= Reaction Molecular Dynamics Comp Lab =

==Exercise 1: H + H2 System ==

In this section we are considering a system of 3 hydrogen atoms all aligned on the same line of motion. By altering the distance between A and B (rAB), the distance between B and C (rAB), the momenta between A and B (pAB) and the momenta between B and C (pBC) we can assign initial conditions that simulate a collision between H2 and a H atom.

[[File:Screenshot_2019-05-15_at_21.15.32.png|centre|]]

=== Representation of the transition state on a potential energy surface: ===

[[File:Screenshot_2019-05-14_at_16.28.01.png|thumb|right|upright=1.5|Figure 1. A surface plot of the H + H2 system with the transition state circled]] On the potential energy surface the transition state is the point in the reaction pathway with the highest potential energy, as shown in figure 1. Mathematically, at this point, both <math>\frac{\partial V}{\partial r_1}</math> and <math>\frac{\partial V}{\partial r_2}</math> will equal 0. However this result is true for all points on the reaction pathway. The transition state is the point on this pathway with the highest value of V (potential energy). It is a maximum in a trough of minima.

More correctly the transition state is represented by a saddle point. A saddle point is mathematically defined as satisfying the below equation:

<math>\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1 \partial r_2})^2 < 0</math>

If the point were a minimum then the above result would not be true. It would be greater than 0.

=== Locating the transition state position: ===

We know that the transition state is going to be symmetric (the new H bond is forming as the old H bond is breaking) and so it can be found by setting rAB = rBC and giving them both no initial momenta (pAB=pBC=0). By doing this the trajectory is 'trapped' on the ridge created by the minimum component of the saddle point that represents the transition state. It can only oscillate at the top of this ridge rather than going down either of the sides. By adjusting the value of rAB and rBC maintaining that rAB = rBC a point can be found where the trajectory doesn't oscillate at all (ie. it is stationary), this will be the transition state position (rts).

rts was found to equal 0.908 Å (3 d.p.)

The 2 graphs shown in Figure 2 and 3 display the results of having rAB = rBC=0.908 Å and pAB=pBC=0. Both confirm that the H atoms are stationary and that rts=0.908 Å.
[[File:Screenshot_2019-05-14_at_16.53.26.png|thumb|left|upright=1.7|Figure 2. Internuclear Distance vs Time graph with rAB = rBC=0.908 Å and pAB=pBC=0. The horizontal lines indicate no change in internuclear distances and that the H atoms are stationary ]] [[File:Screenshot_2019-05-14_at_16.52.19.png|thumb|right|upright=1.7|Figure 3. Contour plot with rAB = rBC=0.908 Å and pAB=pBC=0. The trajectory is represented by a single point, again indicating that the H atoms are stationary ]]

=== The minimum energy path: ===

The minimum energy path way is a special type of trajectory that resets all the momenta to 0 for every step of the simulation. By doing this the inertial motion of the gaseous H atoms is removed, resulting in a trajectory that exactly follows the bottom of the reaction pathway, hence the name minimum energy pathway. In order to simulate this the MEP setting is chosen and rAB is set to rts (0.908 Å), rBC is set to rts + δ (0.910 Å) and both momenta are set to 0. This produces the 2D contour plot shown in figure 4.

When the same simulation is run using the Dynamic setting instead of the MEP setting the inertial motion is now taken into account, the particles now have momentum/kinetic energy and are therefore seen to oscillate. This is shown in figure 5, the trajectory follows the reaction pathway but oscillates up and down the sides.

[[File:Screenshot_2019-05-15_at_19.24.43.png|left|thumb|upright=1.7|Figure 4. A 2D contour plot showing the MEP. (rAB = 0.908 Å, rBC = 0.910 Å and pAB=pBC=0]] [[File:Screenshot_2019-05-15_at_19.25.36.png|right|upright=1.7|thumb|Figure 5. A 2D contour plot that is simulated with all the same settings as figure 4 except it is using the Dynamic setting rather than MEP]]

=== Reactive or Unreactive? ===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.018 || yes ||The system has sufficient momentum for the approach of atom C to the H2 molecule BA to result in the formation of the H2 molecule BC and the free atom A (The system has sufficient momentum to overcome the barrier created by the transition state). ||[[File:Screenshot_2019-05-16_at_14.26.02.png]]
|-
| -1.5 || -2.0 ||-100.456 || no ||The system doesn't have sufficient energy to overcome the transition state barrier. No new products are formed. ||[[File:Screenshot_2019-05-16_at_14.29.21.png]]
|-
| -1.5 || -2.5 ||-98.956 || yes || Similar to the trajectory at the top of the table. The system has sufficient momentum to overcome the transition state and the new products H2 molecule BC and the free atom A are formed. ||[[File:Screenshot_2019-05-16_at_14.39.21.png]]
|-
| -2.5 || -5.0 ||-84.956 ||no || The system has a much greater momentum compared to the previous examples and can overcome the transition state but the high energy system results in the reactants being reformed. ||[[File:Screenshot_2019-05-16_at_14.42.07.png]]
|-
| -2.5 || -5.2 ||-83.416 ||yes ||This is also a high energy system and overcomes the transition state barrier. The reactants are then reformed but these overcome the transition state barrier again to give the product. ||[[File:Screenshot_2019-05-16_at_14.44.38.png]]
|}

This table highlights 2 key ideas:

1) By increasing the momentum (and therefore energy) of the system the transition state barrier can be overcome.

2) For higher energy systems the transition state barrier will be overcome but there is a chance the reactants can reform.

=== Transition State Theory ===

The transition state theory attempts to explain rates of reaction using 3 key assumptions[1]:

1) Once reactants have reached the transition state they cannot return back to being reactants.

2) The distribution of the particles potential energies follows the Boltzmann distribution.

3) The reactants are in an equilibrium with the activated complex (transition state)

My previous results show an obvious discrepancy with the 1st assumption as trajectories have passed back and forth over the transition state. There is also a discrepancy with the 3rd assumption, all previous trajectories have ended by passing over the transition state (in a forwards or backwards direction relative to the reaction itself) and travelling away from the transition state, not returning for any sort of equilibrium.

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

=== PES Inspection ===

==== Classifying the reactions: ====

F + H2

[[File:Screenshot_2019-05-16_at_16.06.02.png]]

This reaction is exothermic.

FH + H

[[File:Screenshot_2019-05-16_at_16.06.52.png]]

This reaction is endothermic.

These conclusions were made by assessing the potential energy surface (PES) for the F-H-H system. Figure 6 shows the PES at an angle that clearly shows how the energy at the bottom of the reaction pathway varies with distance BC (the distance between the 2 H atoms). Effectively forming a 2D representation of potential energy against distance BC. It shows that when BC increases (ie the H-H bond dissociates) the energy of the system becomes more negative, it is stabilised. This is characteristic of an exothermic reaction. In figure 7 the PES is orientated in a way that shows the relationship between the distance AB (the distance between F and HB). Now we see that the energy is raised as the distance AB is increased (ie the F-H bond dissociates). This raising of energy is characteristic of an endothermic reaction, energy needs to be put in for the reaction to occur.

These results are confirmed by the relative strengths of the H-H and H-F bonds. The H-F bond is stronger (565 kJmol-1) than the H-H bond (432 kJmol-1)[2]. In the F + H2 reaction the weaker H-H bond is broken and a stronger H-F bond is formed having a net release of energy therefore making it exothermic (ΔG = -133 kJmol-1). In the reverse direction the opposite occurs making it endothermic.

[[File:Screenshot_2019-05-16_at_16.43.32.png|left|thumb|upright=1.7|Figure 6. Potential energy surface for the F-H-H system orientated to clearly exemplify the relationship between the potential energy and the distance between the 2 H atoms (distance BC)]] [[File:Screenshot_2019-05-16_at_16.48.31.png|right|thumb|upright=1.7|Figure 7. Potential energy surface for the F-H-H system orientated to clearly exemplify the relationship between the potential energy and the distance between the F atom an central H atom (distance AB)]]

==== Location of the transition state: ====

The PES can also be used to find the transition state position of the reaction using a method similar to that of the H + H2 system. Its not quite as simple as though since the transition state will no longer be symmetric as all the atoms aren't the same. Both the momenta are set to 0 and the correct distances between bonds can be approximated by looking for the characteristics of the transition state described in [[#Representation of the transition state on a potential energy surface:|Representation of the transition state on a potential energy surface]] on the PES, these distances can then be adjusted until the trajectory doesn't move. The Hammond postulate can also be used to help approximate the position of the transition state. Considering F + H2 → F-H + H, we have seen that this process is exothermic and it therefore has an early transition state and Hammonds postulate states that similar energy species will be similar shapes[3], we can therefore conclude that the transition state is similar in shape to the reactants. This means that the H-H separation in the transition state will be similar to the H-H bond length (74 pm or 0.74 Å[4]).

Once found the transition state position can be displayed by all horizontal lines on an Internuclear distances vs Time graph (figure 8) or on a 2D contour graph that shows no movement of the trajectory (figure 9).

The F-H separation in the transition state was found to be 1.808 Å (3 d.p.)

The H-H separation in the transition state was found to be 0.745 Å (3 d.p.) (note similarity to the H-H bond length)

[[File:Screenshot_2019-05-16_at_20.38.09.png|left|upright=1.7|thumb|Figure 8. An Internuclear distances vs Time graph showing that with momenta set to 0, F-H separation set to 1.808 Å and H-H separation set to 0.745 Å there is no movement of the atoms, they are all stationary. Indicating that this is the position of the transition state. ]] [[File:Screenshot_2019-05-16_at_20.37.50.png|right|upright=1.7|thumb|Figure 9. A 2D contour plot showing that with momenta set to 0, F-H separation set to 1.808 Å and H-H separation set to 0.745 Å the trajectory is represented by a single point, indicating that this is the transition state position. ]]

=== Calculating activation energies: ===

The 2 activation energies can be found by running an MEP, setting all momenta to 0 and adjusting the size of the H-H separation to the distance in the transition state (0.745 Å) and the F-H separation slightly displaced from its transition state value (1.798 Å and 1.818 Å). By running these 2 MEPs the trajectory can flow down either side of the transition state. Its the Energy vs Time graphs that these simulations produce that are crucial, using these we can find the energy difference between the transition state and the reactants/products by calculating the difference between the highest (ie transition state energy) and lowest energies (Ie reactant/products energies) on the Energy vs Time graphs. The Energy vs Time graphs and their 2 respective 2D contour plot MEPs are shown in figure 10 and figure 11.

[[File:Screenshot 2019-05-17 at 12.45.21.png|centre|upright=3.5|thumb| Figure 10. The Energy vs Time graph and 2D contour plot of the MEP simulated with F-H separation equal to 1.798 Å, the H-H separation set to 0.745 Å and momenta set to 0. This can be used to find the activation energy of the F-H + H → F + H2 reaction. This was found to be 30.1 (2 d.p.).]]

Using the Energy vs Time graph shown above:

Ea for F-H + H → F + H2 = (-103.752)-(-133.868) = 30.1 (2 d.p.)

(Exact values for energies found by zooming in on the graph on the lepsgui.py software)

[[File:Screenshot 2019-05-17 at 13.21.13.png|centre|upright=3.5|thumb| Figure 11. The Energy vs Time graph and 2D contour plot of the MEP simulated with F-H separation equal to 1.818 Å, the H-H separation set to 0.745 Å and momenta set to 0. This can be used to find the activation energy of the F + H2 → F-H + H reaction. This was found to be 0.23 (2 d.p.).]]

Using the Energy vs Time graph shown above:

Ea for F + H2 → F-H + H = (-103.752)-(-103.928) = 0.23 (2 d.p.)

(Exact values for energies again found by zooming in on the graph on the lepsgui.py software)

=== Reaction dynamics: ===

1 set of successful F + H2 → F-H + H reaction conditions are rFH = 2.3 Å, rHH = 0.74 Å, pFH = -2.5 and pHH = -1.5. The results of running a simulation with these conditions are shown below in figure 12.

[[File:Screenshot 2019-05-17 at 14.01.54.png|cetnre|upright=4|thumb|Figure 12. 2D contour plot, Momenta vs Time graph and Energy vs Time graph for a successful F + H2 → F-H + H reaction (rFH = 2.3 Å, rHH = 0.74 Å, pFH = -2.5 and pHH = -1.5). ]]

Considering the Momenta vs Time graph in figure 12, the orange line is initially oscillating with a negative momentum, representative of the H2 molecule possessing vibrational and translational kinetic energy and travelling towards the F atom. Once the reaction is complete (~ at t = 2.0) the orange line becomes positive and flat, representative of the now displaced H atom only possessing translational kinetic energy and travelling in the opposite direction. The blue line also starts of with a negative momentum, the F atom is actually travelling away from the H2 molecule but the F atom is ~ 19 times larger than the H atom and so having a momentum that is marginally larger than that of the H2 molecule means the H2 molecule easily catches up/collides with the F atom. It too is flat, showing that the F atom only possesses translational kinetic energy at this point. When the reaction is complete the blue line then starts to oscillate with a large amplitude, indicative of the large vibrational energy possessed by the F-H bond. The reaction energy has been released as increased vibrational energy. Note the total energy of the system is constant throughout, there is a balance between kinetic and potential energy.

