MRD:ZN 01510257

2020-06-03T17:53:41Z

== Molecular Reaction Dynamics: Applications to Triatomic systems ==
This is a report written for the Molecular Reaction Dynamics Lab, due 22/05/2020. This is done by Zaeem Najeeb, CID : 01510257.

== EXERCISE 1: H + H2 system ==

=== 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? ===

The transition state is defined as the saddle point for a potential energy surface. This is a stationary point in a multivariate function (in our case where the variables are the AB and BC bond distances, with the z-axis defined as the potential energy of the system) where the gradient with respect to each orthogonal variable is zero. However, the curvatures (the second differential with respect to the same variable) are neither all positive or negative.
[[File:ZN 01510257 Q1 Figure 1.png|centre|thumb|A saddle point.[http://www.offconvex.org/2016/03/22/saddlepoints/]]]

In general, the transition state can be found by analysing the hessian matrix. For our case, the hessian matrix is a 2 by 2 symmetric matrix comprised of the second differential with respect to each variable. If we then take the determinant of this matrix, denoted as D, we can then analyse the nature of the stationary point.
[[File:ZN 01510257 Q1 Figure 2.png|centre|thumb|The determinant of a 2 by 2 hessian matrix. The variables are x and y in this case, where (x0,y0) denote the co-ordinates of the stationary point.[https://www.math24.net/multivariable-functions/]]]
When the value of D is calculated, a simple test for the nature stationary point is as follows:
* If D>0 and fxx>0, then (x0,y0) is a point of local minimum.
* If D>0 and fxx<0, then (x0,y0) is a point of local maximum.
* If D<0, then (x0,y0) is a saddle point.
* If D=0, then this test fails.
We can use this test to determine the transition state as the only case where the test could fail would be if D=0. This would not be possible in our physical system as this would require no curvature for the second partial derivatives to be 0. Given that the atoms are close together, there is no physical case where D=0 occurs so this test is sufficient for determining where the transition state is.

{{fontcolor1|green|Good description. More clarity would be gained if you did analyse the matrix by showing how to calculate the sum of the matrix. [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 17:53, 3 June 2020 (BST)}}

=== 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. ===

The best estimate for the transition state position, rts, is rts≈ 90.775 pm.

This was determined by first considering a trajectory that resulted in a reaction. Due to the symmetric nature of the potential surface, as the energy level of the products and reactions is the same. This means that there is neither a late or early transition state, so by analysis using Hammond's postulate, the form of the transition state molecule gains equal contribution from the reactants and products. This leads to the AB and BC bond being equal when at the transition state, which is therefore also equal to the transition state position.
[[File:ZN 01510257 Q2 Figure 1.png|centre|thumb|An internuclear distances vs time plot for a successful reaction of H2 + H → H + H2. Here AB is the initial H2 molecule which react with C. The variables used were AB distance = 74 pm, AB momentum = -1 g.mol-1pm.fs-1, BC distance = 200 pm, BC momentum = -5 g.mol-1.pm.fs-1. The step size is 0.1 fs.]]
Using this plot, we are able to see that rts≈ 90 pm, however this can then be further improved by setting the AB and BC bond distances to 90 pm and the momenta of both as zero.
[[File:ZN 01510257 Q2 Figure 2.png|centre|thumb|An internuclear distances vs time plot where AB distance = BC distance = 90 pm. Both the AB and BC momenta are set to zero. The step size is 0.1 fs. As the AB and BC bond distances are equal, their curves overlap, appearing as a single orange curve.]]
Given that the gradient at the transition state is zero, there should be no force at the stationary point. This is due to the force being equal to the negative of the gradient of a potential. By altering the value of rts, we can monitor how close we are to the transition state by seeing whether the calculated force is zero, which would also be when the triatomic molecule would no longer be moving. This occurs to 3 decimal places (the limit of the model used in this lab) at 90.775 pm, and can be seen by how the AB and BC bond distances no longer vary, suggesting that they are no longer oscillating.
[[File:ZN 01510257 Q2 Figure 3.png|centre|thumb|An internuclear distances vs time plot where AB distance = BC distance = 90.775 pm. Both the AB and BC momenta are set to zero. The step size is 0.1 fs. Here, no oscillations are observed as the bond distances are constant. As the AB and BC bond distances are equal, their curves overlap, appearing as a single orange curve.]]

{{fontcolor1|green|Excellent description but even though the figures are in line with the text you should still refer to them within the text because the figures are easily ignored when reading through the text. Also, where did you get the information on Hammond's postulate? Reference needed [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 18:02, 3 June 2020 (BST)}}

=== Q3) Comment on how the ''mep'' and the trajectory you just calculated differ. ===

An mep calculation shows the trajectory of the system at infinitely slow time steps. This would mean that the momenta of the atoms is zero, so there should be no form of kinetic energy present. This can be seen by how the mep trajectory follows along the minimum potential energy value as the bond distance for BC does not vary (after forming H2). This is done for the case of AB = rts + 1 pm, BC = rts.
[[File:ZN 01510257 Q3 Figure 1.png|centre|thumb|Mep trajectory of the system when displaced from the transition state. The initial conditions are AB distance = 91.775 pm, BC distance = 90.775 pm. Both the AB and BC momenta are set to zero. The step size is 0.1 fs.]]
A dynamic calculation of this same system will the inertial motion of the atoms such that, when displaced in the same manner as the mep trajectory, oscillations of the BC bond length is observed. This is due to the potential energy being converted to kinetic energy in the form of vibrational motion.
[[File:ZN 01510257 Q3 Figure 2.png|centre|thumb|Dynamic trajectory of the system when displaced from the transition state. The initial conditions are the same as the mep trajectory.]]
Comparing the 2, we would expect the dynamic trajectory to approach the mep trajectory as the temperature decreases. This would be due to the system approaching a state of no molecular vibrations when the temperature approaches 0 K. As the momenta are always zero in the mep trajectory, we can neglect kinetic energy and see how the potential energy changes. This would be useful for determining activation energies of the system by perturbing it from the transition state in a particular direction.
[[File:ZN 01510257 Q3 Figure 3.png|centre|thumb|Mep calculation of the energy against time. As the kinetic energy is zero, the total energy of the system is the same as that of the potential energy. We can see a drop in this energy, where the difference between the initial and final state energies is that activation energy. This is approximately 17.681 kJ.mol-1]]
Another difference can be seen if we attempt to get the final positions of the trajectory and reverse both momenta. For the dynamic trajectory, this will effectively retrace the trajectory it took and return to the transition state. It will never overcome the transition state as, due to conservation of energy, the total energy of the system is lower than the transition state. This is due to the initial conditions placing the system slightly below the transition state with no momenta.
[[File:ZN 01510257 Q3 Figure 4.png|centre|thumb|Dynamic trajectory of the system when the final position and momenta of the previous system (displaced from the transition state) is used as the new initial conditions. The momenta values are multiplied by a factor of -1.]]

However, this fails for an mep calculation as there is no momenta at any point in time. The energy that would have been converted to kinetic energy is effectively lost in the process of moving from the transition state due to the momenta constantly being zeroed. This highlights how mep is unrealistic as this would break the conservation of energy.

If we were to now instead displace via is done for the case of AB = rts , BC = rts+ 1 pm then the same trajectories will occur, but instead the AB molecule will form. This would appear as a mirror image of the previous displacements about the y=x axis. This is due to the symmetric nature of this system.
[[File:ZN 01510257 Q3 Figure 5.png|centre|thumb|Dynamic trajectory of the system when displaced from the transition state, where AB = rts , BC = rts+ 1 pm. This is symmetric about the y=x axis to the previously calculated dynamic trajectory.]]

{{fontcolor1|green|Very good discussion but again refer to graphs, especially when you talk about the lack of momenta in mep calculation, state that it can been seen in figure X, from the zero kinetic energy [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 18:09, 3 June 2020 (BST)}}

=== 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? ===

Below is a table recording the observations of a H2 + H reaction, where each H is denoted as either A, B or C respectively. This therefore is equivalent to a AB + C reaction. The initial position of each reaction was AB distance = 74 pm and BC distance = 200 pm. The time step used in the trajectories is at 0.1 fs.

{| 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>
|Yes
|Here the non-vibrating AB molecule collides with C, forming a vibrating BC molecule and A which moves away at a greater velocity than when in the AB molecule.
|[[File:ZN 01510257 Q4 Figure 1.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.08</nowiki>
|No
|The vibrating AB approaches C, however the collision occurs such that the transition state does not form. The vibrating AB and C move apart.
|[[File:ZN 01510257 Q4 Figure 2.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.97</nowiki>
|Yes
|This is a similar case to the 1st situation, except now AB has enough momenta to have significant vibration energy, and so is vibrating. The form BC is also vibrating, with A moving away.
|[[File:ZN 01510257 Q4 Figure 3.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.10</nowiki>
|No
|AB approaches C, leading to to a successful collision which produces a strongly vibrating BC. This BC manages to then interact with A, leading to a second collision where AB is reformed and now strongly vibrating and C moves away fast enough to prevent further recrossing.
|[[File:ZN 01510257 Q4 Figure 4.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.48</nowiki>
|Yes
|This is a similar case to the 4th system. However, C does not move away fast enough so a second recrossing occurs due to the interaction between the strongly vibrating AB molecule and C. This leads to products forming.
|[[File:ZN 01510257 Q4 Figure 5.png|thumb]]
|}
From this table we can see that even though the kinetic energy of the system was sufficient to successfully react, as seen in the 1st trajectory, the different momenta effect whether phenomena such as recrossing, or even reaching the transition state is possible. This could be due to the different bond momenta leading to different phases of vibration upon collision. This would explain why the 1st and 3rd case differ in success, as if there is significant vibration, the point of reaction is no longer the same. This would also suggest an explanation for the final 2 cases of recrossing, where the slight difference in momenta altered the vibrations enough that one trajectory was able to recross again whilst the other was stuck back at the reactants.

