2019-05-30T21:09:55Z

Pu12:

= 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 momentum between A and B (pAB) and the momentum 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 {{fontcolor1|red|So the entire reaction pathway has no gradient??. [[User:Pu12|Pu12]] ([[User talk:Pu12|talk]]) 22:09, 30 May 2019 (BST)}}. 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. {{fontcolor1|red|The transition state is both a maximum and a minimum depending on direction considered, unlike a local minimum which is always a minimum. It looks like you copied the equation from somewhere, it's correct but, it's easier to consider the TS point as somewhere where the second derivative of potential energy with respect to time is <0 in one direction and >0 in the other. [[User:Pu12|Pu12]] ([[User talk:Pu12|talk]]) 22:09, 30 May 2019 (BST)}}

=== 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 ]]

{{fontcolor1|red|Good. [[User:Pu12|Pu12]] ([[User talk:Pu12|talk]]) 22:09, 30 May 2019 (BST)}}

=== 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 (instead of Dynamic) 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]]
|}

{{fontcolor1|red|Good discussions but you should also mention the vibrations. [[User:Pu12|Pu12]] ([[User talk:Pu12|talk]]) 22:09, 30 May 2019 (BST)}}
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 after this.

=== 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 a constant 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, where there is significant recrossing over the transition state barrier it is reasonable to approximate an equilibrium between reactants and the transition state but all trajectories end by travelling infinitely far away from the transition state. In these examples the equilibrium will never be regained.

{{fontcolor1|red|Good, there is also an assumption about quantum tunnelling but its effect is minuscule. The discrepancy you mention with the 3rd reaction is a quirk of the computations, I think that the question means in-lab obtained values for experimental values where the reaction would be contained in a vessel (although this is unclear so your answer is fine). [[User:Pu12|Pu12]] ([[User talk:Pu12|talk]]) 22:09, 30 May 2019 (BST)}}

== 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)]]

{{fontcolor1|red|Good explanation. [[User:Pu12|Pu12]] ([[User talk:Pu12|talk]]) 22:09, 30 May 2019 (BST)}}

==== 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. F + H2 → F-H + H is an exothermic process and therefore has an early transition state. 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 respective MEP 2D contour plots 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.)
{{fontcolor1|red|Correct, but units are in kcal/mol. [[User:Pu12|Pu12]] ([[User talk:Pu12|talk]]) 22:09, 30 May 2019 (BST)}}

(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)

The larger Ea for F-H + H → F + H2 makes sense as this process is endothermic and therefore has a late transition state, the kinetic barrier is high (F + H2 → F-H + H is exothermic and has an early transition state).

=== 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 ~ 10 times larger than the H2 molecule 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.
{{fontcolor1|red|Good, calorimetry could also be used because the vibrational energy would be picked up as heat and this would likely be an easier method. [[User:Pu12|Pu12]] ([[User talk:Pu12|talk]]) 22:09, 30 May 2019 (BST)}}

=== Polanyi’s rules: ===

Polanyi’s rules state that vibrational energy will overcome a late transition state barrier more efficiently than translational energy 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 with minimal recrossing. 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 F-H + H reaction can be observed. The plot on the left has higher vibrational energy (pFH = -6.674) compared to the simulation on the right (pFH = 1.108), it is also less efficient in crossing the transition state barrier (the translational energies were similar pHH = -1.016 on the left and pHH = -1.171 on the right). This actually contradicts Polanyi's rules as it suggests the lower vibrational energy system is more efficient at crossing the late transition state.

[[File:Screenshot 2019-05-17 at 17.30.58.png|centre|upright=3.5|thumb|Figure 14. The 2 2D plots acquired using the inversion of momentum method, the left having higher vibrational energy than the right.]]

Acquiring simulations that clearly depicted Polanyi's rule for late transition states proved difficult as the inversion of momentum procedure dictated which starting conditions were feasible. This limited control over how much relative vibrational and translational energy you could give the system.