Since the reaction energy is released as vibrational energy IR spectroscopy could be used to monitor this.

=== Polanyi’s rules: ===

Polanyi’s rules state that vibrational energy is better for overcoming a late transition state barrier more efficiently whereas translational energy is better for an early transition state barrier[5].

By varying the initial conditions and utilising the fact that the F + H2 reaction has an early transition state and the F-H + H reaction has a late transition state these rules can be explored.

==== Early transition state ====
Figure 13 displays the effects on varying the amount of translational and vibrational energy in the system on the efficiency of the F + H2 reaction with an early transition state. Both trajectories start at the same position, rFH = 2 and rHH = 0.74. The 2D contour plot on the left has a much higher H-H bond vibrational energy (pFH = -0.5 and pHH = 2.4). In this case the reaction is successful but there is significant recrossing, the reaction isn't very efficient. The 2D contour plot on the right of figure 13 shows what happens when some of the vibrational energy of the H2 molecule is converted into translational energy of the F atom, the magnitude of pFH is increased (from -0.5 to -0.8) but the magnitude of pHH is significantly decreased (from 2.4 to 0.1). In this case the reaction is seen to be successful. These results align with Polanyi’s rule that translational energy overcomes an early transition state more efficiently compared to vibrational energy.

[[File:Screenshot 2019-05-17 at 17.12.23.png|centre|upright=3.5|thumb|Figure 13. The 2D contour plot on the left shows a less efficient trajectory due to considerable H-H vibrational energy and less translational energy of the F atom. The 2D contour plot on the right shows a more efficient trajectory, this time with less H-H vibrational energy and more translational energy of the F atom. Both trajectories start at the same position. ]]

==== Late transition state ====

The trajectories shown in figure 14 were found using the inversion of momentum procedure on the 2 trajectories above in figure 13. By doing this the effects on varying the amount of translational and vibrational energy in the system on the efficiency of the

[[File:Screenshot 2019-05-17 at 17.30.58.png]]

== References ==

1 - Steinfeld, Jeffrey I., Francisco, Joseph S, and Hase, William L. Chemical Kinetics and Dynamics. Englewood Cliffs, N.J.: Prentice Hall, 1989.

2 - Darwent, B. National Standard Reference Data Series. No. 31 31, (1970).

3 - Hammond, G. S. A Correlation of Reaction Rates. J. Am. Chem. Soc. 77, 334–338 (1955).

4 - CRC Handbook of Chemistry and Physics. (Boca Raton, 2013).

5 - Polanyi, J. C. Concepts in reaction dynamics. Acc. Chem. Res. 5, 161–168 (1972).

File:Screenshot 2019-05-17 at 17.30.58.png

2019-05-17T16:31:26Z

Oja16:

MRD:OA010298

2019-05-17T16:24:12Z

Oja16: /* Early transition state */

= Reaction Molecular Dynamics Comp Lab =

==Exercise 1: H + H2 System ==

In this section we are considering a system of 3 hydrogen atoms all aligned on the same line of motion. By altering the distance between A and B (rAB), the distance between B and C (rAB), the momenta between A and B (pAB) and the momenta between B and C (pBC) we can assign initial conditions that simulate a collision between H2 and a H atom.

[[File:Screenshot_2019-05-15_at_21.15.32.png|centre|]]

=== Representation of the transition state on a potential energy surface: ===

[[File:Screenshot_2019-05-14_at_16.28.01.png|thumb|right|upright=1.5|Figure 1. A surface plot of the H + H2 system with the transition state circled]] On the potential energy surface the transition state is the point in the reaction pathway with the highest potential energy, as shown in figure 1. Mathematically, at this point, both <math>\frac{\partial V}{\partial r_1}</math> and <math>\frac{\partial V}{\partial r_2}</math> will equal 0. However this result is true for all points on the reaction pathway. The transition state is the point on this pathway with the highest value of V (potential energy). It is a maximum in a trough of minima.

More correctly the transition state is represented by a saddle point. A saddle point is mathematically defined as satisfying the below equation:

<math>\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1 \partial r_2})^2 < 0</math>

If the point were a minimum then the above result would not be true. It would be greater than 0.

=== Locating the transition state position: ===

We know that the transition state is going to be symmetric (the new H bond is forming as the old H bond is breaking) and so it can be found by setting rAB = rBC and giving them both no initial momenta (pAB=pBC=0). By doing this the trajectory is 'trapped' on the ridge created by the minimum component of the saddle point that represents the transition state. It can only oscillate at the top of this ridge rather than going down either of the sides. By adjusting the value of rAB and rBC maintaining that rAB = rBC a point can be found where the trajectory doesn't oscillate at all (ie. it is stationary), this will be the transition state position (rts).

rts was found to equal 0.908 Å (3 d.p.)

The 2 graphs shown in Figure 2 and 3 display the results of having rAB = rBC=0.908 Å and pAB=pBC=0. Both confirm that the H atoms are stationary and that rts=0.908 Å.
[[File:Screenshot_2019-05-14_at_16.53.26.png|thumb|left|upright=1.7|Figure 2. Internuclear Distance vs Time graph with rAB = rBC=0.908 Å and pAB=pBC=0. The horizontal lines indicate no change in internuclear distances and that the H atoms are stationary ]] [[File:Screenshot_2019-05-14_at_16.52.19.png|thumb|right|upright=1.7|Figure 3. Contour plot with rAB = rBC=0.908 Å and pAB=pBC=0. The trajectory is represented by a single point, again indicating that the H atoms are stationary ]]

=== The minimum energy path: ===

The minimum energy path way is a special type of trajectory that resets all the momenta to 0 for every step of the simulation. By doing this the inertial motion of the gaseous H atoms is removed, resulting in a trajectory that exactly follows the bottom of the reaction pathway, hence the name minimum energy pathway. In order to simulate this the MEP setting is chosen and rAB is set to rts (0.908 Å), rBC is set to rts + δ (0.910 Å) and both momenta are set to 0. This produces the 2D contour plot shown in figure 4.

When the same simulation is run using the Dynamic setting instead of the MEP setting the inertial motion is now taken into account, the particles now have momentum/kinetic energy and are therefore seen to oscillate. This is shown in figure 5, the trajectory follows the reaction pathway but oscillates up and down the sides.

[[File:Screenshot_2019-05-15_at_19.24.43.png|left|thumb|upright=1.7|Figure 4. A 2D contour plot showing the MEP. (rAB = 0.908 Å, rBC = 0.910 Å and pAB=pBC=0]] [[File:Screenshot_2019-05-15_at_19.25.36.png|right|upright=1.7|thumb|Figure 5. A 2D contour plot that is simulated with all the same settings as figure 4 except it is using the Dynamic setting rather than MEP]]

=== Reactive or Unreactive? ===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.018 || yes ||The system has sufficient momentum for the approach of atom C to the H2 molecule BA to result in the formation of the H2 molecule BC and the free atom A (The system has sufficient momentum to overcome the barrier created by the transition state). ||[[File:Screenshot_2019-05-16_at_14.26.02.png]]
|-
| -1.5 || -2.0 ||-100.456 || no ||The system doesn't have sufficient energy to overcome the transition state barrier. No new products are formed. ||[[File:Screenshot_2019-05-16_at_14.29.21.png]]
|-
| -1.5 || -2.5 ||-98.956 || yes || Similar to the trajectory at the top of the table. The system has sufficient momentum to overcome the transition state and the new products H2 molecule BC and the free atom A are formed. ||[[File:Screenshot_2019-05-16_at_14.39.21.png]]
|-
| -2.5 || -5.0 ||-84.956 ||no || The system has a much greater momentum compared to the previous examples and can overcome the transition state but the high energy system results in the reactants being reformed. ||[[File:Screenshot_2019-05-16_at_14.42.07.png]]
|-
| -2.5 || -5.2 ||-83.416 ||yes ||This is also a high energy system and overcomes the transition state barrier. The reactants are then reformed but these overcome the transition state barrier again to give the product. ||[[File:Screenshot_2019-05-16_at_14.44.38.png]]
|}

This table highlights 2 key ideas:

1) By increasing the momentum (and therefore energy) of the system the transition state barrier can be overcome.

2) For higher energy systems the transition state barrier will be overcome but there is a chance the reactants can reform.

=== Transition State Theory ===

The transition state theory attempts to explain rates of reaction using 3 key assumptions[1]:

1) Once reactants have reached the transition state they cannot return back to being reactants.

2) The distribution of the particles potential energies follows the Boltzmann distribution.

3) The reactants are in an equilibrium with the activated complex (transition state)

My previous results show an obvious discrepancy with the 1st assumption as trajectories have passed back and forth over the transition state. There is also a discrepancy with the 3rd assumption, all previous trajectories have ended by passing over the transition state (in a forwards or backwards direction relative to the reaction itself) and travelling away from the transition state, not returning for any sort of equilibrium.

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

=== PES Inspection ===

==== Classifying the reactions: ====

F + H2

[[File:Screenshot_2019-05-16_at_16.06.02.png]]

This reaction is exothermic.

FH + H

[[File:Screenshot_2019-05-16_at_16.06.52.png]]

This reaction is endothermic.

These conclusions were made by assessing the potential energy surface (PES) for the F-H-H system. Figure 6 shows the PES at an angle that clearly shows how the energy at the bottom of the reaction pathway varies with distance BC (the distance between the 2 H atoms). Effectively forming a 2D representation of potential energy against distance BC. It shows that when BC increases (ie the H-H bond dissociates) the energy of the system becomes more negative, it is stabilised. This is characteristic of an exothermic reaction. In figure 7 the PES is orientated in a way that shows the relationship between the distance AB (the distance between F and HB). Now we see that the energy is raised as the distance AB is increased (ie the F-H bond dissociates). This raising of energy is characteristic of an endothermic reaction, energy needs to be put in for the reaction to occur.

These results are confirmed by the relative strengths of the H-H and H-F bonds. The H-F bond is stronger (565 kJmol-1) than the H-H bond (432 kJmol-1)[2]. In the F + H2 reaction the weaker H-H bond is broken and a stronger H-F bond is formed having a net release of energy therefore making it exothermic (ΔG = -133 kJmol-1). In the reverse direction the opposite occurs making it endothermic.

[[File:Screenshot_2019-05-16_at_16.43.32.png|left|thumb|upright=1.7|Figure 6. Potential energy surface for the F-H-H system orientated to clearly exemplify the relationship between the potential energy and the distance between the 2 H atoms (distance BC)]] [[File:Screenshot_2019-05-16_at_16.48.31.png|right|thumb|upright=1.7|Figure 7. Potential energy surface for the F-H-H system orientated to clearly exemplify the relationship between the potential energy and the distance between the F atom an central H atom (distance AB)]]

==== Location of the transition state: ====

The PES can also be used to find the transition state position of the reaction using a method similar to that of the H + H2 system. Its not quite as simple as though since the transition state will no longer be symmetric as all the atoms aren't the same. Both the momenta are set to 0 and the correct distances between bonds can be approximated by looking for the characteristics of the transition state described in [[#Representation of the transition state on a potential energy surface:|Representation of the transition state on a potential energy surface]] on the PES, these distances can then be adjusted until the trajectory doesn't move. The Hammond postulate can also be used to help approximate the position of the transition state. Considering F + H2 → F-H + H, we have seen that this process is exothermic and it therefore has an early transition state and Hammonds postulate states that similar energy species will be similar shapes[3], we can therefore conclude that the transition state is similar in shape to the reactants. This means that the H-H separation in the transition state will be similar to the H-H bond length (74 pm or 0.74 Å[4]).

Once found the transition state position can be displayed by all horizontal lines on an Internuclear distances vs Time graph (figure 8) or on a 2D contour graph that shows no movement of the trajectory (figure 9).

The F-H separation in the transition state was found to be 1.808 Å (3 d.p.)

The H-H separation in the transition state was found to be 0.745 Å (3 d.p.) (note similarity to the H-H bond length)

[[File:Screenshot_2019-05-16_at_20.38.09.png|left|upright=1.7|thumb|Figure 8. An Internuclear distances vs Time graph showing that with momenta set to 0, F-H separation set to 1.808 Å and H-H separation set to 0.745 Å there is no movement of the atoms, they are all stationary. Indicating that this is the position of the transition state. ]] [[File:Screenshot_2019-05-16_at_20.37.50.png|right|upright=1.7|thumb|Figure 9. A 2D contour plot showing that with momenta set to 0, F-H separation set to 1.808 Å and H-H separation set to 0.745 Å the trajectory is represented by a single point, indicating that this is the transition state position. ]]

=== Calculating activation energies: ===

The 2 activation energies can be found by running an MEP, setting all momenta to 0 and adjusting the size of the H-H separation to the distance in the transition state (0.745 Å) and the F-H separation slightly displaced from its transition state value (1.798 Å and 1.818 Å). By running these 2 MEPs the trajectory can flow down either side of the transition state. Its the Energy vs Time graphs that these simulations produce that are crucial, using these we can find the energy difference between the transition state and the reactants/products by calculating the difference between the highest (ie transition state energy) and lowest energies (Ie reactant/products energies) on the Energy vs Time graphs. The Energy vs Time graphs and their 2 respective 2D contour plot MEPs are shown in figure 10 and figure 11.

