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MRD:ZN 01510257

2020-05-22T22:58:01Z

Zn4318: /* 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. */

== 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.

=== 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.]]

=== 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.]]

=== 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.

=== 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.

== 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.

=== 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]]

=== 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.

=== 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].

=== 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.

MRD:ZN 01510257

2020-05-22T22:53:44Z

Zn4318: /* Q8) Report the activation energy for both reactions. */

== 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.

=== 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.]]

=== 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.]]

=== 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.

=== 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.

== 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.

=== 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]]

=== 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.

=== 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 fundamental peak) and the now significantly populated 1st to 2nd level (a 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].

=== 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.

MRD:ZN 01510257

2020-05-22T22:47:09Z

Zn4318: /* 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 */

== 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.

=== 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.]]

=== 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.]]

=== 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.

=== 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.

== 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.

=== 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]]

=== 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.

=== 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 fundamental peak) and the now significantly populated 1st to 2nd level (a 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].

=== 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.

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 effeciency.

MRD:ZN 01510257

2020-05-22T22:43:03Z

== 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.

=== 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.]]

=== 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.]]

=== 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.

=== 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.

== 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.

=== 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]]

=== 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.

=== 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 fundamental peak) and the now significantly populated 1st to 2nd level (a 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].

=== 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.

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.

MRD:ZN 01510257

2020-05-22T22:29:01Z

== 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.

=== 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.]]

=== 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.]]

=== 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.

=== 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.

== 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.

=== 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]]

=== 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.

=== 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 fundamental peak) and the now significantly populated 1st to 2nd level (a 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].

=== 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 +H

F=A

r1=200, p1=-1, r2=200, t=0.1fs, 4000 steps
{| 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 rule:

MRD:ZN 01510257

2020-05-22T22:26:44Z

Zn4318: /* 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. */

== 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.

=== 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.]]

=== 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.]]

=== 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.

=== 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.

== 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.

=== 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]]

=== 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.

=== 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 fundamental peak) and the now significantly populated 1st to 2nd level (a 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].

=== 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 ===
F=A

r1=200, p1=-1, r2=200, t=0.1fs, 4000 steps
{| 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 rule:

File:ZN 01510257 Q9 Figure 4.png

2020-05-22T22:07:56Z

Zn4318: Energy vs time plot of successful HF formation

Energy vs time plot of successful HF formation

File:ZN 01510257 Q9 Figure 2.png

2020-05-22T22:04:13Z

Zn4318: Surface plot of successful F + H2 reaction

Surface plot of successful F + H2 reaction

File:ZN 01510257 Q9 Figure 1.png

2020-05-22T21:58:13Z

Zn4318: Contour plot for a successful F + H2 trajectory.

Contour plot for a successful F + H2 trajectory.

MRD:ZN 01510257

2020-05-22T21:50:24Z

== 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.

=== 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.]]

=== 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.]]

=== 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.

=== 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.

== 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.

=== 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]]

=== 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.

=== 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. ===
r1 = 200, p1=-1

r2=74, p2=-5

t=0.1fs, 2500 steps

successful reaction after several recrossing

show contour plot.

See that upon successful reaction, the kinetic energy of the products is significantly larger than that of the reactants, Due to a drop in PE via exothermic reaction to for HF, with the PE being converted to vibrational KE.

can see on fig2, shown in animation1

Confirm large increase via momenta vs time graph to see the large increase in momenta. Can see in fig 3 to see the oscillation between PE and KE , depicting the release of reaction energy.

can confirm via Calorimetry to find translation

in IR expect to see another smaller peak at smaller wavenumber for 2 to ground state due to higher vibrational states being occupied. Intensity of side band will get smaller over time as the product relaxes, causiong fundamental to increase proportionally-

Or observe the emmision and notice a new peak representing the 1 to ground state emmision when product forms initially, slowly decreases in size as energy lost via other processes.- Technique called ___

=== 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 ===
F=A

r1=200, p1=-1, r2=200, t=0.1fs, 4000 steps
{| 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 rule:

File:ZN 01510257 Q10 Figure7 2.gif

2020-05-22T21:49:32Z

Zn4318: The 7th gif in the table

The 7th gif in the table

File:ZN 01510257 Q10 Figure6 2.gif

2020-05-22T21:47:10Z

Zn4318: The 6th gif in the table

The 6th gif in the table

File:ZN 01510257 Q10 Figure5 2.gif

2020-05-22T21:46:29Z

Zn4318: The 5th gif in the table

The 5th gif in the table

File:ZN 01510257 Q10 Figure4 2.gif

2020-05-22T21:45:42Z

Zn4318: The 4th gif in the table

The 4th gif in the table

File:ZN 01510257 Q10 Figure3 2.gif

2020-05-22T21:44:57Z

Zn4318: The 3rd gif in the table

The 3rd gif in the table

File:ZN 01510257 Q10 Figure2 2.gif

2020-05-22T21:44:11Z

Zn4318: The second gif in the table

The second gif in the table

File:ZN 01510257 Q10 Figure1 2.gif

2020-05-22T21:43:22Z

Zn4318: The 1st gif in the table

The 1st gif in the table

File:ZN 01510257 Q10 Figure 7 1.png

2020-05-22T21:42:25Z

Zn4318: The 7th contour plot in the table

The 7th contour plot in the table

File:ZN 01510257 Q10 Figure 6 1.png

2020-05-22T21:41:39Z

Zn4318: The 6th contour plot in the table

The 6th contour plot in the table

File:ZN 01510257 Q10 Figure 5 1.png

2020-05-22T21:40:49Z

Zn4318: The fifth contour plot in the table

The fifth contour plot in the table

File:ZN 01510257 Q10 Figure 4 1.png

2020-05-22T21:40:09Z

Zn4318: The 4th contour plot in the table

The 4th contour plot in the table

File:ZN 01510257 Q10 Figure 3 1.png

2020-05-22T21:39:17Z

Zn4318: The third contour plot in the table

The third contour plot in the table

File:ZN 01510257 Q10 Figure 2 1.png

2020-05-22T21:38:33Z

Zn4318: The second contour plot in the table

The second contour plot in the table

File:ZN 01510257 Q10 Figure 1 1.png

2020-05-22T21:37:29Z

Zn4318: The first contour plot in the table

The first contour plot in the table

MRD:ZN 01510257

2020-05-22T21:35:56Z

Zn4318: /* Q8) Report the activation energy for both reactions. */

== 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.

=== 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.]]

=== 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.]]

=== 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.

=== 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.

== 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.

=== 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]]

=== 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.

=== 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. ===
r1 = 200, p1=-1

r2=74, p2=-5

t=0.1fs, 2500 steps

successful reaction after several recrossing

show contour plot.

See that upon successful reaction, the kinetic energy of the products is significantly larger than that of the reactants, Due to a drop in PE via exothermic reaction to for HF, with the PE being converted to vibrational KE.

can see on fig2, shown in animation1

Confirm large increase via momenta vs time graph to see the large increase in momenta. Can see in fig 3 to see the oscillation between PE and KE , depicting the release of reaction energy.

can confirm via Calorimetry to find translation

in IR expect to see another smaller peak at smaller wavenumber for 2 to ground state due to higher vibrational states being occupied. Intensity of side band will get smaller over time as the product relaxes, causiong fundamental to increase proportionally-

Or observe the emmision and notice a new peak representing the 1 to ground state emmision when product forms initially, slowly decreases in size as energy lost via other processes.- Technique called ___

=== 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 ===
F=A

r1=200, p1=-1, r2=200, t=0.1fs, 4000 steps
{| class="wikitable"
!p2/ g.mol-1.pm.fs-1
!Reaction successful?
!Animation of trajectory
!Illustration of the trajectory
|-
|5
|No
|
|
|-
|3
|Yes
|
|
|-
|1
|Yes
|
|
|-
|0
|Yes
|
|
|-
|<nowiki>-1</nowiki>
|Yes
|
|
|-
|<nowiki>-3</nowiki>
|No
|
|
|-
|<nowiki>-5</nowiki>
|Yes
|
|
|}

Polanyi's rule:

MRD:ZN 01510257

2020-05-22T21:18:18Z

Zn4318: /* Q8) Report the activation energy for both reactions. */

== 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.

=== 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.]]

=== 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.]]

=== 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.

=== 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.

== 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.

=== 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]]

=== 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.
{| 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
|-
|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
|-
|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
|}

=== 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. ===
r1 = 200, p1=-1

r2=74, p2=-5

t=0.1fs, 2500 steps

successful reaction after several recrossing

show contour plot.

See that upon successful reaction, the kinetic energy of the products is significantly larger than that of the reactants, Due to a drop in PE via exothermic reaction to for HF, with the PE being converted to vibrational KE.

can see on fig2, shown in animation1

Confirm large increase via momenta vs time graph to see the large increase in momenta. Can see in fig 3 to see the oscillation between PE and KE , depicting the release of reaction energy.

can confirm via Calorimetry to find translation

in IR expect to see another smaller peak at smaller wavenumber for 2 to ground state due to higher vibrational states being occupied. Intensity of side band will get smaller over time as the product relaxes, causiong fundamental to increase proportionally-

Or observe the emmision and notice a new peak representing the 1 to ground state emmision when product forms initially, slowly decreases in size as energy lost via other processes.- Technique called ___

=== 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 ===
F=A

r1=200, p1=-1, r2=200, t=0.1fs, 4000 steps
{| class="wikitable"
!p2/ g.mol-1.pm.fs-1
!Reaction successful?
!Animation of trajectory
!Illustration of the trajectory
|-
|5
|No
|
|
|-
|3
|Yes
|
|
|-
|1
|Yes
|
|
|-
|0
|Yes
|
|
|-
|<nowiki>-1</nowiki>
|Yes
|
|
|-
|<nowiki>-3</nowiki>
|No
|
|
|-
|<nowiki>-5</nowiki>
|Yes
|
|
|}

Polanyi's rule:

File:ZN 01510257 Q8 Figure 2 2.png

2020-05-22T21:13:04Z

Zn4318: Energy vs time plot of 2nd situation zoomed in

Energy vs time plot of 2nd situation zoomed in

File:ZN 01510257 Q8 Figure 2 1.png

2020-05-22T21:10:57Z

Zn4318: Zn4318 uploaded a new version of File:ZN 01510257 Q8 Figure 2 1.png

Energy vs time plot for 2nd situation

File:ZN 01510257 Q8 Figure 2 1.png

2020-05-22T21:10:38Z

Zn4318: Energy vs time plot for 2nd situation

Energy vs time plot for 2nd situation

File:ZN 01510257 Q8 Figure 1 2.png

2020-05-22T21:09:38Z

Zn4318: Energy vs time plot zoomed in for first situation

Energy vs time plot zoomed in for first situation

File:ZN 01510257 Q8 Figure 1 1.png

2020-05-22T21:08:20Z

Zn4318: Energy vs time plot for the first system.

Energy vs time plot for the first system.

MRD:ZN 01510257

2020-05-22T20:42:45Z

Zn4318: /* Q7) Locate the approximate position of the transition state. */

== 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.

=== 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.]]

=== 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.]]

=== 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.

=== 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.

== 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.

=== 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]]

=== Q8) Report the activation energy for both reactions. ===

for H2+F--->H + HF (HF+H reaction)

r1=r1ts-1

r2=r2ts

40,000 steps, t=0.1fs

Ets=-433.981

Eend=-560.290

asdasdasdasdasdasdasdasdasdasdasdasda

for HF+H--->F + H2 (F + H2 reaction)

r1=r1ts+1

r2=r2ts

40,000 steps, t=0.1fs

Ets=-433.981

Eend=-435.013

=== 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. ===
r1 = 200, p1=-1

r2=74, p2=-5

t=0.1fs, 2500 steps

successful reaction after several recrossing

show contour plot.

See that upon successful reaction, the kinetic energy of the products is significantly larger than that of the reactants, Due to a drop in PE via exothermic reaction to for HF, with the PE being converted to vibrational KE.

can see on fig2, shown in animation1

Confirm large increase via momenta vs time graph to see the large increase in momenta. Can see in fig 3 to see the oscillation between PE and KE , depicting the release of reaction energy.

can confirm via Calorimetry to find translation

in IR expect to see another smaller peak at smaller wavenumber for 2 to ground state due to higher vibrational states being occupied. Intensity of side band will get smaller over time as the product relaxes, causiong fundamental to increase proportionally-

Or observe the emmision and notice a new peak representing the 1 to ground state emmision when product forms initially, slowly decreases in size as energy lost via other processes.- Technique called ___

=== 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 ===
F=A

r1=200, p1=-1, r2=200, t=0.1fs, 4000 steps
{| class="wikitable"
!p2/ g.mol-1.pm.fs-1
!Reaction successful?
!Animation of trajectory
!Illustration of the trajectory
|-
|5
|No
|
|
|-
|3
|Yes
|
|
|-
|1
|Yes
|
|
|-
|0
|Yes
|
|
|-
|<nowiki>-1</nowiki>
|Yes
|
|
|-
|<nowiki>-3</nowiki>
|No
|
|
|-
|<nowiki>-5</nowiki>
|Yes
|
|
|}

Polanyi's rule:

MRD:ZN 01510257

2020-05-22T20:41:10Z

Zn4318: /* Q7) Locate the approximate position of the transition state. */

== 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.

=== 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.]]

=== 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.]]

=== 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.

=== 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.

== 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.

=== 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. We can then confirm this is the case by looking at the internuclear distance against time plot to see no oscillations as done with the H2 + H system. Internuclear distance vs time plot of final approximation of the TS.
[[File:ZN 01510257 Q7 Figure 3.png|left|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.]]
[[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.]]

=== Q8) Report the activation energy for both reactions. ===

for H2+F--->H + HF (HF+H reaction)

r1=r1ts-1

r2=r2ts

40,000 steps, t=0.1fs

Ets=-433.981

Eend=-560.290

asdasdasdasdasdasdasdasdasdasdasdasda

for HF+H--->F + H2 (F + H2 reaction)

r1=r1ts+1

r2=r2ts

40,000 steps, t=0.1fs

Ets=-433.981

Eend=-435.013

=== 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. ===
r1 = 200, p1=-1

r2=74, p2=-5

t=0.1fs, 2500 steps

successful reaction after several recrossing

show contour plot.

See that upon successful reaction, the kinetic energy of the products is significantly larger than that of the reactants, Due to a drop in PE via exothermic reaction to for HF, with the PE being converted to vibrational KE.

can see on fig2, shown in animation1

Confirm large increase via momenta vs time graph to see the large increase in momenta. Can see in fig 3 to see the oscillation between PE and KE , depicting the release of reaction energy.

can confirm via Calorimetry to find translation

in IR expect to see another smaller peak at smaller wavenumber for 2 to ground state due to higher vibrational states being occupied. Intensity of side band will get smaller over time as the product relaxes, causiong fundamental to increase proportionally-

Or observe the emmision and notice a new peak representing the 1 to ground state emmision when product forms initially, slowly decreases in size as energy lost via other processes.- Technique called ___

=== 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 ===
F=A

r1=200, p1=-1, r2=200, t=0.1fs, 4000 steps
{| class="wikitable"
!p2/ g.mol-1.pm.fs-1
!Reaction successful?
!Animation of trajectory
!Illustration of the trajectory
|-
|5
|No
|
|
|-
|3
|Yes
|
|
|-
|1
|Yes
|
|
|-
|0
|Yes
|
|
|-
|<nowiki>-1</nowiki>
|Yes
|
|
|-
|<nowiki>-3</nowiki>
|No
|
|
|-
|<nowiki>-5</nowiki>
|Yes
|
|
|}

Polanyi's rule:

File:ZN 01510257 Q7 Figure 4.png

2020-05-22T20:36:09Z

Zn4318: Internuclear distance vs time plot of final approximation of the TS

Internuclear distance vs time plot of final approximation of the TS

File:ZN 01510257 Q7 Figure 3.png

2020-05-22T20:35:00Z

Zn4318: Final approximation for the TS for a FH + H system by ZN

Final approximation for the TS for a FH + H system by ZN

File:ZN 01510257 Q7 Figure 2.png

2020-05-22T20:26:22Z

Zn4318: Second approximation for the transition state position.

Second approximation for the transition state position.

File:ZN 01510257 Q7 Figure 1.png

2020-05-22T20:21:19Z

Zn4318: Dynamic calculation of first approximation of the HF + H TS.

Dynamic calculation of first approximation of the HF + H TS.

MRD:ZN 01510257

2020-05-22T20:11:37Z

Zn4318:

MRD:ZN 01510257

2020-05-22T20:10:55Z

Zn4318: /* EXERCISE 2: F - H - H system */

MUST ANSWER ANY AND EVERY Question

contour plot is the PES, with trajectory overlay-ed

Skew plot shows new coordinate system designed such that force acting on trajectory is orthogonal to contours.

surface plot: PES in r1 and r2 coords (wavy trail=vibrating molecule)
ALWAYS OSCILLATING BETWEEN SAME CONTOUR LINES (conservation of E)

Internuclear-distance with t: see r2 (colliding atom with molecule) decreases and levels off/ vibrates. r1 starts vibrating but then increases. r1=r2 at TS (not always true)

Internuclear-velocity with t: Oscilation about 0 is vibration, average positive and negative values show the 2 atoms distancing or coming together

E vs t: Constantly swapping between KE and PE, sum is the same. if total E seems to decay, use smaller time step. Bottom of trough of oscillation, PE is low and KE is high. KE at minimum here at TS.

momenta vs t: momenta corresponds to bond stretching displacements of r1 and r2
atom 2 momentum effects r1 and r2 momentum. So to collide 3 with molecule 1-2, by giving some p2, such that it does not vibrate, then some p1 must by given as well.

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

== 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.

=== 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.]]

=== 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.]]

=== 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.

=== 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.

== 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.

=== Q7) Locate the approximate position of the transition state. ===
first assume the H2 bond distance, 74pm and HF (AB) was 100. Done as Late TS so hammond postulate so TS should resemble H2 and a F atom

Fig 1, BC slightly to small as seems to have gone up contours, AB much too small. Start by changing AB. When the force along AB was approximately 0 to 2 dp, BC was adjusted.

Fig 2 r1=178.5 (F is now 0 along AB), r2=74 (F is still large)

In this region, the BC bond and AB bond effect each-others force so from further adjusting both parameters, a final value of

for A=F, B, C=H

r1=181.040

r2=74.490

as shown in fig 3.

insert fig 4 to show constant internuclear distance

=== Q8) Report the activation energy for both reactions. ===

for H2+F--->H + HF (HF+H reaction)

r1=r1ts-1

r2=r2ts

40,000 steps, t=0.1fs

Ets=-433.981

Eend=-560.290

asdasdasdasdasdasdasdasdasdasdasdasda

for HF+H--->F + H2 (F + H2 reaction)

r1=r1ts+1

r2=r2ts

40,000 steps, t=0.1fs

Ets=-433.981

Eend=-435.013

=== 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. ===
r1 = 200, p1=-1

r2=74, p2=-5

t=0.1fs, 2500 steps

successful reaction after several recrossing

show contour plot.

See that upon successful reaction, the kinetic energy of the products is significantly larger than that of the reactants, Due to a drop in PE via exothermic reaction to for HF, with the PE being converted to vibrational KE.

can see on fig2, shown in animation1

Confirm large increase via momenta vs time graph to see the large increase in momenta. Can see in fig 3 to see the oscillation between PE and KE , depicting the release of reaction energy.

can confirm via Calorimetry to find translation

in IR expect to see another smaller peak at smaller wavenumber for 2 to ground state due to higher vibrational states being occupied. Intensity of side band will get smaller over time as the product relaxes, causiong fundamental to increase proportionally-

Or observe the emmision and notice a new peak representing the 1 to ground state emmision when product forms initially, slowly decreases in size as energy lost via other processes.- Technique called ___

=== 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 ===
F=A

r1=200, p1=-1, r2=200, t=0.1fs, 4000 steps
{| class="wikitable"
!p2/ g.mol-1.pm.fs-1
!Reaction successful?
!Animation of trajectory
!Illustration of the trajectory
|-
|5
|No
|
|
|-
|3
|Yes
|
|
|-
|1
|Yes
|
|
|-
|0
|Yes
|
|
|-
|<nowiki>-1</nowiki>
|Yes
|
|
|-
|<nowiki>-3</nowiki>
|No
|
|
|-
|<nowiki>-5</nowiki>
|Yes
|
|
|}

Polanyi's rule:

File:ZN 01510257 Q6 Figure 1.png

2020-05-22T19:53:38Z

Zn4318: Zn4318 uploaded a new version of File:ZN 01510257 Q6 Figure 1.png

PES of FH + H system

File:ZN 01510257 Q6 Figure 1.png

2020-05-22T19:53:08Z

Zn4318: PES of FH + H system

PES of FH + H system

MRD:ZN 01510257

2020-05-22T19:42:39Z

Zn4318: /* 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? */

MUST ANSWER ANY AND EVERY Question

contour plot is the PES, with trajectory overlay-ed

Skew plot shows new coordinate system designed such that force acting on trajectory is orthogonal to contours.

surface plot: PES in r1 and r2 coords (wavy trail=vibrating molecule)
ALWAYS OSCILLATING BETWEEN SAME CONTOUR LINES (conservation of E)

Internuclear-distance with t: see r2 (colliding atom with molecule) decreases and levels off/ vibrates. r1 starts vibrating but then increases. r1=r2 at TS (not always true)

Internuclear-velocity with t: Oscilation about 0 is vibration, average positive and negative values show the 2 atoms distancing or coming together

E vs t: Constantly swapping between KE and PE, sum is the same. if total E seems to decay, use smaller time step. Bottom of trough of oscillation, PE is low and KE is high. KE at minimum here at TS.

momenta vs t: momenta corresponds to bond stretching displacements of r1 and r2
atom 2 momentum effects r1 and r2 momentum. So to collide 3 with molecule 1-2, by giving some p2, such that it does not vibrate, then some p1 must by given as well.

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

== 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.

=== 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.]]

=== 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.]]

=== 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.

=== 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.

== EXERCISE 2: F - H - H system ==
If the system is non equivalent in energies, e.g. endo or exo, so

HF is lower in E as ionic contributions so H +HF - HF should be exothermic so TS resembles HH

=== 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? ===

HF is lower in E as ionic contributions so H +HF -- H2 + F should be endothermic so TS resembles HH

AB bond distance constant, lower in E so HF more stable that HH. etc. compare to bond dissociation E

insert fig showing exothermic E via Z axis

=== Q7) Locate the approximate position of the transition state. ===
first assume the H2 bond distance, 74pm and HF (AB) was 100. Done as Late TS so hammond postulate so TS should resemble H2 and a F atom

Fig 1, BC slightly to small as seems to have gone up contours, AB much too small. Start by changing AB. When the force along AB was approximately 0 to 2 dp, BC was adjusted.

Fig 2 r1=178.5 (F is now 0 along AB), r2=74 (F is still large)

In this region, the BC bond and AB bond effect each-others force so from further adjusting both parameters, a final value of

for A=F, B, C=H

r1=181.040

r2=74.490

as shown in fig 3.

insert fig 4 to show constant internuclear distance

=== Q8) Report the activation energy for both reactions. ===

for H2+F--->H + HF (HF+H reaction)

r1=r1ts-1

r2=r2ts

40,000 steps, t=0.1fs

Ets=-433.981

Eend=-560.290

asdasdasdasdasdasdasdasdasdasdasdasda

for HF+H--->F + H2 (F + H2 reaction)

r1=r1ts+1

r2=r2ts

40,000 steps, t=0.1fs

Ets=-433.981

Eend=-435.013

=== 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. ===
r1 = 200, p1=-1

r2=74, p2=-5

t=0.1fs, 2500 steps

successful reaction after several recrossing

show contour plot.

See that upon successful reaction, the kinetic energy of the products is significantly larger than that of the reactants, Due to a drop in PE via exothermic reaction to for HF, with the PE being converted to vibrational KE.

can see on fig2, shown in animation1

Confirm large increase via momenta vs time graph to see the large increase in momenta. Can see in fig 3 to see the oscillation between PE and KE , depicting the release of reaction energy.

can confirm via Calorimetry to find translation

in IR expect to see another smaller peak at smaller wavenumber for 2 to ground state due to higher vibrational states being occupied. Intensity of side band will get smaller over time as the product relaxes, causiong fundamental to increase proportionally-

Or observe the emmision and notice a new peak representing the 1 to ground state emmision when product forms initially, slowly decreases in size as energy lost via other processes.- Technique called ___

=== 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 ===
F=A

r1=200, p1=-1, r2=200, t=0.1fs, 4000 steps
{| class="wikitable"
!p2/ g.mol-1.pm.fs-1
!Reaction successful?
!Animation of trajectory
!Illustration of the trajectory
|-
|5
|No
|
|
|-
|3
|Yes
|
|
|-
|1
|Yes
|
|
|-
|0
|Yes
|
|
|-
|<nowiki>-1</nowiki>
|Yes
|
|
|-
|<nowiki>-3</nowiki>
|No
|
|
|-
|<nowiki>-5</nowiki>
|Yes
|
|
|}

Polanyi's rule:

MRD:ZN 01510257

2020-05-22T19:12:40Z

MUST ANSWER ANY AND EVERY Question

contour plot is the PES, with trajectory overlay-ed

Skew plot shows new coordinate system designed such that force acting on trajectory is orthogonal to contours.

surface plot: PES in r1 and r2 coords (wavy trail=vibrating molecule)
ALWAYS OSCILLATING BETWEEN SAME CONTOUR LINES (conservation of E)

Internuclear-distance with t: see r2 (colliding atom with molecule) decreases and levels off/ vibrates. r1 starts vibrating but then increases. r1=r2 at TS (not always true)

Internuclear-velocity with t: Oscilation about 0 is vibration, average positive and negative values show the 2 atoms distancing or coming together

E vs t: Constantly swapping between KE and PE, sum is the same. if total E seems to decay, use smaller time step. Bottom of trough of oscillation, PE is low and KE is high. KE at minimum here at TS.

momenta vs t: momenta corresponds to bond stretching displacements of r1 and r2
atom 2 momentum effects r1 and r2 momentum. So to collide 3 with molecule 1-2, by giving some p2, such that it does not vibrate, then some p1 must by given as well.

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

== 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.

=== 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.]]

=== 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.]]

=== 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? ===

r1=74 pm and r2= 200 pm

{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|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]]
|}

=== Q5) Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===

TS assumes:

transition state to products, no reverse. Clearly false as seen by recrossing.(so overestimates)

assumes no tunneling - considers systerm as purely classical(so underestimates-relative contrivbution?)

Assumes all reactions with sufficient energy will form products, false as seen in table. (so overestimates reaction rate)

== EXERCISE 2: F - H - H system ==
If the system is non equivalent in energies, e.g. endo or exo, so

HF is lower in E as ionic contributions so H +HF - HF should be exothermic so TS resembles HH

=== 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? ===

HF is lower in E as ionic contributions so H +HF -- H2 + F should be endothermic so TS resembles HH

AB bond distance constant, lower in E so HF more stable that HH. etc. compare to bond dissociation E

insert fig showing exothermic E via Z axis

=== Q7) Locate the approximate position of the transition state. ===
first assume the H2 bond distance, 74pm and HF (AB) was 100. Done as Late TS so hammond postulate so TS should resemble H2 and a F atom

Fig 1, BC slightly to small as seems to have gone up contours, AB much too small. Start by changing AB. When the force along AB was approximately 0 to 2 dp, BC was adjusted.

Fig 2 r1=178.5 (F is now 0 along AB), r2=74 (F is still large)

In this region, the BC bond and AB bond effect each-others force so from further adjusting both parameters, a final value of

for A=F, B, C=H

r1=181.040

r2=74.490

as shown in fig 3.

insert fig 4 to show constant internuclear distance

=== Q8) Report the activation energy for both reactions. ===

for H2+F--->H + HF (HF+H reaction)

r1=r1ts-1

r2=r2ts

40,000 steps, t=0.1fs

Ets=-433.981

Eend=-560.290

asdasdasdasdasdasdasdasdasdasdasdasda

for HF+H--->F + H2 (F + H2 reaction)

r1=r1ts+1

r2=r2ts

40,000 steps, t=0.1fs

Ets=-433.981

Eend=-435.013

=== 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. ===
r1 = 200, p1=-1

r2=74, p2=-5

t=0.1fs, 2500 steps

successful reaction after several recrossing

show contour plot.

See that upon successful reaction, the kinetic energy of the products is significantly larger than that of the reactants, Due to a drop in PE via exothermic reaction to for HF, with the PE being converted to vibrational KE.

can see on fig2, shown in animation1

Confirm large increase via momenta vs time graph to see the large increase in momenta. Can see in fig 3 to see the oscillation between PE and KE , depicting the release of reaction energy.

can confirm via Calorimetry to find translation

in IR expect to see another smaller peak at smaller wavenumber for 2 to ground state due to higher vibrational states being occupied. Intensity of side band will get smaller over time as the product relaxes, causiong fundamental to increase proportionally-

Or observe the emmision and notice a new peak representing the 1 to ground state emmision when product forms initially, slowly decreases in size as energy lost via other processes.- Technique called ___

=== 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 ===
F=A

r1=200, p1=-1, r2=200, t=0.1fs, 4000 steps
{| class="wikitable"
!p2/ g.mol-1.pm.fs-1
!Reaction successful?
!Animation of trajectory
!Illustration of the trajectory
|-
|5
|No
|
|
|-
|3
|Yes
|
|
|-
|1
|Yes
|
|
|-
|0
|Yes
|
|
|-
|<nowiki>-1</nowiki>
|Yes
|
|
|-
|<nowiki>-3</nowiki>
|No
|
|
|-
|<nowiki>-5</nowiki>
|Yes
|
|
|}

Polanyi's rule:

MRD:ZN 01510257

2020-05-22T18:57:23Z

Zn4318: /* Q3) Comment on how the mep and the trajectory you just calculated differ. */

MUST ANSWER ANY AND EVERY Question

contour plot is the PES, with trajectory overlay-ed

Skew plot shows new coordinate system designed such that force acting on trajectory is orthogonal to contours.

surface plot: PES in r1 and r2 coords (wavy trail=vibrating molecule)
ALWAYS OSCILLATING BETWEEN SAME CONTOUR LINES (conservation of E)

Internuclear-distance with t: see r2 (colliding atom with molecule) decreases and levels off/ vibrates. r1 starts vibrating but then increases. r1=r2 at TS (not always true)

Internuclear-velocity with t: Oscilation about 0 is vibration, average positive and negative values show the 2 atoms distancing or coming together

E vs t: Constantly swapping between KE and PE, sum is the same. if total E seems to decay, use smaller time step. Bottom of trough of oscillation, PE is low and KE is high. KE at minimum here at TS.

momenta vs t: momenta corresponds to bond stretching displacements of r1 and r2
atom 2 momentum effects r1 and r2 momentum. So to collide 3 with molecule 1-2, by giving some p2, such that it does not vibrate, then some p1 must by given as well.

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

== 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.

=== 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.]]

=== 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.]]

=== 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? ===

Q4) Complete the table above by adding the total energy, whether the trajectory is reactive or nonreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

r1=74 pm and r2= 200 pm

{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Yes
|
|[[File:ZN 01510257 Q4 Figure 1.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.08</nowiki>
|No
|
|[[File:ZN 01510257 Q4 Figure 2.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.97</nowiki>
|Yes
|
|[[File:ZN 01510257 Q4 Figure 3.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.10</nowiki>
|No
|
|[[File:ZN 01510257 Q4 Figure 4.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.48</nowiki>
|Yes
|
|[[File:ZN 01510257 Q4 Figure 5.png|thumb]]
|}

=== Q5) Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===

TS assumes:

transition state to products, no reverse. Clearly false as seen by recrossing.(so overestimates)

assumes no tunneling - considers systerm as purely classical(so underestimates-relative contrivbution?)

Assumes all reactions with sufficient energy will form products, false as seen in table. (so overestimates reaction rate)

== EXERCISE 2: F - H - H system ==
If the system is non equivalent in energies, e.g. endo or exo, so

HF is lower in E as ionic contributions so H +HF - HF should be exothermic so TS resembles HH

=== 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? ===

HF is lower in E as ionic contributions so H +HF -- H2 + F should be endothermic so TS resembles HH

AB bond distance constant, lower in E so HF more stable that HH. etc. compare to bond dissociation E

insert fig showing exothermic E via Z axis

=== Q7) Locate the approximate position of the transition state. ===
first assume the H2 bond distance, 74pm and HF (AB) was 100. Done as Late TS so hammond postulate so TS should resemble H2 and a F atom

Fig 1, BC slightly to small as seems to have gone up contours, AB much too small. Start by changing AB. When the force along AB was approximately 0 to 2 dp, BC was adjusted.

Fig 2 r1=178.5 (F is now 0 along AB), r2=74 (F is still large)

In this region, the BC bond and AB bond effect each-others force so from further adjusting both parameters, a final value of

for A=F, B, C=H

r1=181.040

r2=74.490

as shown in fig 3.

insert fig 4 to show constant internuclear distance

=== Q8) Report the activation energy for both reactions. ===

for H2+F--->H + HF (HF+H reaction)

r1=r1ts-1

r2=r2ts

40,000 steps, t=0.1fs

Ets=-433.981

Eend=-560.290

asdasdasdasdasdasdasdasdasdasdasdasda

for HF+H--->F + H2 (F + H2 reaction)

r1=r1ts+1

r2=r2ts

40,000 steps, t=0.1fs

Ets=-433.981

Eend=-435.013

=== 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. ===
r1 = 200, p1=-1

r2=74, p2=-5

t=0.1fs, 2500 steps

successful reaction after several recrossing

show contour plot.

See that upon successful reaction, the kinetic energy of the products is significantly larger than that of the reactants, Due to a drop in PE via exothermic reaction to for HF, with the PE being converted to vibrational KE.

can see on fig2, shown in animation1

Confirm large increase via momenta vs time graph to see the large increase in momenta. Can see in fig 3 to see the oscillation between PE and KE , depicting the release of reaction energy.

can confirm via Calorimetry to find translation

in IR expect to see another smaller peak at smaller wavenumber for 2 to ground state due to higher vibrational states being occupied. Intensity of side band will get smaller over time as the product relaxes, causiong fundamental to increase proportionally-

Or observe the emmision and notice a new peak representing the 1 to ground state emmision when product forms initially, slowly decreases in size as energy lost via other processes.- Technique called ___

=== 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 ===
F=A

r1=200, p1=-1, r2=200, t=0.1fs, 4000 steps
{| class="wikitable"
!p2/ g.mol-1.pm.fs-1
!Reaction successful?
!Animation of trajectory
!Illustration of the trajectory
|-
|5
|No
|
|
|-
|3
|Yes
|
|
|-
|1
|Yes
|
|
|-
|0
|Yes
|
|
|-
|<nowiki>-1</nowiki>
|Yes
|
|
|-
|<nowiki>-3</nowiki>
|No
|
|
|-
|<nowiki>-5</nowiki>
|Yes
|
|
|}

Polanyi's rule:

File:ZN 01510257 Q3 Figure 5.png

2020-05-22T18:52:18Z

Zn4318: Dynamic calculated trajectory of the H2 + H system when displaced about r2 from TS.

Dynamic calculated trajectory of the H2 + H system when displaced about r2 from TS.

MRD:ZN 01510257

2020-05-22T18:31:30Z

Zn4318:

MUST ANSWER ANY AND EVERY Question

contour plot is the PES, with trajectory overlay-ed

Skew plot shows new coordinate system designed such that force acting on trajectory is orthogonal to contours.

surface plot: PES in r1 and r2 coords (wavy trail=vibrating molecule)
ALWAYS OSCILLATING BETWEEN SAME CONTOUR LINES (conservation of E)

Internuclear-distance with t: see r2 (colliding atom with molecule) decreases and levels off/ vibrates. r1 starts vibrating but then increases. r1=r2 at TS (not always true)

Internuclear-velocity with t: Oscilation about 0 is vibration, average positive and negative values show the 2 atoms distancing or coming together

E vs t: Constantly swapping between KE and PE, sum is the same. if total E seems to decay, use smaller time step. Bottom of trough of oscillation, PE is low and KE is high. KE at minimum here at TS.

momenta vs t: momenta corresponds to bond stretching displacements of r1 and r2
atom 2 momentum effects r1 and r2 momentum. So to collide 3 with molecule 1-2, by giving some p2, such that it does not vibrate, then some p1 must by given as well.

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

== 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.

=== 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.]]

=== 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).
[[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]]

just follows the valley that is the BC bond with A atom- no motion via vibration etc

dynamic version shows motion via vibration etc

if you do r1=90.775 and r2=91.775, AB bond forms instead of BC

for dynamic reverse:

at t=2500 fs

r1=1874.29

r2=74.92

p1=5.07

p2=2.56

PRESS GET LAST GEOMETRY- WILL RETRACE PATH

Returns to TS, unable to overcome barrier. shows this as it is essentially retracing it's steps, and as we have placed the molecule to one side of the TS, they do not have sufficient energy to overcome this barrier so the collision should fail. If done more accurately, the calc should end right at TS. Can't check this via mep as momenta are 0 at every step by definition.

=== 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? ===

Q4) Complete the table above by adding the total energy, whether the trajectory is reactive or nonreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

r1=74 pm and r2= 200 pm

{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Yes
|
|[[File:ZN 01510257 Q4 Figure 1.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.08</nowiki>
|No
|
|[[File:ZN 01510257 Q4 Figure 2.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.97</nowiki>
|Yes
|
|[[File:ZN 01510257 Q4 Figure 3.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.10</nowiki>
|No
|
|[[File:ZN 01510257 Q4 Figure 4.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.48</nowiki>
|Yes
|
|[[File:ZN 01510257 Q4 Figure 5.png|thumb]]
|}

=== Q5) Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===

TS assumes:

transition state to products, no reverse. Clearly false as seen by recrossing.(so overestimates)

assumes no tunneling - considers systerm as purely classical(so underestimates-relative contrivbution?)

Assumes all reactions with sufficient energy will form products, false as seen in table. (so overestimates reaction rate)

== EXERCISE 2: F - H - H system ==
If the system is non equivalent in energies, e.g. endo or exo, so

HF is lower in E as ionic contributions so H +HF - HF should be exothermic so TS resembles HH

=== 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? ===

HF is lower in E as ionic contributions so H +HF -- H2 + F should be endothermic so TS resembles HH

AB bond distance constant, lower in E so HF more stable that HH. etc. compare to bond dissociation E

insert fig showing exothermic E via Z axis

=== Q7) Locate the approximate position of the transition state. ===
first assume the H2 bond distance, 74pm and HF (AB) was 100. Done as Late TS so hammond postulate so TS should resemble H2 and a F atom

Fig 1, BC slightly to small as seems to have gone up contours, AB much too small. Start by changing AB. When the force along AB was approximately 0 to 2 dp, BC was adjusted.

Fig 2 r1=178.5 (F is now 0 along AB), r2=74 (F is still large)

In this region, the BC bond and AB bond effect each-others force so from further adjusting both parameters, a final value of

for A=F, B, C=H

r1=181.040

r2=74.490

as shown in fig 3.

insert fig 4 to show constant internuclear distance

=== Q8) Report the activation energy for both reactions. ===

for H2+F--->H + HF (HF+H reaction)

r1=r1ts-1

r2=r2ts

40,000 steps, t=0.1fs

Ets=-433.981

Eend=-560.290

asdasdasdasdasdasdasdasdasdasdasdasda

for HF+H--->F + H2 (F + H2 reaction)

r1=r1ts+1

r2=r2ts

40,000 steps, t=0.1fs

Ets=-433.981

Eend=-435.013

=== 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. ===
r1 = 200, p1=-1

r2=74, p2=-5

t=0.1fs, 2500 steps

successful reaction after several recrossing

show contour plot.

See that upon successful reaction, the kinetic energy of the products is significantly larger than that of the reactants, Due to a drop in PE via exothermic reaction to for HF, with the PE being converted to vibrational KE.

can see on fig2, shown in animation1

Confirm large increase via momenta vs time graph to see the large increase in momenta. Can see in fig 3 to see the oscillation between PE and KE , depicting the release of reaction energy.

can confirm via Calorimetry to find translation

in IR expect to see another smaller peak at smaller wavenumber for 2 to ground state due to higher vibrational states being occupied. Intensity of side band will get smaller over time as the product relaxes, causiong fundamental to increase proportionally-

Or observe the emmision and notice a new peak representing the 1 to ground state emmision when product forms initially, slowly decreases in size as energy lost via other processes.- Technique called ___

=== 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 ===
F=A

r1=200, p1=-1, r2=200, t=0.1fs, 4000 steps
{| class="wikitable"
!p2/ g.mol-1.pm.fs-1
!Reaction successful?
!Animation of trajectory
!Illustration of the trajectory
|-
|5
|No
|
|
|-
|3
|Yes
|
|
|-
|1
|Yes
|
|
|-
|0
|Yes
|
|
|-
|<nowiki>-1</nowiki>
|Yes
|
|
|-
|<nowiki>-3</nowiki>
|No
|
|
|-
|<nowiki>-5</nowiki>
|Yes
|
|
|}

Polanyi's rule:

File:ZN 01510257 Q3 Figure 4.png

2020-05-22T18:31:08Z

Zn4318: Reversed dynamic calculation of the H2 + H system slightly displaced from the TS.

Reversed dynamic calculation of the H2 + H system slightly displaced from the TS.

File:ZN 01510257 Q4 Figure 4.png

2020-05-22T18:30:04Z

Zn4318: Zn4318 uploaded a new version of File:ZN 01510257 Q4 Figure 4.png

Surface plot of the 4th trajectory as shown in the table.

File:ZN 01510257 Q4 Figure 4.png

2020-05-22T18:29:21Z

Zn4318: Zn4318 uploaded a new version of File:ZN 01510257 Q4 Figure 4.png

Surface plot of the 4th trajectory as shown in the table.

File:ZN 01510257 Q3 Figure 3.png

2020-05-22T18:19:56Z

Zn4318: mep calculated energy vs time graph of H2+H reaction

mep calculated energy vs time graph of H2+H reaction