{{fontcolor1|green|Good table layout but your conclusion is a little hard to follow. When talking about vibrations it's best to label p1 and p2 with respect to the AB and BC distances. Also, label the "cases" in your table [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 18:19, 3 June 2020 (BST)}}

=== Q5) Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
The three main assumptions in transition state theory are:
# Systems that have surmounted the transition state barrier cannot go back to reform the reactants
# The energy distribution among the reactants is in accordance with the Maxwell-Boltzmann distribution
# A chemical reaction can be treated in terms of classical motion. Quantum effects can be ignored.[https://contentstore.cla.co.uk/EReader/Index?p=RDpcU2l0ZXNccHJvZHVjdGlvblxUZW1wXERDUy0zYzczN2JlOS01OTEwLTRlYWUtYWE2Yi0yOTUyMTBmODJlMTEucGRm&o=JnB1Ymxpc2hlZENvbnRlbnRfSWQ9OTcyMjYyJmNvbnRlbnRSZXF1ZXN0X0lkPTEwNjE3NzMmZG9jdW1lbnRMaW5rPTRkY2E2MmVmLTJlODktZWExMS04MGNkLTAwNTA1NmFmNDA5OSZjb250ZW50SXRlbV9JZD0zMDE5Njc=&id=4dca62ef-2e89-ea11-80cd-005056af4099]
When comparing this to our system, we see that the first assumption is clearly false in the form of recrossing at specific momenta. This leads to transition state theory overestimating the reaction rate as more products are assumed to be formed when the bulk system is considered.

However, we can also see that the 3rd assumption is also false as in reality energy the vibrational modes are quantised is energy. Similarly effects such as quantum tunnelling are ignored which would allow reactants of lower energy to form the products. This would increase the reaction rate.

Therefore, transitions state theory fails to account for phenomena such as quantum tunnelling and recrossing, leading to this model failing to represent the experimental values.

{{fontcolor1|green|Very good and clear description. Where did you get the information on transition state theory? [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 18:22, 3 June 2020 (BST)}}

== EXERCISE 2: F - H - H system ==
For the following systems, A=F atom and B,C = H atom.

=== 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? ===
[[File:ZN 01510257 Q6 Figure 1.png|centre|thumb|Potential energy surface of a FH + H → F + H2 system. A=F atom, B, C=H atom]]
Looking at the potential energy surface we can see that the AB bond is more stable than the BC bond. This is shown with how the potential energy of the AB bond is much lower than that of the BC bond when in a stable state. This matches literature values where the H-F bond dissociation energy is 565 kJ.mol-1 and the H-H bond dissociation energy is 432 kJ.mol-1 and is due to large difference in electronegativities between H and F producing significant ionic effects that contribute to the bonding[http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html]. This also shows how the reaction of FH + H → F + H2 must be endothermic, with the reverse being exothermic. We can see this on the potential energy surface as the process of going from BC to AB includes a large drop in potential energy, showing that the formation of HF is favoured.

{{fontcolor1|green|Good. Your link to the website for the bond energies shows a references at the bottom of the page for the information on it which your could of used as your own reference [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 18:27, 3 June 2020 (BST)}}

=== Q7) Locate the approximate position of the transition state. ===
The approximate position for the transition state was located at AB = 181.04 pm and BC = 74.49 pm, where A= F atom and B, C are H atoms.

To locate the approximate position for the transition state for an asymmetric system, we can no longer assume that AB distance = BC distance when at the transition state. Instead, by knowing that the formation of HF is highly exothermic, the transition state can be assumed to resemble the reactants. This is according to Hammond's postulate where an exothermic reaction will have an early transition state. The first assumption therefore would be a BC distance of 74 pm and an AB distance of 100 pm as the transition state should closely resemble a H2 + F system (The momenta are kept as zero).
[[File:ZN 01510257 Q7 Figure 1.png|centre|thumb|Dynamic calculation of the trajectory for an FH + H system. AB = 100 pm, BC = 74 pm. The momenta are kept as zero. ]]
Looking at the result of the first approximation, we can visually see that the AB distance is too small. This can also be seen by the relatively large force along the AB bond. The AB bond distance was therefore increased until the force along the AB bond was considered zero to 2 decimal places by the interface to produce the second approximation where AB = 178.5 pm and BC = 74 pm.
[[File:ZN 01510257 Q7 Figure 2.png|centre|thumb|Second approximation for the transition state position. The AB bond was adjusted independently until they were both approximately zero. Here AB = 178.5 pm and BC = 74 pm.]]
At this point, there is still forces present along the AB and BC bonds, however these forces are no longer independent of each other. Therefore, to produce the final approximation for the transition state both bonds were altered until the force along them were zero to 3 decimal places. This occurred when AB = 181.04 pm and BC = 74.49 pm.
[[File:ZN 01510257 Q7 Figure 3.png|thumb|Final approximation for the transition state position. The AB and BC bonds were adjusted until they were both approximately zero. Here AB = 181.04 pm and BC = 74.49 pm.|centre]]
We can then confirm this is the case by looking at the internuclear distance against time plot to see no oscillations occur, as done with the H2 + H system.
[[File:ZN 01510257 Q7 Figure 4.png|thumb|Internuclear distance against time plot for the FH + H system. No oscillations are observed in the system, as shown by the lack of variation in the kinetic and potential energies.|centre]]

{{fontcolor1|green|Very good description on how you find the transition state distances [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 18:31, 3 June 2020 (BST)}}

=== Q8) Report the activation energy for both reactions. ===
Given that the coordinates for the transition state are known, we can perturb the system slightly to induce the the reactants to fail to overcome this transition energy and return to the reactants. This was done for both the exothermic and endothermic reaction pathways. The activation energy can therefore be calculated by taking the difference between the transition state energy, ETS ,with the energy at equilibrium, EEnd. This can be seen by looking at the change in energy of the system in an mep calculated trajectory. A large number of steps were considered to ensure that the potential energy had reached a fixed value.

The reaction pathway is referring to the direction the reaction would have occurred if the reaction was successful. As they are unsuccessful, this will show the energy difference between the reactants and transition state energy. This highlights the relatively small barrier of HF formation from H2.
{| class="wikitable"
!Reaction pathway
!ETS /kJmol-1
!EEnd /kJmol-1
!Activation energy/kJmol-1
!Energy vs time plot
!Energy vs time plot zoomed in
!Initial conditions used
!Percentage error
|-
|FH + H → F + H2
|<nowiki>-433.981</nowiki>
|<nowiki>-560.290</nowiki>
|126.309
|[[File:ZN 01510257 Q8 Figure 1 1.png|thumb]]
|[[File:ZN 01510257 Q8 Figure 1 2.png|thumb]]
|AB = 180.04 pm
BC = 74.49 pm

Both momenta are 0

Time step of 0.1 fs
|5.84 %
|-
|H2 + F → HF + H
|<nowiki>-433.981</nowiki>
|<nowiki>-435.013</nowiki>
|1.032
|[[File:ZN 01510257 Q8 Figure 2 1.png|thumb]]
|[[File:ZN 01510257 Q8 Figure 2 2.png|thumb]]
|AB = 182.04 pm
BC = 74.49 pm

Both momenta are 0

Time step of 0.1 fs
|71.32 %
|}
The activation energies were compared with experimental values to find a relatively small error for H2 formation compared to the error for HF formation [https://aip.scitation.org/doi/10.1063/1.1899125]. This is likely due to the systems resolution being too large to accurately approximate the much smaller activation energy. When the activation energy is large however, the model worked within an acceptable range.

{{fontcolor1|green|Very good. Your reference is not linked properly to give a reference list, details on this was in the reference section of the "wiki page instructions". How was your error calculated? Also, there is no need to speculate why the values are different because the theoretical study will explain why they may be different based on the use of different models to calculate the values [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 18:43, 3 June 2020 (BST)}}

=== 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. ===
A set of initial conditions for a reactive trajectory of the F + H2 reaction, where A= F atom and B, C are the H atoms was used to demonstrate the release of reaction energy. This was done with AB distance = 200 pm, BC distance = 74 pm, AB momentum = -1 g.mol-1.pm.fs-1 and the BC momentum = -5 g.mol-1.pm.fs-1. A time step of 0.1 fs was used and the reaction was observed for 2500 steps.
[[File:ZN 01510257 Q9 Figure 1.png|centre|thumb|Contour plot for a successful reaction trajectory of H2 + F.]]
This initial conditions show a successful reaction after multiple recrossings occur. As conservation of energy applies to this model, and the reaction is exothermic, the loss in potential energy due to a successful reaction must still be present in the system. This potential energy is converted to kinetic energy in the form of vibrations.
[[File:ZN 01510257 Q9 Figure 2.png|centre|thumb|Surface plot for a successful reaction trajectory of H2 + F. The greater kinetic energy of the HF molecule can be seen more clearly here as the HF molecule has a much greater change in potential energy.]]
We can further confirm this by looking at an energy against time plot, where we can see how the kinetic energy has a much greater variation than before the reaction at roughly 60 fs. This also shows how the kinetic energy has increased on average, and therefore the lost potential energy due to the exothermic reaction must have been converted into some form of kinetic energy.
[[File:ZN 01510257 Q9 Figure 4.png|centre|thumb|Energy against time plot for a successful reaction trajectory of H2 + F. The kinetic energy can clearly be seen to have increased on average whilst the potential energy has decreased, with the total energy remaining constant.]]
We can experimentally confirm that the system has released potential energy as vibrational kinetic energy via absorption IR. This is because higher vibarational modes would become excited upon successful reaction. Therefore, unlike the relaxed system which is primarily composed of molecules in the ground state vibrational modes, the 1st excited state would now also be significantly populated. This would allow for absorption of IR wavelengths from the 0 to 1st level (the 0 to 1 fundamental peak) and the now significantly populated 1st to 2nd level (a 1 to 2 hot band). Experimentally, this would appear as a smaller peak to the left of the main vibrational peak. If measured over a given time period, the smaller peak would be expected to decrease in size whilst the main peak would increase. This is due to the higher vibrational states relaxing over time and losing energy via several means [https://journals.aps.org/pr/abstract/10.1103/PhysRev.41.304].