[[File:Screenshot 2019-05-17 at 12.45.21.png|centre|upright=3.5|thumb| Figure 10. The Energy vs Time graph and 2D contour plot of the MEP simulated with F-H separation equal to 1.798 Å, the H-H separation set to 0.745 Å and momenta set to 0. This can be used to find the activation energy of the F-H + H → F + H2 reaction. This was found to be 30.1 (2 d.p.).]]

Using the Energy vs Time graph shown above:

Ea for F-H + H → F + H2 = (-103.752)-(-133.868) = 30.1 (2 d.p.)

(Exact values for energies found by zooming in on the graph on the lepsgui.py software)

[[File:Screenshot 2019-05-17 at 13.21.13.png|centre|upright=3.5|thumb| Figure 11. The Energy vs Time graph and 2D contour plot of the MEP simulated with F-H separation equal to 1.818 Å, the H-H separation set to 0.745 Å and momenta set to 0. This can be used to find the activation energy of the F + H2 → F-H + H reaction. This was found to be 0.23 (2 d.p.).]]

Using the Energy vs Time graph shown above:

Ea for F + H2 → F-H + H = (-103.752)-(-103.928) = 0.23 (2 d.p.)

(Exact values for energies again found by zooming in on the graph on the lepsgui.py software)

=== Reaction dynamics: ===

1 set of successful F + H2 → F-H + H reaction conditions are rFH = 2.3 Å, rHH = 0.74 Å, pFH = -2.5 and pHH = -1.5. The results of running a simulation with these conditions are shown below in figure 12.

[[File:Screenshot 2019-05-17 at 14.01.54.png|cetnre|upright=4|thumb|Figure 12. 2D contour plot, Momenta vs Time graph and Energy vs Time graph for a successful F + H2 → F-H + H reaction (rFH = 2.3 Å, rHH = 0.74 Å, pFH = -2.5 and pHH = -1.5). ]]

Considering the Momenta vs Time graph in figure 12, the orange line is initially oscillating with a negative momentum, representative of the H2 molecule possessing vibrational and translational kinetic energy and travelling towards the F atom. Once the reaction is complete (~ at t = 2.0) the orange line becomes positive and flat, representative of the now displaced H atom only possessing translational kinetic energy and travelling in the opposite direction. The blue line also starts of with a negative momentum, the F atom is actually travelling away from the H2 molecule but the F atom is ~ 19 times larger than the H atom and so having a momentum that is marginally larger than that of the H2 molecule means the H2 molecule easily catches up/collides with the F atom. It too is flat, showing that the F atom only possesses translational kinetic energy at this point. When the reaction is complete the blue line then starts to oscillate with a large amplitude, indicative of the large vibrational energy possessed by the F-H bond. The reaction energy has been released as increased vibrational energy. Note the total energy of the system is constant throughout, there is a balance between kinetic and potential energy.

Since the reaction energy is released as vibrational energy IR spectroscopy could be used to monitor this.

=== Polanyi’s rules: ===

Polanyi’s rules state that vibrational energy is better for overcoming a late transition state barrier more efficiently whereas translational energy is better for an early transition state barrier[5].

By varying the initial conditions and utilising the fact that the F + H2 reaction has an early transition state and the F-H + H reaction has a late transition state these rules can be explored.

==== Early transition state ====
Figure 13 displays the effects on varying the amount of translational and vibrational energy in the system on the efficiency of the F + H2 reaction with an early transition state. Both trajectories start at the same position, rFH = 2 and rHH = 0.74. The 2D contour plot on the left has a much higher H-H bond vibrational energy (pFH = -0.5 and pHH = 2.4). In this case the reaction is successful but there is significant recrossing, the reaction isn't very efficient. The 2D contour plot on the right of figure 13 shows what happens when some of the vibrational energy of the H2 molecule is converted into translational energy of the F atom, the magnitude of pFH is increased (from -0.5 to -0.8) but the magnitude of pHH is significantly decreased (from 2.4 to 0.1). In this case the reaction is seen to be successful. These results align with Polanyi’s rule that translational energy overcomes an early transition state more efficiently compared to vibrational energy.

[[File:Screenshot 2019-05-17 at 17.12.23.png|centre|upright=3.5|thumb|Figure 13. The 2D contour plot on the left shows a less efficient trajectory due to considerable H-H vibrational energy and less translational energy of the F atom. The 2D contour plot on the right shows a more efficient trajectory, this time with less H-H vibrational energy and more translational energy of the F atom. Both trajectories start at the same position. ]]

==== Late transition state ====

== References ==

1 - Steinfeld, Jeffrey I., Francisco, Joseph S, and Hase, William L. Chemical Kinetics and Dynamics. Englewood Cliffs, N.J.: Prentice Hall, 1989.

2 - Darwent, B. National Standard Reference Data Series. No. 31 31, (1970).

3 - Hammond, G. S. A Correlation of Reaction Rates. J. Am. Chem. Soc. 77, 334–338 (1955).

4 - CRC Handbook of Chemistry and Physics. (Boca Raton, 2013).

5 - Polanyi, J. C. Concepts in reaction dynamics. Acc. Chem. Res. 5, 161–168 (1972).

MRD:OA010298

2019-05-17T16:23:07Z

Oja16: /* Early transition state */

= Reaction Molecular Dynamics Comp Lab =

==Exercise 1: H + H2 System ==

In this section we are considering a system of 3 hydrogen atoms all aligned on the same line of motion. By altering the distance between A and B (rAB), the distance between B and C (rAB), the momenta between A and B (pAB) and the momenta between B and C (pBC) we can assign initial conditions that simulate a collision between H2 and a H atom.

[[File:Screenshot_2019-05-15_at_21.15.32.png|centre|]]

=== Representation of the transition state on a potential energy surface: ===

[[File:Screenshot_2019-05-14_at_16.28.01.png|thumb|right|upright=1.5|Figure 1. A surface plot of the H + H2 system with the transition state circled]] On the potential energy surface the transition state is the point in the reaction pathway with the highest potential energy, as shown in figure 1. Mathematically, at this point, both <math>\frac{\partial V}{\partial r_1}</math> and <math>\frac{\partial V}{\partial r_2}</math> will equal 0. However this result is true for all points on the reaction pathway. The transition state is the point on this pathway with the highest value of V (potential energy). It is a maximum in a trough of minima.

More correctly the transition state is represented by a saddle point. A saddle point is mathematically defined as satisfying the below equation:

<math>\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1 \partial r_2})^2 < 0</math>

If the point were a minimum then the above result would not be true. It would be greater than 0.

=== Locating the transition state position: ===

We know that the transition state is going to be symmetric (the new H bond is forming as the old H bond is breaking) and so it can be found by setting rAB = rBC and giving them both no initial momenta (pAB=pBC=0). By doing this the trajectory is 'trapped' on the ridge created by the minimum component of the saddle point that represents the transition state. It can only oscillate at the top of this ridge rather than going down either of the sides. By adjusting the value of rAB and rBC maintaining that rAB = rBC a point can be found where the trajectory doesn't oscillate at all (ie. it is stationary), this will be the transition state position (rts).

rts was found to equal 0.908 Å (3 d.p.)

The 2 graphs shown in Figure 2 and 3 display the results of having rAB = rBC=0.908 Å and pAB=pBC=0. Both confirm that the H atoms are stationary and that rts=0.908 Å.
[[File:Screenshot_2019-05-14_at_16.53.26.png|thumb|left|upright=1.7|Figure 2. Internuclear Distance vs Time graph with rAB = rBC=0.908 Å and pAB=pBC=0. The horizontal lines indicate no change in internuclear distances and that the H atoms are stationary ]] [[File:Screenshot_2019-05-14_at_16.52.19.png|thumb|right|upright=1.7|Figure 3. Contour plot with rAB = rBC=0.908 Å and pAB=pBC=0. The trajectory is represented by a single point, again indicating that the H atoms are stationary ]]

=== The minimum energy path: ===

The minimum energy path way is a special type of trajectory that resets all the momenta to 0 for every step of the simulation. By doing this the inertial motion of the gaseous H atoms is removed, resulting in a trajectory that exactly follows the bottom of the reaction pathway, hence the name minimum energy pathway. In order to simulate this the MEP setting is chosen and rAB is set to rts (0.908 Å), rBC is set to rts + δ (0.910 Å) and both momenta are set to 0. This produces the 2D contour plot shown in figure 4.

When the same simulation is run using the Dynamic setting instead of the MEP setting the inertial motion is now taken into account, the particles now have momentum/kinetic energy and are therefore seen to oscillate. This is shown in figure 5, the trajectory follows the reaction pathway but oscillates up and down the sides.

[[File:Screenshot_2019-05-15_at_19.24.43.png|left|thumb|upright=1.7|Figure 4. A 2D contour plot showing the MEP. (rAB = 0.908 Å, rBC = 0.910 Å and pAB=pBC=0]] [[File:Screenshot_2019-05-15_at_19.25.36.png|right|upright=1.7|thumb|Figure 5. A 2D contour plot that is simulated with all the same settings as figure 4 except it is using the Dynamic setting rather than MEP]]

=== Reactive or Unreactive? ===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.018 || yes ||The system has sufficient momentum for the approach of atom C to the H2 molecule BA to result in the formation of the H2 molecule BC and the free atom A (The system has sufficient momentum to overcome the barrier created by the transition state). ||[[File:Screenshot_2019-05-16_at_14.26.02.png]]
|-
| -1.5 || -2.0 ||-100.456 || no ||The system doesn't have sufficient energy to overcome the transition state barrier. No new products are formed. ||[[File:Screenshot_2019-05-16_at_14.29.21.png]]
|-
| -1.5 || -2.5 ||-98.956 || yes || Similar to the trajectory at the top of the table. The system has sufficient momentum to overcome the transition state and the new products H2 molecule BC and the free atom A are formed. ||[[File:Screenshot_2019-05-16_at_14.39.21.png]]
|-
| -2.5 || -5.0 ||-84.956 ||no || The system has a much greater momentum compared to the previous examples and can overcome the transition state but the high energy system results in the reactants being reformed. ||[[File:Screenshot_2019-05-16_at_14.42.07.png]]
|-
| -2.5 || -5.2 ||-83.416 ||yes ||This is also a high energy system and overcomes the transition state barrier. The reactants are then reformed but these overcome the transition state barrier again to give the product. ||[[File:Screenshot_2019-05-16_at_14.44.38.png]]
|}

This table highlights 2 key ideas:

1) By increasing the momentum (and therefore energy) of the system the transition state barrier can be overcome.

2) For higher energy systems the transition state barrier will be overcome but there is a chance the reactants can reform.

=== Transition State Theory ===

The transition state theory attempts to explain rates of reaction using 3 key assumptions[1]:

1) Once reactants have reached the transition state they cannot return back to being reactants.

2) The distribution of the particles potential energies follows the Boltzmann distribution.

3) The reactants are in an equilibrium with the activated complex (transition state)

My previous results show an obvious discrepancy with the 1st assumption as trajectories have passed back and forth over the transition state. There is also a discrepancy with the 3rd assumption, all previous trajectories have ended by passing over the transition state (in a forwards or backwards direction relative to the reaction itself) and travelling away from the transition state, not returning for any sort of equilibrium.

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

=== PES Inspection ===

==== Classifying the reactions: ====

F + H2

[[File:Screenshot_2019-05-16_at_16.06.02.png]]

This reaction is exothermic.

FH + H

[[File:Screenshot_2019-05-16_at_16.06.52.png]]

This reaction is endothermic.

These conclusions were made by assessing the potential energy surface (PES) for the F-H-H system. Figure 6 shows the PES at an angle that clearly shows how the energy at the bottom of the reaction pathway varies with distance BC (the distance between the 2 H atoms). Effectively forming a 2D representation of potential energy against distance BC. It shows that when BC increases (ie the H-H bond dissociates) the energy of the system becomes more negative, it is stabilised. This is characteristic of an exothermic reaction. In figure 7 the PES is orientated in a way that shows the relationship between the distance AB (the distance between F and HB). Now we see that the energy is raised as the distance AB is increased (ie the F-H bond dissociates). This raising of energy is characteristic of an endothermic reaction, energy needs to be put in for the reaction to occur.

These results are confirmed by the relative strengths of the H-H and H-F bonds. The H-F bond is stronger (565 kJmol-1) than the H-H bond (432 kJmol-1)[2]. In the F + H2 reaction the weaker H-H bond is broken and a stronger H-F bond is formed having a net release of energy therefore making it exothermic (ΔG = -133 kJmol-1). In the reverse direction the opposite occurs making it endothermic.

[[File:Screenshot_2019-05-16_at_16.43.32.png|left|thumb|upright=1.7|Figure 6. Potential energy surface for the F-H-H system orientated to clearly exemplify the relationship between the potential energy and the distance between the 2 H atoms (distance BC)]] [[File:Screenshot_2019-05-16_at_16.48.31.png|right|thumb|upright=1.7|Figure 7. Potential energy surface for the F-H-H system orientated to clearly exemplify the relationship between the potential energy and the distance between the F atom an central H atom (distance AB)]]

==== Location of the transition state: ====

The PES can also be used to find the transition state position of the reaction using a method similar to that of the H + H2 system. Its not quite as simple as though since the transition state will no longer be symmetric as all the atoms aren't the same. Both the momenta are set to 0 and the correct distances between bonds can be approximated by looking for the characteristics of the transition state described in [[#Representation of the transition state on a potential energy surface:|Representation of the transition state on a potential energy surface]] on the PES, these distances can then be adjusted until the trajectory doesn't move. The Hammond postulate can also be used to help approximate the position of the transition state. Considering F + H2 → F-H + H, we have seen that this process is exothermic and it therefore has an early transition state and Hammonds postulate states that similar energy species will be similar shapes[3], we can therefore conclude that the transition state is similar in shape to the reactants. This means that the H-H separation in the transition state will be similar to the H-H bond length (74 pm or 0.74 Å[4]).