{{fontcolor1|red|Good discussion, it's difficult to get contour plots to demonstrate the efficiency of reaction clearly, you don't need to use them for this. [[User:Pu12|Pu12]] ([[User talk:Pu12|talk]]) 22:09, 30 May 2019 (BST)}}

{{fontcolor1|red|Good report overall. [[User:Pu12|Pu12]] ([[User talk:Pu12|talk]]) 22:09, 30 May 2019 (BST)}}
= 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:HNA4117

2019-05-30T20:18:21Z

Pu12:

== H + H2 ''' '''system ==

=== Transition State on the PES ===
The potential energy surface has a transition state that is a saddle point - it can be both a maxima and a minima relative to the direction you are looking at (shown in the pictures below). Whether it is a maxima or a minima, its gradient is always 0. The TS can be distinguished from a local minima on the PES using a second derivative of the gradient.
{{fontcolor1|red|How do you use the second derivative of the gradient? [[User:Pu12|Pu12]] ([[User talk:Pu12|talk]]) 21:18, 30 May 2019 (BST)}}

[[File:Hna4117_Maxima.PNG]] [[File:Minima_hna4117.PNG]]
{{fontcolor1|red|I'm not sure what you mean by this diagram. [[User:Pu12|Pu12]] ([[User talk:Pu12|talk]]) 21:18, 30 May 2019 (BST)}}

=== TS value - locating the TS ===
0.91 seems to be a good estimate of the bond distance of TS where the atoms don't seem to fall off. When increased or decreased slightly to 0.90 or to 0.92 respectively, it seems to be inclined to one side or the other with the potential to fall off.
[[File:Hna4117_0.91.PNG]] [[File:Hna4117_0.91(2).PNG]]

The pictures above a contour plot and a surface plot when r1=r2=0.91
{{fontcolor1|red|Correct, but you should include an intermolecular distances vs time plot as the question asks. [[User:Pu12|Pu12]] ([[User talk:Pu12|talk]]) 21:18, 30 May 2019 (BST)}}

[[File:0.90_hna4117.PNG]] [[File:0.90_hna4117(2).PNG]]

The pictures above a contour plot and a surface plot when r1=r2=0.90. The atoms seems to be inclined to fall off on one side of the ridge.

=== Calculating the reaction path (MEP vs Dynamic) ===
[[File:Hna4117_reactionpath1.PNG]] [[File:Hna4117_reactionpath2.PNG]]

The picture on the left shows the trajectory of atoms run on a dynamic energy path and the right picture, run on MEP. MEP is the lowest energy path and it shows a straight line trajectory which deviates from the true energy path. When calculating this, r1 was maintained at the TS length while r2 was increased by +0.01 to 0.92. The MEP produces a straight line because it disregards the interactions (vibrations) between the atoms.

{{fontcolor1|red|What causes the MEP to disregard vibrations between the atoms? You should mention how re-setting the momentum to 0 after each step in the MEP causes the path to stay on the valley floor (no vibration). [[User:Pu12|Pu12]] ([[User talk:Pu12|talk]]) 21:18, 30 May 2019 (BST)}}

[[File:Hna4117_intvstime.PNG]]

The picture above is the internuclear distance vs time plot.
For the very first few steps, the bond distances remained constant. Then, the internuclear distance between B-C quickly decreases then plateaus indicating that bond is formed and remained that way. The bond distances between A-B and A-C increases at the same rate showing bonds are broken and they are moving away from each other at the same rate.

