Jas213: /* Polanyi's empirical rules */

==Transition state==
The transition state of a reaction is a transient structure that exists at a saddle point on a potential energy surface. The gradient of the total energy equals zero and is the maximum of the minimum energy path of the potential energy surface. The minimum point of the surface also has a gradient of 0 and so in order to determine which critical point is a saddle or minimum point is to take the second derivative of the potential energy surface. If the second derivative is less than zero than it is a saddle point/transition state and if it is greater than zero it is a minimum.

∂2V('''ri''')/∂2'''ri'''> 0 'Minimum'

∂2V('''ri''')/∂2'''ri'''< 0 'Saddle point'

==Locating The Transition state==
At the transition state there is no force acting on the hydrogen molecule as -∂V(ri)/∂ri=0. Therefore, if you start a trajectory exactly at the transition state with no momentum then it will stay there forever. As Hydrogen is symmetry we can set r1=r2 and set p1=p2=0 to estimate a value for the transition state.
By adjusting the bond distance in response to oscillation on the Inter-nuclear distance Vs Time graph, the transition bond length estimated to be a value of 0.9076 Å. This is because no change in the inter nuclear distance over time shows that the atoms are being held in that position. Also, the kinetic energy equals 0 and so this also indicates the transition state had been reached.

[[File:PG_KE_T_H.png|281x281px|thumb|right|Figure 3 - A plot inter-nuclear distance vs. Time for the H + H2 system]]

[[File:PG_ID_T_H.png|323x323px|thumb|left|Figure 1 - A plot of the inter-nuclear distance vs. Time]]

[[File:PG_ID_T_H_H.png|313x313px|thumb|centre|Figure 2 - A plot of the internuclear distance for the system, where the points intersect is the transition state]]

==Calculating the reaction path==

The reaction path ( minimum energy path) is a trajectory that requires the minimum energy to form the products. This can be mapped from the transition state and can be seen to follow the valley floor. Both this and the dynamic trajectory were run by changing one of the bond distances to 0.9176. The black line seen is the mapped trajectory. The dynamic surface plot shows oscillations of potential energy surface showing that the molecule is vibrating and this agrees with an inertial trajectory whereas the mep trajectory does not show any oscillation which is not an accurate representation of what is actually occurring. Also, the dynamic path shows the full movement hydrogen atom away in the gas phase whereas the mep shows no further movement to get the lowest energy which isn't a realistic account of the motion and is a very slow rate.

[[File:PG_SP_MEP.png|298x298px|thumb|right|Figure 4 - Surface plot using MEP]]
[[File:PG_SP_D.png|298x298px|thumb|left|Figure 5 - A surface plot using dynamic trajectory]]

[[User:Jas213|Jas213]] ([[User talk:Jas213|talk]]) 00:51, 29 May 2018 (BST) If you had increased the number of steps you would have obtained a longer MEP. Why does the MEP show no vibrations energy? Because the momentum is reset to zero at each step. 

==Reactive and unreactive trajectories==

It can be concluded that trajectories within the range '''r1''' = 0.74, '''r2''' = 2.0, with -1.5 < '''p1''' < -0.8 and '''p2''' = -2.5 are reactive. We can also assume that any trajectory with a momenta over the activation barrier will reach completion as they have enough kinetic energy to overcome the transition state and reach product formation. This was tested by using the initial positions '''r1''' = 0.74 and '''r2''' = 2.0 and changing the momenta.

[[User:Jas213|Jas213]] ([[User talk:Jas213|talk]]) 00:53, 29 May 2018 (BST) Nice intro to your table. 

{| class="wikitable" border="1"
|+ Effect of inertia (Dynamic calculation type)
! p1 !! p2 !! ETotal !! React/Non!!Contour plot !! Discussion
|-
| -1.25 || -2.5 || -99.199 || Reactive || [[File:PG_M_1.png|315px]] ||Prior to the transition state, the reaction trajectory is in a straight line and so shows that the BC molecule does not oscillate. The A molecule then attacks the diatomic molecule when it is stationary. The new AB diatomic molecule formed however does oscillate and so does vibrate. ||
|-
| -1.5 || -2.0 || -100.456 || Unreactive || [[File:PG_M_2.png|315px]] || This is an unreactive path as the trajectory does not reach the product channel. This is because the momentum is not high enough to overcome the activation barrier and so the reaction does not reach completion.
|-
| -1.5 || -2.5 || -98.956 || Reactive || [[File:PG_M_3.png|315px]] || The BC diatomic molecule oscillates slightly as A approaches but it seems to look like the BC bond length increases closer to when A approaches at the transition state compared to the other reactive trajectory's.
|-
| -2.5 || -5.0 || -84.956 || Unreactive || [[File:PG_M_4.png|315px]] || Initially the BC diatomic molecule is not vibrating but once A approaches there is heavy vibration and a triatomic complex appears to form and an AB bond forms. However this doesn't have enough kinetic energy to overcome the barrier and the complex dissociates and A disperses away from the now vibrating BC diatomic and reverts back down the reactant channel .||
|-
| -2.5 || -5.2 || -83.416 || Reactive || [[File:PG_M_5.png|315px]] || The reaction path again starts with A approaching a non oscillating BC molecule and then another complex is formed but this time it has enough energy to form the products and and the trajectory moves into the product channel. The AB diatomic molecule then has a large amount of oscillation.
||
|}

From this investigation we can then see that the previous assumption is not always correct as when p gave a value of -2.5 and -5.2 the reaction became unreactive.

===Transition State Theory===

Transition state theory is used to explain the rates of reactions and qualitatively how they take place. The three main assumptions are:

1) The reactants and the activated complex are in equilibrium ( quasi equilibrium ) but not the products
2) The reaction trajectory will pass through the saddle point/transition state on the potential energy surface.
3) The reactants nuclei adhere to classical mechanics

Assumption three, assumes that the nuclei obey newtons law's of motion and do not take into account tunneling or the fact that molecular vibrations are quantized as these are quantum mechanical effects and it also doesn't take into account barrier re crossing that can occur as was seen in the table. The complex formed and was high energy but did not produce products, this shows that not all energetic collisions result in product formation but TST assumes this. So, the rates of reaction assumed by TST would be faster than those determined experimentally.

[[User:Jas213|Jas213]] ([[User talk:Jas213|talk]]) 00:55, 29 May 2018 (BST) Correct, where are your references for TST?

==F-H-H system==
===Reaction energetics===
The F + H2 reaction is exothermic because the HF bond is lower in energy ( lower enthalpy) than the H-H bond. Therefore, when the new bond is formed energy is released into the surroundings making the reaction exothermic. Therefore H + HF is endothermic as the H-H bond is higher in energy ( higher enthalpy) than the H-F bond and so when the new H-H bond it has to take in energy from the surroundings making the reaction endothermic.

[[User:Jas213|Jas213]] ([[User talk:Jas213|talk]]) 00:56, 29 May 2018 (BST) Justifying one being endothermic because the other is exothermic could have been better. You refer to the bond energies, you could have stated explicit values or described the relative height of the channels in the surface plot. 

===Transition state of reactions===
Locating the transition state in this scenario is a little different as the system is no longer symmetric and so we cannot assume that r1=r2. But, the momentum (kinetic energy) is still 0. In an exothermic reaction the transition state most closely resembles the reactants and in an endothermic reaction the transition state more closely resembles the product. So for F + H-H the values weren't moved very far from the original values and the momentum was set to zero and it was found that the transition state was found at H-H 0.7463 and F-H 1.810 angstroms. For the H + HF system, the transition state was found by moving the bond lengths closer to the products which is the same as the reactants in the previous reaction. The same transition state was found as the same complex is formed in both cases; this is a good representation of Hammond's postulate. As can be seen the kinetic energy is zero showing that the transition state has been reached.

[[File:PG_F_T.png|300x300px|thumb|Figure 6 - A plot of Energy vs. Time for the F-H-H system]][[File:PG_T_F_H.png|298x298px|thumb|center|Figure 7 - A Surface plot of the transition state for the F + H2 reaction]]

[[User:Jas213|Jas213]] ([[User talk:Jas213|talk]]) 00:58, 29 May 2018 (BST) Nice, good answer. 

===Activation energy of the reaction===
The activation energy is the energy barrier that the reactants have to reach in order to form products. This barrier is difference between the energy of the transition state and the minimum of the reactants. The energy of the transition state was found to be -103.751. the F + H2 reaction was studied. MEP trajectory was undertaken and the bond length was changed towards the formation of FH just slightly (1.80) this gave an energy minima for the reaction to be -133.927, and for the HF + H reaction the same was done and the energy minima was found to be -104.028

Activation energy = Energy of transition state - Energy of reactants

For the F + H2 reaction : -103.751 -(-104.028) = 0.277 Kcal/mol

For the HF + H reaction : -103.751 -(-133.927) = 30.176 Kcal/mol

This makes sense as the HF bond is lower in energy than the H-H bond and therefore the reaction to form the HF bond would have a lower activation energy as the reactant bond (H-H) is of higher energy and so has less energy to gain to overcome the activation barrier. This is the reverse for the other reaction.

[[User:Jas213|Jas213]] ([[User talk:Jas213|talk]]) 00:59, 29 May 2018 (BST) This is all correct, but not using the MEP approach that was asked for. Consequently, no energy vs. time plots were provided.

===Reaction Dynamics===
The conditions found to get a reactive trajectory of F + H2 reaction were:

'''rF-H = 1.81, rH-H = 0.75, pF-H = -0.5 , pH-H = 0.8'''
This is an exothermic reaction and as discussed previously therefore releases energy into the surroundings, usually as heat energy if we assume no work on the surroundings is being done. This heat energy is released to the surroundings through vibrational motion. We can see this increase in vibrational motion on the contour map clearly as when the HF bond is made it has a high vibrational movement. On the energy Vs time plot we can clearly see that the energy is being conserved as as the potential energy decreases the kinetic energy rises. This could be analyzed through IR spectroscopy to see the high overtone bands produced by molecules with higher vibrational states

[[File:PG_F_H_C.png | thumb | 300px | left |Figure 8 Reaction trajectory]]
[[File:PG_F_H_E.png | thumb | 300px | centre |Figure 9 Energy vs Time]]
[[File:PG_F_H_M.png| thumb | 300px | left |Figure 10 Momentum vs Time]]

===Polanyi's empirical rules===

The reactants in a reaction must have enough energy to overcome the activation barrier of a reaction. This energy can be in the form of vibrational or translational. Polanyi's rules state that vibrational energy is better at getting your reactants to overcome a late transition state and they say that translational energy is better at getting your reactants to overcome an early transition state. Therefore for the exothermic reaction, F + H2 , translational energy in the molecule is more effective and for the H + HF reaction, vibrational energy is more effective. (p1=vibrational energy, p2 = translational energy)

[[User:Jas213|Jas213]] ([[User talk:Jas213|talk]]) 01:02, 29 May 2018 (BST) Where's your citation for polanyi's rules? Would have been better if you had said pHH and pHF, with one and two I can't tell if it's correct, since in the previous example you used pHF and pHH and I don't know which way round you entered the atom labels in the software. 

The H + HF reaction is endothermic, therefore according to Hammond's postulate has an early transition state, and so an increase in vibrational energy over translational is more effective. The momenta was set to (p1=1.5, p2 = -0.55). This was then reversed. As can be seen this system does obey Polanyi's rules and a reaction occurs when the vibrational energy is high but not when it is low in respect to the translational energy. This coincides with what it should be for an endothermic reaction.

[[User:Jas213|Jas213]] ([[User talk:Jas213|talk]]) 01:04, 29 May 2018 (BST) As can be seen where? Which ones of your plots are you referring to? According to your labels they both show the other reaction. 

[[File:PG_H_F_C.png|284x284px|thumb|left|Figure 11 - A surface plot for the F + H2 reaction of High vibrational energy]]
[[File:PG_H_F_C_2.png|289x289px|thumb|right|Figure 12 - A surface plot for the F + H2 reaction of low vibrational energy]]

2018-05-28T23:13:09Z

Jas213: /* F-H-H System */

=== Potential Energy Surface, Transition State and Minima ===

At the transition state of the reaction, the potential energy surface is at a maximum following increase in energy from from the reactants' energy, equal to the activation energy. At this ridge point, ∂V(ri)/∂ri=0 . There is a point where the potential energy graph direction changes direction to be orthogonal to the direction of the reactants, the point where the graph changes direction is at the transition state. Likewise, at a minimum in the potential energy surface ∂V(ri)/∂ri=0.

[[User:Jas213|Jas213]] ([[User talk:Jas213|talk]]) 22:26, 28 May 2018 (BST)What about the second derivative?? You need it to differentiate between the two. 

=== Position of the Transition State, rts ===
Figure 1 shows the plot of internuclear distance against time with the given parameters of r1= 0.74, r2= 2.30, p1= 0, p2= -2.7. The transition state is symmetrical, and occurs where the distances r1 = r2 = 0.908 Angstroms- at this internuclear distance, ∂V(ri)/∂ri=0 are satisfied. To locate the radii of the transition state, the momenta of AB and BC were set to zero in order to remove the tendency of the atoms to be carried by their momenta away from one another once reaching the transition state radii. The graph of internuclear distance vs. time, Figure 2, thereby shows a straight line of zero gradient, illustrating that the internuclear distance is not changing with time.

[[File:DistancePlot1.png|thumb|left|Figure 1: Plot of Internuclear Distance vs. Time]]
[[File:PlotDistanceTime.png|thumb|right|Figure 2: Internuclear Distance vs. Time at the Transition State]]

=== Dynamics and Minimum Energy Pathway, MEP ===
The minimum energy reaction pathway given by the system with infinitely slow molecular motion and zero moment of inertia. Oscillation due to vibrational energy is observed in the 'dynamics' plot, however in the MEP plot there is no observed vibrational oscillation. Figure 3 shows the internuclear distance vs. time MEP at a slight deviation of r1 from 0.908 at the transition state to 0.918. Momenta remained at zero and r2 remained at the radius for the transition state rts= 0.908.

[[File:InitialMEP.png|thumb|centre|Figure 3: MEP Internuclear Distance vs. Time]]

With the radii r1 and r2 interchanged so that now r2= 0.918, r1= rts, the graph of internuclear distance vs. time shows an exchange in the trends for AB and BC. Conversely to before, now the distance between A and B decreases with time, but the distance between nuclei B and C increases with time.

[[File:Oppositeplot.png|thumb|centre|Figure 4: Reversed MEP Internuclear Distance vs. Time]]

Setting the initial radii values to the final values of the trajectory above, and reversing the momenta vectors shows

[[User:Jas213|Jas213]] ([[User talk:Jas213|talk]]) 22:27, 28 May 2018 (BST)Units units units. An additional contour or surface plot would have illustrated this better. You don't see any vibrations in the MEP because the vibrational energy is set to zero at each step. A momentum vs time plot would have illustrated this further. 

== Reaction Trajectories: Reactive and Unreactive ==
The energy profile of a reaction, regardless of whether the free energy change is endothermic or exothermic, will pass over an energy barrier known as the activation energy, and reach a saddle point (the derivative of the graph equals zero here) corresponding to the energy of the transition state. In order to react, the reactants must collide with reactive orientation, and sufficient energy to react- the reactants possess this energy in a number of forms- here we are primarily concerned with kinetic, translational, and vibrational energy. If there is insufficient energy to react upon collision, the system will revert back to the reactants. It is possible for the system to pass the transition state energy and still revert back to reactants without reaction completion. This is known as barrier recrossing <ref name="barrier recrossing" />, and is illustrated with the penultimate parameters.

{| class="wikitable" border="1"
|+ Effect of momenta on reactivity
! p1 !! p2 !! Total Energy !! Reactivity !! Surface Plot !! Description
|-
| -1.25 || -2.5 || -99.119 || Reactive || [[File:Reactiontraj1.png|400px]] || This trajectory is reactive. The reaction starts with no A-B oscillation, however once the product B-C is formed there is significant vibrational oscillation. The fact that there is no A-B oscillation before reaction shows that the atom C attacks A-B to form the triatomic transition state.
|-
| -1.5 || -2.0 || -100.456 || Unreactive ||[[File:Reactiontraj2.png|400px]] || The A-B bond has significant vibrational energy, as shown by the oscillation along the A-B distance axis, however this energy is insufficient for overcoming the energy barrier of the reaction activation energy. Therefore the reaction does not continue towards the products and this trajectory is said to be unreactive.
|-
| -1.5 || -2.5 || -98.956 || Reactive ||[[File:Reactiontraj3.png|400px]] || The momenta in this case yield sufficient energy to overcome the activation energy barrier to reaction, and the reaction occurs. Both the approach interaction B-C and the product A-B molecule show significant vibrational oscillation. As expected with theory, the additional energy in the A-B molecule is used in breaking the A-B bond- a process which begins by elongation of the bond which is illustrated in this surface plot.
|-
| -2.5 || -5.0 || -84.956 || Unreactive || [[File:Reactiontraj4.png|400px]] || This trajectory is unreactive. The A-B bond is shown not to oscillate initially. The A-B molecule and atom C then collide and begin to form the triatomic A-B-C transition state. The reaction path, however, does not proceed to products- this must be due to the forming transition state not reaching the saddle point of the minimum energy pathway due to insufficient energy in the complex. The symmetrical transition state does not form and therefore the reaction does not proceed to completion.
|-
| -2.5 || -5.2 || -83.416 || Reactive || [[File:Reactiontraj5.png|400px]] || The reaction path here is shown to be reactive. The A-B molecule again shows not to oscillate upon approach of the attacking C atom. Conversely, the product B-C is formed and oscillates very strongly, suggesting significant energy localised in vibrational modes.
|}

Theoretically, if the reactants begin at the same positions but with greater momentum, they will possess greater kinetic energy. The starting positions of the reactants were kept constant throughout the stimulations, at r1= 0.74 and r2= 2.0 Generally this trend is followed, as seen between stimulations 2 and 3, where increasing the momentum by (-)0.5 produces a reaction that previously did not occur. However, some cases above show that this is not necessarily always true. For example, stimulation 4 is with greater momentum for both A-B and C compared to trajectories that previously, produced a reaction, however they did not react. This proves that there is more to a successful collision than having enough kinetic energy.
The main assumptions of Transition State Theory<ref name="tst" /> are:
<pre>
1. The reaction will follow the minimum energy pathway (mep).
2. Classical behaviour must be obeyed.
3. Quasi-equilibrium assumption: The activated complex is in equilibrium with the reactants. The activated complex to products is irreversible.
</pre>
Assumption one is not always true, since reactions can proceed via pathways other than the minimum energy pathway. An example could when the activation energy barrier is surpassed via activation photochemically. Assumption three is clearly broken in the example previously discussed where barrier crossing occurs. In this case, if reaction does not go to completion it must be because the activated complex is in equilibrium with the products as well as the reactants- hence giving reversibility. With regards to assumption two, classical behaviour is not always obeyed since quantum mechanical behaviour, such as tunnelling<ref name= "tunnelling"/>, are possible.

[[User:Jas213|Jas213]] ([[User talk:Jas213|talk]]) 22:33, 28 May 2018 (BST) I like the smooth transition from table to TST as in a proper report. The term you mean for the penultimate case is called "recrossing"

== F-H-H System ==
<pre>
This stimulation concerns the following reaction:
F + H2 ⇌ H + HF
</pre>
The reaction is reversible, exothermic in the forwards direction. This is expected due to the formation of a strongly ionic H-F bond, more than compensating for the breaking of a covalent H-H bond, despite this bond being strong.
In this stimulation, atom A is fluorine, atoms B and C are hydrogen. The transition state for the reaction between fluorine and diatomic hydrogen is observed at distances A-B= 1.814 Angstroms, B-C= 0.741 Angstroms, as shown in figure 5 below:

[[User:Jas213|Jas213]] ([[User talk:Jas213|talk]]) 00:12, 29 May 2018 (BST) IF the forwards reaction is exothermic then what does this mean for the backwards reaction? The HF bond is not strongly ionic, it's strongly dipolar covalent. Ionic bonds only exist between anions and cations. Why is the HF bond stronger than the HH bond? Here you could have given the EN, the bond energy or described the difference in energy illustrating the endothermic and exothermic trajectory using the relative energies of the PES surface plot. 

[[File:HHFeqm.png|thumb|left|Figure 5: Plot of Internuclear Distance vs. Time for F-H-H]]

The method for finding the transition state was 'dynamics' with momenta set to zero to prevent deviation of the atom positions from the transition state once it has been reached. Figure 5 shows no change in interatomic distances with time and no vibrational oscillations- true for a system at rest.

[[User:Jas213|Jas213]] ([[User talk:Jas213|talk]]) How did you obtain the TS? what was your method? Did you think about hammond's postulate?

==== Activation Energy ====
In order to find the activation energy the calculation method was changed fromm dynamics to MEP. The distance of the approaching fluorine was displaced from that of the transition state. The system, when displaced from the transition state, follows the reaction pathway either back to the reactants or to the products. By Hammond's Postulate, at small deviation in energy on the potential surface there will be a structurally similar system to that at the original point, since they are only separated by a small amount of energy. For exothermic reactions, since the energy profile shows a drop below the energy of the reactants to reach the products, show reactants that are more structurally similar to the transition state than the products are. Therefore it is more sensible to vary the distance A-B to find the activation energy than to vary the B-C distance.

The A-B distance was varied from 1.814 to 1.914 Angstroms. It was necessary to increase the number of steps from 500 to 10,000 since the activation energy is so small.
[[File:Transitionstatehhf.png|thumb|right|Figure 6: Surface Plot for Calculation of Activation Energy]]

The energy of the transition state was found to be -103.869 kcal/mol. The activation energy was +0.355 kcal/mol.

The reverse reaction, however, is endothermic. The energy of the products was -133.404 kcal/mol. Taking into account the energy of the transition state, the activation energy for the backwards reaction was therefore +29.535 kcal/mol. The potential energy against time graph for HF + H is shown in figure 7.
[[File:Surface_Plot-2.png|thumb|right|Figure 7: Potential Energy vs. Time for the Backwards Reaction]]

=== Reaction Dynamics ===
<pre>
For atom A= F, atom B= H, atom C= H, and the following parameters:
Radii: A-B= 1.91Å, B-C= 0.7455 Å
Momenta: AB = -1.5
BC = 1.0
</pre>
[[File:Exothermplot.png|thumb|right|Figure 7: Potential Energy vs. Time for the Backwards Reaction]]
The reaction of H-H + F to HF + H appears to be exothermic since vibrational oscillation increases significantly following reaction at around 1ns. A release in energy would be expected to increase vibrational activity in the products. Figure 8 shows that the HF product has significantly more vibrational energy than the reactants, as shown by very high frequency amplitude of oscillation. Therefore this is related to the conservation of energy because the exothermic reaction energy is taken up into the vibrational modes of the product.

[[File:Surface_Plot-3.png|thumb|centre|Figure 8: Internuclear Momenta vs. Time for Formation of HF]]
This reaction could be measured experimentally by bomb calorimetry. Using the equation of heat flow equal to the enthalpy of the reaction at constant pressure, enthalpy changes can be measured.

==== Effect of Varying Momentum ====
With F-H distance: 1.91 Angstroms, H-H distance: 0.74555, FH momentum: -0.5 kg⋅m/s, H-H momentum: -3.0 kg⋅m/s, the following contour surface is given:
[[File:Surface_Plot-5.png|thumb|centre|Figure 9: Contour Plot at Diatomic Hydrogen Momentum= -3 kg⋅m/s]]
The contour plot shows that at these values of momenta the reaction pathway does not proceed to products.

[[File:Surface_Plot-6.png|thumb|centre|Figure 10: Contour Plot at Diatomic Hydrogen Momentum= -0.5kg⋅m/s]]
This Reaction pathway shows completion to products.
[[File:Surface_Plot-7.png|thumb|centre|Figure 11: Contour Plot at Diatomic Hydrogen Momentum= +3 kg⋅m/s]]
This reaction pathway shows an unreactive trajectory.

Figure 9 where diatomic hydrogen momentum is -3 kg⋅m/s s does not lead to a reaction. Instead, the energy is dissipated as translational and vibrational motion in hydrogen fluoride. Figure 10 shows a complete reaction pathway, the vibrational energy here is in the product of diatomic hydrogen. The vibrational energy is significant, since the amplitude of oscillation is high. Figure 11 again shows failure of the reaction pathway to move towards the products, and instead the energy is dissipated in vibrational and translational energy in hydrogen fluoride.

==== Polanyi's Rules ====

Polanyi's rules describe the effects of the distribution of the total energy of a molecule into its different modes on the transition state of a reaction. The modes concerned are translational and vibrational. One rule is that vibrational energy is more significant for promoting a late activation energy in a reaction profile than translational energy is. Conversely, translational energy promotes an early reaction barrier more significantly than vibrational energy does. In the first case the reaction is exothermic, the second, endothermic.

The forward reaction: F + H2 → FH + H is exothermic and so it can be predicted that a high proportion of the total energy of the molecule being located in translational modes would promote reaction. This is illustrated in looking at the above plots of momenta and energy against time.
In the case of the reverse reaction, which is endothermic, a high contribution of vibrational energy to the total energy of the reactant promotes completion of the reaction.

=== References ===
<references>
<ref name="barrier recrossing"> Reaction Rate Theory, http://www.acmm.nl/ensing/thesis/node9.html, ( accessed May 2018) </ref>

<ref name="tst"> E. Anslyn, A. Dennis, Transition State Theory and Related Topics. In Modern Physical Organic Chemistry University Science Books, 2006, p. 365–373 </ref>

<ref name="tunnelling"> R. Gurney, E. Condon, 1929, Quantum Mechanics and Radioactive Disintegration, Phys. Rev. 33, p. 127–140. <ref/>

</references>

MRD:01235058220518

2018-05-28T23:12:14Z

Jas213: /* F-H-H System */

=== Potential Energy Surface, Transition State and Minima ===

At the transition state of the reaction, the potential energy surface is at a maximum following increase in energy from from the reactants' energy, equal to the activation energy. At this ridge point, ∂V(ri)/∂ri=0 . There is a point where the potential energy graph direction changes direction to be orthogonal to the direction of the reactants, the point where the graph changes direction is at the transition state. Likewise, at a minimum in the potential energy surface ∂V(ri)/∂ri=0.

[[User:Jas213|Jas213]] ([[User talk:Jas213|talk]]) 22:26, 28 May 2018 (BST)What about the second derivative?? You need it to differentiate between the two. 

=== Position of the Transition State, rts ===
Figure 1 shows the plot of internuclear distance against time with the given parameters of r1= 0.74, r2= 2.30, p1= 0, p2= -2.7. The transition state is symmetrical, and occurs where the distances r1 = r2 = 0.908 Angstroms- at this internuclear distance, ∂V(ri)/∂ri=0 are satisfied. To locate the radii of the transition state, the momenta of AB and BC were set to zero in order to remove the tendency of the atoms to be carried by their momenta away from one another once reaching the transition state radii. The graph of internuclear distance vs. time, Figure 2, thereby shows a straight line of zero gradient, illustrating that the internuclear distance is not changing with time.

[[File:DistancePlot1.png|thumb|left|Figure 1: Plot of Internuclear Distance vs. Time]]
[[File:PlotDistanceTime.png|thumb|right|Figure 2: Internuclear Distance vs. Time at the Transition State]]

=== Dynamics and Minimum Energy Pathway, MEP ===
The minimum energy reaction pathway given by the system with infinitely slow molecular motion and zero moment of inertia. Oscillation due to vibrational energy is observed in the 'dynamics' plot, however in the MEP plot there is no observed vibrational oscillation. Figure 3 shows the internuclear distance vs. time MEP at a slight deviation of r1 from 0.908 at the transition state to 0.918. Momenta remained at zero and r2 remained at the radius for the transition state rts= 0.908.

[[File:InitialMEP.png|thumb|centre|Figure 3: MEP Internuclear Distance vs. Time]]

With the radii r1 and r2 interchanged so that now r2= 0.918, r1= rts, the graph of internuclear distance vs. time shows an exchange in the trends for AB and BC. Conversely to before, now the distance between A and B decreases with time, but the distance between nuclei B and C increases with time.

[[File:Oppositeplot.png|thumb|centre|Figure 4: Reversed MEP Internuclear Distance vs. Time]]

Setting the initial radii values to the final values of the trajectory above, and reversing the momenta vectors shows

[[User:Jas213|Jas213]] ([[User talk:Jas213|talk]]) 22:27, 28 May 2018 (BST)Units units units. An additional contour or surface plot would have illustrated this better. You don't see any vibrations in the MEP because the vibrational energy is set to zero at each step. A momentum vs time plot would have illustrated this further. 

== Reaction Trajectories: Reactive and Unreactive ==
The energy profile of a reaction, regardless of whether the free energy change is endothermic or exothermic, will pass over an energy barrier known as the activation energy, and reach a saddle point (the derivative of the graph equals zero here) corresponding to the energy of the transition state. In order to react, the reactants must collide with reactive orientation, and sufficient energy to react- the reactants possess this energy in a number of forms- here we are primarily concerned with kinetic, translational, and vibrational energy. If there is insufficient energy to react upon collision, the system will revert back to the reactants. It is possible for the system to pass the transition state energy and still revert back to reactants without reaction completion. This is known as barrier recrossing <ref name="barrier recrossing" />, and is illustrated with the penultimate parameters.

{| class="wikitable" border="1"
|+ Effect of momenta on reactivity
! p1 !! p2 !! Total Energy !! Reactivity !! Surface Plot !! Description
|-
| -1.25 || -2.5 || -99.119 || Reactive || [[File:Reactiontraj1.png|400px]] || This trajectory is reactive. The reaction starts with no A-B oscillation, however once the product B-C is formed there is significant vibrational oscillation. The fact that there is no A-B oscillation before reaction shows that the atom C attacks A-B to form the triatomic transition state.
|-
| -1.5 || -2.0 || -100.456 || Unreactive ||[[File:Reactiontraj2.png|400px]] || The A-B bond has significant vibrational energy, as shown by the oscillation along the A-B distance axis, however this energy is insufficient for overcoming the energy barrier of the reaction activation energy. Therefore the reaction does not continue towards the products and this trajectory is said to be unreactive.
|-
| -1.5 || -2.5 || -98.956 || Reactive ||[[File:Reactiontraj3.png|400px]] || The momenta in this case yield sufficient energy to overcome the activation energy barrier to reaction, and the reaction occurs. Both the approach interaction B-C and the product A-B molecule show significant vibrational oscillation. As expected with theory, the additional energy in the A-B molecule is used in breaking the A-B bond- a process which begins by elongation of the bond which is illustrated in this surface plot.
|-
| -2.5 || -5.0 || -84.956 || Unreactive || [[File:Reactiontraj4.png|400px]] || This trajectory is unreactive. The A-B bond is shown not to oscillate initially. The A-B molecule and atom C then collide and begin to form the triatomic A-B-C transition state. The reaction path, however, does not proceed to products- this must be due to the forming transition state not reaching the saddle point of the minimum energy pathway due to insufficient energy in the complex. The symmetrical transition state does not form and therefore the reaction does not proceed to completion.
|-
| -2.5 || -5.2 || -83.416 || Reactive || [[File:Reactiontraj5.png|400px]] || The reaction path here is shown to be reactive. The A-B molecule again shows not to oscillate upon approach of the attacking C atom. Conversely, the product B-C is formed and oscillates very strongly, suggesting significant energy localised in vibrational modes.
|}

Theoretically, if the reactants begin at the same positions but with greater momentum, they will possess greater kinetic energy. The starting positions of the reactants were kept constant throughout the stimulations, at r1= 0.74 and r2= 2.0 Generally this trend is followed, as seen between stimulations 2 and 3, where increasing the momentum by (-)0.5 produces a reaction that previously did not occur. However, some cases above show that this is not necessarily always true. For example, stimulation 4 is with greater momentum for both A-B and C compared to trajectories that previously, produced a reaction, however they did not react. This proves that there is more to a successful collision than having enough kinetic energy.
The main assumptions of Transition State Theory<ref name="tst" /> are:
<pre>
1. The reaction will follow the minimum energy pathway (mep).
2. Classical behaviour must be obeyed.
3. Quasi-equilibrium assumption: The activated complex is in equilibrium with the reactants. The activated complex to products is irreversible.
</pre>
Assumption one is not always true, since reactions can proceed via pathways other than the minimum energy pathway. An example could when the activation energy barrier is surpassed via activation photochemically. Assumption three is clearly broken in the example previously discussed where barrier crossing occurs. In this case, if reaction does not go to completion it must be because the activated complex is in equilibrium with the products as well as the reactants- hence giving reversibility. With regards to assumption two, classical behaviour is not always obeyed since quantum mechanical behaviour, such as tunnelling<ref name= "tunnelling"/>, are possible.

[[User:Jas213|Jas213]] ([[User talk:Jas213|talk]]) 22:33, 28 May 2018 (BST) I like the smooth transition from table to TST as in a proper report. The term you mean for the penultimate case is called "recrossing"

== F-H-H System ==
<pre>
This stimulation concerns the following reaction:
F + H2 ⇌ H + HF
</pre>
The reaction is reversible, exothermic in the forwards direction. This is expected due to the formation of a strongly ionic H-F bond, more than compensating for the breaking of a covalent H-H bond, despite this bond being strong.
In this stimulation, atom A is fluorine, atoms B and C are hydrogen. The transition state for the reaction between fluorine and diatomic hydrogen is observed at distances A-B= 1.814 Angstroms, B-C= 0.741 Angstroms, as shown in figure 5 below:

[[User:Jas213|Jas213]] ([[User talk:Jas213|talk]]) 00:12, 29 May 2018 (BST) IF the forwards reaction is exothermic then what does this mean for the backwards reaction? The HF bond is not strongly ionic, it's strongly dipolar covalent. Ionic bonds only exist between anions and cations. Why is the HF bond stronger than the HH bond? Here you could have given the EN, the bond energy or described the difference in energy illustrating the endothermic and exothermic trajectory using the relative energies of the PES surface plot. 

[[File:HHFeqm.png|thumb|left|Figure 5: Plot of Internuclear Distance vs. Time for F-H-H]]

The method for finding the transition state was 'dynamics' with momenta set to zero to prevent deviation of the atom positions from the transition state once it has been reached. Figure 5 shows no change in interatomic distances with time and no vibrational oscillations- true for a system at rest.

==== Activation Energy ====
In order to find the activation energy the calculation method was changed fromm dynamics to MEP. The distance of the approaching fluorine was displaced from that of the transition state. The system, when displaced from the transition state, follows the reaction pathway either back to the reactants or to the products. By Hammond's Postulate, at small deviation in energy on the potential surface there will be a structurally similar system to that at the original point, since they are only separated by a small amount of energy. For exothermic reactions, since the energy profile shows a drop below the energy of the reactants to reach the products, show reactants that are more structurally similar to the transition state than the products are. Therefore it is more sensible to vary the distance A-B to find the activation energy than to vary the B-C distance.

The A-B distance was varied from 1.814 to 1.914 Angstroms. It was necessary to increase the number of steps from 500 to 10,000 since the activation energy is so small.
[[File:Transitionstatehhf.png|thumb|right|Figure 6: Surface Plot for Calculation of Activation Energy]]

The energy of the transition state was found to be -103.869 kcal/mol. The activation energy was +0.355 kcal/mol.

The reverse reaction, however, is endothermic. The energy of the products was -133.404 kcal/mol. Taking into account the energy of the transition state, the activation energy for the backwards reaction was therefore +29.535 kcal/mol. The potential energy against time graph for HF + H is shown in figure 7.
[[File:Surface_Plot-2.png|thumb|right|Figure 7: Potential Energy vs. Time for the Backwards Reaction]]

=== Reaction Dynamics ===
<pre>
For atom A= F, atom B= H, atom C= H, and the following parameters:
Radii: A-B= 1.91Å, B-C= 0.7455 Å
Momenta: AB = -1.5
BC = 1.0
</pre>
[[File:Exothermplot.png|thumb|right|Figure 7: Potential Energy vs. Time for the Backwards Reaction]]
The reaction of H-H + F to HF + H appears to be exothermic since vibrational oscillation increases significantly following reaction at around 1ns. A release in energy would be expected to increase vibrational activity in the products. Figure 8 shows that the HF product has significantly more vibrational energy than the reactants, as shown by very high frequency amplitude of oscillation. Therefore this is related to the conservation of energy because the exothermic reaction energy is taken up into the vibrational modes of the product.

[[File:Surface_Plot-3.png|thumb|centre|Figure 8: Internuclear Momenta vs. Time for Formation of HF]]
This reaction could be measured experimentally by bomb calorimetry. Using the equation of heat flow equal to the enthalpy of the reaction at constant pressure, enthalpy changes can be measured.

==== Effect of Varying Momentum ====
With F-H distance: 1.91 Angstroms, H-H distance: 0.74555, FH momentum: -0.5 kg⋅m/s, H-H momentum: -3.0 kg⋅m/s, the following contour surface is given:
[[File:Surface_Plot-5.png|thumb|centre|Figure 9: Contour Plot at Diatomic Hydrogen Momentum= -3 kg⋅m/s]]
The contour plot shows that at these values of momenta the reaction pathway does not proceed to products.

[[File:Surface_Plot-6.png|thumb|centre|Figure 10: Contour Plot at Diatomic Hydrogen Momentum= -0.5kg⋅m/s]]
This Reaction pathway shows completion to products.
[[File:Surface_Plot-7.png|thumb|centre|Figure 11: Contour Plot at Diatomic Hydrogen Momentum= +3 kg⋅m/s]]
This reaction pathway shows an unreactive trajectory.

Figure 9 where diatomic hydrogen momentum is -3 kg⋅m/s s does not lead to a reaction. Instead, the energy is dissipated as translational and vibrational motion in hydrogen fluoride. Figure 10 shows a complete reaction pathway, the vibrational energy here is in the product of diatomic hydrogen. The vibrational energy is significant, since the amplitude of oscillation is high. Figure 11 again shows failure of the reaction pathway to move towards the products, and instead the energy is dissipated in vibrational and translational energy in hydrogen fluoride.

==== Polanyi's Rules ====

Polanyi's rules describe the effects of the distribution of the total energy of a molecule into its different modes on the transition state of a reaction. The modes concerned are translational and vibrational. One rule is that vibrational energy is more significant for promoting a late activation energy in a reaction profile than translational energy is. Conversely, translational energy promotes an early reaction barrier more significantly than vibrational energy does. In the first case the reaction is exothermic, the second, endothermic.

The forward reaction: F + H2 → FH + H is exothermic and so it can be predicted that a high proportion of the total energy of the molecule being located in translational modes would promote reaction. This is illustrated in looking at the above plots of momenta and energy against time.
In the case of the reverse reaction, which is endothermic, a high contribution of vibrational energy to the total energy of the reactant promotes completion of the reaction.

=== References ===
<references>
<ref name="barrier recrossing"> Reaction Rate Theory, http://www.acmm.nl/ensing/thesis/node9.html, ( accessed May 2018) </ref>

<ref name="tst"> E. Anslyn, A. Dennis, Transition State Theory and Related Topics. In Modern Physical Organic Chemistry University Science Books, 2006, p. 365–373 </ref>

<ref name="tunnelling"> R. Gurney, E. Condon, 1929, Quantum Mechanics and Radioactive Disintegration, Phys. Rev. 33, p. 127–140. <ref/>

</references>