Once found the transition state position can be displayed by all horizontal lines on an Internuclear distances vs Time graph (figure 8) or on a 2D contour graph that shows no movement of the trajectory (figure 9).

The F-H separation in the transition state was found to be 1.808 Å (3 d.p.)

The H-H separation in the transition state was found to be 0.745 Å (3 d.p.) (note similarity to the H-H bond length)

[[File:Screenshot_2019-05-16_at_20.38.09.png|left|upright=1.7|thumb|Figure 8. An Internuclear distances vs Time graph showing that with momenta set to 0, F-H separation set to 1.808 Å and H-H separation set to 0.745 Å there is no movement of the atoms, they are all stationary. Indicating that this is the position of the transition state. ]] [[File:Screenshot_2019-05-16_at_20.37.50.png|right|upright=1.7|thumb|Figure 9. A 2D contour plot showing that with momenta set to 0, F-H separation set to 1.808 Å and H-H separation set to 0.745 Å the trajectory is represented by a single point, indicating that this is the transition state position. ]]

=== Calculating activation energies: ===

The 2 activation energies can be found by running an MEP, setting all momenta to 0 and adjusting the size of the H-H separation to the distance in the transition state (0.745 Å) and the F-H separation slightly displaced from its transition state value (1.798 Å and 1.818 Å). By running these 2 MEPs the trajectory can flow down either side of the transition state. Its the Energy vs Time graphs that these simulations produce that are crucial, using these we can find the energy difference between the transition state and the reactants/products by calculating the difference between the highest (ie transition state energy) and lowest energies (Ie reactant/products energies) on the Energy vs Time graphs. The Energy vs Time graphs and their 2 respective 2D contour plot MEPs are shown in figure 10 and figure 11.

[[File:Screenshot 2019-05-17 at 12.45.21.png|centre|upright=3.5|thumb| Figure 10. The Energy vs Time graph and 2D contour plot of the MEP simulated with F-H separation equal to 1.798 Å, the H-H separation set to 0.745 Å and momenta set to 0. This can be used to find the activation energy of the F-H + H → F + H2 reaction. This was found to be 30.1 (2 d.p.).]]

Using the Energy vs Time graph shown above:

Ea for F-H + H → F + H2 = (-103.752)-(-133.868) = 30.1 (2 d.p.)

(Exact values for energies found by zooming in on the graph on the lepsgui.py software)

[[File:Screenshot 2019-05-17 at 13.21.13.png|centre|upright=3.5|thumb| Figure 11. The Energy vs Time graph and 2D contour plot of the MEP simulated with F-H separation equal to 1.818 Å, the H-H separation set to 0.745 Å and momenta set to 0. This can be used to find the activation energy of the F + H2 → F-H + H reaction. This was found to be 0.23 (2 d.p.).]]

Using the Energy vs Time graph shown above:

Ea for F + H2 → F-H + H = (-103.752)-(-103.928) = 0.23 (2 d.p.)

(Exact values for energies again found by zooming in on the graph on the lepsgui.py software)

=== Reaction dynamics: ===

1 set of successful F + H2 → F-H + H reaction conditions are rFH = 2.3 Å, rHH = 0.74 Å, pFH = -2.5 and pHH = -1.5. The results of running a simulation with these conditions are shown below in figure 12.

[[File:Screenshot 2019-05-17 at 14.01.54.png|cetnre|upright=4|thumb|Figure 12. 2D contour plot, Momenta vs Time graph and Energy vs Time graph for a successful F + H2 → F-H + H reaction (rFH = 2.3 Å, rHH = 0.74 Å, pFH = -2.5 and pHH = -1.5). ]]

Considering the Momenta vs Time graph in figure 12, the orange line is initially oscillating with a negative momentum, representative of the H2 molecule possessing vibrational and translational kinetic energy and travelling towards the F atom. Once the reaction is complete (~ at t = 2.0) the orange line becomes positive and flat, representative of the now displaced H atom only possessing translational kinetic energy and travelling in the opposite direction. The blue line also starts of with a negative momentum, the F atom is actually travelling away from the H2 molecule but the F atom is ~ 19 times larger than the H atom and so having a momentum that is marginally larger than that of the H2 molecule means the H2 molecule easily catches up/collides with the F atom. It too is flat, showing that the F atom only possesses translational kinetic energy at this point. When the reaction is complete the blue line then starts to oscillate with a large amplitude, indicative of the large vibrational energy possessed by the F-H bond. The reaction energy has been released as increased vibrational energy. Note the total energy of the system is constant throughout, there is a balance between kinetic and potential energy.

Since the reaction energy is released as vibrational energy IR spectroscopy could be used to monitor this.

=== Polanyi’s rules: ===

Polanyi’s rules state that vibrational energy is better for overcoming a late transition state barrier more efficiently whereas translational energy is better for an early transition state barrier[5].

By varying the initial conditions and utilising the fact that the F + H2 reaction has an early transition state and the F-H + H reaction has a late transition state these rules can be explored.

==== Early transition state ====
Figure 13 displays the effects on varying the amount of translational and vibrational energy in the system on the efficiency of the F + H2 reaction with an early transition state. Both trajectories start at the same position, rFH = 2.3 and rHH = 0.74. The 2D contour plot on the left has a much higher H-H bond vibrational energy (pFH = -0.5 and pHH = 2.4). In this case the reaction is successful but there is significant recrossing, the reaction isn't very efficient. The 2D contour plot on the right of figure 13 shows what happens when some of the vibrational energy of the H2 molecule is converted into translational energy of the F atom, the magnitude of pFH is increased (from -0.5 to -0.8) but the magnitude of pHH is significantly decreased (from 2.4 to 0.1). In this case the reaction is seen to be successful. These results align with Polanyi’s rule that translational energy overcomes an early transition state more efficiently compared to vibrational energy.

[[File:Screenshot 2019-05-17 at 17.12.23.png|centre|upright=3.5|thumb|Figure 13. The 2D contour plot on the left shows a less efficient trajectory due to considerable H-H vibrational energy and less translational energy of the F atom. The 2D contour plot on the right shows a more efficient trajectory, this time with less H-H vibrational energy and more translational energy of the F atom. Both trajectories start at the same position. ]]

==== Late transition state ====

== References ==

1 - Steinfeld, Jeffrey I., Francisco, Joseph S, and Hase, William L. Chemical Kinetics and Dynamics. Englewood Cliffs, N.J.: Prentice Hall, 1989.

2 - Darwent, B. National Standard Reference Data Series. No. 31 31, (1970).

3 - Hammond, G. S. A Correlation of Reaction Rates. J. Am. Chem. Soc. 77, 334–338 (1955).

4 - CRC Handbook of Chemistry and Physics. (Boca Raton, 2013).

5 - Polanyi, J. C. Concepts in reaction dynamics. Acc. Chem. Res. 5, 161–168 (1972).

MRD:OA010298

2019-05-17T16:15:04Z

Oja16: /* Early transition state */

= Reaction Molecular Dynamics Comp Lab =

==Exercise 1: H + H2 System ==

In this section we are considering a system of 3 hydrogen atoms all aligned on the same line of motion. By altering the distance between A and B (rAB), the distance between B and C (rAB), the momenta between A and B (pAB) and the momenta between B and C (pBC) we can assign initial conditions that simulate a collision between H2 and a H atom.

[[File:Screenshot_2019-05-15_at_21.15.32.png|centre|]]

=== Representation of the transition state on a potential energy surface: ===

[[File:Screenshot_2019-05-14_at_16.28.01.png|thumb|right|upright=1.5|Figure 1. A surface plot of the H + H2 system with the transition state circled]] On the potential energy surface the transition state is the point in the reaction pathway with the highest potential energy, as shown in figure 1. Mathematically, at this point, both <math>\frac{\partial V}{\partial r_1}</math> and <math>\frac{\partial V}{\partial r_2}</math> will equal 0. However this result is true for all points on the reaction pathway. The transition state is the point on this pathway with the highest value of V (potential energy). It is a maximum in a trough of minima.

More correctly the transition state is represented by a saddle point. A saddle point is mathematically defined as satisfying the below equation:

<math>\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1 \partial r_2})^2 < 0</math>

If the point were a minimum then the above result would not be true. It would be greater than 0.

=== Locating the transition state position: ===

We know that the transition state is going to be symmetric (the new H bond is forming as the old H bond is breaking) and so it can be found by setting rAB = rBC and giving them both no initial momenta (pAB=pBC=0). By doing this the trajectory is 'trapped' on the ridge created by the minimum component of the saddle point that represents the transition state. It can only oscillate at the top of this ridge rather than going down either of the sides. By adjusting the value of rAB and rBC maintaining that rAB = rBC a point can be found where the trajectory doesn't oscillate at all (ie. it is stationary), this will be the transition state position (rts).

rts was found to equal 0.908 Å (3 d.p.)

The 2 graphs shown in Figure 2 and 3 display the results of having rAB = rBC=0.908 Å and pAB=pBC=0. Both confirm that the H atoms are stationary and that rts=0.908 Å.
[[File:Screenshot_2019-05-14_at_16.53.26.png|thumb|left|upright=1.7|Figure 2. Internuclear Distance vs Time graph with rAB = rBC=0.908 Å and pAB=pBC=0. The horizontal lines indicate no change in internuclear distances and that the H atoms are stationary ]] [[File:Screenshot_2019-05-14_at_16.52.19.png|thumb|right|upright=1.7|Figure 3. Contour plot with rAB = rBC=0.908 Å and pAB=pBC=0. The trajectory is represented by a single point, again indicating that the H atoms are stationary ]]

=== The minimum energy path: ===

The minimum energy path way is a special type of trajectory that resets all the momenta to 0 for every step of the simulation. By doing this the inertial motion of the gaseous H atoms is removed, resulting in a trajectory that exactly follows the bottom of the reaction pathway, hence the name minimum energy pathway. In order to simulate this the MEP setting is chosen and rAB is set to rts (0.908 Å), rBC is set to rts + δ (0.910 Å) and both momenta are set to 0. This produces the 2D contour plot shown in figure 4.

When the same simulation is run using the Dynamic setting instead of the MEP setting the inertial motion is now taken into account, the particles now have momentum/kinetic energy and are therefore seen to oscillate. This is shown in figure 5, the trajectory follows the reaction pathway but oscillates up and down the sides.

[[File:Screenshot_2019-05-15_at_19.24.43.png|left|thumb|upright=1.7|Figure 4. A 2D contour plot showing the MEP. (rAB = 0.908 Å, rBC = 0.910 Å and pAB=pBC=0]] [[File:Screenshot_2019-05-15_at_19.25.36.png|right|upright=1.7|thumb|Figure 5. A 2D contour plot that is simulated with all the same settings as figure 4 except it is using the Dynamic setting rather than MEP]]

=== Reactive or Unreactive? ===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.018 || yes ||The system has sufficient momentum for the approach of atom C to the H2 molecule BA to result in the formation of the H2 molecule BC and the free atom A (The system has sufficient momentum to overcome the barrier created by the transition state). ||[[File:Screenshot_2019-05-16_at_14.26.02.png]]
|-
| -1.5 || -2.0 ||-100.456 || no ||The system doesn't have sufficient energy to overcome the transition state barrier. No new products are formed. ||[[File:Screenshot_2019-05-16_at_14.29.21.png]]
|-
| -1.5 || -2.5 ||-98.956 || yes || Similar to the trajectory at the top of the table. The system has sufficient momentum to overcome the transition state and the new products H2 molecule BC and the free atom A are formed. ||[[File:Screenshot_2019-05-16_at_14.39.21.png]]
|-
| -2.5 || -5.0 ||-84.956 ||no || The system has a much greater momentum compared to the previous examples and can overcome the transition state but the high energy system results in the reactants being reformed. ||[[File:Screenshot_2019-05-16_at_14.42.07.png]]
|-
| -2.5 || -5.2 ||-83.416 ||yes ||This is also a high energy system and overcomes the transition state barrier. The reactants are then reformed but these overcome the transition state barrier again to give the product. ||[[File:Screenshot_2019-05-16_at_14.44.38.png]]
|}

This table highlights 2 key ideas:

1) By increasing the momentum (and therefore energy) of the system the transition state barrier can be overcome.

2) For higher energy systems the transition state barrier will be overcome but there is a chance the reactants can reform.

=== Transition State Theory ===

The transition state theory attempts to explain rates of reaction using 3 key assumptions[1]:

1) Once reactants have reached the transition state they cannot return back to being reactants.

2) The distribution of the particles potential energies follows the Boltzmann distribution.

3) The reactants are in an equilibrium with the activated complex (transition state)

My previous results show an obvious discrepancy with the 1st assumption as trajectories have passed back and forth over the transition state. There is also a discrepancy with the 3rd assumption, all previous trajectories have ended by passing over the transition state (in a forwards or backwards direction relative to the reaction itself) and travelling away from the transition state, not returning for any sort of equilibrium.

