B are bonded to each other and atom C is the approaching atom. R1 remains almost constant as r2 begins to decrease. r1 begins to increase until both r1 and r2 are equal at the transition state. From this point r2 remains almost constant whilst r1 increases. Atom A and B have disassociated whilst a new B-C bond has been formed. This bond has kinetic energy hence the vibrations.
|[[File:1.yes.png|200px|thumb|left|Figure 8 - Trajectory set 1]]
|-
|2
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.1</nowiki>
|No
|From the initial positions, r1 oscillates around 74 pm. Distance r2, intially decreases as C approaches B. The trajectory does not go through the TS instead r2 begins to increase again. In this trajectory A and B remain bonded and atom C is repelled from atom B. This is because the momentum p2 is not large enough for a reaction to occur.
|[[File:2.no.png|200px|thumb|left|Figure 9 - Trajectory set 2]]
|-
|3
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.0</nowiki>
|Yes
|From the initial position, r2 decreases with little change in r1. r1 beings to increase until the atoms are in the transition state. r1 increases suggesting the breaking of the A-B bond. r2 continues to oscilate due to internal vibration of the B-C bond.
|[[File:3.yes.png|200px|thumb|left|Figure 10 - Trajectory set 3]]
|-
|4
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.3</nowiki>
|No
|This trajectory does not go through the transition state.Atom C approaches with too much momentum. The A-B bond breaks and the B-C bond forms. B-C vibrates with a large amount of momentum and kinetic energy. B-C vibrates with increasing r2 until passing back through to the reactant channel with the formation of A-B and the disassociation of B-C This is a case of barrier recrossing.Therefore there is no overall reaction.
|[[File:4.no.png|200px|thumb|left|Figure 11 - Trajectory set 4]]
|-
|5
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.5</nowiki>
|Yes
|Atom B and C approach each other with a large amount of momentum causing the A-B bond to break and the formation of the B-C bond.r2 begins to increase agin while the A-B bond reforms. r1 increaes again while r2 decreases with the trajectory returning to the product channel leading to an overall reaction.
|[[File:5.yes.png|200px|thumb|left|Figure 12 - Trajectory set 5]]
|}
''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?''

From this table we can conclude, trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are always reactive. However trajectories higher momenta and therefore higher kinetic energy does not necessarily mean a successful reaction.This is evident by trajectory set 4 having a kinetic energy of 76.5 kJ/mol and resulting in no reaction and set trajectory 3 having only 19.8 kJ/mol and being successful. Increasing the momentum can lead to barrier crossing which if there isn't sufficient energy to cross back to the product channel can lead to overall no reaction.

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

Transition state theory assumes quasi-equilibrium between the reactants and the activated transition state complex and only requires details bout the PES. The Born-Oppenheimer approximation is used and quantum tunnelling affects are assumed to be negligible. The energies of the reactants also follow the Boltzman distribution curve.<ref>T. Bligaard, J.K. Nørskov, in Chemical Bonding at Surfaces and Interfaces, 2008</ref> From the table above we can see that quantum tunneling and barrier recrossing can occur, As TST does not account for that the value of rate calculated would be an overestimate compared to the experimental value. However error due to this is only marginal.<ref>PetersBaron , in Reaction Rate Theory and Rare Events Simulations, 2017</ref> {{fontcolor1|red| Think about this - barrier recrossing we do see in the table of results above and does mean that overall TST overestimates rate. However quantum tunnelling (not visible in your simulations above) leads to an underestimate in the rate predicted by TST because this is when the system can tunnel through an energy barrier - therefore reaching the products faster than we expect... [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:45, 29 May 2020 (BST)}}

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

=== PES inspection ===
[[File:Fdiagram.png|200px|thumb|left|H2 + F]]
F + H2 ----> HF + H (1)

Initial conditions: p1= p2= 0 g.mol-1.pm.fs-1, r1=74.1 pm, r2= 400 pm

HF + H -----> H2 + F (2)

<nowiki> </nowiki>Initial conditions: p1=p2=0 g.mol-1.pm.fs-1 r1=400 pm, r2= 91.7 pm
[[File:FH2.png|200px|thumb|left|Figure 13 - PES of H2 + F]]
[[File:HfH.png|200px|thumb|left|Figure 14 - PES of HF + H]]
From figure we can conclude this reaction is exothermic as the formation of the H-F bond releases energy greater than the energy required to break the H-H bond. In the PES diagram as p2 decreases p1 increases the trajectory falls into the product channel at a lower energy than the products. This is reflective of the bonds being broken and formed. Reaction 2 is endothermic as the H-H bond being formed releases less energy than required to break the H-F bond. From this we can conclude the H-F bond is stronger that the H-H bond.

H-H bond energy = 435.7799 kJ/mol<ref name=":0">Luo, Yu-Ran. ''Comprehensive handbook of chemical bond energies''. CRC press, 2007.</ref>

H-F bond energy = 569.680 kJ/mol<ref name=":0" />

''Locate the approximate position of the transition state.''
[[File:TScontuor.png|200px|thumb|Figure 15 - Transition state of H2 + F]]

Fluorine is highly electronegative and so attacks the hydrogen early on in the reaction this is fitting of the exothermic characteristic of the reaction. Using the Hammond postulate, this early transition state therefore means that the structure of the transition state resembles that of the reactants. Using the liturature value of H-H bond length, r1 was set to 74.1 pm as an initial estimate.<ref>Huber, K.P.; Herzberg, G., Molecular Spectra and Molecular Structure. IV. Constants of Diatomic Molecules,, Van Nostrand Reinhold Co., 1979</ref> Trial and error was performed in order to calculate the r1 and r2 values that give the closest to 0 kJ/mol.pm of force along the axis.

Transition state position - r1= 74.5 pm r2= 181.1 pm

''Report the activation energy for both reactions.''

To calculate the activation energy of both reactions the potential energy at transition state was recorded. The energy of the reactants of the forwards reaction was measured by setting r1 as the bond length of H2, 74.1 pm and r2 being set as 500 pm. This allows us to view the energy of the reactants when they are not reacting with each other. The same was repeated for the backward reaction using the bond length of HF, 91.7 pm as r2.<ref>NIST Diatomic Spectral Database (www.physics.nist.gov/PhysRefData/MolSpec/Diatomic/index.html)</ref>

<nowiki> </nowiki>Transition state energy = -434.0 kJ/mol

Energy of Reactants H2 + F = -435.1 kJ/mol Ea1 = 1.1 kJ/mol

Energy of Reactants HF + H = -560.7 kJ/mol Ea2 = 125.6 kJ/mol

{{fontcolor1|red| Nice work. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:22, 29 May 2020 (BST)}}

=== Reaction dynamics ===
''Set of initial conditions that results in a reactive trajectory for the F + H2'': r1= 74.5 pm, r2= 130 pm, p1= -1g.mol-1.pm.fs-1 p2= -3 g.mol-1.pm.fs-1 

''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.''

Translational energy is required in order for the reactants to collide with one another with enough energy to react. However large amounts of excess translational energy can lead to the reacts rebounding off of each other.
{{fontcolor1|red| There is another question to answer here - look into Polanyis rules which talks about the trade off between translational and vibrational energy coming into a transition state. This can be related to exothermic and endothermic reactions favouring either vibrational or translational kinetic energy of reactants. A good report but not complete. See a couple of my comments. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:25, 29 May 2020 (BST)}}

MRD:M4H1M4

2020-05-29T18:22:56Z

Mak214: /* PES inspection */

MRD:YX8818CX

2020-05-29T18:20:12Z

Mak214: /* Part 2: F - H - H system */

== Part 1: H + H2 system ==

=== Dynamics from the transition state region ===
'''In part 1, r1 = rAB, r2 = rBC.'''

{{fontcolor|blue|Q1: 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?}}

[[File:Saddle point.svg|thumb|right|300px|A saddle point (in red) on the graph of z=x2−y2 Reference: Wikipedia: Saddle point https://https://en.wikipedia.org/wiki/Saddle_point]]

On a potential energy (PE) surface, the transition state is a saddle point. At that point the gradient of the potential is zero. Mathematically speaking, a saddle point is a point on a surface of a function graph where the slopes (derivatives) in orthogonal directions are all zero, but which is not a local maximum or minimum of the function (as illustrated in the figure<ref>''Wikipedia, The Free Encyclopedia'', s.v. "Saddle point," (accessed May 20, 2020), https://en.wikipedia.org/wiki/Saddle_point</ref> on the right).

The saddle point can be determined mathematically by calculating the Hessian matrix for the function at that point.<ref>Smith, Colin M. "How to find a saddle point." ''International journal of quantum chemistry'' 37.6 (1990): 773-783.</ref> For a function f(x,y) of two variables, a point is a saddle point if:

If fx = 0, fy = 0, and fxxfyy - f2xy < 0. Where fx and fxx are the first and second derivative of f with respect to x respectively.

The transition state (the saddle point) can be identified in a three-dimensional PE plot by looking at the surface at both direcions (r1 and r2 ). A saddle point will be a maximum in one direction and a minimum in the other direction. It can be easily distinguished from a local minimum, as a local minimum will be a minimum in all directions.<ref>Henkelman, Graeme, Gísli Jóhannesson, and Hannes Jónsson. "Methods for finding saddle points and minimum energy paths." ''Theoretical methods in condensed phase chemistry''. Springer, Dordrecht, 2002. 269-302.</ref> {{fontcolor1|red| Great! [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:09, 29 May 2020 (BST)}}

==== Trajectories from r1 = r2: locating the transition state ====
{{fontcolor|blue|Q2: 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.}}

My best estimate of the TS position rts is 90.775 pm.

As p1 = p2 = 0, there is no initial momentum, hence no kinetic energy. So if the H + H2 surface start at the transition state, there will be no oscillation seen in the Distance vs Time plot, thus rAB and rBC will be constant. When r1 = r2 = 90.775 pm, there is negligible oscillatory behaviour displayed in the Distance vs Time plot and zero force along AB and BC, as shown in Figures. This indicates the transition state position ('''rts''') is very close to this value. {{fontcolor1|red| Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:09, 29 May 2020 (BST)}}
{| class="wikitable"
|[[File:TS_force_zero_YX8818.PNG|thumb|center|400px|Settings for rAB and rBC when forces is zero]]
|[[File:Q2_Distance_vs_time_plot_TS_YX8818.png|thumb|center|400px|Internuclear Distances vs Time Plot]]
|}

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

==== Trajectories from r1 = rts+δ, r2 = rts ====
{{fontcolor1|blue|Q3: Comment on how the ''mep'' and the trajectory you just calculated differ.}}

1 decimal place of the transition state position here was used for simplicity, i.e. rts = 91.8 pm

Using r1 = rts+1 = 91.8 pm, r2 = rts = 90.8 pm

As shown from the contour and surface plots below, The minimum energy path (''mep'') corresponded to infinitely slow motions, with velocities / momenta reset to zero for each step. Thus, the trajectory corresponds to infinitely slow motion, and simply follows the valley floor, no oscilatiory behaviour is visible. While the mep plot is useful in characterising the reaction, it does not give an realistic account of the atomic motions.

In comparison, the dynamics surface plot gave a more accurate description of atomic motions (inertial) during the reaction. Oscillations of the potential for the H atoms were observed for dynamics plots. {{fontcolor1|red| Yes. A clear response, thank you. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:09, 29 May 2020 (BST)}}

{| class="wikitable"
|-
! Header 1
! Contour plot
! Surface plot
|-
| Dynamics
| [[File:Q3_contour_plot_dynamics_YX8818.png|thumb|center|300px|Contour plot using Dynamics calculation type]]
| [[File:Q3_surface_plot_dynamics_YX8818.png|thumb|center|300px|Surface plot using Dynamics calculation type]]
|-
| MEP
| [[File:Q3_contour_plot_MEP_YX8818.png|thumb|center|300px|Contour plot using MEP calculation type]]
| [[File:Q3_surface_plot_MEP_YX8818.png|thumb|center|300px|Surface plot using MEP calculation type]]
|}

==== Additional Questions: ====
'''A1'': Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions r1 = rts and r2 = rts+1 pm instead?'''''

As shown in the “Internuclear Distances vs Time” and “Momenta vs Time” plots below, when r1 = rts+1, r2 = rts, the reaction proceed to form the products, and H2 molecule BC and H atom A were formed. This is because the transition state was displaced slightly towards the products.

On the other hand, when r1 = rts, r2 = rts+1, the reaction proceed to form the reactants, and molecule AB and atom C were formed.This is because the transition state was displaced slightly towards the reactants.

{| class="wikitable"
|-
!initial conditions
!r1 = rts+1, r2 = rts
!r1 = rts, r2 = rts+1
|-
|Internuclear Distances vs Time Plot
| [[File:internuclear_vs_time_r2TS_YX8818.png|300px]]
|[[File:internuclear_vs_time_r1TS_YX8818.png|300px]]
|-
|Momenta vs Time Plot
| [[File:Momenta_vs_time_r2TS_YX8818.png|300px]]
| [[File:Momenta_vs_time_r1TS_YX8818.png|300px]]
|}

'''A2: Note final values of the positions r1(t) r2(t) and p1(t) p2(t) for your trajectory for large enough t .'''

'''t = 100 fs '''was chosen, and for r1 = rts, r2 = rts+1, ''' r1'''(t) = 731 pm, '''r2'''(t) = 75.5 pm, '''p1'''(t) = 5.1 g.mol-1.pm.fs-1, '''p2'''(t) =2.5 g.mol-1.pm.fs-1.

'''A3'': ''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. What do you observe?'''

When momentum was reversed, the trajectory traced back to the original position (slightly towards the product from the TS). But as it does not have enough energy to overcome the energy barrier at the TS, it traced back to the product, H2 molecules BC and H atoms A.

[[File:A3_momentum_reversed_YX8818.png|thumb|center|350px|Contour plot when monenta were reversed]]
''<nowiki/>''

=== Reactive and unreactive trajectories ===
The following simulations were investigated with different reaction conditions (momentum) to see whether they resulted in reactive trajectory.
* Initial positions '''r1''' = 74 pm and '''r2''' = 200 pm.

{| class="wikitable"
!No
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot ''/''kJ. mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory:
contour plot
!Illustration of the trajectory:

surface plot
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|The distance rBC decreases as the molecule AB approaches H atom C, and rBC reaches minium as the system reaches the transition state (TS). The system has sufficient energy to overcome the energy barrier, and the products BC molecules is formed (indicated by relatively constant small rBC) and increasing rAB as the H atom A moves away.
|[[File:Q4_1_Contour_YX8818.png|250px]]
|[[File:Q4_1_Surface_YX8818.png|300px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The distance rBC decreases as the molecule AB approaches H atom C, but upon collision there is insufficient energy to overcome the energy barrier at TS. Hence, the molecule AB remains, and atom C moves away (rBC increases)
|[[File:Q4_2_Contour_YX8818.png|250px]]
|[[File:Q4_2_Surface_YX8818.png|300px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The distance rBC decreases as the molecule AB approaches H atom C and rBC reaches minium as the system reaches the transition state (TS). The system has sufficient energy to overcome the energy barrier, the products BC molecules is formed (indicated by relatively constant small rBC) and increasing rAB as the H atom A moves away.
|[[File:Q4_3_Contour_YX8818.png|250px]]
|[[File:Q4_3_Surface_YX8818.png|300px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|The distance rBC decreases at first when the molecule AB approaches H atom C and reaches the transition state (TS). As the system has sufficient energy, it was able to cross the TS and from the product BC molecule and atom A. However, due to the excess kinetic energy / momentum the system possess, the system recrossed the energy barrier at the TS again, and refromed the reactants molecule AB, and atom C.
|[[File:Q4_4_Contour_YX8818.png|250px]]
|[[File:Q4_4_Surface_YX8818.png|300px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|The excess kinetic energy / momentum of the system allows it to cross the energy barrier at the TS to the product side, recrossed to the reactant side, and then crossed the energy barrier again to finallay form the refromed the products, molecule BC, and atom A.
|[[File:Q4_5_Contour_YX8818.png|250px]]
|[[File:Q4_5_Surface_YX8818.png|300px]]
|}
{{fontcolor1|blue|Q4: 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 hypthesis tested with the above simulation was: For a reactive trajectory, all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive, as they have enough kinetic energy to overcome the activation barrier.

From the simulations of the trajectory, it is found that the hypothesis is false. This is supported by simulation 4, which have momenta higher than some reactive trajectories, but was not reactive. This shows that not all all trajectories starting with the same positions but with higher values of momenta would be reactive. The reactivity of a trajectory thus depend on not just the total energy in the system but also on how the energy is distributed.

==== Transition State Theory ====
{{fontcolor1|blue|Q5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

During a molecular collision, we must consider two molecules to form a single quantum-mechanical entity, which is called a supermolecule.<ref name=":1" />

The Transition State Theory (TST)<ref name=":1">Daniels, Farrington, J. Howard Mathews, and John Warren Williams. ''Experimental physical chemistry''. No. 541 D35 1962. New York: McGraw-Hill, 1962, chap 22.</ref> <ref>Henriksen, Niels E., and Flemming Y. Hansen. ''Theories of molecular reaction dynamics: the microscopic foundation of chemical kinetics''. Oxford University Press, 2018.</ref>chooses a boundary surface located between the reactant and product regions and assumes that all supermolecules that cross this boundary surface become products. The boundary surface, called the (critical) dividing surface, is taken to pass through the saddle point of the potential-energy surface.

TST has the following three main assumptions:

1. All supermolecules that cross the critical dividing surface from the reactant side becomes product.

2. During the reaction, the Boltzmann distribution of energy is maintained for the reactant molecules.

3. The supermolecules crossing the critical surface from the reactant side have a Boltzmann distribution of energy corresponding to the temperature of the reacting system.

Experimental results may violate the assumption 1, as in reality reactants and products are in equilibrium, thus allowing both forward and backward reactions to take place, so not all supermolecules that crossed the critical dividing surface from the reactant side will becomes product, causing experimental rates to be lower than TST prediction.This is supported by the simulations findings above.

It was found that majority of the results obtained (Set 1, 2, 3, and 5) agrees with the Transition State Theory, as the supermolecules with sufficient momenta to cross the critical dividing surface from the reactant side becomes product, and those that did not crossed the critical dividing surface remain as reactants. However, the results from Set 4, where the supermolecule recrossed the boundary, contradicted Assumption 1 of TST, showing that some supermolecules that crossed the critical dividing surface containing the transition state, can rebound and reformed the reactants. This shows that TST prediction for reaction rate values might not accurately agrees with experimental values for some cases.

Furthermore, as TST is a classical theory, it does not factor in the quantumn tunneling effect, which could result in the experimental rates to be higher the theory.

Note: the simulations above does not allow investigation in to the Boltzmann distribution of energy in the TST assumption as the system investigated consist of only two reactants, hence is only a small part on the Boltzmann distribution. {{fontcolor1|red| Great [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:46, 29 May 2020 (BST)}}

== Part 2: F - H - H system ==
'''In part 1, rHF = r1 = rAB, rHH = r2 = rBC.'''

=== PES inspection ===
{{fontcolor1|blue|Q6: 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?}}

The potential energy surface for F + H2 → HF + H prcess is shown below:
[[File:Q6_PE_surface_F_H2_YX8818.png|thumb|right|350px|Potential energy surface for F + H2 → H + HF prcess ]]

F + H2 → HF + H : Exothermic.

This can be shown from the lowering of potential energy going from reactant to products.

HF + H → F + H2: Endothermic.

This can be shown from the increase in energy going from reactants to products going in the reverse direction.

Bond strength of HF = 565 kJ.mol-1<ref name=":0">Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958; B. deB. Darwent, "National Standard Reference Data Series," National Bureau of Standards, No. 31, Washington, DC, 1970; S.W. Benson, J. Chem. Educ., 42, 502 (1965).</ref>

Bond strength of H2 = 432 kJ.mol-1<ref name=":0" />

The stronger bond strength of HF bond is a result of the high electronegativity of the F atom, which results in high bond polarisation in the HF bond and a strong ionic / electrostatic attraction between H and F atoms.

F + H2 → HF + H, ΔH = -565 + 432 = -133 kJ.mol-1, which supports it being an exothermic process as energy is being released overall. This is because the energy required to break the H-H bond is less than the energy released in the formation of H-F bond.

HF + H → F + H2 , ΔH = 565 - 432 = 133 kJ.mol-1, which supports it being an enothermic process as energy is being absorbed overall. This is because the energy required to break the H-F bond is more than the energy released in the formation of H-H bond.

{{fontcolor1|red| Exemplary answer. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:20, 29 May 2020 (BST)}}

{{fontcolor1|blue|Q7: Locate the approximate position of the transition state.}}

By Hammond postulate, an exothermic reaction of F + H2 to HF + H has an early transition state that resemble the reactant.

Thus, the distance rAB between F and H2 molecule should be large, while the H2 molecule distance rBC should be close the H2 bond distance.

Using the same method in Q2, the transition state position was found to be as follows:

F-H distance, rAB = 181.13 pm.

H-H distance, rBC = 74.45 pm.

{| class="wikitable"
|[[File:Q7_TS_zero_force_YX8818.png|thumb|center|350px|Settings for rAB and rBC when forces is zero]]
|[[File:Q7_TS_distance_vs_Time_YX8818.png|thumb|center|350px|Internuclear Distances vs Time Plot]]
|[[File:Q7_TS_contour_plot_YX8818.png|thumb|350px|Contour plot]]
|}

{{fontcolor1|blue|Q8: Report the activation energy for both reactions.}}

The activation energy of the forward and backward reactions where found by locating the energy of the transition state, reactants, and product, then determining the difference. The reactant and product formation trajectories were obtained by perturbing the system slightly away from the transition state. Results are summarised in the tables below:

{| class="wikitable"
!system
!rHF / pm
!rHH / pm
!Total Energy / kJ.mol-1
!Activation EnergyTotal Energy / kJ.mol-1
|-
|transition state: F-H-H
|181.13
|74.45
|<nowiki>-433.978</nowiki>
|<nowiki>---</nowiki>
|-
|reactant to product:
F + H2 → HF + H
|180.13
|74.45
|<nowiki>-560.328</nowiki>
|<nowiki>-433.978 - (-560.328) = 126.350</nowiki>
|-
|product to reactant:
HF + H → F + H2
|182.95
|74.45
|<nowiki>-435.015</nowiki>
|<nowiki>-433.978 - (-435.015) = 1.037</nowiki>
|}
{| class="wikitable"
!system
!Energy vs time plot (full scale)
!Energy vs time plot (zoomed)
!Contour plot
|-
|transition state: F-H-H
|<nowiki>---</nowiki>
|<nowiki>---</nowiki>
|<nowiki>---</nowiki>
|-
|reactant to product:
F + H2 → HF + H
|[[File:Q8_forming_HF_Activation_energy_E_vs_time_YX8818.png|300px]]
|[[File:Q8_zoomed_HF_Activation_energy_E_vs_time_YX8818.png|300px]]
|[[File:Q8_forming_HF_Contour_Plot_YX8818.png|300px]]
|-
|product to reactant:
HF + H → F + H2
|[[File:Q8_original_forming_reactant_H2_Activation_energy_E_vs_time_YX8818.png|300px]]
|[[File:Q8_forming_reactant_H2_Activation_energy_E_vs_time_YX8818.png|300px]]
|[[File:Q8_forming_reactant_H2_contour_YX8818.png|300px]]
|}

{{fontcolor1|red| Good work well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:20, 29 May 2020 (BST)}}

=== Reaction dynamics ===

The following is a set of reaction conditions that resulted in a reactive trajectory for F + H2
{| class="wikitable"
!rFH/pm
!rHH / pm
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Illustration of the trajectory
|-
|200
|75
|<nowiki>-1.5</nowiki>
|0.5
|[[File:Reactive_HFH_conditions_YX8818.png|350px]]
|}
{| class="wikitable"
!Energy vs time plot
!Momentum vs time plot
|-
|[[File:Reactive_HFH_energy_vs_time_YX8818.png|400px]]
|[[File:Reactive_HFH_momenta_vs_time_YX8818.png|400px]]
|}

{{fontcolor1|blue|Q9: 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.}}

Before the reaction took place (t = 0 to 60 s), the F atom only has translational energy, while H2 molecule has both translational and vibrational energy (as shown from the oscillatory behavior in momentum vs time diagram). Before the reaction, the total kinetic and potential energy were relatively constant (at 0 kJ.mol-1 and -433 kJ.mol-1 respectively).

After the exothermic reaction (at around t = 60 s), F + H2 → HF + H, some of the potential energy was converted into kinetic energy of the system, with the total energy remained constant (conservation of energy), as shown in the Energy vs time plot. The kinetic energy could be in the form of translational and vibrational (rotational and electronic energies were ignored here). For the HF molecule formed, its kinetic energy / momentum was in both translation and vibrational; while for the H atom, only translation energy was left. The HF molecule was observed to have a larg bond oscillation along the H-F bond, which resulted in larger oscillations in kinetic and potential energy after the reaction.

The overall release of energy / heat of this exothermic reaction could be monitored using calorimetry.<ref>Sunner, Stig, and Margaret Mansson. "Experimental chemical thermodynamics. Volume I. Combustion calorimetry." (1979).</ref> For a reaction involving gaseous reactants, constant-volume calorimeters, such as the bomb calorimeter, is suitable. The reactants can be placed inside a steel vessel with know heat capacity and sealed within an insulated container with known amount of water. The heat released could then be calculated from the temperature change.

In particular, the vibrational energy can be measured using IR spectroscopy, where absorption spectrum can be monitored over time. Before the reaction, the reactants mainly occupied the ground state (lowest vibrational energy level), there would essentially be one absorption peak from 0 to 1. During the exothermic reaction, the molecules will gain vibrational kinetic energy and be excited to occupy the first excited state, allowing overtones (1 to 2) to be observed at lower wavenumbers, thus resulting in two absorption peaks. As the reaction takes place, the intensity of the overtone will be come larger, and the intensity of the fundamental would decrease. After the reaction is complete, the vibrational modes will gradually relax back to the ground state, causing the overtones to disappear, and the fundamental to increase.

Emission of light as a result of this reaction, chemiluminescence,<ref>Dodeigne, C., L. Thunus, and R. Lejeune. "Chemiluminescence as diagnostic tool. A review." ''Talanta'' 51.3 (2000): 415-439.</ref> could also be used to monitor this process. The process monitored here would be the decay of the excited state to the lower energy level (ground state). {{fontcolor1|red| Nice. Thank you for doing some extra reading to give an interesting answer. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:20, 29 May 2020 (BST)}}

=== Further studies ===
Further invesitgations were conducted to investigate the effect of distribution of energy among translational and vibrational modes on the efficiency of the reaction:
* F + H2 → HF + H. Initial conditions: '''r1''' = rHF = 200 pm, '''r2''' = rHH = 74 pm, p1 = pFH = -1.0 g.mol-1.pm.fs-1, pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 .
* '''Table 1:'''
{| class="wikitable"
!pFH/ g.mol-1.pm.fs-1
!pHH/ g.mol-1.pm.fs-1
!Reactive?
|-
|<nowiki>-1.0 </nowiki>
|<nowiki>-6.0</nowiki>
|No
|-
|<nowiki>-1.0 </nowiki>
|<nowiki>-3</nowiki>
|No
|-
|<nowiki>-1.0 </nowiki>
|<nowiki>-2</nowiki>
|No
|-
|<nowiki>-1.0</nowiki>
|<nowiki>-1</nowiki>
|Yes
|-
|<nowiki>-1.0</nowiki>
|0
|Yes
|-
|<nowiki>-1.0</nowiki>
|1
|Yes
|-
|<nowiki>-1.0</nowiki>
|2
|No
|-
|<nowiki>-1.0 </nowiki>
|4
|No
|-
|<nowiki>-1.0</nowiki>
|5.5
|No
|}
* F + H2 → HF + H. For the same initial positions, simulations below were conducted.
* '''Table 2:'''
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Reactive?
|-
|<nowiki>-1.5</nowiki>
|0.2
|Yes
|-
|<nowiki>-1.6</nowiki>
|0.2
|Yes
|-
|<nowiki>-6.0</nowiki>
|0.2
|Yes
|}

* For the reverse reaction, HF + H → F + H2. '''r1''' = rHF = 180 pm and '''r2''' = rHH = 74 pm, the following simulations were conducted.
* '''Table 3:'''
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Reactive?
|-
|<nowiki>-8.0</nowiki>
|1
|Yes
|-
|<nowiki>-4.0</nowiki>
|1
|Yes
|-
|<nowiki>-1.0</nowiki>
|1
|No
|-
|<nowiki>-1.0</nowiki>
|<nowiki>-3.0</nowiki>
|No
|}

{{fontcolor1|blue|Q10: 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.}}

The effect of energy distribution on reaction efficiency can be described by '''Polanyi's empirical rules.'''<ref>D. M. Hirst Potential Energy Surfaces: Molecular Structure and Reaction Dynamics Taylor and Francis, 1985, Chap 6.3. </ref>

Polanyi's empirical rules describes how different forms of energy affect the rates of reactions. The rules state that for a given amount of energy / momentum, the vibrational energy is more efficient than translational energy in activating an late barrier / endothermic reactio; whereas translational energy is more efficient than vibrational energy for an early barrier / exothermic reaction.

'''F + H2 → HF + H is an exothermic reaction with an early barrier, thus the trajectory with high translational energy (pFH) and low vibrational energy (pHH) is reactive.'''

This is suported by the simulations in Tables 1 and 2, where trajectories with high pFH and low pHH are generally reactive, whereas trajectories with low pFH and high pHH are generally unreactive. This is because the momentum in F atom (pFH) is almost entirely translational, whereas the energy in H2 molecules (pHH) is both translational and vibrational, and the vibrational energy can be thought of as proportional to the total energy present. The trends in the table largely suports Polanyi's rules that translational energy is more efficient in early barrier.

'''HF + H → F + H2 is an endothermic reaction, thus the trajectory with high vibrational energy (pFH) and low translational energy(pHH) is reactive.'''

Trends shown in Table 3 also largely supports Polanyi's rules that vibrational energy is more efficient for late barriers. As when higher vibrational energy (pFH) and lower translational energy(pHH) tend to give reactive trajectories.

However, its worthy to note that in the above simulations, the total energy was not kept constants while varying the proportions of the translational and vibrational energy. And the trajectories of the simulations are quite sensitives, sometimes giving conflicting results for certain values of initial conditions. Further explorations could be considered by keeping the total energy / momentum in the one-dimensional trajectory constant, perhaps by varying the collison angle or using more sophisticated simulation package. {{fontcolor1|red| Well done. A thorough, thoughtful report thank you. It was interesting to see everything that you did to explore the program you were given. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:20, 29 May 2020 (BST)}}

= References =

MRD:xmh01513932

2020-05-29T18:16:19Z

Mak214: /* 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. */

== 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? ==
On a potential energy surface diagram, the transition state is identified as the maximum on the minimum potential energy path between the reactants and products.

To explain the transition state mathematically, several derivatives have to be addressed first,
1. The first derivative of potential energy (V) with respect to either atomic separation (r1, r2) is zero, i.e. ∂V/∂r1 = 0 and ∂V/∂r2 = 0;
2. The second derivative D = ∂2V/∂r1∂r1 × ∂2V/∂r2∂r2 – ∂2V/∂r1∂r2 = 0;
3. If D > 0, ∂2V(r1)/∂r1∂r1 > 0, then the point with this particular (r1,r2) coordinates is a local minimum point; ∂2V(r1)/∂r1∂r1 < 0 for a local maximum point;
4. If D < 0, then this point is a saddle point, i.e. the transition state. Simply put, the saddle point is simultaneously a minimum and a maximum along two orthogonal directions (i.e. the two atomic separation axes in this case). {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:11, 29 May 2020 (BST)}}
[[File:Figure 1.log|250px|thumb|center|Figure 1. Contour Plot.]]
[[File:xmhFigure 2.png|250px|thumb|center|Figure 2. Surface Plot.]]

{{fontcolor1|red| What did you want to show with these plots? Use figures to illustrate what you are saying in the text, and make sure you discuss every figure you include in your text. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:12, 29 May 2020 (BST)}}

== 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. ==
For a symmetric system like H + H2, the internuclear distances of r1 = r2 is expected in the transition state. To test out the best input values for r1 and r2, there would be some expected features on various plots, as shown below,
1． In Figure 3., the kinetic energy of the system is constant at 0. This is explained by the fact that, at the transition state, the net force acting on the system F = dp/dt = ∂V/∂r1 = ∂V/∂r2 = 0 (as defined), the triatomic system is at a state of equilibrium with no change in motion. The same idea is conveyed in Figure 4., where the internuclear distances are expected to have a variation as small as possible, the atoms are “frozen” in space, moreover, rAB(r2) = rBC(r1).
2． In Figure 5., the transition state position (rts) is also checked by the fact that the input r1 and r2 values (denoted by the red cross) map onto the circle which gives the coordinates of the exact rts. {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:13, 29 May 2020 (BST)}}
The final value of r1 = r2 = 90.7 pm is chosen to satisfy the above requirements.
[[File:xmhFigure 3.png|250px|thumb|center|Figure 3. Energy vs Time Plot.]]
[[File:xmhFigure 4.png|250px|thumb|center|Figure 4. Internuclear Distances vs Time Plot.]]
[[File:xmhFigure 5.png|250px|thumb|center|Figure 5. Contour Plot]]

== Comment on how the mep and the trajectory you just calculated differ. ==
Two paths were generated under the calculation types of MEP and Dynamics respectively, with initial conditions of r1 = rts+1=91.7 and r2 = 90.7 and p1 = p2 = 0 g.mol-1.pm.fs-1.
The dynamic graph shows a wavy line (shown in Figure 6.). After passing though rts (now r1 < r2), HAHB molecule is formed, detaching HC. This case takes into account of the intramolecular vibration of the new H2 molecule, since there is an oscillating trajectory at around a fixed rAB value with an increasing rBC value which indicates detaching of HC.
[[File:xmhFigure 6.png|250px|thumb|center|Figure 6. Contour Plot.]][[File:xmhFigure 7.png|250px|thumb|center|Figure 7. Energy vs Time Plot.]]
[[File:xmhFigure 8.png|250px|thumb|center|Figure 8. Momentum vs Time Plot]][[File:xmhFigure 8.png|250px|thumb|center|Figure 9. Internuclear Distances vs Time Plot]]
The mep (the minimum energy path) corresponds to a trajectory with infinitely slow motion, i.e. the system’s velocities are reset to zero in each time step, consequently no momentum or kinetic energy, as reflected in Figure 11 & 12. The reaction path way as a smooth curve (shown in Figure .) without any information of the intrinsic vibration within the H2 molecule.
[[File:xmhFigure 10.png|250px|thumb|center|thumb|Figure 10. Contour Plot.]]
[[File:xmhFigure 11.png|250px|thumb|center|Figure 11. Energy vs Time Plot.]]
[[File:xmhFigure 12.png|250px|thumb|center|Figure 12. Momentum vs Time Plot]]
[[File:xmhFigure 13.png|250px|thumb|center|Figure 13. Internuclear Distances vs Time Plot]]
If initial conditions are changed to r2 = rts+1=91.7 and r1 = 90.7 and p1 = p2 = 0 g.mol-1.pm.fs-1, same trends in both the Internuclear Distances vs Time graph and the Momenta vs Time graph are observed as for the previous conditions. Since the system is symmetric, hence the differences between these 2 sets of conditions are just attributed to the fact that whether HAHB or HBHC dissociates. {{fontcolor1|red| Great. Good use of figures here. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:14, 29 May 2020 (BST)}}

== 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? ==
{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Reactive || HA approaches HBHC molecule from a distance of rAB = 200 (starting from the bottom-right of the figure), with rBC being constant since HB and HC still remain bonded. After passing through rts, the new HAHB molecule is formed as shown by the constant rAB and increasing rBC as HC leaves(at the top left of the figure). || [[File:xmhA.png|250px]]
|-
| -3.1 || -4.1 || -420.077 || Unreactive || HA approaches HBHc (bottom-right), however it does not quite get to rts due to a reduced amount of kinetic energy compared to the first case, hence HA bounces off, there is no interaction with the hydrogen molecule. ||[[File:xmhB.png|250px]]
|-
| -3.1 || -5.1 || 413.977 || Reactive || Same as the first case except for the fact that p1 has a larger value, hence the system is more energetic to go over the activation barrier, and due to energy conservation, after the collision, the new molecule has a greater vibration. ||[[File:xmhC.png|250px]]
|-
| -5.1 || -10.1 || -357.277 || Unreactive || Barrier recrossing - the system is energetic enough to go over the transition state region and form the new HAHB, since rBC increases from its original bond length, indicating the leaving of HC, and the Etot is also greater than the first case (a reactive case). However, rBC drops down again to the original bond length and rAB increases as the system reverts back to HBHC. ||[[File:xmhD.png|250px]]
|-
| -5.1 || -10.6 || -349.477 || Reactive || The system HBHC converts to HAHB after reaching rts, but reverts given the excess momentum in p2 as rAB increases and rBC decreases, the system crosses the transition region again to form the final HAHB molecule as rBC continue increasing and rAB stays oscillating around a fixed value. ||[[File:xmhE.png|250px]]
|}
From the first sight, it seems reasonable to encourage a successful collision by adding more momentum into the system since this increases the kinetic energy of the atoms to overcome the activation barrier. However, as concluded from the table, the excess momentum and energy could give rise to greater vibration within the molecule and hence cause the bond to dissociate. {{fontcolor1|red| Great. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:46, 29 May 2020 (BST)}}

== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ==
In the transition state theory, the average transmission rate/frequency is estimated based on the assumption that all trajectories with some kinetic energy greater than the activation energy will be reactive. From the empirical data of the last section, it was shown that if momentum is provided such that the vibration is so great that the bond dissociates again, i.e. barrier recrossing occurs, this reduces the rate of a particular conversion, e.g. from HBHC and HA to HAHB and HC. This would overestimate the transition state theory rate constant (and hence the rate).
The transition state theory also consider the reaction pathway through classical mechanic collision (for macroscopic system). However, wavefunction (for microscopic system) of the atoms should be considered. Quantum tunneling may take place, it is not necessary to jump over the activation barrier, hence less kinetic energy is required for a successful collision, rate is higher than the theory prediction. {{fontcolor1|red| Yes. Include references. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:47, 29 May 2020 (BST)}}

== 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? ==
In GUI, it is set up that the atom A approaches and collide with the molecule HBHC.
In the F+H2 system, F is considered as atom A, and the 2 H atoms as atoms B and C. Figure 14(A) shows the entrance channel for the reactants F+H2, since rBC (the molecule’s bond length) is constant and rAB is big; Figure 14(B) shows the exit channel for the products (FH+H), since rAB is now constant and rBC is big. Comparing the V for the 2 channels, the reactants are at a higher V than the products, hence this is an exothermic reaction.
[[File:xmhFigure 14(A).png|250px|thumb|center|Figure 14(A). Surface Plot.]]
[[File:Figure 14(B)xmh.png|250px|thumb|center|Figure 14(B). Surface Plot.]]
For the H+FH system, the same idea applies, from Figure 15(A). & (B)., the reactants are at a lower V than the products, hence it is an endothermic reaction.
[[File:Figure 15(A)mh.png|250px|thumb|center|Figure 15(A). Surface Plot.]]
[[File:Figure 15(B)xmh.png|250px|thumb|center|Figure 15(B). Surface Plot.]]
This correctly reflects the result obtained from qualitative bond-strength analysis. F-H is a very strong bond (compared to other possible bonds in the system) due to the different electronegativities and hence ionic contribution to the nature of the bond. Hence breaking this bond requires much energy, and forming it releases much energy. In the F+H2 reaction, a pure covalent bond is broken to form the strong F-H bond, this is expected to be exothermic; in the H+FH system, the energy released from forming H2 does not compensate for the energy required to break F-H first, hence it is expected to be endothermic. {{fontcolor1|red| Nice discussion and well illustrated. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:12, 29 May 2020 (BST)}}

== Locate the approximate position of the transition state. ==
The approximate position of the transition state was investigated in the case of F+H2, where F is considered as atom A, and the 2 H atoms as atoms B and C. It was found that rAB = 181.1 and rBC = 74.5. This is reflected by Figure 20., where the internuclear distances are constant between all 3 atoms, corresponding to equilibrium at the transition state. Etot = - 433.980.
For the case of H+HF, same rts was found, since this reaction is just the reverse reaction of the above case, and the same geometry is found at the transition state. {{fontcolor1|red| Great. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:14, 29 May 2020 (BST)}}
[[File:xmhFigure 16.png|250px|thumb|center|Figure 16. Contour Plot.]]
[[File:Figure 17xmh.png|250px|thumb|center|Figure 17. Internuclear Distancesvs Time Plot.]]

== Report the activation energy for both reactions. ==
In order to find the reactants’ energy for the F+H2 reaction, initial conditions were set as rAB = 1000 and rBC = 74.5. The value of rAB was chosen so that there is no interaction in the system since the F atom is at a very distant away from the H2 molecule, as reflected by Figure 20., where the only momentum is given by the oscillation between the bonded molecule. Etot = - 435.057.
Hence the activation energy = Etot(transition state) – Etot(reactants) = - 433.980 + 435.057 = 1.077.
[[File:xmhFigure 18.png|250px|thumb|center|Figure 18. Momentum vs Time Plot.]]
Same method applies to the H+FH system, the activation energy = 125.322.
{{fontcolor1|red| Good, well done. I think a different type of plot would have been more illustrative here, personally. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:13, 29 May 2020 (BST)}}

== 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. ==
There are 3 types of energies associated with the system: potential energy due to the interactions between the atom and the molecule, vibrational kinetic energy and translational kinetic energy. After a successful collision, these energies would redistribute based on the mass and bond strength of the atom and molecule, some energy might excite the new molecule into a higher vibrational state. However, the molecule would eventually drop down to the ground state, releasing energy.
The IR overtone region could be used to confirm the presence of the transition between higher states, the peaks appear at higher wavelength (higher energy). Consequently, this confirms the following release of energy. {{fontcolor1|red| Yes. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:14, 29 May 2020 (BST)}}

== 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. ==
According to Polanyi's rules, the reactants' translational energy activates them more effectively to the activation barrier and hence increases rate for exothermic reactions; conversely, the reactants' vibrational energy activates more effectively for endothermic reactions.
For the exothermic F+H2 system, the translational energy is more important to give a successful collision.
For the endothermic H+HF system, the vibrational energy is more important. {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:16, 29 May 2020 (BST)}}

{{fontcolor1|red| Overall a nice report but missing references. Get into the habit of using literature references to support your discussion in your work. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:16, 29 May 2020 (BST)}}

MRD:xmh01513932

2020-05-29T18:14:55Z

Mak214: /* 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. */

MRD:xmh01513932

2020-05-29T18:14:09Z

Mak214: /* Locate the approximate position of the transition state. */

== 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? ==
On a potential energy surface diagram, the transition state is identified as the maximum on the minimum potential energy path between the reactants and products.

To explain the transition state mathematically, several derivatives have to be addressed first,
1. The first derivative of potential energy (V) with respect to either atomic separation (r1, r2) is zero, i.e. ∂V/∂r1 = 0 and ∂V/∂r2 = 0;
2. The second derivative D = ∂2V/∂r1∂r1 × ∂2V/∂r2∂r2 – ∂2V/∂r1∂r2 = 0;
3. If D > 0, ∂2V(r1)/∂r1∂r1 > 0, then the point with this particular (r1,r2) coordinates is a local minimum point; ∂2V(r1)/∂r1∂r1 < 0 for a local maximum point;
4. If D < 0, then this point is a saddle point, i.e. the transition state. Simply put, the saddle point is simultaneously a minimum and a maximum along two orthogonal directions (i.e. the two atomic separation axes in this case). {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:11, 29 May 2020 (BST)}}
[[File:Figure 1.log|250px|thumb|center|Figure 1. Contour Plot.]]
[[File:xmhFigure 2.png|250px|thumb|center|Figure 2. Surface Plot.]]

{{fontcolor1|red| What did you want to show with these plots? Use figures to illustrate what you are saying in the text, and make sure you discuss every figure you include in your text. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:12, 29 May 2020 (BST)}}

== 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. ==
For a symmetric system like H + H2, the internuclear distances of r1 = r2 is expected in the transition state. To test out the best input values for r1 and r2, there would be some expected features on various plots, as shown below,
1． In Figure 3., the kinetic energy of the system is constant at 0. This is explained by the fact that, at the transition state, the net force acting on the system F = dp/dt = ∂V/∂r1 = ∂V/∂r2 = 0 (as defined), the triatomic system is at a state of equilibrium with no change in motion. The same idea is conveyed in Figure 4., where the internuclear distances are expected to have a variation as small as possible, the atoms are “frozen” in space, moreover, rAB(r2) = rBC(r1).
2． In Figure 5., the transition state position (rts) is also checked by the fact that the input r1 and r2 values (denoted by the red cross) map onto the circle which gives the coordinates of the exact rts. {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:13, 29 May 2020 (BST)}}
The final value of r1 = r2 = 90.7 pm is chosen to satisfy the above requirements.
[[File:xmhFigure 3.png|250px|thumb|center|Figure 3. Energy vs Time Plot.]]
[[File:xmhFigure 4.png|250px|thumb|center|Figure 4. Internuclear Distances vs Time Plot.]]
[[File:xmhFigure 5.png|250px|thumb|center|Figure 5. Contour Plot]]

== Comment on how the mep and the trajectory you just calculated differ. ==
Two paths were generated under the calculation types of MEP and Dynamics respectively, with initial conditions of r1 = rts+1=91.7 and r2 = 90.7 and p1 = p2 = 0 g.mol-1.pm.fs-1.
The dynamic graph shows a wavy line (shown in Figure 6.). After passing though rts (now r1 < r2), HAHB molecule is formed, detaching HC. This case takes into account of the intramolecular vibration of the new H2 molecule, since there is an oscillating trajectory at around a fixed rAB value with an increasing rBC value which indicates detaching of HC.
[[File:xmhFigure 6.png|250px|thumb|center|Figure 6. Contour Plot.]][[File:xmhFigure 7.png|250px|thumb|center|Figure 7. Energy vs Time Plot.]]
[[File:xmhFigure 8.png|250px|thumb|center|Figure 8. Momentum vs Time Plot]][[File:xmhFigure 8.png|250px|thumb|center|Figure 9. Internuclear Distances vs Time Plot]]
The mep (the minimum energy path) corresponds to a trajectory with infinitely slow motion, i.e. the system’s velocities are reset to zero in each time step, consequently no momentum or kinetic energy, as reflected in Figure 11 & 12. The reaction path way as a smooth curve (shown in Figure .) without any information of the intrinsic vibration within the H2 molecule.
[[File:xmhFigure 10.png|250px|thumb|center|thumb|Figure 10. Contour Plot.]]
[[File:xmhFigure 11.png|250px|thumb|center|Figure 11. Energy vs Time Plot.]]
[[File:xmhFigure 12.png|250px|thumb|center|Figure 12. Momentum vs Time Plot]]
[[File:xmhFigure 13.png|250px|thumb|center|Figure 13. Internuclear Distances vs Time Plot]]
If initial conditions are changed to r2 = rts+1=91.7 and r1 = 90.7 and p1 = p2 = 0 g.mol-1.pm.fs-1, same trends in both the Internuclear Distances vs Time graph and the Momenta vs Time graph are observed as for the previous conditions. Since the system is symmetric, hence the differences between these 2 sets of conditions are just attributed to the fact that whether HAHB or HBHC dissociates. {{fontcolor1|red| Great. Good use of figures here. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:14, 29 May 2020 (BST)}}

== 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? ==
{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Reactive || HA approaches HBHC molecule from a distance of rAB = 200 (starting from the bottom-right of the figure), with rBC being constant since HB and HC still remain bonded. After passing through rts, the new HAHB molecule is formed as shown by the constant rAB and increasing rBC as HC leaves(at the top left of the figure). || [[File:xmhA.png|250px]]
|-
| -3.1 || -4.1 || -420.077 || Unreactive || HA approaches HBHc (bottom-right), however it does not quite get to rts due to a reduced amount of kinetic energy compared to the first case, hence HA bounces off, there is no interaction with the hydrogen molecule. ||[[File:xmhB.png|250px]]
|-
| -3.1 || -5.1 || 413.977 || Reactive || Same as the first case except for the fact that p1 has a larger value, hence the system is more energetic to go over the activation barrier, and due to energy conservation, after the collision, the new molecule has a greater vibration. ||[[File:xmhC.png|250px]]
|-
| -5.1 || -10.1 || -357.277 || Unreactive || Barrier recrossing - the system is energetic enough to go over the transition state region and form the new HAHB, since rBC increases from its original bond length, indicating the leaving of HC, and the Etot is also greater than the first case (a reactive case). However, rBC drops down again to the original bond length and rAB increases as the system reverts back to HBHC. ||[[File:xmhD.png|250px]]
|-
| -5.1 || -10.6 || -349.477 || Reactive || The system HBHC converts to HAHB after reaching rts, but reverts given the excess momentum in p2 as rAB increases and rBC decreases, the system crosses the transition region again to form the final HAHB molecule as rBC continue increasing and rAB stays oscillating around a fixed value. ||[[File:xmhE.png|250px]]
|}
From the first sight, it seems reasonable to encourage a successful collision by adding more momentum into the system since this increases the kinetic energy of the atoms to overcome the activation barrier. However, as concluded from the table, the excess momentum and energy could give rise to greater vibration within the molecule and hence cause the bond to dissociate. {{fontcolor1|red| Great. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:46, 29 May 2020 (BST)}}

== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ==
In the transition state theory, the average transmission rate/frequency is estimated based on the assumption that all trajectories with some kinetic energy greater than the activation energy will be reactive. From the empirical data of the last section, it was shown that if momentum is provided such that the vibration is so great that the bond dissociates again, i.e. barrier recrossing occurs, this reduces the rate of a particular conversion, e.g. from HBHC and HA to HAHB and HC. This would overestimate the transition state theory rate constant (and hence the rate).
The transition state theory also consider the reaction pathway through classical mechanic collision (for macroscopic system). However, wavefunction (for microscopic system) of the atoms should be considered. Quantum tunneling may take place, it is not necessary to jump over the activation barrier, hence less kinetic energy is required for a successful collision, rate is higher than the theory prediction. {{fontcolor1|red| Yes. Include references. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:47, 29 May 2020 (BST)}}

== 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? ==
In GUI, it is set up that the atom A approaches and collide with the molecule HBHC.
In the F+H2 system, F is considered as atom A, and the 2 H atoms as atoms B and C. Figure 14(A) shows the entrance channel for the reactants F+H2, since rBC (the molecule’s bond length) is constant and rAB is big; Figure 14(B) shows the exit channel for the products (FH+H), since rAB is now constant and rBC is big. Comparing the V for the 2 channels, the reactants are at a higher V than the products, hence this is an exothermic reaction.
[[File:xmhFigure 14(A).png|250px|thumb|center|Figure 14(A). Surface Plot.]]
[[File:Figure 14(B)xmh.png|250px|thumb|center|Figure 14(B). Surface Plot.]]
For the H+FH system, the same idea applies, from Figure 15(A). & (B)., the reactants are at a lower V than the products, hence it is an endothermic reaction.
[[File:Figure 15(A)mh.png|250px|thumb|center|Figure 15(A). Surface Plot.]]
[[File:Figure 15(B)xmh.png|250px|thumb|center|Figure 15(B). Surface Plot.]]
This correctly reflects the result obtained from qualitative bond-strength analysis. F-H is a very strong bond (compared to other possible bonds in the system) due to the different electronegativities and hence ionic contribution to the nature of the bond. Hence breaking this bond requires much energy, and forming it releases much energy. In the F+H2 reaction, a pure covalent bond is broken to form the strong F-H bond, this is expected to be exothermic; in the H+FH system, the energy released from forming H2 does not compensate for the energy required to break F-H first, hence it is expected to be endothermic. {{fontcolor1|red| Nice discussion and well illustrated. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:12, 29 May 2020 (BST)}}

== Locate the approximate position of the transition state. ==
The approximate position of the transition state was investigated in the case of F+H2, where F is considered as atom A, and the 2 H atoms as atoms B and C. It was found that rAB = 181.1 and rBC = 74.5. This is reflected by Figure 20., where the internuclear distances are constant between all 3 atoms, corresponding to equilibrium at the transition state. Etot = - 433.980.
For the case of H+HF, same rts was found, since this reaction is just the reverse reaction of the above case, and the same geometry is found at the transition state. {{fontcolor1|red| Great. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:14, 29 May 2020 (BST)}}
[[File:xmhFigure 16.png|250px|thumb|center|Figure 16. Contour Plot.]]
[[File:Figure 17xmh.png|250px|thumb|center|Figure 17. Internuclear Distancesvs Time Plot.]]

== Report the activation energy for both reactions. ==
In order to find the reactants’ energy for the F+H2 reaction, initial conditions were set as rAB = 1000 and rBC = 74.5. The value of rAB was chosen so that there is no interaction in the system since the F atom is at a very distant away from the H2 molecule, as reflected by Figure 20., where the only momentum is given by the oscillation between the bonded molecule. Etot = - 435.057.
Hence the activation energy = Etot(transition state) – Etot(reactants) = - 433.980 + 435.057 = 1.077.
[[File:xmhFigure 18.png|250px|thumb|center|Figure 18. Momentum vs Time Plot.]]
Same method applies to the H+FH system, the activation energy = 125.322.
{{fontcolor1|red| Good, well done. I think a different type of plot would have been more illustrative here, personally. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:13, 29 May 2020 (BST)}}

== 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. ==
There are 3 types of energies associated with the system: potential energy due to the interactions between the atom and the molecule, vibrational kinetic energy and translational kinetic energy. After a successful collision, these energies would redistribute based on the mass and bond strength of the atom and molecule, some energy might excite the new molecule into a higher vibrational state. However, the molecule would eventually drop down to the ground state, releasing energy.
The IR overtone region could be used to confirm the presence of the transition between higher states, the peaks appear at higher wavelength (higher energy). Consequently, this confirms the following release of energy.

== 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. ==
According to Polanyi's rules, the reactants' translational energy activates them more effectively to the activation barrier and hence increases rate for exothermic reactions; conversely, the reactants' vibrational energy activates more effectively for endothermic reactions.
For the exothermic F+H2 system, the translational energy is more important to give a successful collision.
For the endothermic H+HF system, the vibrational energy is more important.

MRD:xmh01513932

2020-05-29T18:13:45Z

Mak214: /* Report the activation energy for both reactions. */

== 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? ==
On a potential energy surface diagram, the transition state is identified as the maximum on the minimum potential energy path between the reactants and products.

To explain the transition state mathematically, several derivatives have to be addressed first,
1. The first derivative of potential energy (V) with respect to either atomic separation (r1, r2) is zero, i.e. ∂V/∂r1 = 0 and ∂V/∂r2 = 0;
2. The second derivative D = ∂2V/∂r1∂r1 × ∂2V/∂r2∂r2 – ∂2V/∂r1∂r2 = 0;
3. If D > 0, ∂2V(r1)/∂r1∂r1 > 0, then the point with this particular (r1,r2) coordinates is a local minimum point; ∂2V(r1)/∂r1∂r1 < 0 for a local maximum point;
4. If D < 0, then this point is a saddle point, i.e. the transition state. Simply put, the saddle point is simultaneously a minimum and a maximum along two orthogonal directions (i.e. the two atomic separation axes in this case). {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:11, 29 May 2020 (BST)}}
[[File:Figure 1.log|250px|thumb|center|Figure 1. Contour Plot.]]
[[File:xmhFigure 2.png|250px|thumb|center|Figure 2. Surface Plot.]]

{{fontcolor1|red| What did you want to show with these plots? Use figures to illustrate what you are saying in the text, and make sure you discuss every figure you include in your text. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:12, 29 May 2020 (BST)}}

== 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. ==
For a symmetric system like H + H2, the internuclear distances of r1 = r2 is expected in the transition state. To test out the best input values for r1 and r2, there would be some expected features on various plots, as shown below,
1． In Figure 3., the kinetic energy of the system is constant at 0. This is explained by the fact that, at the transition state, the net force acting on the system F = dp/dt = ∂V/∂r1 = ∂V/∂r2 = 0 (as defined), the triatomic system is at a state of equilibrium with no change in motion. The same idea is conveyed in Figure 4., where the internuclear distances are expected to have a variation as small as possible, the atoms are “frozen” in space, moreover, rAB(r2) = rBC(r1).
2． In Figure 5., the transition state position (rts) is also checked by the fact that the input r1 and r2 values (denoted by the red cross) map onto the circle which gives the coordinates of the exact rts. {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:13, 29 May 2020 (BST)}}
The final value of r1 = r2 = 90.7 pm is chosen to satisfy the above requirements.
[[File:xmhFigure 3.png|250px|thumb|center|Figure 3. Energy vs Time Plot.]]
[[File:xmhFigure 4.png|250px|thumb|center|Figure 4. Internuclear Distances vs Time Plot.]]
[[File:xmhFigure 5.png|250px|thumb|center|Figure 5. Contour Plot]]

== Comment on how the mep and the trajectory you just calculated differ. ==
Two paths were generated under the calculation types of MEP and Dynamics respectively, with initial conditions of r1 = rts+1=91.7 and r2 = 90.7 and p1 = p2 = 0 g.mol-1.pm.fs-1.
The dynamic graph shows a wavy line (shown in Figure 6.). After passing though rts (now r1 < r2), HAHB molecule is formed, detaching HC. This case takes into account of the intramolecular vibration of the new H2 molecule, since there is an oscillating trajectory at around a fixed rAB value with an increasing rBC value which indicates detaching of HC.
[[File:xmhFigure 6.png|250px|thumb|center|Figure 6. Contour Plot.]][[File:xmhFigure 7.png|250px|thumb|center|Figure 7. Energy vs Time Plot.]]
[[File:xmhFigure 8.png|250px|thumb|center|Figure 8. Momentum vs Time Plot]][[File:xmhFigure 8.png|250px|thumb|center|Figure 9. Internuclear Distances vs Time Plot]]
The mep (the minimum energy path) corresponds to a trajectory with infinitely slow motion, i.e. the system’s velocities are reset to zero in each time step, consequently no momentum or kinetic energy, as reflected in Figure 11 & 12. The reaction path way as a smooth curve (shown in Figure .) without any information of the intrinsic vibration within the H2 molecule.
[[File:xmhFigure 10.png|250px|thumb|center|thumb|Figure 10. Contour Plot.]]
[[File:xmhFigure 11.png|250px|thumb|center|Figure 11. Energy vs Time Plot.]]
[[File:xmhFigure 12.png|250px|thumb|center|Figure 12. Momentum vs Time Plot]]
[[File:xmhFigure 13.png|250px|thumb|center|Figure 13. Internuclear Distances vs Time Plot]]
If initial conditions are changed to r2 = rts+1=91.7 and r1 = 90.7 and p1 = p2 = 0 g.mol-1.pm.fs-1, same trends in both the Internuclear Distances vs Time graph and the Momenta vs Time graph are observed as for the previous conditions. Since the system is symmetric, hence the differences between these 2 sets of conditions are just attributed to the fact that whether HAHB or HBHC dissociates. {{fontcolor1|red| Great. Good use of figures here. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:14, 29 May 2020 (BST)}}

== 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? ==
{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Reactive || HA approaches HBHC molecule from a distance of rAB = 200 (starting from the bottom-right of the figure), with rBC being constant since HB and HC still remain bonded. After passing through rts, the new HAHB molecule is formed as shown by the constant rAB and increasing rBC as HC leaves(at the top left of the figure). || [[File:xmhA.png|250px]]
|-
| -3.1 || -4.1 || -420.077 || Unreactive || HA approaches HBHc (bottom-right), however it does not quite get to rts due to a reduced amount of kinetic energy compared to the first case, hence HA bounces off, there is no interaction with the hydrogen molecule. ||[[File:xmhB.png|250px]]
|-
| -3.1 || -5.1 || 413.977 || Reactive || Same as the first case except for the fact that p1 has a larger value, hence the system is more energetic to go over the activation barrier, and due to energy conservation, after the collision, the new molecule has a greater vibration. ||[[File:xmhC.png|250px]]
|-
| -5.1 || -10.1 || -357.277 || Unreactive || Barrier recrossing - the system is energetic enough to go over the transition state region and form the new HAHB, since rBC increases from its original bond length, indicating the leaving of HC, and the Etot is also greater than the first case (a reactive case). However, rBC drops down again to the original bond length and rAB increases as the system reverts back to HBHC. ||[[File:xmhD.png|250px]]
|-
| -5.1 || -10.6 || -349.477 || Reactive || The system HBHC converts to HAHB after reaching rts, but reverts given the excess momentum in p2 as rAB increases and rBC decreases, the system crosses the transition region again to form the final HAHB molecule as rBC continue increasing and rAB stays oscillating around a fixed value. ||[[File:xmhE.png|250px]]
|}
From the first sight, it seems reasonable to encourage a successful collision by adding more momentum into the system since this increases the kinetic energy of the atoms to overcome the activation barrier. However, as concluded from the table, the excess momentum and energy could give rise to greater vibration within the molecule and hence cause the bond to dissociate. {{fontcolor1|red| Great. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:46, 29 May 2020 (BST)}}

== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ==
In the transition state theory, the average transmission rate/frequency is estimated based on the assumption that all trajectories with some kinetic energy greater than the activation energy will be reactive. From the empirical data of the last section, it was shown that if momentum is provided such that the vibration is so great that the bond dissociates again, i.e. barrier recrossing occurs, this reduces the rate of a particular conversion, e.g. from HBHC and HA to HAHB and HC. This would overestimate the transition state theory rate constant (and hence the rate).
The transition state theory also consider the reaction pathway through classical mechanic collision (for macroscopic system). However, wavefunction (for microscopic system) of the atoms should be considered. Quantum tunneling may take place, it is not necessary to jump over the activation barrier, hence less kinetic energy is required for a successful collision, rate is higher than the theory prediction. {{fontcolor1|red| Yes. Include references. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:47, 29 May 2020 (BST)}}

== 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? ==
In GUI, it is set up that the atom A approaches and collide with the molecule HBHC.
In the F+H2 system, F is considered as atom A, and the 2 H atoms as atoms B and C. Figure 14(A) shows the entrance channel for the reactants F+H2, since rBC (the molecule’s bond length) is constant and rAB is big; Figure 14(B) shows the exit channel for the products (FH+H), since rAB is now constant and rBC is big. Comparing the V for the 2 channels, the reactants are at a higher V than the products, hence this is an exothermic reaction.
[[File:xmhFigure 14(A).png|250px|thumb|center|Figure 14(A). Surface Plot.]]
[[File:Figure 14(B)xmh.png|250px|thumb|center|Figure 14(B). Surface Plot.]]
For the H+FH system, the same idea applies, from Figure 15(A). & (B)., the reactants are at a lower V than the products, hence it is an endothermic reaction.
[[File:Figure 15(A)mh.png|250px|thumb|center|Figure 15(A). Surface Plot.]]
[[File:Figure 15(B)xmh.png|250px|thumb|center|Figure 15(B). Surface Plot.]]
This correctly reflects the result obtained from qualitative bond-strength analysis. F-H is a very strong bond (compared to other possible bonds in the system) due to the different electronegativities and hence ionic contribution to the nature of the bond. Hence breaking this bond requires much energy, and forming it releases much energy. In the F+H2 reaction, a pure covalent bond is broken to form the strong F-H bond, this is expected to be exothermic; in the H+FH system, the energy released from forming H2 does not compensate for the energy required to break F-H first, hence it is expected to be endothermic. {{fontcolor1|red| Nice discussion and well illustrated. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:12, 29 May 2020 (BST)}}

== Locate the approximate position of the transition state. ==
The approximate position of the transition state was investigated in the case of F+H2, where F is considered as atom A, and the 2 H atoms as atoms B and C. It was found that rAB = 181.1 and rBC = 74.5. This is reflected by Figure 20., where the internuclear distances are constant between all 3 atoms, corresponding to equilibrium at the transition state. Etot = - 433.980.
For the case of H+HF, same rts was found, since this reaction is just the reverse reaction of the above case, and the same geometry is found at the transition state.
[[File:xmhFigure 16.png|250px|thumb|center|Figure 16. Contour Plot.]]
[[File:Figure 17xmh.png|250px|thumb|center|Figure 17. Internuclear Distancesvs Time Plot.]]

== Report the activation energy for both reactions. ==
In order to find the reactants’ energy for the F+H2 reaction, initial conditions were set as rAB = 1000 and rBC = 74.5. The value of rAB was chosen so that there is no interaction in the system since the F atom is at a very distant away from the H2 molecule, as reflected by Figure 20., where the only momentum is given by the oscillation between the bonded molecule. Etot = - 435.057.
Hence the activation energy = Etot(transition state) – Etot(reactants) = - 433.980 + 435.057 = 1.077.
[[File:xmhFigure 18.png|250px|thumb|center|Figure 18. Momentum vs Time Plot.]]
Same method applies to the H+FH system, the activation energy = 125.322.
{{fontcolor1|red| Good, well done. I think a different type of plot would have been more illustrative here, personally. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:13, 29 May 2020 (BST)}}

== 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. ==
There are 3 types of energies associated with the system: potential energy due to the interactions between the atom and the molecule, vibrational kinetic energy and translational kinetic energy. After a successful collision, these energies would redistribute based on the mass and bond strength of the atom and molecule, some energy might excite the new molecule into a higher vibrational state. However, the molecule would eventually drop down to the ground state, releasing energy.
The IR overtone region could be used to confirm the presence of the transition between higher states, the peaks appear at higher wavelength (higher energy). Consequently, this confirms the following release of energy.

== 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. ==
According to Polanyi's rules, the reactants' translational energy activates them more effectively to the activation barrier and hence increases rate for exothermic reactions; conversely, the reactants' vibrational energy activates more effectively for endothermic reactions.
For the exothermic F+H2 system, the translational energy is more important to give a successful collision.
For the endothermic H+HF system, the vibrational energy is more important.

MRD:xmh01513932

2020-05-29T18:12:37Z

Mak214: /* 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? */

== 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? ==
On a potential energy surface diagram, the transition state is identified as the maximum on the minimum potential energy path between the reactants and products.

To explain the transition state mathematically, several derivatives have to be addressed first,
1. The first derivative of potential energy (V) with respect to either atomic separation (r1, r2) is zero, i.e. ∂V/∂r1 = 0 and ∂V/∂r2 = 0;
2. The second derivative D = ∂2V/∂r1∂r1 × ∂2V/∂r2∂r2 – ∂2V/∂r1∂r2 = 0;
3. If D > 0, ∂2V(r1)/∂r1∂r1 > 0, then the point with this particular (r1,r2) coordinates is a local minimum point; ∂2V(r1)/∂r1∂r1 < 0 for a local maximum point;
4. If D < 0, then this point is a saddle point, i.e. the transition state. Simply put, the saddle point is simultaneously a minimum and a maximum along two orthogonal directions (i.e. the two atomic separation axes in this case). {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:11, 29 May 2020 (BST)}}
[[File:Figure 1.log|250px|thumb|center|Figure 1. Contour Plot.]]
[[File:xmhFigure 2.png|250px|thumb|center|Figure 2. Surface Plot.]]

{{fontcolor1|red| What did you want to show with these plots? Use figures to illustrate what you are saying in the text, and make sure you discuss every figure you include in your text. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:12, 29 May 2020 (BST)}}

== 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. ==
For a symmetric system like H + H2, the internuclear distances of r1 = r2 is expected in the transition state. To test out the best input values for r1 and r2, there would be some expected features on various plots, as shown below,
1． In Figure 3., the kinetic energy of the system is constant at 0. This is explained by the fact that, at the transition state, the net force acting on the system F = dp/dt = ∂V/∂r1 = ∂V/∂r2 = 0 (as defined), the triatomic system is at a state of equilibrium with no change in motion. The same idea is conveyed in Figure 4., where the internuclear distances are expected to have a variation as small as possible, the atoms are “frozen” in space, moreover, rAB(r2) = rBC(r1).
2． In Figure 5., the transition state position (rts) is also checked by the fact that the input r1 and r2 values (denoted by the red cross) map onto the circle which gives the coordinates of the exact rts. {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:13, 29 May 2020 (BST)}}
The final value of r1 = r2 = 90.7 pm is chosen to satisfy the above requirements.
[[File:xmhFigure 3.png|250px|thumb|center|Figure 3. Energy vs Time Plot.]]
[[File:xmhFigure 4.png|250px|thumb|center|Figure 4. Internuclear Distances vs Time Plot.]]
[[File:xmhFigure 5.png|250px|thumb|center|Figure 5. Contour Plot]]

== Comment on how the mep and the trajectory you just calculated differ. ==
Two paths were generated under the calculation types of MEP and Dynamics respectively, with initial conditions of r1 = rts+1=91.7 and r2 = 90.7 and p1 = p2 = 0 g.mol-1.pm.fs-1.
The dynamic graph shows a wavy line (shown in Figure 6.). After passing though rts (now r1 < r2), HAHB molecule is formed, detaching HC. This case takes into account of the intramolecular vibration of the new H2 molecule, since there is an oscillating trajectory at around a fixed rAB value with an increasing rBC value which indicates detaching of HC.
[[File:xmhFigure 6.png|250px|thumb|center|Figure 6. Contour Plot.]][[File:xmhFigure 7.png|250px|thumb|center|Figure 7. Energy vs Time Plot.]]
[[File:xmhFigure 8.png|250px|thumb|center|Figure 8. Momentum vs Time Plot]][[File:xmhFigure 8.png|250px|thumb|center|Figure 9. Internuclear Distances vs Time Plot]]
The mep (the minimum energy path) corresponds to a trajectory with infinitely slow motion, i.e. the system’s velocities are reset to zero in each time step, consequently no momentum or kinetic energy, as reflected in Figure 11 & 12. The reaction path way as a smooth curve (shown in Figure .) without any information of the intrinsic vibration within the H2 molecule.
[[File:xmhFigure 10.png|250px|thumb|center|thumb|Figure 10. Contour Plot.]]
[[File:xmhFigure 11.png|250px|thumb|center|Figure 11. Energy vs Time Plot.]]
[[File:xmhFigure 12.png|250px|thumb|center|Figure 12. Momentum vs Time Plot]]
[[File:xmhFigure 13.png|250px|thumb|center|Figure 13. Internuclear Distances vs Time Plot]]
If initial conditions are changed to r2 = rts+1=91.7 and r1 = 90.7 and p1 = p2 = 0 g.mol-1.pm.fs-1, same trends in both the Internuclear Distances vs Time graph and the Momenta vs Time graph are observed as for the previous conditions. Since the system is symmetric, hence the differences between these 2 sets of conditions are just attributed to the fact that whether HAHB or HBHC dissociates. {{fontcolor1|red| Great. Good use of figures here. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:14, 29 May 2020 (BST)}}

== 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? ==
{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Reactive || HA approaches HBHC molecule from a distance of rAB = 200 (starting from the bottom-right of the figure), with rBC being constant since HB and HC still remain bonded. After passing through rts, the new HAHB molecule is formed as shown by the constant rAB and increasing rBC as HC leaves(at the top left of the figure). || [[File:xmhA.png|250px]]
|-
| -3.1 || -4.1 || -420.077 || Unreactive || HA approaches HBHc (bottom-right), however it does not quite get to rts due to a reduced amount of kinetic energy compared to the first case, hence HA bounces off, there is no interaction with the hydrogen molecule. ||[[File:xmhB.png|250px]]
|-
| -3.1 || -5.1 || 413.977 || Reactive || Same as the first case except for the fact that p1 has a larger value, hence the system is more energetic to go over the activation barrier, and due to energy conservation, after the collision, the new molecule has a greater vibration. ||[[File:xmhC.png|250px]]
|-
| -5.1 || -10.1 || -357.277 || Unreactive || Barrier recrossing - the system is energetic enough to go over the transition state region and form the new HAHB, since rBC increases from its original bond length, indicating the leaving of HC, and the Etot is also greater than the first case (a reactive case). However, rBC drops down again to the original bond length and rAB increases as the system reverts back to HBHC. ||[[File:xmhD.png|250px]]
|-
| -5.1 || -10.6 || -349.477 || Reactive || The system HBHC converts to HAHB after reaching rts, but reverts given the excess momentum in p2 as rAB increases and rBC decreases, the system crosses the transition region again to form the final HAHB molecule as rBC continue increasing and rAB stays oscillating around a fixed value. ||[[File:xmhE.png|250px]]
|}
From the first sight, it seems reasonable to encourage a successful collision by adding more momentum into the system since this increases the kinetic energy of the atoms to overcome the activation barrier. However, as concluded from the table, the excess momentum and energy could give rise to greater vibration within the molecule and hence cause the bond to dissociate. {{fontcolor1|red| Great. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:46, 29 May 2020 (BST)}}

== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ==
In the transition state theory, the average transmission rate/frequency is estimated based on the assumption that all trajectories with some kinetic energy greater than the activation energy will be reactive. From the empirical data of the last section, it was shown that if momentum is provided such that the vibration is so great that the bond dissociates again, i.e. barrier recrossing occurs, this reduces the rate of a particular conversion, e.g. from HBHC and HA to HAHB and HC. This would overestimate the transition state theory rate constant (and hence the rate).
The transition state theory also consider the reaction pathway through classical mechanic collision (for macroscopic system). However, wavefunction (for microscopic system) of the atoms should be considered. Quantum tunneling may take place, it is not necessary to jump over the activation barrier, hence less kinetic energy is required for a successful collision, rate is higher than the theory prediction. {{fontcolor1|red| Yes. Include references. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:47, 29 May 2020 (BST)}}

== 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? ==
In GUI, it is set up that the atom A approaches and collide with the molecule HBHC.
In the F+H2 system, F is considered as atom A, and the 2 H atoms as atoms B and C. Figure 14(A) shows the entrance channel for the reactants F+H2, since rBC (the molecule’s bond length) is constant and rAB is big; Figure 14(B) shows the exit channel for the products (FH+H), since rAB is now constant and rBC is big. Comparing the V for the 2 channels, the reactants are at a higher V than the products, hence this is an exothermic reaction.
[[File:xmhFigure 14(A).png|250px|thumb|center|Figure 14(A). Surface Plot.]]
[[File:Figure 14(B)xmh.png|250px|thumb|center|Figure 14(B). Surface Plot.]]
For the H+FH system, the same idea applies, from Figure 15(A). & (B)., the reactants are at a lower V than the products, hence it is an endothermic reaction.
[[File:Figure 15(A)mh.png|250px|thumb|center|Figure 15(A). Surface Plot.]]
[[File:Figure 15(B)xmh.png|250px|thumb|center|Figure 15(B). Surface Plot.]]
This correctly reflects the result obtained from qualitative bond-strength analysis. F-H is a very strong bond (compared to other possible bonds in the system) due to the different electronegativities and hence ionic contribution to the nature of the bond. Hence breaking this bond requires much energy, and forming it releases much energy. In the F+H2 reaction, a pure covalent bond is broken to form the strong F-H bond, this is expected to be exothermic; in the H+FH system, the energy released from forming H2 does not compensate for the energy required to break F-H first, hence it is expected to be endothermic. {{fontcolor1|red| Nice discussion and well illustrated. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:12, 29 May 2020 (BST)}}

== Locate the approximate position of the transition state. ==
The approximate position of the transition state was investigated in the case of F+H2, where F is considered as atom A, and the 2 H atoms as atoms B and C. It was found that rAB = 181.1 and rBC = 74.5. This is reflected by Figure 20., where the internuclear distances are constant between all 3 atoms, corresponding to equilibrium at the transition state. Etot = - 433.980.
For the case of H+HF, same rts was found, since this reaction is just the reverse reaction of the above case, and the same geometry is found at the transition state.
[[File:xmhFigure 16.png|250px|thumb|center|Figure 16. Contour Plot.]]
[[File:Figure 17xmh.png|250px|thumb|center|Figure 17. Internuclear Distancesvs Time Plot.]]

== Report the activation energy for both reactions. ==
In order to find the reactants’ energy for the F+H2 reaction, initial conditions were set as rAB = 1000 and rBC = 74.5. The value of rAB was chosen so that there is no interaction in the system since the F atom is at a very distant away from the H2 molecule, as reflected by Figure 20., where the only momentum is given by the oscillation between the bonded molecule. Etot = - 435.057.
Hence the activation energy = Etot(transition state) – Etot(reactants) = - 433.980 + 435.057 = 1.077.
[[File:xmhFigure 18.png|250px|thumb|center|Figure 18. Momentum vs Time Plot.]]
Same method applies to the H+FH system, the activation energy = 125.322.

== 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. ==
There are 3 types of energies associated with the system: potential energy due to the interactions between the atom and the molecule, vibrational kinetic energy and translational kinetic energy. After a successful collision, these energies would redistribute based on the mass and bond strength of the atom and molecule, some energy might excite the new molecule into a higher vibrational state. However, the molecule would eventually drop down to the ground state, releasing energy.
The IR overtone region could be used to confirm the presence of the transition between higher states, the peaks appear at higher wavelength (higher energy). Consequently, this confirms the following release of energy.

== 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. ==
According to Polanyi's rules, the reactants' translational energy activates them more effectively to the activation barrier and hence increases rate for exothermic reactions; conversely, the reactants' vibrational energy activates more effectively for endothermic reactions.
For the exothermic F+H2 system, the translational energy is more important to give a successful collision.
For the endothermic H+HF system, the vibrational energy is more important.

MRD:01493832

2020-05-29T18:10:34Z

Mak214: /* Reaction dynamics */

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

== The transition state ==
On a potential energy surface diagram, the transition state is mathematically defined as a stationary point where the derivative of the potential energy curve, which is a force {{fontcolor1|red| I'm not sure exactly what you mean here? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:17, 29 May 2020 (BST)}}, is zero. This means that there are no movement at the transition state.

The transition state is different from a local minimum of a potential energy surface in a sense that it is a minimum in one direction yet a maximum in another.1 Therefore, it is a stationary point but not a point of inflection. From the calculated forces, we are able to confirm that this is a stationary point and from the given corresponding eigenvalues/vectors, the positive and negative combination confirms that this is a saddle point. If we move away from the stationary point, the eigenvalues/vectors will be either all positive or all negative. (For a local minimum, the eigenvalues/vectors will be positive in all directions.) {{fontcolor1|red| IOK. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:17, 29 May 2020 (BST)}}

=== H + H2 : Locating the transition state ===
In the case of H + H2, AB and BC distances must be equal in the transition state because all atoms are identical so the transition state is symmetrical and its position can be identified by using a trial and error process, where the AB and BC distances in the initial conditions are varied until the corresponding forces become as close to zero as possible. For 3 Hydrogen atoms, AB and BC distances (rts) are found to be 90.775 pm.

[[File:Nl3818-H TS dist-time.jpg|200px|thumb|left|Fig. 1 - Internuclear Distances vs Time plot for 3 Hydrogen atoms]]

From Fig.1, it is evidently clear that at rts = 90.775 pm, the distances are constant throughout confirming that it is a stationary point (i.e. the nuclei do not move) and that distances AB = BC. {{fontcolor1|red| I Good. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:17, 29 May 2020 (BST)}}

== Calculating the reaction path ==
There are two types of calculation that can be done when considering the trajectory of the Hydrogen molecule: MEP (minimum energy path) and Dynamics. In MEP, atoms are assumed to be in an infinitely slow motion which corresponds to zero momenta at each time step. {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:20, 29 May 2020 (BST)}}Both calculations can be done by changing the initial conditions to r1 = rts+δ and r2=rts in order to visualise the trajectories. The difference between MEP and Dynamics are shown in two contour plots below. From the MEP contour plot, we can see that the black trajectory line is rather smooth as it approaches the transition state position marked with a red cross compared to the trajectory shown in Dynamics contour plot. The MEP calculated trajectory also starts off at a higher potential energy position than the one from Dynamics calculation. {{fontcolor1|red| Hmm.. they should start with the same energy if you put in the same settings. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:20, 29 May 2020 (BST)}} Because in MEP calculation, atoms are moving very slowly so the vibration between the reactant HA-HB is so small that it does not display an oscillating behaviour as in the Dynamics calculation. {{fontcolor1|red| Not exactly, the MEP calculation there should be no vibration of the bond whatsoever since there is no momentum to move the atoms away from the equilibrium bond distance which is a valley in the potential energy surface. We only see vibration in the dynamics type calculation because there is residual momentum which allows the atoms in the molecule to vibrate. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:20, 29 May 2020 (BST)}}

{| class="wikitable" border=1
|-
|[[File:Dynamic_nl3818.png|200px|thumb|left|Fig.2 - Contour plot of triatomic Hydrogen system using Dynamics calculation]]||
[[File:MEP_nl3818.png|200px|thumb|left|Fig.3 - Contour plot of triatomic Hydrogen system using MEP calculation]]
|}

=== Change in initial conditions ===
If we change the initial condition so that r1 = rts and r2=rts+δ instead, the result is still identical to previous result but with atom A moving away at large t and atom B bonding with C instead. By investigating the “Internuclear Distances vs Time” and “Momenta vs Time” plots and taking note of values for r1(t) r2(t) and p1(t) p2(t) at very large t (50 fs), another set of calculations can be done with using these values for the initial positions and momentum (with reversed sign). The observation is for “Internuclear Distances vs Time” plot, the result is a reflection in the y-axis of Fig. 4 meaning, the bonded BC H2 molecule and the previously isolated HA atom are now approaching each other and their positions at large t corresponds to the transition state position. With reversed sign of the momenta, this is a reverse process of what was previously done, instead of atoms moving apart, they are now moving towards each other with the same kinetic energy. The “Momenta vs Time” is now reflected in the x-axis of Fig. 5, which can be reasoned by the repulsion that the atoms experienced from coming together resulting in positive gradient which is gradual at first then steepens at high t as the three atoms are now very close.

{| class="wikitable" border=1
|-
|[[File:Dtnl3818new.png|200px|thumb|left|Fig.4 - Dist. vs Time plot of H atoms moving apart]]||
[[File:Momentanewnl3818.png|200px|thumb|left|Fig.5 - Momenta vs Time plot of H atoms moving apart]]
|}

== Reactive and unreactive trajectories ==

To test whether trajectories starting at the same position but with higher momenta will all react, a table has been constructed shown below, where r1=74 pm and r2=200 pm.

{| class="wikitable" border=1
!p1/ g.mol-1.pm.fs-1!!p2/ g.mol-1.pm.fs-1!!Etot/kJ.mol-1!!Reactive?!!Description of the dynamics!!Illustration of the trajectory using Dist. vs Time plot
|-
| -2.56 || -5.1 || -414 || YES || This is a reactive trajectory because as the incoming H approaches, its AB distance decreases until it displays an oscillating behaviour which corresponds to the new bond vibration while the initial BC distance increases exponentially showing that the bond is broken and the leaving atom is moving further apart. ||[[File:Nl3818table1.png|200px|thumb|left]]
|-
| -3.1 || -4.1 || -420 || NO || This is not a reactive trajectory since although the AB distance decreases, it increases again while the initial BC distance retains its oscillating behaviour throughout meaning the bond is not broken. The incoming H approaches then moves away, no bond is broken in this case. ||[[File:Nl3818table2.png|200px|thumb|left]]
|-
| -3.1 || -5.1 || -412 || YES || This displays the same behaviour as the first case and thus, same explanation applies. ||[[File:Nl3818table3.png|200px|thumb|left]]
|-
| -5.1 || -10.1 || -357 || NO || In this case, the initial bond is broken but then reforms again, therefore, overall there is no reaction. ||[[File:Nl3818table4.png|200px|thumb|left]]
|-
|-5.1 || -10.6 || -349 || YES || The approaching H forms a new bond then bounces back but did not break the new bond so overall, this trajectory is reactive. ||[[File:Nl3818table5.png|200px|thumb|left]]
|}

[[File:Nl3818table5.png|200px|thumb|left]]
The last plot on the left corresponds to p1 = -5.1 g.mol-1.pm.fs-1, p2 = -10.6 g.mol-1.pm.fs-1 and Etot = -349 kJ.mol-1. This is a reactive trajectory since the approaching H forms a new bond then bounces back but did not break so overall, this trajectory is reactive.

*The last line of the table is not showing so I had to insert it like this.
{{fontcolor1|red| Well done. It would be nice to have a line or two here concluding what you have learned from these trajectories. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:49, 29 May 2020 (BST)}}

==Transition State Theory==

'''The transition state theory is a classical consideration which assumes:'''2

'''1) The reactants and transition state are in a quasi-equilibrium.''' This means that once the reactants reaches the transition state, they {{fontcolor1|red| cannot ? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:48, 29 May 2020 (BST)}} collapse back to form the reactants again when in reality they can. This cause the values that the transition state theory predicts to be overestimated than the actual experimental value.

'''2) Quantum tunnelling is ignored.''' Because the reactants that overcome the barrier by moving across is ignored, this causes the values predicted by transition state theory to be underestimated than the actual experimental values.

'''3) All reactants that have enough energy to overcome the barrier will successfully form the product and the step that involves the formation of the product from the transition state is the rate-determing step.''' This causes an overestimation of the actual rate because in reality, although the reactants have high energy, but if the energy is not located in the right place (i.e. in the bond that needs to be broken) then the reaction will not happen. {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:48, 29 May 2020 (BST)}}

== The F-H-H System==
=== PES Inspection===
===='''F + H2'''====
This is an unsymmetrical system which results in unsymmetrical potential wells in the surface plot.

From the surface plot, we are able to tell that this is an exothermic reaction. BC is the H-H distance of the reactant while AB is the H-F distance, the z-axis of the surface plot represents energy and in this case, the H-H is deeper in energy compared to H-F which corresponds to the energy profile of an exothermic reaction and thus, explains the greater H-F bond strength compared to H-H. {{fontcolor1|red| HF is deeper (lower E) because it is the stronger bond.. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:06, 29 May 2020 (BST)}}

The approximate position of the transition state can be found by the same method that was used for three Hydrogen atoms: trial and error where the corresponding forces are as close to zero as possible. This method gives rF-HA = 180 pm and rHA-HB=74.5 pm. These positions are confirmed to be the transition state by the opposite signs of the eigenvalues/vectors. At this position, the energy of the transition state is -434 kJmol-1. {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:06, 29 May 2020 (BST)}}

In order to find the activation energy, the position of the system is displaced slightly towards the reactant side from the transition state (rF-HA = 179 pm) and MEP calculation was used to obtain the "Energy vs Time" below.

From the graph, the energy of the reactants is -561 kJmol-1. The difference between the reactant state and the transition state energy is the activation energy which is the energy required to cross the barrier to form the product, in this case is +127 kJmol-1. {{fontcolor1|red| Yes. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:06, 29 May 2020 (BST)}}

{| class="wikitable" border=1
|-
|[[File:H2andHFsurface.png|200px|thumb|left|Fig.6 - Surface plot for F+H2]]||[[File:FH2Ea.png|200px|thumb|left|Fig.7 - Energy vs Time plot for F+H2]]
|}

==== H + HF ====
The surface plot of this reaction is a mirrored image of the above surface plot from the previous reaction. This shows that this is an endothermic reaction where the reactants have higher energy than the product. This correlates to the stronger H-F bond that requires input energy to break to form the H-H product. {{fontcolor1|red| Same mistake as above, I think you were looking at this the wrong way around. The HH + F is the higher energy and the HF + H is the lower energy on the PES - reorient yourself with these axes so that this makes more sense. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:09, 29 May 2020 (BST)}}

The approximate transition state can be found in the same manner as the previous reaction and the positions are rF-HB = 180 pm and rHB-HC=74.5 pm, corresponding to total energy of -434 kJmol-1. The activation energy in this case is very small is reported to be +0.930 kJmol-1. Because of such small activation energy, it is easier to locate the position of the transition state by using the Hammond postulate. From the surface plot, we have deduced that this reaction is endothermic therefore it must have a late transition state meaning the structure of the transition state will resemble more of the product.3 As a result, shorter H-H distance and long F-H were predicted for the position of the transition state. {{fontcolor1|red| Really good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:09, 29 May 2020 (BST)}}

{| class="wikitable" border=1
|-
|[[File:HandHFsurface.png|200px|thumb|left|Fig.8 - Surface plot for H + HF]]||[[File:FHHEa.png|200px|thumb|left|Fig.9 - Energy vs Time plot for H + HF]]
|}

===Reaction dynamics===

One of the ways to confirm the mechanism of the release of reaction energy is by taking an Infrared absorption spectrum of the product. In this case, we can measure the IR spectrum of HF gas sample. When the reaction just happened and IR light is shown through, we are exciting the molecules that were once thermally relaxed and exclusively occupying the ground state to its first vibrational excited state and at early times (when the reaction just happened), we may also excitation from first to second vibrational excited state.

As a result, what we can observe in the IR spectrum is two peaks present, one is the fundamental peak and the other is an overtone appearing at a lower wavenumber. This is because of decreasing energy gap between each levels due to anharmonicity.4 By measuring the intensity of the overtone over time, we can extract the number of molecules that are vibrationally excited over time so we can expect the intensity of overtone to be smaller as energy is emitted as radiation and the intensity of the fundamental peak would increase as time goes on.

The other way to confirm the mechanism of energy release experimentally is by measuring the IR emitted directly, instead of probing the reaction using IR spectroscopy, as the molecules fall back from vibrational excited state to the ground state. This method can be done by using the Infrared Emission Spectroscopy (IES).5 {{fontcolor1|red| Great. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:10, 29 May 2020 (BST)}}

====F + H2====

By setting up a calculation with rHH=74 pm, pFH=-1.0 g.mol-1.pm.fs-1 and pHH varied between -6.1 and 6.1 g.mol-1.pm.fs-1, it is visible from the contour plot that the trajectory only successfully rolled over to the product with pHH between -2.7 and 1.2 g.mol-1.pm.fs-1. Values outside of this range results in the trajectory hitting the wall and bouncing back.

{| class="wikitable" border=1
|-
|[[File:H2-2.7.png|200px|thumb|left|Fig.10 - H2 trajectory at pHH = -2.7 g.mol-1.pm.fs-1]]||
[[File:H20nl3818.png|200px|thumb|left|Fig.11 - H2 trajectory at pHH = 0.0 g.mol-1.pm.fs-1]]
|[[File:H21.2nl3818.png|200px|thumb|left|Fig.12 - H2 trajectory at pHH = 1.2 g.mol-1.pm.fs-1]]||
[[File:H2bouncingbacknl3818.png|200px|thumb|left|Fig.13 - trajectory at pHH = 0.2 g.mol-1.pm.fs-1, pFH = -1.6 g.mol-1.pm.fs-1]]
|}

====H + HF====
A reactive trajectory was obtained with a combination of pFH = 1 g.mol-1.pm.fs-1 and pHH = -7 g.mol-1.pm.fs-1. Decreasing the momentum of the incoming H further results in the trajectory hitting the wall and bouncing back while increasing the energy of H-F vibration causes the trajectory to go in the other direction rather than towards the product.

====Polanyi's Empirical Rule====
The above investigation illustrates the Polanyi's empirical rule. In an exothermic reaction, as demonstrated by F + H2, translational energy is favoured over vibrational energy because the trajectory can fall down into the low energy region of the PES whereas the vibrational energy is in a different direction and can cause the trajectory to bounce back near the transition state and cross back to the reactant state.

On the other hand, vibrational energy is favoured over translational energy in an endothermic reaction6 as illustrated by H + HF. This is because the trajectory has the same directionality with the vibrational energy that goes to the product. {{fontcolor1|red| Nice report - see my comments. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:10, 29 May 2020 (BST)}}

==Bibliography==
1 Maxima, minima, and saddle points (article). https://www.khanacademy.org/math/multivariable-calculus/applications-of-multivariable-derivatives/optimizing-multivariable-functions/a/maximums-minimums-and-saddle-points (accessed May 22, 2020).

2 Peters, B. Transition State Theory. Reaction Rate Theory and Rare Events Simulations 2017, 227–271.

3 Libretexts. Hammond's Postulate. https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hammond’s_Postulate (accessed May 22, 2020).

4 Libretexts. 13.5: Vibrational Overtones. https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/13:_Molecular_Spectroscopy/13.05:_Vibrational_Overtones (accessed May 22, 2020).

5 Keresztury, G. Emission Spectroscopy, Infrared. Encyclopedia of Analytical Chemistry 2006.

6 Zhang, Z.; Zhou, Y.; Zhang, D. H.; Czakó, G.; Bowman, J. M. Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl CHD3 Reaction. The Journal of Physical Chemistry Letters 2012, 3 (23), 3416–3419.

MRD:01493832

2020-05-29T18:09:06Z

Mak214: /* H + HF */

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

== The transition state ==
On a potential energy surface diagram, the transition state is mathematically defined as a stationary point where the derivative of the potential energy curve, which is a force {{fontcolor1|red| I'm not sure exactly what you mean here? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:17, 29 May 2020 (BST)}}, is zero. This means that there are no movement at the transition state.

The transition state is different from a local minimum of a potential energy surface in a sense that it is a minimum in one direction yet a maximum in another.1 Therefore, it is a stationary point but not a point of inflection. From the calculated forces, we are able to confirm that this is a stationary point and from the given corresponding eigenvalues/vectors, the positive and negative combination confirms that this is a saddle point. If we move away from the stationary point, the eigenvalues/vectors will be either all positive or all negative. (For a local minimum, the eigenvalues/vectors will be positive in all directions.) {{fontcolor1|red| IOK. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:17, 29 May 2020 (BST)}}

=== H + H2 : Locating the transition state ===
In the case of H + H2, AB and BC distances must be equal in the transition state because all atoms are identical so the transition state is symmetrical and its position can be identified by using a trial and error process, where the AB and BC distances in the initial conditions are varied until the corresponding forces become as close to zero as possible. For 3 Hydrogen atoms, AB and BC distances (rts) are found to be 90.775 pm.

[[File:Nl3818-H TS dist-time.jpg|200px|thumb|left|Fig. 1 - Internuclear Distances vs Time plot for 3 Hydrogen atoms]]

From Fig.1, it is evidently clear that at rts = 90.775 pm, the distances are constant throughout confirming that it is a stationary point (i.e. the nuclei do not move) and that distances AB = BC. {{fontcolor1|red| I Good. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:17, 29 May 2020 (BST)}}

== Calculating the reaction path ==
There are two types of calculation that can be done when considering the trajectory of the Hydrogen molecule: MEP (minimum energy path) and Dynamics. In MEP, atoms are assumed to be in an infinitely slow motion which corresponds to zero momenta at each time step. {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:20, 29 May 2020 (BST)}}Both calculations can be done by changing the initial conditions to r1 = rts+δ and r2=rts in order to visualise the trajectories. The difference between MEP and Dynamics are shown in two contour plots below. From the MEP contour plot, we can see that the black trajectory line is rather smooth as it approaches the transition state position marked with a red cross compared to the trajectory shown in Dynamics contour plot. The MEP calculated trajectory also starts off at a higher potential energy position than the one from Dynamics calculation. {{fontcolor1|red| Hmm.. they should start with the same energy if you put in the same settings. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:20, 29 May 2020 (BST)}} Because in MEP calculation, atoms are moving very slowly so the vibration between the reactant HA-HB is so small that it does not display an oscillating behaviour as in the Dynamics calculation. {{fontcolor1|red| Not exactly, the MEP calculation there should be no vibration of the bond whatsoever since there is no momentum to move the atoms away from the equilibrium bond distance which is a valley in the potential energy surface. We only see vibration in the dynamics type calculation because there is residual momentum which allows the atoms in the molecule to vibrate. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:20, 29 May 2020 (BST)}}

{| class="wikitable" border=1
|-
|[[File:Dynamic_nl3818.png|200px|thumb|left|Fig.2 - Contour plot of triatomic Hydrogen system using Dynamics calculation]]||
[[File:MEP_nl3818.png|200px|thumb|left|Fig.3 - Contour plot of triatomic Hydrogen system using MEP calculation]]
|}

=== Change in initial conditions ===
If we change the initial condition so that r1 = rts and r2=rts+δ instead, the result is still identical to previous result but with atom A moving away at large t and atom B bonding with C instead. By investigating the “Internuclear Distances vs Time” and “Momenta vs Time” plots and taking note of values for r1(t) r2(t) and p1(t) p2(t) at very large t (50 fs), another set of calculations can be done with using these values for the initial positions and momentum (with reversed sign). The observation is for “Internuclear Distances vs Time” plot, the result is a reflection in the y-axis of Fig. 4 meaning, the bonded BC H2 molecule and the previously isolated HA atom are now approaching each other and their positions at large t corresponds to the transition state position. With reversed sign of the momenta, this is a reverse process of what was previously done, instead of atoms moving apart, they are now moving towards each other with the same kinetic energy. The “Momenta vs Time” is now reflected in the x-axis of Fig. 5, which can be reasoned by the repulsion that the atoms experienced from coming together resulting in positive gradient which is gradual at first then steepens at high t as the three atoms are now very close.

{| class="wikitable" border=1
|-
|[[File:Dtnl3818new.png|200px|thumb|left|Fig.4 - Dist. vs Time plot of H atoms moving apart]]||
[[File:Momentanewnl3818.png|200px|thumb|left|Fig.5 - Momenta vs Time plot of H atoms moving apart]]
|}

== Reactive and unreactive trajectories ==

To test whether trajectories starting at the same position but with higher momenta will all react, a table has been constructed shown below, where r1=74 pm and r2=200 pm.

{| class="wikitable" border=1
!p1/ g.mol-1.pm.fs-1!!p2/ g.mol-1.pm.fs-1!!Etot/kJ.mol-1!!Reactive?!!Description of the dynamics!!Illustration of the trajectory using Dist. vs Time plot
|-
| -2.56 || -5.1 || -414 || YES || This is a reactive trajectory because as the incoming H approaches, its AB distance decreases until it displays an oscillating behaviour which corresponds to the new bond vibration while the initial BC distance increases exponentially showing that the bond is broken and the leaving atom is moving further apart. ||[[File:Nl3818table1.png|200px|thumb|left]]
|-
| -3.1 || -4.1 || -420 || NO || This is not a reactive trajectory since although the AB distance decreases, it increases again while the initial BC distance retains its oscillating behaviour throughout meaning the bond is not broken. The incoming H approaches then moves away, no bond is broken in this case. ||[[File:Nl3818table2.png|200px|thumb|left]]
|-
| -3.1 || -5.1 || -412 || YES || This displays the same behaviour as the first case and thus, same explanation applies. ||[[File:Nl3818table3.png|200px|thumb|left]]
|-
| -5.1 || -10.1 || -357 || NO || In this case, the initial bond is broken but then reforms again, therefore, overall there is no reaction. ||[[File:Nl3818table4.png|200px|thumb|left]]
|-
|-5.1 || -10.6 || -349 || YES || The approaching H forms a new bond then bounces back but did not break the new bond so overall, this trajectory is reactive. ||[[File:Nl3818table5.png|200px|thumb|left]]
|}

[[File:Nl3818table5.png|200px|thumb|left]]
The last plot on the left corresponds to p1 = -5.1 g.mol-1.pm.fs-1, p2 = -10.6 g.mol-1.pm.fs-1 and Etot = -349 kJ.mol-1. This is a reactive trajectory since the approaching H forms a new bond then bounces back but did not break so overall, this trajectory is reactive.

*The last line of the table is not showing so I had to insert it like this.
{{fontcolor1|red| Well done. It would be nice to have a line or two here concluding what you have learned from these trajectories. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:49, 29 May 2020 (BST)}}

==Transition State Theory==

'''The transition state theory is a classical consideration which assumes:'''2

'''1) The reactants and transition state are in a quasi-equilibrium.''' This means that once the reactants reaches the transition state, they {{fontcolor1|red| cannot ? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:48, 29 May 2020 (BST)}} collapse back to form the reactants again when in reality they can. This cause the values that the transition state theory predicts to be overestimated than the actual experimental value.

'''2) Quantum tunnelling is ignored.''' Because the reactants that overcome the barrier by moving across is ignored, this causes the values predicted by transition state theory to be underestimated than the actual experimental values.

'''3) All reactants that have enough energy to overcome the barrier will successfully form the product and the step that involves the formation of the product from the transition state is the rate-determing step.''' This causes an overestimation of the actual rate because in reality, although the reactants have high energy, but if the energy is not located in the right place (i.e. in the bond that needs to be broken) then the reaction will not happen. {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:48, 29 May 2020 (BST)}}

== The F-H-H System==
=== PES Inspection===
===='''F + H2'''====
This is an unsymmetrical system which results in unsymmetrical potential wells in the surface plot.

From the surface plot, we are able to tell that this is an exothermic reaction. BC is the H-H distance of the reactant while AB is the H-F distance, the z-axis of the surface plot represents energy and in this case, the H-H is deeper in energy compared to H-F which corresponds to the energy profile of an exothermic reaction and thus, explains the greater H-F bond strength compared to H-H. {{fontcolor1|red| HF is deeper (lower E) because it is the stronger bond.. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:06, 29 May 2020 (BST)}}

The approximate position of the transition state can be found by the same method that was used for three Hydrogen atoms: trial and error where the corresponding forces are as close to zero as possible. This method gives rF-HA = 180 pm and rHA-HB=74.5 pm. These positions are confirmed to be the transition state by the opposite signs of the eigenvalues/vectors. At this position, the energy of the transition state is -434 kJmol-1. {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:06, 29 May 2020 (BST)}}

In order to find the activation energy, the position of the system is displaced slightly towards the reactant side from the transition state (rF-HA = 179 pm) and MEP calculation was used to obtain the "Energy vs Time" below.

From the graph, the energy of the reactants is -561 kJmol-1. The difference between the reactant state and the transition state energy is the activation energy which is the energy required to cross the barrier to form the product, in this case is +127 kJmol-1. {{fontcolor1|red| Yes. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:06, 29 May 2020 (BST)}}

{| class="wikitable" border=1
|-
|[[File:H2andHFsurface.png|200px|thumb|left|Fig.6 - Surface plot for F+H2]]||[[File:FH2Ea.png|200px|thumb|left|Fig.7 - Energy vs Time plot for F+H2]]
|}

==== H + HF ====
The surface plot of this reaction is a mirrored image of the above surface plot from the previous reaction. This shows that this is an endothermic reaction where the reactants have higher energy than the product. This correlates to the stronger H-F bond that requires input energy to break to form the H-H product. {{fontcolor1|red| Same mistake as above, I think you were looking at this the wrong way around. The HH + F is the higher energy and the HF + H is the lower energy on the PES - reorient yourself with these axes so that this makes more sense. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:09, 29 May 2020 (BST)}}

The approximate transition state can be found in the same manner as the previous reaction and the positions are rF-HB = 180 pm and rHB-HC=74.5 pm, corresponding to total energy of -434 kJmol-1. The activation energy in this case is very small is reported to be +0.930 kJmol-1. Because of such small activation energy, it is easier to locate the position of the transition state by using the Hammond postulate. From the surface plot, we have deduced that this reaction is endothermic therefore it must have a late transition state meaning the structure of the transition state will resemble more of the product.3 As a result, shorter H-H distance and long F-H were predicted for the position of the transition state. {{fontcolor1|red| Really good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:09, 29 May 2020 (BST)}}

{| class="wikitable" border=1
|-
|[[File:HandHFsurface.png|200px|thumb|left|Fig.8 - Surface plot for H + HF]]||[[File:FHHEa.png|200px|thumb|left|Fig.9 - Energy vs Time plot for H + HF]]
|}

===Reaction dynamics===

One of the ways to confirm the mechanism of the release of reaction energy is by taking an Infrared absorption spectrum of the product. In this case, we can measure the IR spectrum of HF gas sample. When the reaction just happened and IR light is shown through, we are exciting the molecules that were once thermally relaxed and exclusively occupying the ground state to its first vibrational excited state and at early times (when the reaction just happened), we may also excitation from first to second vibrational excited state.

As a result, what we can observe in the IR spectrum is two peaks present, one is the fundamental peak and the other is an overtone appearing at a lower wavenumber. This is because of decreasing energy gap between each levels due to anharmonicity.4 By measuring the intensity of the overtone over time, we can extract the number of molecules that are vibrationally excited over time so we can expect the intensity of overtone to be smaller as energy is emitted as radiation and the intensity of the fundamental peak would increase as time goes on.

The other way to confirm the mechanism of energy release experimentally is by measuring the IR emitted directly, instead of probing the reaction using IR spectroscopy, as the molecules fall back from vibrational excited state to the ground state. This method can be done by using the Infrared Emission Spectroscopy (IES).5

====F + H2====

By setting up a calculation with rHH=74 pm, pFH=-1.0 g.mol-1.pm.fs-1 and pHH varied between -6.1 and 6.1 g.mol-1.pm.fs-1, it is visible from the contour plot that the trajectory only successfully rolled over to the product with pHH between -2.7 and 1.2 g.mol-1.pm.fs-1. Values outside of this range results in the trajectory hitting the wall and bouncing back.

{| class="wikitable" border=1
|-
|[[File:H2-2.7.png|200px|thumb|left|Fig.10 - H2 trajectory at pHH = -2.7 g.mol-1.pm.fs-1]]||
[[File:H20nl3818.png|200px|thumb|left|Fig.11 - H2 trajectory at pHH = 0.0 g.mol-1.pm.fs-1]]
|[[File:H21.2nl3818.png|200px|thumb|left|Fig.12 - H2 trajectory at pHH = 1.2 g.mol-1.pm.fs-1]]||
[[File:H2bouncingbacknl3818.png|200px|thumb|left|Fig.13 - trajectory at pHH = 0.2 g.mol-1.pm.fs-1, pFH = -1.6 g.mol-1.pm.fs-1]]
|}

====H + HF====
A reactive trajectory was obtained with a combination of pFH = 1 g.mol-1.pm.fs-1 and pHH = -7 g.mol-1.pm.fs-1. Decreasing the momentum of the incoming H further results in the trajectory hitting the wall and bouncing back while increasing the energy of H-F vibration causes the trajectory to go in the other direction rather than towards the product.

====Polanyi's Empirical Rule====
The above investigation illustrates the Polanyi's empirical rule. In an exothermic reaction, as demonstrated by F + H2, translational energy is favoured over vibrational energy because the trajectory can fall down into the low energy region of the PES whereas the vibrational energy is in a different direction and can cause the trajectory to bounce back near the transition state and cross back to the reactant state.

On the other hand, vibrational energy is favoured over translational energy in an endothermic reaction6 as illustrated by H + HF. This is because the trajectory has the same directionality with the vibrational energy that goes to the product.

==Bibliography==
1 Maxima, minima, and saddle points (article). https://www.khanacademy.org/math/multivariable-calculus/applications-of-multivariable-derivatives/optimizing-multivariable-functions/a/maximums-minimums-and-saddle-points (accessed May 22, 2020).

2 Peters, B. Transition State Theory. Reaction Rate Theory and Rare Events Simulations 2017, 227–271.

3 Libretexts. Hammond's Postulate. https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hammond’s_Postulate (accessed May 22, 2020).

4 Libretexts. 13.5: Vibrational Overtones. https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/13:_Molecular_Spectroscopy/13.05:_Vibrational_Overtones (accessed May 22, 2020).

5 Keresztury, G. Emission Spectroscopy, Infrared. Encyclopedia of Analytical Chemistry 2006.

6 Zhang, Z.; Zhou, Y.; Zhang, D. H.; Czakó, G.; Bowman, J. M. Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl CHD3 Reaction. The Journal of Physical Chemistry Letters 2012, 3 (23), 3416–3419.

MRD:01493832

2020-05-29T18:06:54Z

Mak214: /* F + H2 */

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

== The transition state ==
On a potential energy surface diagram, the transition state is mathematically defined as a stationary point where the derivative of the potential energy curve, which is a force {{fontcolor1|red| I'm not sure exactly what you mean here? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:17, 29 May 2020 (BST)}}, is zero. This means that there are no movement at the transition state.

The transition state is different from a local minimum of a potential energy surface in a sense that it is a minimum in one direction yet a maximum in another.1 Therefore, it is a stationary point but not a point of inflection. From the calculated forces, we are able to confirm that this is a stationary point and from the given corresponding eigenvalues/vectors, the positive and negative combination confirms that this is a saddle point. If we move away from the stationary point, the eigenvalues/vectors will be either all positive or all negative. (For a local minimum, the eigenvalues/vectors will be positive in all directions.) {{fontcolor1|red| IOK. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:17, 29 May 2020 (BST)}}

=== H + H2 : Locating the transition state ===
In the case of H + H2, AB and BC distances must be equal in the transition state because all atoms are identical so the transition state is symmetrical and its position can be identified by using a trial and error process, where the AB and BC distances in the initial conditions are varied until the corresponding forces become as close to zero as possible. For 3 Hydrogen atoms, AB and BC distances (rts) are found to be 90.775 pm.

[[File:Nl3818-H TS dist-time.jpg|200px|thumb|left|Fig. 1 - Internuclear Distances vs Time plot for 3 Hydrogen atoms]]

From Fig.1, it is evidently clear that at rts = 90.775 pm, the distances are constant throughout confirming that it is a stationary point (i.e. the nuclei do not move) and that distances AB = BC. {{fontcolor1|red| I Good. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:17, 29 May 2020 (BST)}}

== Calculating the reaction path ==
There are two types of calculation that can be done when considering the trajectory of the Hydrogen molecule: MEP (minimum energy path) and Dynamics. In MEP, atoms are assumed to be in an infinitely slow motion which corresponds to zero momenta at each time step. {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:20, 29 May 2020 (BST)}}Both calculations can be done by changing the initial conditions to r1 = rts+δ and r2=rts in order to visualise the trajectories. The difference between MEP and Dynamics are shown in two contour plots below. From the MEP contour plot, we can see that the black trajectory line is rather smooth as it approaches the transition state position marked with a red cross compared to the trajectory shown in Dynamics contour plot. The MEP calculated trajectory also starts off at a higher potential energy position than the one from Dynamics calculation. {{fontcolor1|red| Hmm.. they should start with the same energy if you put in the same settings. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:20, 29 May 2020 (BST)}} Because in MEP calculation, atoms are moving very slowly so the vibration between the reactant HA-HB is so small that it does not display an oscillating behaviour as in the Dynamics calculation. {{fontcolor1|red| Not exactly, the MEP calculation there should be no vibration of the bond whatsoever since there is no momentum to move the atoms away from the equilibrium bond distance which is a valley in the potential energy surface. We only see vibration in the dynamics type calculation because there is residual momentum which allows the atoms in the molecule to vibrate. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:20, 29 May 2020 (BST)}}

{| class="wikitable" border=1
|-
|[[File:Dynamic_nl3818.png|200px|thumb|left|Fig.2 - Contour plot of triatomic Hydrogen system using Dynamics calculation]]||
[[File:MEP_nl3818.png|200px|thumb|left|Fig.3 - Contour plot of triatomic Hydrogen system using MEP calculation]]
|}

=== Change in initial conditions ===
If we change the initial condition so that r1 = rts and r2=rts+δ instead, the result is still identical to previous result but with atom A moving away at large t and atom B bonding with C instead. By investigating the “Internuclear Distances vs Time” and “Momenta vs Time” plots and taking note of values for r1(t) r2(t) and p1(t) p2(t) at very large t (50 fs), another set of calculations can be done with using these values for the initial positions and momentum (with reversed sign). The observation is for “Internuclear Distances vs Time” plot, the result is a reflection in the y-axis of Fig. 4 meaning, the bonded BC H2 molecule and the previously isolated HA atom are now approaching each other and their positions at large t corresponds to the transition state position. With reversed sign of the momenta, this is a reverse process of what was previously done, instead of atoms moving apart, they are now moving towards each other with the same kinetic energy. The “Momenta vs Time” is now reflected in the x-axis of Fig. 5, which can be reasoned by the repulsion that the atoms experienced from coming together resulting in positive gradient which is gradual at first then steepens at high t as the three atoms are now very close.

{| class="wikitable" border=1
|-
|[[File:Dtnl3818new.png|200px|thumb|left|Fig.4 - Dist. vs Time plot of H atoms moving apart]]||
[[File:Momentanewnl3818.png|200px|thumb|left|Fig.5 - Momenta vs Time plot of H atoms moving apart]]
|}

== Reactive and unreactive trajectories ==

To test whether trajectories starting at the same position but with higher momenta will all react, a table has been constructed shown below, where r1=74 pm and r2=200 pm.

{| class="wikitable" border=1
!p1/ g.mol-1.pm.fs-1!!p2/ g.mol-1.pm.fs-1!!Etot/kJ.mol-1!!Reactive?!!Description of the dynamics!!Illustration of the trajectory using Dist. vs Time plot
|-
| -2.56 || -5.1 || -414 || YES || This is a reactive trajectory because as the incoming H approaches, its AB distance decreases until it displays an oscillating behaviour which corresponds to the new bond vibration while the initial BC distance increases exponentially showing that the bond is broken and the leaving atom is moving further apart. ||[[File:Nl3818table1.png|200px|thumb|left]]
|-
| -3.1 || -4.1 || -420 || NO || This is not a reactive trajectory since although the AB distance decreases, it increases again while the initial BC distance retains its oscillating behaviour throughout meaning the bond is not broken. The incoming H approaches then moves away, no bond is broken in this case. ||[[File:Nl3818table2.png|200px|thumb|left]]
|-
| -3.1 || -5.1 || -412 || YES || This displays the same behaviour as the first case and thus, same explanation applies. ||[[File:Nl3818table3.png|200px|thumb|left]]
|-
| -5.1 || -10.1 || -357 || NO || In this case, the initial bond is broken but then reforms again, therefore, overall there is no reaction. ||[[File:Nl3818table4.png|200px|thumb|left]]
|-
|-5.1 || -10.6 || -349 || YES || The approaching H forms a new bond then bounces back but did not break the new bond so overall, this trajectory is reactive. ||[[File:Nl3818table5.png|200px|thumb|left]]
|}

[[File:Nl3818table5.png|200px|thumb|left]]
The last plot on the left corresponds to p1 = -5.1 g.mol-1.pm.fs-1, p2 = -10.6 g.mol-1.pm.fs-1 and Etot = -349 kJ.mol-1. This is a reactive trajectory since the approaching H forms a new bond then bounces back but did not break so overall, this trajectory is reactive.

*The last line of the table is not showing so I had to insert it like this.
{{fontcolor1|red| Well done. It would be nice to have a line or two here concluding what you have learned from these trajectories. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:49, 29 May 2020 (BST)}}

==Transition State Theory==

'''The transition state theory is a classical consideration which assumes:'''2

'''1) The reactants and transition state are in a quasi-equilibrium.''' This means that once the reactants reaches the transition state, they {{fontcolor1|red| cannot ? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:48, 29 May 2020 (BST)}} collapse back to form the reactants again when in reality they can. This cause the values that the transition state theory predicts to be overestimated than the actual experimental value.

'''2) Quantum tunnelling is ignored.''' Because the reactants that overcome the barrier by moving across is ignored, this causes the values predicted by transition state theory to be underestimated than the actual experimental values.

'''3) All reactants that have enough energy to overcome the barrier will successfully form the product and the step that involves the formation of the product from the transition state is the rate-determing step.''' This causes an overestimation of the actual rate because in reality, although the reactants have high energy, but if the energy is not located in the right place (i.e. in the bond that needs to be broken) then the reaction will not happen. {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:48, 29 May 2020 (BST)}}

== The F-H-H System==
=== PES Inspection===
===='''F + H2'''====
This is an unsymmetrical system which results in unsymmetrical potential wells in the surface plot.

From the surface plot, we are able to tell that this is an exothermic reaction. BC is the H-H distance of the reactant while AB is the H-F distance, the z-axis of the surface plot represents energy and in this case, the H-H is deeper in energy compared to H-F which corresponds to the energy profile of an exothermic reaction and thus, explains the greater H-F bond strength compared to H-H. {{fontcolor1|red| HF is deeper (lower E) because it is the stronger bond.. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:06, 29 May 2020 (BST)}}

The approximate position of the transition state can be found by the same method that was used for three Hydrogen atoms: trial and error where the corresponding forces are as close to zero as possible. This method gives rF-HA = 180 pm and rHA-HB=74.5 pm. These positions are confirmed to be the transition state by the opposite signs of the eigenvalues/vectors. At this position, the energy of the transition state is -434 kJmol-1. {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:06, 29 May 2020 (BST)}}

In order to find the activation energy, the position of the system is displaced slightly towards the reactant side from the transition state (rF-HA = 179 pm) and MEP calculation was used to obtain the "Energy vs Time" below.

From the graph, the energy of the reactants is -561 kJmol-1. The difference between the reactant state and the transition state energy is the activation energy which is the energy required to cross the barrier to form the product, in this case is +127 kJmol-1. {{fontcolor1|red| Yes. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:06, 29 May 2020 (BST)}}

{| class="wikitable" border=1
|-
|[[File:H2andHFsurface.png|200px|thumb|left|Fig.6 - Surface plot for F+H2]]||[[File:FH2Ea.png|200px|thumb|left|Fig.7 - Energy vs Time plot for F+H2]]
|}

==== H + HF ====
The surface plot of this reaction is a mirrored image of the above surface plot from the previous reaction. This shows that this is an endothermic reaction where the reactants have higher energy than the product. This correlates to the stronger H-F bond that requires input energy to break to form the H-H product.

The approximate transition state can be found in the same manner as the previous reaction and the positions are rF-HB = 180 pm and rHB-HC=74.5 pm, corresponding to total energy of -434 kJmol-1. The activation energy in this case is very small is reported to be +0.930 kJmol-1. Because of such small activation energy, it is easier to locate the position of the transition state by using the Hammond postulate. From the surface plot, we have deduced that this reaction is endothermic therefore it must have a late transition state meaning the structure of the transition state will resemble more of the product.3 As a result, shorter H-H distance and long F-H were predicted for the position of the transition state.

{| class="wikitable" border=1
|-
|[[File:HandHFsurface.png|200px|thumb|left|Fig.8 - Surface plot for H + HF]]||[[File:FHHEa.png|200px|thumb|left|Fig.9 - Energy vs Time plot for H + HF]]
|}

===Reaction dynamics===

One of the ways to confirm the mechanism of the release of reaction energy is by taking an Infrared absorption spectrum of the product. In this case, we can measure the IR spectrum of HF gas sample. When the reaction just happened and IR light is shown through, we are exciting the molecules that were once thermally relaxed and exclusively occupying the ground state to its first vibrational excited state and at early times (when the reaction just happened), we may also excitation from first to second vibrational excited state.

As a result, what we can observe in the IR spectrum is two peaks present, one is the fundamental peak and the other is an overtone appearing at a lower wavenumber. This is because of decreasing energy gap between each levels due to anharmonicity.4 By measuring the intensity of the overtone over time, we can extract the number of molecules that are vibrationally excited over time so we can expect the intensity of overtone to be smaller as energy is emitted as radiation and the intensity of the fundamental peak would increase as time goes on.

The other way to confirm the mechanism of energy release experimentally is by measuring the IR emitted directly, instead of probing the reaction using IR spectroscopy, as the molecules fall back from vibrational excited state to the ground state. This method can be done by using the Infrared Emission Spectroscopy (IES).5

====F + H2====

By setting up a calculation with rHH=74 pm, pFH=-1.0 g.mol-1.pm.fs-1 and pHH varied between -6.1 and 6.1 g.mol-1.pm.fs-1, it is visible from the contour plot that the trajectory only successfully rolled over to the product with pHH between -2.7 and 1.2 g.mol-1.pm.fs-1. Values outside of this range results in the trajectory hitting the wall and bouncing back.

{| class="wikitable" border=1
|-
|[[File:H2-2.7.png|200px|thumb|left|Fig.10 - H2 trajectory at pHH = -2.7 g.mol-1.pm.fs-1]]||
[[File:H20nl3818.png|200px|thumb|left|Fig.11 - H2 trajectory at pHH = 0.0 g.mol-1.pm.fs-1]]
|[[File:H21.2nl3818.png|200px|thumb|left|Fig.12 - H2 trajectory at pHH = 1.2 g.mol-1.pm.fs-1]]||
[[File:H2bouncingbacknl3818.png|200px|thumb|left|Fig.13 - trajectory at pHH = 0.2 g.mol-1.pm.fs-1, pFH = -1.6 g.mol-1.pm.fs-1]]
|}

====H + HF====
A reactive trajectory was obtained with a combination of pFH = 1 g.mol-1.pm.fs-1 and pHH = -7 g.mol-1.pm.fs-1. Decreasing the momentum of the incoming H further results in the trajectory hitting the wall and bouncing back while increasing the energy of H-F vibration causes the trajectory to go in the other direction rather than towards the product.

====Polanyi's Empirical Rule====
The above investigation illustrates the Polanyi's empirical rule. In an exothermic reaction, as demonstrated by F + H2, translational energy is favoured over vibrational energy because the trajectory can fall down into the low energy region of the PES whereas the vibrational energy is in a different direction and can cause the trajectory to bounce back near the transition state and cross back to the reactant state.

On the other hand, vibrational energy is favoured over translational energy in an endothermic reaction6 as illustrated by H + HF. This is because the trajectory has the same directionality with the vibrational energy that goes to the product.

==Bibliography==
1 Maxima, minima, and saddle points (article). https://www.khanacademy.org/math/multivariable-calculus/applications-of-multivariable-derivatives/optimizing-multivariable-functions/a/maximums-minimums-and-saddle-points (accessed May 22, 2020).

2 Peters, B. Transition State Theory. Reaction Rate Theory and Rare Events Simulations 2017, 227–271.

3 Libretexts. Hammond's Postulate. https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hammond’s_Postulate (accessed May 22, 2020).

4 Libretexts. 13.5: Vibrational Overtones. https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/13:_Molecular_Spectroscopy/13.05:_Vibrational_Overtones (accessed May 22, 2020).

5 Keresztury, G. Emission Spectroscopy, Infrared. Encyclopedia of Analytical Chemistry 2006.

6 Zhang, Z.; Zhou, Y.; Zhang, D. H.; Czakó, G.; Bowman, J. M. Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl CHD3 Reaction. The Journal of Physical Chemistry Letters 2012, 3 (23), 3416–3419.

MRD:ql2018

2020-05-29T18:04:19Z

Mak214: /* Exercise 2ː F - H - H System */

== Molecular Reaction Dynamics Lab ==

=== Exercise 1- (H+H2 System). ===
{{fontcolor|blue|Q1ː 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?}}
A transition state can be found as a saddle point on the potential energy surface diagram. Mathematically, the transition state is defined as the maximum on the minimum energy path linking reactants and the products. To distinguish the transition state from local minima,this can be done by viewing the potential energy surface from different angles. The transition state is the only point that is a minimum point from one angle(Figure 1)but a maximum from second angle(Figure 2). {{fontcolor1|red| IOK. Could you use some equations to define this also? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:23, 29 May 2020 (BST)}}
{{multiple image
| align = center
| direction = vertical
| width = 400
| header = Transition State at maximum and minimum
| image1 = Surface_Plot1_qw.png
| caption1 = Transition state is a maximum from one angle.
| image2 = Surface_Plot2_qw.png
| caption2 = Transition state is a minimum from another angle.
}}

{{fontcolor|blue|Q1ː 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.}}
My estimation for the transition state position rts is 90.7pm.Because for the transition state, the potential energies of both p1 and p2 are zero.There is no change of the energy which means that the gradients of potential energy surface is zero and there is no force acting on atoms.So r1 and r2 will keep constant.The graph corresponds should be a straight line since there is no oscillation. {{fontcolor1|red| Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:23, 29 May 2020 (BST)}}

[[File:Surface_Plot3_qw.png|thumb|center|Internuclear distances against time graph at r1=r2=90.7pm with zero momenta.]]

{{fontcolor|blue|Q3ːComment on how the mep and the trajectory you just calculated differ.}}
As the two graphs shown below, the difference is obvious that the trajectory on the contour plot of mep is shorter than that of dynamics. And there is also no oscillation compared with the graph generated by dynamics. The reason for this is because that there is zero kinetic energy for mep due to zero velocity and momentum, there will be no gain in vibtational energy as a result. Therefore the trajectory shows no oscillation. Also the change in total energy is different. Since there is no gain in kinetic energy, the total energy will drop as potential energy losses. {{fontcolor1|red| Yes. in MEP the momenta are reset to zero at every timestep so the system falls into the nearest local minimum. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:23, 29 May 2020 (BST)}}

For dynamics, the total energy is conserved as there is gain in kinetic energy when H2 formed.And due to this bond forming, the vibrational energy results in the oscillations on the contour plot.

{{multiple image
| align = left
| direction = vertical
| width = 300
| header = MEP and Dynamics
| image1 = Surface_Plot4_qw.png
| caption1 = MEP contour plot
| image2 = Surface_Plot5_qw.png
| caption2 = Dynamics contour plot
}}

{{fontcolor|blue|Q4ːComplete 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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot/ kJ.mol-1!! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes || This reaction started with small AB distance and large BC distance. It has enough momentum to pass the transition state region. After that region, the bond between BC formed and the bond between AB broken. The trajectry shows no fluctuation before BC bond formed, which is because the energy is mostly transitional energy. When BC bond formed, the energy is mostly vibrational energy which causes oscillation. ||[[File:Surface_Plot6_qw.png|thumb|upright=0.8]]
|-
| -3.1 || -4.1 || -420.077 || No || This reaction started with small AB distance and large BC distance. However, the momentum is not large enough to pass the transition state region. There is no break in bond between AB and no form in bond between BC. Molecule AB moves away forom C and the bond between AB keeps vibrating due to the kinetic energy. ||[[File:Surface_Plot7_qw.png|thumb|upright=0.8]]
|-
| -3.1 || -5.1 || -413.977 || Yes || This time the BC momentum is large enough for the system to pass transition state. AB bond vibrates before bond breaking due to kinetic energy. The vibration is even larger for BC due to larger momentum after the bond forming. ||[[File:Surface_Plot8_qw.png|thumb|upright=0.8]]
|-
| -5.1 || -10.1 || -357.277 || No || The energy is too large for this system. Even the AB bond breaks and the trajectory passes the transition state region, the bond formed between BC vibrates too strongly due to large energy that the bond fromed eventually breaks. The trajectory goes back and recrosses the transition state region. AB forms a bond again and as AB moves away from C, there is no vibration between AB bond. ||[[File:Surface_Plot9_qw.png|thumb|upright=0.8]]
|-
| -5.1 || -10.6 || -349.477 || Yes || The energy for this system is also very large that the trajectory passes through the transition state region for a couple of times. Although the bond formed between BC after the trajectory passes through the transition state region for the first time. The bond of BC breaks due to large energy and recrosses the transition state. However, because the system has more energy than the previous one.it had the energy to collide for a third time, the bond between BC forms again and became stable as BC vibrates and moves away from A. ||[[File:Surface_Plot10_qw.png|thumb|upright=0.8]]
|}

{{fontcolor1|red| Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:23, 29 May 2020 (BST)}}
{{fontcolor|blue|Q5ːGiven the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The transition state theory assumes that once the reactants have enough kinetic energy to pass through the transition state. The product must formed and the reaction would not go to reverse. However, the given experimental results indicate that it was not true. The products can recross the transition state and reform reactants. As a result, the transition state theory overestimates the reaction rate compared to experimental results. Also, due to the quantum tunneling effects, atoms with energy lower than activation energy can still pass through the barrier, which leads to a underestimation of reaction rate. {{fontcolor1|red| Good discussion - you should make reference to the relevant literature here to support what you are saying. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:51, 29 May 2020 (BST)}}

== Exercise 2ː F - H - H System ==
{{fontcolor|blue|Q6ː 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?}}

According to the potential energy surface graph F, H2 to H and HFis an exothermic reaction since it has higher potential energy for F and H2. Therefore H + HF to F, H2 is an endothermic reaction. This shows that the H-F bond is stronger than H-H bond. Because the formation of H-F bond releases more energy than the breaking of H-H bond. {{fontcolor1|red| Good. Could talk about the nature of HH vs HF to bulk out this answer a bit.[[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:03, 29 May 2020 (BST)}}

[[File:Surface_Plot11_qw.png|thumb|center|Potential energy surface of F-H-H system]]

{{fontcolor|blue|Q7ː Locate the approximate position of the transition state.}}
The approximate position of transition state for this reaction is when F-H isr1 = 181.300 pm and H-H isr2 = 74.483 pm. The momentum at this point is zero. {{fontcolor1|red| Yes. well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:03, 29 May 2020 (BST)}}
[[File:Surface_Plot12_qw.png|thumb|center|approximate position of the transition state of F-H-H system]]

{{fontcolor|blue|Q8ː Report the activation energy for both reactions.}}
The potential energy of the reactants F + H2 is -434.625 kJ.mol-1. The potential energy of the transition state is -433.981 kJ.mol-1. The activation energy is equal to the energy of transition state minus the energy of the reactants, which is 0.644 kJ.mol-1.

The potential energy of the products H + HF is -556.231 kJ.mol-1. The potential energy of the transition state is -433.981 kJ.mol-1. The activation energy of this reverse reaction is equal to the energy of transition state minus the energy of the products, which is 122.25 kJ.mol-1.

{{fontcolor1|red| Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:03, 29 May 2020 (BST)}}

{{multiple image
| align = center
| direction = vertical
| width = 300
| header = Activation Energies calculation
| image1 = Surface_Plot13_qw.png
| caption1 = Energy against time r1 = 184 pm, r2 = 74.483 pm
| image2 = Surface_Plot14_qw.png
| caption2 = Energy against time r1 = 176.000 pm, r2 = 74.483 pm
}}

{{fontcolor|blue|Q9ːIn light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.}}

The conditions are set that the the bond distance is 184pm between F and H and the bond distance is 75 between H and H. p1=-1 and p2= -2. From the graph, it can be seen that the product has greater vibration. This is because the potential energy transfer to kinetic energy and transitional energy.

The energy transfer can be confirmed by using IR spectroscopy. Because the vibration of HF bond can be detected by IR. Calorimetry can also be used to detect the heat generated during the reaction. However, this heat includes both vibrational and transitional energy. {{fontcolor1|red| Yes. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:03, 29 May 2020 (BST)}}

{| class="wikitable" border="1"
! Momentum vs Time
|- !! Contour Plot
|-
| [[File:Surface_Plot15_qw.png|400px|thumb|left]] || [[File:Surface_Plot16_qw.png|400px|thumb|right]]
|}
{{fontcolor|blue|Q10ːDiscuss 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.}}

From Polanyi's rule, It is known that the transitional energy can be more efficient for a exothermic reaction as it helps the reactants to pass through early transition state. In endothermic reaction, it has late transition state that the vibrational energy is more effective. It helps the reactants to pass the late transition state.1

{{fontcolor1|red| Nice report - would have great to have some further discussion, especially in the second part where you talk about polanyi's rules to show that you have thought about how this relates to what you have seen in your own simulations. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:03, 29 May 2020 (BST)}}

==Reference==
1.J.C. Polanyi. Some Concepts in Reaction Dynamics. Science 1987, 236 (4802): 680-690. doi:10.1126/science.236.4802.680

MRD:ql2018

2020-05-29T18:03:52Z

Mak214: /* Exercise 2ː F - H - H System */

== Molecular Reaction Dynamics Lab ==

=== Exercise 1- (H+H2 System). ===
{{fontcolor|blue|Q1ː 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?}}
A transition state can be found as a saddle point on the potential energy surface diagram. Mathematically, the transition state is defined as the maximum on the minimum energy path linking reactants and the products. To distinguish the transition state from local minima,this can be done by viewing the potential energy surface from different angles. The transition state is the only point that is a minimum point from one angle(Figure 1)but a maximum from second angle(Figure 2). {{fontcolor1|red| IOK. Could you use some equations to define this also? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:23, 29 May 2020 (BST)}}
{{multiple image
| align = center
| direction = vertical
| width = 400
| header = Transition State at maximum and minimum
| image1 = Surface_Plot1_qw.png
| caption1 = Transition state is a maximum from one angle.
| image2 = Surface_Plot2_qw.png
| caption2 = Transition state is a minimum from another angle.
}}

{{fontcolor|blue|Q1ː 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.}}
My estimation for the transition state position rts is 90.7pm.Because for the transition state, the potential energies of both p1 and p2 are zero.There is no change of the energy which means that the gradients of potential energy surface is zero and there is no force acting on atoms.So r1 and r2 will keep constant.The graph corresponds should be a straight line since there is no oscillation. {{fontcolor1|red| Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:23, 29 May 2020 (BST)}}

[[File:Surface_Plot3_qw.png|thumb|center|Internuclear distances against time graph at r1=r2=90.7pm with zero momenta.]]

{{fontcolor|blue|Q3ːComment on how the mep and the trajectory you just calculated differ.}}
As the two graphs shown below, the difference is obvious that the trajectory on the contour plot of mep is shorter than that of dynamics. And there is also no oscillation compared with the graph generated by dynamics. The reason for this is because that there is zero kinetic energy for mep due to zero velocity and momentum, there will be no gain in vibtational energy as a result. Therefore the trajectory shows no oscillation. Also the change in total energy is different. Since there is no gain in kinetic energy, the total energy will drop as potential energy losses. {{fontcolor1|red| Yes. in MEP the momenta are reset to zero at every timestep so the system falls into the nearest local minimum. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:23, 29 May 2020 (BST)}}

For dynamics, the total energy is conserved as there is gain in kinetic energy when H2 formed.And due to this bond forming, the vibrational energy results in the oscillations on the contour plot.

{{multiple image
| align = left
| direction = vertical
| width = 300
| header = MEP and Dynamics
| image1 = Surface_Plot4_qw.png
| caption1 = MEP contour plot
| image2 = Surface_Plot5_qw.png
| caption2 = Dynamics contour plot
}}

{{fontcolor|blue|Q4ːComplete 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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot/ kJ.mol-1!! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes || This reaction started with small AB distance and large BC distance. It has enough momentum to pass the transition state region. After that region, the bond between BC formed and the bond between AB broken. The trajectry shows no fluctuation before BC bond formed, which is because the energy is mostly transitional energy. When BC bond formed, the energy is mostly vibrational energy which causes oscillation. ||[[File:Surface_Plot6_qw.png|thumb|upright=0.8]]
|-
| -3.1 || -4.1 || -420.077 || No || This reaction started with small AB distance and large BC distance. However, the momentum is not large enough to pass the transition state region. There is no break in bond between AB and no form in bond between BC. Molecule AB moves away forom C and the bond between AB keeps vibrating due to the kinetic energy. ||[[File:Surface_Plot7_qw.png|thumb|upright=0.8]]
|-
| -3.1 || -5.1 || -413.977 || Yes || This time the BC momentum is large enough for the system to pass transition state. AB bond vibrates before bond breaking due to kinetic energy. The vibration is even larger for BC due to larger momentum after the bond forming. ||[[File:Surface_Plot8_qw.png|thumb|upright=0.8]]
|-
| -5.1 || -10.1 || -357.277 || No || The energy is too large for this system. Even the AB bond breaks and the trajectory passes the transition state region, the bond formed between BC vibrates too strongly due to large energy that the bond fromed eventually breaks. The trajectory goes back and recrosses the transition state region. AB forms a bond again and as AB moves away from C, there is no vibration between AB bond. ||[[File:Surface_Plot9_qw.png|thumb|upright=0.8]]
|-
| -5.1 || -10.6 || -349.477 || Yes || The energy for this system is also very large that the trajectory passes through the transition state region for a couple of times. Although the bond formed between BC after the trajectory passes through the transition state region for the first time. The bond of BC breaks due to large energy and recrosses the transition state. However, because the system has more energy than the previous one.it had the energy to collide for a third time, the bond between BC forms again and became stable as BC vibrates and moves away from A. ||[[File:Surface_Plot10_qw.png|thumb|upright=0.8]]
|}

{{fontcolor1|red| Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:23, 29 May 2020 (BST)}}
{{fontcolor|blue|Q5ːGiven the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The transition state theory assumes that once the reactants have enough kinetic energy to pass through the transition state. The product must formed and the reaction would not go to reverse. However, the given experimental results indicate that it was not true. The products can recross the transition state and reform reactants. As a result, the transition state theory overestimates the reaction rate compared to experimental results. Also, due to the quantum tunneling effects, atoms with energy lower than activation energy can still pass through the barrier, which leads to a underestimation of reaction rate. {{fontcolor1|red| Good discussion - you should make reference to the relevant literature here to support what you are saying. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:51, 29 May 2020 (BST)}}

== Exercise 2ː F - H - H System ==
{{fontcolor|blue|Q6ː 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?}}

According to the potential energy surface graph F, H2 to H and HFis an exothermic reaction since it has higher potential energy for F and H2. Therefore H + HF to F, H2 is an endothermic reaction. This shows that the H-F bond is stronger than H-H bond. Because the formation of H-F bond releases more energy than the breaking of H-H bond. {{fontcolor1|red| Good. Could talk about the nature of HH vs HF to bulk out this answer a bit.[[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:03, 29 May 2020 (BST)}}

[[File:Surface_Plot11_qw.png|thumb|center|Potential energy surface of F-H-H system]]

{{fontcolor|blue|Q7ː Locate the approximate position of the transition state.}}
The approximate position of transition state for this reaction is when F-H isr1 = 181.300 pm and H-H isr2 = 74.483 pm. The momentum at this point is zero. {{fontcolor1|red| Yes. well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:03, 29 May 2020 (BST)}}
[[File:Surface_Plot12_qw.png|thumb|center|approximate position of the transition state of F-H-H system]]

{{fontcolor|blue|Q8ː Report the activation energy for both reactions.}}
The potential energy of the reactants F + H2 is -434.625 kJ.mol-1. The potential energy of the transition state is -433.981 kJ.mol-1. The activation energy is equal to the energy of transition state minus the energy of the reactants, which is 0.644 kJ.mol-1.

The potential energy of the products H + HF is -556.231 kJ.mol-1. The potential energy of the transition state is -433.981 kJ.mol-1. The activation energy of this reverse reaction is equal to the energy of transition state minus the energy of the products, which is 122.25 kJ.mol-1.

{{fontcolor1|red| Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:03, 29 May 2020 (BST)}}

{{multiple image
| align = center
| direction = vertical
| width = 300
| header = Activation Energies calculation
| image1 = Surface_Plot13_qw.png
| caption1 = Energy against time r1 = 184 pm, r2 = 74.483 pm
| image2 = Surface_Plot14_qw.png
| caption2 = Energy against time r1 = 176.000 pm, r2 = 74.483 pm
}}

{{fontcolor|blue|Q9ːIn light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.}}

The conditions are set that the the bond distance is 184pm between F and H and the bond distance is 75 between H and H. p1=-1 and p2= -2. From the graph, it can be seen that the product has greater vibration. This is because the potential energy transfer to kinetic energy and transitional energy.

The energy transfer can be confirmed by using IR spectroscopy. Because the vibration of HF bond can be detected by IR. Calorimetry can also be used to detect the heat generated during the reaction. However, this heat includes both vibrational and transitional energy. {{fontcolor1|red| Yes. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:03, 29 May 2020 (BST)}}

{| class="wikitable" border="1"
! Momentum vs Time
|- !! Contour Plot
|-
| [[File:Surface_Plot15_qw.png|400px|thumb|left]] || [[File:Surface_Plot16_qw.png|400px|thumb|right]]
|}
{{fontcolor|blue|Q10ːDiscuss 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.}}

From Polanyi's rule, It is known that the transitional energy can be more efficient for a exothermic reaction as it helps the reactants to pass through early transition state. In endothermic reaction, it has late transition state that the vibrational energy is more effective. It helps the reactants to pass the late transition state.1

{{fontcolor1|red| Nice report - would have been nice to have some references and some further discussion, especially in the second part where you talk about polanyi's rules to show that you have thought about how this relates to what you have seen in your own simulations. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 19:03, 29 May 2020 (BST)}}

==Reference==
1.J.C. Polanyi. Some Concepts in Reaction Dynamics. Science 1987, 236 (4802): 680-690. doi:10.1126/science.236.4802.680

MRD：01508610

2020-05-29T17:59:26Z

Mak214: /* Reaction dynamics */

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

We will be looking at collision and reaction between diatomic molecule of hydrogen and an atom of H in a linear configuration in the gas phase. In simple terms, the atom A collides with the molecule B-C and forms a new molecule with B, while C is detached as a separate atom.
The interatomic distances between A and B, B and C are labelled as rAB and rBC respectively. {{fontcolor1|red| Nice! I like these diagrams you have drawn. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:25, 29 May 2020 (BST)}}

[[File:initial_describe_collision.png]]

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

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 molecular geometry and chemical reaction dynamics are analysed with a potential energy surface (PES) where the necessary points can be evaluated and classified according to first and second derivatives of the energy with respect to position, V(r), which respectively are the gradient and the curvature. Stationary points are those with zero gradient (∂V(ri)/∂ri=0) and have physical meaning: energy minima (or local minima of the PES) correspond to physically stable chemical species (such as reactant and product) and saddle points representing transition states, the maximum on the lowest energy pathway (or minimum energy path) that linking reactant and product.

• Mathematically, a saddle point is a point on the surface of a function where the slopes in orthogonal directions are all zero, but which is not a local extremum of the function. The transition state is a first-order saddle point. In this H + H2 system, the surface is symmetrical, which means that the transition state must be exactly half way between the reactants and products. In another words, the transition state lies somewhere on the diagonal line where the distances rAB and rBC are equal.

• The hydrogen atoms will have no movement at the saddle point since the force on atoms (i.e. the negative derivative of PES with respect to the coordinate ri where the force acting on) will be zero at transition state. Consequently, if one starts a trajectory exactly at the transition state, with no initial momentum, it will remain there forever. Therefore, the method to identify the transition state position from a PES is to change the values of rAB and rBC until the forces along AB and BC are both zero. In fact, forces are negative on one side of the PES and positive one the other side. The transition state is located where forces across zero. {{fontcolor1|red| Great. Thank you. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:25, 29 May 2020 (BST)}}

===Trajectories from rAB = rBC: locating the transition state===

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

• By testing different initial conditions with rAB = rBC, and p1 = p2 = 0.0 g.mol-1.pm.fs-1, the transition state position (rts) is found to be the point where distances rAB and rBC are equal to 90.78 pm. At that position, forces are both -0.001 KJ.mol-1.pm-1 (close to zero).

• Since the H + H2 surface is symmetrical, the transition state must have rAB = rBC. If we start a trajectory on the ridge rAB = rBC there is no gradient in the direction at right angles to the ridge, thus the trajectory will oscillate on the ridge and never fall off.

[[File:threeatomcolliderabrbc.png|600px]]

• To verify the transition state position, “Internuclear Distances vs Time” plots for some relevant rAB = rBC trajectories are shown below. In these graphical display the values of rAB(t), rBC(t) and rAC(t) are given (Y axis) against time t (X axis). At transition state, distance rAB and rBC (and thus rAC) are constant since all three atoms are stationary at the saddle point. The potential energy is maximized while the kinetic energy is zero. This geometry is transition structure.

[[File:90pm_90pm_5000number_gjs.png|thumb|center|A “Internuclear Distances vs Time” plot at transition state.]]

• In contrast, when initial conditions increased to rAB = rBC =100 pm, it shows an oscillatory behaviour for the whole trajectory corresponding to the vibration of atom A and atom C, since there are both potential and kinetic energy. The distance rAB and rBC are overlapped, since they have the same amplitude of vibration and the vibrations are symmetrical around atom B. The distance rAC) has doubled the amplitude of vibration (i.e. the sum or superposition of the amplitude). The oscillatory behaviour is changed after the transition state, at t≈270 fs, to a dissociation pattern where rAB (and thus rAC) keeps increasing with a slight oscillatory behaviour while rBC remained at the same level with a higher vibrational frequency. This means that the atom A leaves away from the molecule B-C with the translational energy while molecule B-C is left with vibrational energy, which resembles the reactant. {{fontcolor1|red| Well done. A clear response. Good use of these figures to show what you did. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:26, 29 May 2020 (BST)}}

[[File:Figure_2_100pm_internuc-time.png|thumb|center|A “Internuclear Distances vs Time” plot at rAB = rBC =100 pm.]]

===Trajectories from rBC = rts+δ, rAB = rts===

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

The reaction path (minimum energy path or mep) is calculated with these initial conditions involving:
• rBC = rts+1 pm = 91.78 pm;
• rAB = rts = 90.78 pm;
• pBC = pAB = 0 g.mol-1.pm.fs-1;
• then change the calculation type from dynamics to MEP;
• increase the number/size of steps to 5000/0.1(fs).

Then repeat the calculation with the same initial conditions used to calculate the reaction path, but change the calculation type back to Dynamics.

The calculated trajectories are shown in contour plots below, which both start from the transition state and simply follows the valley floor to HC + HA-HB. The mep is a smooth curve with no oscillatory behaviour, while the dynamic trajectory is an oscillatory curve.

[[File:Reaction_path_mep_jsg_.png|thumb|center|The contour plot of the reaction path calculated with mep.]]

[[File:Reaction_path_dynamic_jsg.png|thumb|center|The contour plot of the reaction path calculated with Dynamics.]]

The mep has the property that any point on the path is at an energy minimum in all directions perpendicular to the path. The mep can also be described as the union of steepest descent paths from the saddle points to the minima (that is why the mep passes through at least one first-order saddle point.) {{fontcolor1|red| Reference is needed here. We reset the momenta to zero at every timestep for the MEP calculation which means the system will always fall into the nearest minimum - If you started on a saddle point and had no momentum you shouldn't go anywhere! [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:31, 29 May 2020 (BST)}}

From the “Internuclear Distances vs Time” diagrams below, it shows a constant increasing rate of rBC (and thus rAC) after the transition state in Dynamic trajectory, whereas it shows a reducing rate of the increase of the rBC (and thus rAC) in the mep. In addition, the molecule A-B has slight vibrational (or oscillatory) behaviour but has no vibration in the mep.

[[File:Animation mep change of rate gjs123.png|thumb|center|The “Internuclear Distances vs Time” plot of the mep with rBC = rts+1 pm = 91.78 pm.]]

[[File:multiple_graph_123_jsg.png|thumb|center|The “Internuclear Distances vs Time” plot of Dynamic trajectory with rBC = rts+1 pm = 91.78 pm.]]

As mentioned before, the variation of momentum p (the forces acting on a given interatomic coordinate ri) will depend on the derivative of the potential energy surface with respect to that coordinate. Simply, the momenta over time, p(t), is the gradient of V(r).

[[File:EquationInnedgjd.PNG|center]]

{{fontcolor1|red| Reference for this equation? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:31, 29 May 2020 (BST)}}

From the “Momenta vs Time” diagrams below, the momenta of all atoms are zero in the mep since the momenta/velocities are always reset to zero in each time step when calculating the mep. while it have various values in Dynamic trajectory: the momentum of A-B decreases to slightly negative values and then rises to ≈2.5 g.mol-1.pm.fs-1 and keeps oscillating; whereas the momentum of B-C, similarly, decreases to negative values and then increases dramatically to and remains at ≈5 g.mol-1.pm.fs-1.

[[File:Mep momentia.vs.time-1000 gjs.png|thumb|center|The “Momenta vs Time” plot of the mep with rBC = rts+1 pm = 91.78 pm.]]

[[File:Dynamics momenta.vs.time 1000 gjs.png|thumb|center|The “Momenta vs Time” plot of Dynamic trajectory with rBC = rts+1 pm = 91.78 pm.]]

If we change the initial conditions to:
• rBC = rts = 90.78 pm;
• rAB = rts + 1 pm = 91.78 pm.
The “Internuclear Distances vs Time” and “Momenta vs Time” plots will look like this:

[[File:Changedinterdis-time-MEP-5000-gjs.png|thumb|center|The “Internuclear Distances vs Time” plot of the mep when rAB = rts + 1 pm.]]

[[File:Multiple graph 123 jsg.png|thumb|center|The “Internuclear Distances vs Time” plot of Dynamic trajectory when rAB = rts + 1 pm.]]

The “Internuclear Distances vs Time”and “Momenta vs Time” plots of both the mep and Dynamic trajectory have exactly the same pattern as those plots before the initial conditions are changed, except that line A-B (representing rAB) and line B-C (representing rBC) are interconverted. This means that the reaction proceeds in the same way but opposite direction (towards the valley floor to HA + HB-HC).

[[File:Mep momentia.vs.time-1000 gjs.png|thumb|center|The “Momenta vs Time” plot of the mep when rAB = rts + 1 pm.]]

[[File:Changesmomentia-vs-time-Dynamics1000-gjs.png|thumb|center|The “Momenta vs Time” plot of Dynamic trajectory when rAB = rts + 1 pm.]]

Take note of the final values of the positions r1(t) r2(t) and p1(t) p2(t) for your trajectory for large enough t.

• step number = 5000

• step size = 0.1 fs

• rBC = 74.77117297476231 pm

• rAB = 3777.013307800053 pm

• pBC = 1.9461593911826354 g.mol-1.pm.fs-1

• pAB = 5.073830709923566 g.mol-1.pm.fs-1

The atom A is extremely far from the molecule B-C and keeps moving away. The energies are:
• kinetic energy = +19.657 KJ.mol-1

• potential energy = -434.999 KJ.mol-1

• total energy = -415.342 KJ.mol-1

Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. What do you observe?

The energy terms are the same as above, but in this case, the reverse sign of momenta results in the reserve of the moving direction of the free atom (atom A in this case). The atom A is very far from the molecule B-C and it keeps approaching the molecule B-C.

===Reactive and unreactive trajectories===

For the initial positions rBC = 74 pm and rAB = 200 pm, run trajectories with the following momenta combination and complete the table.
Complete the table below 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?

{| class="wikitable" border="1"
|+ caption
! pBC/ g.mol-1.pm.fs-1 !! pAB/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes || The reaction can occurred. The atom A approaches the molecule B-C and forms a new molecule A-B which leaves away with most vibrational energy and a small amount of kinetic energy (KE), while atom C is detached as a separate atom which moves away with most KE. || [[File:R1=-2.5;r2=-5.1 jsg lastday.png|thumb|200px]]
|-
| -3.1 || -4.1 || -420.077 || No || The reaction doesn't occur. The atom A approaches the molecule B-C and return back without colliding with the molecule. || [[File:-3.1and-4.1_jsg_lastday.png|thumb|200px]]
|-
| -3.1 || -5.1 || -413.977 || Yes || The reaction can occur. The atom A approaches the molecule B-C and forms a new molecule A-B which leaves away with most vibrational energy and a small amount of kinetic energy (KE), while atom C is detached as a separate atom which moves away with most KE. || [[File:-3.1and-5.1 jsg lastday.png|thumb|200px]]
|-
| -5.1 || -10.1 || -357.277 || No || A case of barrier recrossing.. The atom A approaches the molecule B-C and forms a temporary new molecule A-B with strong vibrational motion and the molecule A-B move back and collide with atom C to form the B-C again.|| [[File:-5.1and-10.1 jsg lastday.png|thumb|200px]]
|-
| -5.1 || -10.6 || -349.477 || Yes || The reaction can occur. The atom A approaches the molecule B-C and break B-C bond and makes atom B bouncing in-between and eventually form a new molecule A-B. || [[File:-5.1and-10.6 jsg+lastday.png|thumb|200px]]
|}

From this table, you can observe that some trajectories have a total energy greater than the activation energy and still do not react. Therefore, for the reaction to proceed, there are two conditions should be met:(1) the total energy is lager than the activation energy; (2)the interaction between atoms should be appropriate for the collision to happen with correct energy and direction. For instance, if the approaching atom A move opposite to the molecule B-C, there will be no collision and the reaction will definitely not occur. {{fontcolor1|red| OK. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:31, 29 May 2020 (BST)}}

* (The transition state is found to be at rAB= 230.0 pm; rBC=74.0 pm; forces along AB and AC = +0.015 and +0.012 respectively; KE = 0; PE = total E = -434.471.)

===Transition State Theory===

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

Transition state theory (TST) explains the reaction rates of elementary chemical reactions. The theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes. These are three properties of TST:

1.Rates of reaction can be studied by examining activated complexes near the saddle point of a potential energy surface. The details of how these complexes are formed are not important. The saddle point itself is called the transition state.

2.The activated complexes are in a special equilibrium (quasi-equilibrium) with the reactant molecules.

3.The activated complexes can convert into products, and kinetic theory can be used to calculate the rate of this conversion.

TST is used primarily to understand qualitatively how chemical reactions take place. Simply, the Transition State Theory (TST) tells whether a reaction will happen if cross the TS barrier.

So the difference here is that our program is working with a single molecule, the TST deals with a bulk system with a Boltzmann speed distribution. The main differences between experimental and predicted rate constant values will be due to (1) ignoring barrier recrossing and (2) ignoring QM behaviour. {{fontcolor1|red| What will the effect of these phenomena be on the real rate of reaction relative to one predicted by TST?? More discussion is needed here. And references please! [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:52, 29 May 2020 (BST)}}

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

F + H2 --> HF + H

A F atom (A) approaches a hydrogen molecule (B-C) to forms a new HF molecule (A-B) and a separate H atom (C).

===PES inspection===

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?

Locate the approximate position of the transition state.

In order to classify the reaction based on energies, the total energies of reactant and that of product should be identified first by using mep calculation (in order to exclude kinetic energy) starting with a small displacement from the TS and with a high enough number of steps. It will go towards the formation of either the products or the reactants and the potential energy will tend to a constant value which is the total energy of the product/reactants depending which direction the displacement was.

The total energies of reactant and that of product are calculated to be:

• reactant: E(H2) = -435.099 KJ.mol-1;

• product: E(HF) = -560.402 KJ.mol-1.

Since the energy of the product is lower than that of reactant, this is an exothermic reaction. This makes sense with the concept of bond strength in which the H-F bond (565 KJ.mol-1) is much stronger than H-H bond (432 KJ.mol-1) and thus there will be more energy released during the formation of H-F bond.

The transition state is still a maximum or a saddle point on the PES, and found to be at:

• rAB = rH-F = 182 pm;

• rBC = rH-H = 74.2 pm;

• pAB = pBC = 0 g.mol-1.pm.fs-1;

• force along AB = +0.003 ≈0 KJ.mol-1.pm-1;

• forces along BC = +0.089 ≈0 KJ.mol-1.pm-1;

• the total energy = -433.969 KJ.mol-1.

{{fontcolor1|red| Good. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:54, 29 May 2020 (BST)}}

Because the activation energy for one of the reactions is so small, it is not easy to locate the transition state immediately. Use the Hammond postulate to guide your search.

Report the activation energy for both reactions.

The activation energy of the forward reaction is so small (1.13 KJ.mol-1), the Hammond postulate is introduced. Hammond postulate is a hypothesis which states that the transition state of a reaction resembles either the reactants or the products, to whichever it is closer in energy. In an exothermic reaction, the TS is closer to the reactants than to the products in energy. Therefore, the TS resembles the reactant in the reaction. In an exothermic reaction, the TS is closer to the products than to the reactants in energy and therefore, the TS resembles the products.

In F-H-H system, the activation energy is very similar to the energy of the reactant and hence the TS resembles reactant in this case. Therefore, this reaction (F + H2 --> HF + H) is exothermic in forward direction and thus endothermic in backward direction.

In order to classify the reaction based on energies, the total energies of reactant and that of product should be identified first by using mep calculation (in order to exclude kinetic energy) starting with a small displacement from the TS and with a high enough number of steps. It will go towards the formation of either the products or the reactants and the potential energy will tend to a constant value which is the total energy of the product/reactants depending which direction the displacement was.

• mep calculation: number of steps = 5000

Since the transition state is found to be at rAB = rH-F = 182 pm and rBC = rH-H = 74.2 pm, we either change rAB or rBC to achieve the formation of either reactants or products. Here is an example:

The reaction reacts towards the products when initial conditions are:

• rAB = rH-F = 165 pm;

• rBC = rH-H = 74.2 pm;

The "contour plot" indicating the reaction direction clearly and "Energy vs time" diagram which shows a decrease (lower in energy) of potential energy (and also the total energy) from -434.194 to -560.402 KJ.mol-1 are illustrated below.

[[File:Surface Plot 165 74.2 MEP gjs.png|thumb|center|The contour plot of FHH system when rAB =165 pm and rBC = 74.2 pm, calculated with mep.]]

[[File:Contour Plot 165 74.2 MEP gjs.png|thumb|center|The Energy.vs.time plot of FHH system when rAB =165 pm and rBC = 74.2 pm, calculated with mep.]]

Based on this mep calculation, the total energies of reactant and that of product are calculated to be:

• reactant: E(H2) = -435.099 KJ.mol-1;

• product: E(HF) = -560.402 KJ.mol-1.

Consequentially, since the TS energy = -433.969 KJ.mol-1, the forward reaction (F + H2 --> HF + H) has an activation energy of [-433.969-(-435.099)]=1.13 KJ.mol-1; the backward reaction (HF + H --> F + H2）has an activation energy of [-433.969-(-560.402)]=126.433 KJ.mol-1. The latter is much larger than the former. {{fontcolor1|red| Very good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:54, 29 May 2020 (BST)}}

===Reaction dynamics===

Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Momenta vs Time”

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.

For example, when initial conditions fulfil rAB = 160 pm and 0 < rBC < 76 pm, the trajectory are reactive.

The released energy is concerted thermally into electromagnetic radiations (photons). The relaxed potential energy well of electrons have photon absorption and emission between ground state E(0) and the first excited state E(1) and most of the electrons are populated in ground state. While the excited potential energy well involves more possible higher excited states, such as E(0) to the second excited state E(2), E(1) to E(2) which is called overtone. The overtone will contribute to a second absorbance peak at lower wavenumbers in IR spectra. {{fontcolor1|red| Most importantly we want to distinguish between vibrational and translational energy. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:59, 29 May 2020 (BST)}}

The emission of photons can be detected by a special type of IR called a Quantum Well Infrared Photodetector (QWIP) which is an infrared photodetector.

Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration.

Setting initial conditions as:
• rHH = 74 pm;
• rFH = 180 pm;
• pFH = -1.0 g.mol-1.pm.fs-1;

The value of pHH is varied from -6.1 to +6.1 and the resulting "Internuclear Distance.vs.Time" diagams used to show the trajectories are displayed below:

[[File:Q2-gjs-rage-6.1-123.png|thumb|center|pHH= -6.1|300px]] [[File:Q2-gjs-rage-6-123.png|thumb|center|pHH= -6|300px]] [[File:Q2-gjs-rage-5-123.png|thumb|center|pHH= -5|300px]] [[File:Q2-gjs-rage-4-123.png|thumb|center|pHH= -4|300px]] [[File:Q2-gjs-rage-3-123.png|thumb|center|pHH= -3|300px]] [[File:Q2-gjs-rage-2-123.png|thumb|center|pHH= -2|300px]] [[File:Q2-gjs-rage-1-123.png|thumb|center|pHH= -1|300px]] [[File:Q2-gjs-rage-0-123.png|thumb|center|pHH= 0|300px]] [[File:Q2-gjs-rage+1-123.png|thumb|center|pHH= 1|300px]] [[File:Q2-gjs-rage+2-123.png|thumb|center|pHH= 2|300px]] [[File:Q2-gjs-rage+3-123.png|thumb|center|pHH= 3|300px]] [[File:Q2-gjs-rage+4-123.png|thumb|center|pHH= 4|300px]] [[File:Q2-gjs-rage+5-123.png|thumb|center|pHH= 5|300px]] [[File:Q2-gjs-rage+6-123.png|thumb|center|pHH= 6|300px]] [[File:Q2-gjs-rage+6.1-123.png|thumb|center|pHH= 6.1|300px]]

In general, the excess of energy put into the system gives rise to the significant increase in the vibrational energy of H-H. when pHH is negative, the smaller the value (more positive) the more barrier recrossing. When pHH is positive, in most cases there is one barrier recrossing only except when p=+1,+5,+6.1. Even though some barrier recrossing occurs (F atom collides with H atom for several times), it doesn't necessarily mean that the reaction will proceed to form the product HF and a separated atom H.

For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?

The new trajectory is showm in the "Internuclear Distance.vs.Time" diagams below. The vibrating molecule H-H moves towards F and collides with it, making the atom B (H atom) bouncing in-between for three times and then return back to the other H atom to reform the vibrating hydrogen molecule.
[[File:3Q-gjs-changed-initials.png|thumb|center|pHH= 0.2|300px]]

Let us now focus on the reverse reaction, H + HF.

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

As mentioned before, the relaxed potential energy well of electrons have photon absorptions and emissions between ground state E(0) and the first excited state E(1) and most of the electrons are populated in ground state. While the excited potential energy well involves more possible higher excited states, such as E(0) to the second excited state E(2), E(1) to E(2) which is called overtone. The overtone will contribute to a second absorbance peak at lower wavenumbers in IR spectra. Each energy state correspond to a translational energy while the vibrational levels are smaller in energy difference and located around translational energy states.

Polanyi's empirical rules states that vibrational energy is more efficient in prompting a late-barrier reaction (that is, a transition state resembling the products) than translational energy, whereas the reverse is true for an early barrier reaction.

Translational energy is better at promoting exothermic reactions, because the trajectory can just fall into the lower energy region. Vibrational motion might make it harder for the trajectory to cross the barrier because it is in a different direction than the mep. If the reactants have vibrational motion, then they will try to move up the valley near the transition state, and just fall back to the reactants state. {{fontcolor1|red| OK. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:59, 29 May 2020 (BST)}}

{{fontcolor1|red| Good report - see a couple of comments. You need to be in the habit of referencing literature throughout your work to back up your discussion. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:59, 29 May 2020 (BST)}}

MRD：01508610

2020-05-29T17:54:49Z

Mak214: /* PES inspection */

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

We will be looking at collision and reaction between diatomic molecule of hydrogen and an atom of H in a linear configuration in the gas phase. In simple terms, the atom A collides with the molecule B-C and forms a new molecule with B, while C is detached as a separate atom.
The interatomic distances between A and B, B and C are labelled as rAB and rBC respectively. {{fontcolor1|red| Nice! I like these diagrams you have drawn. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:25, 29 May 2020 (BST)}}

[[File:initial_describe_collision.png]]

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

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 molecular geometry and chemical reaction dynamics are analysed with a potential energy surface (PES) where the necessary points can be evaluated and classified according to first and second derivatives of the energy with respect to position, V(r), which respectively are the gradient and the curvature. Stationary points are those with zero gradient (∂V(ri)/∂ri=0) and have physical meaning: energy minima (or local minima of the PES) correspond to physically stable chemical species (such as reactant and product) and saddle points representing transition states, the maximum on the lowest energy pathway (or minimum energy path) that linking reactant and product.

• Mathematically, a saddle point is a point on the surface of a function where the slopes in orthogonal directions are all zero, but which is not a local extremum of the function. The transition state is a first-order saddle point. In this H + H2 system, the surface is symmetrical, which means that the transition state must be exactly half way between the reactants and products. In another words, the transition state lies somewhere on the diagonal line where the distances rAB and rBC are equal.

• The hydrogen atoms will have no movement at the saddle point since the force on atoms (i.e. the negative derivative of PES with respect to the coordinate ri where the force acting on) will be zero at transition state. Consequently, if one starts a trajectory exactly at the transition state, with no initial momentum, it will remain there forever. Therefore, the method to identify the transition state position from a PES is to change the values of rAB and rBC until the forces along AB and BC are both zero. In fact, forces are negative on one side of the PES and positive one the other side. The transition state is located where forces across zero. {{fontcolor1|red| Great. Thank you. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:25, 29 May 2020 (BST)}}

===Trajectories from rAB = rBC: locating the transition state===

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

• By testing different initial conditions with rAB = rBC, and p1 = p2 = 0.0 g.mol-1.pm.fs-1, the transition state position (rts) is found to be the point where distances rAB and rBC are equal to 90.78 pm. At that position, forces are both -0.001 KJ.mol-1.pm-1 (close to zero).

• Since the H + H2 surface is symmetrical, the transition state must have rAB = rBC. If we start a trajectory on the ridge rAB = rBC there is no gradient in the direction at right angles to the ridge, thus the trajectory will oscillate on the ridge and never fall off.

[[File:threeatomcolliderabrbc.png|600px]]

• To verify the transition state position, “Internuclear Distances vs Time” plots for some relevant rAB = rBC trajectories are shown below. In these graphical display the values of rAB(t), rBC(t) and rAC(t) are given (Y axis) against time t (X axis). At transition state, distance rAB and rBC (and thus rAC) are constant since all three atoms are stationary at the saddle point. The potential energy is maximized while the kinetic energy is zero. This geometry is transition structure.

[[File:90pm_90pm_5000number_gjs.png|thumb|center|A “Internuclear Distances vs Time” plot at transition state.]]

• In contrast, when initial conditions increased to rAB = rBC =100 pm, it shows an oscillatory behaviour for the whole trajectory corresponding to the vibration of atom A and atom C, since there are both potential and kinetic energy. The distance rAB and rBC are overlapped, since they have the same amplitude of vibration and the vibrations are symmetrical around atom B. The distance rAC) has doubled the amplitude of vibration (i.e. the sum or superposition of the amplitude). The oscillatory behaviour is changed after the transition state, at t≈270 fs, to a dissociation pattern where rAB (and thus rAC) keeps increasing with a slight oscillatory behaviour while rBC remained at the same level with a higher vibrational frequency. This means that the atom A leaves away from the molecule B-C with the translational energy while molecule B-C is left with vibrational energy, which resembles the reactant. {{fontcolor1|red| Well done. A clear response. Good use of these figures to show what you did. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:26, 29 May 2020 (BST)}}

[[File:Figure_2_100pm_internuc-time.png|thumb|center|A “Internuclear Distances vs Time” plot at rAB = rBC =100 pm.]]

===Trajectories from rBC = rts+δ, rAB = rts===

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

The reaction path (minimum energy path or mep) is calculated with these initial conditions involving:
• rBC = rts+1 pm = 91.78 pm;
• rAB = rts = 90.78 pm;
• pBC = pAB = 0 g.mol-1.pm.fs-1;
• then change the calculation type from dynamics to MEP;
• increase the number/size of steps to 5000/0.1(fs).

Then repeat the calculation with the same initial conditions used to calculate the reaction path, but change the calculation type back to Dynamics.

The calculated trajectories are shown in contour plots below, which both start from the transition state and simply follows the valley floor to HC + HA-HB. The mep is a smooth curve with no oscillatory behaviour, while the dynamic trajectory is an oscillatory curve.

[[File:Reaction_path_mep_jsg_.png|thumb|center|The contour plot of the reaction path calculated with mep.]]

[[File:Reaction_path_dynamic_jsg.png|thumb|center|The contour plot of the reaction path calculated with Dynamics.]]

The mep has the property that any point on the path is at an energy minimum in all directions perpendicular to the path. The mep can also be described as the union of steepest descent paths from the saddle points to the minima (that is why the mep passes through at least one first-order saddle point.) {{fontcolor1|red| Reference is needed here. We reset the momenta to zero at every timestep for the MEP calculation which means the system will always fall into the nearest minimum - If you started on a saddle point and had no momentum you shouldn't go anywhere! [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:31, 29 May 2020 (BST)}}

From the “Internuclear Distances vs Time” diagrams below, it shows a constant increasing rate of rBC (and thus rAC) after the transition state in Dynamic trajectory, whereas it shows a reducing rate of the increase of the rBC (and thus rAC) in the mep. In addition, the molecule A-B has slight vibrational (or oscillatory) behaviour but has no vibration in the mep.

[[File:Animation mep change of rate gjs123.png|thumb|center|The “Internuclear Distances vs Time” plot of the mep with rBC = rts+1 pm = 91.78 pm.]]

[[File:multiple_graph_123_jsg.png|thumb|center|The “Internuclear Distances vs Time” plot of Dynamic trajectory with rBC = rts+1 pm = 91.78 pm.]]

As mentioned before, the variation of momentum p (the forces acting on a given interatomic coordinate ri) will depend on the derivative of the potential energy surface with respect to that coordinate. Simply, the momenta over time, p(t), is the gradient of V(r).

[[File:EquationInnedgjd.PNG|center]]

{{fontcolor1|red| Reference for this equation? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:31, 29 May 2020 (BST)}}

From the “Momenta vs Time” diagrams below, the momenta of all atoms are zero in the mep since the momenta/velocities are always reset to zero in each time step when calculating the mep. while it have various values in Dynamic trajectory: the momentum of A-B decreases to slightly negative values and then rises to ≈2.5 g.mol-1.pm.fs-1 and keeps oscillating; whereas the momentum of B-C, similarly, decreases to negative values and then increases dramatically to and remains at ≈5 g.mol-1.pm.fs-1.

[[File:Mep momentia.vs.time-1000 gjs.png|thumb|center|The “Momenta vs Time” plot of the mep with rBC = rts+1 pm = 91.78 pm.]]

[[File:Dynamics momenta.vs.time 1000 gjs.png|thumb|center|The “Momenta vs Time” plot of Dynamic trajectory with rBC = rts+1 pm = 91.78 pm.]]

If we change the initial conditions to:
• rBC = rts = 90.78 pm;
• rAB = rts + 1 pm = 91.78 pm.
The “Internuclear Distances vs Time” and “Momenta vs Time” plots will look like this:

[[File:Changedinterdis-time-MEP-5000-gjs.png|thumb|center|The “Internuclear Distances vs Time” plot of the mep when rAB = rts + 1 pm.]]

[[File:Multiple graph 123 jsg.png|thumb|center|The “Internuclear Distances vs Time” plot of Dynamic trajectory when rAB = rts + 1 pm.]]

The “Internuclear Distances vs Time”and “Momenta vs Time” plots of both the mep and Dynamic trajectory have exactly the same pattern as those plots before the initial conditions are changed, except that line A-B (representing rAB) and line B-C (representing rBC) are interconverted. This means that the reaction proceeds in the same way but opposite direction (towards the valley floor to HA + HB-HC).

[[File:Mep momentia.vs.time-1000 gjs.png|thumb|center|The “Momenta vs Time” plot of the mep when rAB = rts + 1 pm.]]

[[File:Changesmomentia-vs-time-Dynamics1000-gjs.png|thumb|center|The “Momenta vs Time” plot of Dynamic trajectory when rAB = rts + 1 pm.]]

Take note of the final values of the positions r1(t) r2(t) and p1(t) p2(t) for your trajectory for large enough t.

• step number = 5000

• step size = 0.1 fs

• rBC = 74.77117297476231 pm

• rAB = 3777.013307800053 pm

• pBC = 1.9461593911826354 g.mol-1.pm.fs-1

• pAB = 5.073830709923566 g.mol-1.pm.fs-1

The atom A is extremely far from the molecule B-C and keeps moving away. The energies are:
• kinetic energy = +19.657 KJ.mol-1

• potential energy = -434.999 KJ.mol-1

• total energy = -415.342 KJ.mol-1

Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. What do you observe?

The energy terms are the same as above, but in this case, the reverse sign of momenta results in the reserve of the moving direction of the free atom (atom A in this case). The atom A is very far from the molecule B-C and it keeps approaching the molecule B-C.

===Reactive and unreactive trajectories===

For the initial positions rBC = 74 pm and rAB = 200 pm, run trajectories with the following momenta combination and complete the table.
Complete the table below 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?

{| class="wikitable" border="1"
|+ caption
! pBC/ g.mol-1.pm.fs-1 !! pAB/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes || The reaction can occurred. The atom A approaches the molecule B-C and forms a new molecule A-B which leaves away with most vibrational energy and a small amount of kinetic energy (KE), while atom C is detached as a separate atom which moves away with most KE. || [[File:R1=-2.5;r2=-5.1 jsg lastday.png|thumb|200px]]
|-
| -3.1 || -4.1 || -420.077 || No || The reaction doesn't occur. The atom A approaches the molecule B-C and return back without colliding with the molecule. || [[File:-3.1and-4.1_jsg_lastday.png|thumb|200px]]
|-
| -3.1 || -5.1 || -413.977 || Yes || The reaction can occur. The atom A approaches the molecule B-C and forms a new molecule A-B which leaves away with most vibrational energy and a small amount of kinetic energy (KE), while atom C is detached as a separate atom which moves away with most KE. || [[File:-3.1and-5.1 jsg lastday.png|thumb|200px]]
|-
| -5.1 || -10.1 || -357.277 || No || A case of barrier recrossing.. The atom A approaches the molecule B-C and forms a temporary new molecule A-B with strong vibrational motion and the molecule A-B move back and collide with atom C to form the B-C again.|| [[File:-5.1and-10.1 jsg lastday.png|thumb|200px]]
|-
| -5.1 || -10.6 || -349.477 || Yes || The reaction can occur. The atom A approaches the molecule B-C and break B-C bond and makes atom B bouncing in-between and eventually form a new molecule A-B. || [[File:-5.1and-10.6 jsg+lastday.png|thumb|200px]]
|}

From this table, you can observe that some trajectories have a total energy greater than the activation energy and still do not react. Therefore, for the reaction to proceed, there are two conditions should be met:(1) the total energy is lager than the activation energy; (2)the interaction between atoms should be appropriate for the collision to happen with correct energy and direction. For instance, if the approaching atom A move opposite to the molecule B-C, there will be no collision and the reaction will definitely not occur. {{fontcolor1|red| OK. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:31, 29 May 2020 (BST)}}

* (The transition state is found to be at rAB= 230.0 pm; rBC=74.0 pm; forces along AB and AC = +0.015 and +0.012 respectively; KE = 0; PE = total E = -434.471.)

===Transition State Theory===

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

Transition state theory (TST) explains the reaction rates of elementary chemical reactions. The theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes. These are three properties of TST:

1.Rates of reaction can be studied by examining activated complexes near the saddle point of a potential energy surface. The details of how these complexes are formed are not important. The saddle point itself is called the transition state.

2.The activated complexes are in a special equilibrium (quasi-equilibrium) with the reactant molecules.

3.The activated complexes can convert into products, and kinetic theory can be used to calculate the rate of this conversion.

TST is used primarily to understand qualitatively how chemical reactions take place. Simply, the Transition State Theory (TST) tells whether a reaction will happen if cross the TS barrier.

So the difference here is that our program is working with a single molecule, the TST deals with a bulk system with a Boltzmann speed distribution. The main differences between experimental and predicted rate constant values will be due to (1) ignoring barrier recrossing and (2) ignoring QM behaviour. {{fontcolor1|red| What will the effect of these phenomena be on the real rate of reaction relative to one predicted by TST?? More discussion is needed here. And references please! [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:52, 29 May 2020 (BST)}}

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

F + H2 --> HF + H

A F atom (A) approaches a hydrogen molecule (B-C) to forms a new HF molecule (A-B) and a separate H atom (C).

===PES inspection===

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?

Locate the approximate position of the transition state.

In order to classify the reaction based on energies, the total energies of reactant and that of product should be identified first by using mep calculation (in order to exclude kinetic energy) starting with a small displacement from the TS and with a high enough number of steps. It will go towards the formation of either the products or the reactants and the potential energy will tend to a constant value which is the total energy of the product/reactants depending which direction the displacement was.

The total energies of reactant and that of product are calculated to be:

• reactant: E(H2) = -435.099 KJ.mol-1;

• product: E(HF) = -560.402 KJ.mol-1.

Since the energy of the product is lower than that of reactant, this is an exothermic reaction. This makes sense with the concept of bond strength in which the H-F bond (565 KJ.mol-1) is much stronger than H-H bond (432 KJ.mol-1) and thus there will be more energy released during the formation of H-F bond.

The transition state is still a maximum or a saddle point on the PES, and found to be at:

• rAB = rH-F = 182 pm;

• rBC = rH-H = 74.2 pm;

• pAB = pBC = 0 g.mol-1.pm.fs-1;

• force along AB = +0.003 ≈0 KJ.mol-1.pm-1;

• forces along BC = +0.089 ≈0 KJ.mol-1.pm-1;

• the total energy = -433.969 KJ.mol-1.

{{fontcolor1|red| Good. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:54, 29 May 2020 (BST)}}

Because the activation energy for one of the reactions is so small, it is not easy to locate the transition state immediately. Use the Hammond postulate to guide your search.

Report the activation energy for both reactions.

The activation energy of the forward reaction is so small (1.13 KJ.mol-1), the Hammond postulate is introduced. Hammond postulate is a hypothesis which states that the transition state of a reaction resembles either the reactants or the products, to whichever it is closer in energy. In an exothermic reaction, the TS is closer to the reactants than to the products in energy. Therefore, the TS resembles the reactant in the reaction. In an exothermic reaction, the TS is closer to the products than to the reactants in energy and therefore, the TS resembles the products.

In F-H-H system, the activation energy is very similar to the energy of the reactant and hence the TS resembles reactant in this case. Therefore, this reaction (F + H2 --> HF + H) is exothermic in forward direction and thus endothermic in backward direction.

In order to classify the reaction based on energies, the total energies of reactant and that of product should be identified first by using mep calculation (in order to exclude kinetic energy) starting with a small displacement from the TS and with a high enough number of steps. It will go towards the formation of either the products or the reactants and the potential energy will tend to a constant value which is the total energy of the product/reactants depending which direction the displacement was.

• mep calculation: number of steps = 5000

Since the transition state is found to be at rAB = rH-F = 182 pm and rBC = rH-H = 74.2 pm, we either change rAB or rBC to achieve the formation of either reactants or products. Here is an example:

The reaction reacts towards the products when initial conditions are:

• rAB = rH-F = 165 pm;

• rBC = rH-H = 74.2 pm;

The "contour plot" indicating the reaction direction clearly and "Energy vs time" diagram which shows a decrease (lower in energy) of potential energy (and also the total energy) from -434.194 to -560.402 KJ.mol-1 are illustrated below.

[[File:Surface Plot 165 74.2 MEP gjs.png|thumb|center|The contour plot of FHH system when rAB =165 pm and rBC = 74.2 pm, calculated with mep.]]

[[File:Contour Plot 165 74.2 MEP gjs.png|thumb|center|The Energy.vs.time plot of FHH system when rAB =165 pm and rBC = 74.2 pm, calculated with mep.]]

Based on this mep calculation, the total energies of reactant and that of product are calculated to be:

• reactant: E(H2) = -435.099 KJ.mol-1;

• product: E(HF) = -560.402 KJ.mol-1.

Consequentially, since the TS energy = -433.969 KJ.mol-1, the forward reaction (F + H2 --> HF + H) has an activation energy of [-433.969-(-435.099)]=1.13 KJ.mol-1; the backward reaction (HF + H --> F + H2）has an activation energy of [-433.969-(-560.402)]=126.433 KJ.mol-1. The latter is much larger than the former. {{fontcolor1|red| Very good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:54, 29 May 2020 (BST)}}

===Reaction dynamics===

Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Momenta vs Time”

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.

For example, when initial conditions fulfil rAB = 160 pm and 0 < rBC < 76 pm, the trajectory are reactive.

The released energy is concerted thermally into electromagnetic radiations (photons). The relaxed potential energy well of electrons have photon absorption and emission between ground state E(0) and the first excited state E(1) and most of the electrons are populated in ground state. While the excited potential energy well involves more possible higher excited states, such as E(0) to the second excited state E(2), E(1) to E(2) which is called overtone. The overtone will contribute to a second absorbance peak at lower wavenumbers in IR spectra.

The emission of photons can be detected by a special type of IR called a Quantum Well Infrared Photodetector (QWIP) which is an infrared photodetector.

Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration.

Setting initial conditions as:
• rHH = 74 pm;
• rFH = 180 pm;
• pFH = -1.0 g.mol-1.pm.fs-1;

The value of pHH is varied from -6.1 to +6.1 and the resulting "Internuclear Distance.vs.Time" diagams used to show the trajectories are displayed below:

[[File:Q2-gjs-rage-6.1-123.png|thumb|center|pHH= -6.1|300px]] [[File:Q2-gjs-rage-6-123.png|thumb|center|pHH= -6|300px]] [[File:Q2-gjs-rage-5-123.png|thumb|center|pHH= -5|300px]] [[File:Q2-gjs-rage-4-123.png|thumb|center|pHH= -4|300px]] [[File:Q2-gjs-rage-3-123.png|thumb|center|pHH= -3|300px]] [[File:Q2-gjs-rage-2-123.png|thumb|center|pHH= -2|300px]] [[File:Q2-gjs-rage-1-123.png|thumb|center|pHH= -1|300px]] [[File:Q2-gjs-rage-0-123.png|thumb|center|pHH= 0|300px]] [[File:Q2-gjs-rage+1-123.png|thumb|center|pHH= 1|300px]] [[File:Q2-gjs-rage+2-123.png|thumb|center|pHH= 2|300px]] [[File:Q2-gjs-rage+3-123.png|thumb|center|pHH= 3|300px]] [[File:Q2-gjs-rage+4-123.png|thumb|center|pHH= 4|300px]] [[File:Q2-gjs-rage+5-123.png|thumb|center|pHH= 5|300px]] [[File:Q2-gjs-rage+6-123.png|thumb|center|pHH= 6|300px]] [[File:Q2-gjs-rage+6.1-123.png|thumb|center|pHH= 6.1|300px]]

In general, the excess of energy put into the system gives rise to the significant increase in the vibrational energy of H-H. when pHH is negative, the smaller the value (more positive) the more barrier recrossing. When pHH is positive, in most cases there is one barrier recrossing only except when p=+1,+5,+6.1. Even though some barrier recrossing occurs (F atom collides with H atom for several times), it doesn't necessarily mean that the reaction will proceed to form the product HF and a separated atom H.

For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?

The new trajectory is showm in the "Internuclear Distance.vs.Time" diagams below. The vibrating molecule H-H moves towards F and collides with it, making the atom B (H atom) bouncing in-between for three times and then return back to the other H atom to reform the vibrating hydrogen molecule.
[[File:3Q-gjs-changed-initials.png|thumb|center|pHH= 0.2|300px]]

Let us now focus on the reverse reaction, H + HF.

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

As mentioned before, the relaxed potential energy well of electrons have photon absorptions and emissions between ground state E(0) and the first excited state E(1) and most of the electrons are populated in ground state. While the excited potential energy well involves more possible higher excited states, such as E(0) to the second excited state E(2), E(1) to E(2) which is called overtone. The overtone will contribute to a second absorbance peak at lower wavenumbers in IR spectra. Each energy state correspond to a translational energy while the vibrational levels are smaller in energy difference and located around translational energy states.

Polanyi's empirical rules states that vibrational energy is more efficient in prompting a late-barrier reaction (that is, a transition state resembling the products) than translational energy, whereas the reverse is true for an early barrier reaction.

Translational energy is better at promoting exothermic reactions, because the trajectory can just fall into the lower energy region. Vibrational motion might make it harder for the trajectory to cross the barrier because it is in a different direction than the mep. If the reactants have vibrational motion, then they will try to move up the valley near the transition state, and just fall back to the reactants state.

MRD：01508610

2020-05-29T17:52:53Z

Mak214: /* Transition State Theory */

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

We will be looking at collision and reaction between diatomic molecule of hydrogen and an atom of H in a linear configuration in the gas phase. In simple terms, the atom A collides with the molecule B-C and forms a new molecule with B, while C is detached as a separate atom.
The interatomic distances between A and B, B and C are labelled as rAB and rBC respectively. {{fontcolor1|red| Nice! I like these diagrams you have drawn. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:25, 29 May 2020 (BST)}}

[[File:initial_describe_collision.png]]

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

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 molecular geometry and chemical reaction dynamics are analysed with a potential energy surface (PES) where the necessary points can be evaluated and classified according to first and second derivatives of the energy with respect to position, V(r), which respectively are the gradient and the curvature. Stationary points are those with zero gradient (∂V(ri)/∂ri=0) and have physical meaning: energy minima (or local minima of the PES) correspond to physically stable chemical species (such as reactant and product) and saddle points representing transition states, the maximum on the lowest energy pathway (or minimum energy path) that linking reactant and product.

• Mathematically, a saddle point is a point on the surface of a function where the slopes in orthogonal directions are all zero, but which is not a local extremum of the function. The transition state is a first-order saddle point. In this H + H2 system, the surface is symmetrical, which means that the transition state must be exactly half way between the reactants and products. In another words, the transition state lies somewhere on the diagonal line where the distances rAB and rBC are equal.

• The hydrogen atoms will have no movement at the saddle point since the force on atoms (i.e. the negative derivative of PES with respect to the coordinate ri where the force acting on) will be zero at transition state. Consequently, if one starts a trajectory exactly at the transition state, with no initial momentum, it will remain there forever. Therefore, the method to identify the transition state position from a PES is to change the values of rAB and rBC until the forces along AB and BC are both zero. In fact, forces are negative on one side of the PES and positive one the other side. The transition state is located where forces across zero. {{fontcolor1|red| Great. Thank you. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:25, 29 May 2020 (BST)}}

===Trajectories from rAB = rBC: locating the transition state===

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

• By testing different initial conditions with rAB = rBC, and p1 = p2 = 0.0 g.mol-1.pm.fs-1, the transition state position (rts) is found to be the point where distances rAB and rBC are equal to 90.78 pm. At that position, forces are both -0.001 KJ.mol-1.pm-1 (close to zero).

• Since the H + H2 surface is symmetrical, the transition state must have rAB = rBC. If we start a trajectory on the ridge rAB = rBC there is no gradient in the direction at right angles to the ridge, thus the trajectory will oscillate on the ridge and never fall off.

[[File:threeatomcolliderabrbc.png|600px]]

• To verify the transition state position, “Internuclear Distances vs Time” plots for some relevant rAB = rBC trajectories are shown below. In these graphical display the values of rAB(t), rBC(t) and rAC(t) are given (Y axis) against time t (X axis). At transition state, distance rAB and rBC (and thus rAC) are constant since all three atoms are stationary at the saddle point. The potential energy is maximized while the kinetic energy is zero. This geometry is transition structure.

[[File:90pm_90pm_5000number_gjs.png|thumb|center|A “Internuclear Distances vs Time” plot at transition state.]]

• In contrast, when initial conditions increased to rAB = rBC =100 pm, it shows an oscillatory behaviour for the whole trajectory corresponding to the vibration of atom A and atom C, since there are both potential and kinetic energy. The distance rAB and rBC are overlapped, since they have the same amplitude of vibration and the vibrations are symmetrical around atom B. The distance rAC) has doubled the amplitude of vibration (i.e. the sum or superposition of the amplitude). The oscillatory behaviour is changed after the transition state, at t≈270 fs, to a dissociation pattern where rAB (and thus rAC) keeps increasing with a slight oscillatory behaviour while rBC remained at the same level with a higher vibrational frequency. This means that the atom A leaves away from the molecule B-C with the translational energy while molecule B-C is left with vibrational energy, which resembles the reactant. {{fontcolor1|red| Well done. A clear response. Good use of these figures to show what you did. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:26, 29 May 2020 (BST)}}

[[File:Figure_2_100pm_internuc-time.png|thumb|center|A “Internuclear Distances vs Time” plot at rAB = rBC =100 pm.]]

===Trajectories from rBC = rts+δ, rAB = rts===

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

The reaction path (minimum energy path or mep) is calculated with these initial conditions involving:
• rBC = rts+1 pm = 91.78 pm;
• rAB = rts = 90.78 pm;
• pBC = pAB = 0 g.mol-1.pm.fs-1;
• then change the calculation type from dynamics to MEP;
• increase the number/size of steps to 5000/0.1(fs).

Then repeat the calculation with the same initial conditions used to calculate the reaction path, but change the calculation type back to Dynamics.

The calculated trajectories are shown in contour plots below, which both start from the transition state and simply follows the valley floor to HC + HA-HB. The mep is a smooth curve with no oscillatory behaviour, while the dynamic trajectory is an oscillatory curve.

[[File:Reaction_path_mep_jsg_.png|thumb|center|The contour plot of the reaction path calculated with mep.]]

[[File:Reaction_path_dynamic_jsg.png|thumb|center|The contour plot of the reaction path calculated with Dynamics.]]

The mep has the property that any point on the path is at an energy minimum in all directions perpendicular to the path. The mep can also be described as the union of steepest descent paths from the saddle points to the minima (that is why the mep passes through at least one first-order saddle point.) {{fontcolor1|red| Reference is needed here. We reset the momenta to zero at every timestep for the MEP calculation which means the system will always fall into the nearest minimum - If you started on a saddle point and had no momentum you shouldn't go anywhere! [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:31, 29 May 2020 (BST)}}

From the “Internuclear Distances vs Time” diagrams below, it shows a constant increasing rate of rBC (and thus rAC) after the transition state in Dynamic trajectory, whereas it shows a reducing rate of the increase of the rBC (and thus rAC) in the mep. In addition, the molecule A-B has slight vibrational (or oscillatory) behaviour but has no vibration in the mep.

[[File:Animation mep change of rate gjs123.png|thumb|center|The “Internuclear Distances vs Time” plot of the mep with rBC = rts+1 pm = 91.78 pm.]]

[[File:multiple_graph_123_jsg.png|thumb|center|The “Internuclear Distances vs Time” plot of Dynamic trajectory with rBC = rts+1 pm = 91.78 pm.]]

As mentioned before, the variation of momentum p (the forces acting on a given interatomic coordinate ri) will depend on the derivative of the potential energy surface with respect to that coordinate. Simply, the momenta over time, p(t), is the gradient of V(r).

[[File:EquationInnedgjd.PNG|center]]

{{fontcolor1|red| Reference for this equation? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:31, 29 May 2020 (BST)}}

From the “Momenta vs Time” diagrams below, the momenta of all atoms are zero in the mep since the momenta/velocities are always reset to zero in each time step when calculating the mep. while it have various values in Dynamic trajectory: the momentum of A-B decreases to slightly negative values and then rises to ≈2.5 g.mol-1.pm.fs-1 and keeps oscillating; whereas the momentum of B-C, similarly, decreases to negative values and then increases dramatically to and remains at ≈5 g.mol-1.pm.fs-1.

[[File:Mep momentia.vs.time-1000 gjs.png|thumb|center|The “Momenta vs Time” plot of the mep with rBC = rts+1 pm = 91.78 pm.]]

[[File:Dynamics momenta.vs.time 1000 gjs.png|thumb|center|The “Momenta vs Time” plot of Dynamic trajectory with rBC = rts+1 pm = 91.78 pm.]]

If we change the initial conditions to:
• rBC = rts = 90.78 pm;
• rAB = rts + 1 pm = 91.78 pm.
The “Internuclear Distances vs Time” and “Momenta vs Time” plots will look like this:

[[File:Changedinterdis-time-MEP-5000-gjs.png|thumb|center|The “Internuclear Distances vs Time” plot of the mep when rAB = rts + 1 pm.]]

[[File:Multiple graph 123 jsg.png|thumb|center|The “Internuclear Distances vs Time” plot of Dynamic trajectory when rAB = rts + 1 pm.]]

The “Internuclear Distances vs Time”and “Momenta vs Time” plots of both the mep and Dynamic trajectory have exactly the same pattern as those plots before the initial conditions are changed, except that line A-B (representing rAB) and line B-C (representing rBC) are interconverted. This means that the reaction proceeds in the same way but opposite direction (towards the valley floor to HA + HB-HC).

[[File:Mep momentia.vs.time-1000 gjs.png|thumb|center|The “Momenta vs Time” plot of the mep when rAB = rts + 1 pm.]]

[[File:Changesmomentia-vs-time-Dynamics1000-gjs.png|thumb|center|The “Momenta vs Time” plot of Dynamic trajectory when rAB = rts + 1 pm.]]

Take note of the final values of the positions r1(t) r2(t) and p1(t) p2(t) for your trajectory for large enough t.

• step number = 5000

• step size = 0.1 fs

• rBC = 74.77117297476231 pm

• rAB = 3777.013307800053 pm

• pBC = 1.9461593911826354 g.mol-1.pm.fs-1

• pAB = 5.073830709923566 g.mol-1.pm.fs-1

The atom A is extremely far from the molecule B-C and keeps moving away. The energies are:
• kinetic energy = +19.657 KJ.mol-1

• potential energy = -434.999 KJ.mol-1

• total energy = -415.342 KJ.mol-1

Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. What do you observe?

The energy terms are the same as above, but in this case, the reverse sign of momenta results in the reserve of the moving direction of the free atom (atom A in this case). The atom A is very far from the molecule B-C and it keeps approaching the molecule B-C.

===Reactive and unreactive trajectories===

For the initial positions rBC = 74 pm and rAB = 200 pm, run trajectories with the following momenta combination and complete the table.
Complete the table below 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?

{| class="wikitable" border="1"
|+ caption
! pBC/ g.mol-1.pm.fs-1 !! pAB/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes || The reaction can occurred. The atom A approaches the molecule B-C and forms a new molecule A-B which leaves away with most vibrational energy and a small amount of kinetic energy (KE), while atom C is detached as a separate atom which moves away with most KE. || [[File:R1=-2.5;r2=-5.1 jsg lastday.png|thumb|200px]]
|-
| -3.1 || -4.1 || -420.077 || No || The reaction doesn't occur. The atom A approaches the molecule B-C and return back without colliding with the molecule. || [[File:-3.1and-4.1_jsg_lastday.png|thumb|200px]]
|-
| -3.1 || -5.1 || -413.977 || Yes || The reaction can occur. The atom A approaches the molecule B-C and forms a new molecule A-B which leaves away with most vibrational energy and a small amount of kinetic energy (KE), while atom C is detached as a separate atom which moves away with most KE. || [[File:-3.1and-5.1 jsg lastday.png|thumb|200px]]
|-
| -5.1 || -10.1 || -357.277 || No || A case of barrier recrossing.. The atom A approaches the molecule B-C and forms a temporary new molecule A-B with strong vibrational motion and the molecule A-B move back and collide with atom C to form the B-C again.|| [[File:-5.1and-10.1 jsg lastday.png|thumb|200px]]
|-
| -5.1 || -10.6 || -349.477 || Yes || The reaction can occur. The atom A approaches the molecule B-C and break B-C bond and makes atom B bouncing in-between and eventually form a new molecule A-B. || [[File:-5.1and-10.6 jsg+lastday.png|thumb|200px]]
|}

From this table, you can observe that some trajectories have a total energy greater than the activation energy and still do not react. Therefore, for the reaction to proceed, there are two conditions should be met:(1) the total energy is lager than the activation energy; (2)the interaction between atoms should be appropriate for the collision to happen with correct energy and direction. For instance, if the approaching atom A move opposite to the molecule B-C, there will be no collision and the reaction will definitely not occur. {{fontcolor1|red| OK. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:31, 29 May 2020 (BST)}}

* (The transition state is found to be at rAB= 230.0 pm; rBC=74.0 pm; forces along AB and AC = +0.015 and +0.012 respectively; KE = 0; PE = total E = -434.471.)

===Transition State Theory===

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

Transition state theory (TST) explains the reaction rates of elementary chemical reactions. The theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes. These are three properties of TST:

1.Rates of reaction can be studied by examining activated complexes near the saddle point of a potential energy surface. The details of how these complexes are formed are not important. The saddle point itself is called the transition state.

2.The activated complexes are in a special equilibrium (quasi-equilibrium) with the reactant molecules.

3.The activated complexes can convert into products, and kinetic theory can be used to calculate the rate of this conversion.

TST is used primarily to understand qualitatively how chemical reactions take place. Simply, the Transition State Theory (TST) tells whether a reaction will happen if cross the TS barrier.

So the difference here is that our program is working with a single molecule, the TST deals with a bulk system with a Boltzmann speed distribution. The main differences between experimental and predicted rate constant values will be due to (1) ignoring barrier recrossing and (2) ignoring QM behaviour. {{fontcolor1|red| What will the effect of these phenomena be on the real rate of reaction relative to one predicted by TST?? More discussion is needed here. And references please! [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:52, 29 May 2020 (BST)}}

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

F + H2 --> HF + H

A F atom (A) approaches a hydrogen molecule (B-C) to forms a new HF molecule (A-B) and a separate H atom (C).

===PES inspection===

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?

Locate the approximate position of the transition state.

In order to classify the reaction based on energies, the total energies of reactant and that of product should be identified first by using mep calculation (in order to exclude kinetic energy) starting with a small displacement from the TS and with a high enough number of steps. It will go towards the formation of either the products or the reactants and the potential energy will tend to a constant value which is the total energy of the product/reactants depending which direction the displacement was.

The total energies of reactant and that of product are calculated to be:

• reactant: E(H2) = -435.099 KJ.mol-1;

• product: E(HF) = -560.402 KJ.mol-1.

Since the energy of the product is lower than that of reactant, this is an exothermic reaction. This makes sense with the concept of bond strength in which the H-F bond (565 KJ.mol-1) is much stronger than H-H bond (432 KJ.mol-1) and thus there will be more energy released during the formation of H-F bond.

The transition state is still a maximum or a saddle point on the PES, and found to be at:

• rAB = rH-F = 182 pm;

• rBC = rH-H = 74.2 pm;

• pAB = pBC = 0 g.mol-1.pm.fs-1;

• force along AB = +0.003 ≈0 KJ.mol-1.pm-1;

• forces along BC = +0.089 ≈0 KJ.mol-1.pm-1;

• the total energy = -433.969 KJ.mol-1.

Because the activation energy for one of the reactions is so small, it is not easy to locate the transition state immediately. Use the Hammond postulate to guide your search.

Report the activation energy for both reactions.

The activation energy of the forward reaction is so small (1.13 KJ.mol-1), the Hammond postulate is introduced. Hammond postulate is a hypothesis which states that the transition state of a reaction resembles either the reactants or the products, to whichever it is closer in energy. In an exothermic reaction, the TS is closer to the reactants than to the products in energy. Therefore, the TS resembles the reactant in the reaction. In an exothermic reaction, the TS is closer to the products than to the reactants in energy and therefore, the TS resembles the products.

In F-H-H system, the activation energy is very similar to the energy of the reactant and hence the TS resembles reactant in this case. Therefore, this reaction (F + H2 --> HF + H) is exothermic in forward direction and thus endothermic in backward direction.

In order to classify the reaction based on energies, the total energies of reactant and that of product should be identified first by using mep calculation (in order to exclude kinetic energy) starting with a small displacement from the TS and with a high enough number of steps. It will go towards the formation of either the products or the reactants and the potential energy will tend to a constant value which is the total energy of the product/reactants depending which direction the displacement was.

• mep calculation: number of steps = 5000

Since the transition state is found to be at rAB = rH-F = 182 pm and rBC = rH-H = 74.2 pm, we either change rAB or rBC to achieve the formation of either reactants or products. Here is an example:

The reaction reacts towards the products when initial conditions are:

• rAB = rH-F = 165 pm;

• rBC = rH-H = 74.2 pm;

The "contour plot" indicating the reaction direction clearly and "Energy vs time" diagram which shows a decrease (lower in energy) of potential energy (and also the total energy) from -434.194 to -560.402 KJ.mol-1 are illustrated below.

[[File:Surface Plot 165 74.2 MEP gjs.png|thumb|center|The contour plot of FHH system when rAB =165 pm and rBC = 74.2 pm, calculated with mep.]]

[[File:Contour Plot 165 74.2 MEP gjs.png|thumb|center|The Energy.vs.time plot of FHH system when rAB =165 pm and rBC = 74.2 pm, calculated with mep.]]

Based on this mep calculation, the total energies of reactant and that of product are calculated to be:

• reactant: E(H2) = -435.099 KJ.mol-1;

• product: E(HF) = -560.402 KJ.mol-1.

Consequentially, since the TS energy = -433.969 KJ.mol-1, the forward reaction (F + H2 --> HF + H) has an activation energy of [-433.969-(-435.099)]=1.13 KJ.mol-1; the backward reaction (HF + H --> F + H2）has an activation energy of [-433.969-(-560.402)]=126.433 KJ.mol-1. The latter is much larger than the former.

===Reaction dynamics===

Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Momenta vs Time”

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.

For example, when initial conditions fulfil rAB = 160 pm and 0 < rBC < 76 pm, the trajectory are reactive.

The released energy is concerted thermally into electromagnetic radiations (photons). The relaxed potential energy well of electrons have photon absorption and emission between ground state E(0) and the first excited state E(1) and most of the electrons are populated in ground state. While the excited potential energy well involves more possible higher excited states, such as E(0) to the second excited state E(2), E(1) to E(2) which is called overtone. The overtone will contribute to a second absorbance peak at lower wavenumbers in IR spectra.

The emission of photons can be detected by a special type of IR called a Quantum Well Infrared Photodetector (QWIP) which is an infrared photodetector.

Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration.

Setting initial conditions as:
• rHH = 74 pm;
• rFH = 180 pm;
• pFH = -1.0 g.mol-1.pm.fs-1;

The value of pHH is varied from -6.1 to +6.1 and the resulting "Internuclear Distance.vs.Time" diagams used to show the trajectories are displayed below:

[[File:Q2-gjs-rage-6.1-123.png|thumb|center|pHH= -6.1|300px]] [[File:Q2-gjs-rage-6-123.png|thumb|center|pHH= -6|300px]] [[File:Q2-gjs-rage-5-123.png|thumb|center|pHH= -5|300px]] [[File:Q2-gjs-rage-4-123.png|thumb|center|pHH= -4|300px]] [[File:Q2-gjs-rage-3-123.png|thumb|center|pHH= -3|300px]] [[File:Q2-gjs-rage-2-123.png|thumb|center|pHH= -2|300px]] [[File:Q2-gjs-rage-1-123.png|thumb|center|pHH= -1|300px]] [[File:Q2-gjs-rage-0-123.png|thumb|center|pHH= 0|300px]] [[File:Q2-gjs-rage+1-123.png|thumb|center|pHH= 1|300px]] [[File:Q2-gjs-rage+2-123.png|thumb|center|pHH= 2|300px]] [[File:Q2-gjs-rage+3-123.png|thumb|center|pHH= 3|300px]] [[File:Q2-gjs-rage+4-123.png|thumb|center|pHH= 4|300px]] [[File:Q2-gjs-rage+5-123.png|thumb|center|pHH= 5|300px]] [[File:Q2-gjs-rage+6-123.png|thumb|center|pHH= 6|300px]] [[File:Q2-gjs-rage+6.1-123.png|thumb|center|pHH= 6.1|300px]]

In general, the excess of energy put into the system gives rise to the significant increase in the vibrational energy of H-H. when pHH is negative, the smaller the value (more positive) the more barrier recrossing. When pHH is positive, in most cases there is one barrier recrossing only except when p=+1,+5,+6.1. Even though some barrier recrossing occurs (F atom collides with H atom for several times), it doesn't necessarily mean that the reaction will proceed to form the product HF and a separated atom H.

For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?

The new trajectory is showm in the "Internuclear Distance.vs.Time" diagams below. The vibrating molecule H-H moves towards F and collides with it, making the atom B (H atom) bouncing in-between for three times and then return back to the other H atom to reform the vibrating hydrogen molecule.
[[File:3Q-gjs-changed-initials.png|thumb|center|pHH= 0.2|300px]]

Let us now focus on the reverse reaction, H + HF.

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

As mentioned before, the relaxed potential energy well of electrons have photon absorptions and emissions between ground state E(0) and the first excited state E(1) and most of the electrons are populated in ground state. While the excited potential energy well involves more possible higher excited states, such as E(0) to the second excited state E(2), E(1) to E(2) which is called overtone. The overtone will contribute to a second absorbance peak at lower wavenumbers in IR spectra. Each energy state correspond to a translational energy while the vibrational levels are smaller in energy difference and located around translational energy states.

Polanyi's empirical rules states that vibrational energy is more efficient in prompting a late-barrier reaction (that is, a transition state resembling the products) than translational energy, whereas the reverse is true for an early barrier reaction.

Translational energy is better at promoting exothermic reactions, because the trajectory can just fall into the lower energy region. Vibrational motion might make it harder for the trajectory to cross the barrier because it is in a different direction than the mep. If the reactants have vibrational motion, then they will try to move up the valley near the transition state, and just fall back to the reactants state.

MRD:ql2018

2020-05-29T17:51:08Z

Mak214: /* Exercise 1- (H+H2 System). */

== Molecular Reaction Dynamics Lab ==

=== Exercise 1- (H+H2 System). ===
{{fontcolor|blue|Q1ː 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?}}
A transition state can be found as a saddle point on the potential energy surface diagram. Mathematically, the transition state is defined as the maximum on the minimum energy path linking reactants and the products. To distinguish the transition state from local minima,this can be done by viewing the potential energy surface from different angles. The transition state is the only point that is a minimum point from one angle(Figure 1)but a maximum from second angle(Figure 2). {{fontcolor1|red| IOK. Could you use some equations to define this also? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:23, 29 May 2020 (BST)}}
{{multiple image
| align = center
| direction = vertical
| width = 400
| header = Transition State at maximum and minimum
| image1 = Surface_Plot1_qw.png
| caption1 = Transition state is a maximum from one angle.
| image2 = Surface_Plot2_qw.png
| caption2 = Transition state is a minimum from another angle.
}}

{{fontcolor|blue|Q1ː 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.}}
My estimation for the transition state position rts is 90.7pm.Because for the transition state, the potential energies of both p1 and p2 are zero.There is no change of the energy which means that the gradients of potential energy surface is zero and there is no force acting on atoms.So r1 and r2 will keep constant.The graph corresponds should be a straight line since there is no oscillation. {{fontcolor1|red| Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:23, 29 May 2020 (BST)}}

[[File:Surface_Plot3_qw.png|thumb|center|Internuclear distances against time graph at r1=r2=90.7pm with zero momenta.]]

{{fontcolor|blue|Q3ːComment on how the mep and the trajectory you just calculated differ.}}
As the two graphs shown below, the difference is obvious that the trajectory on the contour plot of mep is shorter than that of dynamics. And there is also no oscillation compared with the graph generated by dynamics. The reason for this is because that there is zero kinetic energy for mep due to zero velocity and momentum, there will be no gain in vibtational energy as a result. Therefore the trajectory shows no oscillation. Also the change in total energy is different. Since there is no gain in kinetic energy, the total energy will drop as potential energy losses. {{fontcolor1|red| Yes. in MEP the momenta are reset to zero at every timestep so the system falls into the nearest local minimum. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:23, 29 May 2020 (BST)}}

For dynamics, the total energy is conserved as there is gain in kinetic energy when H2 formed.And due to this bond forming, the vibrational energy results in the oscillations on the contour plot.

{{multiple image
| align = left
| direction = vertical
| width = 300
| header = MEP and Dynamics
| image1 = Surface_Plot4_qw.png
| caption1 = MEP contour plot
| image2 = Surface_Plot5_qw.png
| caption2 = Dynamics contour plot
}}

{{fontcolor|blue|Q4ːComplete 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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot/ kJ.mol-1!! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes || This reaction started with small AB distance and large BC distance. It has enough momentum to pass the transition state region. After that region, the bond between BC formed and the bond between AB broken. The trajectry shows no fluctuation before BC bond formed, which is because the energy is mostly transitional energy. When BC bond formed, the energy is mostly vibrational energy which causes oscillation. ||[[File:Surface_Plot6_qw.png|thumb|upright=0.8]]
|-
| -3.1 || -4.1 || -420.077 || No || This reaction started with small AB distance and large BC distance. However, the momentum is not large enough to pass the transition state region. There is no break in bond between AB and no form in bond between BC. Molecule AB moves away forom C and the bond between AB keeps vibrating due to the kinetic energy. ||[[File:Surface_Plot7_qw.png|thumb|upright=0.8]]
|-
| -3.1 || -5.1 || -413.977 || Yes || This time the BC momentum is large enough for the system to pass transition state. AB bond vibrates before bond breaking due to kinetic energy. The vibration is even larger for BC due to larger momentum after the bond forming. ||[[File:Surface_Plot8_qw.png|thumb|upright=0.8]]
|-
| -5.1 || -10.1 || -357.277 || No || The energy is too large for this system. Even the AB bond breaks and the trajectory passes the transition state region, the bond formed between BC vibrates too strongly due to large energy that the bond fromed eventually breaks. The trajectory goes back and recrosses the transition state region. AB forms a bond again and as AB moves away from C, there is no vibration between AB bond. ||[[File:Surface_Plot9_qw.png|thumb|upright=0.8]]
|-
| -5.1 || -10.6 || -349.477 || Yes || The energy for this system is also very large that the trajectory passes through the transition state region for a couple of times. Although the bond formed between BC after the trajectory passes through the transition state region for the first time. The bond of BC breaks due to large energy and recrosses the transition state. However, because the system has more energy than the previous one.it had the energy to collide for a third time, the bond between BC forms again and became stable as BC vibrates and moves away from A. ||[[File:Surface_Plot10_qw.png|thumb|upright=0.8]]
|}

{{fontcolor1|red| Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:23, 29 May 2020 (BST)}}
{{fontcolor|blue|Q5ːGiven the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The transition state theory assumes that once the reactants have enough kinetic energy to pass through the transition state. The product must formed and the reaction would not go to reverse. However, the given experimental results indicate that it was not true. The products can recross the transition state and reform reactants. As a result, the transition state theory overestimates the reaction rate compared to experimental results. Also, due to the quantum tunneling effects, atoms with energy lower than activation energy can still pass through the barrier, which leads to a underestimation of reaction rate. {{fontcolor1|red| Good discussion - you should make reference to the relevant literature here to support what you are saying. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:51, 29 May 2020 (BST)}}

== Exercise 2ː F - H - H System ==
{{fontcolor|blue|Q6ː 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?}}

According to the potential energy surface graph F, H2 to H and HFis an exothermic reaction since it has higher potential energy for F and H2. Therefore H + HF to F, H2 is an endothermic reaction. This shows that the H-F bond is stronger than H-H bond. Because the formation of H-F bond releases more energy than the breaking of H-H bond.

[[File:Surface_Plot11_qw.png|thumb|center|Potential energy surface of F-H-H system]]

{{fontcolor|blue|Q7ː Locate the approximate position of the transition state.}}
The approximate position of transition state for this reaction is when F-H isr1 = 181.300 pm and H-H isr2 = 74.483 pm. The momentum at this point is zero.
[[File:Surface_Plot12_qw.png|thumb|center|approximate position of the transition state of F-H-H system]]

{{fontcolor|blue|Q8ː Report the activation energy for both reactions.}}
The potential energy of the reactants F + H2 is -434.625 kJ.mol-1. The potential energy of the transition state is -433.981 kJ.mol-1. The activation energy is equal to the energy of transition state minus the energy of the reactants, which is 0.644 kJ.mol-1.

The potential energy of the products H + HF is -556.231 kJ.mol-1. The potential energy of the transition state is -433.981 kJ.mol-1. The activation energy of this reverse reaction is equal to the energy of transition state minus the energy of the products, which is 122.25 kJ.mol-1.

{{multiple image
| align = center
| direction = vertical
| width = 300
| header = Activation Energies calculation
| image1 = Surface_Plot13_qw.png
| caption1 = Energy against time r1 = 184 pm, r2 = 74.483 pm
| image2 = Surface_Plot14_qw.png
| caption2 = Energy against time r1 = 176.000 pm, r2 = 74.483 pm
}}

{{fontcolor|blue|Q9ːIn light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.}}

The conditions are set that the the bond distance is 184pm between F and H and the bond distance is 75 between H and H. p1=-1 and p2= -2. From the graph, it can be seen that the product has greater vibration. This is because the potential energy transfer to kinetic energy and transitional energy.

The energy transfer can be confirmed by using IR spectroscopy. Because the vibration of HF bond can be detected by IR. Calorimetry can also be used to detect the heat generated during the reaction. However, this heat includes both vibrational and transitional energy.

{| class="wikitable" border="1"
! Momentum vs Time
|- !! Contour Plot
|-
| [[File:Surface_Plot15_qw.png|400px|thumb|left]] || [[File:Surface_Plot16_qw.png|400px|thumb|right]]
|}
{{fontcolor|blue|Q10ːDiscuss 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.}}

From Polanyi's rule, It is known that the transitional energy can be more efficient for a exothermic reaction as it helps the reactants to pass through early transition state. In endothermic reaction, it has late transition state that the vibrational energy is more effective. It helps the reactants to pass the late transition state.1

==Reference==
1.J.C. Polanyi. Some Concepts in Reaction Dynamics. Science 1987, 236 (4802): 680-690. doi:10.1126/science.236.4802.680

MRD:01493832

2020-05-29T17:49:38Z

Mak214: /* Reactive and unreactive trajectories */

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

== The transition state ==
On a potential energy surface diagram, the transition state is mathematically defined as a stationary point where the derivative of the potential energy curve, which is a force {{fontcolor1|red| I'm not sure exactly what you mean here? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:17, 29 May 2020 (BST)}}, is zero. This means that there are no movement at the transition state.

The transition state is different from a local minimum of a potential energy surface in a sense that it is a minimum in one direction yet a maximum in another.1 Therefore, it is a stationary point but not a point of inflection. From the calculated forces, we are able to confirm that this is a stationary point and from the given corresponding eigenvalues/vectors, the positive and negative combination confirms that this is a saddle point. If we move away from the stationary point, the eigenvalues/vectors will be either all positive or all negative. (For a local minimum, the eigenvalues/vectors will be positive in all directions.) {{fontcolor1|red| IOK. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:17, 29 May 2020 (BST)}}

=== H + H2 : Locating the transition state ===
In the case of H + H2, AB and BC distances must be equal in the transition state because all atoms are identical so the transition state is symmetrical and its position can be identified by using a trial and error process, where the AB and BC distances in the initial conditions are varied until the corresponding forces become as close to zero as possible. For 3 Hydrogen atoms, AB and BC distances (rts) are found to be 90.775 pm.

[[File:Nl3818-H TS dist-time.jpg|200px|thumb|left|Fig. 1 - Internuclear Distances vs Time plot for 3 Hydrogen atoms]]

From Fig.1, it is evidently clear that at rts = 90.775 pm, the distances are constant throughout confirming that it is a stationary point (i.e. the nuclei do not move) and that distances AB = BC. {{fontcolor1|red| I Good. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:17, 29 May 2020 (BST)}}

== Calculating the reaction path ==
There are two types of calculation that can be done when considering the trajectory of the Hydrogen molecule: MEP (minimum energy path) and Dynamics. In MEP, atoms are assumed to be in an infinitely slow motion which corresponds to zero momenta at each time step. {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:20, 29 May 2020 (BST)}}Both calculations can be done by changing the initial conditions to r1 = rts+δ and r2=rts in order to visualise the trajectories. The difference between MEP and Dynamics are shown in two contour plots below. From the MEP contour plot, we can see that the black trajectory line is rather smooth as it approaches the transition state position marked with a red cross compared to the trajectory shown in Dynamics contour plot. The MEP calculated trajectory also starts off at a higher potential energy position than the one from Dynamics calculation. {{fontcolor1|red| Hmm.. they should start with the same energy if you put in the same settings. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:20, 29 May 2020 (BST)}} Because in MEP calculation, atoms are moving very slowly so the vibration between the reactant HA-HB is so small that it does not display an oscillating behaviour as in the Dynamics calculation. {{fontcolor1|red| Not exactly, the MEP calculation there should be no vibration of the bond whatsoever since there is no momentum to move the atoms away from the equilibrium bond distance which is a valley in the potential energy surface. We only see vibration in the dynamics type calculation because there is residual momentum which allows the atoms in the molecule to vibrate. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:20, 29 May 2020 (BST)}}

{| class="wikitable" border=1
|-
|[[File:Dynamic_nl3818.png|200px|thumb|left|Fig.2 - Contour plot of triatomic Hydrogen system using Dynamics calculation]]||
[[File:MEP_nl3818.png|200px|thumb|left|Fig.3 - Contour plot of triatomic Hydrogen system using MEP calculation]]
|}

=== Change in initial conditions ===
If we change the initial condition so that r1 = rts and r2=rts+δ instead, the result is still identical to previous result but with atom A moving away at large t and atom B bonding with C instead. By investigating the “Internuclear Distances vs Time” and “Momenta vs Time” plots and taking note of values for r1(t) r2(t) and p1(t) p2(t) at very large t (50 fs), another set of calculations can be done with using these values for the initial positions and momentum (with reversed sign). The observation is for “Internuclear Distances vs Time” plot, the result is a reflection in the y-axis of Fig. 4 meaning, the bonded BC H2 molecule and the previously isolated HA atom are now approaching each other and their positions at large t corresponds to the transition state position. With reversed sign of the momenta, this is a reverse process of what was previously done, instead of atoms moving apart, they are now moving towards each other with the same kinetic energy. The “Momenta vs Time” is now reflected in the x-axis of Fig. 5, which can be reasoned by the repulsion that the atoms experienced from coming together resulting in positive gradient which is gradual at first then steepens at high t as the three atoms are now very close.

{| class="wikitable" border=1
|-
|[[File:Dtnl3818new.png|200px|thumb|left|Fig.4 - Dist. vs Time plot of H atoms moving apart]]||
[[File:Momentanewnl3818.png|200px|thumb|left|Fig.5 - Momenta vs Time plot of H atoms moving apart]]
|}

== Reactive and unreactive trajectories ==

To test whether trajectories starting at the same position but with higher momenta will all react, a table has been constructed shown below, where r1=74 pm and r2=200 pm.

{| class="wikitable" border=1
!p1/ g.mol-1.pm.fs-1!!p2/ g.mol-1.pm.fs-1!!Etot/kJ.mol-1!!Reactive?!!Description of the dynamics!!Illustration of the trajectory using Dist. vs Time plot
|-
| -2.56 || -5.1 || -414 || YES || This is a reactive trajectory because as the incoming H approaches, its AB distance decreases until it displays an oscillating behaviour which corresponds to the new bond vibration while the initial BC distance increases exponentially showing that the bond is broken and the leaving atom is moving further apart. ||[[File:Nl3818table1.png|200px|thumb|left]]
|-
| -3.1 || -4.1 || -420 || NO || This is not a reactive trajectory since although the AB distance decreases, it increases again while the initial BC distance retains its oscillating behaviour throughout meaning the bond is not broken. The incoming H approaches then moves away, no bond is broken in this case. ||[[File:Nl3818table2.png|200px|thumb|left]]
|-
| -3.1 || -5.1 || -412 || YES || This displays the same behaviour as the first case and thus, same explanation applies. ||[[File:Nl3818table3.png|200px|thumb|left]]
|-
| -5.1 || -10.1 || -357 || NO || In this case, the initial bond is broken but then reforms again, therefore, overall there is no reaction. ||[[File:Nl3818table4.png|200px|thumb|left]]
|-
|-5.1 || -10.6 || -349 || YES || The approaching H forms a new bond then bounces back but did not break the new bond so overall, this trajectory is reactive. ||[[File:Nl3818table5.png|200px|thumb|left]]
|}

[[File:Nl3818table5.png|200px|thumb|left]]
The last plot on the left corresponds to p1 = -5.1 g.mol-1.pm.fs-1, p2 = -10.6 g.mol-1.pm.fs-1 and Etot = -349 kJ.mol-1. This is a reactive trajectory since the approaching H forms a new bond then bounces back but did not break so overall, this trajectory is reactive.

*The last line of the table is not showing so I had to insert it like this.
{{fontcolor1|red| Well done. It would be nice to have a line or two here concluding what you have learned from these trajectories. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:49, 29 May 2020 (BST)}}

==Transition State Theory==

'''The transition state theory is a classical consideration which assumes:'''2

'''1) The reactants and transition state are in a quasi-equilibrium.''' This means that once the reactants reaches the transition state, they {{fontcolor1|red| cannot ? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:48, 29 May 2020 (BST)}} collapse back to form the reactants again when in reality they can. This cause the values that the transition state theory predicts to be overestimated than the actual experimental value.

'''2) Quantum tunnelling is ignored.''' Because the reactants that overcome the barrier by moving across is ignored, this causes the values predicted by transition state theory to be underestimated than the actual experimental values.

'''3) All reactants that have enough energy to overcome the barrier will successfully form the product and the step that involves the formation of the product from the transition state is the rate-determing step.''' This causes an overestimation of the actual rate because in reality, although the reactants have high energy, but if the energy is not located in the right place (i.e. in the bond that needs to be broken) then the reaction will not happen. {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:48, 29 May 2020 (BST)}}

== The F-H-H System==
=== PES Inspection===
===='''F + H2'''====
This is an unsymmetrical system which results in unsymmetrical potential wells in the surface plot.

From the surface plot, we are able to tell that this is an exothermic reaction. BC is the H-H distance of the reactant while AB is the H-F distance, the z-axis of the surface plot represents energy and in this case, the H-H is deeper in energy compared to H-F which corresponds to the energy profile of an exothermic reaction and thus, explains the greater H-F bond strength compared to H-H.

The approximate position of the transition state can be found by the same method that was used for three Hydrogen atoms: trial and error where the corresponding forces are as close to zero as possible. This method gives rF-HA = 180 pm and rHA-HB=74.5 pm. These positions are confirmed to be the transition state by the opposite signs of the eigenvalues/vectors. At this position, the energy of the transition state is -434 kJmol-1.

In order to find the activation energy, the position of the system is displaced slightly towards the reactant side from the transition state (rF-HA = 179 pm) and MEP calculation was used to obtain the "Energy vs Time" below.

From the graph, the energy of the reactants is -561 kJmol-1. The difference between the reactant state and the transition state energy is the activation energy which is the energy required to cross the barrier to form the product, in this case is +127 kJmol-1.

{| class="wikitable" border=1
|-
|[[File:H2andHFsurface.png|200px|thumb|left|Fig.6 - Surface plot for F+H2]]||[[File:FH2Ea.png|200px|thumb|left|Fig.7 - Energy vs Time plot for F+H2]]
|}

==== H + HF ====
The surface plot of this reaction is a mirrored image of the above surface plot from the previous reaction. This shows that this is an endothermic reaction where the reactants have higher energy than the product. This correlates to the stronger H-F bond that requires input energy to break to form the H-H product.

The approximate transition state can be found in the same manner as the previous reaction and the positions are rF-HB = 180 pm and rHB-HC=74.5 pm, corresponding to total energy of -434 kJmol-1. The activation energy in this case is very small is reported to be +0.930 kJmol-1. Because of such small activation energy, it is easier to locate the position of the transition state by using the Hammond postulate. From the surface plot, we have deduced that this reaction is endothermic therefore it must have a late transition state meaning the structure of the transition state will resemble more of the product.3 As a result, shorter H-H distance and long F-H were predicted for the position of the transition state.

{| class="wikitable" border=1
|-
|[[File:HandHFsurface.png|200px|thumb|left|Fig.8 - Surface plot for H + HF]]||[[File:FHHEa.png|200px|thumb|left|Fig.9 - Energy vs Time plot for H + HF]]
|}

===Reaction dynamics===

One of the ways to confirm the mechanism of the release of reaction energy is by taking an Infrared absorption spectrum of the product. In this case, we can measure the IR spectrum of HF gas sample. When the reaction just happened and IR light is shown through, we are exciting the molecules that were once thermally relaxed and exclusively occupying the ground state to its first vibrational excited state and at early times (when the reaction just happened), we may also excitation from first to second vibrational excited state.

As a result, what we can observe in the IR spectrum is two peaks present, one is the fundamental peak and the other is an overtone appearing at a lower wavenumber. This is because of decreasing energy gap between each levels due to anharmonicity.4 By measuring the intensity of the overtone over time, we can extract the number of molecules that are vibrationally excited over time so we can expect the intensity of overtone to be smaller as energy is emitted as radiation and the intensity of the fundamental peak would increase as time goes on.

The other way to confirm the mechanism of energy release experimentally is by measuring the IR emitted directly, instead of probing the reaction using IR spectroscopy, as the molecules fall back from vibrational excited state to the ground state. This method can be done by using the Infrared Emission Spectroscopy (IES).5

====F + H2====

By setting up a calculation with rHH=74 pm, pFH=-1.0 g.mol-1.pm.fs-1 and pHH varied between -6.1 and 6.1 g.mol-1.pm.fs-1, it is visible from the contour plot that the trajectory only successfully rolled over to the product with pHH between -2.7 and 1.2 g.mol-1.pm.fs-1. Values outside of this range results in the trajectory hitting the wall and bouncing back.

{| class="wikitable" border=1
|-
|[[File:H2-2.7.png|200px|thumb|left|Fig.10 - H2 trajectory at pHH = -2.7 g.mol-1.pm.fs-1]]||
[[File:H20nl3818.png|200px|thumb|left|Fig.11 - H2 trajectory at pHH = 0.0 g.mol-1.pm.fs-1]]
|[[File:H21.2nl3818.png|200px|thumb|left|Fig.12 - H2 trajectory at pHH = 1.2 g.mol-1.pm.fs-1]]||
[[File:H2bouncingbacknl3818.png|200px|thumb|left|Fig.13 - trajectory at pHH = 0.2 g.mol-1.pm.fs-1, pFH = -1.6 g.mol-1.pm.fs-1]]
|}

====H + HF====
A reactive trajectory was obtained with a combination of pFH = 1 g.mol-1.pm.fs-1 and pHH = -7 g.mol-1.pm.fs-1. Decreasing the momentum of the incoming H further results in the trajectory hitting the wall and bouncing back while increasing the energy of H-F vibration causes the trajectory to go in the other direction rather than towards the product.

====Polanyi's Empirical Rule====
The above investigation illustrates the Polanyi's empirical rule. In an exothermic reaction, as demonstrated by F + H2, translational energy is favoured over vibrational energy because the trajectory can fall down into the low energy region of the PES whereas the vibrational energy is in a different direction and can cause the trajectory to bounce back near the transition state and cross back to the reactant state.

On the other hand, vibrational energy is favoured over translational energy in an endothermic reaction6 as illustrated by H + HF. This is because the trajectory has the same directionality with the vibrational energy that goes to the product.

==Bibliography==
1 Maxima, minima, and saddle points (article). https://www.khanacademy.org/math/multivariable-calculus/applications-of-multivariable-derivatives/optimizing-multivariable-functions/a/maximums-minimums-and-saddle-points (accessed May 22, 2020).

2 Peters, B. Transition State Theory. Reaction Rate Theory and Rare Events Simulations 2017, 227–271.

3 Libretexts. Hammond's Postulate. https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hammond’s_Postulate (accessed May 22, 2020).

4 Libretexts. 13.5: Vibrational Overtones. https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/13:_Molecular_Spectroscopy/13.05:_Vibrational_Overtones (accessed May 22, 2020).

5 Keresztury, G. Emission Spectroscopy, Infrared. Encyclopedia of Analytical Chemistry 2006.

6 Zhang, Z.; Zhou, Y.; Zhang, D. H.; Czakó, G.; Bowman, J. M. Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl CHD3 Reaction. The Journal of Physical Chemistry Letters 2012, 3 (23), 3416–3419.

MRD:01493832

2020-05-29T17:48:56Z

Mak214: /* Transition State Theory */

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

== The transition state ==
On a potential energy surface diagram, the transition state is mathematically defined as a stationary point where the derivative of the potential energy curve, which is a force {{fontcolor1|red| I'm not sure exactly what you mean here? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:17, 29 May 2020 (BST)}}, is zero. This means that there are no movement at the transition state.

The transition state is different from a local minimum of a potential energy surface in a sense that it is a minimum in one direction yet a maximum in another.1 Therefore, it is a stationary point but not a point of inflection. From the calculated forces, we are able to confirm that this is a stationary point and from the given corresponding eigenvalues/vectors, the positive and negative combination confirms that this is a saddle point. If we move away from the stationary point, the eigenvalues/vectors will be either all positive or all negative. (For a local minimum, the eigenvalues/vectors will be positive in all directions.) {{fontcolor1|red| IOK. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:17, 29 May 2020 (BST)}}

=== H + H2 : Locating the transition state ===
In the case of H + H2, AB and BC distances must be equal in the transition state because all atoms are identical so the transition state is symmetrical and its position can be identified by using a trial and error process, where the AB and BC distances in the initial conditions are varied until the corresponding forces become as close to zero as possible. For 3 Hydrogen atoms, AB and BC distances (rts) are found to be 90.775 pm.

[[File:Nl3818-H TS dist-time.jpg|200px|thumb|left|Fig. 1 - Internuclear Distances vs Time plot for 3 Hydrogen atoms]]

From Fig.1, it is evidently clear that at rts = 90.775 pm, the distances are constant throughout confirming that it is a stationary point (i.e. the nuclei do not move) and that distances AB = BC. {{fontcolor1|red| I Good. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:17, 29 May 2020 (BST)}}

== Calculating the reaction path ==
There are two types of calculation that can be done when considering the trajectory of the Hydrogen molecule: MEP (minimum energy path) and Dynamics. In MEP, atoms are assumed to be in an infinitely slow motion which corresponds to zero momenta at each time step. {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:20, 29 May 2020 (BST)}}Both calculations can be done by changing the initial conditions to r1 = rts+δ and r2=rts in order to visualise the trajectories. The difference between MEP and Dynamics are shown in two contour plots below. From the MEP contour plot, we can see that the black trajectory line is rather smooth as it approaches the transition state position marked with a red cross compared to the trajectory shown in Dynamics contour plot. The MEP calculated trajectory also starts off at a higher potential energy position than the one from Dynamics calculation. {{fontcolor1|red| Hmm.. they should start with the same energy if you put in the same settings. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:20, 29 May 2020 (BST)}} Because in MEP calculation, atoms are moving very slowly so the vibration between the reactant HA-HB is so small that it does not display an oscillating behaviour as in the Dynamics calculation. {{fontcolor1|red| Not exactly, the MEP calculation there should be no vibration of the bond whatsoever since there is no momentum to move the atoms away from the equilibrium bond distance which is a valley in the potential energy surface. We only see vibration in the dynamics type calculation because there is residual momentum which allows the atoms in the molecule to vibrate. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:20, 29 May 2020 (BST)}}

{| class="wikitable" border=1
|-
|[[File:Dynamic_nl3818.png|200px|thumb|left|Fig.2 - Contour plot of triatomic Hydrogen system using Dynamics calculation]]||
[[File:MEP_nl3818.png|200px|thumb|left|Fig.3 - Contour plot of triatomic Hydrogen system using MEP calculation]]
|}

=== Change in initial conditions ===
If we change the initial condition so that r1 = rts and r2=rts+δ instead, the result is still identical to previous result but with atom A moving away at large t and atom B bonding with C instead. By investigating the “Internuclear Distances vs Time” and “Momenta vs Time” plots and taking note of values for r1(t) r2(t) and p1(t) p2(t) at very large t (50 fs), another set of calculations can be done with using these values for the initial positions and momentum (with reversed sign). The observation is for “Internuclear Distances vs Time” plot, the result is a reflection in the y-axis of Fig. 4 meaning, the bonded BC H2 molecule and the previously isolated HA atom are now approaching each other and their positions at large t corresponds to the transition state position. With reversed sign of the momenta, this is a reverse process of what was previously done, instead of atoms moving apart, they are now moving towards each other with the same kinetic energy. The “Momenta vs Time” is now reflected in the x-axis of Fig. 5, which can be reasoned by the repulsion that the atoms experienced from coming together resulting in positive gradient which is gradual at first then steepens at high t as the three atoms are now very close.

{| class="wikitable" border=1
|-
|[[File:Dtnl3818new.png|200px|thumb|left|Fig.4 - Dist. vs Time plot of H atoms moving apart]]||
[[File:Momentanewnl3818.png|200px|thumb|left|Fig.5 - Momenta vs Time plot of H atoms moving apart]]
|}

== Reactive and unreactive trajectories ==

To test whether trajectories starting at the same position but with higher momenta will all react, a table has been constructed shown below, where r1=74 pm and r2=200 pm.

{| class="wikitable" border=1
!p1/ g.mol-1.pm.fs-1!!p2/ g.mol-1.pm.fs-1!!Etot/kJ.mol-1!!Reactive?!!Description of the dynamics!!Illustration of the trajectory using Dist. vs Time plot
|-
| -2.56 || -5.1 || -414 || YES || This is a reactive trajectory because as the incoming H approaches, its AB distance decreases until it displays an oscillating behaviour which corresponds to the new bond vibration while the initial BC distance increases exponentially showing that the bond is broken and the leaving atom is moving further apart. ||[[File:Nl3818table1.png|200px|thumb|left]]
|-
| -3.1 || -4.1 || -420 || NO || This is not a reactive trajectory since although the AB distance decreases, it increases again while the initial BC distance retains its oscillating behaviour throughout meaning the bond is not broken. The incoming H approaches then moves away, no bond is broken in this case. ||[[File:Nl3818table2.png|200px|thumb|left]]
|-
| -3.1 || -5.1 || -412 || YES || This displays the same behaviour as the first case and thus, same explanation applies. ||[[File:Nl3818table3.png|200px|thumb|left]]
|-
| -5.1 || -10.1 || -357 || NO || In this case, the initial bond is broken but then reforms again, therefore, overall there is no reaction. ||[[File:Nl3818table4.png|200px|thumb|left]]
|-
|-5.1 || -10.6 || -349 || YES || The approaching H forms a new bond then bounces back but did not break the new bond so overall, this trajectory is reactive. ||[[File:Nl3818table5.png|200px|thumb|left]]
|}

[[File:Nl3818table5.png|200px|thumb|left]]
The last plot on the left corresponds to p1 = -5.1 g.mol-1.pm.fs-1, p2 = -10.6 g.mol-1.pm.fs-1 and Etot = -349 kJ.mol-1. This is a reactive trajectory since the approaching H forms a new bond then bounces back but did not break so overall, this trajectory is reactive.

*The last line of the table is not showing so I had to insert it like this.

==Transition State Theory==

'''The transition state theory is a classical consideration which assumes:'''2

'''1) The reactants and transition state are in a quasi-equilibrium.''' This means that once the reactants reaches the transition state, they {{fontcolor1|red| cannot ? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:48, 29 May 2020 (BST)}} collapse back to form the reactants again when in reality they can. This cause the values that the transition state theory predicts to be overestimated than the actual experimental value.

'''2) Quantum tunnelling is ignored.''' Because the reactants that overcome the barrier by moving across is ignored, this causes the values predicted by transition state theory to be underestimated than the actual experimental values.

'''3) All reactants that have enough energy to overcome the barrier will successfully form the product and the step that involves the formation of the product from the transition state is the rate-determing step.''' This causes an overestimation of the actual rate because in reality, although the reactants have high energy, but if the energy is not located in the right place (i.e. in the bond that needs to be broken) then the reaction will not happen. {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:48, 29 May 2020 (BST)}}

== The F-H-H System==
=== PES Inspection===
===='''F + H2'''====
This is an unsymmetrical system which results in unsymmetrical potential wells in the surface plot.

From the surface plot, we are able to tell that this is an exothermic reaction. BC is the H-H distance of the reactant while AB is the H-F distance, the z-axis of the surface plot represents energy and in this case, the H-H is deeper in energy compared to H-F which corresponds to the energy profile of an exothermic reaction and thus, explains the greater H-F bond strength compared to H-H.

The approximate position of the transition state can be found by the same method that was used for three Hydrogen atoms: trial and error where the corresponding forces are as close to zero as possible. This method gives rF-HA = 180 pm and rHA-HB=74.5 pm. These positions are confirmed to be the transition state by the opposite signs of the eigenvalues/vectors. At this position, the energy of the transition state is -434 kJmol-1.

In order to find the activation energy, the position of the system is displaced slightly towards the reactant side from the transition state (rF-HA = 179 pm) and MEP calculation was used to obtain the "Energy vs Time" below.

From the graph, the energy of the reactants is -561 kJmol-1. The difference between the reactant state and the transition state energy is the activation energy which is the energy required to cross the barrier to form the product, in this case is +127 kJmol-1.

{| class="wikitable" border=1
|-
|[[File:H2andHFsurface.png|200px|thumb|left|Fig.6 - Surface plot for F+H2]]||[[File:FH2Ea.png|200px|thumb|left|Fig.7 - Energy vs Time plot for F+H2]]
|}

==== H + HF ====
The surface plot of this reaction is a mirrored image of the above surface plot from the previous reaction. This shows that this is an endothermic reaction where the reactants have higher energy than the product. This correlates to the stronger H-F bond that requires input energy to break to form the H-H product.

The approximate transition state can be found in the same manner as the previous reaction and the positions are rF-HB = 180 pm and rHB-HC=74.5 pm, corresponding to total energy of -434 kJmol-1. The activation energy in this case is very small is reported to be +0.930 kJmol-1. Because of such small activation energy, it is easier to locate the position of the transition state by using the Hammond postulate. From the surface plot, we have deduced that this reaction is endothermic therefore it must have a late transition state meaning the structure of the transition state will resemble more of the product.3 As a result, shorter H-H distance and long F-H were predicted for the position of the transition state.

{| class="wikitable" border=1
|-
|[[File:HandHFsurface.png|200px|thumb|left|Fig.8 - Surface plot for H + HF]]||[[File:FHHEa.png|200px|thumb|left|Fig.9 - Energy vs Time plot for H + HF]]
|}

===Reaction dynamics===

One of the ways to confirm the mechanism of the release of reaction energy is by taking an Infrared absorption spectrum of the product. In this case, we can measure the IR spectrum of HF gas sample. When the reaction just happened and IR light is shown through, we are exciting the molecules that were once thermally relaxed and exclusively occupying the ground state to its first vibrational excited state and at early times (when the reaction just happened), we may also excitation from first to second vibrational excited state.

As a result, what we can observe in the IR spectrum is two peaks present, one is the fundamental peak and the other is an overtone appearing at a lower wavenumber. This is because of decreasing energy gap between each levels due to anharmonicity.4 By measuring the intensity of the overtone over time, we can extract the number of molecules that are vibrationally excited over time so we can expect the intensity of overtone to be smaller as energy is emitted as radiation and the intensity of the fundamental peak would increase as time goes on.

The other way to confirm the mechanism of energy release experimentally is by measuring the IR emitted directly, instead of probing the reaction using IR spectroscopy, as the molecules fall back from vibrational excited state to the ground state. This method can be done by using the Infrared Emission Spectroscopy (IES).5

====F + H2====

By setting up a calculation with rHH=74 pm, pFH=-1.0 g.mol-1.pm.fs-1 and pHH varied between -6.1 and 6.1 g.mol-1.pm.fs-1, it is visible from the contour plot that the trajectory only successfully rolled over to the product with pHH between -2.7 and 1.2 g.mol-1.pm.fs-1. Values outside of this range results in the trajectory hitting the wall and bouncing back.

{| class="wikitable" border=1
|-
|[[File:H2-2.7.png|200px|thumb|left|Fig.10 - H2 trajectory at pHH = -2.7 g.mol-1.pm.fs-1]]||
[[File:H20nl3818.png|200px|thumb|left|Fig.11 - H2 trajectory at pHH = 0.0 g.mol-1.pm.fs-1]]
|[[File:H21.2nl3818.png|200px|thumb|left|Fig.12 - H2 trajectory at pHH = 1.2 g.mol-1.pm.fs-1]]||
[[File:H2bouncingbacknl3818.png|200px|thumb|left|Fig.13 - trajectory at pHH = 0.2 g.mol-1.pm.fs-1, pFH = -1.6 g.mol-1.pm.fs-1]]
|}

====H + HF====
A reactive trajectory was obtained with a combination of pFH = 1 g.mol-1.pm.fs-1 and pHH = -7 g.mol-1.pm.fs-1. Decreasing the momentum of the incoming H further results in the trajectory hitting the wall and bouncing back while increasing the energy of H-F vibration causes the trajectory to go in the other direction rather than towards the product.

====Polanyi's Empirical Rule====
The above investigation illustrates the Polanyi's empirical rule. In an exothermic reaction, as demonstrated by F + H2, translational energy is favoured over vibrational energy because the trajectory can fall down into the low energy region of the PES whereas the vibrational energy is in a different direction and can cause the trajectory to bounce back near the transition state and cross back to the reactant state.

On the other hand, vibrational energy is favoured over translational energy in an endothermic reaction6 as illustrated by H + HF. This is because the trajectory has the same directionality with the vibrational energy that goes to the product.

==Bibliography==
1 Maxima, minima, and saddle points (article). https://www.khanacademy.org/math/multivariable-calculus/applications-of-multivariable-derivatives/optimizing-multivariable-functions/a/maximums-minimums-and-saddle-points (accessed May 22, 2020).

2 Peters, B. Transition State Theory. Reaction Rate Theory and Rare Events Simulations 2017, 227–271.

3 Libretexts. Hammond's Postulate. https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hammond’s_Postulate (accessed May 22, 2020).

4 Libretexts. 13.5: Vibrational Overtones. https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/13:_Molecular_Spectroscopy/13.05:_Vibrational_Overtones (accessed May 22, 2020).

5 Keresztury, G. Emission Spectroscopy, Infrared. Encyclopedia of Analytical Chemistry 2006.

6 Zhang, Z.; Zhou, Y.; Zhang, D. H.; Czakó, G.; Bowman, J. M. Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl CHD3 Reaction. The Journal of Physical Chemistry Letters 2012, 3 (23), 3416–3419.

MRD:xmh01513932

2020-05-29T17:47:24Z

Mak214: /* Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? */

== 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? ==
On a potential energy surface diagram, the transition state is identified as the maximum on the minimum potential energy path between the reactants and products.

To explain the transition state mathematically, several derivatives have to be addressed first,
1. The first derivative of potential energy (V) with respect to either atomic separation (r1, r2) is zero, i.e. ∂V/∂r1 = 0 and ∂V/∂r2 = 0;
2. The second derivative D = ∂2V/∂r1∂r1 × ∂2V/∂r2∂r2 – ∂2V/∂r1∂r2 = 0;
3. If D > 0, ∂2V(r1)/∂r1∂r1 > 0, then the point with this particular (r1,r2) coordinates is a local minimum point; ∂2V(r1)/∂r1∂r1 < 0 for a local maximum point;
4. If D < 0, then this point is a saddle point, i.e. the transition state. Simply put, the saddle point is simultaneously a minimum and a maximum along two orthogonal directions (i.e. the two atomic separation axes in this case). {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:11, 29 May 2020 (BST)}}
[[File:Figure 1.log|250px|thumb|center|Figure 1. Contour Plot.]]
[[File:xmhFigure 2.png|250px|thumb|center|Figure 2. Surface Plot.]]

{{fontcolor1|red| What did you want to show with these plots? Use figures to illustrate what you are saying in the text, and make sure you discuss every figure you include in your text. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:12, 29 May 2020 (BST)}}

== 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. ==
For a symmetric system like H + H2, the internuclear distances of r1 = r2 is expected in the transition state. To test out the best input values for r1 and r2, there would be some expected features on various plots, as shown below,
1． In Figure 3., the kinetic energy of the system is constant at 0. This is explained by the fact that, at the transition state, the net force acting on the system F = dp/dt = ∂V/∂r1 = ∂V/∂r2 = 0 (as defined), the triatomic system is at a state of equilibrium with no change in motion. The same idea is conveyed in Figure 4., where the internuclear distances are expected to have a variation as small as possible, the atoms are “frozen” in space, moreover, rAB(r2) = rBC(r1).
2． In Figure 5., the transition state position (rts) is also checked by the fact that the input r1 and r2 values (denoted by the red cross) map onto the circle which gives the coordinates of the exact rts. {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:13, 29 May 2020 (BST)}}
The final value of r1 = r2 = 90.7 pm is chosen to satisfy the above requirements.
[[File:xmhFigure 3.png|250px|thumb|center|Figure 3. Energy vs Time Plot.]]
[[File:xmhFigure 4.png|250px|thumb|center|Figure 4. Internuclear Distances vs Time Plot.]]
[[File:xmhFigure 5.png|250px|thumb|center|Figure 5. Contour Plot]]

== Comment on how the mep and the trajectory you just calculated differ. ==
Two paths were generated under the calculation types of MEP and Dynamics respectively, with initial conditions of r1 = rts+1=91.7 and r2 = 90.7 and p1 = p2 = 0 g.mol-1.pm.fs-1.
The dynamic graph shows a wavy line (shown in Figure 6.). After passing though rts (now r1 < r2), HAHB molecule is formed, detaching HC. This case takes into account of the intramolecular vibration of the new H2 molecule, since there is an oscillating trajectory at around a fixed rAB value with an increasing rBC value which indicates detaching of HC.
[[File:xmhFigure 6.png|250px|thumb|center|Figure 6. Contour Plot.]][[File:xmhFigure 7.png|250px|thumb|center|Figure 7. Energy vs Time Plot.]]
[[File:xmhFigure 8.png|250px|thumb|center|Figure 8. Momentum vs Time Plot]][[File:xmhFigure 8.png|250px|thumb|center|Figure 9. Internuclear Distances vs Time Plot]]
The mep (the minimum energy path) corresponds to a trajectory with infinitely slow motion, i.e. the system’s velocities are reset to zero in each time step, consequently no momentum or kinetic energy, as reflected in Figure 11 & 12. The reaction path way as a smooth curve (shown in Figure .) without any information of the intrinsic vibration within the H2 molecule.
[[File:xmhFigure 10.png|250px|thumb|center|thumb|Figure 10. Contour Plot.]]
[[File:xmhFigure 11.png|250px|thumb|center|Figure 11. Energy vs Time Plot.]]
[[File:xmhFigure 12.png|250px|thumb|center|Figure 12. Momentum vs Time Plot]]
[[File:xmhFigure 13.png|250px|thumb|center|Figure 13. Internuclear Distances vs Time Plot]]
If initial conditions are changed to r2 = rts+1=91.7 and r1 = 90.7 and p1 = p2 = 0 g.mol-1.pm.fs-1, same trends in both the Internuclear Distances vs Time graph and the Momenta vs Time graph are observed as for the previous conditions. Since the system is symmetric, hence the differences between these 2 sets of conditions are just attributed to the fact that whether HAHB or HBHC dissociates. {{fontcolor1|red| Great. Good use of figures here. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:14, 29 May 2020 (BST)}}

== 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? ==
{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Reactive || HA approaches HBHC molecule from a distance of rAB = 200 (starting from the bottom-right of the figure), with rBC being constant since HB and HC still remain bonded. After passing through rts, the new HAHB molecule is formed as shown by the constant rAB and increasing rBC as HC leaves(at the top left of the figure). || [[File:xmhA.png|250px]]
|-
| -3.1 || -4.1 || -420.077 || Unreactive || HA approaches HBHc (bottom-right), however it does not quite get to rts due to a reduced amount of kinetic energy compared to the first case, hence HA bounces off, there is no interaction with the hydrogen molecule. ||[[File:xmhB.png|250px]]
|-
| -3.1 || -5.1 || 413.977 || Reactive || Same as the first case except for the fact that p1 has a larger value, hence the system is more energetic to go over the activation barrier, and due to energy conservation, after the collision, the new molecule has a greater vibration. ||[[File:xmhC.png|250px]]
|-
| -5.1 || -10.1 || -357.277 || Unreactive || Barrier recrossing - the system is energetic enough to go over the transition state region and form the new HAHB, since rBC increases from its original bond length, indicating the leaving of HC, and the Etot is also greater than the first case (a reactive case). However, rBC drops down again to the original bond length and rAB increases as the system reverts back to HBHC. ||[[File:xmhD.png|250px]]
|-
| -5.1 || -10.6 || -349.477 || Reactive || The system HBHC converts to HAHB after reaching rts, but reverts given the excess momentum in p2 as rAB increases and rBC decreases, the system crosses the transition region again to form the final HAHB molecule as rBC continue increasing and rAB stays oscillating around a fixed value. ||[[File:xmhE.png|250px]]
|}
From the first sight, it seems reasonable to encourage a successful collision by adding more momentum into the system since this increases the kinetic energy of the atoms to overcome the activation barrier. However, as concluded from the table, the excess momentum and energy could give rise to greater vibration within the molecule and hence cause the bond to dissociate. {{fontcolor1|red| Great. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:46, 29 May 2020 (BST)}}

== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ==
In the transition state theory, the average transmission rate/frequency is estimated based on the assumption that all trajectories with some kinetic energy greater than the activation energy will be reactive. From the empirical data of the last section, it was shown that if momentum is provided such that the vibration is so great that the bond dissociates again, i.e. barrier recrossing occurs, this reduces the rate of a particular conversion, e.g. from HBHC and HA to HAHB and HC. This would overestimate the transition state theory rate constant (and hence the rate).
The transition state theory also consider the reaction pathway through classical mechanic collision (for macroscopic system). However, wavefunction (for microscopic system) of the atoms should be considered. Quantum tunneling may take place, it is not necessary to jump over the activation barrier, hence less kinetic energy is required for a successful collision, rate is higher than the theory prediction. {{fontcolor1|red| Yes. Include references. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:47, 29 May 2020 (BST)}}

== 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? ==
In GUI, it is set up that the atom A approaches and collide with the molecule HBHC.
In the F+H2 system, F is considered as atom A, and the 2 H atoms as atoms B and C. Figure 14(A) shows the entrance channel for the reactants F+H2, since rBC (the molecule’s bond length) is constant and rAB is big; Figure 14(B) shows the exit channel for the products (FH+H), since rAB is now constant and rBC is big. Comparing the V for the 2 channels, the reactants are at a higher V than the products, hence this is an exothermic reaction.
[[File:xmhFigure 14(A).png|250px|thumb|center|Figure 14(A). Surface Plot.]]
[[File:Figure 14(B)xmh.png|250px|thumb|center|Figure 14(B). Surface Plot.]]
For the H+FH system, the same idea applies, from Figure 15(A). & (B)., the reactants are at a lower V than the products, hence it is an endothermic reaction.
[[File:Figure 15(A)mh.png|250px|thumb|center|Figure 15(A). Surface Plot.]]
[[File:Figure 15(B)xmh.png|250px|thumb|center|Figure 15(B). Surface Plot.]]
This correctly reflects the result obtained from qualitative bond-strength analysis. F-H is a very strong bond (compared to other possible bonds in the system) due to the different electronegativities and hence ionic contribution to the nature of the bond. Hence breaking this bond requires much energy, and forming it releases much energy. In the F+H2 reaction, a pure covalent bond is broken to form the strong F-H bond, this is expected to be exothermic; in the H+FH system, the energy released from forming H2 does not compensate for the energy required to break F-H first, hence it is expected to be endothermic.

== Locate the approximate position of the transition state. ==
The approximate position of the transition state was investigated in the case of F+H2, where F is considered as atom A, and the 2 H atoms as atoms B and C. It was found that rAB = 181.1 and rBC = 74.5. This is reflected by Figure 20., where the internuclear distances are constant between all 3 atoms, corresponding to equilibrium at the transition state. Etot = - 433.980.
For the case of H+HF, same rts was found, since this reaction is just the reverse reaction of the above case, and the same geometry is found at the transition state.
[[File:xmhFigure 16.png|250px|thumb|center|Figure 16. Contour Plot.]]
[[File:Figure 17xmh.png|250px|thumb|center|Figure 17. Internuclear Distancesvs Time Plot.]]

== Report the activation energy for both reactions. ==
In order to find the reactants’ energy for the F+H2 reaction, initial conditions were set as rAB = 1000 and rBC = 74.5. The value of rAB was chosen so that there is no interaction in the system since the F atom is at a very distant away from the H2 molecule, as reflected by Figure 20., where the only momentum is given by the oscillation between the bonded molecule. Etot = - 435.057.
Hence the activation energy = Etot(transition state) – Etot(reactants) = - 433.980 + 435.057 = 1.077.
[[File:xmhFigure 18.png|250px|thumb|center|Figure 18. Momentum vs Time Plot.]]
Same method applies to the H+FH system, the activation energy = 125.322.

== 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. ==
There are 3 types of energies associated with the system: potential energy due to the interactions between the atom and the molecule, vibrational kinetic energy and translational kinetic energy. After a successful collision, these energies would redistribute based on the mass and bond strength of the atom and molecule, some energy might excite the new molecule into a higher vibrational state. However, the molecule would eventually drop down to the ground state, releasing energy.
The IR overtone region could be used to confirm the presence of the transition between higher states, the peaks appear at higher wavelength (higher energy). Consequently, this confirms the following release of energy.

== 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. ==
According to Polanyi's rules, the reactants' translational energy activates them more effectively to the activation barrier and hence increases rate for exothermic reactions; conversely, the reactants' vibrational energy activates more effectively for endothermic reactions.
For the exothermic F+H2 system, the translational energy is more important to give a successful collision.
For the endothermic H+HF system, the vibrational energy is more important.

MRD:xmh01513932

2020-05-29T17:46:40Z

Mak214: /* 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? */

== 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? ==
On a potential energy surface diagram, the transition state is identified as the maximum on the minimum potential energy path between the reactants and products.

To explain the transition state mathematically, several derivatives have to be addressed first,
1. The first derivative of potential energy (V) with respect to either atomic separation (r1, r2) is zero, i.e. ∂V/∂r1 = 0 and ∂V/∂r2 = 0;
2. The second derivative D = ∂2V/∂r1∂r1 × ∂2V/∂r2∂r2 – ∂2V/∂r1∂r2 = 0;
3. If D > 0, ∂2V(r1)/∂r1∂r1 > 0, then the point with this particular (r1,r2) coordinates is a local minimum point; ∂2V(r1)/∂r1∂r1 < 0 for a local maximum point;
4. If D < 0, then this point is a saddle point, i.e. the transition state. Simply put, the saddle point is simultaneously a minimum and a maximum along two orthogonal directions (i.e. the two atomic separation axes in this case). {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:11, 29 May 2020 (BST)}}
[[File:Figure 1.log|250px|thumb|center|Figure 1. Contour Plot.]]
[[File:xmhFigure 2.png|250px|thumb|center|Figure 2. Surface Plot.]]

{{fontcolor1|red| What did you want to show with these plots? Use figures to illustrate what you are saying in the text, and make sure you discuss every figure you include in your text. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:12, 29 May 2020 (BST)}}

== 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. ==
For a symmetric system like H + H2, the internuclear distances of r1 = r2 is expected in the transition state. To test out the best input values for r1 and r2, there would be some expected features on various plots, as shown below,
1． In Figure 3., the kinetic energy of the system is constant at 0. This is explained by the fact that, at the transition state, the net force acting on the system F = dp/dt = ∂V/∂r1 = ∂V/∂r2 = 0 (as defined), the triatomic system is at a state of equilibrium with no change in motion. The same idea is conveyed in Figure 4., where the internuclear distances are expected to have a variation as small as possible, the atoms are “frozen” in space, moreover, rAB(r2) = rBC(r1).
2． In Figure 5., the transition state position (rts) is also checked by the fact that the input r1 and r2 values (denoted by the red cross) map onto the circle which gives the coordinates of the exact rts. {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:13, 29 May 2020 (BST)}}
The final value of r1 = r2 = 90.7 pm is chosen to satisfy the above requirements.
[[File:xmhFigure 3.png|250px|thumb|center|Figure 3. Energy vs Time Plot.]]
[[File:xmhFigure 4.png|250px|thumb|center|Figure 4. Internuclear Distances vs Time Plot.]]
[[File:xmhFigure 5.png|250px|thumb|center|Figure 5. Contour Plot]]

== Comment on how the mep and the trajectory you just calculated differ. ==
Two paths were generated under the calculation types of MEP and Dynamics respectively, with initial conditions of r1 = rts+1=91.7 and r2 = 90.7 and p1 = p2 = 0 g.mol-1.pm.fs-1.
The dynamic graph shows a wavy line (shown in Figure 6.). After passing though rts (now r1 < r2), HAHB molecule is formed, detaching HC. This case takes into account of the intramolecular vibration of the new H2 molecule, since there is an oscillating trajectory at around a fixed rAB value with an increasing rBC value which indicates detaching of HC.
[[File:xmhFigure 6.png|250px|thumb|center|Figure 6. Contour Plot.]][[File:xmhFigure 7.png|250px|thumb|center|Figure 7. Energy vs Time Plot.]]
[[File:xmhFigure 8.png|250px|thumb|center|Figure 8. Momentum vs Time Plot]][[File:xmhFigure 8.png|250px|thumb|center|Figure 9. Internuclear Distances vs Time Plot]]
The mep (the minimum energy path) corresponds to a trajectory with infinitely slow motion, i.e. the system’s velocities are reset to zero in each time step, consequently no momentum or kinetic energy, as reflected in Figure 11 & 12. The reaction path way as a smooth curve (shown in Figure .) without any information of the intrinsic vibration within the H2 molecule.
[[File:xmhFigure 10.png|250px|thumb|center|thumb|Figure 10. Contour Plot.]]
[[File:xmhFigure 11.png|250px|thumb|center|Figure 11. Energy vs Time Plot.]]
[[File:xmhFigure 12.png|250px|thumb|center|Figure 12. Momentum vs Time Plot]]
[[File:xmhFigure 13.png|250px|thumb|center|Figure 13. Internuclear Distances vs Time Plot]]
If initial conditions are changed to r2 = rts+1=91.7 and r1 = 90.7 and p1 = p2 = 0 g.mol-1.pm.fs-1, same trends in both the Internuclear Distances vs Time graph and the Momenta vs Time graph are observed as for the previous conditions. Since the system is symmetric, hence the differences between these 2 sets of conditions are just attributed to the fact that whether HAHB or HBHC dissociates. {{fontcolor1|red| Great. Good use of figures here. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:14, 29 May 2020 (BST)}}

== 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? ==
{| class="wikitable" border="1"
|+ caption
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Reactive || HA approaches HBHC molecule from a distance of rAB = 200 (starting from the bottom-right of the figure), with rBC being constant since HB and HC still remain bonded. After passing through rts, the new HAHB molecule is formed as shown by the constant rAB and increasing rBC as HC leaves(at the top left of the figure). || [[File:xmhA.png|250px]]
|-
| -3.1 || -4.1 || -420.077 || Unreactive || HA approaches HBHc (bottom-right), however it does not quite get to rts due to a reduced amount of kinetic energy compared to the first case, hence HA bounces off, there is no interaction with the hydrogen molecule. ||[[File:xmhB.png|250px]]
|-
| -3.1 || -5.1 || 413.977 || Reactive || Same as the first case except for the fact that p1 has a larger value, hence the system is more energetic to go over the activation barrier, and due to energy conservation, after the collision, the new molecule has a greater vibration. ||[[File:xmhC.png|250px]]
|-
| -5.1 || -10.1 || -357.277 || Unreactive || Barrier recrossing - the system is energetic enough to go over the transition state region and form the new HAHB, since rBC increases from its original bond length, indicating the leaving of HC, and the Etot is also greater than the first case (a reactive case). However, rBC drops down again to the original bond length and rAB increases as the system reverts back to HBHC. ||[[File:xmhD.png|250px]]
|-
| -5.1 || -10.6 || -349.477 || Reactive || The system HBHC converts to HAHB after reaching rts, but reverts given the excess momentum in p2 as rAB increases and rBC decreases, the system crosses the transition region again to form the final HAHB molecule as rBC continue increasing and rAB stays oscillating around a fixed value. ||[[File:xmhE.png|250px]]
|}
From the first sight, it seems reasonable to encourage a successful collision by adding more momentum into the system since this increases the kinetic energy of the atoms to overcome the activation barrier. However, as concluded from the table, the excess momentum and energy could give rise to greater vibration within the molecule and hence cause the bond to dissociate. {{fontcolor1|red| Great. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:46, 29 May 2020 (BST)}}

== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ==
In the transition state theory, the average transmission rate/frequency is estimated based on the assumption that all trajectories with some kinetic energy greater than the activation energy will be reactive. From the empirical data of the last section, it was shown that if momentum is provided such that the vibration is so great that the bond dissociates again, i.e. barrier recrossing occurs, this reduces the rate of a particular conversion, e.g. from HBHC and HA to HAHB and HC. This would overestimate the transition state theory rate constant (and hence the rate).
The transition state theory also consider the reaction pathway through classical mechanic collision (for macroscopic system). However, wavefunction (for microscopic system) of the atoms should be considered. Quantum tunneling may take place, it is not necessary to jump over the activation barrier, hence less kinetic energy is required for a successful collision, rate is higher than the theory prediction.

== 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? ==
In GUI, it is set up that the atom A approaches and collide with the molecule HBHC.
In the F+H2 system, F is considered as atom A, and the 2 H atoms as atoms B and C. Figure 14(A) shows the entrance channel for the reactants F+H2, since rBC (the molecule’s bond length) is constant and rAB is big; Figure 14(B) shows the exit channel for the products (FH+H), since rAB is now constant and rBC is big. Comparing the V for the 2 channels, the reactants are at a higher V than the products, hence this is an exothermic reaction.
[[File:xmhFigure 14(A).png|250px|thumb|center|Figure 14(A). Surface Plot.]]
[[File:Figure 14(B)xmh.png|250px|thumb|center|Figure 14(B). Surface Plot.]]
For the H+FH system, the same idea applies, from Figure 15(A). & (B)., the reactants are at a lower V than the products, hence it is an endothermic reaction.
[[File:Figure 15(A)mh.png|250px|thumb|center|Figure 15(A). Surface Plot.]]
[[File:Figure 15(B)xmh.png|250px|thumb|center|Figure 15(B). Surface Plot.]]
This correctly reflects the result obtained from qualitative bond-strength analysis. F-H is a very strong bond (compared to other possible bonds in the system) due to the different electronegativities and hence ionic contribution to the nature of the bond. Hence breaking this bond requires much energy, and forming it releases much energy. In the F+H2 reaction, a pure covalent bond is broken to form the strong F-H bond, this is expected to be exothermic; in the H+FH system, the energy released from forming H2 does not compensate for the energy required to break F-H first, hence it is expected to be endothermic.

== Locate the approximate position of the transition state. ==
The approximate position of the transition state was investigated in the case of F+H2, where F is considered as atom A, and the 2 H atoms as atoms B and C. It was found that rAB = 181.1 and rBC = 74.5. This is reflected by Figure 20., where the internuclear distances are constant between all 3 atoms, corresponding to equilibrium at the transition state. Etot = - 433.980.
For the case of H+HF, same rts was found, since this reaction is just the reverse reaction of the above case, and the same geometry is found at the transition state.
[[File:xmhFigure 16.png|250px|thumb|center|Figure 16. Contour Plot.]]
[[File:Figure 17xmh.png|250px|thumb|center|Figure 17. Internuclear Distancesvs Time Plot.]]

== Report the activation energy for both reactions. ==
In order to find the reactants’ energy for the F+H2 reaction, initial conditions were set as rAB = 1000 and rBC = 74.5. The value of rAB was chosen so that there is no interaction in the system since the F atom is at a very distant away from the H2 molecule, as reflected by Figure 20., where the only momentum is given by the oscillation between the bonded molecule. Etot = - 435.057.
Hence the activation energy = Etot(transition state) – Etot(reactants) = - 433.980 + 435.057 = 1.077.
[[File:xmhFigure 18.png|250px|thumb|center|Figure 18. Momentum vs Time Plot.]]
Same method applies to the H+FH system, the activation energy = 125.322.

== 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. ==
There are 3 types of energies associated with the system: potential energy due to the interactions between the atom and the molecule, vibrational kinetic energy and translational kinetic energy. After a successful collision, these energies would redistribute based on the mass and bond strength of the atom and molecule, some energy might excite the new molecule into a higher vibrational state. However, the molecule would eventually drop down to the ground state, releasing energy.
The IR overtone region could be used to confirm the presence of the transition between higher states, the peaks appear at higher wavelength (higher energy). Consequently, this confirms the following release of energy.

== 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. ==
According to Polanyi's rules, the reactants' translational energy activates them more effectively to the activation barrier and hence increases rate for exothermic reactions; conversely, the reactants' vibrational energy activates more effectively for endothermic reactions.
For the exothermic F+H2 system, the translational energy is more important to give a successful collision.
For the endothermic H+HF system, the vibrational energy is more important.

MRD:YX8818CX

2020-05-29T17:46:14Z

Mak214: /* Transition State Theory */

== Part 1: H + H2 system ==

=== Dynamics from the transition state region ===
'''In part 1, r1 = rAB, r2 = rBC.'''

{{fontcolor|blue|Q1: 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?}}

[[File:Saddle point.svg|thumb|right|300px|A saddle point (in red) on the graph of z=x2−y2 Reference: Wikipedia: Saddle point https://https://en.wikipedia.org/wiki/Saddle_point]]

On a potential energy (PE) surface, the transition state is a saddle point. At that point the gradient of the potential is zero. Mathematically speaking, a saddle point is a point on a surface of a function graph where the slopes (derivatives) in orthogonal directions are all zero, but which is not a local maximum or minimum of the function (as illustrated in the figure<ref>''Wikipedia, The Free Encyclopedia'', s.v. "Saddle point," (accessed May 20, 2020), https://en.wikipedia.org/wiki/Saddle_point</ref> on the right).

The saddle point can be determined mathematically by calculating the Hessian matrix for the function at that point.<ref>Smith, Colin M. "How to find a saddle point." ''International journal of quantum chemistry'' 37.6 (1990): 773-783.</ref> For a function f(x,y) of two variables, a point is a saddle point if:

If fx = 0, fy = 0, and fxxfyy - f2xy < 0. Where fx and fxx are the first and second derivative of f with respect to x respectively.

The transition state (the saddle point) can be identified in a three-dimensional PE plot by looking at the surface at both direcions (r1 and r2 ). A saddle point will be a maximum in one direction and a minimum in the other direction. It can be easily distinguished from a local minimum, as a local minimum will be a minimum in all directions.<ref>Henkelman, Graeme, Gísli Jóhannesson, and Hannes Jónsson. "Methods for finding saddle points and minimum energy paths." ''Theoretical methods in condensed phase chemistry''. Springer, Dordrecht, 2002. 269-302.</ref> {{fontcolor1|red| Great! [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:09, 29 May 2020 (BST)}}

==== Trajectories from r1 = r2: locating the transition state ====
{{fontcolor|blue|Q2: 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.}}

My best estimate of the TS position rts is 90.775 pm.

As p1 = p2 = 0, there is no initial momentum, hence no kinetic energy. So if the H + H2 surface start at the transition state, there will be no oscillation seen in the Distance vs Time plot, thus rAB and rBC will be constant. When r1 = r2 = 90.775 pm, there is negligible oscillatory behaviour displayed in the Distance vs Time plot and zero force along AB and BC, as shown in Figures. This indicates the transition state position ('''rts''') is very close to this value. {{fontcolor1|red| Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:09, 29 May 2020 (BST)}}
{| class="wikitable"
|[[File:TS_force_zero_YX8818.PNG|thumb|center|400px|Settings for rAB and rBC when forces is zero]]
|[[File:Q2_Distance_vs_time_plot_TS_YX8818.png|thumb|center|400px|Internuclear Distances vs Time Plot]]
|}

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

==== Trajectories from r1 = rts+δ, r2 = rts ====
{{fontcolor1|blue|Q3: Comment on how the ''mep'' and the trajectory you just calculated differ.}}

1 decimal place of the transition state position here was used for simplicity, i.e. rts = 91.8 pm

Using r1 = rts+1 = 91.8 pm, r2 = rts = 90.8 pm

As shown from the contour and surface plots below, The minimum energy path (''mep'') corresponded to infinitely slow motions, with velocities / momenta reset to zero for each step. Thus, the trajectory corresponds to infinitely slow motion, and simply follows the valley floor, no oscilatiory behaviour is visible. While the mep plot is useful in characterising the reaction, it does not give an realistic account of the atomic motions.

In comparison, the dynamics surface plot gave a more accurate description of atomic motions (inertial) during the reaction. Oscillations of the potential for the H atoms were observed for dynamics plots. {{fontcolor1|red| Yes. A clear response, thank you. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:09, 29 May 2020 (BST)}}

{| class="wikitable"
|-
! Header 1
! Contour plot
! Surface plot
|-
| Dynamics
| [[File:Q3_contour_plot_dynamics_YX8818.png|thumb|center|300px|Contour plot using Dynamics calculation type]]
| [[File:Q3_surface_plot_dynamics_YX8818.png|thumb|center|300px|Surface plot using Dynamics calculation type]]
|-
| MEP
| [[File:Q3_contour_plot_MEP_YX8818.png|thumb|center|300px|Contour plot using MEP calculation type]]
| [[File:Q3_surface_plot_MEP_YX8818.png|thumb|center|300px|Surface plot using MEP calculation type]]
|}

==== Additional Questions: ====
'''A1'': Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. What would change if we used the initial conditions r1 = rts and r2 = rts+1 pm instead?'''''

As shown in the “Internuclear Distances vs Time” and “Momenta vs Time” plots below, when r1 = rts+1, r2 = rts, the reaction proceed to form the products, and H2 molecule BC and H atom A were formed. This is because the transition state was displaced slightly towards the products.

On the other hand, when r1 = rts, r2 = rts+1, the reaction proceed to form the reactants, and molecule AB and atom C were formed.This is because the transition state was displaced slightly towards the reactants.

{| class="wikitable"
|-
!initial conditions
!r1 = rts+1, r2 = rts
!r1 = rts, r2 = rts+1
|-
|Internuclear Distances vs Time Plot
| [[File:internuclear_vs_time_r2TS_YX8818.png|300px]]
|[[File:internuclear_vs_time_r1TS_YX8818.png|300px]]
|-
|Momenta vs Time Plot
| [[File:Momenta_vs_time_r2TS_YX8818.png|300px]]
| [[File:Momenta_vs_time_r1TS_YX8818.png|300px]]
|}

'''A2: Note final values of the positions r1(t) r2(t) and p1(t) p2(t) for your trajectory for large enough t .'''

'''t = 100 fs '''was chosen, and for r1 = rts, r2 = rts+1, ''' r1'''(t) = 731 pm, '''r2'''(t) = 75.5 pm, '''p1'''(t) = 5.1 g.mol-1.pm.fs-1, '''p2'''(t) =2.5 g.mol-1.pm.fs-1.

'''A3'': ''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. What do you observe?'''

When momentum was reversed, the trajectory traced back to the original position (slightly towards the product from the TS). But as it does not have enough energy to overcome the energy barrier at the TS, it traced back to the product, H2 molecules BC and H atoms A.

[[File:A3_momentum_reversed_YX8818.png|thumb|center|350px|Contour plot when monenta were reversed]]
''<nowiki/>''

=== Reactive and unreactive trajectories ===
The following simulations were investigated with different reaction conditions (momentum) to see whether they resulted in reactive trajectory.
* Initial positions '''r1''' = 74 pm and '''r2''' = 200 pm.

{| class="wikitable"
!No
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot ''/''kJ. mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory:
contour plot
!Illustration of the trajectory:

surface plot
|-
|'''1'''
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|The distance rBC decreases as the molecule AB approaches H atom C, and rBC reaches minium as the system reaches the transition state (TS). The system has sufficient energy to overcome the energy barrier, and the products BC molecules is formed (indicated by relatively constant small rBC) and increasing rAB as the H atom A moves away.
|[[File:Q4_1_Contour_YX8818.png|250px]]
|[[File:Q4_1_Surface_YX8818.png|300px]]
|-
|'''2'''
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The distance rBC decreases as the molecule AB approaches H atom C, but upon collision there is insufficient energy to overcome the energy barrier at TS. Hence, the molecule AB remains, and atom C moves away (rBC increases)
|[[File:Q4_2_Contour_YX8818.png|250px]]
|[[File:Q4_2_Surface_YX8818.png|300px]]
|-
|'''3'''
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The distance rBC decreases as the molecule AB approaches H atom C and rBC reaches minium as the system reaches the transition state (TS). The system has sufficient energy to overcome the energy barrier, the products BC molecules is formed (indicated by relatively constant small rBC) and increasing rAB as the H atom A moves away.
|[[File:Q4_3_Contour_YX8818.png|250px]]
|[[File:Q4_3_Surface_YX8818.png|300px]]
|-
|'''4'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|The distance rBC decreases at first when the molecule AB approaches H atom C and reaches the transition state (TS). As the system has sufficient energy, it was able to cross the TS and from the product BC molecule and atom A. However, due to the excess kinetic energy / momentum the system possess, the system recrossed the energy barrier at the TS again, and refromed the reactants molecule AB, and atom C.
|[[File:Q4_4_Contour_YX8818.png|250px]]
|[[File:Q4_4_Surface_YX8818.png|300px]]
|-
|'''5'''
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|The excess kinetic energy / momentum of the system allows it to cross the energy barrier at the TS to the product side, recrossed to the reactant side, and then crossed the energy barrier again to finallay form the refromed the products, molecule BC, and atom A.
|[[File:Q4_5_Contour_YX8818.png|250px]]
|[[File:Q4_5_Surface_YX8818.png|300px]]
|}
{{fontcolor1|blue|Q4: 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 hypthesis tested with the above simulation was: For a reactive trajectory, all trajectories starting with the same positions but with higher values of momenta (higher kinetic energy) would be reactive, as they have enough kinetic energy to overcome the activation barrier.

From the simulations of the trajectory, it is found that the hypothesis is false. This is supported by simulation 4, which have momenta higher than some reactive trajectories, but was not reactive. This shows that not all all trajectories starting with the same positions but with higher values of momenta would be reactive. The reactivity of a trajectory thus depend on not just the total energy in the system but also on how the energy is distributed.

==== Transition State Theory ====
{{fontcolor1|blue|Q5: Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

During a molecular collision, we must consider two molecules to form a single quantum-mechanical entity, which is called a supermolecule.<ref name=":1" />

The Transition State Theory (TST)<ref name=":1">Daniels, Farrington, J. Howard Mathews, and John Warren Williams. ''Experimental physical chemistry''. No. 541 D35 1962. New York: McGraw-Hill, 1962, chap 22.</ref> <ref>Henriksen, Niels E., and Flemming Y. Hansen. ''Theories of molecular reaction dynamics: the microscopic foundation of chemical kinetics''. Oxford University Press, 2018.</ref>chooses a boundary surface located between the reactant and product regions and assumes that all supermolecules that cross this boundary surface become products. The boundary surface, called the (critical) dividing surface, is taken to pass through the saddle point of the potential-energy surface.

TST has the following three main assumptions:

1. All supermolecules that cross the critical dividing surface from the reactant side becomes product.

2. During the reaction, the Boltzmann distribution of energy is maintained for the reactant molecules.

3. The supermolecules crossing the critical surface from the reactant side have a Boltzmann distribution of energy corresponding to the temperature of the reacting system.

Experimental results may violate the assumption 1, as in reality reactants and products are in equilibrium, thus allowing both forward and backward reactions to take place, so not all supermolecules that crossed the critical dividing surface from the reactant side will becomes product, causing experimental rates to be lower than TST prediction.This is supported by the simulations findings above.

It was found that majority of the results obtained (Set 1, 2, 3, and 5) agrees with the Transition State Theory, as the supermolecules with sufficient momenta to cross the critical dividing surface from the reactant side becomes product, and those that did not crossed the critical dividing surface remain as reactants. However, the results from Set 4, where the supermolecule recrossed the boundary, contradicted Assumption 1 of TST, showing that some supermolecules that crossed the critical dividing surface containing the transition state, can rebound and reformed the reactants. This shows that TST prediction for reaction rate values might not accurately agrees with experimental values for some cases.

Furthermore, as TST is a classical theory, it does not factor in the quantumn tunneling effect, which could result in the experimental rates to be higher the theory.

Note: the simulations above does not allow investigation in to the Boltzmann distribution of energy in the TST assumption as the system investigated consist of only two reactants, hence is only a small part on the Boltzmann distribution. {{fontcolor1|red| Great [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:46, 29 May 2020 (BST)}}

== Part 2: F - H - H system ==
'''In part 1, rHF = r1 = rAB, rHH = r2 = rBC.'''

=== PES inspection ===
{{fontcolor1|blue|Q6: 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?}}

The potential energy surface for F + H2 → HF + H prcess is shown below:
[[File:Q6_PE_surface_F_H2_YX8818.png|thumb|right|350px|Potential energy surface for F + H2 → H + HF prcess ]]

F + H2 → HF + H : Exothermic.

This can be shown from the lowering of potential energy going from reactant to products.

HF + H → F + H2: Endothermic.

This can be shown from the increase in energy going from reactants to products going in the reverse direction.

Bond strength of HF = 565 kJ.mol-1<ref name=":0">Huheey, pps. A-21 to A-34; T.L. Cottrell, "The Strengths of Chemical Bonds," 2nd ed., Butterworths, London, 1958; B. deB. Darwent, "National Standard Reference Data Series," National Bureau of Standards, No. 31, Washington, DC, 1970; S.W. Benson, J. Chem. Educ., 42, 502 (1965).</ref>

Bond strength of H2 = 432 kJ.mol-1<ref name=":0" />

The stronger bond strength of HF bond is a result of the high electronegativity of the F atom, which results in high bond polarisation in the HF bond and a strong ionic / electrostatic attraction between H and F atoms.

F + H2 → HF + H, ΔH = -565 + 432 = -133 kJ.mol-1, which supports it being an exothermic process as energy is being released overall. This is because the energy required to break the H-H bond is less than the energy released in the formation of H-F bond.

HF + H → F + H2 , ΔH = 565 - 432 = 133 kJ.mol-1, which supports it being an enothermic process as energy is being absorbed overall. This is because the energy required to break the H-F bond is more than the energy released in the formation of H-H bond.

{{fontcolor1|blue|Q7: Locate the approximate position of the transition state.}}

By Hammond postulate, an exothermic reaction of F + H2 to HF + H has an early transition state that resemble the reactant.

Thus, the distance rAB between F and H2 molecule should be large, while the H2 molecule distance rBC should be close the H2 bond distance.

Using the same method in Q2, the transition state position was found to be as follows:

F-H distance, rAB = 181.13 pm.

H-H distance, rBC = 74.45 pm.

{| class="wikitable"
|[[File:Q7_TS_zero_force_YX8818.png|thumb|center|350px|Settings for rAB and rBC when forces is zero]]
|[[File:Q7_TS_distance_vs_Time_YX8818.png|thumb|center|350px|Internuclear Distances vs Time Plot]]
|[[File:Q7_TS_contour_plot_YX8818.png|thumb|350px|Contour plot]]
|}

{{fontcolor1|blue|Q8: Report the activation energy for both reactions.}}

The activation energy of the forward and backward reactions where found by locating the energy of the transition state, reactants, and product, then determining the difference. The reactant and product formation trajectories were obtained by perturbing the system slightly away from the transition state. Results are summarised in the tables below:

{| class="wikitable"
!system
!rHF / pm
!rHH / pm
!Total Energy / kJ.mol-1
!Activation EnergyTotal Energy / kJ.mol-1
|-
|transition state: F-H-H
|181.13
|74.45
|<nowiki>-433.978</nowiki>
|<nowiki>---</nowiki>
|-
|reactant to product:
F + H2 → HF + H
|180.13
|74.45
|<nowiki>-560.328</nowiki>
|<nowiki>-433.978 - (-560.328) = 126.350</nowiki>
|-
|product to reactant:
HF + H → F + H2
|182.95
|74.45
|<nowiki>-435.015</nowiki>
|<nowiki>-433.978 - (-435.015) = 1.037</nowiki>
|}
{| class="wikitable"
!system
!Energy vs time plot (full scale)
!Energy vs time plot (zoomed)
!Contour plot
|-
|transition state: F-H-H
|<nowiki>---</nowiki>
|<nowiki>---</nowiki>
|<nowiki>---</nowiki>
|-
|reactant to product:
F + H2 → HF + H
|[[File:Q8_forming_HF_Activation_energy_E_vs_time_YX8818.png|300px]]
|[[File:Q8_zoomed_HF_Activation_energy_E_vs_time_YX8818.png|300px]]
|[[File:Q8_forming_HF_Contour_Plot_YX8818.png|300px]]
|-
|product to reactant:
HF + H → F + H2
|[[File:Q8_original_forming_reactant_H2_Activation_energy_E_vs_time_YX8818.png|300px]]
|[[File:Q8_forming_reactant_H2_Activation_energy_E_vs_time_YX8818.png|300px]]
|[[File:Q8_forming_reactant_H2_contour_YX8818.png|300px]]
|}

=== Reaction dynamics ===

The following is a set of reaction conditions that resulted in a reactive trajectory for F + H2
{| class="wikitable"
!rFH/pm
!rHH / pm
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Illustration of the trajectory
|-
|200
|75
|<nowiki>-1.5</nowiki>
|0.5
|[[File:Reactive_HFH_conditions_YX8818.png|350px]]
|}
{| class="wikitable"
!Energy vs time plot
!Momentum vs time plot
|-
|[[File:Reactive_HFH_energy_vs_time_YX8818.png|400px]]
|[[File:Reactive_HFH_momenta_vs_time_YX8818.png|400px]]
|}

{{fontcolor1|blue|Q9: 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.}}

Before the reaction took place (t = 0 to 60 s), the F atom only has translational energy, while H2 molecule has both translational and vibrational energy (as shown from the oscillatory behavior in momentum vs time diagram). Before the reaction, the total kinetic and potential energy were relatively constant (at 0 kJ.mol-1 and -433 kJ.mol-1 respectively).

After the exothermic reaction (at around t = 60 s), F + H2 → HF + H, some of the potential energy was converted into kinetic energy of the system, with the total energy remained constant (conservation of energy), as shown in the Energy vs time plot. The kinetic energy could be in the form of translational and vibrational (rotational and electronic energies were ignored here). For the HF molecule formed, its kinetic energy / momentum was in both translation and vibrational; while for the H atom, only translation energy was left. The HF molecule was observed to have a larg bond oscillation along the H-F bond, which resulted in larger oscillations in kinetic and potential energy after the reaction.

The overall release of energy / heat of this exothermic reaction could be monitored using calorimetry.<ref>Sunner, Stig, and Margaret Mansson. "Experimental chemical thermodynamics. Volume I. Combustion calorimetry." (1979).</ref> For a reaction involving gaseous reactants, constant-volume calorimeters, such as the bomb calorimeter, is suitable. The reactants can be placed inside a steel vessel with know heat capacity and sealed within an insulated container with known amount of water. The heat released could then be calculated from the temperature change.

In particular, the vibrational energy can be measured using IR spectroscopy, where absorption spectrum can be monitored over time. Before the reaction, the reactants mainly occupied the ground state (lowest vibrational energy level), there would essentially be one absorption peak from 0 to 1. During the exothermic reaction, the molecules will gain vibrational kinetic energy and be excited to occupy the first excited state, allowing overtones (1 to 2) to be observed at lower wavenumbers, thus resulting in two absorption peaks. As the reaction takes place, the intensity of the overtone will be come larger, and the intensity of the fundamental would decrease. After the reaction is complete, the vibrational modes will gradually relax back to the ground state, causing the overtones to disappear, and the fundamental to increase.

Emission of light as a result of this reaction, chemiluminescence,<ref>Dodeigne, C., L. Thunus, and R. Lejeune. "Chemiluminescence as diagnostic tool. A review." ''Talanta'' 51.3 (2000): 415-439.</ref> could also be used to monitor this process. The process monitored here would be the decay of the excited state to the lower energy level (ground state).

=== Further studies ===
Further invesitgations were conducted to investigate the effect of distribution of energy among translational and vibrational modes on the efficiency of the reaction:
* F + H2 → HF + H. Initial conditions: '''r1''' = rHF = 200 pm, '''r2''' = rHH = 74 pm, p1 = pFH = -1.0 g.mol-1.pm.fs-1, pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 .
* '''Table 1:'''
{| class="wikitable"
!pFH/ g.mol-1.pm.fs-1
!pHH/ g.mol-1.pm.fs-1
!Reactive?
|-
|<nowiki>-1.0 </nowiki>
|<nowiki>-6.0</nowiki>
|No
|-
|<nowiki>-1.0 </nowiki>
|<nowiki>-3</nowiki>
|No
|-
|<nowiki>-1.0 </nowiki>
|<nowiki>-2</nowiki>
|No
|-
|<nowiki>-1.0</nowiki>
|<nowiki>-1</nowiki>
|Yes
|-
|<nowiki>-1.0</nowiki>
|0
|Yes
|-
|<nowiki>-1.0</nowiki>
|1
|Yes
|-
|<nowiki>-1.0</nowiki>
|2
|No
|-
|<nowiki>-1.0 </nowiki>
|4
|No
|-
|<nowiki>-1.0</nowiki>
|5.5
|No
|}
* F + H2 → HF + H. For the same initial positions, simulations below were conducted.
* '''Table 2:'''
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Reactive?
|-
|<nowiki>-1.5</nowiki>
|0.2
|Yes
|-
|<nowiki>-1.6</nowiki>
|0.2
|Yes
|-
|<nowiki>-6.0</nowiki>
|0.2
|Yes
|}

* For the reverse reaction, HF + H → F + H2. '''r1''' = rHF = 180 pm and '''r2''' = rHH = 74 pm, the following simulations were conducted.
* '''Table 3:'''
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Reactive?
|-
|<nowiki>-8.0</nowiki>
|1
|Yes
|-
|<nowiki>-4.0</nowiki>
|1
|Yes
|-
|<nowiki>-1.0</nowiki>
|1
|No
|-
|<nowiki>-1.0</nowiki>
|<nowiki>-3.0</nowiki>
|No
|}

{{fontcolor1|blue|Q10: 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.}}

The effect of energy distribution on reaction efficiency can be described by '''Polanyi's empirical rules.'''<ref>D. M. Hirst Potential Energy Surfaces: Molecular Structure and Reaction Dynamics Taylor and Francis, 1985, Chap 6.3. </ref>

Polanyi's empirical rules describes how different forms of energy affect the rates of reactions. The rules state that for a given amount of energy / momentum, the vibrational energy is more efficient than translational energy in activating an late barrier / endothermic reactio; whereas translational energy is more efficient than vibrational energy for an early barrier / exothermic reaction.

'''F + H2 → HF + H is an exothermic reaction with an early barrier, thus the trajectory with high translational energy (pFH) and low vibrational energy (pHH) is reactive.'''

This is suported by the simulations in Tables 1 and 2, where trajectories with high pFH and low pHH are generally reactive, whereas trajectories with low pFH and high pHH are generally unreactive. This is because the momentum in F atom (pFH) is almost entirely translational, whereas the energy in H2 molecules (pHH) is both translational and vibrational, and the vibrational energy can be thought of as proportional to the total energy present. The trends in the table largely suports Polanyi's rules that translational energy is more efficient in early barrier.

'''HF + H → F + H2 is an endothermic reaction, thus the trajectory with high vibrational energy (pFH) and low translational energy(pHH) is reactive.'''

Trends shown in Table 3 also largely supports Polanyi's rules that vibrational energy is more efficient for late barriers. As when higher vibrational energy (pFH) and lower translational energy(pHH) tend to give reactive trajectories.

However, its worthy to note that in the above simulations, the total energy was not kept constants while varying the proportions of the translational and vibrational energy. And the trajectories of the simulations are quite sensitives, sometimes giving conflicting results for certain values of initial conditions. Further explorations could be considered by keeping the total energy / momentum in the one-dimensional trajectory constant, perhaps by varying the collison angle or using more sophisticated simulation package.

= References =

MRD:M4H1M4

2020-05-29T17:45:13Z

Mak214: /* Transition State Theory */

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

=== Dynamics from the transition state region ===
[[File:Hdiagram.png|200px|thumb|left|H + H2]]
''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 first order saddle point on the potential energy surface (PES) diagram. The derivative of potential energy is equal to the force acting on the atoms/molecules. At this point the gradient is equal to 0 and therefore no force is acting on the particles. This allows us to use a constrained one dimensional optimisation to calculate transition state. In order to verify this stationary point as the saddle transition state the Hessian eigenvectors can be used. Only one of the second derivatives must be negative and the other positive so that (fxxfyy - f2xy) < 0. Therefore the determinent of the Hessian matrix less than 0.

This verifies the point as a saddle rather than a local minimum of the potential energy surface. {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:06, 29 May 2020 (BST)}}

==== Trajectories from r1 = r2: locating the transition state ====
''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.''

The the H + H2 ---> H2 + H surface is symmetrical, neither H2 or H atom are favoured.Therefore r1 is equal to r2 in the transition state. This restriction allows us to calculate the transition state position through a process of trial and error.

[[File:R1=R2.png|600px|thumb|left|Figure 1 - PES of transition state of reaction H + H2]]

[[File:Hessian.png|600px|thumb|Figure 2 - Hessian matrix and determinant for H + H2 reaction]]

[[File:Dvst01499529.png|600px|thumb|left|Figure 3 - Displacement vs Time for reaction at transition state]]Transition state position r1= r2 = 90.8 pm

As shown in figure 1, at these positions the forces action along AB and BC are approximately 0 suggesting a stationary point. Figure 3 shows two horizontal lines with gradient 0 which therefore means 0 acceleration and 0 forces acting on the atoms.

Looking at the Hessian matrix we can verify this as a saddle point. To further confirm this, trajectories were calculated with either r1 or r2 being slightly displaced from equilibrium. Figure 4 shows that increasing r2 by 1 pm pushing the reaction to the reactant side figure 5 shows increasing r2 by 1 pm favours the formation of the products. {{fontcolor1|red| Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:06, 29 May 2020 (BST)}}

[[File:Reactantchannel.01499529.png|200px|thumb|Figure 4 - Trajectory in reactant channel at '''r2''' = '''rts'''+1, '''r1''' = '''rts p1''' = '''p2''' = 0 g.mol-1.pm.fs-1]]

[[File:Productchannel.01499529.png|200px|thumb|Figure 5 - Trajectory in product channel at '''r1''' = '''rts'''+1 pm, '''r2''' = '''rts, p1''' = '''p2''' = 0 g.mol-1.pm.fs-1]]

==== Trajectories from r1 = rts+δ, r2 = rts ====
''Comment on how the mep and the trajectory you just calculated differ.''

Figure 6 shows a straight line suggesting the velocity and momentum do not change with time. Figure 7, the dynamic trajectory, accounts for the oscillation of the bonds. {{fontcolor1|red| Why? What are the conditions for the MEP calculation that lead you to see this. More discussion is needed here. MEP calculations involve resetting the momenta to zero at every time step so we end up finding the nearest local minimum.. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:06, 29 May 2020 (BST)}}

Final values of the positions '''r1'''(t) '''r2'''(t) and '''p1'''(t) '''p2'''(t):

r1=74.0 pm r2= 352.6 pm p1= 3.2 g.mol-1.pm.fs-1 p2= 5.1 g.mol-1.pm.fs-1
* ''Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.''
''What do you observe?''

The reaction does not occur atom C reaches a constant velocity and atom B and A vibrate with constant acceleration.
[[File:MEPMT.png|200px|thumb|middle|Figure 6 - Momentum vs Time MEP]]
[[File:DYNMT.png|200px|thumb|middle|Figure 7 - Momentum vs Time dynamic]]

=== Reactive and unreactive trajectories ===
{| 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
|-
|1
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.3</nowiki>
|Yes
|Initially r1 < r2, and therefore atoms A and B are bonded to each other and atom C is the approaching atom. R1 remains almost constant as r2 begins to decrease. r1 begins to increase until both r1 and r2 are equal at the transition state. From this point r2 remains almost constant whilst r1 increases. Atom A and B have disassociated whilst a new B-C bond has been formed. This bond has kinetic energy hence the vibrations.
|[[File:1.yes.png|200px|thumb|left|Figure 8 - Trajectory set 1]]
|-
|2
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.1</nowiki>
|No
|From the initial positions, r1 oscillates around 74 pm. Distance r2, intially decreases as C approaches B. The trajectory does not go through the TS instead r2 begins to increase again. In this trajectory A and B remain bonded and atom C is repelled from atom B. This is because the momentum p2 is not large enough for a reaction to occur.
|[[File:2.no.png|200px|thumb|left|Figure 9 - Trajectory set 2]]
|-
|3
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.0</nowiki>
|Yes
|From the initial position, r2 decreases with little change in r1. r1 beings to increase until the atoms are in the transition state. r1 increases suggesting the breaking of the A-B bond. r2 continues to oscilate due to internal vibration of the B-C bond.
|[[File:3.yes.png|200px|thumb|left|Figure 10 - Trajectory set 3]]
|-
|4
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.3</nowiki>
|No
|This trajectory does not go through the transition state.Atom C approaches with too much momentum. The A-B bond breaks and the B-C bond forms. B-C vibrates with a large amount of momentum and kinetic energy. B-C vibrates with increasing r2 until passing back through to the reactant channel with the formation of A-B and the disassociation of B-C This is a case of barrier recrossing.Therefore there is no overall reaction.
|[[File:4.no.png|200px|thumb|left|Figure 11 - Trajectory set 4]]
|-
|5
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.5</nowiki>
|Yes
|Atom B and C approach each other with a large amount of momentum causing the A-B bond to break and the formation of the B-C bond.r2 begins to increase agin while the A-B bond reforms. r1 increaes again while r2 decreases with the trajectory returning to the product channel leading to an overall reaction.
|[[File:5.yes.png|200px|thumb|left|Figure 12 - Trajectory set 5]]
|}
''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?''

From this table we can conclude, trajectories with initial conditions in the range '''r1''' =74 pm, '''r2''' = 200 pm, with -3.1 < '''p1'''/ g.mol-1.pm.fs-1 < -1.6 and '''p2''' = -5.1 g.mol-1.pm.fs-1 are always reactive. However trajectories higher momenta and therefore higher kinetic energy does not necessarily mean a successful reaction.This is evident by trajectory set 4 having a kinetic energy of 76.5 kJ/mol and resulting in no reaction and set trajectory 3 having only 19.8 kJ/mol and being successful. Increasing the momentum can lead to barrier crossing which if there isn't sufficient energy to cross back to the product channel can lead to overall no reaction.

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

Transition state theory assumes quasi-equilibrium between the reactants and the activated transition state complex and only requires details bout the PES. The Born-Oppenheimer approximation is used and quantum tunnelling affects are assumed to be negligible. The energies of the reactants also follow the Boltzman distribution curve.<ref>T. Bligaard, J.K. Nørskov, in Chemical Bonding at Surfaces and Interfaces, 2008</ref> From the table above we can see that quantum tunneling and barrier recrossing can occur, As TST does not account for that the value of rate calculated would be an overestimate compared to the experimental value. However error due to this is only marginal.<ref>PetersBaron , in Reaction Rate Theory and Rare Events Simulations, 2017</ref> {{fontcolor1|red| Think about this - barrier recrossing we do see in the table of results above and does mean that overall TST overestimates rate. However quantum tunnelling (not visible in your simulations above) leads to an underestimate in the rate predicted by TST because this is when the system can tunnel through an energy barrier - therefore reaching the products faster than we expect... [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:45, 29 May 2020 (BST)}}

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

=== PES inspection ===
[[File:Fdiagram.png|200px|thumb|left|H2 + F]]
F + H2 ----> HF + H (1)

Initial conditions: p1= p2= 0 g.mol-1.pm.fs-1, r1=74.1 pm, r2= 400 pm

HF + H -----> H2 + F (2)

<nowiki> </nowiki>Initial conditions: p1=p2=0 g.mol-1.pm.fs-1 r1=400 pm, r2= 91.7 pm
[[File:FH2.png|200px|thumb|left|Figure 13 - PES of H2 + F]]
[[File:HfH.png|200px|thumb|left|Figure 14 - PES of HF + H]]
From figure we can conclude this reaction is exothermic as the formation of the H-F bond releases energy greater than the energy required to break the H-H bond. In the PES diagram as p2 decreases p1 increases the trajectory falls into the product channel at a lower energy than the products. This is reflective of the bonds being broken and formed. Reaction 2 is endothermic as the H-H bond being formed releases less energy than required to break the H-F bond. From this we can conclude the H-F bond is stronger that the H-H bond.

H-H bond energy = 435.7799 kJ/mol<ref name=":0">Luo, Yu-Ran. ''Comprehensive handbook of chemical bond energies''. CRC press, 2007.</ref>

H-F bond energy = 569.680 kJ/mol<ref name=":0" />

''Locate the approximate position of the transition state.''
[[File:TScontuor.png|200px|thumb|Figure 15 - Transition state of H2 + F]]

Fluorine is highly electronegative and so attacks the hydrogen early on in the reaction this is fitting of the exothermic characteristic of the reaction. Using the Hammond postulate, this early transition state therefore means that the structure of the transition state resembles that of the reactants. Using the liturature value of H-H bond length, r1 was set to 74.1 pm as an initial estimate.<ref>Huber, K.P.; Herzberg, G., Molecular Spectra and Molecular Structure. IV. Constants of Diatomic Molecules,, Van Nostrand Reinhold Co., 1979</ref> Trial and error was performed in order to calculate the r1 and r2 values that give the closest to 0 kJ/mol.pm of force along the axis.

Transition state position - r1= 74.5 pm r2= 181.1 pm

''Report the activation energy for both reactions.''

To calculate the activation energy of both reactions the potential energy at transition state was recorded. The energy of the reactants of the forwards reaction was measured by setting r1 as the bond length of H2, 74.1 pm and r2 being set as 500 pm. This allows us to view the energy of the reactants when they are not reacting with each other. The same was repeated for the backward reaction using the bond length of HF, 91.7 pm as r2.<ref>NIST Diatomic Spectral Database (www.physics.nist.gov/PhysRefData/MolSpec/Diatomic/index.html)</ref>

<nowiki> </nowiki>Transition state energy = -434.0 kJ/mol

Energy of Reactants H2 + F = -435.1 kJ/mol Ea1 = 1.1 kJ/mol

Energy of Reactants HF + H = -560.7 kJ/mol Ea2 = 125.6 kJ/mol

=== Reaction dynamics ===
''Set of initial conditions that results in a reactive trajectory for the F + H2'': r1= 74.5 pm, r2= 130 pm, p1= -1g.mol-1.pm.fs-1 p2= -3 g.mol-1.pm.fs-1 

''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.''

Translational energy is required in order for the reactants to collide with one another with enough energy to react. However large amounts of excess translational energy can lead to the reacts rebounding off of each other.

MRD:mb7418

2020-05-29T17:41:39Z

Mak214: /* Transition state theory */

== EXERCISE 1: H + H2 system ==

===Dynamics from the transition state region===
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? {{fontcolor1|red| This text is repeated below which makes this section difficult to read. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:03, 29 May 2020 (BST)}} Potential energy surface diagrams show the potential energy the system has according to the coordinates of the atoms in the system.<ref>Chemistry LibreTexts. 2020. ''2.6: Potential Energy Surfaces''. [online] Available at: <<nowiki>https://chem.libretexts.org/Courses/University_of_California_Davis/UCD_Chem_107B%3A_Physical_Chemistry_for_Life_Scientists/Chapters/2%3A_Chemical_Kinetics/2.06%3A_Potential_Energy_Surfaces</nowiki>> [Accessed 19 May 2020].</ref> Figure 1 shows a potential energy surface diagram representing the reaction, equation 1: A + BC → AB + C.
{| class="wikitable"
|+Table 1. Molecules distances (r) and momenta (p)
!
!r / ppm
!p/ g.mol-1.pm.fs-1
|-
|r2=r(AB)
|230
|<nowiki>-5.2</nowiki>
|-
|r1=r(BC)
|74
|0
|}
In this case r(1)=r(BC) and r(2)=r(AB). BC is a hydrogen molecule consisting of two hydrogen atoms, B and C. Atom A is a hydrogen atom which has energy equal to or more than the activation energy, thus reaching the transition state. The transition state structure consists of atom A forming a bond with molecule BC while the BC bond is slightly dissociating. After the transition state, the bond between AB would form making a new molecule and the BC bond would be fully dissociated.
* Q1. 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?
[[File:PE with TS.png|left|thumb|[[Figure]] 1. Potential energy surface diagram with transition state labelled ]]
In a potential energy surface diagram, the transition state is mathematically defined as the first-order saddle point<ref>Chemistry LibreTexts. 2020. ''Hammond’S Postulate''. [online] Available at: <<nowiki>https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hammond%E2%80%99s_Postulate</nowiki>> [Accessed 22 May 2020].</ref>:

∂V(r1)/∂(r1)=0 and ∂V(r2)/∂r2=0

∂2V(r)/∂(r1)2 > 0 and ∂2V(r)/∂(r2)2 < 0

[∂2V(r)/∂(r1)2]*[ ∂2V(r)/∂(r2)2] - [ ∂2V/∂(r1)(r2)]2 < 0

Hessian matrix is shown in Table 2, resulting in DET = -0.999698 < 0
{| class="wikitable"
|+Table 1. Hessian matrix for the transition state
!
!f(x)
!f(y)
|-
|f(x)
|<nowiki>+0.707</nowiki>
|<nowiki>-0.707</nowiki>
|-
|f(y)
|<nowiki>-0.707</nowiki>
|<nowiki>-0.707</nowiki>
|}
In Figure 1, the black wavy lines shows that the molecule is vibrating. This is the minimum energy path required for the reactants to react and turn into the products via the transition state.<ref>Atkins, de Paula, Keeler: Physical Chemistry, 11th Edition. Section 18D.3 Potential Energy Surfaces (PESs)</ref> This suggests that the transition state is the peak in the minimum energy path. The transition state is the minimum energy required for the reactants to convert into the products. Hence once the transition state is reached, if the reaction was successful the products would be formed. Figure 1 shows the location of the transition state in the potential energy diagram.

The transition state can be distinguished from a local minimum of the potential energy surface because it is on the diagonal axis. {{fontcolor1|red|Because it is also a maximum (minimum AND maximum = saddle point) therefore distinguishable from a local minimum. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:03, 29 May 2020 (BST)}} It is the saddle point.This is when everything in the system is stationary when there is no momentum.

===Locating the transition state r1=r2===
* Q2. 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.

<nowiki> </nowiki>When the transition state is formed, due to all the atoms being hydrogen, this suggest that the transition state is symmetrical. This means that the bond distance between A and B is the same as the bond distance of B and C. This can be used to locate the transition state when the momentum is 0. Table 3 shows an estimate of the transition state position (rts).
{| class="wikitable"
|+Table 3. Distances and momenta to locate the transition state
!
!r / ppm
!p/ g.mol-1.pm.fs-1
|-
|r2=r(AB)
|90.75
|0
|-
|r1=r(BC)
|90.75
|0
|}
[[File:Mb7418 dis vs time ts.png|thumb|[[Figure 2]]. Distance vs time plot of the transition state ]]
Figure 2 shows an Intermolecular distance vs time graph resulted from the initial conditions from Table 2. The line is flat, this suggest that everything in the system is stationary. Whereas, if it wasn't the transition state, in the intermolecular distance vs time graph you'd see oscillations representing the vibrational movements. The more oscillations would indicate the more vibrational movements. Hence the transition state is a flat line as there isn't any. {{fontcolor1|red| OK. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:33, 29 May 2020 (BST)}}
[[File:Mb7418 ts contour.png|thumb|[[Figure 3]]. Potential energy diagram as a contour plot of the transition state]]
The contour plot shown in Figure 3 shows the transition state as a red cross. This is the saddle point of the potential surface. Whenever r1 and r2 are the same, the point would be anywhere across the diagonal axis. This contour plot helped estimate the transition state position as the trajectory shows that the AB and BC bond lengths are the same. This is the point at which reactants have to overcome to react into the products.

The minimum energy path can differ depending on the relative momentum of the atoms/molecules and their vibrational excitation energy. The transition state of the reaction won't change its position. The reactants require sufficient translational kinetic energy and vibrational energy to overcome the transition state (saddle point) in order to react and turn into the products.

===Trajectories from r1 = rts+δ, r2 = rts ===
* Q3. Comment on how the ''mep'' and the trajectory you just calculated differ. '''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''' and the momenta '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1
The trajectory of the minimum energy path can be visualised slightly near the transition state, this is when atom A is approaching molecule BC to form a bond with it, A + BC.
{| class="wikitable"
|+Table 4. Initial conditions slightly displaced from transition state
!
!r /ppm
!p/ g.mol-1.pm.fs-1
|-
|r2=r(AB)
|90.75
|0
|-
|r1=r(BC)
|91.75
|0
|}
The initial conditions shown in Table 4 gives the potential energy diagram in Figure 4 and 5 when the calculation type is MEP. The minimum energy path shown by the black line isn't wavy. [[File:Mb7418 mep.png|left|thumb|[[Figure 4]]. MEP calculation diagram]]
[[File:Mb7418 mep 1.png|thumb|234x234px|[[Figure 5]]. MEP calculation diagram bottom view]]However, when the same initial conditions are applied when the calculation type is Dynamic, the minimum energy path is wavy shown in Figure 6. This shows that the MEP calculation gives a different result of the reaction trajectory. The molecule is constantly moving at a velocity since there is no external force acting upon the reactants. This is called inertia motion.<ref>Evans, M.G. and Polanyi, M., 1938. Inertia and driving force of chemical reactions. ''Transactions of the Faraday Society'', ''34'', pp.11-24.</ref> Whereas the potential energy diagram produced from the Dynamics calculation takes into account the reactants inertia by representing the pathway as non-linear. Additionally, the MEP calculation diagram has a smaller trajectory than the Dynamics calculation.The reason for this is beacause the MEP calculation sets the momentum to zero at every step whereas the dynamic calculation the momenta is developed classically<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., 8.3.1, Prentice-Hall, 1998.</ref>. The MEP calculation only takes the gradient into account. The amount of steps there are corresponds to the number of times it is calculating what the gradient is at that point. The only thing relative to the MEP calculation is the gradient. Whereas with dynamics, it takes into account how long it takes for the gradient to change. The momentum (velocity) is changing each time instead of setting it to zero every step. Hence it's longer as after the products are formed, they are still have motion such as vibrational and kinetic. {{fontcolor1|red| Yes. This does show an understanding of MEP vs Dynamics. Just to clarify - the amount of steps is the number of times the system moves during the simulation - this was a bit confused in your response. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:36, 29 May 2020 (BST)}} [[File:Mb7418 dyn 1.png|thumb|[[Figure 6]]. Dynamics calculation bottom view]]

== Reactive and unreactive trajectories ==

===Interpreting trajectories===

* Q4. For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, run trajectories with the following momenta combination and complete the table.

{| class="wikitable"
|+Table 5. Reactive and unreactive trajectories
!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>
|Reactive
|The starting point (represented by the red cross) starts at 200 pm between atoms A and B. Then the distance AB decreases. Atoms A and B are approaching each other and once it reaches the transition state, the AB distance and BC distance are the same and all the atoms are stationary. Then the AB bond is formed and atom C repels from atom B. The molecule AB is constantly vibrating. This could suggest that some translational kinetic energy got converted into vibrational energy after the transition state. Hence after the trajectory has oscillations after it passes the transition state.
|[[File:Mb7418 6.png|centre|thumb|200x200px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The trajectory has some oscillations, showing that the reactants are constantly vibrating. Hence it has vibrational energy. However, it doesn't go over the transition state so it doesn't react into the products. This suggests that the reactants approach each other but then they repel. While they do that they oscillate as they get closer. It ends up with a high AB distance (essentially the starting point of the reaction), this shows that the reactants haven't reacted due to the AB molecule not forming. This could be due to not having enough translational kinetic energy.
|[[File:Mb7418 2.png|centre|thumb|200x200px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The starting point starts at 200 pm between atoms A and B. Then the the atoms got closer and reaches the transition state. This is when the distance AB and BC are equal. Then the AB distance remains continuous while the BC distance begins to increase. The BC bond breaks and the AB bond forms. It only crosses the transition state once. It ends up in the products as the distance AB is short (molecule containing atoms A and B bonding) and BC is distance is long (reactant molecule BC bond has been broken).
|[[File:Mb7418 3.png|centre|thumb|200x200px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Unreactive
|The whole trajectory shows that it's unreactive because the starting point and ending point are similar. It started at a high AB distance value and also ended at a high AB distance value. This suggests this is end unreactive.
At the starting point, AB are far apart. Then they begin approaching each other and reach the transition state. However, they repel from each other and then start approaching each other again. This happens again and then the distance AB begins to increase again. The BC bond is formed again, resulting in the reactant molecule BC at the end.
|[[File:Mb7418 4.png|centre|thumb|200x200px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Reactive
|This crosses the transition point more than once. It's reactive as it's end point is at a high BC distance and low AB distance, suggesting that the BC bond broke and the AB bond has formed resulting in AB + B.

From the starting point, the AB distance starts to decrease. This suggests atom A and B approach each other, but then the AB distance begins to increase again so the A and B atoms repel. There is a moment when the distance between atoms A and B is the same as the distance between B and C (transition state) Then the B and C atoms approach each other and then repel B continuously. While atoms A and B get closer forming a bond. The molecule AB vibrates while atom C continuously repels away from it.
|[[File:Mb7418 5.png|centre|thumb|200x200px]]
|}

{{fontcolor1|red| OK. What do you conclude from the table? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:40, 29 May 2020 (BST)}}

=== Transition state theory ===
* Q5. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?
The three assumptions of the transition state theory<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Chapter 10 pg. 20, Prentice-Hall, 1998</ref> are:

1) The transition state will only be crossed over once. Once the transition state it reached, the products will be formed.

2) During the transition state, any motion along the reaction coordinate can be separate from the other motions and treated classically as a translation. The concentration of the transition state is in equilibrium with the reactants. (concentration constant relating them both, there's a Maxwell-Boltzmann distribution between the reactants and the transition state at any given time) This is defined as the Quasi equilibrium. It's assumed that the kinetic energy is predicted by the Boltzmann distribution. The transition states that are turning into products are distributed among their states according to the Maxwell-Boltzmann Laws<ref>Rowlinson*, J.S., 2005. The Maxwell–Boltzmann distribution. ''Molecular Physics'', ''103''(21-23), pp.2821-2828.</ref> even in absence of an equilibrium between the reactant and product molecules.

'''3)''' There's minimum effects of quantum tunneling.

The quantum tunneling effect<ref>Abyss.uoregon.edu. 2020. ''Quantum Tunneling''. [online] Available at: <<nowiki>http://abyss.uoregon.edu/~js/glossary/quantum_tunneling.html</nowiki>> [Accessed 22 May 2020].</ref> is when the reactants can overcome the barrier by moving across it to turn into the products. The transition state theory doesn't include quantum tunneling. If quantum tunneling was included, it would increase the rate of reaction as more reactants will turn into products. There are only classical motions in the transition state theory. Hence it is suggested that quantum tunneling isn't included in the model used in Table 5 because it's assumed there's a faster alternative route used.

Additionally, in the Table 5 model the transition state theory wasn't true because the transition state was crossed over more than once in the last two reactions in Table 5, this could decrease the rate of reaction due to the products being able to cross the transition state again and go back into the reactants. Therefore, the transition state theory is an over-estimate in comparison to the model used. {{fontcolor1|red| Yes. Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:41, 29 May 2020 (BST)}}

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

=== PES inspection ===
* Q6. 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?
{| class="wikitable"
|+Table 6. Initial conditions for Figure 7 (F + H2 → HF + H)
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|500
|0
|-
|r(BC)
|70
|0
|}

[[File:Mb7418 F + H2 exo edit CORRECT.png|left|thumb|[[Figure 7]]. F + H2 -> HF + H exothermic potential surface diagram]]
In the following equations, A = atom F, B = atom H, C = atom H.

The initial conditions in Table 6 gives the potential energy diagram for Figure 7. The reason for this is because the F and H bond is very far apart, suggesting that it's not bonded whereas the hydrogen atoms are closer together, suggesting a F atom and H2 molecule are in the initial conditions.

Equation 2: F + H2 → HF + H is an exothermic reaction<ref name=":0">Duff, J.W. and Truhlar, D.G., 1975. Effect of curvature of the reaction path on dynamic effects in endothermic chemical reactions and product energies in exothermic reactions. ''The Journal of Chemical Physics'', ''62''(6), pp.2477-2491.</ref> shown in Figure 7. The energy of the reactants is higher than the energy of the products. This shows that this is an exothermic reaction due to the process releasing energy. This suggests that the bond strength of H2 is weaker than the bond strength of HF as no additional energy was required to break its bond. This reaction has an early transition state<ref>Chemistry LibreTexts. 2020. ''Hammond’S Postulate''. [online] Available at: <<nowiki>https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hammond%E2%80%99s_Postulate</nowiki>> [Accessed 22 May 2020].</ref> due to the the fluorine atom being highly electro-negative so it attacks the hydrogen molecule very quickly.
{| class="wikitable"
|+Table 7. Initial conditions for Figure 8 (HF + H → F + H2)
!
!r / ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|70
|0
|-
|r(BC)
|500
|0
|}
[[File:Mb7418 NEW CORRECT ENDO.png|left|thumb|[[Figure 8]]. HF + H -> F + H2 endothermic potential surface diagram]]
The initial conditions in Table 7 gives the potential energy diagram for Figure 8. The reason for this is because the F and H atoms are very close together compared to both the hydrogens. This suggests that a HF molecule and one H atom are in the initial conditions.

Equation 3: HF + H → F + H2 is an endothermic reaction<ref name=":0" /> shown in Figure 8. The energy of the reactants is lower than the energy of the products. This suggests that this reaction is endothermic as the process used additional energy to break the H-F bond to form the H-H bond. Therefore, in the potential energy surface diagram there will be a lot of vibrational energy before the transition state compared to after. This also suggest that the HF bond is stronger than the HH bond as it requires additional energy to break its bond. This will give a late transition state.The translational energy could get converted into vibrational energy in order to break the bond between the strong H-F bond<ref name=":1">Polanyi, J.C. and Tardy, D., 1969. Energy Distribution in the Exothermic Reaction F+ H2 and the Endothermic Reaction HF+ H. ''The Journal of Chemical Physics'', ''51''(12), pp.5717-5719.</ref>. This favours the endothermic reaction because the reactants position will be aligned with the direction of the minimum energy path.''''' '''''

* Q7. Locate the approximate position of the transition state.
{| class="wikitable"
|+Table 8. Transition state position
!
!r /ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|180
|0
|-
|r(BC)
|74.5
|0
|}
The transition state position shown in Table 8 was estimated by looking at the initial conditions in Table 6 for the exothermic F + H2 → HF + H reaction. Due to exothermic reactions having early transition states, this suggested that the bond distance BC (the hydrogen atoms) would be small as it's the reactant. The initial bond distance of AB was reduced until the hessian matrix, shown in Table 9, gave a value of Det= -1.0 <0 which proved it was a saddle point. Additionally, there was no force. This also suggested it was the transition state.
{| class="wikitable"
|+Table 9. Hessian matrix for Transition state
|AB direction
|<nowiki>+1.000</nowiki>
|<nowiki>-0.024</nowiki>
|-
|BC direction
|<nowiki>-0.024</nowiki>
|<nowiki>-1.000</nowiki>
|}
[[File:Mb7418 dis vs time ts ex2.png|thumb|[[Figure 9]]. Intermolecular distance vs time graph for transition state for equations two and three]]
Additionally, Figure 9 shows the intermolecular distance vs time graph. This is a flat line, this could be due to everything in the system being stationary. There are no oscillations showing that there are any vibrational motions. This suggest that it's at the initial conditions in Table 8 represent the transition state position.

* Q8. Report the activation energy for both reactions.
Equation 2: F + H2 → HF + H the activation energy value is 1.0 kJ/mol.

Equation 3: HF + H → F + H2 the activation energy value is 126.7 kJ/mol.

The activation energy for equation 2 was calculated by getting the potential energy for transition state by using the initial conditions in Table 8. Then this was subtracted by the potential energy obtained by using the initial conditions in Table 10. Activation energy = TS(potential energy) - Reactants(potential energy) = (-433.981)-(435.056) = 1.075 kJ/mol
{| class="wikitable"
|+Table 10. Initial conditions for equation 2
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|500
|0
|-
|r(BC)
|74.5
|0
|}
The activation energy for equation 3 was calculated by getting the potential energy for for transition state by using the initial conditions in Table 8. Then this was subtracted by the potential energy obtained by using the initial conditions in Table 11. r(AB) which is the HF bond distance = 92 ppm<ref>Cccbdb.nist.gov. 2020. ''CCCBDB Listing Of Experimental Data Page 2''. [online] Available at: <<nowiki>https://cccbdb.nist.gov/exp2x.asp?casno=7664393&charge=0</nowiki>> [Accessed 22 May 2020].</ref>'''. '''The r(BC) bond which is the HH bond distance was set very high because the hydrogen atoms aren't bonded in the reactants.

Activation energy = TS(potential energy) - Reactants(potential energy) = (-433.981)-(560.698) = 126.7 kJ/mol
{| class="wikitable"
|+Table 11. Initial conditions for equation 3
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|92
|0
|-
|r(BC)
|500
|0
|}

=== Reaction dynamics ===
* Q9. 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.
{| class="wikitable"
|+Table 12. Initial conditions for reactive trajectory
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|175
|<nowiki>-1.6</nowiki>
|-
|r(BC)
|74
|0.3
|}
[[File:Mb7418 THIS reactive.png|left|thumb|[[Figure 10]]. Reactive trajectory]]
The step number was increased to 4000 and the initial conditions in Table 12 gives the potential energy surface diagram in Figure 10. This shows a reactive trajectory for the exothermic reaction F + H2 → HF + H as it passes the transition state as well as having many oscillations. The end point is also far away from the starting point.

The starting point (reactants) has a high energy level compared to the products due to being exothermic<ref name=":0" />. Hence the potential energy converts into translational kinetic energy. Therefore, there are many oscillations in Figure 10 due to energy being conserved.

This can be explained by the difference in distribution of the energy levels between an excited and relaxed state. The excited state will have an occupied ground state (v=0) as well as an occupied first state (v=1) where v=vibrational number. Whereas in the relaxed state only the ground state will be occupied. This approach uses the harmonic oscillator<ref>Cccbdb.nist.gov. 2020. ''CCCBDB Listing Of N2 Experimental Data Page 2''. [online] Available at: <<nowiki>https://cccbdb.nist.gov/exp2x.asp?casno=7727379&charge=0</nowiki>> [Accessed 18 May 2020].</ref>. When energy is released, in the excited state there will be two transitions, v(0)→v(1) and v(1)→v(2), the fundamental peak and overtone. In the ground state it will only have transitions from v(0)→v(1).

Therefore, in an IR spectrum<ref>Chemistry LibreTexts. 2020. ''5.5: The Harmonic Oscillator And Infrared Spectra''. [online] Available at: <<nowiki>https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.05%3A_The_Harmonic_Oscillator_and_Infrared_Spectra</nowiki>> [Accessed 22 May 2020].</ref>, for the excited state there will be two peaks corresponding to the overtone and fundamental peak. The overtone peak intensity will be smaller than the fundamental's peak intensity. Hence there are many oscillations due to the overtone decreasing and then increasing in intensity while the fundamental increases and then decreases in intensity.

Therefore, this could be confirmed experimentally by Infrared spectroscopy (IR).
* Q10. 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.
[[File:Mb7418 last Q.png|left|thumb|[[Figure 11]]. Reactive potential energy surface contour plot]]
The position of the transition state affects whether the reaction is either exothermic or endothermic. An endothermic reaction will have a late transition state, suggesting that the vibrational energy efficiency will be higher<ref>Guo, H. and Liu, K., 2020. ''Control Of Chemical Reactivity By Transition-State And Beyond''.</ref>. The reason for this is because if it had a higher translational kinetic energy, the reactants wouldn't follow the minimum energy path and will just collide with each other but then immediately repel resulting in being back at the starting point. Hence, the translational kinetic energy converts into vibrational energy<ref name=":1" /> instead so the reactants collide and are aligned with the minimum energy pathway. The direction of the path taken is shown in Figure 11. The initial conditions in Table 12 was used to give the reactive trajectory. This suggests that endothermic reactions have higher vibrational energy to make their reactions more efficient so it's able to overcome the transition state. Hence there are a lot of oscillations before the transition state is reached because it requires more efficient vibrational energy in order to reach the transition state.

Whereas if the reaction has an early transition state, it will be an exothermic reaction. This suggest that there will be a higher translational kinetic energy compared to vibrational energy. This means that there will be less oscillations before it reaches the transition state as shown in Figure 11.

Overall, an efficient endothermic reaction will have a higher vibrational energy whereas an efficient exothermic reaction will have a higher translational kinetic energy.

MRD:mb7418

2020-05-29T17:40:40Z

Mak214: /* Interpreting trajectories */

== EXERCISE 1: H + H2 system ==

===Dynamics from the transition state region===
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? {{fontcolor1|red| This text is repeated below which makes this section difficult to read. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:03, 29 May 2020 (BST)}} Potential energy surface diagrams show the potential energy the system has according to the coordinates of the atoms in the system.<ref>Chemistry LibreTexts. 2020. ''2.6: Potential Energy Surfaces''. [online] Available at: <<nowiki>https://chem.libretexts.org/Courses/University_of_California_Davis/UCD_Chem_107B%3A_Physical_Chemistry_for_Life_Scientists/Chapters/2%3A_Chemical_Kinetics/2.06%3A_Potential_Energy_Surfaces</nowiki>> [Accessed 19 May 2020].</ref> Figure 1 shows a potential energy surface diagram representing the reaction, equation 1: A + BC → AB + C.
{| class="wikitable"
|+Table 1. Molecules distances (r) and momenta (p)
!
!r / ppm
!p/ g.mol-1.pm.fs-1
|-
|r2=r(AB)
|230
|<nowiki>-5.2</nowiki>
|-
|r1=r(BC)
|74
|0
|}
In this case r(1)=r(BC) and r(2)=r(AB). BC is a hydrogen molecule consisting of two hydrogen atoms, B and C. Atom A is a hydrogen atom which has energy equal to or more than the activation energy, thus reaching the transition state. The transition state structure consists of atom A forming a bond with molecule BC while the BC bond is slightly dissociating. After the transition state, the bond between AB would form making a new molecule and the BC bond would be fully dissociated.
* Q1. 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?
[[File:PE with TS.png|left|thumb|[[Figure]] 1. Potential energy surface diagram with transition state labelled ]]
In a potential energy surface diagram, the transition state is mathematically defined as the first-order saddle point<ref>Chemistry LibreTexts. 2020. ''Hammond’S Postulate''. [online] Available at: <<nowiki>https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hammond%E2%80%99s_Postulate</nowiki>> [Accessed 22 May 2020].</ref>:

∂V(r1)/∂(r1)=0 and ∂V(r2)/∂r2=0

∂2V(r)/∂(r1)2 > 0 and ∂2V(r)/∂(r2)2 < 0

[∂2V(r)/∂(r1)2]*[ ∂2V(r)/∂(r2)2] - [ ∂2V/∂(r1)(r2)]2 < 0

Hessian matrix is shown in Table 2, resulting in DET = -0.999698 < 0
{| class="wikitable"
|+Table 1. Hessian matrix for the transition state
!
!f(x)
!f(y)
|-
|f(x)
|<nowiki>+0.707</nowiki>
|<nowiki>-0.707</nowiki>
|-
|f(y)
|<nowiki>-0.707</nowiki>
|<nowiki>-0.707</nowiki>
|}
In Figure 1, the black wavy lines shows that the molecule is vibrating. This is the minimum energy path required for the reactants to react and turn into the products via the transition state.<ref>Atkins, de Paula, Keeler: Physical Chemistry, 11th Edition. Section 18D.3 Potential Energy Surfaces (PESs)</ref> This suggests that the transition state is the peak in the minimum energy path. The transition state is the minimum energy required for the reactants to convert into the products. Hence once the transition state is reached, if the reaction was successful the products would be formed. Figure 1 shows the location of the transition state in the potential energy diagram.

The transition state can be distinguished from a local minimum of the potential energy surface because it is on the diagonal axis. {{fontcolor1|red|Because it is also a maximum (minimum AND maximum = saddle point) therefore distinguishable from a local minimum. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:03, 29 May 2020 (BST)}} It is the saddle point.This is when everything in the system is stationary when there is no momentum.

===Locating the transition state r1=r2===
* Q2. 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.

<nowiki> </nowiki>When the transition state is formed, due to all the atoms being hydrogen, this suggest that the transition state is symmetrical. This means that the bond distance between A and B is the same as the bond distance of B and C. This can be used to locate the transition state when the momentum is 0. Table 3 shows an estimate of the transition state position (rts).
{| class="wikitable"
|+Table 3. Distances and momenta to locate the transition state
!
!r / ppm
!p/ g.mol-1.pm.fs-1
|-
|r2=r(AB)
|90.75
|0
|-
|r1=r(BC)
|90.75
|0
|}
[[File:Mb7418 dis vs time ts.png|thumb|[[Figure 2]]. Distance vs time plot of the transition state ]]
Figure 2 shows an Intermolecular distance vs time graph resulted from the initial conditions from Table 2. The line is flat, this suggest that everything in the system is stationary. Whereas, if it wasn't the transition state, in the intermolecular distance vs time graph you'd see oscillations representing the vibrational movements. The more oscillations would indicate the more vibrational movements. Hence the transition state is a flat line as there isn't any. {{fontcolor1|red| OK. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:33, 29 May 2020 (BST)}}
[[File:Mb7418 ts contour.png|thumb|[[Figure 3]]. Potential energy diagram as a contour plot of the transition state]]
The contour plot shown in Figure 3 shows the transition state as a red cross. This is the saddle point of the potential surface. Whenever r1 and r2 are the same, the point would be anywhere across the diagonal axis. This contour plot helped estimate the transition state position as the trajectory shows that the AB and BC bond lengths are the same. This is the point at which reactants have to overcome to react into the products.

The minimum energy path can differ depending on the relative momentum of the atoms/molecules and their vibrational excitation energy. The transition state of the reaction won't change its position. The reactants require sufficient translational kinetic energy and vibrational energy to overcome the transition state (saddle point) in order to react and turn into the products.

===Trajectories from r1 = rts+δ, r2 = rts ===
* Q3. Comment on how the ''mep'' and the trajectory you just calculated differ. '''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''' and the momenta '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1
The trajectory of the minimum energy path can be visualised slightly near the transition state, this is when atom A is approaching molecule BC to form a bond with it, A + BC.
{| class="wikitable"
|+Table 4. Initial conditions slightly displaced from transition state
!
!r /ppm
!p/ g.mol-1.pm.fs-1
|-
|r2=r(AB)
|90.75
|0
|-
|r1=r(BC)
|91.75
|0
|}
The initial conditions shown in Table 4 gives the potential energy diagram in Figure 4 and 5 when the calculation type is MEP. The minimum energy path shown by the black line isn't wavy. [[File:Mb7418 mep.png|left|thumb|[[Figure 4]]. MEP calculation diagram]]
[[File:Mb7418 mep 1.png|thumb|234x234px|[[Figure 5]]. MEP calculation diagram bottom view]]However, when the same initial conditions are applied when the calculation type is Dynamic, the minimum energy path is wavy shown in Figure 6. This shows that the MEP calculation gives a different result of the reaction trajectory. The molecule is constantly moving at a velocity since there is no external force acting upon the reactants. This is called inertia motion.<ref>Evans, M.G. and Polanyi, M., 1938. Inertia and driving force of chemical reactions. ''Transactions of the Faraday Society'', ''34'', pp.11-24.</ref> Whereas the potential energy diagram produced from the Dynamics calculation takes into account the reactants inertia by representing the pathway as non-linear. Additionally, the MEP calculation diagram has a smaller trajectory than the Dynamics calculation.The reason for this is beacause the MEP calculation sets the momentum to zero at every step whereas the dynamic calculation the momenta is developed classically<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., 8.3.1, Prentice-Hall, 1998.</ref>. The MEP calculation only takes the gradient into account. The amount of steps there are corresponds to the number of times it is calculating what the gradient is at that point. The only thing relative to the MEP calculation is the gradient. Whereas with dynamics, it takes into account how long it takes for the gradient to change. The momentum (velocity) is changing each time instead of setting it to zero every step. Hence it's longer as after the products are formed, they are still have motion such as vibrational and kinetic. {{fontcolor1|red| Yes. This does show an understanding of MEP vs Dynamics. Just to clarify - the amount of steps is the number of times the system moves during the simulation - this was a bit confused in your response. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:36, 29 May 2020 (BST)}} [[File:Mb7418 dyn 1.png|thumb|[[Figure 6]]. Dynamics calculation bottom view]]

== Reactive and unreactive trajectories ==

===Interpreting trajectories===

* Q4. For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, run trajectories with the following momenta combination and complete the table.

{| class="wikitable"
|+Table 5. Reactive and unreactive trajectories
!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>
|Reactive
|The starting point (represented by the red cross) starts at 200 pm between atoms A and B. Then the distance AB decreases. Atoms A and B are approaching each other and once it reaches the transition state, the AB distance and BC distance are the same and all the atoms are stationary. Then the AB bond is formed and atom C repels from atom B. The molecule AB is constantly vibrating. This could suggest that some translational kinetic energy got converted into vibrational energy after the transition state. Hence after the trajectory has oscillations after it passes the transition state.
|[[File:Mb7418 6.png|centre|thumb|200x200px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The trajectory has some oscillations, showing that the reactants are constantly vibrating. Hence it has vibrational energy. However, it doesn't go over the transition state so it doesn't react into the products. This suggests that the reactants approach each other but then they repel. While they do that they oscillate as they get closer. It ends up with a high AB distance (essentially the starting point of the reaction), this shows that the reactants haven't reacted due to the AB molecule not forming. This could be due to not having enough translational kinetic energy.
|[[File:Mb7418 2.png|centre|thumb|200x200px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The starting point starts at 200 pm between atoms A and B. Then the the atoms got closer and reaches the transition state. This is when the distance AB and BC are equal. Then the AB distance remains continuous while the BC distance begins to increase. The BC bond breaks and the AB bond forms. It only crosses the transition state once. It ends up in the products as the distance AB is short (molecule containing atoms A and B bonding) and BC is distance is long (reactant molecule BC bond has been broken).
|[[File:Mb7418 3.png|centre|thumb|200x200px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Unreactive
|The whole trajectory shows that it's unreactive because the starting point and ending point are similar. It started at a high AB distance value and also ended at a high AB distance value. This suggests this is end unreactive.
At the starting point, AB are far apart. Then they begin approaching each other and reach the transition state. However, they repel from each other and then start approaching each other again. This happens again and then the distance AB begins to increase again. The BC bond is formed again, resulting in the reactant molecule BC at the end.
|[[File:Mb7418 4.png|centre|thumb|200x200px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Reactive
|This crosses the transition point more than once. It's reactive as it's end point is at a high BC distance and low AB distance, suggesting that the BC bond broke and the AB bond has formed resulting in AB + B.

From the starting point, the AB distance starts to decrease. This suggests atom A and B approach each other, but then the AB distance begins to increase again so the A and B atoms repel. There is a moment when the distance between atoms A and B is the same as the distance between B and C (transition state) Then the B and C atoms approach each other and then repel B continuously. While atoms A and B get closer forming a bond. The molecule AB vibrates while atom C continuously repels away from it.
|[[File:Mb7418 5.png|centre|thumb|200x200px]]
|}

{{fontcolor1|red| OK. What do you conclude from the table? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:40, 29 May 2020 (BST)}}

=== Transition state theory ===
* Q5. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?
The three assumptions of the transition state theory<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Chapter 10 pg. 20, Prentice-Hall, 1998</ref> are:

1) The transition state will only be crossed over once. Once the transition state it reached, the products will be formed.

2) During the transition state, any motion along the reaction coordinate can be separate from the other motions and treated classically as a translation. The concentration of the transition state is in equilibrium with the reactants. (concentration constant relating them both, there's a Maxwell-Boltzmann distribution between the reactants and the transition state at any given time) This is defined as the Quasi equilibrium. It's assumed that the kinetic energy is predicted by the Boltzmann distribution. The transition states that are turning into products are distributed among their states according to the Maxwell-Boltzmann Laws<ref>Rowlinson*, J.S., 2005. The Maxwell–Boltzmann distribution. ''Molecular Physics'', ''103''(21-23), pp.2821-2828.</ref> even in absence of an equilibrium between the reactant and product molecules.

'''3)''' There's minimum effects of quantum tunneling.

The quantum tunneling effect<ref>Abyss.uoregon.edu. 2020. ''Quantum Tunneling''. [online] Available at: <<nowiki>http://abyss.uoregon.edu/~js/glossary/quantum_tunneling.html</nowiki>> [Accessed 22 May 2020].</ref> is when the reactants can overcome the barrier by moving across it to turn into the products. The transition state theory doesn't include quantum tunneling. If quantum tunneling was included, it would increase the rate of reaction as more reactants will turn into products. There are only classical motions in the transition state theory. Hence it is suggested that quantum tunneling isn't included in the model used in Table 5 because it's assumed there's a faster alternative route used.

Additionally, in the Table 5 model the transition state theory wasn't true because the transition state was crossed over more than once in the last two reactions in Table 5, this could decrease the rate of reaction due to the products being able to cross the transition state again and go back into the reactants. Therefore, the transition state theory is an over-estimate in comparison to the model used.

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

=== PES inspection ===
* Q6. 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?
{| class="wikitable"
|+Table 6. Initial conditions for Figure 7 (F + H2 → HF + H)
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|500
|0
|-
|r(BC)
|70
|0
|}

[[File:Mb7418 F + H2 exo edit CORRECT.png|left|thumb|[[Figure 7]]. F + H2 -> HF + H exothermic potential surface diagram]]
In the following equations, A = atom F, B = atom H, C = atom H.

The initial conditions in Table 6 gives the potential energy diagram for Figure 7. The reason for this is because the F and H bond is very far apart, suggesting that it's not bonded whereas the hydrogen atoms are closer together, suggesting a F atom and H2 molecule are in the initial conditions.

Equation 2: F + H2 → HF + H is an exothermic reaction<ref name=":0">Duff, J.W. and Truhlar, D.G., 1975. Effect of curvature of the reaction path on dynamic effects in endothermic chemical reactions and product energies in exothermic reactions. ''The Journal of Chemical Physics'', ''62''(6), pp.2477-2491.</ref> shown in Figure 7. The energy of the reactants is higher than the energy of the products. This shows that this is an exothermic reaction due to the process releasing energy. This suggests that the bond strength of H2 is weaker than the bond strength of HF as no additional energy was required to break its bond. This reaction has an early transition state<ref>Chemistry LibreTexts. 2020. ''Hammond’S Postulate''. [online] Available at: <<nowiki>https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hammond%E2%80%99s_Postulate</nowiki>> [Accessed 22 May 2020].</ref> due to the the fluorine atom being highly electro-negative so it attacks the hydrogen molecule very quickly.
{| class="wikitable"
|+Table 7. Initial conditions for Figure 8 (HF + H → F + H2)
!
!r / ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|70
|0
|-
|r(BC)
|500
|0
|}
[[File:Mb7418 NEW CORRECT ENDO.png|left|thumb|[[Figure 8]]. HF + H -> F + H2 endothermic potential surface diagram]]
The initial conditions in Table 7 gives the potential energy diagram for Figure 8. The reason for this is because the F and H atoms are very close together compared to both the hydrogens. This suggests that a HF molecule and one H atom are in the initial conditions.

Equation 3: HF + H → F + H2 is an endothermic reaction<ref name=":0" /> shown in Figure 8. The energy of the reactants is lower than the energy of the products. This suggests that this reaction is endothermic as the process used additional energy to break the H-F bond to form the H-H bond. Therefore, in the potential energy surface diagram there will be a lot of vibrational energy before the transition state compared to after. This also suggest that the HF bond is stronger than the HH bond as it requires additional energy to break its bond. This will give a late transition state.The translational energy could get converted into vibrational energy in order to break the bond between the strong H-F bond<ref name=":1">Polanyi, J.C. and Tardy, D., 1969. Energy Distribution in the Exothermic Reaction F+ H2 and the Endothermic Reaction HF+ H. ''The Journal of Chemical Physics'', ''51''(12), pp.5717-5719.</ref>. This favours the endothermic reaction because the reactants position will be aligned with the direction of the minimum energy path.''''' '''''

* Q7. Locate the approximate position of the transition state.
{| class="wikitable"
|+Table 8. Transition state position
!
!r /ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|180
|0
|-
|r(BC)
|74.5
|0
|}
The transition state position shown in Table 8 was estimated by looking at the initial conditions in Table 6 for the exothermic F + H2 → HF + H reaction. Due to exothermic reactions having early transition states, this suggested that the bond distance BC (the hydrogen atoms) would be small as it's the reactant. The initial bond distance of AB was reduced until the hessian matrix, shown in Table 9, gave a value of Det= -1.0 <0 which proved it was a saddle point. Additionally, there was no force. This also suggested it was the transition state.
{| class="wikitable"
|+Table 9. Hessian matrix for Transition state
|AB direction
|<nowiki>+1.000</nowiki>
|<nowiki>-0.024</nowiki>
|-
|BC direction
|<nowiki>-0.024</nowiki>
|<nowiki>-1.000</nowiki>
|}
[[File:Mb7418 dis vs time ts ex2.png|thumb|[[Figure 9]]. Intermolecular distance vs time graph for transition state for equations two and three]]
Additionally, Figure 9 shows the intermolecular distance vs time graph. This is a flat line, this could be due to everything in the system being stationary. There are no oscillations showing that there are any vibrational motions. This suggest that it's at the initial conditions in Table 8 represent the transition state position.

* Q8. Report the activation energy for both reactions.
Equation 2: F + H2 → HF + H the activation energy value is 1.0 kJ/mol.

Equation 3: HF + H → F + H2 the activation energy value is 126.7 kJ/mol.

The activation energy for equation 2 was calculated by getting the potential energy for transition state by using the initial conditions in Table 8. Then this was subtracted by the potential energy obtained by using the initial conditions in Table 10. Activation energy = TS(potential energy) - Reactants(potential energy) = (-433.981)-(435.056) = 1.075 kJ/mol
{| class="wikitable"
|+Table 10. Initial conditions for equation 2
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|500
|0
|-
|r(BC)
|74.5
|0
|}
The activation energy for equation 3 was calculated by getting the potential energy for for transition state by using the initial conditions in Table 8. Then this was subtracted by the potential energy obtained by using the initial conditions in Table 11. r(AB) which is the HF bond distance = 92 ppm<ref>Cccbdb.nist.gov. 2020. ''CCCBDB Listing Of Experimental Data Page 2''. [online] Available at: <<nowiki>https://cccbdb.nist.gov/exp2x.asp?casno=7664393&charge=0</nowiki>> [Accessed 22 May 2020].</ref>'''. '''The r(BC) bond which is the HH bond distance was set very high because the hydrogen atoms aren't bonded in the reactants.

Activation energy = TS(potential energy) - Reactants(potential energy) = (-433.981)-(560.698) = 126.7 kJ/mol
{| class="wikitable"
|+Table 11. Initial conditions for equation 3
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|92
|0
|-
|r(BC)
|500
|0
|}

=== Reaction dynamics ===
* Q9. 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.
{| class="wikitable"
|+Table 12. Initial conditions for reactive trajectory
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|175
|<nowiki>-1.6</nowiki>
|-
|r(BC)
|74
|0.3
|}
[[File:Mb7418 THIS reactive.png|left|thumb|[[Figure 10]]. Reactive trajectory]]
The step number was increased to 4000 and the initial conditions in Table 12 gives the potential energy surface diagram in Figure 10. This shows a reactive trajectory for the exothermic reaction F + H2 → HF + H as it passes the transition state as well as having many oscillations. The end point is also far away from the starting point.

The starting point (reactants) has a high energy level compared to the products due to being exothermic<ref name=":0" />. Hence the potential energy converts into translational kinetic energy. Therefore, there are many oscillations in Figure 10 due to energy being conserved.

This can be explained by the difference in distribution of the energy levels between an excited and relaxed state. The excited state will have an occupied ground state (v=0) as well as an occupied first state (v=1) where v=vibrational number. Whereas in the relaxed state only the ground state will be occupied. This approach uses the harmonic oscillator<ref>Cccbdb.nist.gov. 2020. ''CCCBDB Listing Of N2 Experimental Data Page 2''. [online] Available at: <<nowiki>https://cccbdb.nist.gov/exp2x.asp?casno=7727379&charge=0</nowiki>> [Accessed 18 May 2020].</ref>. When energy is released, in the excited state there will be two transitions, v(0)→v(1) and v(1)→v(2), the fundamental peak and overtone. In the ground state it will only have transitions from v(0)→v(1).

Therefore, in an IR spectrum<ref>Chemistry LibreTexts. 2020. ''5.5: The Harmonic Oscillator And Infrared Spectra''. [online] Available at: <<nowiki>https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.05%3A_The_Harmonic_Oscillator_and_Infrared_Spectra</nowiki>> [Accessed 22 May 2020].</ref>, for the excited state there will be two peaks corresponding to the overtone and fundamental peak. The overtone peak intensity will be smaller than the fundamental's peak intensity. Hence there are many oscillations due to the overtone decreasing and then increasing in intensity while the fundamental increases and then decreases in intensity.

Therefore, this could be confirmed experimentally by Infrared spectroscopy (IR).
* Q10. 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.
[[File:Mb7418 last Q.png|left|thumb|[[Figure 11]]. Reactive potential energy surface contour plot]]
The position of the transition state affects whether the reaction is either exothermic or endothermic. An endothermic reaction will have a late transition state, suggesting that the vibrational energy efficiency will be higher<ref>Guo, H. and Liu, K., 2020. ''Control Of Chemical Reactivity By Transition-State And Beyond''.</ref>. The reason for this is because if it had a higher translational kinetic energy, the reactants wouldn't follow the minimum energy path and will just collide with each other but then immediately repel resulting in being back at the starting point. Hence, the translational kinetic energy converts into vibrational energy<ref name=":1" /> instead so the reactants collide and are aligned with the minimum energy pathway. The direction of the path taken is shown in Figure 11. The initial conditions in Table 12 was used to give the reactive trajectory. This suggests that endothermic reactions have higher vibrational energy to make their reactions more efficient so it's able to overcome the transition state. Hence there are a lot of oscillations before the transition state is reached because it requires more efficient vibrational energy in order to reach the transition state.

Whereas if the reaction has an early transition state, it will be an exothermic reaction. This suggest that there will be a higher translational kinetic energy compared to vibrational energy. This means that there will be less oscillations before it reaches the transition state as shown in Figure 11.

Overall, an efficient endothermic reaction will have a higher vibrational energy whereas an efficient exothermic reaction will have a higher translational kinetic energy.

MRD:mb7418

2020-05-29T17:36:47Z

Mak214: /* Trajectories from r1 = rts+δ, r2 = rts */

== EXERCISE 1: H + H2 system ==

===Dynamics from the transition state region===
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? {{fontcolor1|red| This text is repeated below which makes this section difficult to read. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:03, 29 May 2020 (BST)}} Potential energy surface diagrams show the potential energy the system has according to the coordinates of the atoms in the system.<ref>Chemistry LibreTexts. 2020. ''2.6: Potential Energy Surfaces''. [online] Available at: <<nowiki>https://chem.libretexts.org/Courses/University_of_California_Davis/UCD_Chem_107B%3A_Physical_Chemistry_for_Life_Scientists/Chapters/2%3A_Chemical_Kinetics/2.06%3A_Potential_Energy_Surfaces</nowiki>> [Accessed 19 May 2020].</ref> Figure 1 shows a potential energy surface diagram representing the reaction, equation 1: A + BC → AB + C.
{| class="wikitable"
|+Table 1. Molecules distances (r) and momenta (p)
!
!r / ppm
!p/ g.mol-1.pm.fs-1
|-
|r2=r(AB)
|230
|<nowiki>-5.2</nowiki>
|-
|r1=r(BC)
|74
|0
|}
In this case r(1)=r(BC) and r(2)=r(AB). BC is a hydrogen molecule consisting of two hydrogen atoms, B and C. Atom A is a hydrogen atom which has energy equal to or more than the activation energy, thus reaching the transition state. The transition state structure consists of atom A forming a bond with molecule BC while the BC bond is slightly dissociating. After the transition state, the bond between AB would form making a new molecule and the BC bond would be fully dissociated.
* Q1. 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?
[[File:PE with TS.png|left|thumb|[[Figure]] 1. Potential energy surface diagram with transition state labelled ]]
In a potential energy surface diagram, the transition state is mathematically defined as the first-order saddle point<ref>Chemistry LibreTexts. 2020. ''Hammond’S Postulate''. [online] Available at: <<nowiki>https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hammond%E2%80%99s_Postulate</nowiki>> [Accessed 22 May 2020].</ref>:

∂V(r1)/∂(r1)=0 and ∂V(r2)/∂r2=0

∂2V(r)/∂(r1)2 > 0 and ∂2V(r)/∂(r2)2 < 0

[∂2V(r)/∂(r1)2]*[ ∂2V(r)/∂(r2)2] - [ ∂2V/∂(r1)(r2)]2 < 0

Hessian matrix is shown in Table 2, resulting in DET = -0.999698 < 0
{| class="wikitable"
|+Table 1. Hessian matrix for the transition state
!
!f(x)
!f(y)
|-
|f(x)
|<nowiki>+0.707</nowiki>
|<nowiki>-0.707</nowiki>
|-
|f(y)
|<nowiki>-0.707</nowiki>
|<nowiki>-0.707</nowiki>
|}
In Figure 1, the black wavy lines shows that the molecule is vibrating. This is the minimum energy path required for the reactants to react and turn into the products via the transition state.<ref>Atkins, de Paula, Keeler: Physical Chemistry, 11th Edition. Section 18D.3 Potential Energy Surfaces (PESs)</ref> This suggests that the transition state is the peak in the minimum energy path. The transition state is the minimum energy required for the reactants to convert into the products. Hence once the transition state is reached, if the reaction was successful the products would be formed. Figure 1 shows the location of the transition state in the potential energy diagram.

The transition state can be distinguished from a local minimum of the potential energy surface because it is on the diagonal axis. {{fontcolor1|red|Because it is also a maximum (minimum AND maximum = saddle point) therefore distinguishable from a local minimum. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:03, 29 May 2020 (BST)}} It is the saddle point.This is when everything in the system is stationary when there is no momentum.

===Locating the transition state r1=r2===
* Q2. 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.

<nowiki> </nowiki>When the transition state is formed, due to all the atoms being hydrogen, this suggest that the transition state is symmetrical. This means that the bond distance between A and B is the same as the bond distance of B and C. This can be used to locate the transition state when the momentum is 0. Table 3 shows an estimate of the transition state position (rts).
{| class="wikitable"
|+Table 3. Distances and momenta to locate the transition state
!
!r / ppm
!p/ g.mol-1.pm.fs-1
|-
|r2=r(AB)
|90.75
|0
|-
|r1=r(BC)
|90.75
|0
|}
[[File:Mb7418 dis vs time ts.png|thumb|[[Figure 2]]. Distance vs time plot of the transition state ]]
Figure 2 shows an Intermolecular distance vs time graph resulted from the initial conditions from Table 2. The line is flat, this suggest that everything in the system is stationary. Whereas, if it wasn't the transition state, in the intermolecular distance vs time graph you'd see oscillations representing the vibrational movements. The more oscillations would indicate the more vibrational movements. Hence the transition state is a flat line as there isn't any. {{fontcolor1|red| OK. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:33, 29 May 2020 (BST)}}
[[File:Mb7418 ts contour.png|thumb|[[Figure 3]]. Potential energy diagram as a contour plot of the transition state]]
The contour plot shown in Figure 3 shows the transition state as a red cross. This is the saddle point of the potential surface. Whenever r1 and r2 are the same, the point would be anywhere across the diagonal axis. This contour plot helped estimate the transition state position as the trajectory shows that the AB and BC bond lengths are the same. This is the point at which reactants have to overcome to react into the products.

The minimum energy path can differ depending on the relative momentum of the atoms/molecules and their vibrational excitation energy. The transition state of the reaction won't change its position. The reactants require sufficient translational kinetic energy and vibrational energy to overcome the transition state (saddle point) in order to react and turn into the products.

===Trajectories from r1 = rts+δ, r2 = rts ===
* Q3. Comment on how the ''mep'' and the trajectory you just calculated differ. '''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''' and the momenta '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1
The trajectory of the minimum energy path can be visualised slightly near the transition state, this is when atom A is approaching molecule BC to form a bond with it, A + BC.
{| class="wikitable"
|+Table 4. Initial conditions slightly displaced from transition state
!
!r /ppm
!p/ g.mol-1.pm.fs-1
|-
|r2=r(AB)
|90.75
|0
|-
|r1=r(BC)
|91.75
|0
|}
The initial conditions shown in Table 4 gives the potential energy diagram in Figure 4 and 5 when the calculation type is MEP. The minimum energy path shown by the black line isn't wavy. [[File:Mb7418 mep.png|left|thumb|[[Figure 4]]. MEP calculation diagram]]
[[File:Mb7418 mep 1.png|thumb|234x234px|[[Figure 5]]. MEP calculation diagram bottom view]]However, when the same initial conditions are applied when the calculation type is Dynamic, the minimum energy path is wavy shown in Figure 6. This shows that the MEP calculation gives a different result of the reaction trajectory. The molecule is constantly moving at a velocity since there is no external force acting upon the reactants. This is called inertia motion.<ref>Evans, M.G. and Polanyi, M., 1938. Inertia and driving force of chemical reactions. ''Transactions of the Faraday Society'', ''34'', pp.11-24.</ref> Whereas the potential energy diagram produced from the Dynamics calculation takes into account the reactants inertia by representing the pathway as non-linear. Additionally, the MEP calculation diagram has a smaller trajectory than the Dynamics calculation.The reason for this is beacause the MEP calculation sets the momentum to zero at every step whereas the dynamic calculation the momenta is developed classically<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., 8.3.1, Prentice-Hall, 1998.</ref>. The MEP calculation only takes the gradient into account. The amount of steps there are corresponds to the number of times it is calculating what the gradient is at that point. The only thing relative to the MEP calculation is the gradient. Whereas with dynamics, it takes into account how long it takes for the gradient to change. The momentum (velocity) is changing each time instead of setting it to zero every step. Hence it's longer as after the products are formed, they are still have motion such as vibrational and kinetic. {{fontcolor1|red| Yes. This does show an understanding of MEP vs Dynamics. Just to clarify - the amount of steps is the number of times the system moves during the simulation - this was a bit confused in your response. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:36, 29 May 2020 (BST)}} [[File:Mb7418 dyn 1.png|thumb|[[Figure 6]]. Dynamics calculation bottom view]]

== Reactive and unreactive trajectories ==

===Interpreting trajectories===

* Q4. For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, run trajectories with the following momenta combination and complete the table.

{| class="wikitable"
|+Table 5. Reactive and unreactive trajectories
!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>
|Reactive
|The starting point (represented by the red cross) starts at 200 pm between atoms A and B. Then the distance AB decreases. Atoms A and B are approaching each other and once it reaches the transition state, the AB distance and BC distance are the same and all the atoms are stationary. Then the AB bond is formed and atom C repels from atom B. The molecule AB is constantly vibrating. This could suggest that some translational kinetic energy got converted into vibrational energy after the transition state. Hence after the trajectory has oscillations after it passes the transition state.
|[[File:Mb7418 6.png|centre|thumb|200x200px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The trajectory has some oscillations, showing that the reactants are constantly vibrating. Hence it has vibrational energy. However, it doesn't go over the transition state so it doesn't react into the products. This suggests that the reactants approach each other but then they repel. While they do that they oscillate as they get closer. It ends up with a high AB distance (essentially the starting point of the reaction), this shows that the reactants haven't reacted due to the AB molecule not forming. This could be due to not having enough translational kinetic energy.
|[[File:Mb7418 2.png|centre|thumb|200x200px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The starting point starts at 200 pm between atoms A and B. Then the the atoms got closer and reaches the transition state. This is when the distance AB and BC are equal. Then the AB distance remains continuous while the BC distance begins to increase. The BC bond breaks and the AB bond forms. It only crosses the transition state once. It ends up in the products as the distance AB is short (molecule containing atoms A and B bonding) and BC is distance is long (reactant molecule BC bond has been broken).
|[[File:Mb7418 3.png|centre|thumb|200x200px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Unreactive
|The whole trajectory shows that it's unreactive because the starting point and ending point are similar. It started at a high AB distance value and also ended at a high AB distance value. This suggests this is end unreactive.
At the starting point, AB are far apart. Then they begin approaching each other and reach the transition state. However, they repel from each other and then start approaching each other again. This happens again and then the distance AB begins to increase again. The BC bond is formed again, resulting in the reactant molecule BC at the end.
|[[File:Mb7418 4.png|centre|thumb|200x200px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Reactive
|This crosses the transition point more than once. It's reactive as it's end point is at a high BC distance and low AB distance, suggesting that the BC bond broke and the AB bond has formed resulting in AB + B.

From the starting point, the AB distance starts to decrease. This suggests atom A and B approach each other, but then the AB distance begins to increase again so the A and B atoms repel. There is a moment when the distance between atoms A and B is the same as the distance between B and C (transition state) Then the B and C atoms approach each other and then repel B continuously. While atoms A and B get closer forming a bond. The molecule AB vibrates while atom C continuously repels away from it.
|[[File:Mb7418 5.png|centre|thumb|200x200px]]
|}
=== Transition state theory ===
* Q5. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?
The three assumptions of the transition state theory<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Chapter 10 pg. 20, Prentice-Hall, 1998</ref> are:

1) The transition state will only be crossed over once. Once the transition state it reached, the products will be formed.

2) During the transition state, any motion along the reaction coordinate can be separate from the other motions and treated classically as a translation. The concentration of the transition state is in equilibrium with the reactants. (concentration constant relating them both, there's a Maxwell-Boltzmann distribution between the reactants and the transition state at any given time) This is defined as the Quasi equilibrium. It's assumed that the kinetic energy is predicted by the Boltzmann distribution. The transition states that are turning into products are distributed among their states according to the Maxwell-Boltzmann Laws<ref>Rowlinson*, J.S., 2005. The Maxwell–Boltzmann distribution. ''Molecular Physics'', ''103''(21-23), pp.2821-2828.</ref> even in absence of an equilibrium between the reactant and product molecules.

'''3)''' There's minimum effects of quantum tunneling.

The quantum tunneling effect<ref>Abyss.uoregon.edu. 2020. ''Quantum Tunneling''. [online] Available at: <<nowiki>http://abyss.uoregon.edu/~js/glossary/quantum_tunneling.html</nowiki>> [Accessed 22 May 2020].</ref> is when the reactants can overcome the barrier by moving across it to turn into the products. The transition state theory doesn't include quantum tunneling. If quantum tunneling was included, it would increase the rate of reaction as more reactants will turn into products. There are only classical motions in the transition state theory. Hence it is suggested that quantum tunneling isn't included in the model used in Table 5 because it's assumed there's a faster alternative route used.

Additionally, in the Table 5 model the transition state theory wasn't true because the transition state was crossed over more than once in the last two reactions in Table 5, this could decrease the rate of reaction due to the products being able to cross the transition state again and go back into the reactants. Therefore, the transition state theory is an over-estimate in comparison to the model used.

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

=== PES inspection ===
* Q6. 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?
{| class="wikitable"
|+Table 6. Initial conditions for Figure 7 (F + H2 → HF + H)
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|500
|0
|-
|r(BC)
|70
|0
|}

[[File:Mb7418 F + H2 exo edit CORRECT.png|left|thumb|[[Figure 7]]. F + H2 -> HF + H exothermic potential surface diagram]]
In the following equations, A = atom F, B = atom H, C = atom H.

The initial conditions in Table 6 gives the potential energy diagram for Figure 7. The reason for this is because the F and H bond is very far apart, suggesting that it's not bonded whereas the hydrogen atoms are closer together, suggesting a F atom and H2 molecule are in the initial conditions.

Equation 2: F + H2 → HF + H is an exothermic reaction<ref name=":0">Duff, J.W. and Truhlar, D.G., 1975. Effect of curvature of the reaction path on dynamic effects in endothermic chemical reactions and product energies in exothermic reactions. ''The Journal of Chemical Physics'', ''62''(6), pp.2477-2491.</ref> shown in Figure 7. The energy of the reactants is higher than the energy of the products. This shows that this is an exothermic reaction due to the process releasing energy. This suggests that the bond strength of H2 is weaker than the bond strength of HF as no additional energy was required to break its bond. This reaction has an early transition state<ref>Chemistry LibreTexts. 2020. ''Hammond’S Postulate''. [online] Available at: <<nowiki>https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hammond%E2%80%99s_Postulate</nowiki>> [Accessed 22 May 2020].</ref> due to the the fluorine atom being highly electro-negative so it attacks the hydrogen molecule very quickly.
{| class="wikitable"
|+Table 7. Initial conditions for Figure 8 (HF + H → F + H2)
!
!r / ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|70
|0
|-
|r(BC)
|500
|0
|}
[[File:Mb7418 NEW CORRECT ENDO.png|left|thumb|[[Figure 8]]. HF + H -> F + H2 endothermic potential surface diagram]]
The initial conditions in Table 7 gives the potential energy diagram for Figure 8. The reason for this is because the F and H atoms are very close together compared to both the hydrogens. This suggests that a HF molecule and one H atom are in the initial conditions.

Equation 3: HF + H → F + H2 is an endothermic reaction<ref name=":0" /> shown in Figure 8. The energy of the reactants is lower than the energy of the products. This suggests that this reaction is endothermic as the process used additional energy to break the H-F bond to form the H-H bond. Therefore, in the potential energy surface diagram there will be a lot of vibrational energy before the transition state compared to after. This also suggest that the HF bond is stronger than the HH bond as it requires additional energy to break its bond. This will give a late transition state.The translational energy could get converted into vibrational energy in order to break the bond between the strong H-F bond<ref name=":1">Polanyi, J.C. and Tardy, D., 1969. Energy Distribution in the Exothermic Reaction F+ H2 and the Endothermic Reaction HF+ H. ''The Journal of Chemical Physics'', ''51''(12), pp.5717-5719.</ref>. This favours the endothermic reaction because the reactants position will be aligned with the direction of the minimum energy path.''''' '''''

* Q7. Locate the approximate position of the transition state.
{| class="wikitable"
|+Table 8. Transition state position
!
!r /ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|180
|0
|-
|r(BC)
|74.5
|0
|}
The transition state position shown in Table 8 was estimated by looking at the initial conditions in Table 6 for the exothermic F + H2 → HF + H reaction. Due to exothermic reactions having early transition states, this suggested that the bond distance BC (the hydrogen atoms) would be small as it's the reactant. The initial bond distance of AB was reduced until the hessian matrix, shown in Table 9, gave a value of Det= -1.0 <0 which proved it was a saddle point. Additionally, there was no force. This also suggested it was the transition state.
{| class="wikitable"
|+Table 9. Hessian matrix for Transition state
|AB direction
|<nowiki>+1.000</nowiki>
|<nowiki>-0.024</nowiki>
|-
|BC direction
|<nowiki>-0.024</nowiki>
|<nowiki>-1.000</nowiki>
|}
[[File:Mb7418 dis vs time ts ex2.png|thumb|[[Figure 9]]. Intermolecular distance vs time graph for transition state for equations two and three]]
Additionally, Figure 9 shows the intermolecular distance vs time graph. This is a flat line, this could be due to everything in the system being stationary. There are no oscillations showing that there are any vibrational motions. This suggest that it's at the initial conditions in Table 8 represent the transition state position.

* Q8. Report the activation energy for both reactions.
Equation 2: F + H2 → HF + H the activation energy value is 1.0 kJ/mol.

Equation 3: HF + H → F + H2 the activation energy value is 126.7 kJ/mol.

The activation energy for equation 2 was calculated by getting the potential energy for transition state by using the initial conditions in Table 8. Then this was subtracted by the potential energy obtained by using the initial conditions in Table 10. Activation energy = TS(potential energy) - Reactants(potential energy) = (-433.981)-(435.056) = 1.075 kJ/mol
{| class="wikitable"
|+Table 10. Initial conditions for equation 2
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|500
|0
|-
|r(BC)
|74.5
|0
|}
The activation energy for equation 3 was calculated by getting the potential energy for for transition state by using the initial conditions in Table 8. Then this was subtracted by the potential energy obtained by using the initial conditions in Table 11. r(AB) which is the HF bond distance = 92 ppm<ref>Cccbdb.nist.gov. 2020. ''CCCBDB Listing Of Experimental Data Page 2''. [online] Available at: <<nowiki>https://cccbdb.nist.gov/exp2x.asp?casno=7664393&charge=0</nowiki>> [Accessed 22 May 2020].</ref>'''. '''The r(BC) bond which is the HH bond distance was set very high because the hydrogen atoms aren't bonded in the reactants.

Activation energy = TS(potential energy) - Reactants(potential energy) = (-433.981)-(560.698) = 126.7 kJ/mol
{| class="wikitable"
|+Table 11. Initial conditions for equation 3
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|92
|0
|-
|r(BC)
|500
|0
|}

=== Reaction dynamics ===
* Q9. 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.
{| class="wikitable"
|+Table 12. Initial conditions for reactive trajectory
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|175
|<nowiki>-1.6</nowiki>
|-
|r(BC)
|74
|0.3
|}
[[File:Mb7418 THIS reactive.png|left|thumb|[[Figure 10]]. Reactive trajectory]]
The step number was increased to 4000 and the initial conditions in Table 12 gives the potential energy surface diagram in Figure 10. This shows a reactive trajectory for the exothermic reaction F + H2 → HF + H as it passes the transition state as well as having many oscillations. The end point is also far away from the starting point.

The starting point (reactants) has a high energy level compared to the products due to being exothermic<ref name=":0" />. Hence the potential energy converts into translational kinetic energy. Therefore, there are many oscillations in Figure 10 due to energy being conserved.

This can be explained by the difference in distribution of the energy levels between an excited and relaxed state. The excited state will have an occupied ground state (v=0) as well as an occupied first state (v=1) where v=vibrational number. Whereas in the relaxed state only the ground state will be occupied. This approach uses the harmonic oscillator<ref>Cccbdb.nist.gov. 2020. ''CCCBDB Listing Of N2 Experimental Data Page 2''. [online] Available at: <<nowiki>https://cccbdb.nist.gov/exp2x.asp?casno=7727379&charge=0</nowiki>> [Accessed 18 May 2020].</ref>. When energy is released, in the excited state there will be two transitions, v(0)→v(1) and v(1)→v(2), the fundamental peak and overtone. In the ground state it will only have transitions from v(0)→v(1).

Therefore, in an IR spectrum<ref>Chemistry LibreTexts. 2020. ''5.5: The Harmonic Oscillator And Infrared Spectra''. [online] Available at: <<nowiki>https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.05%3A_The_Harmonic_Oscillator_and_Infrared_Spectra</nowiki>> [Accessed 22 May 2020].</ref>, for the excited state there will be two peaks corresponding to the overtone and fundamental peak. The overtone peak intensity will be smaller than the fundamental's peak intensity. Hence there are many oscillations due to the overtone decreasing and then increasing in intensity while the fundamental increases and then decreases in intensity.

Therefore, this could be confirmed experimentally by Infrared spectroscopy (IR).
* Q10. 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.
[[File:Mb7418 last Q.png|left|thumb|[[Figure 11]]. Reactive potential energy surface contour plot]]
The position of the transition state affects whether the reaction is either exothermic or endothermic. An endothermic reaction will have a late transition state, suggesting that the vibrational energy efficiency will be higher<ref>Guo, H. and Liu, K., 2020. ''Control Of Chemical Reactivity By Transition-State And Beyond''.</ref>. The reason for this is because if it had a higher translational kinetic energy, the reactants wouldn't follow the minimum energy path and will just collide with each other but then immediately repel resulting in being back at the starting point. Hence, the translational kinetic energy converts into vibrational energy<ref name=":1" /> instead so the reactants collide and are aligned with the minimum energy pathway. The direction of the path taken is shown in Figure 11. The initial conditions in Table 12 was used to give the reactive trajectory. This suggests that endothermic reactions have higher vibrational energy to make their reactions more efficient so it's able to overcome the transition state. Hence there are a lot of oscillations before the transition state is reached because it requires more efficient vibrational energy in order to reach the transition state.

Whereas if the reaction has an early transition state, it will be an exothermic reaction. This suggest that there will be a higher translational kinetic energy compared to vibrational energy. This means that there will be less oscillations before it reaches the transition state as shown in Figure 11.

Overall, an efficient endothermic reaction will have a higher vibrational energy whereas an efficient exothermic reaction will have a higher translational kinetic energy.

MRD:mb7418

2020-05-29T17:33:32Z

Mak214: /* Locating the transition state r1=r2 */

== EXERCISE 1: H + H2 system ==

===Dynamics from the transition state region===
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? {{fontcolor1|red| This text is repeated below which makes this section difficult to read. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:03, 29 May 2020 (BST)}} Potential energy surface diagrams show the potential energy the system has according to the coordinates of the atoms in the system.<ref>Chemistry LibreTexts. 2020. ''2.6: Potential Energy Surfaces''. [online] Available at: <<nowiki>https://chem.libretexts.org/Courses/University_of_California_Davis/UCD_Chem_107B%3A_Physical_Chemistry_for_Life_Scientists/Chapters/2%3A_Chemical_Kinetics/2.06%3A_Potential_Energy_Surfaces</nowiki>> [Accessed 19 May 2020].</ref> Figure 1 shows a potential energy surface diagram representing the reaction, equation 1: A + BC → AB + C.
{| class="wikitable"
|+Table 1. Molecules distances (r) and momenta (p)
!
!r / ppm
!p/ g.mol-1.pm.fs-1
|-
|r2=r(AB)
|230
|<nowiki>-5.2</nowiki>
|-
|r1=r(BC)
|74
|0
|}
In this case r(1)=r(BC) and r(2)=r(AB). BC is a hydrogen molecule consisting of two hydrogen atoms, B and C. Atom A is a hydrogen atom which has energy equal to or more than the activation energy, thus reaching the transition state. The transition state structure consists of atom A forming a bond with molecule BC while the BC bond is slightly dissociating. After the transition state, the bond between AB would form making a new molecule and the BC bond would be fully dissociated.
* Q1. 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?
[[File:PE with TS.png|left|thumb|[[Figure]] 1. Potential energy surface diagram with transition state labelled ]]
In a potential energy surface diagram, the transition state is mathematically defined as the first-order saddle point<ref>Chemistry LibreTexts. 2020. ''Hammond’S Postulate''. [online] Available at: <<nowiki>https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hammond%E2%80%99s_Postulate</nowiki>> [Accessed 22 May 2020].</ref>:

∂V(r1)/∂(r1)=0 and ∂V(r2)/∂r2=0

∂2V(r)/∂(r1)2 > 0 and ∂2V(r)/∂(r2)2 < 0

[∂2V(r)/∂(r1)2]*[ ∂2V(r)/∂(r2)2] - [ ∂2V/∂(r1)(r2)]2 < 0

Hessian matrix is shown in Table 2, resulting in DET = -0.999698 < 0
{| class="wikitable"
|+Table 1. Hessian matrix for the transition state
!
!f(x)
!f(y)
|-
|f(x)
|<nowiki>+0.707</nowiki>
|<nowiki>-0.707</nowiki>
|-
|f(y)
|<nowiki>-0.707</nowiki>
|<nowiki>-0.707</nowiki>
|}
In Figure 1, the black wavy lines shows that the molecule is vibrating. This is the minimum energy path required for the reactants to react and turn into the products via the transition state.<ref>Atkins, de Paula, Keeler: Physical Chemistry, 11th Edition. Section 18D.3 Potential Energy Surfaces (PESs)</ref> This suggests that the transition state is the peak in the minimum energy path. The transition state is the minimum energy required for the reactants to convert into the products. Hence once the transition state is reached, if the reaction was successful the products would be formed. Figure 1 shows the location of the transition state in the potential energy diagram.

The transition state can be distinguished from a local minimum of the potential energy surface because it is on the diagonal axis. {{fontcolor1|red|Because it is also a maximum (minimum AND maximum = saddle point) therefore distinguishable from a local minimum. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:03, 29 May 2020 (BST)}} It is the saddle point.This is when everything in the system is stationary when there is no momentum.

===Locating the transition state r1=r2===
* Q2. 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.

<nowiki> </nowiki>When the transition state is formed, due to all the atoms being hydrogen, this suggest that the transition state is symmetrical. This means that the bond distance between A and B is the same as the bond distance of B and C. This can be used to locate the transition state when the momentum is 0. Table 3 shows an estimate of the transition state position (rts).
{| class="wikitable"
|+Table 3. Distances and momenta to locate the transition state
!
!r / ppm
!p/ g.mol-1.pm.fs-1
|-
|r2=r(AB)
|90.75
|0
|-
|r1=r(BC)
|90.75
|0
|}
[[File:Mb7418 dis vs time ts.png|thumb|[[Figure 2]]. Distance vs time plot of the transition state ]]
Figure 2 shows an Intermolecular distance vs time graph resulted from the initial conditions from Table 2. The line is flat, this suggest that everything in the system is stationary. Whereas, if it wasn't the transition state, in the intermolecular distance vs time graph you'd see oscillations representing the vibrational movements. The more oscillations would indicate the more vibrational movements. Hence the transition state is a flat line as there isn't any. {{fontcolor1|red| OK. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:33, 29 May 2020 (BST)}}
[[File:Mb7418 ts contour.png|thumb|[[Figure 3]]. Potential energy diagram as a contour plot of the transition state]]
The contour plot shown in Figure 3 shows the transition state as a red cross. This is the saddle point of the potential surface. Whenever r1 and r2 are the same, the point would be anywhere across the diagonal axis. This contour plot helped estimate the transition state position as the trajectory shows that the AB and BC bond lengths are the same. This is the point at which reactants have to overcome to react into the products.

The minimum energy path can differ depending on the relative momentum of the atoms/molecules and their vibrational excitation energy. The transition state of the reaction won't change its position. The reactants require sufficient translational kinetic energy and vibrational energy to overcome the transition state (saddle point) in order to react and turn into the products.

===Trajectories from r1 = rts+δ, r2 = rts ===
* Q3. Comment on how the ''mep'' and the trajectory you just calculated differ. '''r1''' = '''rts'''+1 pm, '''r2''' = '''rts''' and the momenta '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1
The trajectory of the minimum energy path can be visualised slightly near the transition state, this is when atom A is approaching molecule BC to form a bond with it, A + BC.
{| class="wikitable"
|+Table 4. Initial conditions slightly displaced from transition state
!
!r /ppm
!p/ g.mol-1.pm.fs-1
|-
|r2=r(AB)
|90.75
|0
|-
|r1=r(BC)
|91.75
|0
|}
The initial conditions shown in Table 4 gives the potential energy diagram in Figure 4 and 5 when the calculation type is MEP. The minimum energy path shown by the black line isn't wavy. [[File:Mb7418 mep.png|left|thumb|[[Figure 4]]. MEP calculation diagram]]
[[File:Mb7418 mep 1.png|thumb|234x234px|[[Figure 5]]. MEP calculation diagram bottom view]]However, when the same initial conditions are applied when the calculation type is Dynamic, the minimum energy path is wavy shown in Figure 6. This shows that the MEP calculation gives a different result of the reaction trajectory. The molecule is constantly moving at a velocity since there is no external force acting upon the reactants. This is called inertia motion.<ref>Evans, M.G. and Polanyi, M., 1938. Inertia and driving force of chemical reactions. ''Transactions of the Faraday Society'', ''34'', pp.11-24.</ref> Whereas the potential energy diagram produced from the Dynamics calculation takes into account the reactants inertia by representing the pathway as non-linear. Additionally, the MEP calculation diagram has a smaller trajectory than the Dynamics calculation.The reason for this is beacause the MEP calculation sets the momentum to zero at every step whereas the dynamic calculation the momenta is developed classically<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., 8.3.1, Prentice-Hall, 1998.</ref>. The MEP calculation only takes the gradient into account. The amount of steps there are corresponds to the number of times it is calculating what the gradient is at that point. The only thing relative to the MEP calculation is the gradient. Whereas with dynamics, it takes into account how long it takes for the gradient to change. The momentum (velocity) is changing each time instead of setting it to zero every step. Hence it's longer as after the products are formed, they are still have motion such as vibrational and kinetic.[[File:Mb7418 dyn 1.png|thumb|[[Figure 6]]. Dynamics calculation bottom view]]

== Reactive and unreactive trajectories ==

===Interpreting trajectories===

* Q4. For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm, run trajectories with the following momenta combination and complete the table.

{| class="wikitable"
|+Table 5. Reactive and unreactive trajectories
!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>
|Reactive
|The starting point (represented by the red cross) starts at 200 pm between atoms A and B. Then the distance AB decreases. Atoms A and B are approaching each other and once it reaches the transition state, the AB distance and BC distance are the same and all the atoms are stationary. Then the AB bond is formed and atom C repels from atom B. The molecule AB is constantly vibrating. This could suggest that some translational kinetic energy got converted into vibrational energy after the transition state. Hence after the trajectory has oscillations after it passes the transition state.
|[[File:Mb7418 6.png|centre|thumb|200x200px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|Unreactive
|The trajectory has some oscillations, showing that the reactants are constantly vibrating. Hence it has vibrational energy. However, it doesn't go over the transition state so it doesn't react into the products. This suggests that the reactants approach each other but then they repel. While they do that they oscillate as they get closer. It ends up with a high AB distance (essentially the starting point of the reaction), this shows that the reactants haven't reacted due to the AB molecule not forming. This could be due to not having enough translational kinetic energy.
|[[File:Mb7418 2.png|centre|thumb|200x200px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Reactive
|The starting point starts at 200 pm between atoms A and B. Then the the atoms got closer and reaches the transition state. This is when the distance AB and BC are equal. Then the AB distance remains continuous while the BC distance begins to increase. The BC bond breaks and the AB bond forms. It only crosses the transition state once. It ends up in the products as the distance AB is short (molecule containing atoms A and B bonding) and BC is distance is long (reactant molecule BC bond has been broken).
|[[File:Mb7418 3.png|centre|thumb|200x200px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|Unreactive
|The whole trajectory shows that it's unreactive because the starting point and ending point are similar. It started at a high AB distance value and also ended at a high AB distance value. This suggests this is end unreactive.
At the starting point, AB are far apart. Then they begin approaching each other and reach the transition state. However, they repel from each other and then start approaching each other again. This happens again and then the distance AB begins to increase again. The BC bond is formed again, resulting in the reactant molecule BC at the end.
|[[File:Mb7418 4.png|centre|thumb|200x200px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Reactive
|This crosses the transition point more than once. It's reactive as it's end point is at a high BC distance and low AB distance, suggesting that the BC bond broke and the AB bond has formed resulting in AB + B.

From the starting point, the AB distance starts to decrease. This suggests atom A and B approach each other, but then the AB distance begins to increase again so the A and B atoms repel. There is a moment when the distance between atoms A and B is the same as the distance between B and C (transition state) Then the B and C atoms approach each other and then repel B continuously. While atoms A and B get closer forming a bond. The molecule AB vibrates while atom C continuously repels away from it.
|[[File:Mb7418 5.png|centre|thumb|200x200px]]
|}
=== Transition state theory ===
* Q5. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?
The three assumptions of the transition state theory<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Chapter 10 pg. 20, Prentice-Hall, 1998</ref> are:

1) The transition state will only be crossed over once. Once the transition state it reached, the products will be formed.

2) During the transition state, any motion along the reaction coordinate can be separate from the other motions and treated classically as a translation. The concentration of the transition state is in equilibrium with the reactants. (concentration constant relating them both, there's a Maxwell-Boltzmann distribution between the reactants and the transition state at any given time) This is defined as the Quasi equilibrium. It's assumed that the kinetic energy is predicted by the Boltzmann distribution. The transition states that are turning into products are distributed among their states according to the Maxwell-Boltzmann Laws<ref>Rowlinson*, J.S., 2005. The Maxwell–Boltzmann distribution. ''Molecular Physics'', ''103''(21-23), pp.2821-2828.</ref> even in absence of an equilibrium between the reactant and product molecules.

'''3)''' There's minimum effects of quantum tunneling.

The quantum tunneling effect<ref>Abyss.uoregon.edu. 2020. ''Quantum Tunneling''. [online] Available at: <<nowiki>http://abyss.uoregon.edu/~js/glossary/quantum_tunneling.html</nowiki>> [Accessed 22 May 2020].</ref> is when the reactants can overcome the barrier by moving across it to turn into the products. The transition state theory doesn't include quantum tunneling. If quantum tunneling was included, it would increase the rate of reaction as more reactants will turn into products. There are only classical motions in the transition state theory. Hence it is suggested that quantum tunneling isn't included in the model used in Table 5 because it's assumed there's a faster alternative route used.

Additionally, in the Table 5 model the transition state theory wasn't true because the transition state was crossed over more than once in the last two reactions in Table 5, this could decrease the rate of reaction due to the products being able to cross the transition state again and go back into the reactants. Therefore, the transition state theory is an over-estimate in comparison to the model used.

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

=== PES inspection ===
* Q6. 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?
{| class="wikitable"
|+Table 6. Initial conditions for Figure 7 (F + H2 → HF + H)
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|500
|0
|-
|r(BC)
|70
|0
|}

[[File:Mb7418 F + H2 exo edit CORRECT.png|left|thumb|[[Figure 7]]. F + H2 -> HF + H exothermic potential surface diagram]]
In the following equations, A = atom F, B = atom H, C = atom H.

The initial conditions in Table 6 gives the potential energy diagram for Figure 7. The reason for this is because the F and H bond is very far apart, suggesting that it's not bonded whereas the hydrogen atoms are closer together, suggesting a F atom and H2 molecule are in the initial conditions.

Equation 2: F + H2 → HF + H is an exothermic reaction<ref name=":0">Duff, J.W. and Truhlar, D.G., 1975. Effect of curvature of the reaction path on dynamic effects in endothermic chemical reactions and product energies in exothermic reactions. ''The Journal of Chemical Physics'', ''62''(6), pp.2477-2491.</ref> shown in Figure 7. The energy of the reactants is higher than the energy of the products. This shows that this is an exothermic reaction due to the process releasing energy. This suggests that the bond strength of H2 is weaker than the bond strength of HF as no additional energy was required to break its bond. This reaction has an early transition state<ref>Chemistry LibreTexts. 2020. ''Hammond’S Postulate''. [online] Available at: <<nowiki>https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hammond%E2%80%99s_Postulate</nowiki>> [Accessed 22 May 2020].</ref> due to the the fluorine atom being highly electro-negative so it attacks the hydrogen molecule very quickly.
{| class="wikitable"
|+Table 7. Initial conditions for Figure 8 (HF + H → F + H2)
!
!r / ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|70
|0
|-
|r(BC)
|500
|0
|}
[[File:Mb7418 NEW CORRECT ENDO.png|left|thumb|[[Figure 8]]. HF + H -> F + H2 endothermic potential surface diagram]]
The initial conditions in Table 7 gives the potential energy diagram for Figure 8. The reason for this is because the F and H atoms are very close together compared to both the hydrogens. This suggests that a HF molecule and one H atom are in the initial conditions.

Equation 3: HF + H → F + H2 is an endothermic reaction<ref name=":0" /> shown in Figure 8. The energy of the reactants is lower than the energy of the products. This suggests that this reaction is endothermic as the process used additional energy to break the H-F bond to form the H-H bond. Therefore, in the potential energy surface diagram there will be a lot of vibrational energy before the transition state compared to after. This also suggest that the HF bond is stronger than the HH bond as it requires additional energy to break its bond. This will give a late transition state.The translational energy could get converted into vibrational energy in order to break the bond between the strong H-F bond<ref name=":1">Polanyi, J.C. and Tardy, D., 1969. Energy Distribution in the Exothermic Reaction F+ H2 and the Endothermic Reaction HF+ H. ''The Journal of Chemical Physics'', ''51''(12), pp.5717-5719.</ref>. This favours the endothermic reaction because the reactants position will be aligned with the direction of the minimum energy path.''''' '''''

* Q7. Locate the approximate position of the transition state.
{| class="wikitable"
|+Table 8. Transition state position
!
!r /ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|180
|0
|-
|r(BC)
|74.5
|0
|}
The transition state position shown in Table 8 was estimated by looking at the initial conditions in Table 6 for the exothermic F + H2 → HF + H reaction. Due to exothermic reactions having early transition states, this suggested that the bond distance BC (the hydrogen atoms) would be small as it's the reactant. The initial bond distance of AB was reduced until the hessian matrix, shown in Table 9, gave a value of Det= -1.0 <0 which proved it was a saddle point. Additionally, there was no force. This also suggested it was the transition state.
{| class="wikitable"
|+Table 9. Hessian matrix for Transition state
|AB direction
|<nowiki>+1.000</nowiki>
|<nowiki>-0.024</nowiki>
|-
|BC direction
|<nowiki>-0.024</nowiki>
|<nowiki>-1.000</nowiki>
|}
[[File:Mb7418 dis vs time ts ex2.png|thumb|[[Figure 9]]. Intermolecular distance vs time graph for transition state for equations two and three]]
Additionally, Figure 9 shows the intermolecular distance vs time graph. This is a flat line, this could be due to everything in the system being stationary. There are no oscillations showing that there are any vibrational motions. This suggest that it's at the initial conditions in Table 8 represent the transition state position.

* Q8. Report the activation energy for both reactions.
Equation 2: F + H2 → HF + H the activation energy value is 1.0 kJ/mol.

Equation 3: HF + H → F + H2 the activation energy value is 126.7 kJ/mol.

The activation energy for equation 2 was calculated by getting the potential energy for transition state by using the initial conditions in Table 8. Then this was subtracted by the potential energy obtained by using the initial conditions in Table 10. Activation energy = TS(potential energy) - Reactants(potential energy) = (-433.981)-(435.056) = 1.075 kJ/mol
{| class="wikitable"
|+Table 10. Initial conditions for equation 2
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|500
|0
|-
|r(BC)
|74.5
|0
|}
The activation energy for equation 3 was calculated by getting the potential energy for for transition state by using the initial conditions in Table 8. Then this was subtracted by the potential energy obtained by using the initial conditions in Table 11. r(AB) which is the HF bond distance = 92 ppm<ref>Cccbdb.nist.gov. 2020. ''CCCBDB Listing Of Experimental Data Page 2''. [online] Available at: <<nowiki>https://cccbdb.nist.gov/exp2x.asp?casno=7664393&charge=0</nowiki>> [Accessed 22 May 2020].</ref>'''. '''The r(BC) bond which is the HH bond distance was set very high because the hydrogen atoms aren't bonded in the reactants.

Activation energy = TS(potential energy) - Reactants(potential energy) = (-433.981)-(560.698) = 126.7 kJ/mol
{| class="wikitable"
|+Table 11. Initial conditions for equation 3
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|92
|0
|-
|r(BC)
|500
|0
|}

=== Reaction dynamics ===
* Q9. 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.
{| class="wikitable"
|+Table 12. Initial conditions for reactive trajectory
!
!r/ ppm
!p/ g.mol-1.pm.fs-1
|-
|r(AB)
|175
|<nowiki>-1.6</nowiki>
|-
|r(BC)
|74
|0.3
|}
[[File:Mb7418 THIS reactive.png|left|thumb|[[Figure 10]]. Reactive trajectory]]
The step number was increased to 4000 and the initial conditions in Table 12 gives the potential energy surface diagram in Figure 10. This shows a reactive trajectory for the exothermic reaction F + H2 → HF + H as it passes the transition state as well as having many oscillations. The end point is also far away from the starting point.

The starting point (reactants) has a high energy level compared to the products due to being exothermic<ref name=":0" />. Hence the potential energy converts into translational kinetic energy. Therefore, there are many oscillations in Figure 10 due to energy being conserved.

This can be explained by the difference in distribution of the energy levels between an excited and relaxed state. The excited state will have an occupied ground state (v=0) as well as an occupied first state (v=1) where v=vibrational number. Whereas in the relaxed state only the ground state will be occupied. This approach uses the harmonic oscillator<ref>Cccbdb.nist.gov. 2020. ''CCCBDB Listing Of N2 Experimental Data Page 2''. [online] Available at: <<nowiki>https://cccbdb.nist.gov/exp2x.asp?casno=7727379&charge=0</nowiki>> [Accessed 18 May 2020].</ref>. When energy is released, in the excited state there will be two transitions, v(0)→v(1) and v(1)→v(2), the fundamental peak and overtone. In the ground state it will only have transitions from v(0)→v(1).

Therefore, in an IR spectrum<ref>Chemistry LibreTexts. 2020. ''5.5: The Harmonic Oscillator And Infrared Spectra''. [online] Available at: <<nowiki>https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/05%3A_The_Harmonic_Oscillator_and_the_Rigid_Rotor/5.05%3A_The_Harmonic_Oscillator_and_Infrared_Spectra</nowiki>> [Accessed 22 May 2020].</ref>, for the excited state there will be two peaks corresponding to the overtone and fundamental peak. The overtone peak intensity will be smaller than the fundamental's peak intensity. Hence there are many oscillations due to the overtone decreasing and then increasing in intensity while the fundamental increases and then decreases in intensity.

Therefore, this could be confirmed experimentally by Infrared spectroscopy (IR).
* Q10. 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.
[[File:Mb7418 last Q.png|left|thumb|[[Figure 11]]. Reactive potential energy surface contour plot]]
The position of the transition state affects whether the reaction is either exothermic or endothermic. An endothermic reaction will have a late transition state, suggesting that the vibrational energy efficiency will be higher<ref>Guo, H. and Liu, K., 2020. ''Control Of Chemical Reactivity By Transition-State And Beyond''.</ref>. The reason for this is because if it had a higher translational kinetic energy, the reactants wouldn't follow the minimum energy path and will just collide with each other but then immediately repel resulting in being back at the starting point. Hence, the translational kinetic energy converts into vibrational energy<ref name=":1" /> instead so the reactants collide and are aligned with the minimum energy pathway. The direction of the path taken is shown in Figure 11. The initial conditions in Table 12 was used to give the reactive trajectory. This suggests that endothermic reactions have higher vibrational energy to make their reactions more efficient so it's able to overcome the transition state. Hence there are a lot of oscillations before the transition state is reached because it requires more efficient vibrational energy in order to reach the transition state.

Whereas if the reaction has an early transition state, it will be an exothermic reaction. This suggest that there will be a higher translational kinetic energy compared to vibrational energy. This means that there will be less oscillations before it reaches the transition state as shown in Figure 11.

Overall, an efficient endothermic reaction will have a higher vibrational energy whereas an efficient exothermic reaction will have a higher translational kinetic energy.

MRD：01508610

2020-05-29T17:31:50Z

Mak214: /* Reactive and unreactive trajectories */

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

We will be looking at collision and reaction between diatomic molecule of hydrogen and an atom of H in a linear configuration in the gas phase. In simple terms, the atom A collides with the molecule B-C and forms a new molecule with B, while C is detached as a separate atom.
The interatomic distances between A and B, B and C are labelled as rAB and rBC respectively. {{fontcolor1|red| Nice! I like these diagrams you have drawn. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:25, 29 May 2020 (BST)}}

[[File:initial_describe_collision.png]]

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

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 molecular geometry and chemical reaction dynamics are analysed with a potential energy surface (PES) where the necessary points can be evaluated and classified according to first and second derivatives of the energy with respect to position, V(r), which respectively are the gradient and the curvature. Stationary points are those with zero gradient (∂V(ri)/∂ri=0) and have physical meaning: energy minima (or local minima of the PES) correspond to physically stable chemical species (such as reactant and product) and saddle points representing transition states, the maximum on the lowest energy pathway (or minimum energy path) that linking reactant and product.

• Mathematically, a saddle point is a point on the surface of a function where the slopes in orthogonal directions are all zero, but which is not a local extremum of the function. The transition state is a first-order saddle point. In this H + H2 system, the surface is symmetrical, which means that the transition state must be exactly half way between the reactants and products. In another words, the transition state lies somewhere on the diagonal line where the distances rAB and rBC are equal.

• The hydrogen atoms will have no movement at the saddle point since the force on atoms (i.e. the negative derivative of PES with respect to the coordinate ri where the force acting on) will be zero at transition state. Consequently, if one starts a trajectory exactly at the transition state, with no initial momentum, it will remain there forever. Therefore, the method to identify the transition state position from a PES is to change the values of rAB and rBC until the forces along AB and BC are both zero. In fact, forces are negative on one side of the PES and positive one the other side. The transition state is located where forces across zero. {{fontcolor1|red| Great. Thank you. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:25, 29 May 2020 (BST)}}

===Trajectories from rAB = rBC: locating the transition state===

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

• By testing different initial conditions with rAB = rBC, and p1 = p2 = 0.0 g.mol-1.pm.fs-1, the transition state position (rts) is found to be the point where distances rAB and rBC are equal to 90.78 pm. At that position, forces are both -0.001 KJ.mol-1.pm-1 (close to zero).

• Since the H + H2 surface is symmetrical, the transition state must have rAB = rBC. If we start a trajectory on the ridge rAB = rBC there is no gradient in the direction at right angles to the ridge, thus the trajectory will oscillate on the ridge and never fall off.

[[File:threeatomcolliderabrbc.png|600px]]

• To verify the transition state position, “Internuclear Distances vs Time” plots for some relevant rAB = rBC trajectories are shown below. In these graphical display the values of rAB(t), rBC(t) and rAC(t) are given (Y axis) against time t (X axis). At transition state, distance rAB and rBC (and thus rAC) are constant since all three atoms are stationary at the saddle point. The potential energy is maximized while the kinetic energy is zero. This geometry is transition structure.

[[File:90pm_90pm_5000number_gjs.png|thumb|center|A “Internuclear Distances vs Time” plot at transition state.]]

• In contrast, when initial conditions increased to rAB = rBC =100 pm, it shows an oscillatory behaviour for the whole trajectory corresponding to the vibration of atom A and atom C, since there are both potential and kinetic energy. The distance rAB and rBC are overlapped, since they have the same amplitude of vibration and the vibrations are symmetrical around atom B. The distance rAC) has doubled the amplitude of vibration (i.e. the sum or superposition of the amplitude). The oscillatory behaviour is changed after the transition state, at t≈270 fs, to a dissociation pattern where rAB (and thus rAC) keeps increasing with a slight oscillatory behaviour while rBC remained at the same level with a higher vibrational frequency. This means that the atom A leaves away from the molecule B-C with the translational energy while molecule B-C is left with vibrational energy, which resembles the reactant. {{fontcolor1|red| Well done. A clear response. Good use of these figures to show what you did. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:26, 29 May 2020 (BST)}}

[[File:Figure_2_100pm_internuc-time.png|thumb|center|A “Internuclear Distances vs Time” plot at rAB = rBC =100 pm.]]

===Trajectories from rBC = rts+δ, rAB = rts===

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

The reaction path (minimum energy path or mep) is calculated with these initial conditions involving:
• rBC = rts+1 pm = 91.78 pm;
• rAB = rts = 90.78 pm;
• pBC = pAB = 0 g.mol-1.pm.fs-1;
• then change the calculation type from dynamics to MEP;
• increase the number/size of steps to 5000/0.1(fs).

Then repeat the calculation with the same initial conditions used to calculate the reaction path, but change the calculation type back to Dynamics.

The calculated trajectories are shown in contour plots below, which both start from the transition state and simply follows the valley floor to HC + HA-HB. The mep is a smooth curve with no oscillatory behaviour, while the dynamic trajectory is an oscillatory curve.

[[File:Reaction_path_mep_jsg_.png|thumb|center|The contour plot of the reaction path calculated with mep.]]

[[File:Reaction_path_dynamic_jsg.png|thumb|center|The contour plot of the reaction path calculated with Dynamics.]]

The mep has the property that any point on the path is at an energy minimum in all directions perpendicular to the path. The mep can also be described as the union of steepest descent paths from the saddle points to the minima (that is why the mep passes through at least one first-order saddle point.) {{fontcolor1|red| Reference is needed here. We reset the momenta to zero at every timestep for the MEP calculation which means the system will always fall into the nearest minimum - If you started on a saddle point and had no momentum you shouldn't go anywhere! [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:31, 29 May 2020 (BST)}}

From the “Internuclear Distances vs Time” diagrams below, it shows a constant increasing rate of rBC (and thus rAC) after the transition state in Dynamic trajectory, whereas it shows a reducing rate of the increase of the rBC (and thus rAC) in the mep. In addition, the molecule A-B has slight vibrational (or oscillatory) behaviour but has no vibration in the mep.

[[File:Animation mep change of rate gjs123.png|thumb|center|The “Internuclear Distances vs Time” plot of the mep with rBC = rts+1 pm = 91.78 pm.]]

[[File:multiple_graph_123_jsg.png|thumb|center|The “Internuclear Distances vs Time” plot of Dynamic trajectory with rBC = rts+1 pm = 91.78 pm.]]

As mentioned before, the variation of momentum p (the forces acting on a given interatomic coordinate ri) will depend on the derivative of the potential energy surface with respect to that coordinate. Simply, the momenta over time, p(t), is the gradient of V(r).

[[File:EquationInnedgjd.PNG|center]]

{{fontcolor1|red| Reference for this equation? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:31, 29 May 2020 (BST)}}

From the “Momenta vs Time” diagrams below, the momenta of all atoms are zero in the mep since the momenta/velocities are always reset to zero in each time step when calculating the mep. while it have various values in Dynamic trajectory: the momentum of A-B decreases to slightly negative values and then rises to ≈2.5 g.mol-1.pm.fs-1 and keeps oscillating; whereas the momentum of B-C, similarly, decreases to negative values and then increases dramatically to and remains at ≈5 g.mol-1.pm.fs-1.

[[File:Mep momentia.vs.time-1000 gjs.png|thumb|center|The “Momenta vs Time” plot of the mep with rBC = rts+1 pm = 91.78 pm.]]

[[File:Dynamics momenta.vs.time 1000 gjs.png|thumb|center|The “Momenta vs Time” plot of Dynamic trajectory with rBC = rts+1 pm = 91.78 pm.]]

If we change the initial conditions to:
• rBC = rts = 90.78 pm;
• rAB = rts + 1 pm = 91.78 pm.
The “Internuclear Distances vs Time” and “Momenta vs Time” plots will look like this:

[[File:Changedinterdis-time-MEP-5000-gjs.png|thumb|center|The “Internuclear Distances vs Time” plot of the mep when rAB = rts + 1 pm.]]

[[File:Multiple graph 123 jsg.png|thumb|center|The “Internuclear Distances vs Time” plot of Dynamic trajectory when rAB = rts + 1 pm.]]

The “Internuclear Distances vs Time”and “Momenta vs Time” plots of both the mep and Dynamic trajectory have exactly the same pattern as those plots before the initial conditions are changed, except that line A-B (representing rAB) and line B-C (representing rBC) are interconverted. This means that the reaction proceeds in the same way but opposite direction (towards the valley floor to HA + HB-HC).

[[File:Mep momentia.vs.time-1000 gjs.png|thumb|center|The “Momenta vs Time” plot of the mep when rAB = rts + 1 pm.]]

[[File:Changesmomentia-vs-time-Dynamics1000-gjs.png|thumb|center|The “Momenta vs Time” plot of Dynamic trajectory when rAB = rts + 1 pm.]]

Take note of the final values of the positions r1(t) r2(t) and p1(t) p2(t) for your trajectory for large enough t.

• step number = 5000

• step size = 0.1 fs

• rBC = 74.77117297476231 pm

• rAB = 3777.013307800053 pm

• pBC = 1.9461593911826354 g.mol-1.pm.fs-1

• pAB = 5.073830709923566 g.mol-1.pm.fs-1

The atom A is extremely far from the molecule B-C and keeps moving away. The energies are:
• kinetic energy = +19.657 KJ.mol-1

• potential energy = -434.999 KJ.mol-1

• total energy = -415.342 KJ.mol-1

Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. What do you observe?

The energy terms are the same as above, but in this case, the reverse sign of momenta results in the reserve of the moving direction of the free atom (atom A in this case). The atom A is very far from the molecule B-C and it keeps approaching the molecule B-C.

===Reactive and unreactive trajectories===

For the initial positions rBC = 74 pm and rAB = 200 pm, run trajectories with the following momenta combination and complete the table.
Complete the table below 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?

{| class="wikitable" border="1"
|+ caption
! pBC/ g.mol-1.pm.fs-1 !! pAB/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes || The reaction can occurred. The atom A approaches the molecule B-C and forms a new molecule A-B which leaves away with most vibrational energy and a small amount of kinetic energy (KE), while atom C is detached as a separate atom which moves away with most KE. || [[File:R1=-2.5;r2=-5.1 jsg lastday.png|thumb|200px]]
|-
| -3.1 || -4.1 || -420.077 || No || The reaction doesn't occur. The atom A approaches the molecule B-C and return back without colliding with the molecule. || [[File:-3.1and-4.1_jsg_lastday.png|thumb|200px]]
|-
| -3.1 || -5.1 || -413.977 || Yes || The reaction can occur. The atom A approaches the molecule B-C and forms a new molecule A-B which leaves away with most vibrational energy and a small amount of kinetic energy (KE), while atom C is detached as a separate atom which moves away with most KE. || [[File:-3.1and-5.1 jsg lastday.png|thumb|200px]]
|-
| -5.1 || -10.1 || -357.277 || No || A case of barrier recrossing.. The atom A approaches the molecule B-C and forms a temporary new molecule A-B with strong vibrational motion and the molecule A-B move back and collide with atom C to form the B-C again.|| [[File:-5.1and-10.1 jsg lastday.png|thumb|200px]]
|-
| -5.1 || -10.6 || -349.477 || Yes || The reaction can occur. The atom A approaches the molecule B-C and break B-C bond and makes atom B bouncing in-between and eventually form a new molecule A-B. || [[File:-5.1and-10.6 jsg+lastday.png|thumb|200px]]
|}

From this table, you can observe that some trajectories have a total energy greater than the activation energy and still do not react. Therefore, for the reaction to proceed, there are two conditions should be met:(1) the total energy is lager than the activation energy; (2)the interaction between atoms should be appropriate for the collision to happen with correct energy and direction. For instance, if the approaching atom A move opposite to the molecule B-C, there will be no collision and the reaction will definitely not occur. {{fontcolor1|red| OK. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:31, 29 May 2020 (BST)}}

* (The transition state is found to be at rAB= 230.0 pm; rBC=74.0 pm; forces along AB and AC = +0.015 and +0.012 respectively; KE = 0; PE = total E = -434.471.)

===Transition State Theory===

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

Transition state theory (TST) explains the reaction rates of elementary chemical reactions. The theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes. These are three properties of TST:

1.Rates of reaction can be studied by examining activated complexes near the saddle point of a potential energy surface. The details of how these complexes are formed are not important. The saddle point itself is called the transition state.

2.The activated complexes are in a special equilibrium (quasi-equilibrium) with the reactant molecules.

3.The activated complexes can convert into products, and kinetic theory can be used to calculate the rate of this conversion.

TST is used primarily to understand qualitatively how chemical reactions take place. Simply, the Transition State Theory (TST) tells whether a reaction will happen if cross the TS barrier.

So the difference here is that our program is working with a single molecule, the TST deals with a bulk system with a Boltzmann speed distribution. The main differences between experimental and predicted rate constant values will be due to (1) ignoring barrier recrossing and (2) ignoring QM behaviour.

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

F + H2 --> HF + H

A F atom (A) approaches a hydrogen molecule (B-C) to forms a new HF molecule (A-B) and a separate H atom (C).

===PES inspection===

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?

Locate the approximate position of the transition state.

In order to classify the reaction based on energies, the total energies of reactant and that of product should be identified first by using mep calculation (in order to exclude kinetic energy) starting with a small displacement from the TS and with a high enough number of steps. It will go towards the formation of either the products or the reactants and the potential energy will tend to a constant value which is the total energy of the product/reactants depending which direction the displacement was.

The total energies of reactant and that of product are calculated to be:

• reactant: E(H2) = -435.099 KJ.mol-1;

• product: E(HF) = -560.402 KJ.mol-1.

Since the energy of the product is lower than that of reactant, this is an exothermic reaction. This makes sense with the concept of bond strength in which the H-F bond (565 KJ.mol-1) is much stronger than H-H bond (432 KJ.mol-1) and thus there will be more energy released during the formation of H-F bond.

The transition state is still a maximum or a saddle point on the PES, and found to be at:

• rAB = rH-F = 182 pm;

• rBC = rH-H = 74.2 pm;

• pAB = pBC = 0 g.mol-1.pm.fs-1;

• force along AB = +0.003 ≈0 KJ.mol-1.pm-1;

• forces along BC = +0.089 ≈0 KJ.mol-1.pm-1;

• the total energy = -433.969 KJ.mol-1.

Because the activation energy for one of the reactions is so small, it is not easy to locate the transition state immediately. Use the Hammond postulate to guide your search.

Report the activation energy for both reactions.

The activation energy of the forward reaction is so small (1.13 KJ.mol-1), the Hammond postulate is introduced. Hammond postulate is a hypothesis which states that the transition state of a reaction resembles either the reactants or the products, to whichever it is closer in energy. In an exothermic reaction, the TS is closer to the reactants than to the products in energy. Therefore, the TS resembles the reactant in the reaction. In an exothermic reaction, the TS is closer to the products than to the reactants in energy and therefore, the TS resembles the products.

In F-H-H system, the activation energy is very similar to the energy of the reactant and hence the TS resembles reactant in this case. Therefore, this reaction (F + H2 --> HF + H) is exothermic in forward direction and thus endothermic in backward direction.

In order to classify the reaction based on energies, the total energies of reactant and that of product should be identified first by using mep calculation (in order to exclude kinetic energy) starting with a small displacement from the TS and with a high enough number of steps. It will go towards the formation of either the products or the reactants and the potential energy will tend to a constant value which is the total energy of the product/reactants depending which direction the displacement was.

• mep calculation: number of steps = 5000

Since the transition state is found to be at rAB = rH-F = 182 pm and rBC = rH-H = 74.2 pm, we either change rAB or rBC to achieve the formation of either reactants or products. Here is an example:

The reaction reacts towards the products when initial conditions are:

• rAB = rH-F = 165 pm;

• rBC = rH-H = 74.2 pm;

The "contour plot" indicating the reaction direction clearly and "Energy vs time" diagram which shows a decrease (lower in energy) of potential energy (and also the total energy) from -434.194 to -560.402 KJ.mol-1 are illustrated below.

[[File:Surface Plot 165 74.2 MEP gjs.png|thumb|center|The contour plot of FHH system when rAB =165 pm and rBC = 74.2 pm, calculated with mep.]]

[[File:Contour Plot 165 74.2 MEP gjs.png|thumb|center|The Energy.vs.time plot of FHH system when rAB =165 pm and rBC = 74.2 pm, calculated with mep.]]

Based on this mep calculation, the total energies of reactant and that of product are calculated to be:

• reactant: E(H2) = -435.099 KJ.mol-1;

• product: E(HF) = -560.402 KJ.mol-1.

Consequentially, since the TS energy = -433.969 KJ.mol-1, the forward reaction (F + H2 --> HF + H) has an activation energy of [-433.969-(-435.099)]=1.13 KJ.mol-1; the backward reaction (HF + H --> F + H2）has an activation energy of [-433.969-(-560.402)]=126.433 KJ.mol-1. The latter is much larger than the former.

===Reaction dynamics===

Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Momenta vs Time”

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.

For example, when initial conditions fulfil rAB = 160 pm and 0 < rBC < 76 pm, the trajectory are reactive.

The released energy is concerted thermally into electromagnetic radiations (photons). The relaxed potential energy well of electrons have photon absorption and emission between ground state E(0) and the first excited state E(1) and most of the electrons are populated in ground state. While the excited potential energy well involves more possible higher excited states, such as E(0) to the second excited state E(2), E(1) to E(2) which is called overtone. The overtone will contribute to a second absorbance peak at lower wavenumbers in IR spectra.

The emission of photons can be detected by a special type of IR called a Quantum Well Infrared Photodetector (QWIP) which is an infrared photodetector.

Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration.

Setting initial conditions as:
• rHH = 74 pm;
• rFH = 180 pm;
• pFH = -1.0 g.mol-1.pm.fs-1;

The value of pHH is varied from -6.1 to +6.1 and the resulting "Internuclear Distance.vs.Time" diagams used to show the trajectories are displayed below:

[[File:Q2-gjs-rage-6.1-123.png|thumb|center|pHH= -6.1|300px]] [[File:Q2-gjs-rage-6-123.png|thumb|center|pHH= -6|300px]] [[File:Q2-gjs-rage-5-123.png|thumb|center|pHH= -5|300px]] [[File:Q2-gjs-rage-4-123.png|thumb|center|pHH= -4|300px]] [[File:Q2-gjs-rage-3-123.png|thumb|center|pHH= -3|300px]] [[File:Q2-gjs-rage-2-123.png|thumb|center|pHH= -2|300px]] [[File:Q2-gjs-rage-1-123.png|thumb|center|pHH= -1|300px]] [[File:Q2-gjs-rage-0-123.png|thumb|center|pHH= 0|300px]] [[File:Q2-gjs-rage+1-123.png|thumb|center|pHH= 1|300px]] [[File:Q2-gjs-rage+2-123.png|thumb|center|pHH= 2|300px]] [[File:Q2-gjs-rage+3-123.png|thumb|center|pHH= 3|300px]] [[File:Q2-gjs-rage+4-123.png|thumb|center|pHH= 4|300px]] [[File:Q2-gjs-rage+5-123.png|thumb|center|pHH= 5|300px]] [[File:Q2-gjs-rage+6-123.png|thumb|center|pHH= 6|300px]] [[File:Q2-gjs-rage+6.1-123.png|thumb|center|pHH= 6.1|300px]]

In general, the excess of energy put into the system gives rise to the significant increase in the vibrational energy of H-H. when pHH is negative, the smaller the value (more positive) the more barrier recrossing. When pHH is positive, in most cases there is one barrier recrossing only except when p=+1,+5,+6.1. Even though some barrier recrossing occurs (F atom collides with H atom for several times), it doesn't necessarily mean that the reaction will proceed to form the product HF and a separated atom H.

For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?

The new trajectory is showm in the "Internuclear Distance.vs.Time" diagams below. The vibrating molecule H-H moves towards F and collides with it, making the atom B (H atom) bouncing in-between for three times and then return back to the other H atom to reform the vibrating hydrogen molecule.
[[File:3Q-gjs-changed-initials.png|thumb|center|pHH= 0.2|300px]]

Let us now focus on the reverse reaction, H + HF.

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

As mentioned before, the relaxed potential energy well of electrons have photon absorptions and emissions between ground state E(0) and the first excited state E(1) and most of the electrons are populated in ground state. While the excited potential energy well involves more possible higher excited states, such as E(0) to the second excited state E(2), E(1) to E(2) which is called overtone. The overtone will contribute to a second absorbance peak at lower wavenumbers in IR spectra. Each energy state correspond to a translational energy while the vibrational levels are smaller in energy difference and located around translational energy states.

Polanyi's empirical rules states that vibrational energy is more efficient in prompting a late-barrier reaction (that is, a transition state resembling the products) than translational energy, whereas the reverse is true for an early barrier reaction.

Translational energy is better at promoting exothermic reactions, because the trajectory can just fall into the lower energy region. Vibrational motion might make it harder for the trajectory to cross the barrier because it is in a different direction than the mep. If the reactants have vibrational motion, then they will try to move up the valley near the transition state, and just fall back to the reactants state.

MRD：01508610

2020-05-29T17:31:07Z

Mak214: /* Trajectories from rBC = rts+δ, rAB = rts */

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

We will be looking at collision and reaction between diatomic molecule of hydrogen and an atom of H in a linear configuration in the gas phase. In simple terms, the atom A collides with the molecule B-C and forms a new molecule with B, while C is detached as a separate atom.
The interatomic distances between A and B, B and C are labelled as rAB and rBC respectively. {{fontcolor1|red| Nice! I like these diagrams you have drawn. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:25, 29 May 2020 (BST)}}

[[File:initial_describe_collision.png]]

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

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 molecular geometry and chemical reaction dynamics are analysed with a potential energy surface (PES) where the necessary points can be evaluated and classified according to first and second derivatives of the energy with respect to position, V(r), which respectively are the gradient and the curvature. Stationary points are those with zero gradient (∂V(ri)/∂ri=0) and have physical meaning: energy minima (or local minima of the PES) correspond to physically stable chemical species (such as reactant and product) and saddle points representing transition states, the maximum on the lowest energy pathway (or minimum energy path) that linking reactant and product.

• Mathematically, a saddle point is a point on the surface of a function where the slopes in orthogonal directions are all zero, but which is not a local extremum of the function. The transition state is a first-order saddle point. In this H + H2 system, the surface is symmetrical, which means that the transition state must be exactly half way between the reactants and products. In another words, the transition state lies somewhere on the diagonal line where the distances rAB and rBC are equal.

• The hydrogen atoms will have no movement at the saddle point since the force on atoms (i.e. the negative derivative of PES with respect to the coordinate ri where the force acting on) will be zero at transition state. Consequently, if one starts a trajectory exactly at the transition state, with no initial momentum, it will remain there forever. Therefore, the method to identify the transition state position from a PES is to change the values of rAB and rBC until the forces along AB and BC are both zero. In fact, forces are negative on one side of the PES and positive one the other side. The transition state is located where forces across zero. {{fontcolor1|red| Great. Thank you. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:25, 29 May 2020 (BST)}}

===Trajectories from rAB = rBC: locating the transition state===

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

• By testing different initial conditions with rAB = rBC, and p1 = p2 = 0.0 g.mol-1.pm.fs-1, the transition state position (rts) is found to be the point where distances rAB and rBC are equal to 90.78 pm. At that position, forces are both -0.001 KJ.mol-1.pm-1 (close to zero).

• Since the H + H2 surface is symmetrical, the transition state must have rAB = rBC. If we start a trajectory on the ridge rAB = rBC there is no gradient in the direction at right angles to the ridge, thus the trajectory will oscillate on the ridge and never fall off.

[[File:threeatomcolliderabrbc.png|600px]]

• To verify the transition state position, “Internuclear Distances vs Time” plots for some relevant rAB = rBC trajectories are shown below. In these graphical display the values of rAB(t), rBC(t) and rAC(t) are given (Y axis) against time t (X axis). At transition state, distance rAB and rBC (and thus rAC) are constant since all three atoms are stationary at the saddle point. The potential energy is maximized while the kinetic energy is zero. This geometry is transition structure.

[[File:90pm_90pm_5000number_gjs.png|thumb|center|A “Internuclear Distances vs Time” plot at transition state.]]

• In contrast, when initial conditions increased to rAB = rBC =100 pm, it shows an oscillatory behaviour for the whole trajectory corresponding to the vibration of atom A and atom C, since there are both potential and kinetic energy. The distance rAB and rBC are overlapped, since they have the same amplitude of vibration and the vibrations are symmetrical around atom B. The distance rAC) has doubled the amplitude of vibration (i.e. the sum or superposition of the amplitude). The oscillatory behaviour is changed after the transition state, at t≈270 fs, to a dissociation pattern where rAB (and thus rAC) keeps increasing with a slight oscillatory behaviour while rBC remained at the same level with a higher vibrational frequency. This means that the atom A leaves away from the molecule B-C with the translational energy while molecule B-C is left with vibrational energy, which resembles the reactant. {{fontcolor1|red| Well done. A clear response. Good use of these figures to show what you did. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:26, 29 May 2020 (BST)}}

[[File:Figure_2_100pm_internuc-time.png|thumb|center|A “Internuclear Distances vs Time” plot at rAB = rBC =100 pm.]]

===Trajectories from rBC = rts+δ, rAB = rts===

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

The reaction path (minimum energy path or mep) is calculated with these initial conditions involving:
• rBC = rts+1 pm = 91.78 pm;
• rAB = rts = 90.78 pm;
• pBC = pAB = 0 g.mol-1.pm.fs-1;
• then change the calculation type from dynamics to MEP;
• increase the number/size of steps to 5000/0.1(fs).

Then repeat the calculation with the same initial conditions used to calculate the reaction path, but change the calculation type back to Dynamics.

The calculated trajectories are shown in contour plots below, which both start from the transition state and simply follows the valley floor to HC + HA-HB. The mep is a smooth curve with no oscillatory behaviour, while the dynamic trajectory is an oscillatory curve.

[[File:Reaction_path_mep_jsg_.png|thumb|center|The contour plot of the reaction path calculated with mep.]]

[[File:Reaction_path_dynamic_jsg.png|thumb|center|The contour plot of the reaction path calculated with Dynamics.]]

The mep has the property that any point on the path is at an energy minimum in all directions perpendicular to the path. The mep can also be described as the union of steepest descent paths from the saddle points to the minima (that is why the mep passes through at least one first-order saddle point.) {{fontcolor1|red| Reference is needed here. We reset the momenta to zero at every timestep for the MEP calculation which means the system will always fall into the nearest minimum - If you started on a saddle point and had no momentum you shouldn't go anywhere! [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:31, 29 May 2020 (BST)}}

From the “Internuclear Distances vs Time” diagrams below, it shows a constant increasing rate of rBC (and thus rAC) after the transition state in Dynamic trajectory, whereas it shows a reducing rate of the increase of the rBC (and thus rAC) in the mep. In addition, the molecule A-B has slight vibrational (or oscillatory) behaviour but has no vibration in the mep.

[[File:Animation mep change of rate gjs123.png|thumb|center|The “Internuclear Distances vs Time” plot of the mep with rBC = rts+1 pm = 91.78 pm.]]

[[File:multiple_graph_123_jsg.png|thumb|center|The “Internuclear Distances vs Time” plot of Dynamic trajectory with rBC = rts+1 pm = 91.78 pm.]]

As mentioned before, the variation of momentum p (the forces acting on a given interatomic coordinate ri) will depend on the derivative of the potential energy surface with respect to that coordinate. Simply, the momenta over time, p(t), is the gradient of V(r).

[[File:EquationInnedgjd.PNG|center]]

{{fontcolor1|red| Reference for this equation? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:31, 29 May 2020 (BST)}}

From the “Momenta vs Time” diagrams below, the momenta of all atoms are zero in the mep since the momenta/velocities are always reset to zero in each time step when calculating the mep. while it have various values in Dynamic trajectory: the momentum of A-B decreases to slightly negative values and then rises to ≈2.5 g.mol-1.pm.fs-1 and keeps oscillating; whereas the momentum of B-C, similarly, decreases to negative values and then increases dramatically to and remains at ≈5 g.mol-1.pm.fs-1.

[[File:Mep momentia.vs.time-1000 gjs.png|thumb|center|The “Momenta vs Time” plot of the mep with rBC = rts+1 pm = 91.78 pm.]]

[[File:Dynamics momenta.vs.time 1000 gjs.png|thumb|center|The “Momenta vs Time” plot of Dynamic trajectory with rBC = rts+1 pm = 91.78 pm.]]

If we change the initial conditions to:
• rBC = rts = 90.78 pm;
• rAB = rts + 1 pm = 91.78 pm.
The “Internuclear Distances vs Time” and “Momenta vs Time” plots will look like this:

[[File:Changedinterdis-time-MEP-5000-gjs.png|thumb|center|The “Internuclear Distances vs Time” plot of the mep when rAB = rts + 1 pm.]]

[[File:Multiple graph 123 jsg.png|thumb|center|The “Internuclear Distances vs Time” plot of Dynamic trajectory when rAB = rts + 1 pm.]]

The “Internuclear Distances vs Time”and “Momenta vs Time” plots of both the mep and Dynamic trajectory have exactly the same pattern as those plots before the initial conditions are changed, except that line A-B (representing rAB) and line B-C (representing rBC) are interconverted. This means that the reaction proceeds in the same way but opposite direction (towards the valley floor to HA + HB-HC).

[[File:Mep momentia.vs.time-1000 gjs.png|thumb|center|The “Momenta vs Time” plot of the mep when rAB = rts + 1 pm.]]

[[File:Changesmomentia-vs-time-Dynamics1000-gjs.png|thumb|center|The “Momenta vs Time” plot of Dynamic trajectory when rAB = rts + 1 pm.]]

Take note of the final values of the positions r1(t) r2(t) and p1(t) p2(t) for your trajectory for large enough t.

• step number = 5000

• step size = 0.1 fs

• rBC = 74.77117297476231 pm

• rAB = 3777.013307800053 pm

• pBC = 1.9461593911826354 g.mol-1.pm.fs-1

• pAB = 5.073830709923566 g.mol-1.pm.fs-1

The atom A is extremely far from the molecule B-C and keeps moving away. The energies are:
• kinetic energy = +19.657 KJ.mol-1

• potential energy = -434.999 KJ.mol-1

• total energy = -415.342 KJ.mol-1

Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. What do you observe?

The energy terms are the same as above, but in this case, the reverse sign of momenta results in the reserve of the moving direction of the free atom (atom A in this case). The atom A is very far from the molecule B-C and it keeps approaching the molecule B-C.

===Reactive and unreactive trajectories===

For the initial positions rBC = 74 pm and rAB = 200 pm, run trajectories with the following momenta combination and complete the table.
Complete the table below 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?

{| class="wikitable" border="1"
|+ caption
! pBC/ g.mol-1.pm.fs-1 !! pAB/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes || The reaction can occurred. The atom A approaches the molecule B-C and forms a new molecule A-B which leaves away with most vibrational energy and a small amount of kinetic energy (KE), while atom C is detached as a separate atom which moves away with most KE. || [[File:R1=-2.5;r2=-5.1 jsg lastday.png|thumb|200px]]
|-
| -3.1 || -4.1 || -420.077 || No || The reaction doesn't occur. The atom A approaches the molecule B-C and return back without colliding with the molecule. || [[File:-3.1and-4.1_jsg_lastday.png|thumb|200px]]
|-
| -3.1 || -5.1 || -413.977 || Yes || The reaction can occur. The atom A approaches the molecule B-C and forms a new molecule A-B which leaves away with most vibrational energy and a small amount of kinetic energy (KE), while atom C is detached as a separate atom which moves away with most KE. || [[File:-3.1and-5.1 jsg lastday.png|thumb|200px]]
|-
| -5.1 || -10.1 || -357.277 || No || A case of barrier recrossing.. The atom A approaches the molecule B-C and forms a temporary new molecule A-B with strong vibrational motion and the molecule A-B move back and collide with atom C to form the B-C again.|| [[File:-5.1and-10.1 jsg lastday.png|thumb|200px]]
|-
| -5.1 || -10.6 || -349.477 || Yes || The reaction can occur. The atom A approaches the molecule B-C and break B-C bond and makes atom B bouncing in-between and eventually form a new molecule A-B. || [[File:-5.1and-10.6 jsg+lastday.png|thumb|200px]]
|}

From this table, you can observe that some trajectories have a total energy greater than the activation energy and still do not react. Therefore, for the reaction to proceed, there are two conditions should be met:(1) the total energy is lager than the activation energy; (2)the interaction between atoms should be appropriate for the collision to happen with correct energy and direction. For instance, if the approaching atom A move opposite to the molecule B-C, there will be no collision and the reaction will definitely not occur.

* (The transition state is found to be at rAB= 230.0 pm; rBC=74.0 pm; forces along AB and AC = +0.015 and +0.012 respectively; KE = 0; PE = total E = -434.471.)

===Transition State Theory===

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

Transition state theory (TST) explains the reaction rates of elementary chemical reactions. The theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes. These are three properties of TST:

1.Rates of reaction can be studied by examining activated complexes near the saddle point of a potential energy surface. The details of how these complexes are formed are not important. The saddle point itself is called the transition state.

2.The activated complexes are in a special equilibrium (quasi-equilibrium) with the reactant molecules.

3.The activated complexes can convert into products, and kinetic theory can be used to calculate the rate of this conversion.

TST is used primarily to understand qualitatively how chemical reactions take place. Simply, the Transition State Theory (TST) tells whether a reaction will happen if cross the TS barrier.

So the difference here is that our program is working with a single molecule, the TST deals with a bulk system with a Boltzmann speed distribution. The main differences between experimental and predicted rate constant values will be due to (1) ignoring barrier recrossing and (2) ignoring QM behaviour.

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

F + H2 --> HF + H

A F atom (A) approaches a hydrogen molecule (B-C) to forms a new HF molecule (A-B) and a separate H atom (C).

===PES inspection===

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?

Locate the approximate position of the transition state.

In order to classify the reaction based on energies, the total energies of reactant and that of product should be identified first by using mep calculation (in order to exclude kinetic energy) starting with a small displacement from the TS and with a high enough number of steps. It will go towards the formation of either the products or the reactants and the potential energy will tend to a constant value which is the total energy of the product/reactants depending which direction the displacement was.

The total energies of reactant and that of product are calculated to be:

• reactant: E(H2) = -435.099 KJ.mol-1;

• product: E(HF) = -560.402 KJ.mol-1.

Since the energy of the product is lower than that of reactant, this is an exothermic reaction. This makes sense with the concept of bond strength in which the H-F bond (565 KJ.mol-1) is much stronger than H-H bond (432 KJ.mol-1) and thus there will be more energy released during the formation of H-F bond.

The transition state is still a maximum or a saddle point on the PES, and found to be at:

• rAB = rH-F = 182 pm;

• rBC = rH-H = 74.2 pm;

• pAB = pBC = 0 g.mol-1.pm.fs-1;

• force along AB = +0.003 ≈0 KJ.mol-1.pm-1;

• forces along BC = +0.089 ≈0 KJ.mol-1.pm-1;

• the total energy = -433.969 KJ.mol-1.

Because the activation energy for one of the reactions is so small, it is not easy to locate the transition state immediately. Use the Hammond postulate to guide your search.

Report the activation energy for both reactions.

The activation energy of the forward reaction is so small (1.13 KJ.mol-1), the Hammond postulate is introduced. Hammond postulate is a hypothesis which states that the transition state of a reaction resembles either the reactants or the products, to whichever it is closer in energy. In an exothermic reaction, the TS is closer to the reactants than to the products in energy. Therefore, the TS resembles the reactant in the reaction. In an exothermic reaction, the TS is closer to the products than to the reactants in energy and therefore, the TS resembles the products.

In F-H-H system, the activation energy is very similar to the energy of the reactant and hence the TS resembles reactant in this case. Therefore, this reaction (F + H2 --> HF + H) is exothermic in forward direction and thus endothermic in backward direction.

In order to classify the reaction based on energies, the total energies of reactant and that of product should be identified first by using mep calculation (in order to exclude kinetic energy) starting with a small displacement from the TS and with a high enough number of steps. It will go towards the formation of either the products or the reactants and the potential energy will tend to a constant value which is the total energy of the product/reactants depending which direction the displacement was.

• mep calculation: number of steps = 5000

Since the transition state is found to be at rAB = rH-F = 182 pm and rBC = rH-H = 74.2 pm, we either change rAB or rBC to achieve the formation of either reactants or products. Here is an example:

The reaction reacts towards the products when initial conditions are:

• rAB = rH-F = 165 pm;

• rBC = rH-H = 74.2 pm;

The "contour plot" indicating the reaction direction clearly and "Energy vs time" diagram which shows a decrease (lower in energy) of potential energy (and also the total energy) from -434.194 to -560.402 KJ.mol-1 are illustrated below.

[[File:Surface Plot 165 74.2 MEP gjs.png|thumb|center|The contour plot of FHH system when rAB =165 pm and rBC = 74.2 pm, calculated with mep.]]

[[File:Contour Plot 165 74.2 MEP gjs.png|thumb|center|The Energy.vs.time plot of FHH system when rAB =165 pm and rBC = 74.2 pm, calculated with mep.]]

Based on this mep calculation, the total energies of reactant and that of product are calculated to be:

• reactant: E(H2) = -435.099 KJ.mol-1;

• product: E(HF) = -560.402 KJ.mol-1.

Consequentially, since the TS energy = -433.969 KJ.mol-1, the forward reaction (F + H2 --> HF + H) has an activation energy of [-433.969-(-435.099)]=1.13 KJ.mol-1; the backward reaction (HF + H --> F + H2）has an activation energy of [-433.969-(-560.402)]=126.433 KJ.mol-1. The latter is much larger than the former.

===Reaction dynamics===

Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Momenta vs Time”

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.

For example, when initial conditions fulfil rAB = 160 pm and 0 < rBC < 76 pm, the trajectory are reactive.

The released energy is concerted thermally into electromagnetic radiations (photons). The relaxed potential energy well of electrons have photon absorption and emission between ground state E(0) and the first excited state E(1) and most of the electrons are populated in ground state. While the excited potential energy well involves more possible higher excited states, such as E(0) to the second excited state E(2), E(1) to E(2) which is called overtone. The overtone will contribute to a second absorbance peak at lower wavenumbers in IR spectra.

The emission of photons can be detected by a special type of IR called a Quantum Well Infrared Photodetector (QWIP) which is an infrared photodetector.

Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration.

Setting initial conditions as:
• rHH = 74 pm;
• rFH = 180 pm;
• pFH = -1.0 g.mol-1.pm.fs-1;

The value of pHH is varied from -6.1 to +6.1 and the resulting "Internuclear Distance.vs.Time" diagams used to show the trajectories are displayed below:

[[File:Q2-gjs-rage-6.1-123.png|thumb|center|pHH= -6.1|300px]] [[File:Q2-gjs-rage-6-123.png|thumb|center|pHH= -6|300px]] [[File:Q2-gjs-rage-5-123.png|thumb|center|pHH= -5|300px]] [[File:Q2-gjs-rage-4-123.png|thumb|center|pHH= -4|300px]] [[File:Q2-gjs-rage-3-123.png|thumb|center|pHH= -3|300px]] [[File:Q2-gjs-rage-2-123.png|thumb|center|pHH= -2|300px]] [[File:Q2-gjs-rage-1-123.png|thumb|center|pHH= -1|300px]] [[File:Q2-gjs-rage-0-123.png|thumb|center|pHH= 0|300px]] [[File:Q2-gjs-rage+1-123.png|thumb|center|pHH= 1|300px]] [[File:Q2-gjs-rage+2-123.png|thumb|center|pHH= 2|300px]] [[File:Q2-gjs-rage+3-123.png|thumb|center|pHH= 3|300px]] [[File:Q2-gjs-rage+4-123.png|thumb|center|pHH= 4|300px]] [[File:Q2-gjs-rage+5-123.png|thumb|center|pHH= 5|300px]] [[File:Q2-gjs-rage+6-123.png|thumb|center|pHH= 6|300px]] [[File:Q2-gjs-rage+6.1-123.png|thumb|center|pHH= 6.1|300px]]

In general, the excess of energy put into the system gives rise to the significant increase in the vibrational energy of H-H. when pHH is negative, the smaller the value (more positive) the more barrier recrossing. When pHH is positive, in most cases there is one barrier recrossing only except when p=+1,+5,+6.1. Even though some barrier recrossing occurs (F atom collides with H atom for several times), it doesn't necessarily mean that the reaction will proceed to form the product HF and a separated atom H.

For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?

The new trajectory is showm in the "Internuclear Distance.vs.Time" diagams below. The vibrating molecule H-H moves towards F and collides with it, making the atom B (H atom) bouncing in-between for three times and then return back to the other H atom to reform the vibrating hydrogen molecule.
[[File:3Q-gjs-changed-initials.png|thumb|center|pHH= 0.2|300px]]

Let us now focus on the reverse reaction, H + HF.

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

As mentioned before, the relaxed potential energy well of electrons have photon absorptions and emissions between ground state E(0) and the first excited state E(1) and most of the electrons are populated in ground state. While the excited potential energy well involves more possible higher excited states, such as E(0) to the second excited state E(2), E(1) to E(2) which is called overtone. The overtone will contribute to a second absorbance peak at lower wavenumbers in IR spectra. Each energy state correspond to a translational energy while the vibrational levels are smaller in energy difference and located around translational energy states.

Polanyi's empirical rules states that vibrational energy is more efficient in prompting a late-barrier reaction (that is, a transition state resembling the products) than translational energy, whereas the reverse is true for an early barrier reaction.

Translational energy is better at promoting exothermic reactions, because the trajectory can just fall into the lower energy region. Vibrational motion might make it harder for the trajectory to cross the barrier because it is in a different direction than the mep. If the reactants have vibrational motion, then they will try to move up the valley near the transition state, and just fall back to the reactants state.

MRD：01508610

2020-05-29T17:26:39Z

Mak214: /* Trajectories from rAB = rBC: locating the transition state */

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

We will be looking at collision and reaction between diatomic molecule of hydrogen and an atom of H in a linear configuration in the gas phase. In simple terms, the atom A collides with the molecule B-C and forms a new molecule with B, while C is detached as a separate atom.
The interatomic distances between A and B, B and C are labelled as rAB and rBC respectively. {{fontcolor1|red| Nice! I like these diagrams you have drawn. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:25, 29 May 2020 (BST)}}

[[File:initial_describe_collision.png]]

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

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 molecular geometry and chemical reaction dynamics are analysed with a potential energy surface (PES) where the necessary points can be evaluated and classified according to first and second derivatives of the energy with respect to position, V(r), which respectively are the gradient and the curvature. Stationary points are those with zero gradient (∂V(ri)/∂ri=0) and have physical meaning: energy minima (or local minima of the PES) correspond to physically stable chemical species (such as reactant and product) and saddle points representing transition states, the maximum on the lowest energy pathway (or minimum energy path) that linking reactant and product.

• Mathematically, a saddle point is a point on the surface of a function where the slopes in orthogonal directions are all zero, but which is not a local extremum of the function. The transition state is a first-order saddle point. In this H + H2 system, the surface is symmetrical, which means that the transition state must be exactly half way between the reactants and products. In another words, the transition state lies somewhere on the diagonal line where the distances rAB and rBC are equal.

• The hydrogen atoms will have no movement at the saddle point since the force on atoms (i.e. the negative derivative of PES with respect to the coordinate ri where the force acting on) will be zero at transition state. Consequently, if one starts a trajectory exactly at the transition state, with no initial momentum, it will remain there forever. Therefore, the method to identify the transition state position from a PES is to change the values of rAB and rBC until the forces along AB and BC are both zero. In fact, forces are negative on one side of the PES and positive one the other side. The transition state is located where forces across zero. {{fontcolor1|red| Great. Thank you. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:25, 29 May 2020 (BST)}}

===Trajectories from rAB = rBC: locating the transition state===

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

• By testing different initial conditions with rAB = rBC, and p1 = p2 = 0.0 g.mol-1.pm.fs-1, the transition state position (rts) is found to be the point where distances rAB and rBC are equal to 90.78 pm. At that position, forces are both -0.001 KJ.mol-1.pm-1 (close to zero).

• Since the H + H2 surface is symmetrical, the transition state must have rAB = rBC. If we start a trajectory on the ridge rAB = rBC there is no gradient in the direction at right angles to the ridge, thus the trajectory will oscillate on the ridge and never fall off.

[[File:threeatomcolliderabrbc.png|600px]]

• To verify the transition state position, “Internuclear Distances vs Time” plots for some relevant rAB = rBC trajectories are shown below. In these graphical display the values of rAB(t), rBC(t) and rAC(t) are given (Y axis) against time t (X axis). At transition state, distance rAB and rBC (and thus rAC) are constant since all three atoms are stationary at the saddle point. The potential energy is maximized while the kinetic energy is zero. This geometry is transition structure.

[[File:90pm_90pm_5000number_gjs.png|thumb|center|A “Internuclear Distances vs Time” plot at transition state.]]

• In contrast, when initial conditions increased to rAB = rBC =100 pm, it shows an oscillatory behaviour for the whole trajectory corresponding to the vibration of atom A and atom C, since there are both potential and kinetic energy. The distance rAB and rBC are overlapped, since they have the same amplitude of vibration and the vibrations are symmetrical around atom B. The distance rAC) has doubled the amplitude of vibration (i.e. the sum or superposition of the amplitude). The oscillatory behaviour is changed after the transition state, at t≈270 fs, to a dissociation pattern where rAB (and thus rAC) keeps increasing with a slight oscillatory behaviour while rBC remained at the same level with a higher vibrational frequency. This means that the atom A leaves away from the molecule B-C with the translational energy while molecule B-C is left with vibrational energy, which resembles the reactant. {{fontcolor1|red| Well done. A clear response. Good use of these figures to show what you did. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:26, 29 May 2020 (BST)}}

[[File:Figure_2_100pm_internuc-time.png|thumb|center|A “Internuclear Distances vs Time” plot at rAB = rBC =100 pm.]]

===Trajectories from rBC = rts+δ, rAB = rts===

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

The reaction path (minimum energy path or mep) is calculated with these initial conditions involving:
• rBC = rts+1 pm = 91.78 pm;
• rAB = rts = 90.78 pm;
• pBC = pAB = 0 g.mol-1.pm.fs-1;
• then change the calculation type from dynamics to MEP;
• increase the number/size of steps to 5000/0.1(fs).

Then repeat the calculation with the same initial conditions used to calculate the reaction path, but change the calculation type back to Dynamics.

The calculated trajectories are shown in contour plots below, which both start from the transition state and simply follows the valley floor to HC + HA-HB. The mep is a smooth curve with no oscillatory behaviour, while the dynamic trajectory is an oscillatory curve.

[[File:Reaction_path_mep_jsg_.png|thumb|center|The contour plot of the reaction path calculated with mep.]]

[[File:Reaction_path_dynamic_jsg.png|thumb|center|The contour plot of the reaction path calculated with Dynamics.]]

The mep has the property that any point on the path is at an energy minimum in all directions perpendicular to the path. The mep can also be described as the union of steepest descent paths from the saddle points to the minima (that is why the mep passes through at least one first-order saddle point.)

From the “Internuclear Distances vs Time” diagrams below, it shows a constant increasing rate of rBC (and thus rAC) after the transition state in Dynamic trajectory, whereas it shows a reducing rate of the increase of the rBC (and thus rAC) in the mep. In addition, the molecule A-B has slight vibrational (or oscillatory) behaviour but has no vibration in the mep.

[[File:Animation mep change of rate gjs123.png|thumb|center|The “Internuclear Distances vs Time” plot of the mep with rBC = rts+1 pm = 91.78 pm.]]

[[File:multiple_graph_123_jsg.png|thumb|center|The “Internuclear Distances vs Time” plot of Dynamic trajectory with rBC = rts+1 pm = 91.78 pm.]]

As mentioned before, the variation of momentum p (the forces acting on a given interatomic coordinate ri) will depend on the derivative of the potential energy surface with respect to that coordinate. Simply, the momenta over time, p(t), is the gradient of V(r).

[[File:EquationInnedgjd.PNG|center]]

From the “Momenta vs Time” diagrams below, the momenta of all atoms are zero in the mep since the momenta/velocities are always reset to zero in each time step when calculating the mep. while it have various values in Dynamic trajectory: the momentum of A-B decreases to slightly negative values and then rises to ≈2.5 g.mol-1.pm.fs-1 and keeps oscillating; whereas the momentum of B-C, similarly, decreases to negative values and then increases dramatically to and remains at ≈5 g.mol-1.pm.fs-1.

[[File:Mep momentia.vs.time-1000 gjs.png|thumb|center|The “Momenta vs Time” plot of the mep with rBC = rts+1 pm = 91.78 pm.]]

[[File:Dynamics momenta.vs.time 1000 gjs.png|thumb|center|The “Momenta vs Time” plot of Dynamic trajectory with rBC = rts+1 pm = 91.78 pm.]]

If we change the initial conditions to:
• rBC = rts = 90.78 pm;
• rAB = rts + 1 pm = 91.78 pm.
The “Internuclear Distances vs Time” and “Momenta vs Time” plots will look like this:

[[File:Changedinterdis-time-MEP-5000-gjs.png|thumb|center|The “Internuclear Distances vs Time” plot of the mep when rAB = rts + 1 pm.]]

[[File:Multiple graph 123 jsg.png|thumb|center|The “Internuclear Distances vs Time” plot of Dynamic trajectory when rAB = rts + 1 pm.]]

The “Internuclear Distances vs Time”and “Momenta vs Time” plots of both the mep and Dynamic trajectory have exactly the same pattern as those plots before the initial conditions are changed, except that line A-B (representing rAB) and line B-C (representing rBC) are interconverted. This means that the reaction proceeds in the same way but opposite direction (towards the valley floor to HA + HB-HC).

[[File:Mep momentia.vs.time-1000 gjs.png|thumb|center|The “Momenta vs Time” plot of the mep when rAB = rts + 1 pm.]]

[[File:Changesmomentia-vs-time-Dynamics1000-gjs.png|thumb|center|The “Momenta vs Time” plot of Dynamic trajectory when rAB = rts + 1 pm.]]

Take note of the final values of the positions r1(t) r2(t) and p1(t) p2(t) for your trajectory for large enough t.

• step number = 5000

• step size = 0.1 fs

• rBC = 74.77117297476231 pm

• rAB = 3777.013307800053 pm

• pBC = 1.9461593911826354 g.mol-1.pm.fs-1

• pAB = 5.073830709923566 g.mol-1.pm.fs-1

The atom A is extremely far from the molecule B-C and keeps moving away. The energies are:
• kinetic energy = +19.657 KJ.mol-1

• potential energy = -434.999 KJ.mol-1

• total energy = -415.342 KJ.mol-1

Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. What do you observe?

The energy terms are the same as above, but in this case, the reverse sign of momenta results in the reserve of the moving direction of the free atom (atom A in this case). The atom A is very far from the molecule B-C and it keeps approaching the molecule B-C.

===Reactive and unreactive trajectories===

For the initial positions rBC = 74 pm and rAB = 200 pm, run trajectories with the following momenta combination and complete the table.
Complete the table below 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?

{| class="wikitable" border="1"
|+ caption
! pBC/ g.mol-1.pm.fs-1 !! pAB/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes || The reaction can occurred. The atom A approaches the molecule B-C and forms a new molecule A-B which leaves away with most vibrational energy and a small amount of kinetic energy (KE), while atom C is detached as a separate atom which moves away with most KE. || [[File:R1=-2.5;r2=-5.1 jsg lastday.png|thumb|200px]]
|-
| -3.1 || -4.1 || -420.077 || No || The reaction doesn't occur. The atom A approaches the molecule B-C and return back without colliding with the molecule. || [[File:-3.1and-4.1_jsg_lastday.png|thumb|200px]]
|-
| -3.1 || -5.1 || -413.977 || Yes || The reaction can occur. The atom A approaches the molecule B-C and forms a new molecule A-B which leaves away with most vibrational energy and a small amount of kinetic energy (KE), while atom C is detached as a separate atom which moves away with most KE. || [[File:-3.1and-5.1 jsg lastday.png|thumb|200px]]
|-
| -5.1 || -10.1 || -357.277 || No || A case of barrier recrossing.. The atom A approaches the molecule B-C and forms a temporary new molecule A-B with strong vibrational motion and the molecule A-B move back and collide with atom C to form the B-C again.|| [[File:-5.1and-10.1 jsg lastday.png|thumb|200px]]
|-
| -5.1 || -10.6 || -349.477 || Yes || The reaction can occur. The atom A approaches the molecule B-C and break B-C bond and makes atom B bouncing in-between and eventually form a new molecule A-B. || [[File:-5.1and-10.6 jsg+lastday.png|thumb|200px]]
|}

From this table, you can observe that some trajectories have a total energy greater than the activation energy and still do not react. Therefore, for the reaction to proceed, there are two conditions should be met:(1) the total energy is lager than the activation energy; (2)the interaction between atoms should be appropriate for the collision to happen with correct energy and direction. For instance, if the approaching atom A move opposite to the molecule B-C, there will be no collision and the reaction will definitely not occur.

* (The transition state is found to be at rAB= 230.0 pm; rBC=74.0 pm; forces along AB and AC = +0.015 and +0.012 respectively; KE = 0; PE = total E = -434.471.)

===Transition State Theory===

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

Transition state theory (TST) explains the reaction rates of elementary chemical reactions. The theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes. These are three properties of TST:

1.Rates of reaction can be studied by examining activated complexes near the saddle point of a potential energy surface. The details of how these complexes are formed are not important. The saddle point itself is called the transition state.

2.The activated complexes are in a special equilibrium (quasi-equilibrium) with the reactant molecules.

3.The activated complexes can convert into products, and kinetic theory can be used to calculate the rate of this conversion.

TST is used primarily to understand qualitatively how chemical reactions take place. Simply, the Transition State Theory (TST) tells whether a reaction will happen if cross the TS barrier.

So the difference here is that our program is working with a single molecule, the TST deals with a bulk system with a Boltzmann speed distribution. The main differences between experimental and predicted rate constant values will be due to (1) ignoring barrier recrossing and (2) ignoring QM behaviour.

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

F + H2 --> HF + H

A F atom (A) approaches a hydrogen molecule (B-C) to forms a new HF molecule (A-B) and a separate H atom (C).

===PES inspection===

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?

Locate the approximate position of the transition state.

In order to classify the reaction based on energies, the total energies of reactant and that of product should be identified first by using mep calculation (in order to exclude kinetic energy) starting with a small displacement from the TS and with a high enough number of steps. It will go towards the formation of either the products or the reactants and the potential energy will tend to a constant value which is the total energy of the product/reactants depending which direction the displacement was.

The total energies of reactant and that of product are calculated to be:

• reactant: E(H2) = -435.099 KJ.mol-1;

• product: E(HF) = -560.402 KJ.mol-1.

Since the energy of the product is lower than that of reactant, this is an exothermic reaction. This makes sense with the concept of bond strength in which the H-F bond (565 KJ.mol-1) is much stronger than H-H bond (432 KJ.mol-1) and thus there will be more energy released during the formation of H-F bond.

The transition state is still a maximum or a saddle point on the PES, and found to be at:

• rAB = rH-F = 182 pm;

• rBC = rH-H = 74.2 pm;

• pAB = pBC = 0 g.mol-1.pm.fs-1;

• force along AB = +0.003 ≈0 KJ.mol-1.pm-1;

• forces along BC = +0.089 ≈0 KJ.mol-1.pm-1;

• the total energy = -433.969 KJ.mol-1.

Because the activation energy for one of the reactions is so small, it is not easy to locate the transition state immediately. Use the Hammond postulate to guide your search.

Report the activation energy for both reactions.

The activation energy of the forward reaction is so small (1.13 KJ.mol-1), the Hammond postulate is introduced. Hammond postulate is a hypothesis which states that the transition state of a reaction resembles either the reactants or the products, to whichever it is closer in energy. In an exothermic reaction, the TS is closer to the reactants than to the products in energy. Therefore, the TS resembles the reactant in the reaction. In an exothermic reaction, the TS is closer to the products than to the reactants in energy and therefore, the TS resembles the products.

In F-H-H system, the activation energy is very similar to the energy of the reactant and hence the TS resembles reactant in this case. Therefore, this reaction (F + H2 --> HF + H) is exothermic in forward direction and thus endothermic in backward direction.

In order to classify the reaction based on energies, the total energies of reactant and that of product should be identified first by using mep calculation (in order to exclude kinetic energy) starting with a small displacement from the TS and with a high enough number of steps. It will go towards the formation of either the products or the reactants and the potential energy will tend to a constant value which is the total energy of the product/reactants depending which direction the displacement was.

• mep calculation: number of steps = 5000

Since the transition state is found to be at rAB = rH-F = 182 pm and rBC = rH-H = 74.2 pm, we either change rAB or rBC to achieve the formation of either reactants or products. Here is an example:

The reaction reacts towards the products when initial conditions are:

• rAB = rH-F = 165 pm;

• rBC = rH-H = 74.2 pm;

The "contour plot" indicating the reaction direction clearly and "Energy vs time" diagram which shows a decrease (lower in energy) of potential energy (and also the total energy) from -434.194 to -560.402 KJ.mol-1 are illustrated below.

[[File:Surface Plot 165 74.2 MEP gjs.png|thumb|center|The contour plot of FHH system when rAB =165 pm and rBC = 74.2 pm, calculated with mep.]]

[[File:Contour Plot 165 74.2 MEP gjs.png|thumb|center|The Energy.vs.time plot of FHH system when rAB =165 pm and rBC = 74.2 pm, calculated with mep.]]

Based on this mep calculation, the total energies of reactant and that of product are calculated to be:

• reactant: E(H2) = -435.099 KJ.mol-1;

• product: E(HF) = -560.402 KJ.mol-1.

Consequentially, since the TS energy = -433.969 KJ.mol-1, the forward reaction (F + H2 --> HF + H) has an activation energy of [-433.969-(-435.099)]=1.13 KJ.mol-1; the backward reaction (HF + H --> F + H2）has an activation energy of [-433.969-(-560.402)]=126.433 KJ.mol-1. The latter is much larger than the former.

===Reaction dynamics===

Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Momenta vs Time”

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.

For example, when initial conditions fulfil rAB = 160 pm and 0 < rBC < 76 pm, the trajectory are reactive.

The released energy is concerted thermally into electromagnetic radiations (photons). The relaxed potential energy well of electrons have photon absorption and emission between ground state E(0) and the first excited state E(1) and most of the electrons are populated in ground state. While the excited potential energy well involves more possible higher excited states, such as E(0) to the second excited state E(2), E(1) to E(2) which is called overtone. The overtone will contribute to a second absorbance peak at lower wavenumbers in IR spectra.

The emission of photons can be detected by a special type of IR called a Quantum Well Infrared Photodetector (QWIP) which is an infrared photodetector.

Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration.

Setting initial conditions as:
• rHH = 74 pm;
• rFH = 180 pm;
• pFH = -1.0 g.mol-1.pm.fs-1;

The value of pHH is varied from -6.1 to +6.1 and the resulting "Internuclear Distance.vs.Time" diagams used to show the trajectories are displayed below:

[[File:Q2-gjs-rage-6.1-123.png|thumb|center|pHH= -6.1|300px]] [[File:Q2-gjs-rage-6-123.png|thumb|center|pHH= -6|300px]] [[File:Q2-gjs-rage-5-123.png|thumb|center|pHH= -5|300px]] [[File:Q2-gjs-rage-4-123.png|thumb|center|pHH= -4|300px]] [[File:Q2-gjs-rage-3-123.png|thumb|center|pHH= -3|300px]] [[File:Q2-gjs-rage-2-123.png|thumb|center|pHH= -2|300px]] [[File:Q2-gjs-rage-1-123.png|thumb|center|pHH= -1|300px]] [[File:Q2-gjs-rage-0-123.png|thumb|center|pHH= 0|300px]] [[File:Q2-gjs-rage+1-123.png|thumb|center|pHH= 1|300px]] [[File:Q2-gjs-rage+2-123.png|thumb|center|pHH= 2|300px]] [[File:Q2-gjs-rage+3-123.png|thumb|center|pHH= 3|300px]] [[File:Q2-gjs-rage+4-123.png|thumb|center|pHH= 4|300px]] [[File:Q2-gjs-rage+5-123.png|thumb|center|pHH= 5|300px]] [[File:Q2-gjs-rage+6-123.png|thumb|center|pHH= 6|300px]] [[File:Q2-gjs-rage+6.1-123.png|thumb|center|pHH= 6.1|300px]]

In general, the excess of energy put into the system gives rise to the significant increase in the vibrational energy of H-H. when pHH is negative, the smaller the value (more positive) the more barrier recrossing. When pHH is positive, in most cases there is one barrier recrossing only except when p=+1,+5,+6.1. Even though some barrier recrossing occurs (F atom collides with H atom for several times), it doesn't necessarily mean that the reaction will proceed to form the product HF and a separated atom H.

For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?

The new trajectory is showm in the "Internuclear Distance.vs.Time" diagams below. The vibrating molecule H-H moves towards F and collides with it, making the atom B (H atom) bouncing in-between for three times and then return back to the other H atom to reform the vibrating hydrogen molecule.
[[File:3Q-gjs-changed-initials.png|thumb|center|pHH= 0.2|300px]]

Let us now focus on the reverse reaction, H + HF.

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

As mentioned before, the relaxed potential energy well of electrons have photon absorptions and emissions between ground state E(0) and the first excited state E(1) and most of the electrons are populated in ground state. While the excited potential energy well involves more possible higher excited states, such as E(0) to the second excited state E(2), E(1) to E(2) which is called overtone. The overtone will contribute to a second absorbance peak at lower wavenumbers in IR spectra. Each energy state correspond to a translational energy while the vibrational levels are smaller in energy difference and located around translational energy states.

Polanyi's empirical rules states that vibrational energy is more efficient in prompting a late-barrier reaction (that is, a transition state resembling the products) than translational energy, whereas the reverse is true for an early barrier reaction.

Translational energy is better at promoting exothermic reactions, because the trajectory can just fall into the lower energy region. Vibrational motion might make it harder for the trajectory to cross the barrier because it is in a different direction than the mep. If the reactants have vibrational motion, then they will try to move up the valley near the transition state, and just fall back to the reactants state.

MRD：01508610

2020-05-29T17:25:50Z

Mak214: /* Dynamics from the transition state region */

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

We will be looking at collision and reaction between diatomic molecule of hydrogen and an atom of H in a linear configuration in the gas phase. In simple terms, the atom A collides with the molecule B-C and forms a new molecule with B, while C is detached as a separate atom.
The interatomic distances between A and B, B and C are labelled as rAB and rBC respectively. {{fontcolor1|red| Nice! I like these diagrams you have drawn. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:25, 29 May 2020 (BST)}}

[[File:initial_describe_collision.png]]

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

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 molecular geometry and chemical reaction dynamics are analysed with a potential energy surface (PES) where the necessary points can be evaluated and classified according to first and second derivatives of the energy with respect to position, V(r), which respectively are the gradient and the curvature. Stationary points are those with zero gradient (∂V(ri)/∂ri=0) and have physical meaning: energy minima (or local minima of the PES) correspond to physically stable chemical species (such as reactant and product) and saddle points representing transition states, the maximum on the lowest energy pathway (or minimum energy path) that linking reactant and product.

• Mathematically, a saddle point is a point on the surface of a function where the slopes in orthogonal directions are all zero, but which is not a local extremum of the function. The transition state is a first-order saddle point. In this H + H2 system, the surface is symmetrical, which means that the transition state must be exactly half way between the reactants and products. In another words, the transition state lies somewhere on the diagonal line where the distances rAB and rBC are equal.

• The hydrogen atoms will have no movement at the saddle point since the force on atoms (i.e. the negative derivative of PES with respect to the coordinate ri where the force acting on) will be zero at transition state. Consequently, if one starts a trajectory exactly at the transition state, with no initial momentum, it will remain there forever. Therefore, the method to identify the transition state position from a PES is to change the values of rAB and rBC until the forces along AB and BC are both zero. In fact, forces are negative on one side of the PES and positive one the other side. The transition state is located where forces across zero. {{fontcolor1|red| Great. Thank you. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:25, 29 May 2020 (BST)}}

===Trajectories from rAB = rBC: locating the transition state===

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

• By testing different initial conditions with rAB = rBC, and p1 = p2 = 0.0 g.mol-1.pm.fs-1, the transition state position (rts) is found to be the point where distances rAB and rBC are equal to 90.78 pm. At that position, forces are both -0.001 KJ.mol-1.pm-1 (close to zero).

• Since the H + H2 surface is symmetrical, the transition state must have rAB = rBC. If we start a trajectory on the ridge rAB = rBC there is no gradient in the direction at right angles to the ridge, thus the trajectory will oscillate on the ridge and never fall off.

[[File:threeatomcolliderabrbc.png|600px]]

• To verify the transition state position, “Internuclear Distances vs Time” plots for some relevant rAB = rBC trajectories are shown below. In these graphical display the values of rAB(t), rBC(t) and rAC(t) are given (Y axis) against time t (X axis). At transition state, distance rAB and rBC (and thus rAC) are constant since all three atoms are stationary at the saddle point. The potential energy is maximized while the kinetic energy is zero. This geometry is transition structure.

[[File:90pm_90pm_5000number_gjs.png|thumb|center|A “Internuclear Distances vs Time” plot at transition state.]]

• In contrast, when initial conditions increased to rAB = rBC =100 pm, it shows an oscillatory behaviour for the whole trajectory corresponding to the vibration of atom A and atom C, since there are both potential and kinetic energy. The distance rAB and rBC are overlapped, since they have the same amplitude of vibration and the vibrations are symmetrical around atom B. The distance rAC) has doubled the amplitude of vibration (i.e. the sum or superposition of the amplitude). The oscillatory behaviour is changed after the transition state, at t≈270 fs, to a dissociation pattern where rAB (and thus rAC) keeps increasing with a slight oscillatory behaviour while rBC remained at the same level with a higher vibrational frequency. This means that the atom A leaves away from the molecule B-C with the translational energy while molecule B-C is left with vibrational energy, which resembles the reactant.

[[File:Figure_2_100pm_internuc-time.png|thumb|center|A “Internuclear Distances vs Time” plot at rAB = rBC =100 pm.]]

===Trajectories from rBC = rts+δ, rAB = rts===

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

The reaction path (minimum energy path or mep) is calculated with these initial conditions involving:
• rBC = rts+1 pm = 91.78 pm;
• rAB = rts = 90.78 pm;
• pBC = pAB = 0 g.mol-1.pm.fs-1;
• then change the calculation type from dynamics to MEP;
• increase the number/size of steps to 5000/0.1(fs).

Then repeat the calculation with the same initial conditions used to calculate the reaction path, but change the calculation type back to Dynamics.

The calculated trajectories are shown in contour plots below, which both start from the transition state and simply follows the valley floor to HC + HA-HB. The mep is a smooth curve with no oscillatory behaviour, while the dynamic trajectory is an oscillatory curve.

[[File:Reaction_path_mep_jsg_.png|thumb|center|The contour plot of the reaction path calculated with mep.]]

[[File:Reaction_path_dynamic_jsg.png|thumb|center|The contour plot of the reaction path calculated with Dynamics.]]

The mep has the property that any point on the path is at an energy minimum in all directions perpendicular to the path. The mep can also be described as the union of steepest descent paths from the saddle points to the minima (that is why the mep passes through at least one first-order saddle point.)

From the “Internuclear Distances vs Time” diagrams below, it shows a constant increasing rate of rBC (and thus rAC) after the transition state in Dynamic trajectory, whereas it shows a reducing rate of the increase of the rBC (and thus rAC) in the mep. In addition, the molecule A-B has slight vibrational (or oscillatory) behaviour but has no vibration in the mep.

[[File:Animation mep change of rate gjs123.png|thumb|center|The “Internuclear Distances vs Time” plot of the mep with rBC = rts+1 pm = 91.78 pm.]]

[[File:multiple_graph_123_jsg.png|thumb|center|The “Internuclear Distances vs Time” plot of Dynamic trajectory with rBC = rts+1 pm = 91.78 pm.]]

As mentioned before, the variation of momentum p (the forces acting on a given interatomic coordinate ri) will depend on the derivative of the potential energy surface with respect to that coordinate. Simply, the momenta over time, p(t), is the gradient of V(r).

[[File:EquationInnedgjd.PNG|center]]

From the “Momenta vs Time” diagrams below, the momenta of all atoms are zero in the mep since the momenta/velocities are always reset to zero in each time step when calculating the mep. while it have various values in Dynamic trajectory: the momentum of A-B decreases to slightly negative values and then rises to ≈2.5 g.mol-1.pm.fs-1 and keeps oscillating; whereas the momentum of B-C, similarly, decreases to negative values and then increases dramatically to and remains at ≈5 g.mol-1.pm.fs-1.

[[File:Mep momentia.vs.time-1000 gjs.png|thumb|center|The “Momenta vs Time” plot of the mep with rBC = rts+1 pm = 91.78 pm.]]

[[File:Dynamics momenta.vs.time 1000 gjs.png|thumb|center|The “Momenta vs Time” plot of Dynamic trajectory with rBC = rts+1 pm = 91.78 pm.]]

If we change the initial conditions to:
• rBC = rts = 90.78 pm;
• rAB = rts + 1 pm = 91.78 pm.
The “Internuclear Distances vs Time” and “Momenta vs Time” plots will look like this:

[[File:Changedinterdis-time-MEP-5000-gjs.png|thumb|center|The “Internuclear Distances vs Time” plot of the mep when rAB = rts + 1 pm.]]

[[File:Multiple graph 123 jsg.png|thumb|center|The “Internuclear Distances vs Time” plot of Dynamic trajectory when rAB = rts + 1 pm.]]

The “Internuclear Distances vs Time”and “Momenta vs Time” plots of both the mep and Dynamic trajectory have exactly the same pattern as those plots before the initial conditions are changed, except that line A-B (representing rAB) and line B-C (representing rBC) are interconverted. This means that the reaction proceeds in the same way but opposite direction (towards the valley floor to HA + HB-HC).

[[File:Mep momentia.vs.time-1000 gjs.png|thumb|center|The “Momenta vs Time” plot of the mep when rAB = rts + 1 pm.]]

[[File:Changesmomentia-vs-time-Dynamics1000-gjs.png|thumb|center|The “Momenta vs Time” plot of Dynamic trajectory when rAB = rts + 1 pm.]]

Take note of the final values of the positions r1(t) r2(t) and p1(t) p2(t) for your trajectory for large enough t.

• step number = 5000

• step size = 0.1 fs

• rBC = 74.77117297476231 pm

• rAB = 3777.013307800053 pm

• pBC = 1.9461593911826354 g.mol-1.pm.fs-1

• pAB = 5.073830709923566 g.mol-1.pm.fs-1

The atom A is extremely far from the molecule B-C and keeps moving away. The energies are:
• kinetic energy = +19.657 KJ.mol-1

• potential energy = -434.999 KJ.mol-1

• total energy = -415.342 KJ.mol-1

Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. What do you observe?

The energy terms are the same as above, but in this case, the reverse sign of momenta results in the reserve of the moving direction of the free atom (atom A in this case). The atom A is very far from the molecule B-C and it keeps approaching the molecule B-C.

===Reactive and unreactive trajectories===

For the initial positions rBC = 74 pm and rAB = 200 pm, run trajectories with the following momenta combination and complete the table.
Complete the table below 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?

{| class="wikitable" border="1"
|+ caption
! pBC/ g.mol-1.pm.fs-1 !! pAB/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes || The reaction can occurred. The atom A approaches the molecule B-C and forms a new molecule A-B which leaves away with most vibrational energy and a small amount of kinetic energy (KE), while atom C is detached as a separate atom which moves away with most KE. || [[File:R1=-2.5;r2=-5.1 jsg lastday.png|thumb|200px]]
|-
| -3.1 || -4.1 || -420.077 || No || The reaction doesn't occur. The atom A approaches the molecule B-C and return back without colliding with the molecule. || [[File:-3.1and-4.1_jsg_lastday.png|thumb|200px]]
|-
| -3.1 || -5.1 || -413.977 || Yes || The reaction can occur. The atom A approaches the molecule B-C and forms a new molecule A-B which leaves away with most vibrational energy and a small amount of kinetic energy (KE), while atom C is detached as a separate atom which moves away with most KE. || [[File:-3.1and-5.1 jsg lastday.png|thumb|200px]]
|-
| -5.1 || -10.1 || -357.277 || No || A case of barrier recrossing.. The atom A approaches the molecule B-C and forms a temporary new molecule A-B with strong vibrational motion and the molecule A-B move back and collide with atom C to form the B-C again.|| [[File:-5.1and-10.1 jsg lastday.png|thumb|200px]]
|-
| -5.1 || -10.6 || -349.477 || Yes || The reaction can occur. The atom A approaches the molecule B-C and break B-C bond and makes atom B bouncing in-between and eventually form a new molecule A-B. || [[File:-5.1and-10.6 jsg+lastday.png|thumb|200px]]
|}

From this table, you can observe that some trajectories have a total energy greater than the activation energy and still do not react. Therefore, for the reaction to proceed, there are two conditions should be met:(1) the total energy is lager than the activation energy; (2)the interaction between atoms should be appropriate for the collision to happen with correct energy and direction. For instance, if the approaching atom A move opposite to the molecule B-C, there will be no collision and the reaction will definitely not occur.

* (The transition state is found to be at rAB= 230.0 pm; rBC=74.0 pm; forces along AB and AC = +0.015 and +0.012 respectively; KE = 0; PE = total E = -434.471.)

===Transition State Theory===

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

Transition state theory (TST) explains the reaction rates of elementary chemical reactions. The theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes. These are three properties of TST:

1.Rates of reaction can be studied by examining activated complexes near the saddle point of a potential energy surface. The details of how these complexes are formed are not important. The saddle point itself is called the transition state.

2.The activated complexes are in a special equilibrium (quasi-equilibrium) with the reactant molecules.

3.The activated complexes can convert into products, and kinetic theory can be used to calculate the rate of this conversion.

TST is used primarily to understand qualitatively how chemical reactions take place. Simply, the Transition State Theory (TST) tells whether a reaction will happen if cross the TS barrier.

So the difference here is that our program is working with a single molecule, the TST deals with a bulk system with a Boltzmann speed distribution. The main differences between experimental and predicted rate constant values will be due to (1) ignoring barrier recrossing and (2) ignoring QM behaviour.

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

F + H2 --> HF + H

A F atom (A) approaches a hydrogen molecule (B-C) to forms a new HF molecule (A-B) and a separate H atom (C).

===PES inspection===

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?

Locate the approximate position of the transition state.

In order to classify the reaction based on energies, the total energies of reactant and that of product should be identified first by using mep calculation (in order to exclude kinetic energy) starting with a small displacement from the TS and with a high enough number of steps. It will go towards the formation of either the products or the reactants and the potential energy will tend to a constant value which is the total energy of the product/reactants depending which direction the displacement was.

The total energies of reactant and that of product are calculated to be:

• reactant: E(H2) = -435.099 KJ.mol-1;

• product: E(HF) = -560.402 KJ.mol-1.

Since the energy of the product is lower than that of reactant, this is an exothermic reaction. This makes sense with the concept of bond strength in which the H-F bond (565 KJ.mol-1) is much stronger than H-H bond (432 KJ.mol-1) and thus there will be more energy released during the formation of H-F bond.

The transition state is still a maximum or a saddle point on the PES, and found to be at:

• rAB = rH-F = 182 pm;

• rBC = rH-H = 74.2 pm;

• pAB = pBC = 0 g.mol-1.pm.fs-1;

• force along AB = +0.003 ≈0 KJ.mol-1.pm-1;

• forces along BC = +0.089 ≈0 KJ.mol-1.pm-1;

• the total energy = -433.969 KJ.mol-1.

Because the activation energy for one of the reactions is so small, it is not easy to locate the transition state immediately. Use the Hammond postulate to guide your search.

Report the activation energy for both reactions.

The activation energy of the forward reaction is so small (1.13 KJ.mol-1), the Hammond postulate is introduced. Hammond postulate is a hypothesis which states that the transition state of a reaction resembles either the reactants or the products, to whichever it is closer in energy. In an exothermic reaction, the TS is closer to the reactants than to the products in energy. Therefore, the TS resembles the reactant in the reaction. In an exothermic reaction, the TS is closer to the products than to the reactants in energy and therefore, the TS resembles the products.

In F-H-H system, the activation energy is very similar to the energy of the reactant and hence the TS resembles reactant in this case. Therefore, this reaction (F + H2 --> HF + H) is exothermic in forward direction and thus endothermic in backward direction.

In order to classify the reaction based on energies, the total energies of reactant and that of product should be identified first by using mep calculation (in order to exclude kinetic energy) starting with a small displacement from the TS and with a high enough number of steps. It will go towards the formation of either the products or the reactants and the potential energy will tend to a constant value which is the total energy of the product/reactants depending which direction the displacement was.

• mep calculation: number of steps = 5000

Since the transition state is found to be at rAB = rH-F = 182 pm and rBC = rH-H = 74.2 pm, we either change rAB or rBC to achieve the formation of either reactants or products. Here is an example:

The reaction reacts towards the products when initial conditions are:

• rAB = rH-F = 165 pm;

• rBC = rH-H = 74.2 pm;

The "contour plot" indicating the reaction direction clearly and "Energy vs time" diagram which shows a decrease (lower in energy) of potential energy (and also the total energy) from -434.194 to -560.402 KJ.mol-1 are illustrated below.

[[File:Surface Plot 165 74.2 MEP gjs.png|thumb|center|The contour plot of FHH system when rAB =165 pm and rBC = 74.2 pm, calculated with mep.]]

[[File:Contour Plot 165 74.2 MEP gjs.png|thumb|center|The Energy.vs.time plot of FHH system when rAB =165 pm and rBC = 74.2 pm, calculated with mep.]]

Based on this mep calculation, the total energies of reactant and that of product are calculated to be:

• reactant: E(H2) = -435.099 KJ.mol-1;

• product: E(HF) = -560.402 KJ.mol-1.

Consequentially, since the TS energy = -433.969 KJ.mol-1, the forward reaction (F + H2 --> HF + H) has an activation energy of [-433.969-(-435.099)]=1.13 KJ.mol-1; the backward reaction (HF + H --> F + H2）has an activation energy of [-433.969-(-560.402)]=126.433 KJ.mol-1. The latter is much larger than the former.

===Reaction dynamics===

Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Momenta vs Time”

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.

For example, when initial conditions fulfil rAB = 160 pm and 0 < rBC < 76 pm, the trajectory are reactive.

The released energy is concerted thermally into electromagnetic radiations (photons). The relaxed potential energy well of electrons have photon absorption and emission between ground state E(0) and the first excited state E(1) and most of the electrons are populated in ground state. While the excited potential energy well involves more possible higher excited states, such as E(0) to the second excited state E(2), E(1) to E(2) which is called overtone. The overtone will contribute to a second absorbance peak at lower wavenumbers in IR spectra.

The emission of photons can be detected by a special type of IR called a Quantum Well Infrared Photodetector (QWIP) which is an infrared photodetector.

Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration.

Setting initial conditions as:
• rHH = 74 pm;
• rFH = 180 pm;
• pFH = -1.0 g.mol-1.pm.fs-1;

The value of pHH is varied from -6.1 to +6.1 and the resulting "Internuclear Distance.vs.Time" diagams used to show the trajectories are displayed below:

[[File:Q2-gjs-rage-6.1-123.png|thumb|center|pHH= -6.1|300px]] [[File:Q2-gjs-rage-6-123.png|thumb|center|pHH= -6|300px]] [[File:Q2-gjs-rage-5-123.png|thumb|center|pHH= -5|300px]] [[File:Q2-gjs-rage-4-123.png|thumb|center|pHH= -4|300px]] [[File:Q2-gjs-rage-3-123.png|thumb|center|pHH= -3|300px]] [[File:Q2-gjs-rage-2-123.png|thumb|center|pHH= -2|300px]] [[File:Q2-gjs-rage-1-123.png|thumb|center|pHH= -1|300px]] [[File:Q2-gjs-rage-0-123.png|thumb|center|pHH= 0|300px]] [[File:Q2-gjs-rage+1-123.png|thumb|center|pHH= 1|300px]] [[File:Q2-gjs-rage+2-123.png|thumb|center|pHH= 2|300px]] [[File:Q2-gjs-rage+3-123.png|thumb|center|pHH= 3|300px]] [[File:Q2-gjs-rage+4-123.png|thumb|center|pHH= 4|300px]] [[File:Q2-gjs-rage+5-123.png|thumb|center|pHH= 5|300px]] [[File:Q2-gjs-rage+6-123.png|thumb|center|pHH= 6|300px]] [[File:Q2-gjs-rage+6.1-123.png|thumb|center|pHH= 6.1|300px]]

In general, the excess of energy put into the system gives rise to the significant increase in the vibrational energy of H-H. when pHH is negative, the smaller the value (more positive) the more barrier recrossing. When pHH is positive, in most cases there is one barrier recrossing only except when p=+1,+5,+6.1. Even though some barrier recrossing occurs (F atom collides with H atom for several times), it doesn't necessarily mean that the reaction will proceed to form the product HF and a separated atom H.

For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?

The new trajectory is showm in the "Internuclear Distance.vs.Time" diagams below. The vibrating molecule H-H moves towards F and collides with it, making the atom B (H atom) bouncing in-between for three times and then return back to the other H atom to reform the vibrating hydrogen molecule.
[[File:3Q-gjs-changed-initials.png|thumb|center|pHH= 0.2|300px]]

Let us now focus on the reverse reaction, H + HF.

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

As mentioned before, the relaxed potential energy well of electrons have photon absorptions and emissions between ground state E(0) and the first excited state E(1) and most of the electrons are populated in ground state. While the excited potential energy well involves more possible higher excited states, such as E(0) to the second excited state E(2), E(1) to E(2) which is called overtone. The overtone will contribute to a second absorbance peak at lower wavenumbers in IR spectra. Each energy state correspond to a translational energy while the vibrational levels are smaller in energy difference and located around translational energy states.

Polanyi's empirical rules states that vibrational energy is more efficient in prompting a late-barrier reaction (that is, a transition state resembling the products) than translational energy, whereas the reverse is true for an early barrier reaction.

Translational energy is better at promoting exothermic reactions, because the trajectory can just fall into the lower energy region. Vibrational motion might make it harder for the trajectory to cross the barrier because it is in a different direction than the mep. If the reactants have vibrational motion, then they will try to move up the valley near the transition state, and just fall back to the reactants state.

MRD：01508610

2020-05-29T17:25:31Z

Mak214: /* Coordinates */

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

We will be looking at collision and reaction between diatomic molecule of hydrogen and an atom of H in a linear configuration in the gas phase. In simple terms, the atom A collides with the molecule B-C and forms a new molecule with B, while C is detached as a separate atom.
The interatomic distances between A and B, B and C are labelled as rAB and rBC respectively. {{fontcolor1|red| Nice! I like these diagrams you have drawn. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:25, 29 May 2020 (BST)}}

[[File:initial_describe_collision.png]]

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

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 molecular geometry and chemical reaction dynamics are analysed with a potential energy surface (PES) where the necessary points can be evaluated and classified according to first and second derivatives of the energy with respect to position, V(r), which respectively are the gradient and the curvature. Stationary points are those with zero gradient (∂V(ri)/∂ri=0) and have physical meaning: energy minima (or local minima of the PES) correspond to physically stable chemical species (such as reactant and product) and saddle points representing transition states, the maximum on the lowest energy pathway (or minimum energy path) that linking reactant and product.

• Mathematically, a saddle point is a point on the surface of a function where the slopes in orthogonal directions are all zero, but which is not a local extremum of the function. The transition state is a first-order saddle point. In this H + H2 system, the surface is symmetrical, which means that the transition state must be exactly half way between the reactants and products. In another words, the transition state lies somewhere on the diagonal line where the distances rAB and rBC are equal.

• The hydrogen atoms will have no movement at the saddle point since the force on atoms (i.e. the negative derivative of PES with respect to the coordinate ri where the force acting on) will be zero at transition state. Consequently, if one starts a trajectory exactly at the transition state, with no initial momentum, it will remain there forever. Therefore, the method to identify the transition state position from a PES is to change the values of rAB and rBC until the forces along AB and BC are both zero. In fact, forces are negative on one side of the PES and positive one the other side. The transition state is located where forces across zero.

===Trajectories from rAB = rBC: locating the transition state===

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

• By testing different initial conditions with rAB = rBC, and p1 = p2 = 0.0 g.mol-1.pm.fs-1, the transition state position (rts) is found to be the point where distances rAB and rBC are equal to 90.78 pm. At that position, forces are both -0.001 KJ.mol-1.pm-1 (close to zero).

• Since the H + H2 surface is symmetrical, the transition state must have rAB = rBC. If we start a trajectory on the ridge rAB = rBC there is no gradient in the direction at right angles to the ridge, thus the trajectory will oscillate on the ridge and never fall off.

[[File:threeatomcolliderabrbc.png|600px]]

• To verify the transition state position, “Internuclear Distances vs Time” plots for some relevant rAB = rBC trajectories are shown below. In these graphical display the values of rAB(t), rBC(t) and rAC(t) are given (Y axis) against time t (X axis). At transition state, distance rAB and rBC (and thus rAC) are constant since all three atoms are stationary at the saddle point. The potential energy is maximized while the kinetic energy is zero. This geometry is transition structure.

[[File:90pm_90pm_5000number_gjs.png|thumb|center|A “Internuclear Distances vs Time” plot at transition state.]]

• In contrast, when initial conditions increased to rAB = rBC =100 pm, it shows an oscillatory behaviour for the whole trajectory corresponding to the vibration of atom A and atom C, since there are both potential and kinetic energy. The distance rAB and rBC are overlapped, since they have the same amplitude of vibration and the vibrations are symmetrical around atom B. The distance rAC) has doubled the amplitude of vibration (i.e. the sum or superposition of the amplitude). The oscillatory behaviour is changed after the transition state, at t≈270 fs, to a dissociation pattern where rAB (and thus rAC) keeps increasing with a slight oscillatory behaviour while rBC remained at the same level with a higher vibrational frequency. This means that the atom A leaves away from the molecule B-C with the translational energy while molecule B-C is left with vibrational energy, which resembles the reactant.

[[File:Figure_2_100pm_internuc-time.png|thumb|center|A “Internuclear Distances vs Time” plot at rAB = rBC =100 pm.]]

===Trajectories from rBC = rts+δ, rAB = rts===

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

The reaction path (minimum energy path or mep) is calculated with these initial conditions involving:
• rBC = rts+1 pm = 91.78 pm;
• rAB = rts = 90.78 pm;
• pBC = pAB = 0 g.mol-1.pm.fs-1;
• then change the calculation type from dynamics to MEP;
• increase the number/size of steps to 5000/0.1(fs).

Then repeat the calculation with the same initial conditions used to calculate the reaction path, but change the calculation type back to Dynamics.

The calculated trajectories are shown in contour plots below, which both start from the transition state and simply follows the valley floor to HC + HA-HB. The mep is a smooth curve with no oscillatory behaviour, while the dynamic trajectory is an oscillatory curve.

[[File:Reaction_path_mep_jsg_.png|thumb|center|The contour plot of the reaction path calculated with mep.]]

[[File:Reaction_path_dynamic_jsg.png|thumb|center|The contour plot of the reaction path calculated with Dynamics.]]

The mep has the property that any point on the path is at an energy minimum in all directions perpendicular to the path. The mep can also be described as the union of steepest descent paths from the saddle points to the minima (that is why the mep passes through at least one first-order saddle point.)

From the “Internuclear Distances vs Time” diagrams below, it shows a constant increasing rate of rBC (and thus rAC) after the transition state in Dynamic trajectory, whereas it shows a reducing rate of the increase of the rBC (and thus rAC) in the mep. In addition, the molecule A-B has slight vibrational (or oscillatory) behaviour but has no vibration in the mep.

[[File:Animation mep change of rate gjs123.png|thumb|center|The “Internuclear Distances vs Time” plot of the mep with rBC = rts+1 pm = 91.78 pm.]]

[[File:multiple_graph_123_jsg.png|thumb|center|The “Internuclear Distances vs Time” plot of Dynamic trajectory with rBC = rts+1 pm = 91.78 pm.]]

As mentioned before, the variation of momentum p (the forces acting on a given interatomic coordinate ri) will depend on the derivative of the potential energy surface with respect to that coordinate. Simply, the momenta over time, p(t), is the gradient of V(r).

[[File:EquationInnedgjd.PNG|center]]

From the “Momenta vs Time” diagrams below, the momenta of all atoms are zero in the mep since the momenta/velocities are always reset to zero in each time step when calculating the mep. while it have various values in Dynamic trajectory: the momentum of A-B decreases to slightly negative values and then rises to ≈2.5 g.mol-1.pm.fs-1 and keeps oscillating; whereas the momentum of B-C, similarly, decreases to negative values and then increases dramatically to and remains at ≈5 g.mol-1.pm.fs-1.

[[File:Mep momentia.vs.time-1000 gjs.png|thumb|center|The “Momenta vs Time” plot of the mep with rBC = rts+1 pm = 91.78 pm.]]

[[File:Dynamics momenta.vs.time 1000 gjs.png|thumb|center|The “Momenta vs Time” plot of Dynamic trajectory with rBC = rts+1 pm = 91.78 pm.]]

If we change the initial conditions to:
• rBC = rts = 90.78 pm;
• rAB = rts + 1 pm = 91.78 pm.
The “Internuclear Distances vs Time” and “Momenta vs Time” plots will look like this:

[[File:Changedinterdis-time-MEP-5000-gjs.png|thumb|center|The “Internuclear Distances vs Time” plot of the mep when rAB = rts + 1 pm.]]

[[File:Multiple graph 123 jsg.png|thumb|center|The “Internuclear Distances vs Time” plot of Dynamic trajectory when rAB = rts + 1 pm.]]

The “Internuclear Distances vs Time”and “Momenta vs Time” plots of both the mep and Dynamic trajectory have exactly the same pattern as those plots before the initial conditions are changed, except that line A-B (representing rAB) and line B-C (representing rBC) are interconverted. This means that the reaction proceeds in the same way but opposite direction (towards the valley floor to HA + HB-HC).

[[File:Mep momentia.vs.time-1000 gjs.png|thumb|center|The “Momenta vs Time” plot of the mep when rAB = rts + 1 pm.]]

[[File:Changesmomentia-vs-time-Dynamics1000-gjs.png|thumb|center|The “Momenta vs Time” plot of Dynamic trajectory when rAB = rts + 1 pm.]]

Take note of the final values of the positions r1(t) r2(t) and p1(t) p2(t) for your trajectory for large enough t.

• step number = 5000

• step size = 0.1 fs

• rBC = 74.77117297476231 pm

• rAB = 3777.013307800053 pm

• pBC = 1.9461593911826354 g.mol-1.pm.fs-1

• pAB = 5.073830709923566 g.mol-1.pm.fs-1

The atom A is extremely far from the molecule B-C and keeps moving away. The energies are:
• kinetic energy = +19.657 KJ.mol-1

• potential energy = -434.999 KJ.mol-1

• total energy = -415.342 KJ.mol-1

Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. What do you observe?

The energy terms are the same as above, but in this case, the reverse sign of momenta results in the reserve of the moving direction of the free atom (atom A in this case). The atom A is very far from the molecule B-C and it keeps approaching the molecule B-C.

===Reactive and unreactive trajectories===

For the initial positions rBC = 74 pm and rAB = 200 pm, run trajectories with the following momenta combination and complete the table.
Complete the table below 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?

{| class="wikitable" border="1"
|+ caption
! pBC/ g.mol-1.pm.fs-1 !! pAB/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes || The reaction can occurred. The atom A approaches the molecule B-C and forms a new molecule A-B which leaves away with most vibrational energy and a small amount of kinetic energy (KE), while atom C is detached as a separate atom which moves away with most KE. || [[File:R1=-2.5;r2=-5.1 jsg lastday.png|thumb|200px]]
|-
| -3.1 || -4.1 || -420.077 || No || The reaction doesn't occur. The atom A approaches the molecule B-C and return back without colliding with the molecule. || [[File:-3.1and-4.1_jsg_lastday.png|thumb|200px]]
|-
| -3.1 || -5.1 || -413.977 || Yes || The reaction can occur. The atom A approaches the molecule B-C and forms a new molecule A-B which leaves away with most vibrational energy and a small amount of kinetic energy (KE), while atom C is detached as a separate atom which moves away with most KE. || [[File:-3.1and-5.1 jsg lastday.png|thumb|200px]]
|-
| -5.1 || -10.1 || -357.277 || No || A case of barrier recrossing.. The atom A approaches the molecule B-C and forms a temporary new molecule A-B with strong vibrational motion and the molecule A-B move back and collide with atom C to form the B-C again.|| [[File:-5.1and-10.1 jsg lastday.png|thumb|200px]]
|-
| -5.1 || -10.6 || -349.477 || Yes || The reaction can occur. The atom A approaches the molecule B-C and break B-C bond and makes atom B bouncing in-between and eventually form a new molecule A-B. || [[File:-5.1and-10.6 jsg+lastday.png|thumb|200px]]
|}

From this table, you can observe that some trajectories have a total energy greater than the activation energy and still do not react. Therefore, for the reaction to proceed, there are two conditions should be met:(1) the total energy is lager than the activation energy; (2)the interaction between atoms should be appropriate for the collision to happen with correct energy and direction. For instance, if the approaching atom A move opposite to the molecule B-C, there will be no collision and the reaction will definitely not occur.

* (The transition state is found to be at rAB= 230.0 pm; rBC=74.0 pm; forces along AB and AC = +0.015 and +0.012 respectively; KE = 0; PE = total E = -434.471.)

===Transition State Theory===

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

Transition state theory (TST) explains the reaction rates of elementary chemical reactions. The theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes. These are three properties of TST:

1.Rates of reaction can be studied by examining activated complexes near the saddle point of a potential energy surface. The details of how these complexes are formed are not important. The saddle point itself is called the transition state.

2.The activated complexes are in a special equilibrium (quasi-equilibrium) with the reactant molecules.

3.The activated complexes can convert into products, and kinetic theory can be used to calculate the rate of this conversion.

TST is used primarily to understand qualitatively how chemical reactions take place. Simply, the Transition State Theory (TST) tells whether a reaction will happen if cross the TS barrier.

So the difference here is that our program is working with a single molecule, the TST deals with a bulk system with a Boltzmann speed distribution. The main differences between experimental and predicted rate constant values will be due to (1) ignoring barrier recrossing and (2) ignoring QM behaviour.

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

F + H2 --> HF + H

A F atom (A) approaches a hydrogen molecule (B-C) to forms a new HF molecule (A-B) and a separate H atom (C).

===PES inspection===

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?

Locate the approximate position of the transition state.

In order to classify the reaction based on energies, the total energies of reactant and that of product should be identified first by using mep calculation (in order to exclude kinetic energy) starting with a small displacement from the TS and with a high enough number of steps. It will go towards the formation of either the products or the reactants and the potential energy will tend to a constant value which is the total energy of the product/reactants depending which direction the displacement was.

The total energies of reactant and that of product are calculated to be:

• reactant: E(H2) = -435.099 KJ.mol-1;

• product: E(HF) = -560.402 KJ.mol-1.

Since the energy of the product is lower than that of reactant, this is an exothermic reaction. This makes sense with the concept of bond strength in which the H-F bond (565 KJ.mol-1) is much stronger than H-H bond (432 KJ.mol-1) and thus there will be more energy released during the formation of H-F bond.

The transition state is still a maximum or a saddle point on the PES, and found to be at:

• rAB = rH-F = 182 pm;

• rBC = rH-H = 74.2 pm;

• pAB = pBC = 0 g.mol-1.pm.fs-1;

• force along AB = +0.003 ≈0 KJ.mol-1.pm-1;

• forces along BC = +0.089 ≈0 KJ.mol-1.pm-1;

• the total energy = -433.969 KJ.mol-1.

Because the activation energy for one of the reactions is so small, it is not easy to locate the transition state immediately. Use the Hammond postulate to guide your search.

Report the activation energy for both reactions.

The activation energy of the forward reaction is so small (1.13 KJ.mol-1), the Hammond postulate is introduced. Hammond postulate is a hypothesis which states that the transition state of a reaction resembles either the reactants or the products, to whichever it is closer in energy. In an exothermic reaction, the TS is closer to the reactants than to the products in energy. Therefore, the TS resembles the reactant in the reaction. In an exothermic reaction, the TS is closer to the products than to the reactants in energy and therefore, the TS resembles the products.

In F-H-H system, the activation energy is very similar to the energy of the reactant and hence the TS resembles reactant in this case. Therefore, this reaction (F + H2 --> HF + H) is exothermic in forward direction and thus endothermic in backward direction.

In order to classify the reaction based on energies, the total energies of reactant and that of product should be identified first by using mep calculation (in order to exclude kinetic energy) starting with a small displacement from the TS and with a high enough number of steps. It will go towards the formation of either the products or the reactants and the potential energy will tend to a constant value which is the total energy of the product/reactants depending which direction the displacement was.

• mep calculation: number of steps = 5000

Since the transition state is found to be at rAB = rH-F = 182 pm and rBC = rH-H = 74.2 pm, we either change rAB or rBC to achieve the formation of either reactants or products. Here is an example:

The reaction reacts towards the products when initial conditions are:

• rAB = rH-F = 165 pm;

• rBC = rH-H = 74.2 pm;

The "contour plot" indicating the reaction direction clearly and "Energy vs time" diagram which shows a decrease (lower in energy) of potential energy (and also the total energy) from -434.194 to -560.402 KJ.mol-1 are illustrated below.

[[File:Surface Plot 165 74.2 MEP gjs.png|thumb|center|The contour plot of FHH system when rAB =165 pm and rBC = 74.2 pm, calculated with mep.]]

[[File:Contour Plot 165 74.2 MEP gjs.png|thumb|center|The Energy.vs.time plot of FHH system when rAB =165 pm and rBC = 74.2 pm, calculated with mep.]]

Based on this mep calculation, the total energies of reactant and that of product are calculated to be:

• reactant: E(H2) = -435.099 KJ.mol-1;

• product: E(HF) = -560.402 KJ.mol-1.

Consequentially, since the TS energy = -433.969 KJ.mol-1, the forward reaction (F + H2 --> HF + H) has an activation energy of [-433.969-(-435.099)]=1.13 KJ.mol-1; the backward reaction (HF + H --> F + H2）has an activation energy of [-433.969-(-560.402)]=126.433 KJ.mol-1. The latter is much larger than the former.

===Reaction dynamics===

Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Momenta vs Time”

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.

For example, when initial conditions fulfil rAB = 160 pm and 0 < rBC < 76 pm, the trajectory are reactive.

The released energy is concerted thermally into electromagnetic radiations (photons). The relaxed potential energy well of electrons have photon absorption and emission between ground state E(0) and the first excited state E(1) and most of the electrons are populated in ground state. While the excited potential energy well involves more possible higher excited states, such as E(0) to the second excited state E(2), E(1) to E(2) which is called overtone. The overtone will contribute to a second absorbance peak at lower wavenumbers in IR spectra.

The emission of photons can be detected by a special type of IR called a Quantum Well Infrared Photodetector (QWIP) which is an infrared photodetector.

Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 74 pm, with a momentum pFH = -1.0 g.mol-1.pm.fs-1, and explore several values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration.

Setting initial conditions as:
• rHH = 74 pm;
• rFH = 180 pm;
• pFH = -1.0 g.mol-1.pm.fs-1;

The value of pHH is varied from -6.1 to +6.1 and the resulting "Internuclear Distance.vs.Time" diagams used to show the trajectories are displayed below:

[[File:Q2-gjs-rage-6.1-123.png|thumb|center|pHH= -6.1|300px]] [[File:Q2-gjs-rage-6-123.png|thumb|center|pHH= -6|300px]] [[File:Q2-gjs-rage-5-123.png|thumb|center|pHH= -5|300px]] [[File:Q2-gjs-rage-4-123.png|thumb|center|pHH= -4|300px]] [[File:Q2-gjs-rage-3-123.png|thumb|center|pHH= -3|300px]] [[File:Q2-gjs-rage-2-123.png|thumb|center|pHH= -2|300px]] [[File:Q2-gjs-rage-1-123.png|thumb|center|pHH= -1|300px]] [[File:Q2-gjs-rage-0-123.png|thumb|center|pHH= 0|300px]] [[File:Q2-gjs-rage+1-123.png|thumb|center|pHH= 1|300px]] [[File:Q2-gjs-rage+2-123.png|thumb|center|pHH= 2|300px]] [[File:Q2-gjs-rage+3-123.png|thumb|center|pHH= 3|300px]] [[File:Q2-gjs-rage+4-123.png|thumb|center|pHH= 4|300px]] [[File:Q2-gjs-rage+5-123.png|thumb|center|pHH= 5|300px]] [[File:Q2-gjs-rage+6-123.png|thumb|center|pHH= 6|300px]] [[File:Q2-gjs-rage+6.1-123.png|thumb|center|pHH= 6.1|300px]]

In general, the excess of energy put into the system gives rise to the significant increase in the vibrational energy of H-H. when pHH is negative, the smaller the value (more positive) the more barrier recrossing. When pHH is positive, in most cases there is one barrier recrossing only except when p=+1,+5,+6.1. Even though some barrier recrossing occurs (F atom collides with H atom for several times), it doesn't necessarily mean that the reaction will proceed to form the product HF and a separated atom H.

For the same initial position, increase slightly the momentum pFH = -1.6 g.mol-1.pm.fs-1, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.2 g.mol-1.pm.fs-1. What do you observe now?

The new trajectory is showm in the "Internuclear Distance.vs.Time" diagams below. The vibrating molecule H-H moves towards F and collides with it, making the atom B (H atom) bouncing in-between for three times and then return back to the other H atom to reform the vibrating hydrogen molecule.
[[File:3Q-gjs-changed-initials.png|thumb|center|pHH= 0.2|300px]]

Let us now focus on the reverse reaction, H + HF.

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

As mentioned before, the relaxed potential energy well of electrons have photon absorptions and emissions between ground state E(0) and the first excited state E(1) and most of the electrons are populated in ground state. While the excited potential energy well involves more possible higher excited states, such as E(0) to the second excited state E(2), E(1) to E(2) which is called overtone. The overtone will contribute to a second absorbance peak at lower wavenumbers in IR spectra. Each energy state correspond to a translational energy while the vibrational levels are smaller in energy difference and located around translational energy states.

Polanyi's empirical rules states that vibrational energy is more efficient in prompting a late-barrier reaction (that is, a transition state resembling the products) than translational energy, whereas the reverse is true for an early barrier reaction.

Translational energy is better at promoting exothermic reactions, because the trajectory can just fall into the lower energy region. Vibrational motion might make it harder for the trajectory to cross the barrier because it is in a different direction than the mep. If the reactants have vibrational motion, then they will try to move up the valley near the transition state, and just fall back to the reactants state.

MRD:ql2018

2020-05-29T17:23:44Z

Mak214: /* Exercise 1- (H+H2 System). */

== Molecular Reaction Dynamics Lab ==

=== Exercise 1- (H+H2 System). ===
{{fontcolor|blue|Q1ː 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?}}
A transition state can be found as a saddle point on the potential energy surface diagram. Mathematically, the transition state is defined as the maximum on the minimum energy path linking reactants and the products. To distinguish the transition state from local minima,this can be done by viewing the potential energy surface from different angles. The transition state is the only point that is a minimum point from one angle(Figure 1)but a maximum from second angle(Figure 2). {{fontcolor1|red| IOK. Could you use some equations to define this also? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:23, 29 May 2020 (BST)}}
{{multiple image
| align = center
| direction = vertical
| width = 400
| header = Transition State at maximum and minimum
| image1 = Surface_Plot1_qw.png
| caption1 = Transition state is a maximum from one angle.
| image2 = Surface_Plot2_qw.png
| caption2 = Transition state is a minimum from another angle.
}}

{{fontcolor|blue|Q1ː 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.}}
My estimation for the transition state position rts is 90.7pm.Because for the transition state, the potential energies of both p1 and p2 are zero.There is no change of the energy which means that the gradients of potential energy surface is zero and there is no force acting on atoms.So r1 and r2 will keep constant.The graph corresponds should be a straight line since there is no oscillation. {{fontcolor1|red| Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:23, 29 May 2020 (BST)}}

[[File:Surface_Plot3_qw.png|thumb|center|Internuclear distances against time graph at r1=r2=90.7pm with zero momenta.]]

{{fontcolor|blue|Q3ːComment on how the mep and the trajectory you just calculated differ.}}
As the two graphs shown below, the difference is obvious that the trajectory on the contour plot of mep is shorter than that of dynamics. And there is also no oscillation compared with the graph generated by dynamics. The reason for this is because that there is zero kinetic energy for mep due to zero velocity and momentum, there will be no gain in vibtational energy as a result. Therefore the trajectory shows no oscillation. Also the change in total energy is different. Since there is no gain in kinetic energy, the total energy will drop as potential energy losses. {{fontcolor1|red| Yes. in MEP the momenta are reset to zero at every timestep so the system falls into the nearest local minimum. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:23, 29 May 2020 (BST)}}

For dynamics, the total energy is conserved as there is gain in kinetic energy when H2 formed.And due to this bond forming, the vibrational energy results in the oscillations on the contour plot.

{{multiple image
| align = left
| direction = vertical
| width = 300
| header = MEP and Dynamics
| image1 = Surface_Plot4_qw.png
| caption1 = MEP contour plot
| image2 = Surface_Plot5_qw.png
| caption2 = Dynamics contour plot
}}

{{fontcolor|blue|Q4ːComplete 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?}}

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot/ kJ.mol-1!! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes || This reaction started with small AB distance and large BC distance. It has enough momentum to pass the transition state region. After that region, the bond between BC formed and the bond between AB broken. The trajectry shows no fluctuation before BC bond formed, which is because the energy is mostly transitional energy. When BC bond formed, the energy is mostly vibrational energy which causes oscillation. ||[[File:Surface_Plot6_qw.png|thumb|upright=0.8]]
|-
| -3.1 || -4.1 || -420.077 || No || This reaction started with small AB distance and large BC distance. However, the momentum is not large enough to pass the transition state region. There is no break in bond between AB and no form in bond between BC. Molecule AB moves away forom C and the bond between AB keeps vibrating due to the kinetic energy. ||[[File:Surface_Plot7_qw.png|thumb|upright=0.8]]
|-
| -3.1 || -5.1 || -413.977 || Yes || This time the BC momentum is large enough for the system to pass transition state. AB bond vibrates before bond breaking due to kinetic energy. The vibration is even larger for BC due to larger momentum after the bond forming. ||[[File:Surface_Plot8_qw.png|thumb|upright=0.8]]
|-
| -5.1 || -10.1 || -357.277 || No || The energy is too large for this system. Even the AB bond breaks and the trajectory passes the transition state region, the bond formed between BC vibrates too strongly due to large energy that the bond fromed eventually breaks. The trajectory goes back and recrosses the transition state region. AB forms a bond again and as AB moves away from C, there is no vibration between AB bond. ||[[File:Surface_Plot9_qw.png|thumb|upright=0.8]]
|-
| -5.1 || -10.6 || -349.477 || Yes || The energy for this system is also very large that the trajectory passes through the transition state region for a couple of times. Although the bond formed between BC after the trajectory passes through the transition state region for the first time. The bond of BC breaks due to large energy and recrosses the transition state. However, because the system has more energy than the previous one.it had the energy to collide for a third time, the bond between BC forms again and became stable as BC vibrates and moves away from A. ||[[File:Surface_Plot10_qw.png|thumb|upright=0.8]]
|}

{{fontcolor1|red| Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:23, 29 May 2020 (BST)}}
{{fontcolor|blue|Q5ːGiven the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?}}

The transition state theory assumes that once the reactants have enough kinetic energy to pass through the transition state. The product must formed and the reaction would not go to reverse. However, the given experimental results indicate that it was not true. The products can recross the transition state and reform reactants. As a result, the transition state theory overestimates the reaction rate compared to experimental results. Also, due to the quantum tunneling effects, atoms with energy lower than activation energy can still pass through the barrier, which leads to a underestimation of reaction rate.

== Exercise 2ː F - H - H System ==
{{fontcolor|blue|Q6ː 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?}}

According to the potential energy surface graph F, H2 to H and HFis an exothermic reaction since it has higher potential energy for F and H2. Therefore H + HF to F, H2 is an endothermic reaction. This shows that the H-F bond is stronger than H-H bond. Because the formation of H-F bond releases more energy than the breaking of H-H bond.

[[File:Surface_Plot11_qw.png|thumb|center|Potential energy surface of F-H-H system]]

{{fontcolor|blue|Q7ː Locate the approximate position of the transition state.}}
The approximate position of transition state for this reaction is when F-H isr1 = 181.300 pm and H-H isr2 = 74.483 pm. The momentum at this point is zero.
[[File:Surface_Plot12_qw.png|thumb|center|approximate position of the transition state of F-H-H system]]

{{fontcolor|blue|Q8ː Report the activation energy for both reactions.}}
The potential energy of the reactants F + H2 is -434.625 kJ.mol-1. The potential energy of the transition state is -433.981 kJ.mol-1. The activation energy is equal to the energy of transition state minus the energy of the reactants, which is 0.644 kJ.mol-1.

The potential energy of the products H + HF is -556.231 kJ.mol-1. The potential energy of the transition state is -433.981 kJ.mol-1. The activation energy of this reverse reaction is equal to the energy of transition state minus the energy of the products, which is 122.25 kJ.mol-1.

{{multiple image
| align = center
| direction = vertical
| width = 300
| header = Activation Energies calculation
| image1 = Surface_Plot13_qw.png
| caption1 = Energy against time r1 = 184 pm, r2 = 74.483 pm
| image2 = Surface_Plot14_qw.png
| caption2 = Energy against time r1 = 176.000 pm, r2 = 74.483 pm
}}

{{fontcolor|blue|Q9ːIn light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.}}

The conditions are set that the the bond distance is 184pm between F and H and the bond distance is 75 between H and H. p1=-1 and p2= -2. From the graph, it can be seen that the product has greater vibration. This is because the potential energy transfer to kinetic energy and transitional energy.

The energy transfer can be confirmed by using IR spectroscopy. Because the vibration of HF bond can be detected by IR. Calorimetry can also be used to detect the heat generated during the reaction. However, this heat includes both vibrational and transitional energy.

{| class="wikitable" border="1"
! Momentum vs Time
|- !! Contour Plot
|-
| [[File:Surface_Plot15_qw.png|400px|thumb|left]] || [[File:Surface_Plot16_qw.png|400px|thumb|right]]
|}
{{fontcolor|blue|Q10ːDiscuss 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.}}

From Polanyi's rule, It is known that the transitional energy can be more efficient for a exothermic reaction as it helps the reactants to pass through early transition state. In endothermic reaction, it has late transition state that the vibrational energy is more effective. It helps the reactants to pass the late transition state.1

==Reference==
1.J.C. Polanyi. Some Concepts in Reaction Dynamics. Science 1987, 236 (4802): 680-690. doi:10.1126/science.236.4802.680

MRD:01493832

2020-05-29T17:20:25Z

Mak214: /* Calculating the reaction path */

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

== The transition state ==
On a potential energy surface diagram, the transition state is mathematically defined as a stationary point where the derivative of the potential energy curve, which is a force {{fontcolor1|red| I'm not sure exactly what you mean here? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:17, 29 May 2020 (BST)}}, is zero. This means that there are no movement at the transition state.

The transition state is different from a local minimum of a potential energy surface in a sense that it is a minimum in one direction yet a maximum in another.1 Therefore, it is a stationary point but not a point of inflection. From the calculated forces, we are able to confirm that this is a stationary point and from the given corresponding eigenvalues/vectors, the positive and negative combination confirms that this is a saddle point. If we move away from the stationary point, the eigenvalues/vectors will be either all positive or all negative. (For a local minimum, the eigenvalues/vectors will be positive in all directions.) {{fontcolor1|red| IOK. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:17, 29 May 2020 (BST)}}

=== H + H2 : Locating the transition state ===
In the case of H + H2, AB and BC distances must be equal in the transition state because all atoms are identical so the transition state is symmetrical and its position can be identified by using a trial and error process, where the AB and BC distances in the initial conditions are varied until the corresponding forces become as close to zero as possible. For 3 Hydrogen atoms, AB and BC distances (rts) are found to be 90.775 pm.

[[File:Nl3818-H TS dist-time.jpg|200px|thumb|left|Fig. 1 - Internuclear Distances vs Time plot for 3 Hydrogen atoms]]

From Fig.1, it is evidently clear that at rts = 90.775 pm, the distances are constant throughout confirming that it is a stationary point (i.e. the nuclei do not move) and that distances AB = BC. {{fontcolor1|red| I Good. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:17, 29 May 2020 (BST)}}

== Calculating the reaction path ==
There are two types of calculation that can be done when considering the trajectory of the Hydrogen molecule: MEP (minimum energy path) and Dynamics. In MEP, atoms are assumed to be in an infinitely slow motion which corresponds to zero momenta at each time step. {{fontcolor1|red| Good. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:20, 29 May 2020 (BST)}}Both calculations can be done by changing the initial conditions to r1 = rts+δ and r2=rts in order to visualise the trajectories. The difference between MEP and Dynamics are shown in two contour plots below. From the MEP contour plot, we can see that the black trajectory line is rather smooth as it approaches the transition state position marked with a red cross compared to the trajectory shown in Dynamics contour plot. The MEP calculated trajectory also starts off at a higher potential energy position than the one from Dynamics calculation. {{fontcolor1|red| Hmm.. they should start with the same energy if you put in the same settings. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:20, 29 May 2020 (BST)}} Because in MEP calculation, atoms are moving very slowly so the vibration between the reactant HA-HB is so small that it does not display an oscillating behaviour as in the Dynamics calculation. {{fontcolor1|red| Not exactly, the MEP calculation there should be no vibration of the bond whatsoever since there is no momentum to move the atoms away from the equilibrium bond distance which is a valley in the potential energy surface. We only see vibration in the dynamics type calculation because there is residual momentum which allows the atoms in the molecule to vibrate. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:20, 29 May 2020 (BST)}}

{| class="wikitable" border=1
|-
|[[File:Dynamic_nl3818.png|200px|thumb|left|Fig.2 - Contour plot of triatomic Hydrogen system using Dynamics calculation]]||
[[File:MEP_nl3818.png|200px|thumb|left|Fig.3 - Contour plot of triatomic Hydrogen system using MEP calculation]]
|}

=== Change in initial conditions ===
If we change the initial condition so that r1 = rts and r2=rts+δ instead, the result is still identical to previous result but with atom A moving away at large t and atom B bonding with C instead. By investigating the “Internuclear Distances vs Time” and “Momenta vs Time” plots and taking note of values for r1(t) r2(t) and p1(t) p2(t) at very large t (50 fs), another set of calculations can be done with using these values for the initial positions and momentum (with reversed sign). The observation is for “Internuclear Distances vs Time” plot, the result is a reflection in the y-axis of Fig. 4 meaning, the bonded BC H2 molecule and the previously isolated HA atom are now approaching each other and their positions at large t corresponds to the transition state position. With reversed sign of the momenta, this is a reverse process of what was previously done, instead of atoms moving apart, they are now moving towards each other with the same kinetic energy. The “Momenta vs Time” is now reflected in the x-axis of Fig. 5, which can be reasoned by the repulsion that the atoms experienced from coming together resulting in positive gradient which is gradual at first then steepens at high t as the three atoms are now very close.

{| class="wikitable" border=1
|-
|[[File:Dtnl3818new.png|200px|thumb|left|Fig.4 - Dist. vs Time plot of H atoms moving apart]]||
[[File:Momentanewnl3818.png|200px|thumb|left|Fig.5 - Momenta vs Time plot of H atoms moving apart]]
|}

== Reactive and unreactive trajectories ==

To test whether trajectories starting at the same position but with higher momenta will all react, a table has been constructed shown below, where r1=74 pm and r2=200 pm.

{| class="wikitable" border=1
!p1/ g.mol-1.pm.fs-1!!p2/ g.mol-1.pm.fs-1!!Etot/kJ.mol-1!!Reactive?!!Description of the dynamics!!Illustration of the trajectory using Dist. vs Time plot
|-
| -2.56 || -5.1 || -414 || YES || This is a reactive trajectory because as the incoming H approaches, its AB distance decreases until it displays an oscillating behaviour which corresponds to the new bond vibration while the initial BC distance increases exponentially showing that the bond is broken and the leaving atom is moving further apart. ||[[File:Nl3818table1.png|200px|thumb|left]]
|-
| -3.1 || -4.1 || -420 || NO || This is not a reactive trajectory since although the AB distance decreases, it increases again while the initial BC distance retains its oscillating behaviour throughout meaning the bond is not broken. The incoming H approaches then moves away, no bond is broken in this case. ||[[File:Nl3818table2.png|200px|thumb|left]]
|-
| -3.1 || -5.1 || -412 || YES || This displays the same behaviour as the first case and thus, same explanation applies. ||[[File:Nl3818table3.png|200px|thumb|left]]
|-
| -5.1 || -10.1 || -357 || NO || In this case, the initial bond is broken but then reforms again, therefore, overall there is no reaction. ||[[File:Nl3818table4.png|200px|thumb|left]]
|-
|-5.1 || -10.6 || -349 || YES || The approaching H forms a new bond then bounces back but did not break the new bond so overall, this trajectory is reactive. ||[[File:Nl3818table5.png|200px|thumb|left]]
|}

[[File:Nl3818table5.png|200px|thumb|left]]
The last plot on the left corresponds to p1 = -5.1 g.mol-1.pm.fs-1, p2 = -10.6 g.mol-1.pm.fs-1 and Etot = -349 kJ.mol-1. This is a reactive trajectory since the approaching H forms a new bond then bounces back but did not break so overall, this trajectory is reactive.

*The last line of the table is not showing so I had to insert it like this.

==Transition State Theory==

'''The transition state theory is a classical consideration which assumes:'''2

'''1) The reactants and transition state are in a quasi-equilibrium.''' This means that once the reactants reaches the transition state, they collapse back to form the reactants again when in reality they can. This cause the values that the transition state theory predicts to be overestimated than the actual experimental value.

'''2) Quantum tunnelling is ignored.''' Because the reactants that overcome the barrier by moving across is ignored, this causes the values predicted by transition state theory to be underestimated than the actual experimental values.

'''3) All reactants that have enough energy to overcome the barrier will successfully form the product and the step that involves the formation of the product from the transition state is the rate-determing step.''' This causes an overestimation of the actual rate because in reality, although the reactants have high energy, but if the energy is not located in the right place (i.e. in the bond that needs to be broken) then the reaction will not happen.

== The F-H-H System==
=== PES Inspection===
===='''F + H2'''====
This is an unsymmetrical system which results in unsymmetrical potential wells in the surface plot.

From the surface plot, we are able to tell that this is an exothermic reaction. BC is the H-H distance of the reactant while AB is the H-F distance, the z-axis of the surface plot represents energy and in this case, the H-H is deeper in energy compared to H-F which corresponds to the energy profile of an exothermic reaction and thus, explains the greater H-F bond strength compared to H-H.

The approximate position of the transition state can be found by the same method that was used for three Hydrogen atoms: trial and error where the corresponding forces are as close to zero as possible. This method gives rF-HA = 180 pm and rHA-HB=74.5 pm. These positions are confirmed to be the transition state by the opposite signs of the eigenvalues/vectors. At this position, the energy of the transition state is -434 kJmol-1.

In order to find the activation energy, the position of the system is displaced slightly towards the reactant side from the transition state (rF-HA = 179 pm) and MEP calculation was used to obtain the "Energy vs Time" below.

From the graph, the energy of the reactants is -561 kJmol-1. The difference between the reactant state and the transition state energy is the activation energy which is the energy required to cross the barrier to form the product, in this case is +127 kJmol-1.

{| class="wikitable" border=1
|-
|[[File:H2andHFsurface.png|200px|thumb|left|Fig.6 - Surface plot for F+H2]]||[[File:FH2Ea.png|200px|thumb|left|Fig.7 - Energy vs Time plot for F+H2]]
|}

==== H + HF ====
The surface plot of this reaction is a mirrored image of the above surface plot from the previous reaction. This shows that this is an endothermic reaction where the reactants have higher energy than the product. This correlates to the stronger H-F bond that requires input energy to break to form the H-H product.

The approximate transition state can be found in the same manner as the previous reaction and the positions are rF-HB = 180 pm and rHB-HC=74.5 pm, corresponding to total energy of -434 kJmol-1. The activation energy in this case is very small is reported to be +0.930 kJmol-1. Because of such small activation energy, it is easier to locate the position of the transition state by using the Hammond postulate. From the surface plot, we have deduced that this reaction is endothermic therefore it must have a late transition state meaning the structure of the transition state will resemble more of the product.3 As a result, shorter H-H distance and long F-H were predicted for the position of the transition state.

{| class="wikitable" border=1
|-
|[[File:HandHFsurface.png|200px|thumb|left|Fig.8 - Surface plot for H + HF]]||[[File:FHHEa.png|200px|thumb|left|Fig.9 - Energy vs Time plot for H + HF]]
|}

===Reaction dynamics===

One of the ways to confirm the mechanism of the release of reaction energy is by taking an Infrared absorption spectrum of the product. In this case, we can measure the IR spectrum of HF gas sample. When the reaction just happened and IR light is shown through, we are exciting the molecules that were once thermally relaxed and exclusively occupying the ground state to its first vibrational excited state and at early times (when the reaction just happened), we may also excitation from first to second vibrational excited state.

As a result, what we can observe in the IR spectrum is two peaks present, one is the fundamental peak and the other is an overtone appearing at a lower wavenumber. This is because of decreasing energy gap between each levels due to anharmonicity.4 By measuring the intensity of the overtone over time, we can extract the number of molecules that are vibrationally excited over time so we can expect the intensity of overtone to be smaller as energy is emitted as radiation and the intensity of the fundamental peak would increase as time goes on.

The other way to confirm the mechanism of energy release experimentally is by measuring the IR emitted directly, instead of probing the reaction using IR spectroscopy, as the molecules fall back from vibrational excited state to the ground state. This method can be done by using the Infrared Emission Spectroscopy (IES).5

====F + H2====

By setting up a calculation with rHH=74 pm, pFH=-1.0 g.mol-1.pm.fs-1 and pHH varied between -6.1 and 6.1 g.mol-1.pm.fs-1, it is visible from the contour plot that the trajectory only successfully rolled over to the product with pHH between -2.7 and 1.2 g.mol-1.pm.fs-1. Values outside of this range results in the trajectory hitting the wall and bouncing back.

{| class="wikitable" border=1
|-
|[[File:H2-2.7.png|200px|thumb|left|Fig.10 - H2 trajectory at pHH = -2.7 g.mol-1.pm.fs-1]]||
[[File:H20nl3818.png|200px|thumb|left|Fig.11 - H2 trajectory at pHH = 0.0 g.mol-1.pm.fs-1]]
|[[File:H21.2nl3818.png|200px|thumb|left|Fig.12 - H2 trajectory at pHH = 1.2 g.mol-1.pm.fs-1]]||
[[File:H2bouncingbacknl3818.png|200px|thumb|left|Fig.13 - trajectory at pHH = 0.2 g.mol-1.pm.fs-1, pFH = -1.6 g.mol-1.pm.fs-1]]
|}

====H + HF====
A reactive trajectory was obtained with a combination of pFH = 1 g.mol-1.pm.fs-1 and pHH = -7 g.mol-1.pm.fs-1. Decreasing the momentum of the incoming H further results in the trajectory hitting the wall and bouncing back while increasing the energy of H-F vibration causes the trajectory to go in the other direction rather than towards the product.

====Polanyi's Empirical Rule====
The above investigation illustrates the Polanyi's empirical rule. In an exothermic reaction, as demonstrated by F + H2, translational energy is favoured over vibrational energy because the trajectory can fall down into the low energy region of the PES whereas the vibrational energy is in a different direction and can cause the trajectory to bounce back near the transition state and cross back to the reactant state.

On the other hand, vibrational energy is favoured over translational energy in an endothermic reaction6 as illustrated by H + HF. This is because the trajectory has the same directionality with the vibrational energy that goes to the product.

==Bibliography==
1 Maxima, minima, and saddle points (article). https://www.khanacademy.org/math/multivariable-calculus/applications-of-multivariable-derivatives/optimizing-multivariable-functions/a/maximums-minimums-and-saddle-points (accessed May 22, 2020).

2 Peters, B. Transition State Theory. Reaction Rate Theory and Rare Events Simulations 2017, 227–271.

3 Libretexts. Hammond's Postulate. https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hammond’s_Postulate (accessed May 22, 2020).

4 Libretexts. 13.5: Vibrational Overtones. https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/13:_Molecular_Spectroscopy/13.05:_Vibrational_Overtones (accessed May 22, 2020).

5 Keresztury, G. Emission Spectroscopy, Infrared. Encyclopedia of Analytical Chemistry 2006.

6 Zhang, Z.; Zhou, Y.; Zhang, D. H.; Czakó, G.; Bowman, J. M. Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl CHD3 Reaction. The Journal of Physical Chemistry Letters 2012, 3 (23), 3416–3419.

MRD:01493832

2020-05-29T17:17:05Z

Mak214: /* The transition state */

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

== The transition state ==
On a potential energy surface diagram, the transition state is mathematically defined as a stationary point where the derivative of the potential energy curve, which is a force {{fontcolor1|red| I'm not sure exactly what you mean here? [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:17, 29 May 2020 (BST)}}, is zero. This means that there are no movement at the transition state.

The transition state is different from a local minimum of a potential energy surface in a sense that it is a minimum in one direction yet a maximum in another.1 Therefore, it is a stationary point but not a point of inflection. From the calculated forces, we are able to confirm that this is a stationary point and from the given corresponding eigenvalues/vectors, the positive and negative combination confirms that this is a saddle point. If we move away from the stationary point, the eigenvalues/vectors will be either all positive or all negative. (For a local minimum, the eigenvalues/vectors will be positive in all directions.) {{fontcolor1|red| IOK. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:17, 29 May 2020 (BST)}}

=== H + H2 : Locating the transition state ===
In the case of H + H2, AB and BC distances must be equal in the transition state because all atoms are identical so the transition state is symmetrical and its position can be identified by using a trial and error process, where the AB and BC distances in the initial conditions are varied until the corresponding forces become as close to zero as possible. For 3 Hydrogen atoms, AB and BC distances (rts) are found to be 90.775 pm.

[[File:Nl3818-H TS dist-time.jpg|200px|thumb|left|Fig. 1 - Internuclear Distances vs Time plot for 3 Hydrogen atoms]]

From Fig.1, it is evidently clear that at rts = 90.775 pm, the distances are constant throughout confirming that it is a stationary point (i.e. the nuclei do not move) and that distances AB = BC. {{fontcolor1|red| I Good. Well done. [[User:Mak214|Mak214]] ([[User talk:Mak214|talk]]) 18:17, 29 May 2020 (BST)}}

== Calculating the reaction path ==
There are two types of calculation that can be done when considering the trajectory of the Hydrogen molecule: MEP (minimum energy path) and Dynamics. In MEP, atoms are assumed to be in an infinitely slow motion which corresponds to zero momenta at each time step. Both calculations can be done by changing the initial conditions to r1 = rts+δ and r2=rts in order to visualise the trajectories. The difference between MEP and Dynamics are shown in two contour plots below. From the MEP contour plot, we can see that the black trajectory line is rather smooth as it approaches the transition state position marked with a red cross compared to the trajectory shown in Dynamics contour plot. The MEP calculated trajectory also starts off at a higher potential energy position than the one from Dynamics calculation. Because in MEP calculation, atoms are moving very slowly so the vibration between the reactant HA-HB is so small that it does not display an oscillating behaviour as in the Dynamics calculation.

{| class="wikitable" border=1
|-
|[[File:Dynamic_nl3818.png|200px|thumb|left|Fig.2 - Contour plot of triatomic Hydrogen system using Dynamics calculation]]||
[[File:MEP_nl3818.png|200px|thumb|left|Fig.3 - Contour plot of triatomic Hydrogen system using MEP calculation]]
|}

=== Change in initial conditions ===
If we change the initial condition so that r1 = rts and r2=rts+δ instead, the result is still identical to previous result but with atom A moving away at large t and atom B bonding with C instead. By investigating the “Internuclear Distances vs Time” and “Momenta vs Time” plots and taking note of values for r1(t) r2(t) and p1(t) p2(t) at very large t (50 fs), another set of calculations can be done with using these values for the initial positions and momentum (with reversed sign). The observation is for “Internuclear Distances vs Time” plot, the result is a reflection in the y-axis of Fig. 4 meaning, the bonded BC H2 molecule and the previously isolated HA atom are now approaching each other and their positions at large t corresponds to the transition state position. With reversed sign of the momenta, this is a reverse process of what was previously done, instead of atoms moving apart, they are now moving towards each other with the same kinetic energy. The “Momenta vs Time” is now reflected in the x-axis of Fig. 5, which can be reasoned by the repulsion that the atoms experienced from coming together resulting in positive gradient which is gradual at first then steepens at high t as the three atoms are now very close.

{| class="wikitable" border=1
|-
|[[File:Dtnl3818new.png|200px|thumb|left|Fig.4 - Dist. vs Time plot of H atoms moving apart]]||
[[File:Momentanewnl3818.png|200px|thumb|left|Fig.5 - Momenta vs Time plot of H atoms moving apart]]
|}

== Reactive and unreactive trajectories ==

To test whether trajectories starting at the same position but with higher momenta will all react, a table has been constructed shown below, where r1=74 pm and r2=200 pm.

{| class="wikitable" border=1
!p1/ g.mol-1.pm.fs-1!!p2/ g.mol-1.pm.fs-1!!Etot/kJ.mol-1!!Reactive?!!Description of the dynamics!!Illustration of the trajectory using Dist. vs Time plot
|-
| -2.56 || -5.1 || -414 || YES || This is a reactive trajectory because as the incoming H approaches, its AB distance decreases until it displays an oscillating behaviour which corresponds to the new bond vibration while the initial BC distance increases exponentially showing that the bond is broken and the leaving atom is moving further apart. ||[[File:Nl3818table1.png|200px|thumb|left]]
|-
| -3.1 || -4.1 || -420 || NO || This is not a reactive trajectory since although the AB distance decreases, it increases again while the initial BC distance retains its oscillating behaviour throughout meaning the bond is not broken. The incoming H approaches then moves away, no bond is broken in this case. ||[[File:Nl3818table2.png|200px|thumb|left]]
|-
| -3.1 || -5.1 || -412 || YES || This displays the same behaviour as the first case and thus, same explanation applies. ||[[File:Nl3818table3.png|200px|thumb|left]]
|-
| -5.1 || -10.1 || -357 || NO || In this case, the initial bond is broken but then reforms again, therefore, overall there is no reaction. ||[[File:Nl3818table4.png|200px|thumb|left]]
|-
|-5.1 || -10.6 || -349 || YES || The approaching H forms a new bond then bounces back but did not break the new bond so overall, this trajectory is reactive. ||[[File:Nl3818table5.png|200px|thumb|left]]
|}

[[File:Nl3818table5.png|200px|thumb|left]]
The last plot on the left corresponds to p1 = -5.1 g.mol-1.pm.fs-1, p2 = -10.6 g.mol-1.pm.fs-1 and Etot = -349 kJ.mol-1. This is a reactive trajectory since the approaching H forms a new bond then bounces back but did not break so overall, this trajectory is reactive.

*The last line of the table is not showing so I had to insert it like this.

==Transition State Theory==

'''The transition state theory is a classical consideration which assumes:'''2

'''1) The reactants and transition state are in a quasi-equilibrium.''' This means that once the reactants reaches the transition state, they collapse back to form the reactants again when in reality they can. This cause the values that the transition state theory predicts to be overestimated than the actual experimental value.

'''2) Quantum tunnelling is ignored.''' Because the reactants that overcome the barrier by moving across is ignored, this causes the values predicted by transition state theory to be underestimated than the actual experimental values.

'''3) All reactants that have enough energy to overcome the barrier will successfully form the product and the step that involves the formation of the product from the transition state is the rate-determing step.''' This causes an overestimation of the actual rate because in reality, although the reactants have high energy, but if the energy is not located in the right place (i.e. in the bond that needs to be broken) then the reaction will not happen.

== The F-H-H System==
=== PES Inspection===
===='''F + H2'''====
This is an unsymmetrical system which results in unsymmetrical potential wells in the surface plot.

From the surface plot, we are able to tell that this is an exothermic reaction. BC is the H-H distance of the reactant while AB is the H-F distance, the z-axis of the surface plot represents energy and in this case, the H-H is deeper in energy compared to H-F which corresponds to the energy profile of an exothermic reaction and thus, explains the greater H-F bond strength compared to H-H.

The approximate position of the transition state can be found by the same method that was used for three Hydrogen atoms: trial and error where the corresponding forces are as close to zero as possible. This method gives rF-HA = 180 pm and rHA-HB=74.5 pm. These positions are confirmed to be the transition state by the opposite signs of the eigenvalues/vectors. At this position, the energy of the transition state is -434 kJmol-1.

In order to find the activation energy, the position of the system is displaced slightly towards the reactant side from the transition state (rF-HA = 179 pm) and MEP calculation was used to obtain the "Energy vs Time" below.

From the graph, the energy of the reactants is -561 kJmol-1. The difference between the reactant state and the transition state energy is the activation energy which is the energy required to cross the barrier to form the product, in this case is +127 kJmol-1.

{| class="wikitable" border=1
|-
|[[File:H2andHFsurface.png|200px|thumb|left|Fig.6 - Surface plot for F+H2]]||[[File:FH2Ea.png|200px|thumb|left|Fig.7 - Energy vs Time plot for F+H2]]
|}

==== H + HF ====
The surface plot of this reaction is a mirrored image of the above surface plot from the previous reaction. This shows that this is an endothermic reaction where the reactants have higher energy than the product. This correlates to the stronger H-F bond that requires input energy to break to form the H-H product.

The approximate transition state can be found in the same manner as the previous reaction and the positions are rF-HB = 180 pm and rHB-HC=74.5 pm, corresponding to total energy of -434 kJmol-1. The activation energy in this case is very small is reported to be +0.930 kJmol-1. Because of such small activation energy, it is easier to locate the position of the transition state by using the Hammond postulate. From the surface plot, we have deduced that this reaction is endothermic therefore it must have a late transition state meaning the structure of the transition state will resemble more of the product.3 As a result, shorter H-H distance and long F-H were predicted for the position of the transition state.

{| class="wikitable" border=1
|-
|[[File:HandHFsurface.png|200px|thumb|left|Fig.8 - Surface plot for H + HF]]||[[File:FHHEa.png|200px|thumb|left|Fig.9 - Energy vs Time plot for H + HF]]
|}

===Reaction dynamics===

One of the ways to confirm the mechanism of the release of reaction energy is by taking an Infrared absorption spectrum of the product. In this case, we can measure the IR spectrum of HF gas sample. When the reaction just happened and IR light is shown through, we are exciting the molecules that were once thermally relaxed and exclusively occupying the ground state to its first vibrational excited state and at early times (when the reaction just happened), we may also excitation from first to second vibrational excited state.

As a result, what we can observe in the IR spectrum is two peaks present, one is the fundamental peak and the other is an overtone appearing at a lower wavenumber. This is because of decreasing energy gap between each levels due to anharmonicity.4 By measuring the intensity of the overtone over time, we can extract the number of molecules that are vibrationally excited over time so we can expect the intensity of overtone to be smaller as energy is emitted as radiation and the intensity of the fundamental peak would increase as time goes on.

The other way to confirm the mechanism of energy release experimentally is by measuring the IR emitted directly, instead of probing the reaction using IR spectroscopy, as the molecules fall back from vibrational excited state to the ground state. This method can be done by using the Infrared Emission Spectroscopy (IES).5

====F + H2====

By setting up a calculation with rHH=74 pm, pFH=-1.0 g.mol-1.pm.fs-1 and pHH varied between -6.1 and 6.1 g.mol-1.pm.fs-1, it is visible from the contour plot that the trajectory only successfully rolled over to the product with pHH between -2.7 and 1.2 g.mol-1.pm.fs-1. Values outside of this range results in the trajectory hitting the wall and bouncing back.

{| class="wikitable" border=1
|-
|[[File:H2-2.7.png|200px|thumb|left|Fig.10 - H2 trajectory at pHH = -2.7 g.mol-1.pm.fs-1]]||
[[File:H20nl3818.png|200px|thumb|left|Fig.11 - H2 trajectory at pHH = 0.0 g.mol-1.pm.fs-1]]
|[[File:H21.2nl3818.png|200px|thumb|left|Fig.12 - H2 trajectory at pHH = 1.2 g.mol-1.pm.fs-1]]||
[[File:H2bouncingbacknl3818.png|200px|thumb|left|Fig.13 - trajectory at pHH = 0.2 g.mol-1.pm.fs-1, pFH = -1.6 g.mol-1.pm.fs-1]]
|}

====H + HF====
A reactive trajectory was obtained with a combination of pFH = 1 g.mol-1.pm.fs-1 and pHH = -7 g.mol-1.pm.fs-1. Decreasing the momentum of the incoming H further results in the trajectory hitting the wall and bouncing back while increasing the energy of H-F vibration causes the trajectory to go in the other direction rather than towards the product.

====Polanyi's Empirical Rule====
The above investigation illustrates the Polanyi's empirical rule. In an exothermic reaction, as demonstrated by F + H2, translational energy is favoured over vibrational energy because the trajectory can fall down into the low energy region of the PES whereas the vibrational energy is in a different direction and can cause the trajectory to bounce back near the transition state and cross back to the reactant state.

On the other hand, vibrational energy is favoured over translational energy in an endothermic reaction6 as illustrated by H + HF. This is because the trajectory has the same directionality with the vibrational energy that goes to the product.

==Bibliography==
1 Maxima, minima, and saddle points (article). https://www.khanacademy.org/math/multivariable-calculus/applications-of-multivariable-derivatives/optimizing-multivariable-functions/a/maximums-minimums-and-saddle-points (accessed May 22, 2020).

2 Peters, B. Transition State Theory. Reaction Rate Theory and Rare Events Simulations 2017, 227–271.

3 Libretexts. Hammond's Postulate. https://chem.libretexts.org/Bookshelves/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Hammond’s_Postulate (accessed May 22, 2020).

4 Libretexts. 13.5: Vibrational Overtones. https://chem.libretexts.org/Courses/Pacific_Union_College/Quantum_Chemistry/13:_Molecular_Spectroscopy/13.05:_Vibrational_Overtones (accessed May 22, 2020).

5 Keresztury, G. Emission Spectroscopy, Infrared. Encyclopedia of Analytical Chemistry 2006.

6 Zhang, Z.; Zhou, Y.; Zhang, D. H.; Czakó, G.; Bowman, J. M. Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl CHD3 Reaction. The Journal of Physical Chemistry Letters 2012, 3 (23), 3416–3419.