=== Reactive and Unreactive Trajectories ===
{| class="wikitable"
!p1
!p2
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.02</nowiki>
|Yes
|The reaction path proceeds towards product formation. The path started with a straight line indicating the diatomic molecule is stationary. After atom A attacks, the newly formed A-B diatomic molecule then vibrates.
|[[File:Hna4117_first.PNG]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.46</nowiki>
|No
|The reaction path does not proceed towards product formation. The atoms do not have enough energy to overcome the activation barrier so reaction does not occur. Atom A and diatom B-C then vibrates away from each other.
|[[File:Hna4117_second.PNG]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.96</nowiki>
|Yes
|The reaction path proceeds toward product formation. The wavy reaction path shows that atom A attacks diatom B-C while the latter is vibrating. After bond B-C break, the newly form A-B bond vibrates.
|[[File:Hna4117_third.PNG]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.96</nowiki>
|No
|The reaction path does not proceed toward product formation. It crosses the TS path which means it forms a TS complex but then did not proceed further and dissociates. Atom A and diatom B-C vibrate away from each other.
|[[File:Hna4117_fourth.PNG]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.42</nowiki>
|Yes
|The reaction path proceeds toward product formation. The extremely wavy reaction path seems like this is a high energy reaction.
|[[File:Hna4117_fifth.PNG]]
|}
{{fontcolor1|red|What can you conclude from the table? [[User:Pu12|Pu12]] ([[User talk:Pu12|talk]]) 21:18, 30 May 2019 (BST)}}
The Transition State Theory assumes that reactant nuclei follow the rules of classical mechanics. As we know now, this is inaccurate. At molecular levels, atoms follow the rules of quantum mechanics. For example, in reaction 4, the atoms have enough energy to form TS but did not proceed. According to classical mechanics, this should've formed product. But the reactants recrosses the activation barrier.
{{fontcolor1|red|What are the specific assumptions of transition state theory? It isn't enough to say that classical mechanics is followed because not all of the assumptions are purely based on mechanics. [[User:Pu12|Pu12]] ([[User talk:Pu12|talk]]) 21:18, 30 May 2019 (BST)}}

== F-H-H system ==

=== PES Inspection ===
F+ H2 ---> F-H + H
The reaction of F + H2 is exothermic. The product, H-F form is more stable hence lower in energy than the reactant. H-F bond is very strong due to the electronegativity difference between H and F. This gives rise to polar bond which is much stronger than a covalent H-H bond. The PES is illustrated below.

[[File:Hna4117_ffs.PNG]]

F-H + H ---> H2 + F
On the other hand, the reaction of H-F + H is endothermic. The strong H-F bond results in highly positive enthalpy which causes the reaction to have net endothermic reaction.
<nowiki> </nowiki>The PES is illustrated below. [[File:HNA4117_FFS3.PNG ‎]]
{{fontcolor1|red|Correct. [[User:Pu12|Pu12]] ([[User talk:Pu12|talk]]) 21:18, 30 May 2019 (BST)}}

{{fontcolor1|red|The last 4 questions are missing. [[User:Pu12|Pu12]] ([[User talk:Pu12|talk]]) 21:18, 30 May 2019 (BST)}}

MRD:bs517-201905

2019-05-24T11:59:20Z

Pu12:

== Units and indexing ==
Throughout this document all energies are expressed in kcal mol<nowiki>-1</nowiki>, distances in Å, time is an arbitrary parametrisation of the trajectory. For each system with atoms A, B and C, r1=rAB, r2=rBC, p1=pAB and p2=pBC.

== Exercise 1: H + H2 system ==

=== Transition states ===
'''On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''
* Maximum of the minimum energy path between the reactants and products
* Saddle point on the potential energy surface; <math>\text{2D: }\nabla i: \frac{\partial V_i}{\partial p_i}=0, D=[V_{xx}V_{yy}-\left(V_{xy}\right)^2]<0</math>.
*Can be located by running simulations starting from initial conditions close to, but not equal to it; these simulations either (for a sufficiently small change in initial conditions) oscillate around the transition state, or more likely, end up on the reactant/product side. A tiny change in initial conditions might change the outcome of the "reaction" completely. Simulations with the exact transition state as initial conditions will stay at the transition state indefinitely (though this cannot be achieved in practice).
*Distinguishing from local minima: simulations starting from sufficiently close to a local minimum will always stay there, oscillating; <math>\text{2D: }\nabla i: \frac{\partial V_i}{\partial p_i}=0, D=[V_{xx}V_{yy}-\left(V_{xy}\right)^2]>0, V_{xx}>0</math>.