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

=== PES Inspection ===

==== Classifying the reactions: ====

F + H2

[[File:Screenshot_2019-05-16_at_16.06.02.png]]

This reaction is exothermic.

FH + H

[[File:Screenshot_2019-05-16_at_16.06.52.png]]

This reaction is endothermic.

These conclusions were made by assessing the potential energy surface (PES) for the F-H-H system. Figure 6 shows the PES at an angle that clearly shows how the energy at the bottom of the reaction pathway varies with distance BC (the distance between the 2 H atoms). Effectively forming a 2D representation of potential energy against distance BC. It shows that when BC increases (ie the H-H bond dissociates) the energy of the system becomes more negative, it is stabilised. This is characteristic of an exothermic reaction. In figure 7 the PES is orientated in a way that shows the relationship between the distance AB (the distance between F and HB). Now we see that the energy is raised as the distance AB is increased (ie the F-H bond dissociates). This raising of energy is characteristic of an endothermic reaction, energy needs to be put in for the reaction to occur.

These results are confirmed by the relative strengths of the H-H and H-F bonds. The H-F bond is stronger (565 kJmol-1) than the H-H bond (432 kJmol-1)[2]. In the F + H2 reaction the weaker H-H bond is broken and a stronger H-F bond is formed having a net release of energy therefore making it exothermic (ΔG = -133 kJmol-1). In the reverse direction the opposite occurs making it endothermic.

[[File:Screenshot_2019-05-16_at_16.43.32.png|left|thumb|upright=1.7|Figure 6. Potential energy surface for the F-H-H system orientated to clearly exemplify the relationship between the potential energy and the distance between the 2 H atoms (distance BC)]] [[File:Screenshot_2019-05-16_at_16.48.31.png|right|thumb|upright=1.7|Figure 7. Potential energy surface for the F-H-H system orientated to clearly exemplify the relationship between the potential energy and the distance between the F atom an central H atom (distance AB)]]

==== Location of the transition state: ====

The PES can also be used to find the transition state position of the reaction using a method similar to that of the H + H2 system. Its not quite as simple as though since the transition state will no longer be symmetric as all the atoms aren't the same. Both the momenta are set to 0 and the correct distances between bonds can be approximated by looking for the characteristics of the transition state described in [[#Representation of the transition state on a potential energy surface:|Representation of the transition state on a potential energy surface]] on the PES, these distances can then be adjusted until the trajectory doesn't move. The Hammond postulate can also be used to help approximate the position of the transition state. Considering F + H2 → F-H + H, we have seen that this process is exothermic and it therefore has an early transition state and Hammonds postulate states that similar energy species will be similar shapes[3], we can therefore conclude that the transition state is similar in shape to the reactants. This means that the H-H separation in the transition state will be similar to the H-H bond length (74 pm or 0.74 Å[4]).

Once found the transition state position can be displayed by all horizontal lines on an Internuclear distances vs Time graph (figure 8) or on a 2D contour graph that shows no movement of the trajectory (figure 9).

The F-H separation in the transition state was found to be 1.808 Å (3 d.p.)

The H-H separation in the transition state was found to be 0.745 Å (3 d.p.) (note similarity to the H-H bond length)

[[File:Screenshot_2019-05-16_at_20.38.09.png|left|upright=1.7|thumb|Figure 8. An Internuclear distances vs Time graph showing that with momenta set to 0, F-H separation set to 1.808 Å and H-H separation set to 0.745 Å there is no movement of the atoms, they are all stationary. Indicating that this is the position of the transition state. ]] [[File:Screenshot_2019-05-16_at_20.37.50.png|right|upright=1.7|thumb|Figure 9. A 2D contour plot showing that with momenta set to 0, F-H separation set to 1.808 Å and H-H separation set to 0.745 Å the trajectory is represented by a single point, indicating that this is the transition state position. ]]

=== Calculating activation energies: ===

The 2 activation energies can be found by running an MEP, setting all momenta to 0 and adjusting the size of the H-H separation to the distance in the transition state (0.745 Å) and the F-H separation slightly displaced from its transition state value (1.798 Å and 1.818 Å). By running these 2 MEPs the trajectory can flow down either side of the transition state. Its the Energy vs Time graphs that these simulations produce that are crucial, using these we can find the energy difference between the transition state and the reactants/products by calculating the difference between the highest (ie transition state energy) and lowest energies (Ie reactant/products energies) on the Energy vs Time graphs. The Energy vs Time graphs and their 2 respective 2D contour plot MEPs are shown in figure 10 and figure 11.

[[File:Screenshot 2019-05-17 at 12.45.21.png|centre|upright=3.5|thumb| Figure 10. The Energy vs Time graph and 2D contour plot of the MEP simulated with F-H separation equal to 1.798 Å, the H-H separation set to 0.745 Å and momenta set to 0. This can be used to find the activation energy of the F-H + H → F + H2 reaction. This was found to be 30.1 (2 d.p.).]]

Using the Energy vs Time graph shown above:

Ea for F-H + H → F + H2 = (-103.752)-(-133.868) = 30.1 (2 d.p.)

(Exact values for energies found by zooming in on the graph on the lepsgui.py software)

[[File:Screenshot 2019-05-17 at 13.21.13.png|centre|upright=3.5|thumb| Figure 11. The Energy vs Time graph and 2D contour plot of the MEP simulated with F-H separation equal to 1.818 Å, the H-H separation set to 0.745 Å and momenta set to 0. This can be used to find the activation energy of the F + H2 → F-H + H reaction. This was found to be 0.23 (2 d.p.).]]

Using the Energy vs Time graph shown above:

Ea for F + H2 → F-H + H = (-103.752)-(-103.928) = 0.23 (2 d.p.)

(Exact values for energies again found by zooming in on the graph on the lepsgui.py software)

=== Reaction dynamics: ===

1 set of successful F + H2 → F-H + H reaction conditions are rFH = 2.3 Å, rHH = 0.74 Å, pFH = -2.5 and pHH = -1.5. The results of running a simulation with these conditions are shown below in figure 12.

[[File:Screenshot 2019-05-17 at 14.01.54.png|cetnre|upright=4|thumb|Figure 12. 2D contour plot, Momenta vs Time graph and Energy vs Time graph for a successful F + H2 → F-H + H reaction (rFH = 2.3 Å, rHH = 0.74 Å, pFH = -2.5 and pHH = -1.5). ]]

Considering the Momenta vs Time graph in figure 12, the orange line is initially oscillating with a negative momentum, representative of the H2 molecule possessing vibrational and translational kinetic energy and travelling towards the F atom. Once the reaction is complete (~ at t = 2.0) the orange line becomes positive and flat, representative of the now displaced H atom only possessing translational kinetic energy and travelling in the opposite direction. The blue line also starts of with a negative momentum, the F atom is actually travelling away from the H2 molecule but the F atom is ~ 19 times larger than the H atom and so having a momentum that is marginally larger than that of the H2 molecule means the H2 molecule easily catches up/collides with the F atom. It too is flat, showing that the F atom only possesses translational kinetic energy at this point. When the reaction is complete the blue line then starts to oscillate with a large amplitude, indicative of the large vibrational energy possessed by the F-H bond. The reaction energy has been released as increased vibrational energy. Note the total energy of the system is constant throughout, there is a balance between kinetic and potential energy.

Since the reaction energy is released as vibrational energy IR spectroscopy could be used to monitor this.

=== Polanyi’s rules: ===

Polanyi’s rules state that vibrational energy is better for overcoming a late transition state barrier more efficiently whereas translational energy is better for an early transition state barrier[5].

By varying the initial conditions and utilising the fact that the F + H2 reaction has an early transition state and the F-H + H reaction has a late transition state these rules can be explored.

==== Early transition state ====
Figure 13 displays the effects on varying the amount of translational and vibrational energy in the system on the efficiency of the F + H2 reaction with an early transition state. Both trajectories start at the same position, rFH = 2.3 and rHH = 0.74. The 2D contour plot on the left has a sufficient energy for the reaction to be feasible but a lot of this is in the vibrational energy of the H-H bond (pFH = -0.5 and pHH = 2.4). In this case the reaction is succesful but there is significant recrossing, the reaction isn't very efficient. The 2D contour plot on the right of figure 13 shows what happens when some of the vibrational energy of the H2 molecule is converted into translational energy of the F atom, the magnitude of pFH is increased (from -0.5 to -0.8) but the magnitude of pHH is significantly decreased (from 2.4 to 0.1). In this case the reaction is seen to be successful. These results validify Polanyi’s rule that translational energy overcomes an early transition state more efficiently compared to vibrational energy.

[[File:Screenshot 2019-05-17 at 17.12.23.png|centre|upright=3.5|thumb|Figure 13. The 2D contour plot on the left shows a less efficient trajectory due to considerable H-H vibrational energy and less translational energy of the F atom. The 2D contour plot on the right shows a more efficient trajectory, this time with less H-H vibrational energy and more translational energy of the F atom. Both trajectories start at the same position. ]]

==== Late transition state ====

== References ==

1 - Steinfeld, Jeffrey I., Francisco, Joseph S, and Hase, William L. Chemical Kinetics and Dynamics. Englewood Cliffs, N.J.: Prentice Hall, 1989.

2 - Darwent, B. National Standard Reference Data Series. No. 31 31, (1970).

3 - Hammond, G. S. A Correlation of Reaction Rates. J. Am. Chem. Soc. 77, 334–338 (1955).

4 - CRC Handbook of Chemistry and Physics. (Boca Raton, 2013).

5 - Polanyi, J. C. Concepts in reaction dynamics. Acc. Chem. Res. 5, 161–168 (1972).

MRD:OA010298

2019-05-17T16:13:17Z

Oja16: /* Early transition state */

= Reaction Molecular Dynamics Comp Lab =

==Exercise 1: H + H2 System ==

In this section we are considering a system of 3 hydrogen atoms all aligned on the same line of motion. By altering the distance between A and B (rAB), the distance between B and C (rAB), the momenta between A and B (pAB) and the momenta between B and C (pBC) we can assign initial conditions that simulate a collision between H2 and a H atom.

[[File:Screenshot_2019-05-15_at_21.15.32.png|centre|]]

=== Representation of the transition state on a potential energy surface: ===

[[File:Screenshot_2019-05-14_at_16.28.01.png|thumb|right|upright=1.5|Figure 1. A surface plot of the H + H2 system with the transition state circled]] On the potential energy surface the transition state is the point in the reaction pathway with the highest potential energy, as shown in figure 1. Mathematically, at this point, both <math>\frac{\partial V}{\partial r_1}</math> and <math>\frac{\partial V}{\partial r_2}</math> will equal 0. However this result is true for all points on the reaction pathway. The transition state is the point on this pathway with the highest value of V (potential energy). It is a maximum in a trough of minima.

More correctly the transition state is represented by a saddle point. A saddle point is mathematically defined as satisfying the below equation:

<math>\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1 \partial r_2})^2 < 0</math>

If the point were a minimum then the above result would not be true. It would be greater than 0.

=== Locating the transition state position: ===

We know that the transition state is going to be symmetric (the new H bond is forming as the old H bond is breaking) and so it can be found by setting rAB = rBC and giving them both no initial momenta (pAB=pBC=0). By doing this the trajectory is 'trapped' on the ridge created by the minimum component of the saddle point that represents the transition state. It can only oscillate at the top of this ridge rather than going down either of the sides. By adjusting the value of rAB and rBC maintaining that rAB = rBC a point can be found where the trajectory doesn't oscillate at all (ie. it is stationary), this will be the transition state position (rts).

rts was found to equal 0.908 Å (3 d.p.)

The 2 graphs shown in Figure 2 and 3 display the results of having rAB = rBC=0.908 Å and pAB=pBC=0. Both confirm that the H atoms are stationary and that rts=0.908 Å.
[[File:Screenshot_2019-05-14_at_16.53.26.png|thumb|left|upright=1.7|Figure 2. Internuclear Distance vs Time graph with rAB = rBC=0.908 Å and pAB=pBC=0. The horizontal lines indicate no change in internuclear distances and that the H atoms are stationary ]] [[File:Screenshot_2019-05-14_at_16.52.19.png|thumb|right|upright=1.7|Figure 3. Contour plot with rAB = rBC=0.908 Å and pAB=pBC=0. The trajectory is represented by a single point, again indicating that the H atoms are stationary ]]

=== The minimum energy path: ===

The minimum energy path way is a special type of trajectory that resets all the momenta to 0 for every step of the simulation. By doing this the inertial motion of the gaseous H atoms is removed, resulting in a trajectory that exactly follows the bottom of the reaction pathway, hence the name minimum energy pathway. In order to simulate this the MEP setting is chosen and rAB is set to rts (0.908 Å), rBC is set to rts + δ (0.910 Å) and both momenta are set to 0. This produces the 2D contour plot shown in figure 4.

When the same simulation is run using the Dynamic setting instead of the MEP setting the inertial motion is now taken into account, the particles now have momentum/kinetic energy and are therefore seen to oscillate. This is shown in figure 5, the trajectory follows the reaction pathway but oscillates up and down the sides.

