= Molecular Reaction Dynamics: Applications to Triatomic systems...........Anjli Suchak =

== Introduction ==
The main aims of this experiment were to study the reactivity of triatomic systems, involving the collision of an atom and a diatomic molecule, by calculating Molecular Dynamic trajectories. The systems studied were the symmetric H-H-H system and the non-symmetric F-H-H system.

== Exercise 1: The H + H2 system ==
The original H2 diatomic is defined as BC with the A H atom colliding to form the AB H2 diatomic molecule.

==== On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ====
[[File:H3 TS contourplot 01524183.png|thumb|292x292px|Figure 1. A contour plot showing the TS of the H-H-H system and the variation of AB and BC distances, with reaction trajectory and eigenvectors shown.]]
On a potential energy surface diagram, the transition state (TS) is mathematically defined as the point at which the gradient of the potential with respect to the internal coordinates is zero (∂V('''ri''')/∂'''ri'''=0). This is because the transition state is the local maximum on the minimum energy paths between the reactants and products. There would be no movement of the H atoms at the TS since the force is a derivative of the potential energy and at the TS this is zero therefore the transition state is a stationary point. However, minima are also stationary points. The TS can be distinguished from a local minimum of the potential energy surface by taking the second differential of the potential function. By doing so, the nature of the stationary point can be determined: if the second differential is greater than zero, the point is a local minimum and if it is less than zero, the point is a local maximum (the TS).

Using the GUI simulation, the TS was identified by trial and error. At the TS the distance AB and BC must be equal (rAB = rBC), as the H + H2 surface is symmetric. Therefore the TS lies somewhere along the diagonal line where AB and BC are equal. As the TS is approached, the size of the forces along AB and BC decreases to zero and so the point at which these forces are close to zero is the TS of the potential energy surface. This maximum point can be distinguished from a local minimum by analysing the omega squared values under the Hessian eigenvalues/vectors. A TS is a maximum in one direction and a minimum in the other direction; it is a saddle point. Therefore at the TS, the omega squared would be positive in one direction and negative in the other direction.

==== 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 transition state position ('''rts''') was found to be '''90.8 pm''' to 1 d.p.[[File:H3 TS distancetime 01524183.png|thumb|Figure 2. A plot of Internuclear Distances vs Time: linear horizontal lines show the constant distances of AB, BC and AC. |359x359px]]This value gave forces of +0.001 kJ.mol-1.pm-1 along AB and BC which are approx. zero. Also, the omega hessian eigenvalues had opposite signs which verified that this point was the TS.

At the TS, the internuclear distances between AB and BC are the same and so the distance between AC is approximately double the AB (BC) distance. This is shown in the plot of "Internuclear distances vs Time" (Figure 2), as two straight lines of constant distances of AB and BC at 90.8 pm and AC at approximately 181.6 pm. This is because at the TS the H atoms are stationary, with one H atom (the B atom in this case) being approximately halfway between the A and C H atoms.

==== Comment on how the ''mep'' and the trajectory you just calculated differ. ====
The minimum energy path (''mep) ''is the lowest energy path from reactant to product; it is a special trajectory that correlates to infinitely slow motion. The ''mep ''was calculated by setting the initial conditions as rBC = 91.8 pm (1 pm displaced from the TS position) and rAB = 90.8 pm (the TS position). The number of steps was also increased to 2500 in order to obtain a complete mep. This calculation was repeated with the calculation type set to 'Dynamics'.
[[File:H3 dynamics mep 01524183.png|thumb|601x601px|Figure 3. A contour plot of BC distance vs AB distance, showing the ''mep ''trajectory: calculation type was set to MEP (left).
Figure 4. A contour plot of BC distance vs AB distance, showing the dynamic'' ''trajectory: calculation type was set to Dynamics (right).
|centre]]
As shown in the plots above, the ''mep ''is a smooth trajectory (left) which indicates that the BC molecule is not vibrating as the system is not oscillating between energy contours. This is because the ''mep'' doesn't provide a realistic view of the motion of atoms during a reaction; it assumes the atoms are moving at an infinitely slow speed. However, there would be vibration occurring between B and C as they are moving further away from each other. This is shown by the wavy trajectory produced from the dynamic calculation (right).

==== Complete the table 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? ====
For the initial positions rBC = 74 pm and rAB = 200 pm, trajectories were run for five different momenta combinations (p1 is pBC and p2 is pAB).

{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot / kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Reactive
|The trajectory crosses the TS therefore it is reactive. The line is wavier after crossing the TS which means the initial diatomic BC has less vibrational energy compared to the forming AB diatomic molecule. However, overall the trajectory is not very wavy therefore there is less vibrational energy compared to translational energy in the whole system.
|[[File:H3 1 01524183.png|thumb|375x375px|Figure 5]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.08</nowiki>
|Unreactive
|The trajectory does not cross the TS therefore it is unreactive. The AB distance is changing as the A H atom is moving towards and away from the BC diatomic molecule, which means it has more translational energy compared to vibrational. This is shown in the plot as the line is not very wavy. However, the BC distance doesn't change as the diatomic molecule remains intact.
|[[File:H3 2 01524183.png|thumb|375x375px|Figure 6]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.98</nowiki>
|Reactive
|The trajectory does cross the TS therefore it is reactive. The line is quite wavy which means there is some vibrational energy but more translational energy. This is quite similar to the plot in Figure 5 as the potential energies are the same therefore the waviness of the trajectories are the same. However, the kinetic energies are a little different, making the total energies different.
|[[File:H3 3 01524183.png|thumb|375x375px|Figure 7]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.28</nowiki>
|Unreactive
|The trajectory just crosses the TS however it does not enter the product channel instead it goes back to the reactant channel. This is called barrier re-crossing. Therefore it is overall unreactive. The trajectory is very wavy which means there is a lot of vibrational energy in the system and less translational energy. In the beginning, the AB and BC distances are changing as the middle B atom is oscillating between the A and C atoms. However, in the end, the molecule BC is reformed which is why the AB distance is large.
|[[File:H3 4 01524183.png|thumb|375x375px|Figure 8]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.48</nowiki>
|Reactive
|The trajectory is reactive as it does cross the TS. The line is very wavy which shows that the system has more vibrational energy compared to translational. As the A atom gets closer to the initial BC molecule, the AB distance decreases and the BC distance increases, however, at the TS the B atom oscillates between the A and C atoms as it is attracted to both. This means there's a lot of vibrational energy near the TS. In the end, it does form the AB product molecule, which is shown by the trajectory increase in BC distance.
|[[File:H3 5 01524183.png|thumb|375x375px|Figure 9]]
|}
From the table, it can be concluded that not all trajectories starting with the same positions but with higher values of momenta would be reactive. For example, increasing the value of momenta to -5.1 g.mol-1.pm.fs-1 for p1 and -10.1 g.mol-1.pm.fs-1 for p2 still gives an unreactive trajectory. Also, low values of momenta can be reactive as is the first momenta combination (Figure 5). This is because vibrational energy also needs to be taken into account, not only the kinetic energy. The degree of vibrational and translational energy is determined by analysing the trajectory line shape: a more wavy line would mean more vibrational energy whereas a more smooth line corresponds to more translational energy.

==== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ====
The TS theory is a theory used to determine absolute reaction rates based on the reactant and transition state structure properties. A few important assumptions this theory makes are<ref name=":0">I. Steinfeld, J. S. Francisco, W. L. Hase, Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998, Ch. 10, pg. 290</ref>:
# Every trajectory that has enough energy to roll over the TS will definitely go to form the products. This means that once the reactants have crossed the energy barrier they can't go back to form reactants.<ref name=":0" /> Therefore the overall rate of reaction of reactants to products increases, overestimating the rate.
# The concentration of the TS is in equilibrium with the reactant concentration; this is known as the Quasi-equilibrium hypothesis ([X]‡ = 0.5 KC‡ [A][B], where A + B → X‡ → P). This means that even though there is no equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws.<ref name=":0" /> This leads to an overestimate of the rate of reaction.
# Quantum tunnelling is not taken into account; the theory is treated purely classical as a translation. If quantum tunnelling was taken into account the rate would increase so in this case the theory underestimates the rate.
As the first assumption is the most crucial, it has a larger weighting effect on the rate compared to quantum tunnelling. Therefore overall the TS theory predictions overestimate the reaction rate values compared to the experimental values.

== Exercise 2: F - H - H system ==
==== By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ====
The F + H2 reaction (forward reaction) is exothermic as the reactants are at a higher energy in comparison to the products therefore energy is released to the surroundings. The H + HF reaction (backward reaction) is endothermic as the products are at a higher energy than the reactants. The exothermicity of the forward reaction indicates that the bond strength of HF is larger than that of HH as the energy required to break the HH bond is less than the energy released upon forming HF. Also, the endothermicity of the backward reaction means that the energy required to break the strong HF bond is greater than the energy released to form the products (F and H2). This is confirmed by the literature values of HF and HH bond strengths of 565 kJmol-1 <ref name=":1">Darwent, B. deB., Bond Dissociation Energies in Simple Molecules. ''Natl. Stand. Ref. data Syst.'' 1970, ''31'', pg. 9-48.</ref> and 436 kJmol-1 <ref name=":1" /> respectively.
[[File:PES FHH 01524183.png|centre|thumb|678x678px|Figure 10. (left) A potential energy surface plot of the F + H2 reaction (A: F, B: H and C: H). The reactant channel (small BC and large AB; the left channel) is at a higher energy value compared to the products (large BC and small AB). Therefore, this forward reaction is exothermic.
Figure 11. (right) A potential energy surface plot of the H + HF reaction. (A: H, B: H and C: F). The reactant channel (small BC and large AB; the left channel) is at a lower energy value compared to the products (large BC and small AB). Therefore, this forward reaction is endothermic.
]]

==== Locate the approximate position of the transition state. ====
[[File:FH2 TS 01524183.png|thumb|Figure 12. A contour plot of the F-H-H system, showing the approximate position of the TS. The point is in the reactant channel as the forward reaction is exothermic. When viewed in terms of the backward reaction the TS is in the product channel as it is endothermic.|326x326px]]
The transition state for both the forward (F + H2) and backward (H + HF) reactions is at the same point. As the system is not symmetric, the TS is not defined as the point where AB = BC, instead Hammond's postulate is used in order to determine the TS point. The Hammond's postulate states that the TS of a reaction resembles the reactants if the reaction is exothermic (an early TS) and it resembles the products if the reaction is endothermic (a late TS). Therefore, the TS would be close to the reactants of the forward reaction and the products of the backward reaction which is the same (F + H2). So by setting the initial conditions such that AB is a larger distance than BC (where A: F, B: H and C: H), the TS was found. The momenta values were set to zero because the nuclei are stationary at the TS as it is a saddle point.

The transition state position ('''rts''') was approximated to be at an AB distance of '''181.4 pm''' and BC distance of '''74.5 pm''' (A: F, B: H and C: H).

These values gave force values along AB and BC of 0.000 kJ.mol-1 and 0.006 kJ.mol-1 respectively, which are close to zero, as expected at the TS point. They also gave Hessian eigenvalues/vectors of opposite signs which also indicated that this was the TS position.
==== Report the activation energy for both reactions. ====
The activation energy is the difference in energy of the TS and the minimum of the reactant channel. This difference is smaller for the forward (F + H2) exothermic reaction compared to the backwards (H + HF) endothermic reaction. The minimum point was found by testing different distance values until the energy value was constant as this is when the plateau has been reached.

The energy of the TS was found to be -433.98 kJ.mol-1.

For the F + H2 reaction, the minimum point (reactants energy) was -435.10 kJ.mol-1. The activation energy was found to be -433.98--435.10 = '''+1.12 kJ.mol-1''' (where A: F, B: H, C: H and AB is a larger distance than the BC bond distance of HH).
[[File:H2 EA 01524183.png|thumb|342x342px|Figure 13. Finding the activation energy using mep. The activation energies could also be found by running mep calculations where the distance values are those of the TS but with the AB distance displaced by approx. 1 pm ( AB = 179 pm and BC = 74.5 pm, where A: F, B: H, C: H).]]
For the H + HF reaction, the minimum point (reactants energy) was -560.70 kJ.mol-1. The activation energy was found to be -433.98--560.70 = '''+126.72 kJ.mol-1''' (where A: F, B: H, C: H and BC is a larger distance than AB bond distance of HF).

The activation energies could also be found by performing a mep calculation (see Figure 13).
==== 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 reactive trajectory of the F + H2 reaction is shown in Figure 14, as a contour plot. Initially, the reactants have more translational kinetic energy therefore the atoms are moving very fast. Whereas, after the reaction has occurred (the region where the BC distance is increasing), there is more vibrational energy as there are more oscillations in the product valley. Therefore, potential energy is being converted to vibrational energy and the system releases the reaction energy as vibrational energy. In this way, the conservation of energy is held true. This could be confirmed experimentally by IR spectroscopy (IR absorption) and infrared chemiluminescence (emission of vibration by IR light).
[[File:01524183 FH2 reaction.png|left|thumb|306x306px|Figure 14. A contour plot of the F + H2 reaction showing the reactive trajectory. Initial conditions were set as AB: 175 pm and -1.6 g.mol-1.pm.fs-1, BC: 74 pm and 0.2 g.mol-1.pm.fs-1 , with the number of steps as 3000. (where A: F, B: H and C: H).]]'''IR spectroscopy '''

For the exothermic F + H2 reaction, initially, all the molecules are in the ground state in the H2 gas. However, as a result of the reaction, a small proportion of the molecules are excited and transfer to a vibrationally first excited state. This can be seen from the plot (Figure 14), as the vibrational energy increases in the product channel. Therefore, the IR spectrum of the thermally relaxed sample would show one absorption peak. However, the excited sample would show a fundamental peak, corresponding to the excitation from the ground state to the first excited state, and also an overtone peak, corresponding to the excitation from the first excited state to the second excited state, at a lower wavenumber. The peaks would be at different wavenumbers due to the anharmonicity of the potential energy surface. By measuring the intensity of the overtone over time, the number of molecules that are vibrationally excited as time goes on can be deduced. The intensity of the overtone would decrease, as energy would be released as radiation due to exothermicity of the reaction, therefore there would be fewer molecules in the first electronically excited state over time. Whereas, the intensity of the fundamental peak would increase as more molecules would be in the ground state which could be excited to the first excited state.

'''Infrared chemiluminescence'''

This is another method for examining the energy distribution in the products. In this case, vibrationally excited molecules emit infrared radiation as they return to their ground states. The populations of the vibrational states of the products may be determined by studying the intensities of the IR emission spectrum.<ref>Atkins, P.; Paula, J. de; Keeler, J. ''Atkins’ Physical Chemistry'', 11th ed.; Oxford University Press, UK, 2018, pg. 804</ref>

==== 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. ====
From Polyani's rules, it can be deduced that for exothermic reactions, translational energy is more effective than vibrational energy at overcoming the TS barrier. Whereas, for endothermic reactions, vibrational energy is more effective than translational energy at passing the TS barrier.<ref>K. J. Laidler, Chemical Kinetics, 3rd ed., Harper-Collins, 1987, Ch. 12, pg. 460-471</ref> Therefore, an early energy barrier favours more highly vibrational products and a late energy barrier favours products with more translational energy.<ref name=":2">J. I. Steinfeld, J. S. Francisco, W. L. Hase, Chemical Kinetic and Dynamics 2nd ed., Prentice-Hall, 1998, Ch. 9, pg. 272-274</ref>

For exothermic reactions the TS is located in the reactant channel (early TS) therefore molecules with most energy in motion would easily be able to surmount the barrier. This is because a vibrationally excited reactant molecule would be too busy oscillating from side to side and so wouldn't have enough energy left to reach the top of the TS barrier. Therefore, the reactant vibrational energy in excess of the TS barrier height may be ineffective for the reaction.<ref name=":2" />

For endothermic reactions, the TS is located in the product channel (late TS), therefore reactant molecules with most vibrational energy would easily be able to cross the barrier. This is because the vibrationally excited reactant molecules would be travelling down the reactant channel; they have to travel further in order to cross the TS barrier. The chances of the reactant molecules with the correct phase finding the TS barrier is higher due to the higher amount of vibrational energy. A reactant molecule with more translational energy, in this case, would simply slam into the repulsive inner wall of the potential surface and bounce back into the entrance channel.<ref name=":2" />

Some example trajectories are given in the table below in order to show this concept.

{| class="wikitable"
|Exothermic F + H2 reaction
(forwards reaction)
|[[File:01524183 FHH 1.png|thumb|381x381px|Figure 15. A contour plot of the F + H2 exothermic reaction, showing that vibrationally excited reactant molecules cannot easily surmount the early TS barrier. The initial conditions were set as, rAB: 175 pm, pAB = -1.0 g.mol-1.pm.fs-1 and rBC: 74 pm, pBC: 5 g.mol-1.pm.fs-1 (where A: F, B: H and C: H). ]]
|[[File:01524183 FHH 2.png|centre|thumb|381x381px|Figure 16. A contour plot of the F + H2 exothermic reaction, showing that reactant molecules with high translational energy can easily surmount the early TS barrier. The initial conditions were set as, rAB: 175 pm, pAB = -1.6 g.mol-1.pm.fs-1 and rBC: 74 pm, pBC: 0.2 g.mol-1.pm.fs-1 (where A: F, B: H and C: H). ]]
|-
|Endothermic H + HF reaction
(backwards reaction)
|[[File:01524183 HHF 1.png|thumb|381x381px|Figure 17. A contour plot of the H + HF endothermic reaction, showing that vibrationally excited reactant molecules can easily surmount the late TS barrier. The initial conditions were set as, rAB: 74 pm, pAB = 0.2 g.mol-1.pm.fs-1 and rBC: 200 pm, pBC: -1.6 g.mol-1.pm.fs-1 (where A: F, B: H and C: H). ]]
|[[File:01524183 HHF 2.png|thumb|382x382px|Figure 18. A contour plot of the H + HF endothermic reaction, showing that reactant molecules with high translational energy cannot easily surmount the late TS barrier. The initial conditions were set as, rAB: 92 pm, pAB = -0.6 g.mol-1.pm.fs-1 and rBC: 200 pm, pBC: -1.6 g.mol-1.pm.fs-1 (where A: F, B: H and C: H). ]]
|}

== Conclusion ==
In conclusion, the outcomes of chemical reactions were studied by analysing transition states, reaction coordinates, contour plots and potential energy surfaces. The systems studied involved the collision and reaction between an atom and a diatomic molecule in a linear configuration in the gas phase. The reactions analysed were H + H2, F + H2 and H + HF. If the reaction had a reactive trajectory, it formed a new diatomic molecule and an atom. Lastly, it was discovered that for a chemical reaction to take place, the energy must be distributed correctly in the molecules; they must be in the right vibrational and translational modes at the right time.

== References ==
<references />

MRD:01524183

2020-05-22T21:55:25Z

2020-05-22T18:33:30Z

Aas1218:

== Introduction ==
The main aims of this experiment were to study the reactivity of triatomic systems, involving the collision of an atom and a diatomic molecule, by calculating Molecular Dynamic trajectories. The systems studied here are the symmetric H-H-H system and the non-symmetric F-H-H system.

== Exercise 1: The H + H2 system ==
The original H2 diatomic is defined as BC with the A H atom colliding to form the AB H2 diatomic molecule.

==== On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ====
[[File:H3 TS contourplot 01524183.png|thumb|329x329px|Figure 1. A contour plot showing the TS of the H-H-H system and the variation of AB and BC distances, with reaction trajectory and eigenvectors shown.]]
On a potential energy surface diagram, the transition state (TS) is mathematically defined as the point at which the gradient of the potential with respect to the internal coordinates is zero (∂V('''ri''')/∂'''ri'''=0). This is because the transition state is the local maximum on the minimum energy paths between the reactants and products. There would be no movement of the H atoms at the TS since the force is a derivative of the potential energy and at the TS this is zero, therefore, the nuclei are stationary; the transition state is a stationary point. However, minima and maxima are stationary points. The TS can be distinguished from a local minimum of the potential energy surface by taking the second differential of the potential function. By doing so, the nature of the stationary point can be determined: if the second differential is greater than zero, the point is a local minimum and if it is less than zero, the point is a local maximum (the TS).

Using the GUI simulation, the TS was identified by trial and error. At the TS the distance AB and BC must be equal (rAB = rBC), as the H + H2 surface is symmetric. Therefore the TS lies somewhere along the diagonal line where AB and BC are equal. As the TS is approached, the size of the forces of AB and BC decrease to zero and so the point at which these forces are close to zero is the TS of the potential energy surface. This maximum point can be distinguished from a local minimum by analysing the omega squared values under the Hessian eigenvalues/vectors. A TS is a maximum in one direction and a minimum in the other direction; it is a saddle point. Therefore the omega squared would be positive in one direction and negative in the other direction.

==== 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 transition state position ('''rts''') was found to be 90.8 pm to 1 d.p.[[File:H3 TS distancetime 01524183.png|thumb|Figure 2. A plot of Internuclear Distances vs Time: linear horizontal lines show the constant distances of AB, BC and AC. |329x329px]]This value gave forces of +0.001 kJ.mol-1.pm-1 along AB and BC which are very close to zero. Also, the omega hessian eigenvalues had opposite signs which verified that this point was the TS.

At the TS, the internuclear distances between AB and BC are the same and so the distance between AC is approximately double the AB (BC) distance. This is shown in the plot of "Internuclear distances vs Time", as two straight lines of constant distances of AB and BC at 90.8 pm and AC at approximately 181.6 pm. This is because at the TS the H atoms are stationary, with one H atom (the B atom in this case) being approx. halfway between the A and C H atoms.

==== Comment on how the ''mep'' and the trajectory you just calculated differ. ====
The minimum energy path (''mep) ''is the lowest energy path from reactant to product; it is a special trajectory that correlates to infinitely slow motion. The ''mep ''was calculated by setting the initial conditions as rBC = 91.8 pm (1 pm displaced from the TS position) and rAB = 90.8 pm (the TS position). The number of steps were also increased to 2500 in order to obtain a complete mep. This calculation was repeated with the calculation type set to Dynamics.
[[File:H3 dynamics mep 01524183.png|thumb|584x584px|Figure 3 and 4. A contour plot of BC distance vs AB distance, showing the ''mep ''trajectory: calculation type was set to MEP (left).
A contour plot of BC distance vs AB distance, showing the dynamic'' ''trajectory: calculation type was set to Dynamics (right).
|centre]]
As shown in the plots above, the ''mep ''is a smooth trajectory (left) which indicates that the BC molecule is not vibrating as the system is not oscillating between energy contours. This is because the ''mep'' doesn't provide a realistic view of the motion of atoms during a reaction; it assumes the atoms are moving at an infinitely slow speed. There would be vibration occurring between B and C as they are moving further away from each other. This is shown by the wavy trajectory produced from the dynamic calculation (right).

==== 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? ====
For the initial positions rBC = 74 pm and rAB = 200 pm, trajectories were run for five different moments combinations. The results of which are given in the table below.

{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot / kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Reactive
|The trajectory crosses the TS, therefore, it is reactive. The line is wavier after crossing the TS, therefore the initial diatomic BC has less vibrational energy than the forming AB diatomic molecule. However, overall the trajectory is not very wavy therefore there is less vibrational energy compared to translational energy in the whole system.
|[[File:H3 1 01524183.png|thumb|375x375px|Figure 5]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.08</nowiki>
|Unreactive
|The trajectory does not cross the TS in this case, therefore it is unreactive. The AB distance is changing as the A H atom is moving towards and away from the BC diatomic molecule, which means it has more translational energy compared to vibrational. This is shown in the plot as the line is not very wavy. However, the BC distance doesn't change as the diatomic molecule remains intact.
|[[File:H3 2 01524183.png|thumb|375x375px|Figure 6]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.98</nowiki>
|Reactive
|The trajectory does cross the TS therefore it is reactive. The lines quite wavy which there is some vibrational energy but more translational energy. This is quite similar to the plot in Figure 5 as the potential energies are the same therefore the waviness of the trajectories are the same. However, the kinetic energies are a little different, making the total energies different.
|[[File:H3 3 01524183.png|thumb|375x375px|Figure 7]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.28</nowiki>
|Unreactive
|The trajectory just crosses the TS however it does not enter the product channel instead it goes back to the reactant channel. This is called barrier re-crossing. Therefore it is overall unreactive. The trajectory is very wavy which means there is a lot of vibrational energy in the system and less translational energy. In the beginning, the AB and BC distances are changing as the middle B atom is oscillating between the A and C atoms. However, in the end, the molecule BC is reformed which is they the AB distance is large.
|[[File:H3 4 01524183.png|thumb|375x375px|Figure 8]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.48</nowiki>
|Reactive
|The trajectory is reactive as it does cross the TS. The line is very wavy which shows that the system has more vibrational energy compared to translational. As the A atom gets closer to the initial BC molecule, the AB distance decreases and the BC distance increases, however, at the TS the B atom oscillates between the A and C atoms as it is attracted to both. This means there's a lot of vibrational energy near the TS. In the end, it does form the AB product molecule, which is shown by the trajectory increase in BC distance.
|[[File:H3 5 01524183.png|thumb|375x375px|Figure 9]]
|}
From the table, it can be concluded that not all trajectories starting with the same positions but with higher values of momenta would be reactive. For example, increasing the value of momenta to -5.1 g.mol-1.pm.fs-1 for p1 and -10.1 g.mol-1.pm.fs-1 for p2 still gives an unreactive trajectory. Also, low values of momenta can be reactive as is the first momenta combination (Figure 5) in the table. This is because vibrational energy also needs to be taken into account, not only the kinetic energy. The degree of vibrational and translational energy is determined by analysing the trajectory line shape: a more wavy line would mean more vibrational energy whereas a more smooth line corresponds to more translational energy.

==== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ====
The TS theory is a theory used to determine absolute reaction rates based on the reactant and transition state structure properties. The assumptions this theory makes are:
# Every trajectory that has enough energy to roll over the TS will definitely go to form the products. This means that once the reactants have crossed the energy barrier they can't go back to reactants. Therefore the overall rate of reaction of reactants to products increases, overestimating the rate.
# The concentration of the TS is in equilibrium with the reactant concentration; this is known as the Quasi-equilibrium hypothesis ([X]‡ = 0.5 KC‡ [A][B], where A + B → X‡ → P). This shows that the overall concentration of TS would be lower than the concentration of the reactants. This is because a small proportion of reactants would be forming the TS from the reactants pool.
# Quantum tunnelling is not taken into account; the theory is treated purely classical as a translation. If quantum tunnelling was taken into account the rate would increase so in this case the theory underestimates the rate.
As the first assumption is the most crucial, it has a larger weighting effect on the rate compared to quantum tunnelling. Therefore overall the TS theory predictions overestimate the reaction rate values compared to the experimental values.

N'''eed to include references here '''

== Exercise 2: F - H - H system ==
==== By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ====
The F + H2 reaction (forward reaction) is exothermic as the reactants are at a higher energy in comparison to the products, therefore, energy is released to the surroundings. The H + HF reaction (backward reaction) is endothermic as the products are at a higher energy than the reactants. Therefore, as shown by the exothermicity of the forward reaction (F + H2), the bond strength of HF is larger than that of HH as the energy required to break the HH bond is less than the energy released upon forming HF. Also, the endothermicity of the backward reaction means that the energy required to break the strong HF bond is greater than the energy released to form the products (F and HH). '''This is confirmed by the literature values of HF and HH bond strengths of:...........................'''
[[File:PES FHH 01524183.png|centre|thumb|678x678px|Figure 10. (left) A potential energy surface plot of the F + H2 reaction (A: F, B: H and C: H). The reactant channel (small BC and large AB; the left channel) is at a higher energy value compared to the products (large BC and small AB). Therefore, this forward reaction is exothermic.
Figure 11. (right) A potential energy surface plot of the H + HF reaction. (A: H, B: H and C: F). The reactant channel (small BC and large AB; the left channel) is at a lower energy value compared to the products (large BC and small AB). Therefore, this forward reaction is endothermic.
]]

==== Locate the approximate position of the transition state. ====
[[File:FH2 TS 01524183.png|thumb|Figure 12. A contour plot of the F-H-H system, showing the TS. There is no trajectory therefore this is the TS. The point is in the reactant channel as expected as the forward reaction is exothermic. When viewed in terms of the backward reaction the TS is in the product channel as it is endothermic.|298x298px]]
The transition state for both the forward (F + H2) and backward (H + HF) reactions is at the same point. As the system is not symmetric, the TS is not defined as the point where AB = BC, instead Hammond's postulate is used in order to determine the TS point. According to Hammonds postulate, the TS of a reaction resembles the reactants if the reaction is exothermic (an early TS) and it resembles the products if the reaction is endothermic (a late TS). Therefore, the TS would be close to the reactants of the forward reaction and the products of the backward reaction which is the same (F + H2). So by setting the initial conditions such that AB is a larger distance than BC, the TS was found (where A: F, B: H and C: H). The momenta values were set to zero because the molecules are stationary at the TS as it is a saddle point.

The transition state position ('''rts''') was found to be at an '''AB distance''' of '''181.4 pm''' and '''BC''' '''distance''' of '''74.5 pm''' (A: F, B: H and C: H).

These values gave force values along AB and BC of 0.000 kJ.mol-1 and 0.006 kJ.mol-1 respectively, which are close to 0 kJ.mol-1, as expected at the TS point. They also gave Hessian eigenvalues/vectors of opposite signs (one negative and one positive) which also indicated that this was the TS position.
==== Report the activation energy for both reactions. ====
The activation energy is the difference in energy of the TS and the minimum of the reactant channel. This difference is smaller for the forward (F + H2) exothermic reaction compared to the backwards (H + HF) endothermic reaction. The minimum point was found by testing different distance values until the energy value was constant; until it plateaued.

The energy of the TS was found to be -433.98 kJ.mol-1.

For the F + H2 reaction, the minimum point (reactants energy) was -435.10 kJ.mol-1. The activation energy was found to be -433.98--435.10 = '''+1.12 kJ.mol-1''' (where A: F, B: H, C: H and AB is a larger distance than the BC bond distance of HH).
[[File:H2 EA 01524183.png|thumb|351x351px|Figure 13. Finding the activation energy using mep. The activation energies could also be found by running mep calculations where the distance values are those of the TS but with the AB distance displaced by approx. 1 pm ( AB = 179 pm and BC = 74.5 pm, where A: F, B: H, C: H).]]
For the H + HF reaction, the minimum point (reactants energy) was -560.70 kJ.mol-1. The activation energy was found to be -433.98--560.70 = '''+126.72 kJ.mol-1''' (where A: F, B: H, C: H and BC is a larger distance than AB bond distance of HF).

The activation energies could also be found by performing a mep (see Figure 13).
==== 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 reactive trajectory of the F + H2 reaction is shown in Figure 14, as a contour plot. Initially, the system (reactants) has more translational kinetic energy therefore the atoms are moving very fast. Whereas, after the reaction has occurred (the region where the BC distance is increasing), there is more vibrational energy as there are more oscillations in the product valley. Therefore potential energy is being converted to vibrational energy and the system releases the reaction energy as vibrational energy. In this way, the conservation of energy is fulfilled. This could be confirmed experimentally by IR spectroscopy (IR absorption) and infrared chemiluminescence (emission of vibration by IR light).
[[File:01524183 FH2 reaction.png|left|thumb|274x274px|Figure 14. A contour plot of the F + H2 reaction showing the reactive trajectory. Initial conditions were set as AB: 175 pm and -1.6 g.mol-1.pm.fs-1, BC: 74 pm and 0.2 g.mol-1.pm.fs-1 , with the number of steps as 3000. (where A: F, B: H and C: H).]]

====== IR spectroscopy ======
Initially, for the exothermic F + H2 reaction, all the molecules are in the ground state in the H2 gas. However, as a result of the reaction, a small proportion of the molecules get excited and transfer to a vibrationally first excited state. This can be seen from the plot (Figure 14), as the vibrational energy increases in the product channel. Therefore, the IR spectrum of the thermally relaxed sample would show one absorption peak. However, the excited sample would show a fundamental peak, corresponding to the excitation from the ground state to the first excited state, and also an overtone peak, corresponding to excitation from the first excited state to the second excited state, at a lower wavenumber. The peaks would be at different wavenumbers due to the anharmonicity of the potential energy surface. By measuring the intensity of the overtone overtime, the number of molecules that are vibrationally excited, as time goes on, can be deduced. This is because the intensity of the overtone would decrease, as energy would be released as radiation (due to exothermicity of the reaction) therefore there would be fewer molecules in the first electronically excited state over time. Whereas, the intensity of the fundamental peak would increase as more molecules would be in the ground state which could be excited to the first excited state.

====== Infrared chemiluminescence ======
This is another method for examining the energy distribution in the products. In this case, vibrationally excited molecules emit infrared radiation as they return to their ground states. The populations of the vibrational states of the products may be determined by studying the intensities of the IR emission spectrum. '''(reference ATKINS PG 804)'''

==== 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. ====
From Polyani's rules, it can be deduced that for exothermic reactions, translational energy is more effective than vibrational energy at overcoming the TS barrier. Whereas, for endothermic reactions, vibrational energy is more effective than translational energy at passing the TS barrier. Therefore, an early energy barrier favours more highly vibrational products and a late energy barrier favours products with more translational energy.

For exothermic reactions the TS is located in the reactant channel (early TS) therefore molecules with the most energy in motion would easily be able to surmount the barrier. This is because a vibrationally excited reactant molecule would be too busy oscillating from side to side and so wouldn't have enough energy left to reach the top of the TS barrier. Therefore, reactant vibrational energy in excess of the TS barrier height may be ineffective for the reaction.

For endothermic reactions, the TS is located in the product channel (late TS), therefore reactant molecules with most vibrational energy would easily be able to cross the barrier. This is because the vibrationally excited reactant molecules would be travelling down the reactant channel; they would have to travel further in order to cross the TS barrier. The chances of the reactant molecules with the correct phase finding the TS barrier is higher due to the higher amount of vibrational energy. A reactant molecule with more translational energy, in this case, would simply slam into the repulsive inner wall of the potential surface and bounce back into the entrance channel.

Some example trajectories are provided in the table below in order to show this concept.

'''(reference here Steinfield - Chemical reactions and dynamics)'''

{| class="wikitable"
|Exothermic F + H2 reaction
(forwards reaction)
|[[File:01524183 FHH 1.png|thumb|381x381px|A contour plot of the F + H2 exothermic reaction, showing that vibrationally excited reactant molecules cannot easily surmount the early TS barrier. The initial conditions were set as, rAB: 175 pm, pAB = -1.0 g.mol-1.pm.fs-1 and rBC: 74 pm, pBC: 5 g.mol-1.pm.fs-1 (where A: F, B: H and C: H). ]]
|[[File:01524183 FHH 2.png|centre|thumb|381x381px|A contour plot of the F + H2 exothermic reaction, showing that reactant molecules with high translational energy can easily surmount the early TS barrier. The initial conditions were set as, rAB: 175 pm, pAB = -1.6 g.mol-1.pm.fs-1 and rBC: 74 pm, pBC: 0.2 g.mol-1.pm.fs-1 (where A: F, B: H and C: H). ]]
|-
|Endothermic H + HF reaction
(backwards reaction)
|[[File:01524183 HHF 1.png|thumb|381x381px|A contour plot of the H + HF endothermic reaction, showing that vibrationally excited reactant molecules can easily surmount the late TS barrier. The initial conditions were set as, rAB: 74 pm, pAB = 0.2 g.mol-1.pm.fs-1 and rBC: 200 pm, pBC: -1.6 g.mol-1.pm.fs-1 (where A: F, B: H and C: H). ]]
|[[File:01524183 HHF 2.png|thumb|382x382px|A contour plot of the H + HF endothermic reaction, showing that reactant molecules with high translational energy cannot easily surmount the late TS barrier. The initial conditions were set as, rAB: 92 pm, pAB = -0.6 g.mol-1.pm.fs-1 and rBC: 200 pm, pBC: -1.6 g.mol-1.pm.fs-1 (where A: F, B: H and C: H). ]]
|}

File:01524183 HHF 2.png

2020-05-22T18:00:10Z

Aas1218:

File:01524183 HHF 1.png

2020-05-22T17:59:27Z

Aas1218:

File:01524183 FHH 2.png

2020-05-22T17:59:05Z

Aas1218:

2020-05-22T17:52:39Z

== Introduction ==
The main aims of this experiment were to study the reactivity of triatomic systems, involving the collision of an atom and a diatomic molecule, by calculating Molecular Dynamic trajectories. The systems studied here are the symmetric H-H-H system and the non-symmetric F-H-H system.

== Exercise 1: The H + H2 system ==
The original H2 diatomic is defined as BC with the A H atom colliding to form the AB H2 diatomic molecule.

==== On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ====
[[File:H3 TS contourplot 01524183.png|thumb|329x329px|Figure 1. A contour plot showing the TS of the H-H-H system and the variation of AB and BC distances, with reaction trajectory and eigenvectors shown.]]
On a potential energy surface diagram, the transition state (TS) is mathematically defined as the point at which the gradient of the potential with respect to the internal coordinates is zero (∂V('''ri''')/∂'''ri'''=0). This is because the transition state is the local maximum on the minimum energy paths between the reactants and products. There would be no movement of the H atoms at the TS since the force is a derivative of the potential energy and at the TS this is zero, therefore, the nuclei are stationary; the transition state is a stationary point. However, minima and maxima are stationary points. The TS can be distinguished from a local minimum of the potential energy surface by taking the second differential of the potential function. By doing so, the nature of the stationary point can be determined: if the second differential is greater than zero, the point is a local minimum and if it is less than zero, the point is a local maximum (the TS).

Using the GUI simulation, the TS was identified by trial and error. At the TS the distance AB and BC must be equal (rAB = rBC), as the H + H2 surface is symmetric. Therefore the TS lies somewhere along the diagonal line where AB and BC are equal. As the TS is approached, the size of the forces of AB and BC decrease to zero and so the point at which these forces are close to zero is the TS of the potential energy surface. This maximum point can be distinguished from a local minimum by analysing the omega squared values under the Hessian eigenvalues/vectors. A TS is a maximum in one direction and a minimum in the other direction; it is a saddle point. Therefore the omega squared would be positive in one direction and negative in the other direction.

==== 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 transition state position ('''rts''') was found to be 90.8 pm to 1 d.p.[[File:H3 TS distancetime 01524183.png|thumb|Figure 2. A plot of Internuclear Distances vs Time: linear horizontal lines show the constant distances of AB, BC and AC. |329x329px]]This value gave forces of +0.001 kJ.mol-1.pm-1 along AB and BC which are very close to zero. Also, the omega hessian eigenvalues had opposite signs which verified that this point was the TS.

At the TS, the internuclear distances between AB and BC are the same and so the distance between AC is approximately double the AB (BC) distance. This is shown in the plot of "Internuclear distances vs Time", as two straight lines of constant distances of AB and BC at 90.8 pm and AC at approximately 181.6 pm. This is because at the TS the H atoms are stationary, with one H atom (the B atom in this case) being approx. halfway between the A and C H atoms.

==== Comment on how the ''mep'' and the trajectory you just calculated differ. ====
The minimum energy path (''mep) ''is the lowest energy path from reactant to product; it is a special trajectory that correlates to infinitely slow motion. The ''mep ''was calculated by setting the initial conditions as rBC = 91.8 pm (1 pm displaced from the TS position) and rAB = 90.8 pm (the TS position). The number of steps were also increased to 2500 in order to obtain a complete mep. This calculation was repeated with the calculation type set to Dynamics.
[[File:H3 dynamics mep 01524183.png|thumb|584x584px|Figure 3 and 4. A contour plot of BC distance vs AB distance, showing the ''mep ''trajectory: calculation type was set to MEP (left).
A contour plot of BC distance vs AB distance, showing the dynamic'' ''trajectory: calculation type was set to Dynamics (right).
|centre]]
As shown in the plots above, the ''mep ''is a smooth trajectory (left) which indicates that the BC molecule is not vibrating as the system is not oscillating between energy contours. This is because the ''mep'' doesn't provide a realistic view of the motion of atoms during a reaction; it assumes the atoms are moving at an infinitely slow speed. There would be vibration occurring between B and C as they are moving further away from each other. This is shown by the wavy trajectory produced from the dynamic calculation (right).

==== 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? ====
For the initial positions rBC = 74 pm and rAB = 200 pm, trajectories were run for five different moments combinations. The results of which are given in the table below.

{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot / kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Reactive
|The trajectory crosses the TS, therefore, it is reactive. The line is wavier after crossing the TS, therefore the initial diatomic BC has less vibrational energy than the forming AB diatomic molecule. However, overall the trajectory is not very wavy therefore there is less vibrational energy compared to translational energy in the whole system.
|[[File:H3 1 01524183.png|thumb|375x375px|Figure 5]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.08</nowiki>
|Unreactive
|The trajectory does not cross the TS in this case, therefore it is unreactive. The AB distance is changing as the A H atom is moving towards and away from the BC diatomic molecule, which means it has more translational energy compared to vibrational. This is shown in the plot as the line is not very wavy. However, the BC distance doesn't change as the diatomic molecule remains intact.
|[[File:H3 2 01524183.png|thumb|375x375px|Figure 6]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.98</nowiki>
|Reactive
|The trajectory does cross the TS therefore it is reactive. The lines quite wavy which there is some vibrational energy but more translational energy. This is quite similar to the plot in Figure 5 as the potential energies are the same therefore the waviness of the trajectories are the same. However, the kinetic energies are a little different, making the total energies different.
|[[File:H3 3 01524183.png|thumb|375x375px|Figure 7]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.28</nowiki>
|Unreactive
|The trajectory just crosses the TS however it does not enter the product channel instead it goes back to the reactant channel. This is called barrier re-crossing. Therefore it is overall unreactive. The trajectory is very wavy which means there is a lot of vibrational energy in the system and less translational energy. In the beginning, the AB and BC distances are changing as the middle B atom is oscillating between the A and C atoms. However, in the end, the molecule BC is reformed which is they the AB distance is large.
|[[File:H3 4 01524183.png|thumb|375x375px|Figure 8]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.48</nowiki>
|Reactive
|The trajectory is reactive as it does cross the TS. The line is very wavy which shows that the system has more vibrational energy compared to translational. As the A atom gets closer to the initial BC molecule, the AB distance decreases and the BC distance increases, however, at the TS the B atom oscillates between the A and C atoms as it is attracted to both. This means there's a lot of vibrational energy near the TS. In the end, it does form the AB product molecule, which is shown by the trajectory increase in BC distance.
|[[File:H3 5 01524183.png|thumb|375x375px|Figure 9]]
|}
From the table, it can be concluded that not all trajectories starting with the same positions but with higher values of momenta would be reactive. For example, increasing the value of momenta to -5.1 g.mol-1.pm.fs-1 for p1 and -10.1 g.mol-1.pm.fs-1 for p2 still gives an unreactive trajectory. Also, low values of momenta can be reactive as is the first momenta combination (Figure 5) in the table. This is because vibrational energy also needs to be taken into account, not only the kinetic energy. The degree of vibrational and translational energy is determined by analysing the trajectory line shape: a more wavy line would mean more vibrational energy whereas a more smooth line corresponds to more translational energy.

==== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ====
The TS theory is a theory used to determine absolute reaction rates based on the reactant and transition state structure properties. The assumptions this theory makes are:
# Every trajectory that has enough energy to roll over the TS will definitely go to form the products. This means that once the reactants have crossed the energy barrier they can't go back to reactants. Therefore the overall rate of reaction of reactants to products increases, overestimating the rate.
# The concentration of the TS is in equilibrium with the reactant concentration; this is known as the Quasi-equilibrium hypothesis ([X]‡ = 0.5 KC‡ [A][B], where A + B → X‡ → P). This shows that the overall concentration of TS would be lower than the concentration of the reactants. This is because a small proportion of reactants would be forming the TS from the reactants pool.
# Quantum tunnelling is not taken into account; the theory is treated purely classical as a translation. If quantum tunnelling was taken into account the rate would increase so in this case the theory underestimates the rate.
As the first assumption is the most crucial, it has a larger weighting effect on the rate compared to quantum tunnelling. Therefore overall the TS theory predictions overestimate the reaction rate values compared to the experimental values.

N'''eed to include references here '''

== Exercise 2: F - H - H system ==
==== By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ====
The F + H2 reaction (forward reaction) is exothermic as the reactants are at a higher energy in comparison to the products, therefore, energy is released to the surroundings. The H + HF reaction (backward reaction) is endothermic as the products are at a higher energy than the reactants. Therefore, as shown by the exothermicity of the forward reaction (F + H2), the bond strength of HF is larger than that of HH as the energy required to break the HH bond is less than the energy released upon forming HF. Also, the endothermicity of the backward reaction means that the energy required to break the strong HF bond is greater than the energy released to form the products (F and HH). '''This is confirmed by the literature values of HF and HH bond strengths of:...........................'''
[[File:PES FHH 01524183.png|centre|thumb|743x743px|Figure 10. (left) A potential energy surface plot of the F + H2 reaction (A: F, B: H and C: H). The reactant channel (small BC and large AB; the left channel) is at a higher energy value compared to the products (large BC and small AB). Therefore, this forward reaction is exothermic.
Figure 11. (right) A potential energy surface plot of the H + HF reaction. (A: H, B: H and C: F). The reactant channel (small BC and large AB; the left channel) is at a lower energy value compared to the products (large BC and small AB). Therefore, this forward reaction is endothermic.
]]

==== Locate the approximate position of the transition state. ====
[[File:FH2 TS 01524183.png|thumb|Figure 12. A contour plot of the F-H-H system, showing the TS. There is no trajectory therefore this is the TS. The point is in the reactant channel as expected as the forward reaction is exothermic. When viewed in terms of the backward reaction the TS is in the product channel as it is endothermic.|338x338px]]
The transition state for both the forward (F + H2) and backward (H + HF) reactions is at the same point. As the system is not symmetric, the TS is not defined as the point where AB = BC, instead Hammond's postulate is used in order to determine the TS point. According to Hammonds postulate, the TS of a reaction resembles the reactants if the reaction is exothermic (an early TS) and it resembles the products if the reaction is endothermic (a late TS). Therefore, the TS would be close to the reactants of the forward reaction and the products of the backward reaction which is the same (F + H2). So by setting the initial conditions such that AB is a larger distance than BC, the TS was found (where A: F, B: H and C: H). The momenta values were set to zero because the molecules are stationary at the TS as it is a saddle point.

The transition state position ('''rts''') was found to be at an '''AB distance''' of '''181.4 pm''' and '''BC''' '''distance''' of '''74.5 pm''' (A: F, B: H and C: H).

These values gave force values along AB and BC of 0.000 kJ.mol-1 and 0.006 kJ.mol-1 respectively, which are close to 0 kJ.mol-1, as expected at the TS point. They also gave Hessian eigenvalues/vectors of opposite signs (one negative and one positive) which also indicated that this was the TS position.
==== Report the activation energy for both reactions. ====
The activation energy is the difference in energy of the TS and the minimum of the reactant channel. This difference is smaller for the forward (F + H2) exothermic reaction compared to the backwards (H + HF) endothermic reaction. The minimum point was found by testing different distance values until the energy value was constant; until it plateaued.

The energy of the TS was found to be -433.98 kJ.mol-1.

For the F + H2 reaction, the minimum point (reactants energy) was -435.10 kJ.mol-1. The activation energy was found to be -433.98--435.10 = '''+1.12 kJ.mol-1''' (where A: F, B: H, C: H and AB is a larger distance than the BC bond distance of HH).
[[File:H2 EA 01524183.png|thumb|340x340px|Figure 13. Finding the activation energy using mep. The activation energies could also be found by running mep calculations where the distance values are those of the TS but with the AB distance displaced by approx. 1 pm ( AB = 179 pm and BC = 74.5 pm, where A: F, B: H, C: H).]]
For the H + HF reaction, the minimum point (reactants energy) was -560.70 kJ.mol-1. The activation energy was found to be -433.98--560.70 = '''+126.72 kJ.mol-1''' (where A: F, B: H, C: H and BC is a larger distance than AB bond distance of HF).

The activation energies could also be found by performing a mep (see Figure 13).
==== 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 reactive trajectory of the F + H2 reaction is shown in Figure 14, as a contour plot. Initially, the system (reactants) has more translational kinetic energy therefore the atoms are moving very fast. Whereas, after the reaction has occurred (the region where the BC distance is increasing), there is more vibrational energy as there are more oscillations in the product valley. Therefore potential energy is being converted to vibrational energy and the system releases the reaction energy as vibrational energy. In this way, the conservation of energy is fulfilled. This could be confirmed experimentally by IR spectroscopy (IR absorption) and Infrared chemiluminescence (emission of vibration by IR light).
[[File:01524183 FH2 reaction.png|left|thumb|273x273px|Figure 14. A contour plot of the F + H2 reaction showing the reactive trajectory. Initial conditions were set as AB: 175 pm and -1.6 g.mol-1.pm.fs-1, BC: 74 pm and 0.2 g.mol-1.pm.fs-1 , with the number of steps as 3000. (where A: F, B: H and C: H).]]

====== IR spectroscopy ======
Initially, for the exothermic F + H2 reaction, all the molecules are in the ground state in the H2 gas. However, as a result of the reaction, a small proportion of the molecules get excited and transfer to a vibrationally first excited state. This can be seen from the plot (Figure 14), as the vibrational energy increases in the product channel. Therefore, the IR spectrum of the thermally relaxed sample would show one absorption peak. However, the excited sample would show a fundamental peak, corresponding to the excitation from the ground state to the first excited state, and also an overtone peak, corresponding to excitation from the first excited state to the second excited state, at a lower wavenumber. The peaks would be at different wavenumbers due to the anharmonicity of the potential energy surface. By measuring the intensity of the overtone overtime, the number of molecules that are vibrationally excited, as time goes on, can be deduced. This is because the intensity of the overtone would decrease, as energy would be released as radiation (due to exothermicity of the reaction) therefore there would be fewer molecules in the first electronically excited state over time. Whereas, the intensity of the fundamental peak would increase as more molecules would be in the ground state which could be excited to the first excited state.

====== Infrared chemiluminescence ======
This is another method for examining the energy distribution in the products. In this case, vibrationally excited molecules emit infrared radiation as they return to their ground states. The populations of the vibrational states of the products may be determined by studying the intensities of the IR emission spectrum. '''(reference ATKINS PG 804)'''

==== 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. ====
From Polyani's rules, it can be deduced that for exothermic reactions, translational energy is more effective than vibrational energy at overcoming the TS barrier. Whereas, for endothermic reactions, vibrational energy is more effective than translational energy at passing the TS barrier. Therefore, an early energy barrier favours more highly vibrational products and a late energy barrier favours products with more translational energy.

For exothermic reactions the TS is located in the reactant channel (early TS) therefore molecules with the most energy in motion would easily be able to surmount the barrier. This is because a vibrationally excited reactant molecule would be too busy oscillating from side to side and so wouldn't have enough energy left to reach the top of the TS barrier. Therefore, reactant vibrational energy in excess of the TS barrier height may be ineffective for the reaction.

For endothermic reactions, the TS is located in the product channel (late TS), therefore reactant molecules with most vibrational energy would easily be able to cross the barrier. This is because the vibrationally excited reactant molecules would be travelling down the reactant channel; they would have to travel further in order to cross the TS barrier. The chances of the reactant molecules with the correct phase finding the TS barrier is higher due to the higher amount of vibrational energy. A reactant molecule with more translational energy, in this case, would simply slam into the repulsive inner wall of the potential surface and bounce back into the entrance channel.

MRD:01524183

2020-05-22T17:47:36Z

Aas1218:

== Introduction ==
The main aims of this experiment were to study the reactivity of triatomic systems, involving the collision of an atom and a diatomic molecule, by calculating Molecular Dynamic trajectories. The systems studied here are the symmetric H-H-H system and the non-symmetric F-H-H system.

== Exercise 1: The H + H2 system ==
The original H2 diatomic is defined as BC with the A H atom colliding to form the AB H2 diatomic molecule.

==== On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ====
[[File:H3 TS contourplot 01524183.png|thumb|329x329px|Figure 1. A contour plot showing the TS of the H-H-H system and the variation of AB and BC distances, with reaction trajectory and eigenvectors shown.]]
On a potential energy surface diagram, the transition state (TS) is mathematically defined as the point at which the gradient of the potential with respect to the internal coordinates is zero (∂V('''ri''')/∂'''ri'''=0). This is because the transition state is the local maximum on the minimum energy paths between the reactants and products. There would be no movement of the H atoms at the TS since the force is a derivative of the potential energy and at the TS this is zero, therefore, the nuclei are stationary; the transition state is a stationary point. However, minima and maxima are stationary points. The TS can be distinguished from a local minimum of the potential energy surface by taking the second differential of the potential function. By doing so, the nature of the stationary point can be determined: if the second differential is greater than zero, the point is a local minimum and if it is less than zero, the point is a local maximum (the TS).

Using the GUI simulation, the TS was identified by trial and error. At the TS the distance AB and BC must be equal (rAB = rBC), as the H + H2 surface is symmetric. Therefore the TS lies somewhere along the diagonal line where AB and BC are equal. As the TS is approached, the size of the forces of AB and BC decrease to zero and so the point at which these forces are close to zero is the TS of the potential energy surface. This maximum point can be distinguished from a local minimum by analysing the omega squared values under the Hessian eigenvalues/vectors. A TS is a maximum in one direction and a minimum in the other direction; it is a saddle point. Therefore the omega squared would be positive in one direction and negative in the other direction.

==== 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 transition state position ('''rts''') was found to be 90.8 pm to 1 d.p.[[File:H3 TS distancetime 01524183.png|thumb|Figure 2. A plot of Internuclear Distances vs Time: linear horizontal lines show the constant distances of AB, BC and AC. |329x329px]]This value gave forces of +0.001 kJ.mol-1.pm-1 along AB and BC which are very close to zero. Also, the omega hessian eigenvalues had opposite signs which verified that this point was the TS.

At the TS, the internuclear distances between AB and BC are the same and so the distance between AC is approximately double the AB (BC) distance. This is shown in the plot of "Internuclear distances vs Time", as two straight lines of constant distances of AB and BC at 90.8 pm and AC at approximately 181.6 pm. This is because at the TS the H atoms are stationary, with one H atom (the B atom in this case) being approx. halfway between the A and C H atoms.

==== Comment on how the ''mep'' and the trajectory you just calculated differ. ====
The minimum energy path (''mep) ''is the lowest energy path from reactant to product; it is a special trajectory that correlates to infinitely slow motion. The ''mep ''was calculated by setting the initial conditions as rBC = 91.8 pm (1 pm displaced from the TS position) and rAB = 90.8 pm (the TS position). The number of steps were also increased to 2500 in order to obtain a complete mep. This calculation was repeated with the calculation type set to Dynamics.
[[File:H3 dynamics mep 01524183.png|thumb|584x584px|Figure 3 and 4. A contour plot of BC distance vs AB distance, showing the ''mep ''trajectory: calculation type was set to MEP (left).
A contour plot of BC distance vs AB distance, showing the dynamic'' ''trajectory: calculation type was set to Dynamics (right).
|centre]]
As shown in the plots above, the ''mep ''is a smooth trajectory (left) which indicates that the BC molecule is not vibrating as the system is not oscillating between energy contours. This is because the ''mep'' doesn't provide a realistic view of the motion of atoms during a reaction; it assumes the atoms are moving at an infinitely slow speed. There would be vibration occurring between B and C as they are moving further away from each other. This is shown by the wavy trajectory produced from the dynamic calculation (right).

==== 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? ====
For the initial positions rBC = 74 pm and rAB = 200 pm, trajectories were run for five different moments combinations. The results of which are given in the table below.

{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot / kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Reactive
|The trajectory crosses the TS, therefore, it is reactive. The line is wavier after crossing the TS, therefore the initial diatomic BC has less vibrational energy than the forming AB diatomic molecule. However, overall the trajectory is not very wavy therefore there is less vibrational energy compared to translational energy in the whole system.
|[[File:H3 1 01524183.png|thumb|375x375px|Figure 5]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.08</nowiki>
|Unreactive
|The trajectory does not cross the TS in this case, therefore it is unreactive. The AB distance is changing as the A H atom is moving towards and away from the BC diatomic molecule, which means it has more translational energy compared to vibrational. This is shown in the plot as the line is not very wavy. However, the BC distance doesn't change as the diatomic molecule remains intact.
|[[File:H3 2 01524183.png|thumb|375x375px|Figure 6]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.98</nowiki>
|Reactive
|The trajectory does cross the TS therefore it is reactive. The lines quite wavy which there is some vibrational energy but more translational energy. This is quite similar to the plot in Figure 5 as the potential energies are the same therefore the waviness of the trajectories are the same. However, the kinetic energies are a little different, making the total energies different.
|[[File:H3 3 01524183.png|thumb|375x375px|Figure 7]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.28</nowiki>
|Unreactive
|The trajectory just crosses the TS however it does not enter the product channel instead it goes back to the reactant channel. This is called barrier re-crossing. Therefore it is overall unreactive. The trajectory is very wavy which means there is a lot of vibrational energy in the system and less translational energy. In the beginning, the AB and BC distances are changing as the middle B atom is oscillating between the A and C atoms. However, in the end, the molecule BC is reformed which is they the AB distance is large.
|[[File:H3 4 01524183.png|thumb|375x375px|Figure 8]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.48</nowiki>
|Reactive
|The trajectory is reactive as it does cross the TS. The line is very wavy which shows that the system has more vibrational energy compared to translational. As the A atom gets closer to the initial BC molecule, the AB distance decreases and the BC distance increases, however, at the TS the B atom oscillates between the A and C atoms as it is attracted to both. This means there's a lot of vibrational energy near the TS. In the end, it does form the AB product molecule, which is shown by the trajectory increase in BC distance.
|[[File:H3 5 01524183.png|thumb|375x375px|Figure 9]]
|}
From the table, it can be concluded that not all trajectories starting with the same positions but with higher values of momenta would be reactive. For example, increasing the value of momenta to -5.1 g.mol-1.pm.fs-1 for p1 and -10.1 g.mol-1.pm.fs-1 for p2 still gives an unreactive trajectory. Also, low values of momenta can be reactive as is the first momenta combination (Figure 5) in the table. This is because vibrational energy also needs to be taken into account, not only the kinetic energy. The degree of vibrational and translational energy is determined by analysing the trajectory line shape: a more wavy line would mean more vibrational energy whereas a more smooth line corresponds to more translational energy.

==== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ====
The TS theory is a theory used to determine absolute reaction rates based on the reactant and transition state structure properties. The assumptions this theory makes are:
# Every trajectory that has enough energy to roll over the TS will definitely go to form the products. This means that once the reactants have crossed the energy barrier they can't go back to reactants. Therefore the overall rate of reaction of reactants to products increases, overestimating the rate.
# The concentration of the TS is in equilibrium with the reactant concentration; this is known as the Quasi-equilibrium hypothesis ([X]‡ = 0.5 KC‡ [A][B], where A + B → X‡ → P). This shows that the overall concentration of TS would be lower than the concentration of the reactants. This is because a small proportion of reactants would be forming the TS from the reactants pool.
# Quantum tunnelling is not taken into account; the theory is treated purely classical as a translation. If quantum tunnelling was taken into account the rate would increase so in this case the theory underestimates the rate.
As the first assumption is the most crucial, it has a larger weighting effect on the rate compared to quantum tunnelling. Therefore overall the TS theory predictions overestimate the reaction rate values compared to the experimental values.

N'''eed to include references here '''

== Exercise 2: F - H - H system ==
==== By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ====
The F + H2 reaction (forward reaction) is exothermic as the reactants are at a higher energy in comparison to the products, therefore, energy is released to the surroundings. The H + HF reaction (backward reaction) is endothermic as the products are at a higher energy than the reactants. Therefore, as shown by the exothermicity of the forward reaction (F + H2), the bond strength of HF is larger than that of HH as the energy required to break the HH bond is less than the energy released upon forming HF. Also, the endothermicity of the backward reaction means that the energy required to break the strong HF bond is greater than the energy released to form the products (F and HH). '''This is confirmed by the literature values of HF and HH bond strengths of:...........................'''
[[File:PES FHH 01524183.png|centre|thumb|743x743px|Figure 10. (left) A potential energy surface plot of the F + H2 reaction (A: F, B: H and C: H). The reactant channel (small BC and large AB; the left channel) is at a higher energy value compared to the products (large BC and small AB). Therefore, this forward reaction is exothermic.
Figure 11. (right) A potential energy surface plot of the H + HF reaction. (A: H, B: H and C: F). The reactant channel (small BC and large AB; the left channel) is at a lower energy value compared to the products (large BC and small AB). Therefore, this forward reaction is endothermic.
]]

==== Locate the approximate position of the transition state. ====
[[File:FH2 TS 01524183.png|thumb|Figure 12. A contour plot of the F-H-H system, showing the TS. There is no trajectory therefore this is the TS. The point is in the reactant channel as expected as the forward reaction is exothermic. When viewed in terms of the backward reaction the TS is in the product channel as it is endothermic.|338x338px]]
The transition state for both the forward (F + H2) and backward (H + HF) reactions is at the same point. As the system is not symmetric, the TS is not defined as the point where AB = BC, instead Hammond's postulate is used in order to determine the TS point. According to Hammonds postulate, the TS of a reaction resembles the reactants if the reaction is exothermic (an early TS) and it resembles the products if the reaction is endothermic (a late TS). Therefore, the TS would be close to the reactants of the forward reaction and the products of the backward reaction which is the same (F + H2). So by setting the initial conditions such that AB is a larger distance than BC, the TS was found (where A: F, B: H and C: H). The momenta values were set to zero because the molecules are stationary at the TS as it is a saddle point.

The transition state position ('''rts''') was found to be at an '''AB distance''' of '''181.4 pm''' and '''BC''' '''distance''' of '''74.5 pm''' (A: F, B: H and C: H).

These values gave force values along AB and BC of 0.000 kJ.mol-1 and 0.006 kJ.mol-1 respectively, which are close to 0 kJ.mol-1, as expected at the TS point. They also gave Hessian eigenvalues/vectors of opposite signs (one negative and one positive) which also indicated that this was the TS position.
==== Report the activation energy for both reactions. ====
The activation energy is the difference in energy of the TS and the minimum of the reactant channel. This difference is smaller for the forward (F + H2) exothermic reaction compared to the backwards (H + HF) endothermic reaction. The minimum point was found by testing different distance values until the energy value was constant; until it plateaued.

The energy of the TS was found to be -433.98 kJ.mol-1.

For the F + H2 reaction, the minimum point (reactants energy) was -435.10 kJ.mol-1. The activation energy was found to be -433.98--435.10 = '''+1.12 kJ.mol-1''' (where A: F, B: H, C: H and AB is a larger distance than the BC bond distance of HH).
[[File:H2 EA 01524183.png|thumb|340x340px|Figure 13. Finding the activation energy using mep. The activation energies could also be found by running mep calculations where the distance values are those of the TS but with the AB distance displaced by approx. 1 pm ( AB = 179 pm and BC = 74.5 pm, where A: F, B: H, C: H).]]
For the H + HF reaction, the minimum point (reactants energy) was -560.70 kJ.mol-1. The activation energy was found to be -433.98--560.70 = '''+126.72 kJ.mol-1''' (where A: F, B: H, C: H and BC is a larger distance than AB bond distance of HF).

The activation energies could also be found by performing a mep (see Figure 13).
==== 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 reactive trajectory of the F + H2 reaction is shown in Figure 14, as a contour plot. Initially, the system (reactants) has more translational kinetic energy therefore the atoms are moving very fast. Whereas, after the reaction has occurred (the region where the BC distance is increasing), there is more vibrational energy as there are more oscillations in the product valley. Therefore potential energy is being converted to vibrational energy and the system releases the reaction energy as vibrational energy. In this way, the conservation of energy is fulfilled. This could be confirmed experimentally by IR spectroscopy (IR absorption) and Infrared chemiluminescence (emission of vibration by IR light).
[[File:01524183 FH2 reaction.png|left|thumb|307x307px|Figure 14. A contour plot of the F + H2 reaction showing the reactive trajectory. Initial conditions were set as AB: 175 pm and -1.6 g.mol-1.pm.fs-1, BC: 74 pm and 0.2 g.mol-1.pm.fs-1 , with the number of steps as 3000. (where A: F, B: H and C: H).]]

====== IR spectroscopy ======
Initially, for the exothermic F + H2 reaction, all the molecules are in the ground state in the H2 gas. However, as a result of the reaction, a small proportion of the molecules get excited and transfer to a vibrationally first excited state. This can be seen from the plot (Figure 14), as the vibrational energy increases in the product channel. Therefore, the IR spectrum of the thermally relaxed sample would show one absorption peak. However, the excited sample would show a fundamental peak, corresponding to the excitation from the ground state to the first excited state, and also an overtone peak, corresponding to excitation from the first excited state to the second excited state, at a lower wavenumber. The peaks would be at different wavenumbers due to the anharmonicity of the potential energy surface. By measuring the intensity of the overtone overtime, the number of molecules that are vibrationally excited, as time goes on, can be deduced. This is because the intensity of the overtone would decrease, as energy would be released as radiation (due to exothermicity of the reaction) therefore there would be fewer molecules in the first electronically excited state over time. Whereas, the intensity of the fundamental peak would increase as more molecules would be in the ground state which could be excited to the first excited state.

====== Infrared chemiluminescence ======
This is another method for examining the energy distribution in the products. In this case, vibrationally excited molecules emit infrared radiation as they return to their ground states. The populations of the vibrational states of the products may be determined by studying the intensities of the IR emission spectrum. '''(reference ATKINS PG 804)'''

==== 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. ====
From Polyani's rules, it can be deduced that for exothermic reactions, translational energy is more effective than vibrational energy at overcoming the TS barrier. Whereas, for endothermic reactions, vibrational energy is more effective than translational energy at passing the TS barrier. Therefore, an early energy barrier favours more highly vibrational products and a late energy barrier favours products with more translational energy.

For exothermic reactions the TS is located in the reactant channel (early TS) therefore molecules with the most energy in motion would easily be able to surmount the barrier. This is because a vibrationally excited reactant molecule would be too busy oscillating from side to side and so wouldn't have enough energy left to reach the top of the TS barrier. Therefore, reactant vibrational energy in excess of the TS barrier height may be ineffective for the reaction.

For endothermic reactions, the TS is located in the product channel (late TS), therefore reactant molecules with most vibrational energy would easily be able to cross the barrier. This is because the vibrationally excited reactant molecules would be travelling down the reactant channel; they would have to travel further in order to cross the TS barrier. The chances of the reactant molecules with the correct phase finding the TS barrier is higher due to the higher amount of vibrational energy. A reactant molecule with more translational energy, in this case, would simply slam into the repulsive inner wall of the potential surface and bounce back into the entrance channel.

MRD:01524183

2020-05-22T15:59:17Z

Aas1218:

== Introduction ==
The main aims of this experiment were to study the reactivity of triatomic systems, involving the collision of an atom and a diatomic molecule, by calculating Molecular Dynamic trajectories. The systems studied here are the symmetric H-H-H system and the non-symmetric F-H-H system.

== Exercise 1: The H + H2 system ==
The original H2 diatomic is defined as BC with the A H atom colliding to form the AB H2 diatomic molecule.

==== On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ====
[[File:H3 TS contourplot 01524183.png|thumb|329x329px|Figure 1. A contour plot showing the TS of the H-H-H system and the variation of AB and BC distances, with reaction trajectory and eigenvectors shown.]]
On a potential energy surface diagram, the transition state (TS) is mathematically defined as the point at which the gradient of the potential with respect to the internal coordinates is zero (∂V('''ri''')/∂'''ri'''=0). This is because the transition state is the local maximum on the minimum energy paths between the reactants and products. There would be no movement of the H atoms at the TS since the force is a derivative of the potential energy and at the TS this is zero, therefore, the nuclei are stationary; the transition state is a stationary point. However, minima and maxima are stationary points. The TS can be distinguished from a local minimum of the potential energy surface by taking the second differential of the potential function. By doing so, the nature of the stationary point can be determined: if the second differential is greater than zero, the point is a local minimum and if it is less than zero, the point is a local maximum (the TS).

Using the GUI simulation, the TS was identified by trial and error. At the TS the distance AB and BC must be equal (rAB = rBC), as the H + H2 surface is symmetric. Therefore the TS lies somewhere along the diagonal line where AB and BC are equal. As the TS is approached, the size of the forces of AB and BC decrease to zero and so the point at which these forces are close to zero is the TS of the potential energy surface. This maximum point can be distinguished from a local minimum by analysing the omega squared values under the Hessian eigenvalues/vectors. A TS is a maximum in one direction and a minimum in the other direction; it is a saddle point. Therefore the omega squared would be positive in one direction and negative in the other direction.

==== 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 transition state position ('''rts''') was found to be 90.8 pm to 1 d.p.[[File:H3 TS distancetime 01524183.png|thumb|Figure 2. A plot of Internuclear Distances vs Time: linear horizontal lines show the constant distances of AB, BC and AC. |329x329px]]This value gave forces of +0.001 kJ.mol-1.pm-1 along AB and BC which are very close to zero. Also, the omega hessian eigenvalues had opposite signs which verified that this point was the TS.

At the TS, the internuclear distances between AB and BC are the same and so the distance between AC is approximately double the AB (BC) distance. This is shown in the plot of "Internuclear distances vs Time", as two straight lines of constant distances of AB and BC at 90.8 pm and AC at approximately 181.6 pm. This is because at the TS the H atoms are stationary, with one H atom (the B atom in this case) being approx. halfway between the A and C H atoms.

==== Comment on how the ''mep'' and the trajectory you just calculated differ. ====
The minimum energy path (''mep) ''is the lowest energy path from reactant to product; it is a special trajectory that correlates to infinitely slow motion. The ''mep ''was calculated by setting the initial conditions as rBC = 91.8 pm (1 pm displaced from the TS position) and rAB = 90.8 pm (the TS position). The number of steps were also increased to 2500 in order to obtain a complete mep. This calculation was repeated with the calculation type set to Dynamics.
[[File:H3 dynamics mep 01524183.png|thumb|584x584px|Figure 3 and 4. A contour plot of BC distance vs AB distance, showing the ''mep ''trajectory: calculation type was set to MEP (left).
A contour plot of BC distance vs AB distance, showing the dynamic'' ''trajectory: calculation type was set to Dynamics (right).
|centre]]
As shown in the plots above, the ''mep ''is a smooth trajectory (left) which indicates that the BC molecule is not vibrating as the system is not oscillating between energy contours. This is because the ''mep'' doesn't provide a realistic view of the motion of atoms during a reaction; it assumes the atoms are moving at an infinitely slow speed. There would be vibration occurring between B and C as they are moving further away from each other. This is shown by the wavy trajectory produced from the dynamic calculation (right).

==== 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? ====
For the initial positions rBC = 74 pm and rAB = 200 pm, trajectories were run for five different moments combinations. The results of which are given in the table below.

{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot / kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Reactive
|The trajectory crosses the TS, therefore, it is reactive. The line is wavier after crossing the TS, therefore the initial diatomic BC has less vibrational energy than the forming AB diatomic molecule. However, overall the trajectory is not very wavy therefore there is less vibrational energy compared to translational energy in the whole system.
|[[File:H3 1 01524183.png|thumb|375x375px|Figure 5]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.08</nowiki>
|Unreactive
|The trajectory does not cross the TS in this case, therefore it is unreactive. The AB distance is changing as the A H atom is moving towards and away from the BC diatomic molecule, which means it has more translational energy compared to vibrational. This is shown in the plot as the line is not very wavy. However, the BC distance doesn't change as the diatomic molecule remains intact.
|[[File:H3 2 01524183.png|thumb|375x375px|Figure 6]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.98</nowiki>
|Reactive
|The trajectory does cross the TS therefore it is reactive. The lines quite wavy which there is some vibrational energy but more translational energy. This is quite similar to the plot in Figure 5 as the potential energies are the same therefore the waviness of the trajectories are the same. However, the kinetic energies are a little different, making the total energies different.
|[[File:H3 3 01524183.png|thumb|375x375px|Figure 7]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.28</nowiki>
|Unreactive
|The trajectory just crosses the TS however it does not enter the product channel instead it goes back to the reactant channel. This is called barrier re-crossing. Therefore it is overall unreactive. The trajectory is very wavy which means there is a lot of vibrational energy in the system and less translational energy. In the beginning, the AB and BC distances are changing as the middle B atom is oscillating between the A and C atoms. However, in the end, the molecule BC is reformed which is they the AB distance is large.
|[[File:H3 4 01524183.png|thumb|375x375px|Figure 8]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.48</nowiki>
|Reactive
|The trajectory is reactive as it does cross the TS. The line is very wavy which shows that the system has more vibrational energy compared to translational. As the A atom gets closer to the initial BC molecule, the AB distance decreases and the BC distance increases, however, at the TS the B atom oscillates between the A and C atoms as it is attracted to both. This means there's a lot of vibrational energy near the TS. In the end, it does form the AB product molecule, which is shown by the trajectory increase in BC distance.
|[[File:H3 5 01524183.png|thumb|375x375px|Figure 9]]
|}
From the table, it can be concluded that not all trajectories starting with the same positions but with higher values of momenta would be reactive. For example, increasing the value of momenta to -5.1 g.mol-1.pm.fs-1 for p1 and -10.1 g.mol-1.pm.fs-1 for p2 still gives an unreactive trajectory. Also, low values of momenta can be reactive as is the first momenta combination (Figure 5) in the table. This is because vibrational energy also needs to be taken into account, not only the kinetic energy. The degree of vibrational and translational energy is determined by analysing the trajectory line shape: a more wavy line would mean more vibrational energy whereas a more smooth line corresponds to more translational energy.

==== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ====
The TS theory is a theory used to determine absolute reaction rates based on the reactant and transition state structure properties. The assumptions this theory makes are:
# Every trajectory that has enough energy to roll over the TS will definitely go to form the products. This means that once the reactants have crossed the energy barrier they can't go back to reactants. Therefore the overall rate of reaction of reactants to products increases, overestimating the rate.
# The concentration of the TS is in equilibrium with the reactant concentration; this is known as the Quasi-equilibrium hypothesis ([X]‡ = 0.5 KC‡ [A][B], where A + B → X‡ → P). This shows that the overall concentration of TS would be lower than the concentration of the reactants. This is because a small proportion of reactants would be forming the TS from the reactants pool.
# Quantum tunnelling is not taken into account; the theory is treated purely classical as a translation. If quantum tunnelling was taken into account the rate would increase so in this case the theory underestimates the rate.
As the first assumption is the most crucial, it has a larger weighting effect on the rate compared to quantum tunnelling. Therefore overall the TS theory predictions overestimate the reaction rate values compared to the experimental values.

N'''eed to include references here '''

== Exercise 2: F - H - H system ==
==== By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ====
The F + H2 reaction (forward reaction) is exothermic as the reactants are at a higher energy in comparison to the products, therefore, energy is released to the surroundings. The H + HF reaction (backward reaction) is endothermic as the products are at a higher energy than the reactants. Therefore, as shown by the exothermicity of the forward reaction (F + H2), the bond strength of HF is larger than that of HH as the energy required to break the HH bond is less than the energy released upon forming HF. Also, the endothermicity of the backward reaction means that the energy required to break the strong HF bond is greater than the energy released to form the products (F and HH). '''This is confirmed by the literature values of HF and HH bond strengths of:...........................'''
[[File:PES FHH 01524183.png|centre|thumb|743x743px|Figure 10. (left) A potential energy surface plot of the F + H2 reaction (A: F, B: H and C: H). The reactant channel (small BC and large AB; the left channel) is at a higher energy value compared to the products (large BC and small AB). Therefore, this forward reaction is exothermic.
Figure 11. (right) A potential energy surface plot of the H + HF reaction. (A: H, B: H and C: F). The reactant channel (small BC and large AB; the left channel) is at a lower energy value compared to the products (large BC and small AB). Therefore, this forward reaction is endothermic.
]]

==== Locate the approximate position of the transition state. ====
[[File:FH2 TS 01524183.png|thumb|Figure 12. A contour plot of the F-H-H system, showing the TS. There is no trajectory therefore this is the TS. The point is in the reactant channel as expected as the forward reaction is exothermic. When viewed in terms of the backward reaction the TS is in the product channel as it is endothermic.|338x338px]]
The transition state for both the forward (F + H2) and backward (H + HF) reactions is at the same point. As the system is not symmetric, the TS is not defined as the point where AB = BC, instead Hammond's postulate is used in order to determine the TS point. According to Hammonds postulate, the TS of a reaction resembles the reactants if the reaction is exothermic (an early TS) and it resembles the products if the reaction is endothermic (a late TS). Therefore, the TS would be close to the reactants of the forward reaction and the products of the backward reaction which is the same (F + H2). So by setting the initial conditions such that AB is a larger distance than BC, the TS was found (where A: F, B: H and C: H). The momenta values were set to zero because the molecules are stationary at the TS as it is a saddle point.

The transition state position ('''rts''') was found to be at an '''AB distance''' of '''181.4 pm''' and '''BC''' '''distance''' of '''74.5 pm''' (A: F, B: H and C: H).

These values gave force values along AB and BC of 0.000 kJ.mol-1 and 0.006 kJ.mol-1 respectively, which are close to 0 kJ.mol-1, as expected at the TS point. They also gave Hessian eigenvalues/vectors of opposite signs (one negative and one positive) which also indicated that this was the TS position.
==== Report the activation energy for both reactions. ====
The activation energy is the difference in energy of the TS and the minimum of the reactant channel. This difference is smaller for the forward (F + H2) exothermic reaction compared to the backwards (H + HF) endothermic reaction. The minimum point was found by testing different distance values until the energy value was constant; until it plateaued.

The energy of the TS was found to be -433.98 kJ.mol-1.

For the F + H2 reaction, the minimum point (reactants energy) was -435.10 kJ.mol-1. The activation energy was found to be -433.98--435.10 = '''+1.12 kJ.mol-1''' (where A: F, B: H, C: H and AB is a larger distance than the BC bond distance of HH).
[[File:H2 EA 01524183.png|thumb|340x340px|Figure 13. Finding the activation energy using mep. The activation energies could also be found by running mep calculations where the distance values are those of the TS but with the AB distance displaced by approx. 1 pm ( AB = 179 pm and BC = 74.5 pm, where A: F, B: H, C: H).]]
For the H + HF reaction, the minimum point (reactants energy) was -560.70 kJ.mol-1. The activation energy was found to be -433.98--560.70 = '''+126.72 kJ.mol-1''' (where A: F, B: H, C: H and BC is a larger distance than AB bond distance of HF).

The activation energies could also be found by performing a mep (see Figure 13).
==== 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 reactive trajectory of the F + H2 reaction is shown in Figure 14, as a contour plot. Initially, the system (reactants) has more translational kinetic energy therefore the atoms are moving very fast. Whereas, after the reaction has occurred (the region where the BC distance is increasing), there is more vibrational energy as there are more oscillations in the product valley. Therefore potential energy is being converted to vibrational energy and the system releases the reaction energy as vibrational energy. In this way, the conservation of energy is fulfilled. This could be confirmed experimentally by IR spectroscopy (IR absorption) and Infrared chemiluminescence (emission of vibration by IR light).
[[File:01524183 FH2 reaction.png|left|thumb|340x340px|Figure 14. A contour plot of the F + H2 reaction showing the reactive trajectory. Initial conditions were set as AB: 175 pm and -1.6 g.mol-1.pm.fs-1, BC: 74 pm and 0.2 g.mol-1.pm.fs-1 , with the number of steps as 3000. (where A: F, B: H and C: H).]]
IR spectroscopy:

Initially, for the exothermic F + H2 reaction, all the molecules are in the ground state in the H2 gas. However, as a result of the reaction, a small proportion of the molecules get excited and transfer to a vibrationally first excited state. This can be seen from the plot (Figure 14), as the vibrational energy increases in the product channel. Therefore, the IR spectrum of the thermally relaxed sample would show one absorption peak. However, the excited sample would show a fundamental peak, corresponding to the excitation from the ground state to the first excited state, and also an overtone peak, corresponding to excitation from the first excited state to the second excited state, at a lower wavenumber. The peaks would be at different wavenumbers due to the anharmonicity of the potential energy surface. By measuring the intensity of the overtone overtime, the number of molecules that are vibrationally excited, as time goes on, can be deduced. This is because the intensity of the overtone would decrease, as energy would be released as radiation (due to exothermicity of the reaction) therefore there would be fewer molecules in the first electronically excited state over time. Whereas, the intensity of the fundamental peak would increase as more molecules would be in the ground state which could be excited to the first excited state.

Infrared chemiluminescence

This is another method for examining the energy distribution in the products. In this case, vibrationally excited molecules emit infrared radiation as they return to their ground states. The populations of the vibrational states of the products may be determined by studying the intensities of the IR emission spectrum. '''(reference ATKINS PG 804)'''

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

File:01524183 FH2 reaction.png

2020-05-22T15:29:56Z

Aas1218:

MRD:01524183

2020-05-22T15:28:14Z

Aas1218: /* Exercise 1: The H + H2 system */

== Introduction ==
The main aims of this experiment were to study the reactivity of triatomic systems, involving the collision of an atom and a diatomic molecule, by calculating Molecular Dynamic trajectories. The systems studied here are the symmetric H-H-H system and the non-symmetric F-H-H system.

== Exercise 1: The H + H2 system ==
The original H2 diatomic is defined as BC with the A H atom colliding to form the AB H2 diatomic molecule.

==== On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ====
[[File:H3 TS contourplot 01524183.png|thumb|329x329px|Figure 1. A contour plot showing the TS of the H-H-H system and the variation of AB and BC distances, with reaction trajectory and eigenvectors shown.]]
On a potential energy surface diagram, the transition state (TS) is mathematically defined as the point at which the gradient of the potential with respect to the internal coordinates is zero (∂V('''ri''')/∂'''ri'''=0). This is because the transition state is the local maximum on the minimum energy paths between the reactants and products. There would be no movement of the H atoms at the TS since the force is a derivative of the potential energy and at the TS this is zero, therefore, the nuclei are stationary; the transition state is a stationary point. However, minima and maxima are stationary points. The TS can be distinguished from a local minimum of the potential energy surface by taking the second differential of the potential function. By doing so, the nature of the stationary point can be determined: if the second differential is greater than zero, the point is a local minimum and if it is less than zero, the point is a local maximum (the TS).

Using the GUI simulation, the TS was identified by trial and error. At the TS the distance AB and BC must be equal (rAB = rBC), as the H + H2 surface is symmetric. Therefore the TS lies somewhere along the diagonal line where AB and BC are equal. As the TS is approached, the size of the forces of AB and BC decrease to zero and so the point at which these forces are close to zero is the TS of the potential energy surface. This maximum point can be distinguished from a local minimum by analysing the omega squared values under the Hessian eigenvalues/vectors. A TS is a maximum in one direction and a minimum in the other direction; it is a saddle point. Therefore the omega squared would be positive in one direction and negative in the other direction.

==== 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 transition state position ('''rts''') was found to be 90.8 pm to 1 d.p.[[File:H3 TS distancetime 01524183.png|thumb|Figure 2. A plot of Internuclear Distances vs Time: linear horizontal lines show the constant distances of AB, BC and AC. |329x329px]]This value gave forces of +0.001 kJ.mol-1.pm-1 along AB and BC which are very close to zero. Also, the omega hessian eigenvalues had opposite signs which verified that this point was the TS.

At the TS, the internuclear distances between AB and BC are the same and so the distance between AC is approximately double the AB (BC) distance. This is shown in the plot of "Internuclear distances vs Time", as two straight lines of constant distances of AB and BC at 90.8 pm and AC at approximately 181.6 pm. This is because at the TS the H atoms are stationary, with one H atom (the B atom in this case) being halfway between the A and C H atoms.

==== Comment on how the ''mep'' and the trajectory you just calculated differ. ====
The minimum energy path (''mep) ''is the lowest energy path from reactant to product; it is a special trajectory that correlates to infinitely slow motion. The ''mep ''was calculated by setting the initial conditions as rBC = 91.8 pm (1 pm displaced from the TS position) and rAB = 90.8 pm (the TS position). The number of steps were also increased to 2500 in order to obtain a complete mep. This calculation was repeated with the calculation type set to Dynamics.
[[File:H3 dynamics mep 01524183.png|thumb|584x584px|Figure 3 and 4. A contour plot of BC distance vs AB distance, showing the ''mep ''trajectory: calculation type was set to MEP (left).
A contour plot of BC distance vs AB distance, showing the dynamic'' ''trajectory: calculation type was set to Dynamics (right).
|centre]]
As shown in the plots above, the ''mep ''is a smooth trajectory (left) which indicates that the BC molecule is not vibrating as the system is not oscillating between energy contours. This is because the ''mep'' doesn't provide a realistic view of the motion of atoms during a reaction; it assumes the atoms are moving at an infinitely slow speed. There would be vibration occurring between B and C as they are moving further away from each other. This is shown by the wavy trajectory produced from the dynamic calculation (right).

==== 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? ====
For the initial positions r1 = 74 pm and r2 = 200 pm, trajectories were run for five different moments combinations. The results of which are given in the table below.

{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot / kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Reactive
|The trajectory crosses the TS, therefore, it is reactive. The line is more wavy after crossing the TS, therefore the initial diatomic BC has less vibrational energy than the forming AB diatomic molecule. However, overall the trajectory is not very wavy therefore there is less vibrational energy compared to translational energy in the whole system.
|[[File:H3 1 01524183.png|thumb|375x375px|Figure 5]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.08</nowiki>
|Unreactive
|The trajectory does not cross the TS in this case, therefore it is unreactive. The AB distance is changing as the A H atom is moving towards and away from the BC diatomic molecule, which means it has more translational energy compared to vibrational. This is shown in the plot as the line is not very wavy. However, the BC distance doesn't change as the diatomic molecule remains intact.
|[[File:H3 2 01524183.png|thumb|375x375px|Figure 6]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.98</nowiki>
|Reactive
|The trajectory does cross the TS therefore it is reactive. The lines quite wavy which there is some vibrational energy but more translational energy. This is quite similar to the plot in Figure 5 as the potential energies are the same therefore the waviness of the trajectories are the same. However, the kinetic energies are a little different, making the total energies different.
|[[File:H3 3 01524183.png|thumb|375x375px|Figure 7]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.28</nowiki>
|Unreactive
|The trajectory just crosses the TS however it does not enter the product channel instead it goes back to the reactant channel. This is called barrier re-crossing. Therefore it is overall unreactive. The trajectory is very wavy which means there is a lot of vibrational energy in the system and less translational energy. In the beginning, the AB and BC distances are changing as the middle B atom is oscillating between the A and C atoms. However, in the end, the molecule BC is reformed which is they the AB distance is large.
|[[File:H3 4 01524183.png|thumb|375x375px|Figure 8]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.48</nowiki>
|Reactive
|The trajectory is reactive as it does cross the TS. The line is very wavy which shows that the system has more vibrational energy compared to translational. As the A atom gets closer to the initial BC molecule, the AB distance decreases and the BC distance increases, however, at the TS the B atom oscillates between the A and C atoms as it is attracted to both. This means there's a lot of vibrational energy near the TS. In the end, it does form the AB product molecule, which is shown by the trajectory increase in BC distance.
|[[File:H3 5 01524183.png|thumb|375x375px|Figure 9]]
|}
From the table, it can be concluded that not all trajectories starting with the same positions but with higher values of momenta would be reactive. For example, increasing the value of momenta to -5.1 g.mol-1.pm.fs-1 for p1 and -10.1 g.mol-1.pm.fs-1 for p2 still gives an unreactive trajectory. Also, low values of momenta can be reactive as is the first momenta combination (Figure 5) in the table. This is because vibrational energy also needs to be taken into account, not only the kinetic energy. The degree of vibrational and translational energy is determined by analysing the trajectory line shape: a more wavy line would mean more vibrational energy whereas a more smooth line corresponds to more translational energy.

==== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ====
The TS theory is a theory used to determine absolute reaction rates based on the reactant and transition state structure properties. The assumptions this theory makes are:
# Every trajectory that has enough energy to roll over the TS will definitely go to form the products. This means that once the reactants have crossed the energy barrier they can't go back to reactants. Therefore the overall rate of reaction of reactants to products increases, overestimating the rate.
# The concentration of the TS is in equilibrium with the reactant concentration; this is known as the Quasi-equilibrium hypothesis ([X]‡ = 0.5 KC‡ [A][B], where A + B → X‡ → P). This shows that the overall concentration of TS would be lower than the concentration of the reactants. This is because a small proportion of reactants would be forming the TS from the reactants pool.
# Quantum tunnelling is not taken into account; the theory is treated purely classical as a translation. If quantum tunnelling was taken into account the rate would increase so in this case the theory underestimates the rate.
As the first assumption is the most crucial, it has a larger weighting effect on the rate compared to quantum tunnelling. Therefore overall the TS theory predictions overestimate the reaction rate values compared with experimental values.

N'''eed to include references here '''

== Exercise 2: F - H - H system ==
==== By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ====
The F + H2 reaction (forward reaction) is exothermic as the reactants are at a higher energy in comparison to the products, therefore, energy is released to the surroundings. The H + HF reaction (backward reaction) is endothermic as the products are at a higher energy than the reactants. Therefore, as shown by the exothermicity of the forward reaction (F + H2), the bond strength of HF is larger than that of HH as the energy required to break the HH bond is less than the energy released upon forming HF. Also, the endothermicity of the backward reaction means that the energy required to break the strong HF bond is greater than the energy released to form the products (F and HH). '''This is confirmed by the literature values of HF and HH bond strengths of:...........................'''
[[File:PES FHH 01524183.png|centre|thumb|827x827px|Figure 10. (left) A potential energy surface plot of the F + H2 reaction (A: F, B: H and C: H). The reactant channel (small BC and large AB; the left channel) is at a higher energy value compared to the products (large BC and small AB). Therefore, this forward reaction is exothermic.
Figure 11. (right) A potential energy surface plot of the H + HF reaction. (A: H, B: H and C: F). The reactant channel (small BC and large AB; the left channel) is at a lower energy value compared to the products (large BC and small AB). Therefore, this forward reaction is endothermic.
]]

==== Locate the approximate position of the transition state. ====
[[File:FH2 TS 01524183.png|thumb|Figure 12. A contour plot of the F-H-H system, showing the TS. There is no trajectory therefore this is the TS. The point is in the reactant channel as expected as the forward reaction is exothermic. When viewed in terms of the backward reaction the TS is in the product channel as it is endothermic.|338x338px]]
The transition state for both the forward (F + H2) and backward (H + HF) reactions is at the same point. As the system is not symmetric, the TS is not defined as the point where AB = BC, instead Hammond's postulate is used in order to determine the TS point. According to Hammonds postulate, the TS of a reaction resembles the reactants if the reaction is exothermic (an early TS) and it resembles the products if the reaction is endothermic (a late TS). Therefore, the TS would be close to the reactants of the forward reaction and the products of the backward reaction which is the same (F + H2). So by setting the initial conditions such that AB is a larger distance than BC, the TS was found (where A: F, B: H and C: H). The momenta values were set to zero because the molecules are stationary at the TS as it is a saddle point.

The transition state position ('''rts''') was found to be at an '''AB distance''' of '''181.4 pm''' and '''BC''' '''distance''' of '''74.5 pm''' (A: F, B: H and C: H).

These values gave force values along AB and BC of 0.000 kJ.mol-1 and 0.006 kJ.mol-1 respectively, which are close to 0 kJ.mol-1, as expected at the TS point. They also gave Hessian eigenvalues/vectors of opposite signs (one negative and one positive) which also indicated that this was the TS position.
==== Report the activation energy for both reactions. ====
The activation energy is the difference in energy of the TS and the minimum of the reactant channel. This difference is smaller for the forward (F + H2) exothermic reaction compared to the backwards (H + HF) endothermic reaction. The minimum point was found by testing different distance values until the energy value was constant; until it plateaued.

The energy of the TS was found to be -433.98 kJ.mol-1.

For the F + H2 reaction, the minimum point (reactants energy) was -435.10 kJ.mol-1. The activation energy was found to be -433.98--435.10 = '''+1.12 kJ.mol-1''' (where A: F, B: H, C: H and AB is a larger distance than the BC bond distance of HH).
[[File:H2 EA 01524183.png|thumb|340x340px|Figure 13. Finding the activation energy using mep. The activation energies could also be found by running mep calculations where the distance values are those of the TS but with the AB distance displaced by approx. 1 pm ( AB = 179 pm and BC = 74.5 pm, where A: F, B: H, C: H).]]
For the H + HF reaction, the minimum point (reactants energy) was -560.70 kJ.mol-1. The activation energy was found to be -433.98--560.70 = '''+126.72 kJ.mol-1''' (where A: F, B: H, C: H and BC is a larger distance than AB bond distance of HF).

The activation energies could also be found by performing a mep (see Figure 13).
==== 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 reactive trajectory of the F + H2 reaction is shown in Figure 14, as a contour plot. Initially, the system (reactants) has more translational kinetic energy therefore the atoms are moving very fast. Whereas, after the reaction has occurred (the region where the BC distance is increasing), there is more vibrational energy as there are more oscillations in the product valley. Therefore potential energy is being converted to vibrational energy and the system releases the reaction energy as vibrational energy. In this way, the conservation of energy is fulfilled. This could be confirmed experimentally by IR spectroscopy (IR absorption) and emission of vibration by IR light '''(need to find the right name). '''

IR spectroscopy:

Initially, for the exothermic F + H2 reaction, all the molecules are in the ground state in the H2 gas. However, as a result of the reaction, a small proportion of the molecules get excited and transfer to a vibrationally first excited state. This can be seen from the plot (Figure 14), as the vibrational energy increases in the product channel. Therefore, the IR spectrum of the thermally relaxed sample would show one absorption peak. However, the excited sample would show a fundamental peak, corresponding to the excitation from the ground state to the first excited state, and also an overtone peak, corresponding to excitation from the first excited state to the second excited state, at a lower wavenumber. The peaks would be at different wavenumbers due to the anharmonicity of the potential energy surface. By measuring the intensity of the overtone overtime, the number of molecules that are vibrationally excited, as time goes on, can be deduced. This is because the intensity of the overtone would decrease, as energy would be released as radiation (due to exothermicity of the reaction) therefore there would be fewer molecules in the first electronically excited state over time. Whereas, the intensity of the fundamental peak would increase as more molecules would be in the ground state which could be excited to the first excited state.

Emission of IR light ('''find name)'''

In this case, the IR light emission is measured when the molecules relax from the excited state to the ground state.

==== 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. ====

MRD:01524183

2020-05-22T09:41:53Z

Aas1218:

MRD:01524183

2020-05-22T09:16:41Z

Aas1218:

== Introduction ==
The main aims of this experiment were to study the reactivity of triatomic systems, involving the collision of an atom and a diatomic molecule, by calculating Molecular Dynamic trajectories. ........... '''say about BC being the original diatomic molecule and A colliding and forming AB. - how the system is initially set up.'''

== Exercise 1: The H + H2 system ==

==== On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ====
[[File:H3 TS contourplot 01524183.png|thumb|278x278px|Figure 1. A contour plot showing the variation of AB and BC distances, with reaction trajectory and eigenvectors shown.]]
On a potential energy surface diagram, the transition state (TS) is mathematically defined as the point at which the gradient of the potential with respect to the internal coordinates is zero (∂V('''ri''')/∂'''ri'''=0). This is because the transition state is the local maximum on the minimum energy paths between the reactants and products. There would be no movement of the H atoms at the TS since the force is a derivative of the potential energy and at the TS this is zero, therefore, the nuclei are stationary; the transition state is a stationary point. However, minima and maxima are stationary points. The TS can be distinguished from a local minimum of the potential energy surface by taking the second differential of the potential function. By doing so, the nature of the stationary point can be determined: if the second differential is greater than zero, the point is a local minimum and if it is less than zero, the point is a local maximum (the TS).

Using the GUI simulation, the TS was identified by trial and error. At the TS the distance AB and BC must be equal (rAB = rBC), as the H + H2 surface is symmetric. Therefore the TS lies somewhere along the diagonal line where AB and BC are equal. As the TS is approached, the size of the forces of AB and BC decrease to zero and so the point at which these forces are close to zero is the TS of the potential energy surface. This maximum point can be distinguished from a local minimum by analysing the omega squared values under the Hessian eigenvalues/vectors. A TS is a maximum in one direction and a minimum in the other direction; it is a saddle point. Therefore the omega squared would be positive in one direction and negative in the other direction.

==== 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 transition state position ('''rts''') was found to be 90.8 pm to 1 d.p.[[File:H3 TS distancetime 01524183.png|thumb|Figure 2. A plot of Internuclear Distances vs Time: linear horizontal lines show the constant distances of AB, BC and AC. |420x420px]]This value gave forces of +0.001 kJ.mol-1.pm-1 along AB and BC which are very close to zero. Also, the omega hessian eigenvalues had opposite signs which verified that this point was the TS.

At the TS, the internuclear distances between AB and BC are the same and so the distance between AC is approximately double the AB (BC) distance. This is shown in the plot of "Internuclear distances vs Time", as two straight lines of constant distances of AB and BC at 90.8 pm and AC at approximately 181.6 pm. This is because at the TS the H atoms are stationary, with one H atom (the B atom in this case) being halfway between the A and C H atoms.

==== Comment on how the ''mep'' and the trajectory you just calculated differ. ====
The minimum energy path (''mep) ''is the lowest energy path from reactant to product; it is a special trajectory that correlates to infinitely slow motion. The ''mep ''was calculated by setting the initial conditions as rBC = 91.8 pm (1 pm displaced from the TS position) and rAB = 90.8 pm (the TS position). The number of steps were also increased to 2500 in order to obtain a complete mep. This calculation was repeated with the calculation type set to Dynamics.
[[File:H3 dynamics mep 01524183.png|thumb|584x584px|Figure 3 and 4. A contour plot of BC distance vs AB distance, showing the ''mep ''trajectory: calculation type was set to MEP (left).
A contour plot of BC distance vs AB distance, showing the dynamic'' ''trajectory: calculation type was set to Dynamics (right).
|centre]]
As shown in the plots above, the ''mep ''is a smooth trajectory (left) which indicates that the BC molecule is not vibrating as the system is not oscillating between energy contours. This is because the ''mep'' doesn't provide a realistic view of the motion of atoms during a reaction; it assumes the atoms are moving at an infinitely slow speed. There would be vibration occurring between B and C as they are moving further away from each other. This is shown by the wavy trajectory produced from the dynamic calculation (right).

==== 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? ====
For the initial positions r1 = 74 pm and r2 = 200 pm, trajectories were run for five different moments combinations. The results of which are given in the table below.

{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot / kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Reactive
|The trajectory crosses the TS, therefore, it is reactive. The line is more wavy after crossing the TS, therefore the initial diatomic BC has less vibrational energy than the forming AB diatomic molecule. However, overall the trajectory is not very wavy therefore there is less vibrational energy compared to translational energy in the whole system.
|[[File:H3 1 01524183.png|thumb|375x375px|Figure 5]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.08</nowiki>
|Unreactive
|The trajectory does not cross the TS in this case, therefore it is unreactive. The AB distance is changing as the A H atom is moving towards and away from the BC diatomic molecule, which means it has more translational energy compared to vibrational. This is shown in the plot as the line is not very wavy. However, the BC distance doesn't change as the diatomic molecule remains intact.
|[[File:H3 2 01524183.png|thumb|375x375px|Figure 6]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.98</nowiki>
|Reactive
|The trajectory does cross the TS therefore it is reactive. The lines quite wavy which there is some vibrational energy but more translational energy. This is quite similar to the plot in Figure 5 as the potential energies are the same therefore the waviness of the trajectories are the same. However, the kinetic energies are a little different, making the total energies different.
|[[File:H3 3 01524183.png|thumb|375x375px|Figure 7]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.28</nowiki>
|Unreactive
|The trajectory just crosses the TS however it does not enter the product channel instead it goes back to the reactant channel. This is called barrier re-crossing. Therefore it is overall unreactive. The trajectory is very wavy which means there is a lot of vibrational energy in the system and less translational energy. In the beginning, the AB and BC distances are changing as the middle B atom is oscillating between the A and C atoms. However, in the end, the molecule BC is reformed which is they the AB distance is large.
|[[File:H3 4 01524183.png|thumb|375x375px|Figure 8]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.48</nowiki>
|Reactive
|The trajectory is reactive as it does cross the TS. The line is very wavy which shows that the system has more vibrational energy compared to translational. As the A atom gets closer to the initial BC molecule, the AB distance decreases and the BC distance increases, however, at the TS the B atom oscillates between the A and C atoms as it is attracted to both. This means there's a lot of vibrational energy near the TS. In the end, it does form the AB product molecule, which is shown by the trajectory increase in BC distance.
|[[File:H3 5 01524183.png|thumb|375x375px|Figure 9]]
|}
From the table, it can be concluded that not all trajectories starting with the same positions but with higher values of momenta would be reactive. For example, increasing the value of momenta to -5.1 g.mol-1.pm.fs-1 for p1 and -10.1 g.mol-1.pm.fs-1 for p2 still gives an unreactive trajectory. Also, low values of momenta can be reactive as is the first momenta combination (Figure 5) in the table. This is because vibrational energy also needs to be taken into account, not only the kinetic energy. The degree of vibrational and translational energy is determined by analysing the trajectory line shape: a more wavy line would mean more vibrational energy whereas a more smooth line corresponds to more translational energy.

==== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ====
The TS theory is a theory used to determine absolute reaction rates based on the reactant and transition state structure properties. The assumptions this theory makes are:
# Every trajectory that has enough energy to roll over the TS will definitely go to form the products. This means that once the reactants have crossed the energy barrier they can't go back to reactants. Therefore the overall rate of reaction of reactants to products increases, overestimating the rate.
# The concentration of the TS is in equilibrium with the reactant concentration; this is known as the Quasi-equilibrium hypothesis ([X]‡ = 0.5 KC‡ [A][B], where A + B → X‡ → P). This shows that the overall concentration of TS would be lower than the concentration of the reactants. This is because a small proportion of reactants would be forming the TS from the reactants pool.
# Quantum tunnelling is not taken into account; the theory is treated purely classical as a translation. If quantum tunnelling was taken into account the rate would increase so in this case the theory underestimates the rate.
As the first assumption is the most crucial, it has a larger weighting effect on the rate compared to quantum tunnelling. Therefore overall the TS theory predictions overestimate the reaction rate values compared with experimental values.

N'''eed to include references here '''

== Exercise 2: F - H - H system ==
* The F - H - H system is defined as A: F, B: H and C: H
* The AB distance is the F to H distance which is initially longer than the BC bond distance of H2 for the forward reaction (F + H2) and initially shorter than BC for the backward reaction (H + HF).

==== By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ====
The F + H2 reaction (forward reaction) is exothermic as the reactants are at a higher energy in comparison to the products, therefore, energy is released to the surroundings. The H + HF reaction (backward reaction) is endothermic as the products are at a higher energy than the reactants. Therefore, as shown by the exothermicity of the forward reaction (F + H2), the bond strength of HF is larger than that of HH as the energy required to break the HH bond is less than the energy released upon forming HF. Also, the endothermicity of the backward reaction means that the energy required to break the strong HF bond is greater than the energy released to form the products (F and HH). '''This is confirmed by the literature values of HF and HH bond strengths of:...........................'''
[[File:F H2 Surface Plot 01524183.png|left|thumb|379x379px|A potential energy surface plot of the F + H2 reaction (A: F, B: H and C: H). The reactant channel (small BC and large AB; the left channel) is at a higher energy value compared to the products (large BC and small AB). Therefore, this forward reaction is exothermic. ]]
[[File:H HF Surface Plot 01524183.png|centre|thumb|379x379px|A potential energy surface plot of the H + HF reaction. (A: H, B: H and C: F). The reactant channel (small BC and large AB; the left channel) is at a lower energy value compared to the products (large BC and small AB). Therefore, this forward reaction is endothermic.]]

==== Locate the approximate position of the transition state. ====
[[File:FH2 TS 01524183.png|thumb|A contour plot of the F-H-H system, showing the TS. There is no trajectory therefore this is the TS. The point is in the reactant channel as expected as the forward reaction is exothermic. When viewed in terms of the backward reaction the TS is in the product channel as it is endothermic.|353x353px]]
The transition state for both the forward (F + H2) and backward (H + HF) reactions is at the same point in terms of energy. As the system is not symmetric anymore, the TS is not defined as the point where AB = BC, instead Hammond's postulate is used in order to determine the TS point. According to Hammonds postulate, the TS of a reaction resembles the reactants if the reaction is exothermic (an early TS) and it resembles the products if the reaction is endothermic (a late TS). Therefore, the TS would be close to the reactants of the forward reaction and the products of the backward reaction which are F + H2. So by setting the initial conditions such that AB is a larger distance than BC, the TS was found. The momenta values were set to zero because the molecules are stationary at the TS as it is a saddle point.

The transition state position ('''rts''') was found to be at an AB distance of 181.4 pm and BC distance of 74.5 pm.

These values gave forces values, along the AB and BC, close to 0 kJ mol-1 which was expected at the TS point. They also gave Hessian eigenvalues/vectors of opposite signs (one negative and one positive) which also indicated that this was the TS position.
==== Report the activation energy for both reactions. ====
The activation energy is the difference in energy of the TS and the minimum of the reactant channel. This difference is smaller for the forward exothermic reaction compared to the backwards endothermic reaction. The minimum point was found by testing different distance values until the energy value was constant; until it plateaued.

The energy of the TS was found to be -433.98 kJmol-1

For the F + H2 reaction, the minimum point (reactants energy) was -435.10 kJ.mol-1. The activation energy was found to be -433.98--435.10 = + 1.12 kJ.mol-1

For the H + HF reaction, the minimum point (reactants energy) was -560.70 kJ.mol-1. The activation energy was found to be -433.98--560.70 = +126.72 kJ.mol-1

==== 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. ====

==== 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. ====

MRD:01524183

2020-05-22T00:00:42Z

Aas1218:

== Introduction ==
The main aims of this experiment were to study the reactivity of triatomic systems, involving the collision of an atom and a diatomic molecule, by calculating Molecular Dynamic trajectories. ........... '''say about BC being the original diatomic molecule and A colliding and forming AB. - how the system is initially set up.'''

== Exercise 1: The H + H2 system ==

==== On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ====
[[File:H3 TS contourplot 01524183.png|thumb|278x278px|Figure 1. A contour plot showing the variation of AB and BC distances, with reaction trajectory and eigenvectors shown.]]
On a potential energy surface diagram, the transition state (TS) is mathematically defined as the point at which the gradient of the potential with respect to the internal coordinates is zero (∂V('''ri''')/∂'''ri'''=0). This is because the transition state is the local maximum on the minimum energy paths between the reactants and products. There would be no movement of the H atoms at the TS since the force is a derivative of the potential energy and at the TS this is zero, therefore, the nuclei are stationary; the transition state is a stationary point. However, minima and maxima are stationary points. The TS can be distinguished from a local minimum of the potential energy surface by taking the second differential of the potential function. By doing so, the nature of the stationary point can be determined: if the second differential is greater than zero, the point is a local minimum and if it is less than zero, the point is a local maximum (the TS).

Using the GUI simulation, the TS was identified by trial and error. At the TS the distance AB and BC must be equal (rAB = rBC), as the H + H2 surface is symmetric. Therefore the TS lies somewhere along the diagonal line where AB and BC are equal. As the TS is approached, the size of the forces of AB and BC decrease to zero and so the point at which these forces are close to zero is the TS of the potential energy surface. This maximum point can be distinguished from a local minimum by analysing the omega squared values under the Hessian eigenvalues/vectors. A TS is a maximum in one direction and a minimum in the other direction; it is a saddle point. Therefore the omega squared would be positive in one direction and negative in the other direction.

==== 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 transition state position ('''rts''') was found to be 90.8 pm to 1 d.p.[[File:H3 TS distancetime 01524183.png|thumb|Figure 2. A plot of Internuclear Distances vs Time: linear horizontal lines show the constant distances of AB, BC and AC. |420x420px]]This value gave forces of +0.001 kJ.mol-1.pm-1 along AB and BC which are very close to zero. Also, the omega hessian eigenvalues had opposite signs which verified that this point was the TS.

At the TS, the internuclear distances between AB and BC are the same and so the distance between AC is approximately double the AB (BC) distance. This is shown in the plot of "Internuclear distances vs Time", as two straight lines of constant distances of AB and BC at 90.8 pm and AC at approximately 181.6 pm. This is because at the TS the H atoms are stationary, with one H atom (the B atom in this case) being halfway between the A and C H atoms.

==== Comment on how the ''mep'' and the trajectory you just calculated differ. ====
The minimum energy path (''mep) ''is the lowest energy path from reactant to product; it is a special trajectory that correlates to infinitely slow motion. The ''mep ''was calculated by setting the initial conditions as rBC = 91.8 pm (1 pm displaced from the TS position) and rAB = 90.8 pm (the TS position). The number of steps were also increased to 2500 in order to obtain a complete mep. This calculation was repeated with the calculation type set to Dynamics.
[[File:H3 dynamics mep 01524183.png|thumb|584x584px|Figure 3 and 4. A contour plot of BC distance vs AB distance, showing the ''mep ''trajectory: calculation type was set to MEP (left).
A contour plot of BC distance vs AB distance, showing the dynamic'' ''trajectory: calculation type was set to Dynamics (right).
|centre]]
As shown in the plots above, the ''mep ''is a smooth trajectory (left) which indicates that the BC molecule is not vibrating as the system is not oscillating between energy contours. This is because the ''mep'' doesn't provide a realistic view of the motion of atoms during a reaction; it assumes the atoms are moving at an infinitely slow speed. There would be vibration occurring between B and C as they are moving further away from each other. This is shown by the wavy trajectory produced from the dynamic calculation (right).

==== 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? ====
For the initial positions r1 = 74 pm and r2 = 200 pm, trajectories were run for five different moments combinations. The results of which are given in the table below.

{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot / kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Reactive
|The trajectory crosses the TS, therefore, it is reactive. The line is more wavy after crossing the TS, therefore the initial diatomic BC has less vibrational energy than the forming AB diatomic molecule. However, overall the trajectory is not very wavy therefore there is less vibrational energy compared to translational energy in the whole system.
|[[File:H3 1 01524183.png|thumb|375x375px|Figure 5]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.08</nowiki>
|Unreactive
|The trajectory does not cross the TS in this case, therefore it is unreactive. The AB distance is changing as the A H atom is moving towards and away from the BC diatomic molecule, which means it has more translational energy compared to vibrational. This is shown in the plot as the line is not very wavy. However, the BC distance doesn't change as the diatomic molecule remains intact.
|[[File:H3 2 01524183.png|thumb|375x375px|Figure 6]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.98</nowiki>
|Reactive
|The trajectory does cross the TS therefore it is reactive. The lines quite wavy which there is some vibrational energy but more translational energy. This is quite similar to the plot in Figure 5 as the potential energies are the same therefore the waviness of the trajectories are the same. However, the kinetic energies are a little different, making the total energies different.
|[[File:H3 3 01524183.png|thumb|375x375px|Figure 7]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.28</nowiki>
|Unreactive
|The trajectory just crosses the TS however it does not enter the product channel instead it goes back to the reactant channel. This is called barrier re-crossing. Therefore it is overall unreactive. The trajectory is very wavy which means there is a lot of vibrational energy in the system and less translational energy. In the beginning, the AB and BC distances are changing as the middle B atom is oscillating between the A and C atoms. However, in the end, the molecule BC is reformed which is they the AB distance is large.
|[[File:H3 4 01524183.png|thumb|375x375px|Figure 8]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.48</nowiki>
|Reactive
|The trajectory is reactive as it does cross the TS. The line is very wavy which shows that the system has more vibrational energy compared to translational. As the A atom gets closer to the initial BC molecule, the AB distance decreases and the BC distance increases, however, at the TS the B atom oscillates between the A and C atoms as it is attracted to both. This means there's a lot of vibrational energy near the TS. In the end, it does form the AB product molecule, which is shown by the trajectory increase in BC distance.
|[[File:H3 5 01524183.png|thumb|375x375px|Figure 9]]
|}
From the table, it can be concluded that not all trajectories starting with the same positions but with higher values of momenta would be reactive. For example, increasing the value of momenta to -5.1 g.mol-1.pm.fs-1 for p1 and -10.1 g.mol-1.pm.fs-1 for p2 still gives an unreactive trajectory. Also, low values of momenta can be reactive as is the first momenta combination (Figure 5) in the table. This is because vibrational energy also needs to be taken into account, not only the kinetic energy. The degree of vibrational and translational energy is determined by analysing the trajectory line shape: a more wavy line would mean more vibrational energy whereas a more smooth line corresponds to more translational energy.

==== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ====
The TS theory is a theory used to determine absolute reaction rates based on the reactant and transition state structure properties. The assumptions this theory makes are:
# Every trajectory that has enough energy to roll over the TS will definitely go to form the products. This means that once the reactants have crossed the energy barrier they can't go back to reactants. Therefore the overall rate of reaction of reactants to products increases, overestimating the rate.
# The concentration of the TS is in equilibrium with the reactant concentration; this is known as the Quasi-equilibrium hypothesis ([X]‡ = 0.5 KC‡ [A][B], where A + B → X‡ → P). This shows that the overall concentration of TS would be lower than the concentration of the reactants. This is because a small proportion of reactants would be forming the TS from the reactants pool.
# Quantum tunnelling is not taken into account; the theory is treated purely classical as a translation. If quantum tunnelling was taken into account the rate would increase so in this case the theory underestimates the rate.
As the first assumption is the most crucial, it has a larger weighting effect on the rate compared to quantum tunnelling. Therefore overall the TS theory predictions overestimate the reaction rate values compared with experimental values.

N'''eed to include references here '''

== Exercise 2: F - H - H system ==
* The F - H - H system is defined as A: F, B: H and C: H.
* The AB distance is the F to H distance which is initially longer than the BC bond distance of H2 for the forward reaction (F + H2) and initially shorter than BC for the backward reaction (H + HF).

==== By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ====
The F + H2 reaction (forward reaction) is exothermic as the reactants are at a higher energy in comparison to the products, therefore, energy is released to the surroundings. The H + HF reaction (backward reaction) is endothermic as the products are at a higher energy than the reactants. Therefore, as shown by the exothermicity of the forward reaction (F + H2), the bond strength of HF is larger than that of HH as the energy required to break the HH bond is less than the energy released upon forming HF. Also, the endothermicity of the backward reaction means that the energy required to break the strong HF bond is greater than the energy released to form the products (F and HH). '''This is confirmed by the literature values of HF and HH bond strengths of:...........................'''
[[File:F H2 Surface Plot 01524183.png|left|thumb|379x379px|A potential energy surface plot of the F + H2 reaction. The reactant channel (small BC and large AB; the left channel) is at a higher energy value compared to the products (large BC and small AB). Therefore, this forward reaction is exothermic. ]]
[[File:H HF Surface Plot 01524183.png|centre|thumb|379x379px|A potential energy surface plot of the H + HF reaction. The reactant channel (small BC and large AB; the left channel) is at a lower energy value compared to the products (large BC and small AB). Therefore, this forward reaction is endothermic.]]

==== Locate the approximate position of the transition state. ====
[[File:FH2 TS 01524183.png|thumb|A contour plot of the F + H2 system. There is no trajectory therefore this is the TS. The point is in the reactant channel as expected as the reaction is exothermic.]]
The transition state for both the forward (F + H2) and backward (H + HF) reactions is at the same point in terms of energy. As the system is not symmetric anymore, the TS is not defined as the point where AB=BC, instead Hammond's postulate is used in order to determine the TS point. According to Hammonds postulate, the TS of a reaction resembles the reactants if the reaction is exothermic (an early TS) and it resembles the products if the reaction is endothermic (a late TS). Therefore, considering the forward reaction (F + H2) which is exothermic, the TS would be close to the reactants and so by setting the initial conditions so that AB is a larger distance than BC (resembling the F + H2 system), the TS was found. The momenta values were set to zero as at the TS, the molecules are stationary as it is a saddle point.

The transition state position ('''rts''') was found to be at an AB distance of 181.4 pm and BC distance of 74.5 pm.

These values gave forces values, along the AB and BC, close to 0 kJ mol-1 which was expected at the TS point. They also gave Hessian eigenvalues/vectors of opposite signs (one negative and one positive) which also indicated that this was the TS position.

'''CONFIRM THIS BIT - HOW ARE WE MEANT TO DEFINE THE TS POSITION (DISTANCES OR ENERGY) AND HOW IS IT NOT THE SAME VALUES - ARE WE MEANT TO CHANGE WHAT WE DEFINED AS A B AND C FOR BOTH THE REACTIONS OR NOT??'''

==== Report the activation energy for both reactions. ====
The activation energy is the difference in energy of the TS and the minimum of the reactant channel. This difference is smaller for the forward exothermic reaction compared to the backwards endothermic reaction. The minimum point was found by testing different distance values until the energy value was constant; until it plateaued.

The energy of the TS was found to be -433.98 kJmol-1

For the F + H2 reaction, the minimum point (reactants energy) was -435.10 kJ.mol-1. The activation energy was found to be -433.98--435.10 = 1.12 kJ.mol-1

For the H + HF reaction, the minimum point (reactants energy) was -560.70 kJ.mol-1. The activation energy was found to be -433.98--560.70 = 126.72 kJ.mol-1

==== 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. ====

==== 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. ====

MRD:01524183

2020-05-21T23:54:16Z

Aas1218:

== Introduction ==
The main aims of this experiment were to study the reactivity of triatomic systems, involving the collision of an atom and a diatomic molecule, by calculating Molecular Dynamic trajectories. ........... '''say about BC being the original diatomic molecule and A colliding and forming AB. - how the system is initially set up.'''

== Exercise 1: The H + H2 system ==

==== On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ====
[[File:H3 TS contourplot 01524183.png|thumb|278x278px|Figure 1. A contour plot showing the variation of AB and BC distances, with reaction trajectory and eigenvectors shown.]]
On a potential energy surface diagram, the transition state (TS) is mathematically defined as the point at which the gradient of the potential with respect to the internal coordinates is zero (∂V('''ri''')/∂'''ri'''=0). This is because the transition state is the local maximum on the minimum energy paths between the reactants and products. There would be no movement of the H atoms at the TS since the force is a derivative of the potential energy and at the TS this is zero, therefore, the nuclei are stationary; the transition state is a stationary point. However, minima and maxima are stationary points. The TS can be distinguished from a local minimum of the potential energy surface by taking the second differential of the potential function. By doing so, the nature of the stationary point can be determined: if the second differential is greater than zero, the point is a local minimum and if it is less than zero, the point is a local maximum (the TS).

Using the GUI simulation, the TS was identified by trial and error. At the TS the distance AB and BC must be equal (rAB = rBC), as the H + H2 surface is symmetric. Therefore the TS lies somewhere along the diagonal line where AB and BC are equal. As the TS is approached, the size of the forces of AB and BC decrease to zero and so the point at which these forces are close to zero is the TS of the potential energy surface. This maximum point can be distinguished from a local minimum by analysing the omega squared values under the Hessian eigenvalues/vectors. A TS is a maximum in one direction and a minimum in the other direction; it is a saddle point. Therefore the omega squared would be positive in one direction and negative in the other direction.

==== 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 transition state position ('''rts''') was found to be 90.8 pm to 1 d.p.[[File:H3 TS distancetime 01524183.png|thumb|Figure 2. A plot of Internuclear Distances vs Time: linear horizontal lines show the constant distances of AB, BC and AC. |420x420px]]This value gave forces of +0.001 kJ.mol-1.pm-1 along AB and BC which are very close to zero. Also, the omega hessian eigenvalues had opposite signs which verified that this point was the TS.

At the TS, the internuclear distances between AB and BC are the same and so the distance between AC is approximately double the AB (BC) distance. This is shown in the plot of "Internuclear distances vs Time", as two straight lines of constant distances of AB and BC at 90.8 pm and AC at approximately 181.6 pm. This is because at the TS the H atoms are stationary, with one H atom (the B atom in this case) being halfway between the A and C H atoms.

==== Comment on how the ''mep'' and the trajectory you just calculated differ. ====
The minimum energy path (''mep) ''is the lowest energy path from reactant to product; it is a special trajectory that correlates to infinitely slow motion. The ''mep ''was calculated by setting the initial conditions as rBC = 91.8 pm (1 pm displaced from the TS position) and rAB = 90.8 pm (the TS position). The number of steps were also increased to 2500 in order to obtain a complete mep. This calculation was repeated with the calculation type set to Dynamics.
[[File:H3 dynamics mep 01524183.png|thumb|584x584px|Figure 3 and 4. A contour plot of BC distance vs AB distance, showing the ''mep ''trajectory: calculation type was set to MEP (left).
A contour plot of BC distance vs AB distance, showing the dynamic'' ''trajectory: calculation type was set to Dynamics (right).
|centre]]
As shown in the plots above, the ''mep ''is a smooth trajectory (left) which indicates that the BC molecule is not vibrating as the system is not oscillating between energy contours. This is because the ''mep'' doesn't provide a realistic view of the motion of atoms during a reaction; it assumes the atoms are moving at an infinitely slow speed. There would be vibration occurring between B and C as they are moving further away from each other. This is shown by the wavy trajectory produced from the dynamic calculation (right).

==== 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? ====
For the initial positions r1 = 74 pm and r2 = 200 pm, trajectories were run for five different moments combinations. The results of which are given in the table below.

{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot / kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Reactive
|The trajectory crosses the TS, therefore, it is reactive. The line is more wavy after crossing the TS, therefore the initial diatomic BC has less vibrational energy than the forming AB diatomic molecule. However, overall the trajectory is not very wavy therefore there is less vibrational energy compared to translational energy in the whole system.
|[[File:H3 1 01524183.png|thumb|375x375px|Figure 5]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.08</nowiki>
|Unreactive
|The trajectory does not cross the TS in this case, therefore it is unreactive. The AB distance is changing as the A H atom is moving towards and away from the BC diatomic molecule, which means it has more translational energy compared to vibrational. This is shown in the plot as the line is not very wavy. However, the BC distance doesn't change as the diatomic molecule remains intact.
|[[File:H3 2 01524183.png|thumb|375x375px|Figure 6]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.98</nowiki>
|Reactive
|The trajectory does cross the TS therefore it is reactive. The lines quite wavy which there is some vibrational energy but more translational energy. This is quite similar to the plot in Figure 5 as the potential energies are the same therefore the waviness of the trajectories are the same. However, the kinetic energies are a little different, making the total energies different.
|[[File:H3 3 01524183.png|thumb|375x375px|Figure 7]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.28</nowiki>
|Unreactive
|The trajectory just crosses the TS however it does not enter the product channel instead it goes back to the reactant channel. This is called barrier re-crossing. Therefore it is overall unreactive. The trajectory is very wavy which means there is a lot of vibrational energy in the system and less translational energy. In the beginning, the AB and BC distances are changing as the middle B atom is oscillating between the A and C atoms. However, in the end, the molecule BC is reformed which is they the AB distance is large.
|[[File:H3 4 01524183.png|thumb|375x375px|Figure 8]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.48</nowiki>
|Reactive
|The trajectory is reactive as it does cross the TS. The line is very wavy which shows that the system has more vibrational energy compared to translational. As the A atom gets closer to the initial BC molecule, the AB distance decreases and the BC distance increases, however, at the TS the B atom oscillates between the A and C atoms as it is attracted to both. This means there's a lot of vibrational energy near the TS. In the end, it does form the AB product molecule, which is shown by the trajectory increase in BC distance.
|[[File:H3 5 01524183.png|thumb|375x375px|Figure 9]]
|}
From the table, it can be concluded that not all trajectories starting with the same positions but with higher values of momenta would be reactive. For example, increasing the value of momenta to -5.1 g.mol-1.pm.fs-1 for p1 and -10.1 g.mol-1.pm.fs-1 for p2 still gives an unreactive trajectory. Also, low values of momenta can be reactive as is the first momenta combination (Figure 5) in the table. This is because vibrational energy also needs to be taken into account, not only the kinetic energy. The degree of vibrational and translational energy is determined by analysing the trajectory line shape: a more wavy line would mean more vibrational energy whereas a more smooth line corresponds to more translational energy.

==== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ====
The TS theory is a theory used to determine absolute reaction rates based on the reactant and transition state structure properties. The assumptions this theory makes are:
# Every trajectory that has enough energy to roll over the TS will definitely go to form the products. This means that once the reactants have crossed the energy barrier they can't go back to reactants. Therefore the overall rate of reaction of reactants to products increases, overestimating the rate.
# The concentration of the TS is in equilibrium with the reactant concentration; this is known as the Quasi-equilibrium hypothesis ([X]‡ = 0.5 KC‡ [A][B], where A + B → X‡ → P). This shows that the overall concentration of TS would be lower than the concentration of the reactants. This is because a small proportion of reactants would be forming the TS from the reactants pool.
# Quantum tunnelling is not taken into account; the theory is treated purely classical as a translation. If quantum tunnelling was taken into account the rate would increase so in this case the theory underestimates the rate.
As the first assumption is the most crucial, it has a larger weighting effect on the rate compared to quantum tunnelling. Therefore overall the TS theory predictions overestimate the reaction rate values compared with experimental values.

N'''eed to include references here '''

== Exercise 2: F - H - H system ==
* The F - H - H system is defined as A: F, B: H and C: H.
* The AB distance is the F to H distance which is initially longer than the BC bond distance of H2 for the forward reaction (F + H2) and initially shorter than BC for the backward reaction (H + HF).

==== By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ====
The F + H2 reaction (forward reaction) is exothermic as the reactants are at a higher energy in comparison to the products, therefore, energy is released to the surroundings. The H + HF reaction (backward reaction) is endothermic as the products are at a higher energy than the reactants. Therefore, as shown by the exothermicity of the forward reaction (F + H2), the bond strength of HF is larger than that of HH as the energy required to break the HH bond is less than the energy released upon forming HF. Also, the endothermicity of the backward reaction means that the energy required to break the strong HF bond is greater than the energy released to form the products (F and HH). '''This is confirmed by the literature values of HF and HH bond strengths of:...........................'''
[[File:F H2 Surface Plot 01524183.png|left|thumb|379x379px|A potential energy surface plot of the F + H2 reaction. The reactant channel (small BC and large AB; the left channel) is at a higher energy value compared to the products (large BC and small AB). Therefore, this forward reaction is exothermic. ]]
[[File:H HF Surface Plot 01524183.png|centre|thumb|379x379px|A potential energy surface plot of the H + HF reaction. The reactant channel (small BC and large AB; the left channel) is at a lower energy value compared to the products (large BC and small AB). Therefore, this forward reaction is endothermic.]]

==== Locate the approximate position of the transition state. ====
[[File:FH2 TS 01524183.png|thumb|A contour plot of the F + H2 system. There is no trajectory therefore this is the TS. The point is in the reactant channel as expected as the reaction is exothermic.]]
The transition state for both the forward (F + H2) and backward (H + HF) reactions is at the same point. As the system is not symmetric anymore, the TS is not defined as the point where AB=BC, instead Hammond's postulate is used in order to determine the TS point. According to Hammonds postulate, the TS of a reaction resembles the reactants if the reaction is exothermic (an early TS) and it resembles the products if the reaction is endothermic (a late TS). Therefore, considering the forward reaction (F + H2) which is exothermic, the TS would be close to the reactants and so by setting the initial conditions so that AB is a larger distance than BC (resembling the F + H2 system), the TS was found. The momenta values were set to zero as at the TS, the molecules are stationary as it is a saddle point.

The transition state position ('''rts''') was found to be at an AB distance of 181.4 pm and BC distance of 74.5 pm.

These values gave forces values, along the AB and BC, close to 0 kJ mol-1 which was expected at the TS point. They also gave Hessian eigenvalues/vectors of opposite signs (one negative and one positive) which also indicated that this was the TS position.

==== Report the activation energy for both reactions. ====
The activation energy is the difference in energy of the TS and the minimum of the reactant channel . This difference is smaller for the forward exothermic reaction compared to the backwards endothermic reaction. The minimum point was found by testing different distance values until the enegy value was constant; until it plateaued.

The energy of the TS was found to be -433.98 kJmol-1

For the F + H2 reaction, the minimum point (reactants energy) was -435.10 kJ.mol-1. The activation energy was found to be -433.98--435.10 = 1.12 kJ.mol-1

For the H + HF reaction, the minimum point (reactants energy) was -560.70 kJ.mol-1. The activation energy was found to be -433.98--560.70 = 126.72 kJ.mol-1

==== 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. ====

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

File:H HF TS 01524183.png

2020-05-21T22:42:00Z

Aas1218:

File:FH2 TS 01524183.png

2020-05-21T22:41:43Z

Aas1218:

MRD:01524183

2020-05-21T22:38:37Z

Aas1218:

== Introduction ==
The main aims of this experiment were to study the reactivity of triatomic systems, involving the collision of an atom and a diatomic molecule, by calculating Molecular Dynamic trajectories. ........... '''say about BC being the original diatomic molecule and A colliding and forming AB. - how the system is initially set up.'''

== Exercise 1: The H + H2 system ==

==== On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ====
[[File:H3 TS contourplot 01524183.png|thumb|278x278px|Figure 1. A contour plot showing the variation of AB and BC distances, with reaction trajectory and eigenvectors shown.]]
On a potential energy surface diagram, the transition state (TS) is mathematically defined as the point at which the gradient of the potential with respect to the internal coordinates is zero (∂V('''ri''')/∂'''ri'''=0). This is because the transition state is the local maximum on the minimum energy paths between the reactants and products. There would be no movement of the H atoms at the TS since the force is a derivative of the potential energy and at the TS this is zero, therefore, the nuclei are stationary; the transition state is a stationary point. However, minima and maxima are stationary points. The TS can be distinguished from a local minimum of the potential energy surface by taking the second differential of the potential function. By doing so, the nature of the stationary point can be determined: if the second differential is greater than zero, the point is a local minimum and if it is less than zero, the point is a local maximum (the TS).

Using the GUI simulation, the TS was identified by trial and error. At the TS the distance AB and BC must be equal (rAB = rBC), as the H + H2 surface is symmetric. Therefore the TS lies somewhere along the diagonal line where AB and BC are equal. As the TS is approached, the size of the forces of AB and BC decrease to zero and so the point at which these forces are close to zero is the TS of the potential energy surface. This maximum point can be distinguished from a local minimum by analysing the omega squared values under the Hessian eigenvalues/vectors. A TS is a maximum in one direction and a minimum in the other direction; it is a saddle point. Therefore the omega squared would be positive in one direction and negative in the other direction.

==== 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 transition state position ('''rts''') was found to be 90.8 pm to 1 d.p.[[File:H3 TS distancetime 01524183.png|thumb|Figure 2. A plot of Internuclear Distances vs Time: linear horizontal lines show the constant distances of AB, BC and AC. |420x420px]]This value gave forces of +0.001 kJ.mol-1.pm-1 along AB and BC which are very close to zero. Also, the omega hessian eigenvalues had opposite signs which verified that this point was the TS.

At the TS, the internuclear distances between AB and BC are the same and so the distance between AC is approximately double the AB (BC) distance. This is shown in the plot of "Internuclear distances vs Time", as two straight lines of constant distances of AB and BC at 90.8 pm and AC at approximately 181.6 pm. This is because at the TS the H atoms are stationary, with one H atom (the B atom in this case) being halfway between the A and C H atoms.

==== Comment on how the ''mep'' and the trajectory you just calculated differ. ====
The minimum energy path (''mep) ''is the lowest energy path from reactant to product; it is a special trajectory that correlates to infinitely slow motion. The ''mep ''was calculated by setting the initial conditions as rBC = 91.8 pm (1 pm displaced from the TS position) and rAB = 90.8 pm (the TS position). The number of steps were also increased to 2500 in order to obtain a complete mep. This calculation was repeated with the calculation type set to Dynamics.
[[File:H3 dynamics mep 01524183.png|thumb|584x584px|Figure 3 and 4. A contour plot of BC distance vs AB distance, showing the ''mep ''trajectory: calculation type was set to MEP (left).
A contour plot of BC distance vs AB distance, showing the dynamic'' ''trajectory: calculation type was set to Dynamics (right).
|centre]]
As shown in the plots above, the ''mep ''is a smooth trajectory (left) which indicates that the BC molecule is not vibrating as the system is not oscillating between energy contours. This is because the ''mep'' doesn't provide a realistic view of the motion of atoms during a reaction; it assumes the atoms are moving at an infinitely slow speed. There would be vibration occurring between B and C as they are moving further away from each other. This is shown by the wavy trajectory produced from the dynamic calculation (right).

==== 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? ====
For the initial positions r1 = 74 pm and r2 = 200 pm, trajectories were run for five different moments combinations. The results of which are given in the table below.

{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot / kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Reactive
|The trajectory crosses the TS, therefore, it is reactive. The line is more wavy after crossing the TS, therefore the initial diatomic BC has less vibrational energy than the forming AB diatomic molecule. However, overall the trajectory is not very wavy therefore there is less vibrational energy compared to translational energy in the whole system.
|[[File:H3 1 01524183.png|thumb|375x375px|Figure 5]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.08</nowiki>
|Unreactive
|The trajectory does not cross the TS in this case, therefore it is unreactive. The AB distance is changing as the A H atom is moving towards and away from the BC diatomic molecule, which means it has more translational energy compared to vibrational. This is shown in the plot as the line is not very wavy. However, the BC distance doesn't change as the diatomic molecule remains intact.
|[[File:H3 2 01524183.png|thumb|375x375px|Figure 6]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.98</nowiki>
|Reactive
|The trajectory does cross the TS therefore it is reactive. The lines quite wavy which there is some vibrational energy but more translational energy. This is quite similar to the plot in Figure 5 as the potential energies are the same therefore the waviness of the trajectories are the same. However, the kinetic energies are a little different, making the total energies different.
|[[File:H3 3 01524183.png|thumb|375x375px|Figure 7]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.28</nowiki>
|Unreactive
|The trajectory just crosses the TS however it does not enter the product channel instead it goes back to the reactant channel. This is called barrier re-crossing. Therefore it is overall unreactive. The trajectory is very wavy which means there is a lot of vibrational energy in the system and less translational energy. In the beginning, the AB and BC distances are changing as the middle B atom is oscillating between the A and C atoms. However, in the end, the molecule BC is reformed which is they the AB distance is large.
|[[File:H3 4 01524183.png|thumb|375x375px|Figure 8]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.48</nowiki>
|Reactive
|The trajectory is reactive as it does cross the TS. The line is very wavy which shows that the system has more vibrational energy compared to translational. As the A atom gets closer to the initial BC molecule, the AB distance decreases and the BC distance increases, however, at the TS the B atom oscillates between the A and C atoms as it is attracted to both. This means there's a lot of vibrational energy near the TS. In the end, it does form the AB product molecule, which is shown by the trajectory increase in BC distance.
|[[File:H3 5 01524183.png|thumb|375x375px|Figure 9]]
|}
From the table, it can be concluded that not all trajectories starting with the same positions but with higher values of momenta would be reactive. For example, increasing the value of momenta to -5.1 g.mol-1.pm.fs-1 for p1 and -10.1 g.mol-1.pm.fs-1 for p2 still gives an unreactive trajectory. Also, low values of momenta can be reactive as is the first momenta combination (Figure 5) in the table. This is because vibrational energy also needs to be taken into account, not only the kinetic energy. The degree of vibrational and translational energy is determined by analysing the trajectory line shape: a more wavy line would mean more vibrational energy whereas a more smooth line corresponds to more translational energy.

==== Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ====
The TS theory is a theory used to determine absolute reaction rates based on the reactant and transition state structure properties. The assumptions this theory makes are:
# Every trajectory that has enough energy to roll over the TS will definitely go to form the products. This means that once the reactants have crossed the energy barrier they can't go back to reactants. Therefore the overall rate of reaction of reactants to products increases, overestimating the rate.
# The concentration of the TS is in equilibrium with the reactant concentration; this is known as the Quasi-equilibrium hypothesis ([X]‡ = 0.5 KC‡ [A][B], where A + B → X‡ → P). This shows that the overall concentration of TS would be lower than the concentration of the reactants. This is because a small proportion of reactants would be forming the TS from the reactants pool.
# Quantum tunnelling is not taken into account; the theory is treated purely classical as a translation. If quantum tunnelling was taken into account the rate would increase so in this case the theory underestimates the rate.
As the first assumption is the most crucial, it has a larger weighting effect on the rate compared to quantum tunnelling. Therefore overall the TS theory predictions overestimate the reaction rate values compared with experimental values.

N'''eed to include references here '''

== Exercise 2: F - H - H system ==
* The F + H2 system is defined as A: F, B: H and C: H. Therefore the AB distance is F to H distance which is initially larger than the BC bond distance of H2.
* The H + HF system is defined as A: H, B: H and C: F. Therefore the AB distance is the H to H distance which is initially larger than the BC bond distance of HF.

==== By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ====
The F + H2 reaction (forward reaction) is exothermic as the reactants are at a higher energy in comparison to the products, therefore, energy is released to the surroundings. The H + HF reaction (backward reaction) is endothermic as the products are at a higher energy than the reactants. Therefore, as shown by the exothermicity of the forward reaction (F + H2), the bond strength of HF is larger than that of HH as the energy required to break the HH bond is less than the energy released upon forming HF. Also, the endothermicity of the backward reaction means that the energy required to break the strong HF bond is greater than the energy released to form the products (F and HH). '''This is confirmed by the literature values of HF and HH bond strengths of:...........................'''
[[File:F H2 Surface Plot 01524183.png|left|thumb|379x379px|A potential energy surface plot of the F + H2 reaction. The reactant channel (small BC and large AB; the left channel) is at a higher energy value compared to the products (large BC and small AB). Therefore, this forward reaction is exothermic. ]]
[[File:H HF Surface Plot 01524183.png|centre|thumb|379x379px|A potential energy surface plot of the H + HF reaction. The reactant channel (small BC and large AB; the left channel) is at a lower energy value compared to the products (large BC and small AB). Therefore, this forward reaction is endothermic.]]

==== Locate the approximate position of the transition state. ====
The transition state for both the forward (F + H2) and backward (H + HF) reactions is at the same point. As the system is not symmetric anymore, the TS is not defined as the point where AB=BC, instead Hammond's postulate is used in order to determine the TS point. According to Hammonds postulate, the TS of a reaction resembles the reactants is the reaction is exothermic (an early TS) and it resembles the products if the reaction is endothermic (a late TS). Therefore, considering the forward reaction (F + H2) which is exothermic, the TS would be close to the reactants and so by setting the initial conditions so that AB is a larger distance than BC (resembling the F + H2 system), the TS was found. The momenta values were set to zero as at the TS the molecules are stationary as it is a saddle point. For the backward reaction, the TS would be same however in this case, the AB would be smaller than BC as the TS would resemble the products which are F and H2.

For the forward reaction, the transition state position ('''rts''') was found to be at an AB distance of 181.4 pm and BC distance of 74.5 pm.

These values gave forces values, along the AB and BC, close to 0 kJ mol-1 which was expected at the TS point. They also gave Hessian eigenvalues/vectors of opposite signs (one negative and one positive) which also indicated that this was the TS position.

For the backward reaction, the transition state position was at an AB distance of 75.5 pm and a BC distance of 74.5 pm.

==== 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. ====

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

File:H HF Surface Plot 01524183.png

2020-05-21T21:16:28Z

Aas1218:

2020-05-20T22:41:52Z

Aas1218:

=== '''Introduction''' ===
The main aims of this experiment were to study the reactivity of triatomic systems, involving the collision of an atom and a diatomic molecule, by calculating Molecular Dynamic trajectories. ........... '''say about BC being the original diatomic molecule and A colliding and forming AB. - how the system is initially set up.'''

==== Transition state (TS) ====
[[File:H3 TS contourplot 01524183.png|thumb|278x278px|Figure 1. A contour plot showing the variation of AB and BC distances, with reaction trajectory and eigenvectors shown.]]
On a potential energy surface diagram, the transition state (TS) is mathematically defined as the point at which the gradient of the potential with respect to the internal coordinates is zero (∂V('''ri''')/∂'''ri'''=0). This is because the transition state is the local maximum on the minimum energy paths between the reactants and products. There would be no movement of the H atoms at the TS since the force is a derivative of the potential energy and at the TS this is zero, therefore, the nuclei are stationary; the transition state is a stationary point. However, minima and maxima are stationary points. The TS can be distinguished from a local minimum of the potential energy surface by taking the second differential of the potential function. By doing so, the nature of the stationary point can be determined: if the second differential is greater than zero, the point is a local minimum and if it is less than zero, the point is a local maximum (the TS).

Using the GUI simulation, the TS was identified by trial and error. At the TS the distance AB and BC must be equal (rAB = rBC), as the H + H2 surface is symmetric. Therefore the TS lies somewhere along the diagonal line where AB and BC are equal. As the TS is approached, the size of the forces of AB and BC decrease to zero and so the point at which these forces are close to zero is the TS of the potential energy surface. This maximum point can be distinguished from a local minimum by analysing the omega squared values under the Hessian eigenvalues/vectors. A TS is a maximum in one direction and a minimum in the other direction; it is a saddle point. Therefore the omega squared would be positive in one direction and negative in the other direction.

The transition state position ('''rts''') was found to be 90.8 pm to 1 d.p.[[File:H3 TS distancetime 01524183.png|thumb|Figure 2. A plot of Internuclear Distances vs Time: linear horizontal lines show the constant distances of AB, BC and AC. |420x420px]]This value gave forces of +0.001 kJ.mol-1.pm-1 along AB and BC which are very close to zero. Also, the omega hessian eigenvalues had opposite signs which verified that this point was the TS.

At the TS, the internuclear distances between AB and BC are the same and so the distance between AC is approximately double the AB (BC) distance. This is shown in the plot of "Internuclear distances vs Time", as two straight lines of constant distances of AB and BC at 90.8 pm and AC at approximately 181.6 pm. This is because at the TS the H atoms are stationary, with one H atom (the B atom in this case) being halfway between the A and C H atoms.

==== Minimum energy path (''mep) ''vs dynamics ====
The minimum energy path (''mep) ''is the lowest energy path from reactant to product; it is a special trajectory that correlates to infinitely slow motion. The ''mep ''was calculated by setting the initial conditions as rBC = 91.8 pm (1 pm displaced from the TS position) and rAB = 90.8 pm (the TS position). The number of steps were also increased to 2500 in order to obtain a complete mep. This calculation was repeated with the calculation type set to Dynamics.
[[File:H3 dynamics mep 01524183.png|thumb|584x584px|Figure 3 and 4. A contour plot of BC distance vs AB distance, showing the ''mep ''trajectory: calculation type was set to MEP (left).
A contour plot of BC distance vs AB distance, showing the dynamic'' ''trajectory: calculation type was set to Dynamics (right).
|centre]]
As shown in the plots above, the ''mep ''is a smooth trajectory (left) which indicates that the BC molecule is not vibrating as the system is not oscillating between energy contours. This is because the ''mep'' doesn't provide a realistic view of the motion of atoms during a reaction; it assumes the atoms are moving at an infinitely slow speed. There would be vibration occurring between B and C as they are moving further away from each other. This is shown by the wavy trajectory produced from the dynamic calculation (right).

==== Reactive and unreactive trajectories ====
For the initial positions r1 = 74 pm and r2 = 200 pm, trajectories were run for five different moments combinations. The results of which are given in the table below.

{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot / kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.28</nowiki>
|Reactive
|The trajectory crosses the TS, therefore, it is reactive. The line is more wavy after crossing the TS, therefore the initial diatomic BC has less vibrational energy than the forming AB diatomic molecule. However, overall the trajectory is not very wavy therefore there is less vibrational energy compared to translational energy in the whole system.
|[[File:H3 1 01524183.png|thumb|375x375px|Figure 5]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.08</nowiki>
|Unreactive
|The trajectory does not cross the TS in this case, therefore it is unreactive. The AB distance is changing as the A H atom is moving towards and away from the BC diatomic molecule, which means it has more translational energy compared to vibrational. This is shown in the plot as the line is not very wavy. However, the BC distance doesn't change as the diatomic molecule remains intact.
|[[File:H3 2 01524183.png|thumb|375x375px|Figure 6]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.98</nowiki>
|Reactive
|The trajectory does cross the TS therefore it is reactive. The lines quite wavy which there is some vibrational energy but more translational energy. This is quite similar to the plot in Figure 5 as the potential energies are the same therefore the waviness of the trajectories are the same. However, the kinetic energies are a little different, making the total energies different.
|[[File:H3 3 01524183.png|thumb|375x375px|Figure 7]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.28</nowiki>
|Unreactive
|The trajectory just crosses the TS however it does not enter the product channel instead it goes back to the reactant channel. This is called barrier re-crossing. Therefore it is overall unreactive. The trajectory is very wavy which means there is a lot of vibrational energy in the system and less translational energy. In the beginning, the AB and BC distances are changing as the middle B atom is oscillating between the A and C atoms. However, in the end, the molecule BC is reformed which is they the AB distance is large.
|[[File:H3 4 01524183.png|thumb|375x375px|Figure 8]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.48</nowiki>
|Reactive
|The trajectory is reactive as it does cross the TS. The line is very wavy which shows that the system has more vibrational energy compared to translational. As the A atom gets closer to the initial BC molecule, the AB distance decreases and the BC distance increases, however, at the TS the B atom oscillates between the A and C atoms as it is attracted to both. This means there's a lot of vibrational energy near the TS. In the end, it does form the AB product molecule, which is shown by the trajectory increase in BC distance.
|[[File:H3 5 01524183.png|thumb|375x375px|Figure 9]]
|}
From the table, it can be concluded that not all trajectories starting with the same positions but with higher values of momenta would be reactive. For example, increasing the value of momenta to -5.1 g.mol-1.pm.fs-1 for p1 and -10.1 g.mol-1.pm.fs-1 for p2 still gives an unreactive trajectory. Also, low values of momenta can be reactive as is the first momenta combination (Figure 5) in the table. This is because vibrational energy also needs to be taken into account, not only the kinetic energy. The degree of vibrational and translational energy is determined by analysing the trajectory line shape: a more wavy line would mean more vibrational energy whereas a more smooth line means more translational energy.