{{fontcolor1|red|This equation looks just copied from somewhere and is not explained, please define the variables. Also it takes the gradient of the potential energy surface with respect to momentum? This doesn't really make sence. A discussion of orthogonal directions having either positive or negative second derivatives of potential energy with respect to interatomic distance would be more appropriate. [[User:Pu12|Pu12]] ([[User talk:Pu12|talk]]) 12:59, 24 May 2019 (BST)}}

=== Locating the transition state ===
'''Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''

The transition state was approximated by simulations with initial conditions r1=r2=cst., p1=p2=0. For initial conditions with internuclear distances sufficiently close to that of the transition state, the atoms stay in the simulation area and an oscillatory behaviour of the atoms was observed. Contrary to the oscillations observed in simulations leading to reactions, these happen in one dimension, i.e. they go thru the transition state as their equilibrium point in every period.

The final approximation of the transition state was reached by changing the initial conditions r1=r2=cst., so the amplitude of the vibrations observed in the "Internuclear Distances vs Time" plot approached zero. The smallest amplitude was observed at r1=r2=0.9077425=rts (Figure 1-3).
[[File:Bs517-201905.1.2.3 0.95.png|thumb|Figure 1: Internuclear Distances vs Time graph of H+H2 from r1=r2=0.95, p1=p2=0. The stoms oscillate at a low amplitude, so we are close to the transition state, but not exactly on it.]]
[[File:Bs517-201905.1.2.1 0.9077.png|thumb|Figure 2: Internuclear Distances vs Time graph of H+H2 from r1=r2=0.9077≈rts, p1=p2=0. Almost no oscillation can be observed on the simulation, we almost exactly on the transition state.]]
[[File:bs517-201905.1.2.4_0.9077425.png|thumb|Figure 3: Internuclear Distances vs Time graph of H+H2 from r1=r2=0.9077=rts p1=p2=0. Vibrations can still be seen, even this close to the transition state, however, only at very high zoom.]]
{{fontcolor1|red|Good use of graphs. [[User:Pu12|Pu12]] ([[User talk:Pu12|talk]]) 16:03, 23 May 2019 (BST)}}
=== Minimum energy path ===

[[File:bs517-201905.1.3.1_0.9177425_mep.png|thumb|Figure 4: Contour plot of the ''mep'' simulation with initial conditions r1=rts+0.01, r2=rts, p1=p2=0.]]'''Comment on how the ''mep'' and the trajectory you just calculated differ.'''

As we can see, the trajectories in Figures 4 and 5 differ in their vibrational states. As ''mep'' simulations reset the momentum of each particle to zero after every step, the particles will always move in the direction of the fastest potential energy descent around them, i.e. <math>\nabla s \in \text{steps}: p_s\parallel -\nabla V</math>. However, when running a ''Dynamics'' simulation the momenta are not reset, so the particles will not travel in the direction of the fastest descent, they will only be accelerated in its direction.{{fontcolor1|red|Good. [[User:Pu12|Pu12]] ([[User talk:Pu12|talk]]) 16:03, 23 May 2019 (BST)}} This retention of their velocities causes them to overshoot the bottom of the well every time, forcing them to vibrate while moving along the surface.
[[File:bs517-201905.1.3.2_0.9177425_dyn.png|thumb|Figure 5: Contour plot of the ''Dynamics'' simulation with initial conditions r1=rts+0.01, r2=rts, p1=p2=0.]]

As we can see from Figures 6 and 7, B and C get closer together and farther apart from A as the reaction progresses. Running the simulation from the final positions of the above trajectory and the final momenta reversed gives us back our original initial conditions at the end of the simulation (if the simulation is done for the same number of frames with the same step size). This supports the theory that when only conservative forces are present all processes could run forwards or backwards in time as well.