[[File:Screenshot_2019-05-15_at_19.24.43.png|left|thumb|upright=1.7|Figure 4. A 2D contour plot showing the MEP. (rAB = 0.908 Å, rBC = 0.910 Å and pAB=pBC=0]] [[File:Screenshot_2019-05-15_at_19.25.36.png|right|upright=1.7|thumb|Figure 5. A 2D contour plot that is simulated with all the same settings as figure 4 except it is using the Dynamic setting rather than MEP]]

=== Reactive or Unreactive? ===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.018 || yes ||The system has sufficient momentum for the approach of atom C to the H2 molecule BA to result in the formation of the H2 molecule BC and the free atom A (The system has sufficient momentum to overcome the barrier created by the transition state). ||[[File:Screenshot_2019-05-16_at_14.26.02.png]]
|-
| -1.5 || -2.0 ||-100.456 || no ||The system doesn't have sufficient energy to overcome the transition state barrier. No new products are formed. ||[[File:Screenshot_2019-05-16_at_14.29.21.png]]
|-
| -1.5 || -2.5 ||-98.956 || yes || Similar to the trajectory at the top of the table. The system has sufficient momentum to overcome the transition state and the new products H2 molecule BC and the free atom A are formed. ||[[File:Screenshot_2019-05-16_at_14.39.21.png]]
|-
| -2.5 || -5.0 ||-84.956 ||no || The system has a much greater momentum compared to the previous examples and can overcome the transition state but the high energy system results in the reactants being reformed. ||[[File:Screenshot_2019-05-16_at_14.42.07.png]]
|-
| -2.5 || -5.2 ||-83.416 ||yes ||This is also a high energy system and overcomes the transition state barrier. The reactants are then reformed but these overcome the transition state barrier again to give the product. ||[[File:Screenshot_2019-05-16_at_14.44.38.png]]
|}

This table highlights 2 key ideas:

1) By increasing the momentum (and therefore energy) of the system the transition state barrier can be overcome.

2) For higher energy systems the transition state barrier will be overcome but there is a chance the reactants can reform.

=== Transition State Theory ===

The transition state theory attempts to explain rates of reaction using 3 key assumptions[1]:

1) Once reactants have reached the transition state they cannot return back to being reactants.

2) The distribution of the particles potential energies follows the Boltzmann distribution.

3) The reactants are in an equilibrium with the activated complex (transition state)

My previous results show an obvious discrepancy with the 1st assumption as trajectories have passed back and forth over the transition state. There is also a discrepancy with the 3rd assumption, all previous trajectories have ended by passing over the transition state (in a forwards or backwards direction relative to the reaction itself) and travelling away from the transition state, not returning for any sort of equilibrium.

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

=== PES Inspection ===

==== Classifying the reactions: ====

F + H2

[[File:Screenshot_2019-05-16_at_16.06.02.png]]

This reaction is exothermic.

FH + H

[[File:Screenshot_2019-05-16_at_16.06.52.png]]

This reaction is endothermic.

These conclusions were made by assessing the potential energy surface (PES) for the F-H-H system. Figure 6 shows the PES at an angle that clearly shows how the energy at the bottom of the reaction pathway varies with distance BC (the distance between the 2 H atoms). Effectively forming a 2D representation of potential energy against distance BC. It shows that when BC increases (ie the H-H bond dissociates) the energy of the system becomes more negative, it is stabilised. This is characteristic of an exothermic reaction. In figure 7 the PES is orientated in a way that shows the relationship between the distance AB (the distance between F and HB). Now we see that the energy is raised as the distance AB is increased (ie the F-H bond dissociates). This raising of energy is characteristic of an endothermic reaction, energy needs to be put in for the reaction to occur.

These results are confirmed by the relative strengths of the H-H and H-F bonds. The H-F bond is stronger (565 kJmol-1) than the H-H bond (432 kJmol-1)[2]. In the F + H2 reaction the weaker H-H bond is broken and a stronger H-F bond is formed having a net release of energy therefore making it exothermic (ΔG = -133 kJmol-1). In the reverse direction the opposite occurs making it endothermic.

[[File:Screenshot_2019-05-16_at_16.43.32.png|left|thumb|upright=1.7|Figure 6. Potential energy surface for the F-H-H system orientated to clearly exemplify the relationship between the potential energy and the distance between the 2 H atoms (distance BC)]] [[File:Screenshot_2019-05-16_at_16.48.31.png|right|thumb|upright=1.7|Figure 7. Potential energy surface for the F-H-H system orientated to clearly exemplify the relationship between the potential energy and the distance between the F atom an central H atom (distance AB)]]

==== Location of the transition state: ====

The PES can also be used to find the transition state position of the reaction using a method similar to that of the H + H2 system. Its not quite as simple as though since the transition state will no longer be symmetric as all the atoms aren't the same. Both the momenta are set to 0 and the correct distances between bonds can be approximated by looking for the characteristics of the transition state described in [[#Representation of the transition state on a potential energy surface:|Representation of the transition state on a potential energy surface]] on the PES, these distances can then be adjusted until the trajectory doesn't move. The Hammond postulate can also be used to help approximate the position of the transition state. Considering F + H2 → F-H + H, we have seen that this process is exothermic and it therefore has an early transition state and Hammonds postulate states that similar energy species will be similar shapes[3], we can therefore conclude that the transition state is similar in shape to the reactants. This means that the H-H separation in the transition state will be similar to the H-H bond length (74 pm or 0.74 Å[4]).

Once found the transition state position can be displayed by all horizontal lines on an Internuclear distances vs Time graph (figure 8) or on a 2D contour graph that shows no movement of the trajectory (figure 9).

The F-H separation in the transition state was found to be 1.808 Å (3 d.p.)

The H-H separation in the transition state was found to be 0.745 Å (3 d.p.) (note similarity to the H-H bond length)

[[File:Screenshot_2019-05-16_at_20.38.09.png|left|upright=1.7|thumb|Figure 8. An Internuclear distances vs Time graph showing that with momenta set to 0, F-H separation set to 1.808 Å and H-H separation set to 0.745 Å there is no movement of the atoms, they are all stationary. Indicating that this is the position of the transition state. ]] [[File:Screenshot_2019-05-16_at_20.37.50.png|right|upright=1.7|thumb|Figure 9. A 2D contour plot showing that with momenta set to 0, F-H separation set to 1.808 Å and H-H separation set to 0.745 Å the trajectory is represented by a single point, indicating that this is the transition state position. ]]

=== Calculating activation energies: ===

The 2 activation energies can be found by running an MEP, setting all momenta to 0 and adjusting the size of the H-H separation to the distance in the transition state (0.745 Å) and the F-H separation slightly displaced from its transition state value (1.798 Å and 1.818 Å). By running these 2 MEPs the trajectory can flow down either side of the transition state. Its the Energy vs Time graphs that these simulations produce that are crucial, using these we can find the energy difference between the transition state and the reactants/products by calculating the difference between the highest (ie transition state energy) and lowest energies (Ie reactant/products energies) on the Energy vs Time graphs. The Energy vs Time graphs and their 2 respective 2D contour plot MEPs are shown in figure 10 and figure 11.

[[File:Screenshot 2019-05-17 at 12.45.21.png|centre|upright=3.5|thumb| Figure 10. The Energy vs Time graph and 2D contour plot of the MEP simulated with F-H separation equal to 1.798 Å, the H-H separation set to 0.745 Å and momenta set to 0. This can be used to find the activation energy of the F-H + H → F + H2 reaction. This was found to be 30.1 (2 d.p.).]]

Using the Energy vs Time graph shown above:

Ea for F-H + H → F + H2 = (-103.752)-(-133.868) = 30.1 (2 d.p.)

(Exact values for energies found by zooming in on the graph on the lepsgui.py software)

[[File:Screenshot 2019-05-17 at 13.21.13.png|centre|upright=3.5|thumb| Figure 11. The Energy vs Time graph and 2D contour plot of the MEP simulated with F-H separation equal to 1.818 Å, the H-H separation set to 0.745 Å and momenta set to 0. This can be used to find the activation energy of the F + H2 → F-H + H reaction. This was found to be 0.23 (2 d.p.).]]

Using the Energy vs Time graph shown above:

Ea for F + H2 → F-H + H = (-103.752)-(-103.928) = 0.23 (2 d.p.)

(Exact values for energies again found by zooming in on the graph on the lepsgui.py software)

=== Reaction dynamics: ===

1 set of successful F + H2 → F-H + H reaction conditions are rFH = 2.3 Å, rHH = 0.74 Å, pFH = -2.5 and pHH = -1.5. The results of running a simulation with these conditions are shown below in figure 12.

[[File:Screenshot 2019-05-17 at 14.01.54.png|cetnre|upright=4|thumb|Figure 12. 2D contour plot, Momenta vs Time graph and Energy vs Time graph for a successful F + H2 → F-H + H reaction (rFH = 2.3 Å, rHH = 0.74 Å, pFH = -2.5 and pHH = -1.5). ]]

Considering the Momenta vs Time graph in figure 12, the orange line is initially oscillating with a negative momentum, representative of the H2 molecule possessing vibrational and translational kinetic energy and travelling towards the F atom. Once the reaction is complete (~ at t = 2.0) the orange line becomes positive and flat, representative of the now displaced H atom only possessing translational kinetic energy and travelling in the opposite direction. The blue line also starts of with a negative momentum, the F atom is actually travelling away from the H2 molecule but the F atom is ~ 19 times larger than the H atom and so having a momentum that is marginally larger than that of the H2 molecule means the H2 molecule easily catches up/collides with the F atom. It too is flat, showing that the F atom only possesses translational kinetic energy at this point. When the reaction is complete the blue line then starts to oscillate with a large amplitude, indicative of the large vibrational energy possessed by the F-H bond. The reaction energy has been released as increased vibrational energy. Note the total energy of the system is constant throughout, there is a balance between kinetic and potential energy.

Since the reaction energy is released as vibrational energy IR spectroscopy could be used to monitor this.

=== Polanyi’s rules: ===

Polanyi’s rules state that vibrational energy is better for overcoming a late transition state barrier more efficiently whereas translational energy is better for an early transition state barrier[5].

By varying the initial conditions and utilising the fact that the F + H2 reaction has an early transition state and the F-H + H reaction has a late transition state these rules can be explored.

==== Early transition state ====
Figure 13 displays the effects on varying the amount of translational and vibrational energy in the system on the success of the F + H2 reaction with an early transition state. Both trajectories start at the same position, rFH = 2.3 and rHH = 0.74. The 2D contour plot on the left has a sufficient energy for the reaction to be feasible but a lot of this is in the vibrational energy of the H-H bond (pFH = -0.5 and pHH = 2.4). In this case the reaction isn’t successful. The 2D contour plot on the right of figure 13 shows what happens when some of the vibrational energy of the H2 molecule is converted into translational energy of the F atom, the magnitude of pFH is increased (from -0.5 to -0.8) but the magnitude of pHH is significantly decreased (from 2.4 to 0.1). In this case the reaction is seen to be successful. These results validify Polanyi’s rule that translational energy is better at successfully overcoming an early transition state compared to vibrational energy.

[[File:Screenshot 2019-05-17 at 17.12.23.png|centre|upright=3.5|thumb|Figure 13. The 2D contour plot on the left shows an unreactive trajectory due to considerable H-H vibrational energy and less translational energy of the F atom. The 2D contour plot on the right shows a reactive trajectory, this time with less H-H vibrational energy and more translational energy of the F atom. Both trajectories start at the same position. ]]

==== Late transition state ====

== References ==

1 - Steinfeld, Jeffrey I., Francisco, Joseph S, and Hase, William L. Chemical Kinetics and Dynamics. Englewood Cliffs, N.J.: Prentice Hall, 1989.

2 - Darwent, B. National Standard Reference Data Series. No. 31 31, (1970).

3 - Hammond, G. S. A Correlation of Reaction Rates. J. Am. Chem. Soc. 77, 334–338 (1955).

4 - CRC Handbook of Chemistry and Physics. (Boca Raton, 2013).

5 - Polanyi, J. C. Concepts in reaction dynamics. Acc. Chem. Res. 5, 161–168 (1972).

2019-05-17T15:35:59Z

Oja16: /* Polanyi’s rules: */

= Reaction Molecular Dynamics Comp Lab =

==Exercise 1: H + H2 System ==

In this section we are considering a system of 3 hydrogen atoms all aligned on the same line of motion. By altering the distance between A and B (rAB), the distance between B and C (rAB), the momenta between A and B (pAB) and the momenta between B and C (pBC) we can assign initial conditions that simulate a collision between H2 and a H atom.

[[File:Screenshot_2019-05-15_at_21.15.32.png|centre|]]

=== Representation of the transition state on a potential energy surface: ===

[[File:Screenshot_2019-05-14_at_16.28.01.png|thumb|right|upright=1.5|Figure 1. A surface plot of the H + H2 system with the transition state circled]] On the potential energy surface the transition state is the point in the reaction pathway with the highest potential energy, as shown in figure 1. Mathematically, at this point, both <math>\frac{\partial V}{\partial r_1}</math> and <math>\frac{\partial V}{\partial r_2}</math> will equal 0. However this result is true for all points on the reaction pathway. The transition state is the point on this pathway with the highest value of V (potential energy). It is a maximum in a trough of minima.