{{fontcolor1|green|Very good. What kind of set up would allow for data collection as the reaction proceeds, for your stated experiment? [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 18:48, 3 June 2020 (BST)}}

=== 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 table below contains the initial conditions used for a calculated dynamic trajectory of the F + H2 reaction where A = F atom and B, C are the H atoms. The AB distance = 200 pm and the AB momentum = -1 g.mol-1.pm.fs-1. The BC distance = 200 pm. The time step is 0.1 fs and each trajectory is calculated for 4000 steps. The BC momenta is changed as shown in the table, with the animation and contour plot shown.
{| class="wikitable"
!p2/ g.mol-1.pm.fs-1
!Reaction successful?
!Animation of trajectory
!Illustration of the trajectory
|-
|5
|No
|[[File:ZN 01510257 Q10 Figure1 2.gif|frameless]]
|[[File:ZN 01510257 Q10 Figure 1 1.png|frameless]]
|-
|3
|Yes
|[[File:ZN 01510257 Q10 Figure2 2.gif|frameless]]
|[[File:ZN 01510257 Q10 Figure 2 1.png|frameless]]
|-
|1
|Yes
|[[File:ZN 01510257 Q10 Figure3 2.gif|frameless]]
|[[File:ZN 01510257 Q10 Figure 3 1.png|frameless]]
|-
|0
|Yes
|[[File:ZN 01510257 Q10 Figure4 2.gif|frameless]]
|[[File:ZN 01510257 Q10 Figure 4 1.png|frameless]]
|-
|<nowiki>-1</nowiki>
|Yes
|[[File:ZN 01510257 Q10 Figure5 2.gif|frameless]]
|[[File:ZN 01510257 Q10 Figure 5 1.png|frameless]]
|-
|<nowiki>-3</nowiki>
|No
|[[File:ZN 01510257 Q10 Figure6 2.gif|frameless]]
|[[File:ZN 01510257 Q10 Figure 6 1.png|right|frameless]]
|-
|<nowiki>-5</nowiki>
|Yes
|[[File:ZN 01510257 Q10 Figure7 2.gif|frameless]]
|[[File:ZN 01510257 Q10 Figure 7 1.png|frameless]]
|}

Polanyi's empirical rules state that an exothermic reaction (early transition state) will favour translational energy over kinetic energy for a successful reaction. This is the opposite when an endothermic reaction (late transition state) is considered [https://contentstore.cla.co.uk/EReader/Index?p=RDpcU2l0ZXNccHJvZHVjdGlvblxUZW1wXERDUy0zYzczN2JlOS01OTEwLTRlYWUtYWE2Yi0yOTUyMTBmODJlMTEucGRm&o=JnB1Ymxpc2hlZENvbnRlbnRfSWQ9OTcyMjYyJmNvbnRlbnRSZXF1ZXN0X0lkPTEwNjE3NzMmZG9jdW1lbnRMaW5rPTRkY2E2MmVmLTJlODktZWExMS04MGNkLTAwNTA1NmFmNDA5OSZjb250ZW50SXRlbV9JZD0zMDE5Njc=&id=4dca62ef-2e89-ea11-80cd-005056af4099]. When considering the exothermic reaction of HF formation, this would mean that systems with smaller kinetic energy are more likely to react successfully, assuming sufficient translational energy is supplied.

This can be explained by how the reactants in an endothermic reaction must go up a steep barrier to react the products on a potential energy surface. If there is no kinetic energy present, then the reactant does not vibrate and therefore the reactants will 'bounce off' the potential energy surface as they have no way to move in the orthogonal direction to reach the product path. Similarly, if the reactants for an exothermic reaction had kinetic energy, they would oscilate in the potential energy surface, hindering the ability to 'fall down' to the products.

A trend seems to appear when considering momenta of BC between 3 and -1 g.mol-1.pm.fs-1, however a discrepancy is seen when the absolute value of BC is 3 or 5 g.mol-1.pm.fs-1. In both cases, although the kinetic energy is the same (as it is proportional to the momentum squared), the success of the reaction differs. This could be due to the phase of the vibrational mode causing recrossing to occur 1 too many times, leading to an unsuccessful reaction.

This suggests that Polanyi's rule works when there is a significant difference in the translation and kinetic energies, as seen by the reactions all being successful when |BC|<3 g.mol-1.pm.fs-1. However, other factors such as the phases of the momenta and potentially even collision angle become significant factors in the determination of the reaction efficiency.

{{fontcolor1|green|There is a mistake in the initial conditions above the table, the AB and BC distances are both written as 200 pm. [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 18:53, 3 June 2020 (BST)}}

2020-06-01T20:45:46Z

Sf3014: /* Polanyi's Empirical Rules */

= '''Molecular Dynamics Lab''' =

== '''Exercise 1: H - H - H System''' ==

=== '''Dynamics from the transition state region''' ===
On a potential energy surface diagram, the transition state is mathematically defined as the maximum on the minimum energy path, so ∂V(ri)/∂ri=0 (the gradient of the potential, or the first derivative, is zero). However, it is not a simple local maximum. The transition state is a saddle point which means that while it is a local maximum in potential energy with respect to reaction coordinates, it is also a local minima with respect to the orthogonal coordinates. Therefore, in order to distinguish the transition state saddle point from a local minimum of maximum, the second partial derivative should be computed: H = fxx(x0,y0)fyy(x0,y0)-fxy(x0,y0)2. If H < 0, then (x0,y0) is a saddle point but if H > 0, then (x0,y0) is either a local maximum or minimum.

{{fontcolor1|green|Good description on the transition state (TS) but there is some confusion when distinguishing between the local minimum and the TS. The TS is the local maximum and you can distinguish between the local maximum and minimum using the second partial derivative of V(r1,r2) [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 20:50, 1 June 2020 (BST)}}

==== Trajectories from r1 = r2 : Locating the Transition State ====
The three atoms are the same (H) and therefore the transition state will be symmetric and the internuclear distances equal (r1 = r2). At the transition state point, as described above, the gradient of the potential is zero, and consequently a trajectory started exactly at the transition state, with no initial momentum, will remain there forever - this can be seen in the Contour Plot, where the trajectory is a single point. This means that the Internuclear Distances will not change (seen in the Internuclear Distance vs Time plot, showing distance as a constant) and the forces along r1 and r2 will be exactly 0 kJ mol-1 pm-1. Therefore to find the transition state position, different initial conditions with r1 = r2 and p1 = p2 = 0 g mol-1 pm fs-1 were tested to meet these provisions. The transition state was found when r1 = r2 = rts = 90.777 pm.
<nowiki> </nowiki><gallery>
File:Ms12218_Ex_1_TS_Contour.png| Figure 1ː Contour Plot showing a trajectory starting exactly at the transition state.
File:Ms12218_Ex_1_TS_IND.png| Figure 2ː Internuclear Distance vs Time for the Transition State
File:Ms12218_Ex_1_TS_A.png| Figure 3ː The Transition State (H - H - H)
</gallery>

{{fontcolor1|green|Good description on locating the TS [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 20:54, 1 June 2020 (BST)}}

==== Trajectories from r1 = rts + 1, r2 = rtsː The Reaction Path ====

The reaction or minimum energy path (mep) is a trajectory that corresponds to infinitely slow motion from the transition state to H1 + H2-H3. However, it does not provide a realistic account of the motion of atoms during a reaction.

With the MEP calculation, the trajectory of r1 is approximately constant, but with the dynamics calculation, it oscillates. This is because MEP does not take account of the vibrational energy of the forming molecule H2-H3 - real atoms do not move in infinitely slow motion, they vibrate.

Also with the MEP calculation, the trajectory of r2 (H1 atom moving away from the H2-H3 molecule) increases to ~ 200 pm, but increases infinitely with the dynamics calculation. This is MEP does not account for the fact that atoms have mass, and as a result, in the gas phase their motion will be inertial - constant motion.

{{fontcolor1|green|Good but refer to your figures in your description. Also, where is the evidence that the mep does not account for mass? Some mention of the momenta in mep calculation before concluding your statement about mass would be clearer to the reader [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 21:03, 1 June 2020 (BST)}}

<gallery>
File:Ms12218_Ex_1_Dynamic.png| Figure 4ː Contour Plot of the Reaction Path (Dynamics)
File:Ms12218_Ex_1_mep.png| Figure 5ː Contour Plot of the Reaction Path (MEP)
</gallery>

If the initial conditions were changed to make r1 = rts, r2 = rts + 1, then the same process would occur, but H1-H2 would form the molecule, with H3 as the atom moving away.

By setting up a dynamics calculation where the initial positions correspond to the final positions of the previous trajectory with the same final momenta values but their signs reversed:
* r1 = 352.598374019175 pm p1 = - 5.06794069218826 g mol-1 pm fs-1 
* r2 = 74.0408880639704 pm p2 = - 3.2038548324798097 g mol-1 pm fs-1
The trajectory then shows the opposite of the reaction path - the formation of the transition state from H1 and H2-H3. The contour plot shows this trajectory clearly, and the Internuclear Distances vs Time plot shows the formation of the transition state as r1 and r2 come to be equal.

<gallery>
File:Ms12218_Ex_1_Reverse_C.png| Figure 6ː Contour Plot of the Reverse of the Reaction Path (Dynamics)
File:Ms12218_Ex_1_Reverse_IND.png| Figure 7ː Internuclear Distances vs Time for the Reverse of the Reaction Path
</gallery>

=== Reactive and Unreactive Trajectories ===
{| class="wikitable"

| '''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
|The trajectory shows the reaction path from H1-H2 + H3, through the transition state, to H1 + H2-H3.
| [[File:Ms12218_Ex_1_T1.png|200px]]

|-

| -3.1
| -4.1
| -420.077
| No
| The trajectory shows H3 approaching H1-H2, however they do not react and H3 moves away again. The transition state is never formed.
| [[File:Ms12218_Ex_1_T2.png|200px]]

|-

| -3.1
| -5.1
| -413.977
| Yes
|The H3 atom has more momenta than the previous example.

The trajectory shows the reaction path from H1-H2 + H3, through the transition state, to H1 + H2-H3.
| [[File:Ms12218_Ex_1_T3.png|200px]]

|-

| -5.1
| -10.1
| -357.277
| No
|The trajectory shows H3 approaching H1-H2 and even forming the transition state. However, though H2 and H3 interact, H1-H2 reforms once more, and H3 moves away again.
| [[File:Ms12218_Ex_1_T4.png|200px]]

|-

| -5.1
| -10.6
| -349.477
| Yes
|The H3 atom has more momenta than the previous example. The trajectory shows the reaction path from H1-H2 + H3, through the transition state (with some interaction between H1 and H2), to H1 + H2-H3.
| [[File:Ms12218_Ex_1_T5.png|200px]]

|}
From the table, it is clear that it is not sufficient for just the momenta (kinetic energy) of the molecule (p1) to increase (i.e. from -3.1 to -5.1), the momenta of the atom (p2) needs to also be sufficiently high for a reaction to occur (i.e. -5.1 vs -5.6 and -10.1 vs -10.6). The atom must also have a larger momentum (more kinetic energy) than the molecule for a reaction to take place.

{{fontcolor1|green|Good table layout. Your conclusion is poor because it doesn't address the trajectories that do not abide by the statements you made in your conclusion, ie in all cases p2 is greater than p1 and the trajectory isn't always successful and p2 doesn't have to be "sufficiently high" (not defined, what is high?) as seen in the first example given in your table [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 21:14, 1 June 2020 (BST)}}

=== Transition State Theory ===

Transition State Theory is a powerful theory to rationalise and calculate the rate of chemical reactions based on the properties of the reactants and the transition state structure. The main assumptions of transition state theory are as follows:
* Molecular systems that have crossed the transition state in the direction of the products cannot turn back and form reactant molecules again.
* The reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann Distribution.
* In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation, quantum effects being ignored.<ref name=":0">J. I. Steinfeld, J. S. Francisco, W. L. Hase, Statistical Approach to Reaction Dynamics: Transition State Theory, ''Chemical Kinetic and Dynamics'', 2, Pentice-Hall, 1998, 287 - 391.</ref><ref name=":1">K. J. Laidler, Conventional Transition-State Theory, Chemical Kinetics, 3, Harper-Collins, 1987, 89-94.</ref>

However, in reality, these assumptions are not always obeyed. Firstly, contrary to transition state theory, there are in fact a number of ways in which a system may re-cross a potential-energy surface. The penultimate set of conditions in the above table (p1 = - 5.1 g mol-1 pm fs-1 and p2 = - 10.1 g mol-1 pm fs-1) show a case of barrier recrossing - the system crosses the transition state region and a bond in the product forms, but then the system reverts back to the reactants. Multiple crossings of the transition state will inevitably decrease the rate of reaction, and thus as a result conventional transition state theory leads to overestimated reaction rates.<ref name=":1" />

Transition state theory assumes that motion across the transition state region can be treated as classical motion, but in reality, quantisation of motion introduces the possibility that the system may tunnel through the potential energy surface. When this occurs, the system does not need to have the activation energy, and therefore quantum tunnelling increases rate of reaction, particularly in reactions involved very light species such as H atoms.<ref name=":1" />

Transition state theory also simplifies motion on the potential energy surface to one-dimensional motion along the reaction coordinate, however, actual motion may include components normal to the reaction coordinate and the reaction path may lie above the minimum energy configuration - as a result, the effective activation energy may be somewhat higher than the barrier height. A larger activation energy will lead to a decreased rate of reaction.<ref name=":0" />

Overall, transition state theory will overestimate the rate because the energy barrier effects have a much larger contribution to the rate of reaction than quantum tunnelling (less significant).

{{fontcolor1|green|Very good and clear description [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 21:23, 1 June 2020 (BST)}}

== '''Exercise 2: F - H - H System''' ==

=== PES Inspection ===
By inspecting the potential energy surfaces, the reactions of a F - H - H system can be classified as follows:
* F + H2 → HF + H is exothermic: the reactants are higher energy than the products.

* HF + H → H2 + F is endothermic: the products are higher energy than the reactants.
From this, it can be concluded that HF is a stronger bond than H2. In the exothermic reaction, more energy is released forming H-F than used to break H-H. In the endothermic reaction, more energy is used to break H-F than is released forming H-H.

<gallery>
File:Ms12218_Ex_2_ExoEndo.png| Figure 8ː Surface Plot of the F-H-H System where A=F, B=C=H.
</gallery>

{{fontcolor1|green|Good but how did you determine which direction was exo- and endo- thermic? refer to your figure 8 and describe how you reached that conclusion. Also, your comment on bond energies would benefit from published experimental values [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 21:27, 1 June 2020 (BST)}}

==== Transition State ====
As the F - H - H system is not symmetric, the transition state is not simply found where r1 = r2. However, it is still the case that at the transition state, the forces along r1 and r2 equal zero. The Internuclear Distances that satisfy these conditions are r1 = 180.1 pm and r2 = 74.488 pm, where r1 is the distance between F and H r2 is the distance between H and H. At this point, with no initial momentum, the trajectory is a single point.

Hammond Postulate:
* Exothermic = early TS, reactant-like.
* Endothermic = late TS, product-like.

<gallery>
File:Ms12218_Ex_2_TS_C.png| Figure 9ː Surface Plot with a trajectory starting exactly at the transition state.
File:Ms12218_Ex_2_TS_IND.png| Figure 10ː Internuclear Distances vs Time for the Transition State
</gallery>

As seen in the surface plot, the transition state point is much closer to the reactants H2 + F. This can be explained with Hammond's Postulate: an exothermic reaction will have an early transition state that is reactant-like, and an endothermic reaction will have a late transition state that is product-like. H2 + F are the reactants of F + H2 → HF + H which is exothermic, and the products of HF + H → H2 + F which is endothermic.

{{fontcolor1|green|Good but how did you find the distances for the transition state? Was it trial and error? What was your process? Did Hammond's postulate help you find the distances? More details required. Also, where did you get the information on Hammond's postulate? Reference [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 21:33, 1 June 2020 (BST)}}

==== Activation Energies ====
The activation energy for each reaction can be found by performing an MEP calculation from a structure neighbouring the transition state to find the trajectory that corresponds to infinitely slow motion from the transition state to the products.

To find the activation energy of the endothermic HF + H → H2 + F:
* r1 = rts - 1, r2 = rts so r1 = 180.1 pm and r2 = 74.488 pm.
* The trajectory shows the reaction path proceeding from the transition state to HF + H.
* By plotting energy against time for this trajectory, the potential energy difference can be found between HF + H (the reactants) and the transition state.
* Transition state potential = 433.981 kJ mol-1.
* Reactants (HF + H) potential = - 560.036 kJ mol-1.
* Therefore the activation energy for HF + H → H2 + F is 126.055 kJ mol-1.
<gallery>
File:Ms12218_Ex_2_HFH_C.png| Figure 11ː Surface Plot with a trajectory of the formation of HF and H from the transition state, A = F, B = C = H.
File:Ms12218_Ex_2_HFH_Ea.png| Figure 12ː Energy vs Time for the formation of HF and H from the transition state.
</gallery>

To find the activation energy of the exothermic H2 + F → HF + H:
* r1 = rts + 1, r2 = rts so r1 = 182.1 pm and r2 = 74.488 pm.
* The trajectory shows the reaction path proceeding from the transition state to H2 + F.
* By plotting energy against time for this trajectory, the potential energy difference can be found between H2 + F (the reactants) and the transition state.
* Transition state potential = 433.981 kJ mol-1.
* Reactants (H2 + F) potential = - 435.013 kJ mol-1.
* Therefore the activation energy for H2 + F → HF + H is 1.032 kJ mol-1.
* This is much smaller than the above reaction () reflecting the fact that this is an exothermic vs an endothermic reaction.
<gallery>
File:Ms12218_Ex_2_H2H_C.png| Figure 13ː Surface Plot with a trajectory showing the formation of H2 and F from the transition state, A = F, B = C = H.
File:Ms12218_Ex_2_H2F_Ea.png| Figure 14ː Energy vs Time, for the formation of H2 and F from the transition state.
File:Ms12218_Ex_2_H2F_EaZ.png| Figure 15ː Energy vs Time for the formation of H2 and F from the transition state.
</gallery>

{{fontcolor1|green|Very good and clear description but refer to your figures [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 21:35, 1 June 2020 (BST)}}

=== Reaction Dynamics ===
==== Mechanism for the release of reaction energy for: F + H2 → HF + H ====
The initial conditions r1 = 175 pm, r2 = 74 pm, p1 = -1.6 g mol-1 pm fs-1, and p2 = 0.2 g mol-1 pm fs-1 resulted in a reactive trajectory for this reaction as seen in the contour plot below.

<gallery>
File:Ms12218_Ex_2_RD_1.png| Figure 16ː Contour Plot with a trajectory showing F + H2 → HF + H, A = F, B = C = H.
</gallery>

Energy in the reaction is conserved, and the reaction energy is released into the vibrational mode of newly formed HF molecule - potential energy is converted into kinetic (vibrational energy) energy. This is clearly seen in the contour plot above by the extent of oscillation in HF which corresponds to vibration of the molecule.

This mechanism of release of reaction energy into vibrational energy can be confirmed experimentally with IR absorption spectroscopy. The formation of the H-F bond and release of reaction energy means that the HF molecule is vibrationally excited: a proportion of the molecules are in the 1st excited level. Therefore, there are two vibrational transitions available: 0 → 1, and 1 → 2, and we expect to see two peaks in the IR spectrum - a fundamental intensity (reflecting the 0 → 1 transition), and a hot band (reflecting the 1 → 2 transition between excited levels) at a lower frequency. The overtone has a lower frequency due to the anharmonicity of the interaction. As the molecule reverts back to the relaxed state, the overtone intensity will become weaker and the fundamental peak will increase in intensity. If the reaction energy was not released as vibrational kinetic energy, the overtone would not be seen and there would be only one peak in the IR spectra.

Another method of confirming the method experimentally is with IR emission spectroscopy (e.g. gas-phase FTIR). As the vibrationally excited molecules revert to the relaxed state, photons are emitted as electrons fall to the ground state - the energy of the photon is equal to the difference in energy levels, and therefore can be used to probe the electronic structure of the molecule (and confirm vibrational excitation - molecules in the ground state will not emit photons).

{{fontcolor1|green|Good. This description on the experiment is missing how the result will be obtained as the reaction proceeds [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 21:41, 1 June 2020 (BST)}}

==== Polanyi's Empirical Rules ====
According to Polanyi's empirical rules, the distribution of energy between different modes (translation and vibration) required for a successful reaction, depends on the position of the transition state.

Exothermic reactions (F + H2 → HF + H) are characterised by an early transition state. An early energy barrier therefore occurs while the reactants are approaching each other. Translational energy is the most effective - a molecule with all its energy in motion along the reaction coordinate can easily overcome the energy barrier. Reactant vibrational energy, on the other hand, may be ineffective for reaction - a vibrationally excited molecule does not have enough energy left to reach the top of its barrier, and the vibrational motion does not allow the atom to simply fall into the reaction channel.<ref name=":2">J. I. Steinfeld, J. S. Francisco, W. L. Hase, Experimental Chemical Dynamics, ''Chemical Kinetic and Dynamics'', 2, Pentice-Hall, 1998, 272 - 274.</ref><ref>K. J. Laidler, Reaction Dynamics, Chemical Kinetics, 3, Harper-Collins, 1987, 460-469</ref>

Endothermic reactions (HF + H → H2 + F) are characterised by a late transition state. Vibrational energy in the reactants will be more effective than translational energy in overcoming the energy barrier. Vibrationally excited molecules with the correct phase can reach the top of the barrier, while molecules with rapid translational motion are repulsed by the inner wall of the potential surface.<ref name=":2" />

This is the core of Polanyi’s Rules - in general, vibrational energy is better at promoting an endothermic reaction and translational energy is better at promoting an exothermic reaction.

Example: F + H2 → HF + H (exothermic).
* Initial conditions: r1 = 225 pm, r2 = 74 pm, A = F, B = C = H.
* In the first trajectory, the molecule (H-H) has a large amount of vibrational energy (large oscillations). The trajectory is not reactive.
* In the second trajectory, the molecule (H-H) has a very small amount of vibrational energy (reduced momentum). The trajectory is reactive.
{| class="wikitable"

| '''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'''

|-

| -1.0
| -6.1
| -402.845
| No
| The trajectory shows F approaching H-H and even forming the transition state. However, H-H reforms once more, and F moves away again.
| [[File:Ms12218_Ex_2_T1.png|200px]]

|-
| -1.6
| -0.2
| -434.482
| Yes
| The trajectory shows the reaction path from F + H-H, through the transition state, to F-H + H.
| [[File:Ms12218_Ex_2_T8.png|200px]]

|}

{{fontcolor1|green|Very good description. What about the endothermic reaction? [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 21:45, 1 June 2020 (BST)}}

== References ==

2020-06-01T16:24:20Z

Sf3014: /* PES Inspection */

== Exercise 1: H2 + H ==
The following plots were produced from observation of the reaction dynamics that occur as a hydrogen atom (A) approaches a hydrogen molecule (BC). The trajectories show how the reaction proceeds from reactants to products. Analysis of these plots allow us to find key parameters, such as the location of the transition state, computationally. [[File:Exe1_contour_plotajdcadjkcbad67867290.PNG|thumb|Contour Plot |centre]]
[[File:Exe1_skew_plotfrgfbgfsgvrvws4718.PNG|thumb|Skew Plot|centre]]
[[File:Animationexe1ws4718.jpeg|thumb|Animation |centre]]
[[File:Exe_1_surface_plotgrgrtgws4718.PNG_‎|thumb|Surface Plot|centre]]
[[File:Exe_1_momentum_v_timews4718.PNG|thumb|Momentum v Time|centre]]
[[File:Invvtws4718.PNG|thumb|Internuclear Velocity v Time |centre]]
[[File:Exe_1_ind_v_timews4718.PNG|thumb|Internuclear Distance v Time|centre]]
[[File:Exe_1_E_v_timews4718.PNG|thumb|Energy v Time |centre]]<blockquote>''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?''</blockquote>A potential energy surface diagram shows how potential energy varies with bond distances. A minimum energy path shows how reactants turn into products and the transition state is the maximum on this path. But as each bond lengths change, the potential energy changes (appearing like a Leonard-Jones potential), where the minima is the most stable energy conformation, at an equilibrium bond length. Making the transition state a saddle point, where the transition state represents a maxima on the reaction pathway (it is higher energy than the reactants/products). But if you assess how the energy varies with bond distance at this maxima, it will be the minima. The saddle point can be distinguished from local minima, because if you take the second derivative of the minima in all directions they will all be positive. But if you take the second derivative of a saddle point , one will be negative and the rest will be positive, showing its dual nature as a maxima from one perspective and a minima from all others.

{{fontcolor1|green|Good but a mathematical definition for the transition state would be when the partial derivative of V(r1,r2) is zero, remember that the derivatives are partial. Also, what are the scales on your animations? [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 16:21, 1 June 2020 (BST)}}

If the trajectory is begun at the transition state, it will remain there as there is no momentum. But geometrical rearrangement to emulate reactants or the products, will push the trajectory in favor of them. To find this state, begin trajectories around this transition state and see in which direction the trajectory follows. In symmetric surfaces, the transition state will also be symmetric. As shown by the following animation.
[[File:_AnimationTS230ws4718.jpeg|centre|thumb|Animation of periodic transition state]]

The transition state can be found by setting bond distances equal to one another and the momentum equal to 0. <blockquote>''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.''</blockquote>[[File:Surface_PlotEXE1TS230ws4718.png|centre|thumb|Potential Energy Surface Plot ]]
[[File:Contour_plot_ts_ws4718.png|centre|thumb|Contour Plot ]]

Analysis of the contour and surface plots show the region where the transition state could be, here the momentum is 0 and the distances are equal (230 pm). It appears the transition state seems to be around 90-100 pm for both AB and BC, as this is where the saddle point seems to be located.
[[File:IDvT_EXE_1_TS̠ws4719.png|centre|thumb|Internuclear distance v Time]]However it is clear that 230 pm is not the the transition state bond length as there is still movement, if it were a transition state it would not have to move to reach it. 90 pm was tried next. There is almost no variation about the lowest energy, but there is still movement shown in the internuclear distance v time graph.
[[File:Surface_Plottsws4718.png|centre|thumb|Contour Plot]]
[[File:Surface_Plotindvtwws4718.png|centre|thumb|Internuclear Distance v Time]]
However when r1 and r2 were set to 90.8 pm, no movement was seen. So this is most likely to be transition state bond length.

{{fontcolor1|green|Good but your distances could be more accurate (to 3 d.p), did you round up your value? If so, state its accuracy and/or zoom in on the distance vs time graphs so it can be shown clearly. Also, if you number your figures at makes it easier to refer to them in your text and make your report clearer. [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 16:29, 1 June 2020 (BST)}}

[[File:Surface_Plotindws471890801549302.png|centre|thumb|Internuclear distance v time : 90.8 pm ]]<blockquote>''Calculating the Reaction Path ː MEP''</blockquote><blockquote>r1 = 91.8 pm
r2 = 90.8 pm</blockquote>[[File:Surface_PlotMEP918908.png|centre|thumb|<blockquote>MEP</blockquote>]]<blockquote>''Trajectories from r1 = rts+δ, r2 = rts''</blockquote>[[File:Surface_Plotdynamics918908ws4718.png|centre|thumb|Surface plot : r1 = 91.8, r2 = 90.8]]
Compared to MEP, the trajectory is oscillating as it represents how the atoms move in real life not in slow motion, where they access vibrational, roatational and translational degrees of freedom. The length of r1 is increased, geometrically becoming more like products, so from the transition state it 'rolls' towards the products.

{{fontcolor1|green|Your comment on the mep calculation seems more like a comment on dynamic calculation. So, you compared dynamic with mep without explaining mep. Comment on your figures, there are so many with no caption or explanation, how do they all fit into your discussion? [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 16:38, 1 June 2020 (BST)}}

[[File:Surface_Plotmomentavtimeexe1ws4718.png|centre|thumb|Internuclear Distance v Time: r1 = 91.8, r2 = 90.8]]
[[File:Surface Plotindvtexe1ws4718.png|centre|thumb|Momenta v Time : r1 = 91.8, r2 = 90.8]]
This is changed if r1 and r2 are switched - r1 = 90.8, r2 = 91.8. This geometrically changes it to be more like the reactants, thus it rolls towards the reactants.
[[File:Surface_Plotbkadbfakjfjja01549302ws4718.png|centre|thumb|Surface Plot : r1 = 90.8, r2=91.8]]
[[File:IntDvTr1918yigpiyg.png|centre|thumb]]
[[File:Momentumreversed.png|centre|thumb|Momentum V Time : r1 = 90.8, r2=91.8]]
For t = 50 s : r1(50) = ~70 pm, r2(50) = ~350 pm : p1(50) = ~3.5 g.mol-1.pm.fs-1, p2(50) = ~5 g.mol-1.pm.fs-1

[[File:Howthesurfaceplotcchangedws4718.png|centre|thumb|Surface Plot]]
[[File:Howthemomentumchangesws4718.png|centre|thumb|Momentum v Time]][[File:Howdistancechangesws4718.png|centre|thumb|Internuclear Distance v Time]]<blockquote>''Reactive and unreactive trajectories''</blockquote>
{| 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>-417</nowiki>
|Reactive
|Trajectory shows the reaction proceeds where they approach and a new H2 molecule is formed with the oncoming H
|[[File:E1ws4718.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-422</nowiki>
|Unreactive
|Trajectory shows that H2 and H move towards each other, but do not react, going back. Shown by the trajectory overlapping with itself.
|[[File:E2ws4718.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-415</nowiki>
|Reactive
|Shows that even though H molecule same momenta, if H2 molecule momenta increases it will react
|[[File:E3ws4718.png ]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-359</nowiki>
|Unreactive
|The molecules approach each other and separate into 3 atoms, H2 tries to form with the original H atom but no permanent bond is made (they seem to rebound off one another) and eventually reforms original H2 molecule
|[[File:E4ws4718.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-351</nowiki>
|Reactive
|There is the same inital approach and split into 3 atoms as above, but the rebounding onto one another results in a H2 molecule formed with the original incoming H
|[[File:E5ws4718.png]]
|}
As you increase the momentum of the H2 molecule, the momentum of the H atom also needs to increase. If this is not sufficient, the reaction will not proceed. In some instances a bond will form but it is not permanent. But a slight increase in momenta will push it towards the products.

{{fontcolor1|green|Good table layout and conclusion. The conclusion would be clearer to follow if p1 and p2 were defined with respect to distances AB and BC. [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 16:59, 1 June 2020 (BST)}}

<blockquote>''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''</blockquote>Transition State Theory''' '''for a bimolecular reaction, Transition State Theory allows us to find the rate constant. However it is based on several assumptionsː

1) It assumes that molecules will have enough kinetic energy to overcome the energy barrier and form products, if it cannot quit make the transition state its treated as a reactant.

2) Once products are formed, the reverse reaction will not proceed and reform reactants.

3) It ignored phenomenon like quantum mechanical tunneling, where instead of going over an energy barrier it will go through it, this is a result of the breakdown of the born-oppenheimer approximation. As such the system here is treated classically, it won't account for this when considering the rate. The rate will therefore be underestimated as tunneling is the lower energy process.

4) When looking at the transition state, the reaction coordinate can be viewed as purely translational motion. But there are many more degrees of freedom available for the molecule, such as vibrational and rotational. It was seen that bond vibrations can result in the formation of bonds and the breaking of them, before permanent bonds form eventually. This can cause a large overestimation of the rate.

{{fontcolor1|green|Good but is point 4 a condition of transition state theory (TST)? Also, it is related to point 1. You would gain more clarity when referring to trajectories in your table above, if you numbered the trajectories and the table. Is the reference below for TST? It is poorly placed and linked poorly to your text. Your reference should be place when the information is stated in your text and linked according to the "general instructions" for wiki's in your lab script. [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 16:59, 1 June 2020 (BST)}}

chapter 10 of J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998

==Exercise 2==
==='''PES Inspection'''===
<blockquote>''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?''</blockquote>'''F + H2:'''

We can conclude the reaction is exothermic, as shown by the BC (reactants) bond being significantly higher in energy with respect to AB (products). The products HF + H would be expected. There is an energy difference between reactants and products of around 124 kJ/mol. So we can assume the HF bond is stronger than HH. This can be seen on the PES.
There is a large oscillation in the products as the bond is vibrating and initially there is bond breaking and reforming with the H and F, but eventually a permanent HF bond is formed.

[[File:PESFHHws4718.png|centre|thumb|F + H2 ]]

'''H + HF:'''

The converse is seen here, where the expected products would be H2 + F. The PES shows the reactants (BC) having a much lower energy than the products, supporting the above and that this is the converse endothermic reaction. Here the trajectory shows that in the following conditions the reaction could not take place.
[[File:PESHHFws4718.png|centre|thumb|H + HF]]

The energy barrier for F + H2 is very large, requiring around 800 K. However interstellar HF has been found to exist at (10-100K) and fast reactions can take place at much lower temperatures providing quantum mechanical tunneling occurs.(1)

{{fontcolor1|green|Good. However, this final statement is a little unclear. Description on bond energies would benefit from published experimental values for the bond energies for HH and HF [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 17:14, 1 June 2020 (BST)}}

(1) Tizniti, M.; Le Picard, S. D.; Lique, F.; Berteloite, C.; Canosa, A.; Alexander, M. H.; Sims, I. R. The Rate of the F + H2 Reaction at Very Low Temperatures. Nat. Chem. 2014, 6 (2), 141–145. https://doi.org/10.1038/nchem.1835.<blockquote>''Locate the approximate position of the transition state''</blockquote>This molecule is not symmetric, so we cannot assume the transition state will be. Using Hammond's postulate, we can gather that in an exothermic reaction the transition state will be closer to the reactants in energy. Therefore will geometrically resemble them granted a small rearrangement, the converse is true for products in an endothermic reaction. As symmetry cannot be used for the F-H-H system, an initial guess would be that the transition state has a H-H bond length very similar to a H2 molecule (74 pm) and the F-H bond would be very large. After trial and error it was found that r1 = 74.54 pm and r2 = 180.62 pm. This is evidenced by the PSE and Internuclear Distance v Time plots.
[[File:F-H-HINDvTws4718.png|centre|thumb|PSE : Transition State F-H-H]]
[[File:F-H-HPSETSws4718.png|centre|thumb|IND v T : Transition State F-H-H]]

For H-H-F, as this system is endothermic, the transition state would most closely represent the products. As this is the reverse reaction of the above, it would be expected the transition state would be in the same place. Where r1 = 74.54 pm and r2 = 180.62 pm. This was evidenced by the PSE and the Internuclear Distance v Time plots.

{{fontcolor1|green|Very good. Where did you get the information on Hammond's postulate and the HH bond distance? [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 17:18, 1 June 2020 (BST)}}

[[File:INDvTHFHTSWS4718.png|centre|thumb|IND v T : Transition State H-H-F]]
[[File:PSEHFHtses4718.png|centre|thumb|PSE : Transition State H-H-F]]<blockquote>''Report the activation energy for both reactions''</blockquote>Transition State Energy (r1 = 74.54 pm, r2 = 180.62 pm) = -433.361 kJ/mol

'''F-H-H:'''

[[File:Etot.png|centre|thumb|MEP : F-H-H Reactants]]
* The activation energy for the reaction was found by an MEP calculation and setting r1=75.54 pm. This 'rolled' the reaction towards the reactants. This averaged the energy of the reactants to a constant.
* E(transition) = -433.361
* Activation Energy = 1.945 kJ/mol

'''H-H-F'''

For the endothermic route the same method was repeated, giving the following MEP.
[[File:EVTHFHws4718.png|centre|thumb|MEP : H-F-H]]
* E(reactants) = -561.804
* E(transition state) = same as F-H-H
* Activation Energy = 128.443 kJ/mol

{{fontcolor1|green|Goog but it isn't clear to how you calculated the activation energies. You could show this more clearly on your energy vs time graphs by zooming in and referring/describing your figures [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 17:21, 1 June 2020 (BST)}}

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

'''F-H-Hː'''

r1 = 74, p1 = -0.2, r2=175, p2= -1.6
[[File:RxndynMTws4718.png|centre|thumb|Momentum v Time : F-H-H]]
[[File:RXNDFHHws4718.jpeg|centre|thumb|Animation : F-H-H]]
<blockquote>''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''</blockquote>The momentum v time and the animation show the product (HF) bond vibrate much more than that of the reactant (H2). This can be seen in the high frequency oscillations of the blue (AB) line, showing fast variations in momentum. Visually this is is seen in the animation as the bond being stretched backwards and forwards. As the reaction is exothermic, energy will be released to compensate the formation of lower energy products. This excess energy could be expressed in the vibrational modes of the molecule, causing the faster oscillations. Spectroscopic techniques such as IR can be used to analyse how this energy is being released. In the IR spectrum of HF, one band would be expected (linear : 3N-5). But as the reaction is exothermic, vibrationally excited states are now available, so we would expect to see an overtone at a much weaker intensity. As time progresses, the overtone becomes weaker as the molecule becomes less excited. Thus it can be shown how the energy is being released into the environment as the reaction proceeds and how it impacts the molecules around it. If energy released had no impact on HF, an overtone would not be seen at all.

Chang, H. C.; Klemperer, W. The
Vibrational Second Overtones of HF Dimer: A Quartet. ''J. Chem. Phys.'' '''1994''',
''100'' (1), 1–14. https://doi.org/10.1063/1.466980.

'''F+H2:'''<blockquote>''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.''</blockquote><blockquote>''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?''</blockquote>
{| class="wikitable"
!p2
!p1
!Contour Plot
!Etot
!Comment
|-
|<nowiki>-6.1</nowiki>
|<nowiki>-1</nowiki>
|[[File:-6.1ws4718.png ]]
|<nowiki>-403</nowiki>
|H2 approaches F, but no permanent bond formed. Unreactive.
|-
|<nowiki>-6</nowiki>
|<nowiki>-1</nowiki>
|[[File:-6ws4718.png ]]
|<nowiki>-403</nowiki>
|large backwards and forwards, but permanent bond formed. Reactive.
|-
|<nowiki>-5.9</nowiki>
|<nowiki>-1</nowiki>
|[[File:-5.9ws4718.png ]]
|<nowiki>-405</nowiki>
|Less backwards and forwards. But no permanent bond formed. Unreactive.
|-
|<nowiki>-4</nowiki>
|<nowiki>-1</nowiki>
|[[File:-4ws4718.png ]]
|<nowiki>-423</nowiki>
|large amount of backwards and forwards but eventually permanent bond formed. Reactive.
|-
|<nowiki>-2</nowiki>
|<nowiki>-1</nowiki>
|[[File:-2ws4718.png ]]
|<nowiki>-435</nowiki>
|Backwards/forwards, but permanent bond. Reactive.
|-
|<nowiki>0</nowiki>
|<nowiki>-1</nowiki>
|[[File:0ws4718.png]]
|<nowiki>-438</nowiki>
|Unreactive
|-
|<nowiki>2</nowiki>
|<nowiki>-1</nowiki>
|[[File:2ws4718.png]]
|<nowiki>-428</nowiki>
|Unreactive.
|-
|<nowiki>4</nowiki>
|<nowiki>-1</nowiki>
|[[File:4ws4718.png]]
|<nowiki>-413</nowiki>
|Unreactive, does not approach F.
|-
|<nowiki>5.9</nowiki>
|<nowiki>-1</nowiki>
|[[File:5.9ws4718.png]]
|<nowiki>-395</nowiki>
|Backwards/forwards, but no permanent bond formed. Unreactive.
|-
|<nowiki>6</nowiki>
|<nowiki>-1</nowiki>
|[[File:6ws4718.png]]
|<nowiki>-394</nowiki>
|Backwards/forwards, but no permanent bond. Unreactive.
|-
|<nowiki>6</nowiki>
|<nowiki>-1</nowiki>
|[[File:6.1ws4718.png]]
|<nowiki>-392</nowiki>
|Backwards/forwards, but permanent bond formed. Reactive.
|-
|<nowiki>0.2</nowiki>
|<nowiki>-1.7</nowiki>
|[[File:0.2ws4718.png]]
|<nowiki>-411</nowiki>
|backwards/forwards, but permanent bond formed. Reactive.
|}
'''H+HF:'''
Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy).
Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.)
{| class="wikitable"
!p1
!p2
!
!
|-
|<nowiki>-0.1</nowiki>
|<nowiki>-6.1</nowiki>
|[[File:Rws4718.png]]
|-
|<nowiki>-3</nowiki>
|<nowiki>-5</nowiki>
|[[File:hws4718.png]]
|-
|<nowiki>0</nowiki>
|<nowiki>8</nowiki>
|[[File:iws4718.png]]
|-
|}
<blockquote>''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.''</blockquote>Polanyi Rules show for an early transition state (exothermic reaction), the product is more likely to possess higher vibrational modes in the product and much less so for late transition states (endothermic). It is seen that for exothermic F+H2, there is a large amount of oscillation in the product HF. This is not seen as much when H2 is formed in H+HF.. Looking at the plots of forwards and reverse reactions, we could analyze the amount of vibrations and assess whether the transition state was likely to be late or early. Helping to establish if the reaction was exothermic or not.

MRD:WS4718

2020-06-01T16:21:45Z

Sf3014: /* PES Inspection */

== Exercise 1: H2 + H ==
The following plots were produced from observation of the reaction dynamics that occur as a hydrogen atom (A) approaches a hydrogen molecule (BC). The trajectories show how the reaction proceeds from reactants to products. Analysis of these plots allow us to find key parameters, such as the location of the transition state, computationally. [[File:Exe1_contour_plotajdcadjkcbad67867290.PNG|thumb|Contour Plot |centre]]
[[File:Exe1_skew_plotfrgfbgfsgvrvws4718.PNG|thumb|Skew Plot|centre]]
[[File:Animationexe1ws4718.jpeg|thumb|Animation |centre]]
[[File:Exe_1_surface_plotgrgrtgws4718.PNG_‎|thumb|Surface Plot|centre]]
[[File:Exe_1_momentum_v_timews4718.PNG|thumb|Momentum v Time|centre]]
[[File:Invvtws4718.PNG|thumb|Internuclear Velocity v Time |centre]]
[[File:Exe_1_ind_v_timews4718.PNG|thumb|Internuclear Distance v Time|centre]]
[[File:Exe_1_E_v_timews4718.PNG|thumb|Energy v Time |centre]]<blockquote>''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?''</blockquote>A potential energy surface diagram shows how potential energy varies with bond distances. A minimum energy path shows how reactants turn into products and the transition state is the maximum on this path. But as each bond lengths change, the potential energy changes (appearing like a Leonard-Jones potential), where the minima is the most stable energy conformation, at an equilibrium bond length. Making the transition state a saddle point, where the transition state represents a maxima on the reaction pathway (it is higher energy than the reactants/products). But if you assess how the energy varies with bond distance at this maxima, it will be the minima. The saddle point can be distinguished from local minima, because if you take the second derivative of the minima in all directions they will all be positive. But if you take the second derivative of a saddle point , one will be negative and the rest will be positive, showing its dual nature as a maxima from one perspective and a minima from all others.

{{fontcolor1|green|Good but a mathematical definition for the transition state would be when the partial derivative of V(r1,r2) is zero, remember that the derivatives are partial. Also, what are the scales on your animations? [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 16:21, 1 June 2020 (BST)}}

If the trajectory is begun at the transition state, it will remain there as there is no momentum. But geometrical rearrangement to emulate reactants or the products, will push the trajectory in favor of them. To find this state, begin trajectories around this transition state and see in which direction the trajectory follows. In symmetric surfaces, the transition state will also be symmetric. As shown by the following animation.
[[File:_AnimationTS230ws4718.jpeg|centre|thumb|Animation of periodic transition state]]

The transition state can be found by setting bond distances equal to one another and the momentum equal to 0. <blockquote>''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.''</blockquote>[[File:Surface_PlotEXE1TS230ws4718.png|centre|thumb|Potential Energy Surface Plot ]]
[[File:Contour_plot_ts_ws4718.png|centre|thumb|Contour Plot ]]

Analysis of the contour and surface plots show the region where the transition state could be, here the momentum is 0 and the distances are equal (230 pm). It appears the transition state seems to be around 90-100 pm for both AB and BC, as this is where the saddle point seems to be located.
[[File:IDvT_EXE_1_TS̠ws4719.png|centre|thumb|Internuclear distance v Time]]However it is clear that 230 pm is not the the transition state bond length as there is still movement, if it were a transition state it would not have to move to reach it. 90 pm was tried next. There is almost no variation about the lowest energy, but there is still movement shown in the internuclear distance v time graph.
[[File:Surface_Plottsws4718.png|centre|thumb|Contour Plot]]
[[File:Surface_Plotindvtwws4718.png|centre|thumb|Internuclear Distance v Time]]
However when r1 and r2 were set to 90.8 pm, no movement was seen. So this is most likely to be transition state bond length.

{{fontcolor1|green|Good but your distances could be more accurate (to 3 d.p), did you round up your value? If so, state its accuracy and/or zoom in on the distance vs time graphs so it can be shown clearly. Also, if you number your figures at makes it easier to refer to them in your text and make your report clearer. [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 16:29, 1 June 2020 (BST)}}

[[File:Surface_Plotindws471890801549302.png|centre|thumb|Internuclear distance v time : 90.8 pm ]]<blockquote>''Calculating the Reaction Path ː MEP''</blockquote><blockquote>r1 = 91.8 pm
r2 = 90.8 pm</blockquote>[[File:Surface_PlotMEP918908.png|centre|thumb|<blockquote>MEP</blockquote>]]<blockquote>''Trajectories from r1 = rts+δ, r2 = rts''</blockquote>[[File:Surface_Plotdynamics918908ws4718.png|centre|thumb|Surface plot : r1 = 91.8, r2 = 90.8]]
Compared to MEP, the trajectory is oscillating as it represents how the atoms move in real life not in slow motion, where they access vibrational, roatational and translational degrees of freedom. The length of r1 is increased, geometrically becoming more like products, so from the transition state it 'rolls' towards the products.

{{fontcolor1|green|Your comment on the mep calculation seems more like a comment on dynamic calculation. So, you compared dynamic with mep without explaining mep. Comment on your figures, there are so many with no caption or explanation, how do they all fit into your discussion? [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 16:38, 1 June 2020 (BST)}}

[[File:Surface_Plotmomentavtimeexe1ws4718.png|centre|thumb|Internuclear Distance v Time: r1 = 91.8, r2 = 90.8]]
[[File:Surface Plotindvtexe1ws4718.png|centre|thumb|Momenta v Time : r1 = 91.8, r2 = 90.8]]
This is changed if r1 and r2 are switched - r1 = 90.8, r2 = 91.8. This geometrically changes it to be more like the reactants, thus it rolls towards the reactants.
[[File:Surface_Plotbkadbfakjfjja01549302ws4718.png|centre|thumb|Surface Plot : r1 = 90.8, r2=91.8]]
[[File:IntDvTr1918yigpiyg.png|centre|thumb]]
[[File:Momentumreversed.png|centre|thumb|Momentum V Time : r1 = 90.8, r2=91.8]]
For t = 50 s : r1(50) = ~70 pm, r2(50) = ~350 pm : p1(50) = ~3.5 g.mol-1.pm.fs-1, p2(50) = ~5 g.mol-1.pm.fs-1

[[File:Howthesurfaceplotcchangedws4718.png|centre|thumb|Surface Plot]]
[[File:Howthemomentumchangesws4718.png|centre|thumb|Momentum v Time]][[File:Howdistancechangesws4718.png|centre|thumb|Internuclear Distance v Time]]<blockquote>''Reactive and unreactive trajectories''</blockquote>
{| 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>-417</nowiki>
|Reactive
|Trajectory shows the reaction proceeds where they approach and a new H2 molecule is formed with the oncoming H
|[[File:E1ws4718.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-422</nowiki>
|Unreactive
|Trajectory shows that H2 and H move towards each other, but do not react, going back. Shown by the trajectory overlapping with itself.
|[[File:E2ws4718.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-415</nowiki>
|Reactive
|Shows that even though H molecule same momenta, if H2 molecule momenta increases it will react
|[[File:E3ws4718.png ]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-359</nowiki>
|Unreactive
|The molecules approach each other and separate into 3 atoms, H2 tries to form with the original H atom but no permanent bond is made (they seem to rebound off one another) and eventually reforms original H2 molecule
|[[File:E4ws4718.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-351</nowiki>
|Reactive
|There is the same inital approach and split into 3 atoms as above, but the rebounding onto one another results in a H2 molecule formed with the original incoming H
|[[File:E5ws4718.png]]
|}
As you increase the momentum of the H2 molecule, the momentum of the H atom also needs to increase. If this is not sufficient, the reaction will not proceed. In some instances a bond will form but it is not permanent. But a slight increase in momenta will push it towards the products.

{{fontcolor1|green|Good table layout and conclusion. The conclusion would be clearer to follow if p1 and p2 were defined with respect to distances AB and BC. [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 16:59, 1 June 2020 (BST)}}

<blockquote>''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''</blockquote>Transition State Theory''' '''for a bimolecular reaction, Transition State Theory allows us to find the rate constant. However it is based on several assumptionsː

1) It assumes that molecules will have enough kinetic energy to overcome the energy barrier and form products, if it cannot quit make the transition state its treated as a reactant.

2) Once products are formed, the reverse reaction will not proceed and reform reactants.

3) It ignored phenomenon like quantum mechanical tunneling, where instead of going over an energy barrier it will go through it, this is a result of the breakdown of the born-oppenheimer approximation. As such the system here is treated classically, it won't account for this when considering the rate. The rate will therefore be underestimated as tunneling is the lower energy process.

4) When looking at the transition state, the reaction coordinate can be viewed as purely translational motion. But there are many more degrees of freedom available for the molecule, such as vibrational and rotational. It was seen that bond vibrations can result in the formation of bonds and the breaking of them, before permanent bonds form eventually. This can cause a large overestimation of the rate.

{{fontcolor1|green|Good but is point 4 a condition of transition state theory (TST)? Also, it is related to point 1. You would gain more clarity when referring to trajectories in your table above, if you numbered the trajectories and the table. Is the reference below for TST? It is poorly placed and linked poorly to your text. Your reference should be place when the information is stated in your text and linked according to the "general instructions" for wiki's in your lab script. [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 16:59, 1 June 2020 (BST)}}

chapter 10 of J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998

==Exercise 2==
==='''PES Inspection'''===
<blockquote>''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?''</blockquote>'''F + H2:'''

We can conclude the reaction is exothermic, as shown by the BC (reactants) bond being significantly higher in energy with respect to AB (products). The products HF + H would be expected. There is an energy difference between reactants and products of around 124 kJ/mol. So we can assume the HF bond is stronger than HH. This can be seen on the PES.
There is a large oscillation in the products as the bond is vibrating and initially there is bond breaking and reforming with the H and F, but eventually a permanent HF bond is formed.

[[File:PESFHHws4718.png|centre|thumb|F + H2 ]]

'''H + HF:'''

The converse is seen here, where the expected products would be H2 + F. The PES shows the reactants (BC) having a much lower energy than the products, supporting the above and that this is the converse endothermic reaction. Here the trajectory shows that in the following conditions the reaction could not take place.
[[File:PESHHFws4718.png|centre|thumb|H + HF]]

The energy barrier for F + H2 is very large, requiring around 800 K. However interstellar HF has been found to exist at (10-100K) and fast reactions can take place at much lower temperatures providing quantum mechanical tunneling occurs.(1)

{{fontcolor1|green|Good. However, this final statement is a little unclear. Description on bond energies would benefit from published experimental values for the bond energies for HH and HF [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 17:14, 1 June 2020 (BST)}}

(1) Tizniti, M.; Le Picard, S. D.; Lique, F.; Berteloite, C.; Canosa, A.; Alexander, M. H.; Sims, I. R. The Rate of the F + H2 Reaction at Very Low Temperatures. Nat. Chem. 2014, 6 (2), 141–145. https://doi.org/10.1038/nchem.1835.<blockquote>''Locate the approximate position of the transition state''</blockquote>This molecule is not symmetric, so we cannot assume the transition state will be. Using Hammond's postulate, we can gather that in an exothermic reaction the transition state will be closer to the reactants in energy. Therefore will geometrically resemble them granted a small rearrangement, the converse is true for products in an endothermic reaction. As symmetry cannot be used for the F-H-H system, an initial guess would be that the transition state has a H-H bond length very similar to a H2 molecule (74 pm) and the F-H bond would be very large. After trial and error it was found that r1 = 74.54 pm and r2 = 180.62 pm. This is evidenced by the PSE and Internuclear Distance v Time plots.
[[File:F-H-HINDvTws4718.png|centre|thumb|PSE : Transition State F-H-H]]
[[File:F-H-HPSETSws4718.png|centre|thumb|IND v T : Transition State F-H-H]]

For H-H-F, as this system is endothermic, the transition state would most closely represent the products. As this is the reverse reaction of the above, it would be expected the transition state would be in the same place. Where r1 = 74.54 pm and r2 = 180.62 pm. This was evidenced by the PSE and the Internuclear Distance v Time plots.

{{fontcolor1|green|Very good. Where did you get the information on Hammond's postulate and the HH bond distance? [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 17:18, 1 June 2020 (BST)}}

[[File:INDvTHFHTSWS4718.png|centre|thumb|IND v T : Transition State H-H-F]]
[[File:PSEHFHtses4718.png|centre|thumb|PSE : Transition State H-H-F]]<blockquote>''Report the activation energy for both reactions''</blockquote>Transition State Energy (r1 = 74.54 pm, r2 = 180.62 pm) = -433.361 kJ/mol

'''F-H-H:'''

[[File:Etot.png|centre|thumb|MEP : F-H-H Reactants]]
* The activation energy for the reaction was found by an MEP calculation and setting r1=75.54 pm. This 'rolled' the reaction towards the reactants. This averaged the energy of the reactants to a constant.
* E(transition) = -433.361
* Activation Energy = 1.945 kJ/mol

'''H-H-F'''

For the endothermic route the same method was repeated, giving the following MEP.
[[File:EVTHFHws4718.png|centre|thumb|MEP : H-F-H]]
* E(reactants) = -561.804
* E(transition state) = same as F-H-H
* Activation Energy = 128.443 kJ/mol

{{fontcolor1|green|Goog but it isn't clear to how you calculated the activation energies. You could show this more clearly on your energy vs time graphs by zooming in [[User:Sf3014|Sf3014]] ([[User talk:Sf3014|talk]]) 17:21, 1 June 2020 (BST)}}

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

'''F-H-Hː'''

r1 = 74, p1 = -0.2, r2=175, p2= -1.6
[[File:RxndynMTws4718.png|centre|thumb|Momentum v Time : F-H-H]]
[[File:RXNDFHHws4718.jpeg|centre|thumb|Animation : F-H-H]]
<blockquote>''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''</blockquote>The momentum v time and the animation show the product (HF) bond vibrate much more than that of the reactant (H2). This can be seen in the high frequency oscillations of the blue (AB) line, showing fast variations in momentum. Visually this is is seen in the animation as the bond being stretched backwards and forwards. As the reaction is exothermic, energy will be released to compensate the formation of lower energy products. This excess energy could be expressed in the vibrational modes of the molecule, causing the faster oscillations. Spectroscopic techniques such as IR can be used to analyse how this energy is being released. In the IR spectrum of HF, one band would be expected (linear : 3N-5). But as the reaction is exothermic, vibrationally excited states are now available, so we would expect to see an overtone at a much weaker intensity. As time progresses, the overtone becomes weaker as the molecule becomes less excited. Thus it can be shown how the energy is being released into the environment as the reaction proceeds and how it impacts the molecules around it. If energy released had no impact on HF, an overtone would not be seen at all.

Chang, H. C.; Klemperer, W. The
Vibrational Second Overtones of HF Dimer: A Quartet. ''J. Chem. Phys.'' '''1994''',
''100'' (1), 1–14. https://doi.org/10.1063/1.466980.

'''F+H2:'''<blockquote>''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.''</blockquote><blockquote>''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?''</blockquote>
{| class="wikitable"
!p2
!p1
!Contour Plot
!Etot
!Comment
|-
|<nowiki>-6.1</nowiki>
|<nowiki>-1</nowiki>
|[[File:-6.1ws4718.png ]]
|<nowiki>-403</nowiki>
|H2 approaches F, but no permanent bond formed. Unreactive.
|-
|<nowiki>-6</nowiki>
|<nowiki>-1</nowiki>
|[[File:-6ws4718.png ]]
|<nowiki>-403</nowiki>
|large backwards and forwards, but permanent bond formed. Reactive.
|-
|<nowiki>-5.9</nowiki>
|<nowiki>-1</nowiki>
|[[File:-5.9ws4718.png ]]
|<nowiki>-405</nowiki>
|Less backwards and forwards. But no permanent bond formed. Unreactive.
|-
|<nowiki>-4</nowiki>
|<nowiki>-1</nowiki>
|[[File:-4ws4718.png ]]
|<nowiki>-423</nowiki>
|large amount of backwards and forwards but eventually permanent bond formed. Reactive.
|-
|<nowiki>-2</nowiki>
|<nowiki>-1</nowiki>
|[[File:-2ws4718.png ]]
|<nowiki>-435</nowiki>
|Backwards/forwards, but permanent bond. Reactive.
|-
|<nowiki>0</nowiki>
|<nowiki>-1</nowiki>
|[[File:0ws4718.png]]
|<nowiki>-438</nowiki>
|Unreactive
|-
|<nowiki>2</nowiki>
|<nowiki>-1</nowiki>
|[[File:2ws4718.png]]
|<nowiki>-428</nowiki>
|Unreactive.
|-
|<nowiki>4</nowiki>
|<nowiki>-1</nowiki>
|[[File:4ws4718.png]]
|<nowiki>-413</nowiki>
|Unreactive, does not approach F.
|-
|<nowiki>5.9</nowiki>
|<nowiki>-1</nowiki>
|[[File:5.9ws4718.png]]
|<nowiki>-395</nowiki>
|Backwards/forwards, but no permanent bond formed. Unreactive.
|-
|<nowiki>6</nowiki>
|<nowiki>-1</nowiki>
|[[File:6ws4718.png]]
|<nowiki>-394</nowiki>
|Backwards/forwards, but no permanent bond. Unreactive.
|-
|<nowiki>6</nowiki>
|<nowiki>-1</nowiki>
|[[File:6.1ws4718.png]]
|<nowiki>-392</nowiki>
|Backwards/forwards, but permanent bond formed. Reactive.
|-
|<nowiki>0.2</nowiki>
|<nowiki>-1.7</nowiki>
|[[File:0.2ws4718.png]]
|<nowiki>-411</nowiki>
|backwards/forwards, but permanent bond formed. Reactive.
|}
'''H+HF:'''
Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy).
Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.)
{| class="wikitable"
!p1
!p2
!
!
|-
|<nowiki>-0.1</nowiki>
|<nowiki>-6.1</nowiki>
|[[File:Rws4718.png]]
|-
|<nowiki>-3</nowiki>
|<nowiki>-5</nowiki>
|[[File:hws4718.png]]
|-
|<nowiki>0</nowiki>
|<nowiki>8</nowiki>
|[[File:iws4718.png]]
|-
|}
<blockquote>''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.''</blockquote>Polanyi Rules show for an early transition state (exothermic reaction), the product is more likely to possess higher vibrational modes in the product and much less so for late transition states (endothermic). It is seen that for exothermic F+H2, there is a large amount of oscillation in the product HF. This is not seen as much when H2 is formed in H+HF.. Looking at the plots of forwards and reverse reactions, we could analyze the amount of vibrations and assess whether the transition state was likely to be late or early. Helping to establish if the reaction was exothermic or not.