Offsetting r2 from the transition state instead of r1 gives similar results, the only difference being that the reaction goes in the other direction towards the reactants.

[[File:bs517-201905.1.3.3_0.9177425_dyn_idvt.png|thumb|Figure 6: Internuclear Distances vs Time plot of the ''Dynamics'' simulation with initial conditions r1=rts+0.01, r2=rts, p1=p2=0.]]

[[File:bs517-201905.1.3.4_0.9177425_dyn_mvt.png|thumb|Figure 7: Momenta vs Time plot of the ''Dynamics'' simulation with initial conditions r1=rts+0.01, r2=rts, p1=p2=0.]]

=== Reactive and unreactive trajectories ===
'''Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Trajectory
|-
| -1.25 || -2.5 ||-99.018||Yes||Weak B-C oscillations after bond has formed || [[File:Bs517-201905.1.3.5.1.png]]
|-
| -1.5 || -2.0 ||-100.456||No||Reaction does not reach TS || [[File:Bs517-201905.1.3.5.2.png]]
|-
| -1.5 || -2.5 ||-98.956||Yes||Weak B-C oscillations after bond has formed || [[File:Bs517-201905.1.3.5.3.png]]
|-
| -2.5 || -5.0 ||-84.956||No||B "bounces" voilently between A and C until finally settling with A again || [[File:Bs517-201905.1.3.5.4.png]]
|-
| -2.5 || -5.2 ||-83.416||Yes||B "bounces" between A and C, but settles on C || [[File:Bs517-201905.1.3.5.5.png]]
|}
{{fontcolor1|red|You should discuss the relative motions of the atoms more in the above descriptions and refer to the crossing of the transition state more. [[User:Pu12|Pu12]] ([[User talk:Pu12|talk]]) 16:03, 23 May 2019 (BST)}}
As we can see from the above table, the distribution of energies is extremely important when determining whether a reaction is going to be successful or not. In the second to last case for example the system enters the product region at a point, but it "bounces back", ultimately ending up as the starting materials again.

'''State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The assumptions of Transition State Theory are as follows:<ref><nowiki>T. Bligaard, J.K. Nørskov, in Chemical Bonding at Surfaces and Interfaces, 2008 (https://www.sciencedirect.com/topics/chemistry/transition-state-theory)</nowiki></ref>
* Born-Oppenheimer approximation
* quantum tunnelling effects are ignored
* Boltzmann distribution of energies in the reactants
* If transition state is reached with a velocity towards product state, system will not come back towards the starting materials
Experimental values are very close to those predicted by TST if all assumptions hold true. TST overestimates reaction rates if, for example, barrier recrossing is significant, and underestimates them if quantum tunnelling pathways are involved in the forwards direction.
{{fontcolor1|red|Good. [[User:Pu12|Pu12]] ([[User talk:Pu12|talk]]) 16:03, 23 May 2019 (BST)}}
== Exercise 2: F-H-H system ==

=== PES inspection ===
'''By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

'''Locate the approximate position of the transition state.'''

The reaction between F + H2 is exothermic and the reverse reaction is endothermic. We can see this from the potential energy surface, as the equipotential lines go lower on the side of the HF species. This is due to the strong H-F bond, that has a high ionic contribution, especially compared to the non-polar H-H bond.

The approximate transition state is r1=1.81069389, r2=0.7449985 (Figure 8, 9). This was found by doing ''Dynamics'' simulations with p1=p2=0, and changing r1 and r2 towards the F + H2 side bit by bit if the reaction went to the HF + H state, and ''vice versa''.

[[File:Bs517-201905.2.1.1 1.81069389.png|thumb|Figure 8: Internuclear Distances vs Time plot of the F-H-H system from the initial conditions r1=1.81069389, r2=0.7449985]]
[[File:Bs517-201905.2.1.2 1.81069389.png|thumb|Figure 9: Contour plot of the F-H-H system from the initial conditions r1=1.81069389, r2=0.7449985]]'''Report the activation energy for both reactions.'''

Transition state energy: -103.752

HF + H energy: -133.910

H2 + F energy: -103.973

EA(F+H2->HF+H)=0.221

EA(HF+H->H2+F)=30.158

{{fontcolor1|red|Correct. [[User:Pu12|Pu12]] ([[User talk:Pu12|talk]]) 16:03, 23 May 2019 (BST)}}
=== Reaction dynamics ===
'''In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''

As we can see from Figure 10, the H-F bond at the end of the reaction vibrates much more, than the H-H bond did at the start of the reaction. This means that throughout the reaction potential energy is converted into heat energy, e.i. the vibration of the atoms. This could be confirmed by for example a simple calorimetry experiment in a bomb calorimeter.

{{fontcolor1|red|Also translational energy is converted into vibrational. [[User:Pu12|Pu12]] ([[User talk:Pu12|talk]]) 16:03, 23 May 2019 (BST)}}

[[File:Bs517-201905.2.1.3.png|thumb|Figure 10: Controur plot of the F+ H2 system from the initial conditions r1=2.424212428049714, r2=0.7399019047604279, p1=-1, p2=-0.4]]

'''Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, rFH = 2.4, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration.'''

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics || Trajectory
|-
| -0.5 || -3 || -96.279 || No || Barrier recrossing || [[File:Bs517-201905.2.2.1 -3.png]]
|-
| -0.5 || -2.8 || -96.816|| No|| Barrier recrossing and re-recrossing || [[File:Bs517-201905.2.2.1 -2.9.png]]
|-
| -0.5 || -2.8 || -97.339 || Yes || Barrier recrossing and re-recrossing || [[File:Bs517-201905.2.2.1 -2.8.png]]
|-
| -0.5 || -2.6 || -98.319|| No|| Barrier recrossing || [[File:Bs517-201905.2.2.1 -2.6.png]]
|-
| -0.5 || -2.4 || 99.219|| Yes|| Barrier recrossing and re-recrossing || [[File:Bs517-201905.2.2.1 -2.4.png]]
|-
| -0.5 || -2.2 || -100.039|| No|| Does not reach TS || [[File:Bs517-201905.2.2.1 -2.2.png]]
|-
| -0.5 || -2.0 || -100.779|| No|| Does not reach TS || [[File:Bs517-201905.2.2.1 -2.0.png]]
|-
| -0.5 || -1.5 || -102.279|| No|| Does not reach TS || [[File:Bs517-201905.2.2.1 -1.5.png]]
|-
| -0.5 || -1 || -103.379|| No|| Does not reach TS|| [[File:Bs517-201905.2.2.1 -1.png]]
|-
| -0.5 || 0 || -103.779|| No|| Does not reach TS|| [[File:Bs517-201905.2.2.1 0.png]]
|-
| -0.5 || 1 || -102.279|| No|| Does not reach TS|| [[File:Bs517-201905.2.2.1 1.png]]
|-
| -0.5 || 1.5 || -100.779|| No|| Does not reach TS || [[File:Bs517-201905.2.2.1 1.5.png]]
|-
| -0.5 || 2 || -98.779|| No|| Barrier recrossing || [[File:Bs517-201905.2.2.1 -2.6.png]]
|-
| -0.5 || 2.2 || -97.839|| No|| Barrier recrossing || [[File:Bs517-201905.2.2.1 2.2.png]]
|-
| -0.5 || 2.4 || -98.819|| No|| Barrier recrossing || [[File:Bs517-201905.2.2.1 2.4.png]]
|-
| -0.5 || 2.6 || -95.719|| No||Barrier recrossing || [[File:Bs517-201905.2.2.1 2.6.png]]
|-
| -0.5 || 2.8 || -94.539|| No|| Multiple barrier recrossings and re-recrossings || [[File:Bs517-201905.2.2.1 2.8.png]]
|-
| -0.5 || 2.9 || -93.919|| No|| Barrier recrossing || [[File:Bs517-201905.2.2.1 2.9.png]]
|-
| -0.5 || 3 || -93.279|| Yes|| Multiple barrier recrossings and re-recrossings || [[File:Bs517-201905.2.2.1 3.png]]
|-
| -0.8 || 0.1 || -103.484|| Yes|| Only one barrier crossing, reaction is successful even though the overall energy is much lower, than in any of the other cases with a successful reaction --> better energy distribution || [[File:Bs517-201905.2.2.2.1.png]]
|}
We can see from the table that the initial energy distribution is even more important for this reaction, than for the reactoin between H and H2. Also, just putting more energy into the system may or may not yield a successful reaction.

Sometimes changing the initial conditions bit a tiny amount changes the outcome of the reaction completely, so though everything here is deterministic, there is an element of chaos theory involved.

'''The cases studied are an illustration of Polanyi's empirical rules. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

Polanyi's empirical rules are based on Hammond's postulate. They state that for endothermic reactions vibrational energy is preferred to translational energy on the reactant side, and the other way around for exothermic reactions. This is because to get up to the top of an endothermic barrier, a system with low vibrational excitation and high translational energy will not be able to pass the potential wall as it just "bounces off", usually a high vibational excitation is required to get up while going back and forth between the two walls. In the exothermic case the system is required to have high translational energy to, instead of just bouncing between the walls and going back to the reactant side, actually pass the barrier and descend to the potential well.<ref name=":0">Steinfeld, J.I., Francisco, Joseph S & Hase, William L, 1989. Chemical kinetics and dynamics, Englewood Cliffs, N.J.: Prentice Hall.</ref>

As we can see from Figures 11-15, for a reaction of HF + F with high initial viabrational energy and low initial translational energy (most of the energy of the reaction is stored in the vibrations of the F-H bond initially), the reaction succeeds even though the total energy of the system is very low (-102.463), meaning that this must be a very efficient trajectory. At first, the system bounces between the walls of the potential well (F-H bond vibrates). Next, the HF hydrogen gets passed back and forth between the F and other H atoms, finally settling with the other H atom. We can see on Figure 13, the H molecule also vibrates at the end of the reaction, but it does so significantly less violently, than the HF molecule did.

{{fontcolor1|red|Good report overall 5/5. [[User:Pu12|Pu12]] ([[User talk:Pu12|talk]]) 12:59, 24 May 2019 (BST)}}

[[File:Bs517-201905.2.3.2 0.925 2.25 -9 -1.5.png|thumb|Figure 11: Contour plot of the reaction of F + H2 initial conditions r1=0.925, r2=2.25, p1=-9, p2=-1.5.]]
[[File:Bs517-201905.2.3.2 0.925 2.25 -9 -1.5_surf.png|thumb|Figure 12: Surface plot of the reaction of F + H2 initial conditions r1=0.925, r2=2.25, p1=-9, p2=-1.5.]]
[[File:Bs517-201905.2.3.3 0.925 2.25 -9 -1.5 idvt.png|thumb|Figure 13: Internuclear distances vs Time plot of the reaction of F + H2 initial conditions r1=0.925, r2=2.25, p1=-9, p2=-1.5.]]
[[File:Bs517-201905.2.3.3 0.925 2.25 -9 -1.5 evt.png|thumb|Figure 14: Energies vs Time plot of the reaction of F + H2 initial conditions r1=0.925, r2=2.25, p1=-9, p2=-1.5.]]
[[File:Bs517-201905.2.3.3 0.925 2.25 -9 -1.5 mvt.png|thumb|Figure 15: Momenta vs Time plot of the reaction of F + H2 initial conditions r1=0.925, r2=2.25, p1=-9, p2=-1.5.]]