More correctly the transition state is represented by a saddle point. A saddle point is mathematically defined as satisfying the below equation:

<math>\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1 \partial r_2})^2 < 0</math>

If the point were a minimum then the above result would not be true. It would be greater than 0.

=== Locating the transition state position: ===

We know that the transition state is going to be symmetric (the new H bond is forming as the old H bond is breaking) and so it can be found by setting rAB = rBC and giving them both no initial momenta (pAB=pBC=0). By doing this the trajectory is 'trapped' on the ridge created by the minimum component of the saddle point that represents the transition state. It can only oscillate at the top of this ridge rather than going down either of the sides. By adjusting the value of rAB and rBC maintaining that rAB = rBC a point can be found where the trajectory doesn't oscillate at all (ie. it is stationary), this will be the transition state position (rts).

rts was found to equal 0.908 Å (3 d.p.)

The 2 graphs shown in Figure 2 and 3 display the results of having rAB = rBC=0.908 Å and pAB=pBC=0. Both confirm that the H atoms are stationary and that rts=0.908 Å.
[[File:Screenshot_2019-05-14_at_16.53.26.png|thumb|left|upright=1.7|Figure 2. Internuclear Distance vs Time graph with rAB = rBC=0.908 Å and pAB=pBC=0. The horizontal lines indicate no change in internuclear distances and that the H atoms are stationary ]] [[File:Screenshot_2019-05-14_at_16.52.19.png|thumb|right|upright=1.7|Figure 3. Contour plot with rAB = rBC=0.908 Å and pAB=pBC=0. The trajectory is represented by a single point, again indicating that the H atoms are stationary ]]

=== The minimum energy path: ===

The minimum energy path way is a special type of trajectory that resets all the momenta to 0 for every step of the simulation. By doing this the inertial motion of the gaseous H atoms is removed, resulting in a trajectory that exactly follows the bottom of the reaction pathway, hence the name minimum energy pathway. In order to simulate this the MEP setting is chosen and rAB is set to rts (0.908 Å), rBC is set to rts + δ (0.910 Å) and both momenta are set to 0. This produces the 2D contour plot shown in figure 4.

When the same simulation is run using the Dynamic setting instead of the MEP setting the inertial motion is now taken into account, the particles now have momentum/kinetic energy and are therefore seen to oscillate. This is shown in figure 5, the trajectory follows the reaction pathway but oscillates up and down the sides.

[[File:Screenshot_2019-05-15_at_19.24.43.png|left|thumb|upright=1.7|Figure 4. A 2D contour plot showing the MEP. (rAB = 0.908 Å, rBC = 0.910 Å and pAB=pBC=0]] [[File:Screenshot_2019-05-15_at_19.25.36.png|right|upright=1.7|thumb|Figure 5. A 2D contour plot that is simulated with all the same settings as figure 4 except it is using the Dynamic setting rather than MEP]]

=== Reactive or Unreactive? ===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.018 || yes ||The system has sufficient momentum for the approach of atom C to the H2 molecule BA to result in the formation of the H2 molecule BC and the free atom A (The system has sufficient momentum to overcome the barrier created by the transition state). ||[[File:Screenshot_2019-05-16_at_14.26.02.png]]
|-
| -1.5 || -2.0 ||-100.456 || no ||The system doesn't have sufficient energy to overcome the transition state barrier. No new products are formed. ||[[File:Screenshot_2019-05-16_at_14.29.21.png]]
|-
| -1.5 || -2.5 ||-98.956 || yes || Similar to the trajectory at the top of the table. The system has sufficient momentum to overcome the transition state and the new products H2 molecule BC and the free atom A are formed. ||[[File:Screenshot_2019-05-16_at_14.39.21.png]]
|-
| -2.5 || -5.0 ||-84.956 ||no || The system has a much greater momentum compared to the previous examples and can overcome the transition state but the high energy system results in the reactants being reformed. ||[[File:Screenshot_2019-05-16_at_14.42.07.png]]
|-
| -2.5 || -5.2 ||-83.416 ||yes ||This is also a high energy system and overcomes the transition state barrier. The reactants are then reformed but these overcome the transition state barrier again to give the product. ||[[File:Screenshot_2019-05-16_at_14.44.38.png]]
|}

This table highlights 2 key ideas:

1) By increasing the momentum (and therefore energy) of the system the transition state barrier can be overcome.

2) For higher energy systems the transition state barrier will be overcome but there is a chance the reactants can reform.

=== Transition State Theory ===

The transition state theory attempts to explain rates of reaction using 3 key assumptions[1]:

1) Once reactants have reached the transition state they cannot return back to being reactants.

2) The distribution of the particles potential energies follows the Boltzmann distribution.

3) The reactants are in an equilibrium with the activated complex (transition state)

My previous results show an obvious discrepancy with the 1st assumption as trajectories have passed back and forth over the transition state. There is also a discrepancy with the 3rd assumption, all previous trajectories have ended by passing over the transition state (in a forwards or backwards direction relative to the reaction itself) and travelling away from the transition state, not returning for any sort of equilibrium.

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

=== PES Inspection ===

==== Classifying the reactions: ====

F + H2

[[File:Screenshot_2019-05-16_at_16.06.02.png]]

This reaction is exothermic.

FH + H

[[File:Screenshot_2019-05-16_at_16.06.52.png]]

This reaction is endothermic.

These conclusions were made by assessing the potential energy surface (PES) for the F-H-H system. Figure 6 shows the PES at an angle that clearly shows how the energy at the bottom of the reaction pathway varies with distance BC (the distance between the 2 H atoms). Effectively forming a 2D representation of potential energy against distance BC. It shows that when BC increases (ie the H-H bond dissociates) the energy of the system becomes more negative, it is stabilised. This is characteristic of an exothermic reaction. In figure 7 the PES is orientated in a way that shows the relationship between the distance AB (the distance between F and HB). Now we see that the energy is raised as the distance AB is increased (ie the F-H bond dissociates). This raising of energy is characteristic of an endothermic reaction, energy needs to be put in for the reaction to occur.

These results are confirmed by the relative strengths of the H-H and H-F bonds. The H-F bond is stronger (565 kJmol-1) than the H-H bond (432 kJmol-1)[2]. In the F + H2 reaction the weaker H-H bond is broken and a stronger H-F bond is formed having a net release of energy therefore making it exothermic (ΔG = -133 kJmol-1). In the reverse direction the opposite occurs making it endothermic.

[[File:Screenshot_2019-05-16_at_16.43.32.png|left|thumb|upright=1.7|Figure 6. Potential energy surface for the F-H-H system orientated to clearly exemplify the relationship between the potential energy and the distance between the 2 H atoms (distance BC)]] [[File:Screenshot_2019-05-16_at_16.48.31.png|right|thumb|upright=1.7|Figure 7. Potential energy surface for the F-H-H system orientated to clearly exemplify the relationship between the potential energy and the distance between the F atom an central H atom (distance AB)]]

==== Location of the transition state: ====

The PES can also be used to find the transition state position of the reaction using a method similar to that of the H + H2 system. Its not quite as simple as though since the transition state will no longer be symmetric as all the atoms aren't the same. Both the momenta are set to 0 and the correct distances between bonds can be approximated by looking for the characteristics of the transition state described in [[#Representation of the transition state on a potential energy surface:|Representation of the transition state on a potential energy surface]] on the PES, these distances can then be adjusted until the trajectory doesn't move. The Hammond postulate can also be used to help approximate the position of the transition state. Considering F + H2 → F-H + H, we have seen that this process is exothermic and it therefore has an early transition state and Hammonds postulate states that similar energy species will be similar shapes[3], we can therefore conclude that the transition state is similar in shape to the reactants. This means that the H-H separation in the transition state will be similar to the H-H bond length (74 pm or 0.74 Å[4]).

Once found the transition state position can be displayed by all horizontal lines on an Internuclear distances vs Time graph (figure 8) or on a 2D contour graph that shows no movement of the trajectory (figure 9).

The F-H separation in the transition state was found to be 1.808 Å (3 d.p.)

The H-H separation in the transition state was found to be 0.745 Å (3 d.p.) (note similarity to the H-H bond length)

[[File:Screenshot_2019-05-16_at_20.38.09.png|left|upright=1.7|thumb|Figure 8. An Internuclear distances vs Time graph showing that with momenta set to 0, F-H separation set to 1.808 Å and H-H separation set to 0.745 Å there is no movement of the atoms, they are all stationary. Indicating that this is the position of the transition state. ]] [[File:Screenshot_2019-05-16_at_20.37.50.png|right|upright=1.7|thumb|Figure 9. A 2D contour plot showing that with momenta set to 0, F-H separation set to 1.808 Å and H-H separation set to 0.745 Å the trajectory is represented by a single point, indicating that this is the transition state position. ]]

=== Calculating activation energies: ===

The 2 activation energies can be found by running an MEP, setting all momenta to 0 and adjusting the size of the H-H separation to the distance in the transition state (0.745 Å) and the F-H separation slightly displaced from its transition state value (1.798 Å and 1.818 Å). By running these 2 MEPs the trajectory can flow down either side of the transition state. Its the Energy vs Time graphs that these simulations produce that are crucial, using these we can find the energy difference between the transition state and the reactants/products by calculating the difference between the highest (ie transition state energy) and lowest energies (Ie reactant/products energies) on the Energy vs Time graphs. The Energy vs Time graphs and their 2 respective 2D contour plot MEPs are shown in figure 10 and figure 11.

[[File:Screenshot 2019-05-17 at 12.45.21.png|centre|upright=3.5|thumb| Figure 10. The Energy vs Time graph and 2D contour plot of the MEP simulated with F-H separation equal to 1.798 Å, the H-H separation set to 0.745 Å and momenta set to 0. This can be used to find the activation energy of the F-H + H → F + H2 reaction. This was found to be 30.1 (2 d.p.).]]

Using the Energy vs Time graph shown above:

Ea for F-H + H → F + H2 = (-103.752)-(-133.868) = 30.1 (2 d.p.)

(Exact values for energies found by zooming in on the graph on the lepsgui.py software)

[[File:Screenshot 2019-05-17 at 13.21.13.png|centre|upright=3.5|thumb| Figure 11. The Energy vs Time graph and 2D contour plot of the MEP simulated with F-H separation equal to 1.818 Å, the H-H separation set to 0.745 Å and momenta set to 0. This can be used to find the activation energy of the F + H2 → F-H + H reaction. This was found to be 0.23 (2 d.p.).]]

Using the Energy vs Time graph shown above:

Ea for F + H2 → F-H + H = (-103.752)-(-103.928) = 0.23 (2 d.p.)

(Exact values for energies again found by zooming in on the graph on the lepsgui.py software)

==== Reaction dynamics: ====

1 set of successful F + H2 → F-H + H reaction conditions are rFH = 2.3 Å, rHH = 0.74 Å, pFH = -2.5 and pHH = -1.5. The results of running a simulation with these conditions are shown below in figure 12.

[[File:Screenshot 2019-05-17 at 14.01.54.png|cetnre|upright=4|thumb|Figure 12. 2D contour plot, Momenta vs Time graph and Energy vs Time graph for a successful F + H2 → F-H + H reaction (rFH = 2.3 Å, rHH = 0.74 Å, pFH = -2.5 and pHH = -1.5). ]]

Considering the Momenta vs Time graph in figure 12, the orange line is initially oscillating with a negative momentum, representative of the H2 molecule possessing vibrational and translational kinetic energy and travelling towards the F atom. Once the reaction is complete (~ at t = 2.0) the orange line becomes positive and flat, representative of the now displaced H atom only possessing translational kinetic energy and travelling in the opposite direction. The blue line also starts of with a negative momentum, the F atom is actually travelling away from the H2 molecule but the F atom is ~ 19 times larger than the H atom and so having a momentum that is marginally larger than that of the H2 molecule means the H2 molecule easily catches up/collides with the F atom. It too is flat, showing that the F atom only possesses translational kinetic energy at this point. When the reaction is complete the blue line then starts to oscillate with a large amplitude, indicative of the large vibrational energy possessed by the F-H bond. The reaction energy has been released as increased vibrational energy. Note the total energy of the system is constant throughout, there is a balance between kinetic and potential energy.

Since the reaction energy is released as vibrational energy IR spectroscopy could be used to monitor this.

==== Polanyi’s rules: ====

Polanyi’s rules state that vibrational energy is better for overcoming a late transition state barrier whereas translational energy is better at overcoming an early transition state barrier[5].

By varying the initial conditions and utilising the fact that the F + H2 reaction has an early transition state and the F-H + H reaction has a late transition state these rules can be explored.

Figure 13 displays the effects on varying the amount of translational and vibrational energy in the system on the success of the F + H2 reaction with an early transition state. Both trajectories start at the same position, rFH = 2 and rHH = 0.74. The 2D contour plot on the left has a sufficient energy for the reaction to be feasible but a lot of this is in the vibrational energy of the H-H bond (pFH = -0.5 and pHH = 2.4). In this case the reaction isn’t successful. The 2D contour plot on the right of figure 13 shows what happens when some of the vibrational energy of the H2 molecule is converted into translational energy of the F atom, the magnitude of pFH is increased (from -0.5 to -0.8) but the magnitude of pHH is significantly decreased (from 2.4 to 0.1). In this case the reaction is seen to be successful. These results validify Polanyi’s rule that translational energy is better at successfully overcoming an early transition state compared to vibrational energy.

[[File:Screenshot 2019-05-17 at 16.02.09.png|centre|upright=3.5|thumb|Figure 13. The 2D contour plot on the left shows an unreactive trajectory due to considerable H-H vibrational energy and less translational energy of the F atom. The 2D contour plot on the right shows a reactive trajectory, this time with less H-H vibrational energy and more translational energy of the F atom. Both trajectories start at the same position. ]]

2019-05-17T13:09:04Z

Oja16: /* Exercise 2: F-H-H System */

= Reaction Molecular Dynamics Comp Lab =

==Exercise 1: H + H2 System ==

In this section we are considering a system of 3 hydrogen atoms all aligned on the same line of motion. By altering the distance between A and B (rAB), the distance between B and C (rAB), the momenta between A and B (pAB) and the momenta between B and C (pBC) we can assign initial conditions that simulate a collision between H2 and a H atom.

[[File:Screenshot_2019-05-15_at_21.15.32.png|centre|]]

=== Representation of the transition state on a potential energy surface: ===

[[File:Screenshot_2019-05-14_at_16.28.01.png|thumb|right|upright=1.5|Figure 1. A surface plot of the H + H2 system with the transition state circled]] On the potential energy surface the transition state is the point in the reaction pathway with the highest potential energy, as shown in figure 1. Mathematically, at this point, both <math>\frac{\partial V}{\partial r_1}</math> and <math>\frac{\partial V}{\partial r_2}</math> will equal 0. However this result is true for all points on the reaction pathway. The transition state is the point on this pathway with the highest value of V (potential energy). It is a maximum in a trough of minima.

More correctly the transition state is represented by a saddle point. A saddle point is mathematically defined as satisfying the below equation:

<math>\frac{\partial^2 V}{\partial r_1^2} \frac{\partial^2 V}{\partial r_2^2} - (\frac{\partial^2 V}{\partial r_1 \partial r_2})^2 < 0</math>

If the point were a minimum then the above result would not be true. It would be greater than 0.

=== Locating the transition state position: ===

We know that the transition state is going to be symmetric (the new H bond is forming as the old H bond is breaking) and so it can be found by setting rAB = rBC and giving them both no initial momenta (pAB=pBC=0). By doing this the trajectory is 'trapped' on the ridge created by the minimum component of the saddle point that represents the transition state. It can only oscillate at the top of this ridge rather than going down either of the sides. By adjusting the value of rAB and rBC maintaining that rAB = rBC a point can be found where the trajectory doesn't oscillate at all (ie. it is stationary), this will be the transition state position (rts).

rts was found to equal 0.908 Å (3 d.p.)

The 2 graphs shown in Figure 2 and 3 display the results of having rAB = rBC=0.908 Å and pAB=pBC=0. Both confirm that the H atoms are stationary and that rts=0.908 Å.
[[File:Screenshot_2019-05-14_at_16.53.26.png|thumb|left|upright=1.7|Figure 2. Internuclear Distance vs Time graph with rAB = rBC=0.908 Å and pAB=pBC=0. The horizontal lines indicate no change in internuclear distances and that the H atoms are stationary ]] [[File:Screenshot_2019-05-14_at_16.52.19.png|thumb|right|upright=1.7|Figure 3. Contour plot with rAB = rBC=0.908 Å and pAB=pBC=0. The trajectory is represented by a single point, again indicating that the H atoms are stationary ]]

=== The minimum energy path: ===

The minimum energy path way is a special type of trajectory that resets all the momenta to 0 for every step of the simulation. By doing this the inertial motion of the gaseous H atoms is removed, resulting in a trajectory that exactly follows the bottom of the reaction pathway, hence the name minimum energy pathway. In order to simulate this the MEP setting is chosen and rAB is set to rts (0.908 Å), rBC is set to rts + δ (0.910 Å) and both momenta are set to 0. This produces the 2D contour plot shown in figure 4.

When the same simulation is run using the Dynamic setting instead of the MEP setting the inertial motion is now taken into account, the particles now have momentum/kinetic energy and are therefore seen to oscillate. This is shown in figure 5, the trajectory follows the reaction pathway but oscillates up and down the sides.

[[File:Screenshot_2019-05-15_at_19.24.43.png|left|thumb|upright=1.7|Figure 4. A 2D contour plot showing the MEP. (rAB = 0.908 Å, rBC = 0.910 Å and pAB=pBC=0]] [[File:Screenshot_2019-05-15_at_19.25.36.png|right|upright=1.7|thumb|Figure 5. A 2D contour plot that is simulated with all the same settings as figure 4 except it is using the Dynamic setting rather than MEP]]

=== Reactive or Unreactive? ===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.018 || yes ||The system has sufficient momentum for the approach of atom C to the H2 molecule BA to result in the formation of the H2 molecule BC and the free atom A (The system has sufficient momentum to overcome the barrier created by the transition state). ||[[File:Screenshot_2019-05-16_at_14.26.02.png]]
|-
| -1.5 || -2.0 ||-100.456 || no ||The system doesn't have sufficient energy to overcome the transition state barrier. No new products are formed. ||[[File:Screenshot_2019-05-16_at_14.29.21.png]]
|-
| -1.5 || -2.5 ||-98.956 || yes || Similar to the trajectory at the top of the table. The system has sufficient momentum to overcome the transition state and the new products H2 molecule BC and the free atom A are formed. ||[[File:Screenshot_2019-05-16_at_14.39.21.png]]
|-
| -2.5 || -5.0 ||-84.956 ||no || The system has a much greater momentum compared to the previous examples and can overcome the transition state but the high energy system results in the reactants being reformed. ||[[File:Screenshot_2019-05-16_at_14.42.07.png]]
|-
| -2.5 || -5.2 ||-83.416 ||yes ||This is also a high energy system and overcomes the transition state barrier. The reactants are then reformed but these overcome the transition state barrier again to give the product. ||[[File:Screenshot_2019-05-16_at_14.44.38.png]]
|}

This table highlights 2 key ideas:

1) By increasing the momentum (and therefore energy) of the system the transition state barrier can be overcome.

2) For higher energy systems the transition state barrier will be overcome but there is a chance the reactants can reform.

=== Transition State Theory ===

The transition state theory attempts to explain rates of reaction using 3 key assumptions[1]:

1) Once reactants have reached the transition state they cannot return back to being reactants.

2) The distribution of the particles potential energies follows the Boltzmann distribution.

3) The reactants are in an equilibrium with the activated complex (transition state)

My previous results show an obvious discrepancy with the 1st assumption as trajectories have passed back and forth over the transition state. There is also a discrepancy with the 3rd assumption, all previous trajectories have ended by passing over the transition state (in a forwards or backwards direction relative to the reaction itself) and travelling away from the transition state, not returning for any sort of equilibrium.

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

=== PES Inspection ===

==== Classifying the reactions: ====

F + H2

[[File:Screenshot_2019-05-16_at_16.06.02.png]]

This reaction is exothermic.

FH + H

[[File:Screenshot_2019-05-16_at_16.06.52.png]]

This reaction is endothermic.

These conclusions were made by assessing the potential energy surface (PES) for the F-H-H system. Figure 6 shows the PES at an angle that clearly shows how the energy at the bottom of the reaction pathway varies with distance BC (the distance between the 2 H atoms). Effectively forming a 2D representation of potential energy against distance BC. It shows that when BC increases (ie the H-H bond dissociates) the energy of the system becomes more negative, it is stabilised. This is characteristic of an exothermic reaction. In figure 7 the PES is orientated in a way that shows the relationship between the distance AB (the distance between F and HB). Now we see that the energy is raised as the distance AB is increased (ie the F-H bond dissociates). This raising of energy is characteristic of an endothermic reaction, energy needs to be put in for the reaction to occur.

These results are confirmed by the relative strengths of the H-H and H-F bonds. The H-F bond is stronger (565 kJmol-1) than the H-H bond (432 kJmol-1)[2]. In the F + H2 reaction the weaker H-H bond is broken and a stronger H-F bond is formed having a net release of energy therefore making it exothermic (ΔG = -133 kJmol-1). In the reverse direction the opposite occurs making it endothermic.

[[File:Screenshot_2019-05-16_at_16.43.32.png|left|thumb|upright=1.7|Figure 6. Potential energy surface for the F-H-H system orientated to clearly exemplify the relationship between the potential energy and the distance between the 2 H atoms (distance BC)]] [[File:Screenshot_2019-05-16_at_16.48.31.png|right|thumb|upright=1.7|Figure 7. Potential energy surface for the F-H-H system orientated to clearly exemplify the relationship between the potential energy and the distance between the F atom an central H atom (distance AB)]]

==== Location of the transition state: ====

The PES can also be used to find the transition state position of the reaction using a method similar to that of the H + H2 system. Its not quite as simple as though since the transition state will no longer be symmetric as all the atoms aren't the same. Both the momenta are set to 0 and the correct distances between bonds can be approximated by looking for the characteristics of the transition state described in [[#Representation of the transition state on a potential energy surface:|Representation of the transition state on a potential energy surface]] on the PES, these distances can then be adjusted until the trajectory doesn't move. The Hammond postulate can also be used to help approximate the position of the transition state. Considering F + H2 → F-H + H, we have seen that this process is exothermic and it therefore has an early transition state and Hammonds postulate states that similar energy species will be similar shapes[3], we can therefore conclude that the transition state is similar in shape to the reactants. This means that the H-H separation in the transition state will be similar to the H-H bond length (74 pm or 0.74 Å[4]).

Once found the transition state position can be displayed by all horizontal lines on an Internuclear distances vs Time graph (figure 8) or on a 2D contour graph that shows no movement of the trajectory (figure 9).

The F-H separation in the transition state was found to be 1.808 Å (3 d.p.)

The H-H separation in the transition state was found to be 0.745 Å (3 d.p.) (note similarity to the H-H bond length)

[[File:Screenshot_2019-05-16_at_20.38.09.png|left|upright=1.7|thumb|Figure 8. An Internuclear distances vs Time graph showing that with momenta set to 0, F-H separation set to 1.808 Å and H-H separation set to 0.745 Å there is no movement of the atoms, they are all stationary. Indicating that this is the position of the transition state. ]] [[File:Screenshot_2019-05-16_at_20.37.50.png|right|upright=1.7|thumb|Figure 9. A 2D contour plot showing that with momenta set to 0, F-H separation set to 1.808 Å and H-H separation set to 0.745 Å the trajectory is represented by a single point, indicating that this is the transition state position. ]]

=== Calculating activation energies: ===

The 2 activation energies can be found by running an MEP, setting all momenta to 0 and adjusting the size of the H-H separation to the distance in the transition state (0.745 Å) and the F-H separation slightly displaced from its transition state value (1.798 Å and 1.818 Å). By running these 2 MEPs the trajectory can flow down either side of the transition state. Its the Energy vs Time graphs that these simulations produce that are crucial, using these we can find the energy difference between the transition state and the reactants/products by calculating the difference between the highest (ie transition state energy) and lowest energies (Ie reactant/products energies) on the Energy vs Time graphs. The Energy vs Time graphs and their 2 respective 2D contour plot MEPs are shown in figure 10 and figure 11.

[[File:Screenshot 2019-05-17 at 12.45.21.png|centre|upright=3.5|thumb| Figure 10. The Energy vs Time graph and 2D contour plot of the MEP simulated with F-H separation equal to 1.798 Å, the H-H separation set to 0.745 Å and momenta set to 0. This can be used to find the activation energy of the F-H + H → F + H2 reaction. This was found to be 30.1 (2 d.p.).]]

Using the Energy vs Time graph shown above:

Ea for F-H + H → F + H2 = (-103.752)-(-133.868) = 30.1 (2 d.p.)

(Exact values for energies found by zooming in on the graph on the lepsgui.py software)

[[File:Screenshot 2019-05-17 at 13.21.13.png|centre|upright=3.5|thumb| Figure 11. The Energy vs Time graph and 2D contour plot of the MEP simulated with F-H separation equal to 1.818 Å, the H-H separation set to 0.745 Å and momenta set to 0. This can be used to find the activation energy of the F + H2 → F-H + H reaction. This was found to be 0.23 (2 d.p.).]]

Using the Energy vs Time graph shown above:

Ea for F + H2 → F-H + H = (-103.752)-(-103.928) = 0.23 (2 d.p.)

(Exact values for energies again found by zooming in on the graph on the lepsgui.py software)

==== Reaction dynamics: ====

[[File:Screenshot 2019-05-17 at 14.01.54.png|cetnre|upright=3.5|thumb|Figure 12. 2D contour plot, Momenta vs Time graph and Energy vs Time graph for a successful F + H2 → F-H + H reaction. rFH = 2.3 Å, rHH = 0.74 Å, pFH = -2.5 and pHH = -1.5. ]]

File:Screenshot 2019-05-17 at 14.01.54.png

2019-05-17T13:02:24Z

Oja16:

MRD:OA010298

2019-05-17T12:26:02Z

Oja16: /* Calculating activation energies: */

MRD:OA010298

2019-05-17T12:25:30Z

Oja16: /* Calculating activation energies: */

MRD:OA010298

2019-05-17T12:25:13Z

Oja16: /* Calculating activation energies: */

File:Screenshot 2019-05-17 at 13.21.13.png

2019-05-17T12:21:32Z

Oja16: