DL3715 Y2

2017-06-02T03:08:52Z

= Molecular Reaction Dynamics: Applications to Triatomic systems =

=== Introduction ===
In this lab, a triatomic system of collision was investigated assuming that the motions of atoms obeyed classical mechanics. The system was investigated with respect to different types of atoms, varying internuclear distances and internuclear momentum by monitoring the molecular dynamic trajectories using MATLAB. Also, the relationship between reaction efficiency and the position of the transition state was demonstrated by the Polanyi's empirical rules.

== EXERCISE 1: H + H2 system ==

=== '''1.1 What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.''' ===
The gradients (first derivative) of the potential energy surface at minimum and at a transition state are both zero. Minima and transition structures can be distinguished by the second derivative of the potential energy surface: at a minimum point, d^^2V/dR^^2>0; whereas at a transition state, d^^2V/dR^^2<0.

{{fontcolor1|red|(This is not necessarily true, depending on your R coordinate. [[User:Lt912|Lt912]] ([[User talk:Lt912|talk]]) 04:08, 2 June 2017 (BST))}}

=== '''1.2 Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.''' ===
The transition state was found at t=0.38 from the internuclear distances vs time plot as shown below.'''<nowiki/>'''
{| class="wikitable"
|-
|[[File:01054770_01.PNG|500px]]
|-
|Internuclear distances vs time (P1=-2.7,P2=0)
|}

The internuclear distance was optimised to 0.90775A when bothe kinetic and potential energy start to oscillate as shown below.
{| class="wikitable"
|-
|[[File:01054770_03.PNG|500px|left]]
|[[File:01054770_04.PNG|500px|left]]
|-
|Oscillating potential energy at TS
|Oscillating kinetic energy at TS
|}

At the transition state (r1=r2=0.90775), the distance between A-B and B-C will always be a constant as illustrated by the internuclear distances vs time plot at P1=P2=0 shown below:
{| class="wikitable"
|-
|[[File:01054770_02.PNG|500px]]
|-
|Internuclear distances vs time (P1=0,P2=0)
|}

=== '''1.3 Comment on how the mep and the trajectory you just calculated differ.''' ===
{| class="wikitable"
|'''MEP'''
|[[File:mepcontourdl3715.PNG|thumb|contour plot]]
|[[File:mepdistancedl3715.PNG|thumb|intermolecular distances vs time]]
|[[File:mepmomentumdl3715.PNG|thumb|momenta vs time]]
|-
|__
|Contour plot in MEP calculation (r1=0.90775, r2=0.91775,p1=p2=0)
|Intermolecular momentum vs time in MEP calculation
|Intermolecular distance vs time in MEP calculation
|-
|'''DYNAMIC'''
|[[File:dyncontourdl3715.PNG|thumb|contour plot]]
|[[File:dyndistancedl3715.PNG|thumb|intermolecular distances vs time]]
|[[File:dynmomentadl3715.PNG|thumb|momenta vs time]]
|-
|__
|Contour plot in Dynamic calculation (r1=0.90775, r2=0.91775,p1=p2=0)
|Intermolecular momentum vs time in Dynamic calculation
|Intermolecular distance vs time in Dynamic calculation
|}

An initial condition was set as : r1=0.91775, r2=0.90775, p1=p2=0 for all plots.
The initial position of r2 was set to deviate from the transitional state position by 0.1. Since A is closer to B than C does, the formation of bond between B-C is favoured over A-B as shown in both of the intermolecular distance vs time plot.
In MEP calculation, there is no internuclear momentum, ie. B-C will not vibrate over time as seen in the contour plot since it is taking the minimum energy at each time point in calculation.The momentum is zero even it is slightly off the transition state.
The trajectory getting from the Dynamics calculation is quite different from the MEP calculation. Dynamic calculation includes all the vibrating situation with the trajectories so when a proton is placed off the transition state, the motion is spontaneous whereas, in MEP, the motion will not take place.

==== '''Calculating the reaction path by using Trajectories from r1 = rts + δ r2 = rts''' ====
At large t (t=5):
the intermolecular distance between A and B is 18.31
the intermolecular distance between B and C is 0.7322
the average intermolecular momentum between B and A is 2.481
the average intermolecular momentum between B and C is 1.237.
When the final state becomes the initial state with inverse sign of momentum, the atoms were expected to be brought back together in the same path as before as shown in the following contour plot.
[[File:01054770_111.PNG|500px|none]]

=== '''1.4 Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.''' ===
{| class="wikitable"
|'''P1'''
|'''P2'''
|'''Contour Plot'''
|'''Intermoleculat distance vs time'''
|'''Momenta vs time'''
|'''Description'''
|-
|'''-1.25'''
|'''-2.5'''
|[[File:g1contourdl3715.PNG|thumb|contour plot]]
|[[File:g1distdl3715.PNG|thumb|intermolecular distances vs time]]
|[[File:g1momentadl3715.PNG|thumb|momenta vs time]]
|Reactive, since the trajectory passes over the transition state which means there is sufficient energy to overcome the activation barrier.
|-
|'''-1.5'''
|'''-2.0'''
|[[File:g2contourdl3715.PNG|thumb|contour plot]]
|[[File:g2distdl3715.PNG|thumb|intermolecular distances vs time]]
|[[File:g2momentadl3715.PNG|thumb|momenta vs time]]
|Unreactive, since the trajectory didn't pass the transition state, instead it reverse back downhill to the A-B side. There is no vibration at B-C(as shown in the intermulcular momentum vs time plot) so there is no bond formation between B-C
|-
|'''-1.5'''
|'''-2.5'''
|[[File:g3contourdl3715.PNG|thumb|contour plot]]
|[[File:g3distdl3715.PNG|thumb|intermolecular distances vs time]]
|[[File:g3momentadl3715.PNG|thumb|momenta vs time]]
|Reactive, since the trajectory passes the transition state so it has sufficient energy to overcome the activation barrier.
|-
|'''-2.5'''
|'''-5.0'''
|[[File:g4contourdl3715.PNG|thumb|contour plot]]
|[[File:g4distdl3715.PNG|thumb|intermolecular distances vs time]]
|[[File:g4momentadl3715.PNG|thumb|momenta vs time]]
|Unreactive, since the trajectory path reverses back to the A-B side. B-C bond was formed shortly but was broken due to the reformation of A-B bond.
|-
|'''-2.5'''
|'''-5.2'''
|[[File:g5contourdl3715.PNG|thumb|contour plot]]
|[[File:g5distdl3715.PNG|thumb|intermolecular distances vs time]]
|[[File:g5momentadl3715.PNG|thumb|momenta vs time]]
|Reactive, since the B-C was oscillating as seen in the intermolecular momentum vs time plot.
|}

The 1st, 3rd and 5th group with Hc-Hb +Ha initial state are all reactive since they have sufficient energy to overcome the activation energy barrier.
While the 2nd and 4th group don't form stable product as they don't have sufficient activation energy. Especially in the 4th group, the product was formed at a high vibration energy but it soon dissociated back to reactants at lower vibration energy.

=== '''1.5 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?''' ===
Transition state theory states that once the system reaches a transition state configuration, the product must be formed. However, in our experimental data of the 4th group, the reactant reaches the transition state but reverses its path after a short while. This dissociation of the already formed product shows that the vibration direction for both reactants is still important. Every reaction has a kinetic energy and a activation barrier. The reaction can occur of the kinetics energy is greater than the activation barrier. However, if the kinetic energy i much larger than the activation energy, it can recross the transition state and cause the reaction to be unreactive. As a result, higher momentum does also give a reactive reaction.

== EXERCISE 2: F - H - H system ==

=== '''2.1 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?''' ===
Since H-F bond strength is much larger than H-H bond, the reaction from F+H2 to H+HF is exothermic (where a fluorine atom is approaching to H2 molecule, H-F bond formation is energetically favoured); whereas the reaction from H-F+H to H2+F is endothermic (where a proton is approaching H-F, H-F bond dissociation is energetically unfavoured)
The first surface plot shows a fluorine atom approaching H2. The reaction proceeds from a higher energy state to a lower energy state so exothermic.
{| class="wikitable"
|-
|[[File:hf1_0831.PNG|400px]]
|-
|surface plot of F + H2
|}

The second surface plot shows a hydrogen atom approaching H-F. The reaction proceeds from a lower energy state to a higher energy state so endothermic.
{| class="wikitable"
|-
|[[File:hf2_0831.PNG|400px]]
|-
|surface plot of H + HF
|}

=== '''2.2 Locate the approximate position of the transition state.''' ===
The transition state was found at r1=1.8107 and r2=0.745 with initial momenta equals to zero.
{| class="wikitable"
|-
|[[File:EofTSdl3715.PNG|400px]]
|-
|transition state surface plot
|}

{| class="wikitable"
|-
|[[File:posnTSdl3715.PNG|400px]]
|-
|transition state contour plot
|}

At transition state, the internuclear distances among all three atoms remain unchanged with time, ie. three atoms are all vibrating at its position.
{| class="wikitable"
|-
|[[File:ts11.PNG|400px]]
|-
|internuclear distances at transition state
|}

=== '''2.3 Report the activation energy for both reactions.''' ===
{| class="wikitable"
|-
|[[File:minima1dl3715.PNG|400px]]
|-
|minimum energy point at -103.3 Kcal/mol
|}

{| class="wikitable"
|-
|[[File:minima2dl3715.PNG|400px]]
|-
|minimum energy point at -133.9 Kcal/mol
|}

The activation energy is the energy difference between the minimum energy point and the transition state point. Therefore:
for energy barrier between the H-H bond and transition state:

the Ea could be calculated as Ea=-103.3-(-103.9)= 0.6 Kcal/mol

for energy barrier between the transition state and the H-F bond:

the Ea could be calculated as Ea=-103.3-(-133.9)= 30.6 Kcal/mol

=== '''2.4 In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?''' ===
'''''Reaction of F + H2'''''
The reaction condition was set as :
r(HF) =2.1å,momentum(HF) =-1 Ns
r(HH) =0.745å, momentum(HF) =0 Ns
The plots were shown in the following table.
{| class="wikitable"
|-
|[[File:last1_0831.PNG|400px|left]]
|[[File:last2_0831.PNG|400px|left]]
|[[File:last3_0831.PNG|400px|left]]
|[[File:last4_0831.PNG|400px|left]]
|-
|Potential energy surface
|Internuclear distance vs time
|Kinetics energy vs time
|Potential energy vs time
|}

The graphs suggested that the mechanism of this reaction can be divided into two steps:
1. The fluorine atom is approaching the vibrating H2 molecule, shown as the oscillating momentum.
2. The fluorine collide with one of the hydrogen atom causing the dissociation of the H2 molecule, shown as the increasing momentum and internuclear distance of B-C. Then, the fluorine atom forms a bond with the hydrogen, shown as the oscillation of A-B internuclear momentum.

Since the energy is always conserved in the reaction, the energy is converting between kinetic energy and potential energy.
The reaction is exothermic so the potential energy of the reactant is higher than that of the product. As a result, the kinetic energy is transformed into potential energy. At the beginning, the change in both energies are small because it was due to the relative small vibration of H2 molecule. The changes became more significant at 2.5s when the H-H bond dissociated and H-F started to form.
This can be experimentally proved by the following two ways:
1. IR spectroscopy. The vibration frequencies of the H-H bond and H-F bond can be distinguished on the spectrum and the heat released was stored as the vibrational energy so can be calculated by E=hv.
2. Thermometer. The energy released can be calculated by Q=mcΔT, where Q is the thermal enegy released, m is the mass of the liquid, c is the specific capacity and ΔT is the change in temperature.

'''''Reaction of H + HF'''''
The initial condition was set that the reactants have low kinetic energy:
r(HH)=2.2 Å
r(HF)=0.91 Å
momentum(HH)=-2.2 kg m s-1
momentum(HF)=-2.0 kg m s-1
The reaction was unreactive as shown below:
[[File:final1_0831.PNG|500px|none]]
The, the momentum of the incoming hydrogen was changed slightly from -2.2 to -2.0; and the moment of H-F was significantly changed from -2.0 to -9.5. The reaction becomes reactive, as shown below.
[[File:final2_0831.PNG|500px|none]]
The H-H bond formation was shown as the oscillation in the internuclear distance plot below.
[[File:final3_0831.PNG|500px|none]]

=== '''2.5 Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.''' ===
The reaction efficiency and transition state were related by the Polanyi's empirical rules. Polanyi's empirical rules state that vibrational energy is more efficient in promoting a late-barrier reaction than translational energy.
In H + HF reaction, the reaction is endothermic so it has a late transition state. According to Polanyi's empirical rules, the vibrational energy would promote the reaction greater than the translational energy. Compared with the data used in 2.4, the reaction becomes reactive when it has a greater momentum (vibration energy) than the translational energy (-9.5 : -2.2). Therefore, the reaction trajectory obeys the Polanyi's empirical rules.

==Conclusion==
In the H + H2 system, the transition state was at 0.9175 Å. According to the Classical Transition State theory, all reactions with sufficient energy to overcome the activation barrier should proceed to product. However, in our experiment, reactions with too high energies recrossed the transition state which leads to a overall unreactive reaction.
In the H-H-F system, two possible reactions behaved differently due to the different strength of bond formed, therefore, they have a different energy profile which related to the reaction efficiency. The H + HF reaction is endothermic with late transition state, as a result, vibrational energy promotes the reactive more effectively than translational energy.

Eo00842229

2017-06-02T03:01:27Z

Lt912: /* Question 9 - Reaction Dynamics - Reaction Energy */

=H + H2 System=

==Question 1 - Gradients at Transition State and Minimum ==

The total value of the potential energy surface at the minimum and at the transition structure will both be zero.

Minima will have a positive value for the second derivative in every axial direction of the energy potential surface whereas a transition state will have a positive value for the second derivative in all axial directions apart from in the reaction pathway.

[[File:Capnew0084.PNG|thumb|centre|''Figure 1'': A surface plot showing the curvature of the minima compared to the transition state]]

Using the reaction pathway as axis helps to tell the difference between these two points as the second derivative of the minima will indicate a minimum whereas that of that of the transition state will indicate a maximum.

==Question 2 - Trajectories from r1 = r2==
To calculate the transition state position, rts, an initial inter-nuclear distance of 1 Å was used with momentum of 0 for both distances AB and BC. ''Figure 2'' shows a vibrating structure about local minimum of the reaction coordinate.

[[File:Cap10084.PNG|thumb|centre|''Figure 2'': A surface plot showing an estimate for the transition state position using the Data Cursor Tool]]

A plot Inter-nuclear Distance vs Time with a value close to the transition state position determined by moving the data cursor up and down to find a minimum inter-nuclear position showed that there was some vibration as the inter-nuclear distances slightly change over time.

[[File:Cap20084.PNG|thumb|centre|''Figure 3'': A plot of Inter-nuclear Distance vs Time for the H-H-H transition state at r1 = r2 = 0.9073 Å showing slight vibrations in the structure.]]

Trying values above and below this initial value for the transition state position allowed to precisely find the position by choosing the value for rts that gave little to no change in inter-nuclear distance with changing time. rts = 0.908 Å (3 d.p.)

[[File:Cap30084.PNG|thumb|centre|''Figure 4'': A plot of Inter-nuclear Distance vs Time for the H-H-H transition state at r1 = r2 = 0.9079 Å showing negligible vribration.]]

==Question 3 - Trajectories from r1 = rts+δ, r2 = rts==
Comparing ''Figure 5'' and ''Figure 6'', it can be seen that the MEP calculation shows a trajectory that does show any vibration whereas the trajectory calculated using the Dynamics calculation does. The MEP calculation also doesn't show any vibration in the bonds whereas the Dynamic calculation does, even with zero momentum for all atoms in the system.

[[File:Cap40084.PNG|thumb|centre|''Figure 5'': A surface plot (MEP) for the H-H-H system with r1 = 0.918 Å and r2 = 0.908 Å.]]

[[File:Cap50084.PNG|thumb|centre|''Figure 6'': A surface plot (Dynamics) for the H-H-H system with r1 = 0.918 Å and r2 = 0.908 Å.]]

This is due to the fact that the MEP resets the velocity to zero for each time step, this therefore doesn't allow the system to transfer kinetic energy into potential energy, keeping the trajectory of the system along the pathway with minimum potential energy. This can be seen when observing ''Figure 7'' and ''Figure 8''.

[[File:Cap70084.PNG|thumb|centre|''Figure 7'': A Potential energy vs. Time for the H-H-H system transition state with r1 = 0.918 Å and r2 = 0.908 Å showing how the system falls to a minimum.]]

[[File:Cap80084.PNG|thumb|centre|''Figure 8'': A Kinetic energy vs. Time plot for the H-H-H system transition state with at r1 = 0.918 Å and r2 = 0.908 Å showing zero kinetic energy over time.]]

==Question 4 - Reactive and unreactive trajectories==
{| class="wikitable" border=1 style="text-align: center;"
! p1
! p2
! Reactive? (Y/N)
!Screenshot
!Description
|-
| -1.25 || -2.5 || Y || [[File:Cap90084.PNG|thumb|centre|]] || The approaching H+ atom, C, comes in with enough momentum to overcome the kinetic barrier for activation and goes to completion. There is little change in the motion of the A-B bond but once the product is formed, there is some vibration in the bond, suggesting that some of the translational energy of the approaching C is transformed into vibrational energy in the new molecule of B-C.
|-
| -1.5 || -2.0 || N || [[File:Cap100084.PNG|thumb|centre|]] || The approaching H+ atom, C, does not have enough kinetic energy to overcome the barrier for the reaction. It initially approaches quite quickly but slows down as it gets closer to the molecule B-C and eventually gets repelled away from the molecule leading to an incomplete reaction (due to an insufficient total momentum). The extra momentum in p1 manifests itself as vibrational motion in the molecule A-B.
|-
| -1.5 || -2.5 || Y || [[File:Cap110084.PNG|thumb|centre|]] || This system is very much similar to the first and also proceeds to completion. The approaching H+ atom, C, has enough kinetic energy to overcome the activation barrier. The only difference is in the greater vibration in B-C due to the large value in its momentum.
|-
| -2.5 || -5.0 || N || [[File:Cap120084.PNG|thumb|centre|]] || The total momentum in this system is very large as can be seen from the much larger vibrational energy in the A-B bond. The approaching H+ atom, C, has a lot of kinetic energy as a result and manages to overcome the activation barrier however it doesn't manage to go to completion despite being in the same proportions as the first system which did. The system crosses the transition state but then recrosses again to finish with an unsuccessful reaction.
|-
| -2.5 || -5.2 || Y || [[File:Cap130084.PNG|thumb|centre|]] || This system is very similar to the previous one, with ''p''2 having a slightly larger value. The approaching H+ atom manages causes and the molecule have enough energy to cross the transition state, recross towards the reactant coordinate and then cross once more to produce a complete reaction.
|-
|}

==Question 5 - Transition State Theory==

The Transition State Theory (TST) assumes that all the molecules obey classical mechanics and works well for heavier molecules but when molecules are very light and quantum mechanical effects are more apparent (i.e. quantum tunneling), TST has its limitations. It also assumes that once the reactants cross the transition state, the reaction will go to completion. In addition, TST assumes that molecules obey the Bolztmann distribution when they occupy energy levels as well as assuming the rate of the reaction rate depends entirely on the transition state. <ref>Moore, John W., ''Kinetics and Mechanism'', A Wiley-Intersection publication, John Wiley & Sons, p. 166, 1981.</ref> <ref>Levine, Ira N., ''Physical Chemistry'', 6th ed, McGraw-Hill, p. 893, 2009.</ref>

Due to the fact that we are dealing with light molecules in this system, TST would not match up very well with experimental values as many of the assumptions would be invalid especially when quantum effects have more of an influence (i.e. quantum tunneling). <ref>Miller, William H., ''Semi-classical limit of quantum mechanical transition state theory for nonseparable systems'', J. Chem. Phys., Vol. 62, No. 5, 1975.</ref>

{{fontcolor1|red|(Would predicted values be higher or lower than expected? Did you notice any barrier recrossing in your simulations? How does that reflect on TST's assumptions? [[User:Lt912|Lt912]] ([[User talk:Lt912|talk]]) 03:58, 2 June 2017 (BST))}}

= F - H - H System=

==Question 6 - PES Inspection==
From ''Figure 9'' it can be seen that the system where the F+ travels towards the H2 molecule to produce H-F as products is an '''exothermic''' as the reactants are higher in potential energy than the products.

[[File:Cap140084.PNG|thumb|centre|''Figure 9'': A surface plot (Dynamics) for the F+ + H2 → HF + H+ system]]

From ''Figure 10'' it can be seen that the system where H+ travels towards the HF molecule to produce H2 as products is an '''endothermic''' as the reactants are lower in potential energy than the products.

[[File:Cap150084.PNG|thumb|centre|''Figure 10'': A surface plot (Dynamics) for the H+ + HF → H2 + F+ system]]

Using tabulated bond enthalpies, the enthalpy of reaction (ΔHR) can be calculated simply by considering the enthalpy of bonds broken (ΔHbroken) and the bonds formed (ΔHformed):

ΔHR = ΔHbroken - ΔHformed

H-F Bond = 569 kJ/mol and H-H Bond = 436 kJ/mol.<ref>T. L. Cottrell, ''The Strengths of Chemical Bonds'', 2nd ed., Butterworth, London, 1958</ref>

For the F+ + H2 → HF + H+ system ΔHR = '''-133 kJ/mol''' (exothermic) and for the H+ + HF → H2 + F+ system ΔHR = '''+133 kJ/mol''' which confirms the prediction of the initial potential surface calculations.

==Question 7 - Approximate Transition State Position==
The approximate position for the transition state is rHaHb = 0.74 Å and rHbF = 1.81 Å (3 s.f.). This was achieved using a similar trial and error method used in ''Question 2'' with the help of the data cursor.

From ''Figure 11'' it can be seen that when this separation is given zero momentum there is virtually no resulting movement and therefore supports that these values are a good approximation for the position of the transition state.

For the opposite reaction, the transition state will also be in the same position for the opposite reaction since they pass through the same transition state.

[[File:Cap160084.PNG|thumb|centre|''Figure 11'': A plot showing Internuclear distance vs Time for the for H2 + F+ → H+ + HF system.]]

==Question 8 - Activation Energies==
The activation energy is defined as the difference in potential energy from the reactants to the products. Using one system only, it is possible to calculate the activation energy for both reactions as they are essentially the same but in reverse, the difference between reactants and the transition state tells you the activation energy of the reverse reaction.

[[File:Cap170084.PNG|thumb|centre|''Figure 12'': A surface plot H2 + F+ → H+ + HF system showing the potential energy for the '''reactants''' kcal/mol (Z axis).]]

[[File:Cap180084.PNG|thumb|centre|''Figure 13'': A surface plot H2 + F+ → H+ + HF system showing the potential energy for the '''transition state''' kcal/mol (Z axis).]]

[[File:Cap190084.PNG|thumb|centre|''Figure 14'': A surface plot H2 + F+ → H+ + HF system showing the potential energy for the '''products''' in kcal/mol (Z axis).]]

The activation energy for H2 + F+ → H+ + HF = (-103.7) - (-103.9) = 0.2 kcal/mol = 0.84 kj/mol.

The activation energy for H+ + HF → H2 + F+ = (-103.7) - (-133.9) = 30.2 kcal/mol = 126 kj/mol.

==Question 9 - Reaction Dynamics - Reaction Energy==
Energy can only be transformed from one form to another. Potential energy can be transformed into vibrational energy and the opposite is also true. for the exothermic reaction of F+ + H2 → HF + H+ system ΔHR, the excess potential energy can be converted to kinetic energy (vibrational energy or translational energy). Some of the energy is also lost as heat to the surroundings.

{{fontcolor1|red|(What surroundings? What is heat? [[User:Lt912|Lt912]] ([[User talk:Lt912|talk]]) 04:01, 2 June 2017 (BST))}}

The extent of reaction can be confirmed experimentally by a calorimetry to measure the transfer of heat to and from the system. The difference in vibrational energy of the products and reactants can be probed using IR spectroscopy methods to give a more detailed distribution of energy conversions between the reactants and products.

{{fontcolor1|red|(How exactly would you use IR to do this? [[User:Lt912|Lt912]] ([[User talk:Lt912|talk]]) 04:01, 2 June 2017 (BST))}}

==Question 10 - Reaction Dynamics - 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.

According to Poliyani's Rules, a reaction that proceeds via an early transition state (exothermic), relative translational energy between the colliding species is the most important factor in overcoming the activation energy. For a reaction with a late transition state (endothermic), relative vibrational energies is more of a governing factor in the efficacy of the reaction.<ref>Guo, Hua, ''Control of chemical reactivity by transition-state and beyond'', Chem. Sci., 7,pp 3992-4003, 2016</ref>

[[File:Cap200084.PNG|thumb|centre|''Figure 15'': A surface plot for the '''exothermic''' H2 + F+ → H+ + HF system showing a '''successful''' reaction.]]

[[File:Cap210084.PNG|thumb|centre|''Figure 16'': A surface plot for the '''exothermic''' H2 + F+ → H+ + HF system showing a '''unsuccessful''' reaction.]]

Comparing ''Figure 15'' and ''Figure 16'' it is clear that there is a larger vibrational contribution (greater amplitude of oscillation in ''Figure 16'') has a significant effect on the reaction. This reaction is exothermic and according to Polanyi's rules, a larger translational contribution would yield a more efficient reaction.

Although ''Figure 15'' shows a successful reaction, a significant amount of vibrational energy is observed in the formation of the products suggesting that there is still significant amount of vibrational energy in the system making the reaction inefficient. This can be seen in the recrossing of the transition state which is quite wasteful as energy is used to arrange the atoms several times over, illustrating that Polanyi's Rules have a profound effect on the efficiency of this reaction.

= References =

Eo00842229

2017-06-02T02:58:23Z

Lt912: /* Question 5 - Transition State Theory */

=H + H2 System=

==Question 1 - Gradients at Transition State and Minimum ==

The total value of the potential energy surface at the minimum and at the transition structure will both be zero.

Minima will have a positive value for the second derivative in every axial direction of the energy potential surface whereas a transition state will have a positive value for the second derivative in all axial directions apart from in the reaction pathway.

[[File:Capnew0084.PNG|thumb|centre|''Figure 1'': A surface plot showing the curvature of the minima compared to the transition state]]

Using the reaction pathway as axis helps to tell the difference between these two points as the second derivative of the minima will indicate a minimum whereas that of that of the transition state will indicate a maximum.

==Question 2 - Trajectories from r1 = r2==
To calculate the transition state position, rts, an initial inter-nuclear distance of 1 Å was used with momentum of 0 for both distances AB and BC. ''Figure 2'' shows a vibrating structure about local minimum of the reaction coordinate.

[[File:Cap10084.PNG|thumb|centre|''Figure 2'': A surface plot showing an estimate for the transition state position using the Data Cursor Tool]]

A plot Inter-nuclear Distance vs Time with a value close to the transition state position determined by moving the data cursor up and down to find a minimum inter-nuclear position showed that there was some vibration as the inter-nuclear distances slightly change over time.

[[File:Cap20084.PNG|thumb|centre|''Figure 3'': A plot of Inter-nuclear Distance vs Time for the H-H-H transition state at r1 = r2 = 0.9073 Å showing slight vibrations in the structure.]]

Trying values above and below this initial value for the transition state position allowed to precisely find the position by choosing the value for rts that gave little to no change in inter-nuclear distance with changing time. rts = 0.908 Å (3 d.p.)

[[File:Cap30084.PNG|thumb|centre|''Figure 4'': A plot of Inter-nuclear Distance vs Time for the H-H-H transition state at r1 = r2 = 0.9079 Å showing negligible vribration.]]

==Question 3 - Trajectories from r1 = rts+δ, r2 = rts==
Comparing ''Figure 5'' and ''Figure 6'', it can be seen that the MEP calculation shows a trajectory that does show any vibration whereas the trajectory calculated using the Dynamics calculation does. The MEP calculation also doesn't show any vibration in the bonds whereas the Dynamic calculation does, even with zero momentum for all atoms in the system.

[[File:Cap40084.PNG|thumb|centre|''Figure 5'': A surface plot (MEP) for the H-H-H system with r1 = 0.918 Å and r2 = 0.908 Å.]]

[[File:Cap50084.PNG|thumb|centre|''Figure 6'': A surface plot (Dynamics) for the H-H-H system with r1 = 0.918 Å and r2 = 0.908 Å.]]

This is due to the fact that the MEP resets the velocity to zero for each time step, this therefore doesn't allow the system to transfer kinetic energy into potential energy, keeping the trajectory of the system along the pathway with minimum potential energy. This can be seen when observing ''Figure 7'' and ''Figure 8''.

[[File:Cap70084.PNG|thumb|centre|''Figure 7'': A Potential energy vs. Time for the H-H-H system transition state with r1 = 0.918 Å and r2 = 0.908 Å showing how the system falls to a minimum.]]

[[File:Cap80084.PNG|thumb|centre|''Figure 8'': A Kinetic energy vs. Time plot for the H-H-H system transition state with at r1 = 0.918 Å and r2 = 0.908 Å showing zero kinetic energy over time.]]

==Question 4 - Reactive and unreactive trajectories==
{| class="wikitable" border=1 style="text-align: center;"
! p1
! p2
! Reactive? (Y/N)
!Screenshot
!Description
|-
| -1.25 || -2.5 || Y || [[File:Cap90084.PNG|thumb|centre|]] || The approaching H+ atom, C, comes in with enough momentum to overcome the kinetic barrier for activation and goes to completion. There is little change in the motion of the A-B bond but once the product is formed, there is some vibration in the bond, suggesting that some of the translational energy of the approaching C is transformed into vibrational energy in the new molecule of B-C.
|-
| -1.5 || -2.0 || N || [[File:Cap100084.PNG|thumb|centre|]] || The approaching H+ atom, C, does not have enough kinetic energy to overcome the barrier for the reaction. It initially approaches quite quickly but slows down as it gets closer to the molecule B-C and eventually gets repelled away from the molecule leading to an incomplete reaction (due to an insufficient total momentum). The extra momentum in p1 manifests itself as vibrational motion in the molecule A-B.
|-
| -1.5 || -2.5 || Y || [[File:Cap110084.PNG|thumb|centre|]] || This system is very much similar to the first and also proceeds to completion. The approaching H+ atom, C, has enough kinetic energy to overcome the activation barrier. The only difference is in the greater vibration in B-C due to the large value in its momentum.
|-
| -2.5 || -5.0 || N || [[File:Cap120084.PNG|thumb|centre|]] || The total momentum in this system is very large as can be seen from the much larger vibrational energy in the A-B bond. The approaching H+ atom, C, has a lot of kinetic energy as a result and manages to overcome the activation barrier however it doesn't manage to go to completion despite being in the same proportions as the first system which did. The system crosses the transition state but then recrosses again to finish with an unsuccessful reaction.
|-
| -2.5 || -5.2 || Y || [[File:Cap130084.PNG|thumb|centre|]] || This system is very similar to the previous one, with ''p''2 having a slightly larger value. The approaching H+ atom manages causes and the molecule have enough energy to cross the transition state, recross towards the reactant coordinate and then cross once more to produce a complete reaction.
|-
|}

==Question 5 - Transition State Theory==

The Transition State Theory (TST) assumes that all the molecules obey classical mechanics and works well for heavier molecules but when molecules are very light and quantum mechanical effects are more apparent (i.e. quantum tunneling), TST has its limitations. It also assumes that once the reactants cross the transition state, the reaction will go to completion. In addition, TST assumes that molecules obey the Bolztmann distribution when they occupy energy levels as well as assuming the rate of the reaction rate depends entirely on the transition state. <ref>Moore, John W., ''Kinetics and Mechanism'', A Wiley-Intersection publication, John Wiley & Sons, p. 166, 1981.</ref> <ref>Levine, Ira N., ''Physical Chemistry'', 6th ed, McGraw-Hill, p. 893, 2009.</ref>

Due to the fact that we are dealing with light molecules in this system, TST would not match up very well with experimental values as many of the assumptions would be invalid especially when quantum effects have more of an influence (i.e. quantum tunneling). <ref>Miller, William H., ''Semi-classical limit of quantum mechanical transition state theory for nonseparable systems'', J. Chem. Phys., Vol. 62, No. 5, 1975.</ref>

{{fontcolor1|red|(Would predicted values be higher or lower than expected? Did you notice any barrier recrossing in your simulations? How does that reflect on TST's assumptions? [[User:Lt912|Lt912]] ([[User talk:Lt912|talk]]) 03:58, 2 June 2017 (BST))}}

= F - H - H System=

==Question 6 - PES Inspection==
From ''Figure 9'' it can be seen that the system where the F+ travels towards the H2 molecule to produce H-F as products is an '''exothermic''' as the reactants are higher in potential energy than the products.

[[File:Cap140084.PNG|thumb|centre|''Figure 9'': A surface plot (Dynamics) for the F+ + H2 → HF + H+ system]]

From ''Figure 10'' it can be seen that the system where H+ travels towards the HF molecule to produce H2 as products is an '''endothermic''' as the reactants are lower in potential energy than the products.

[[File:Cap150084.PNG|thumb|centre|''Figure 10'': A surface plot (Dynamics) for the H+ + HF → H2 + F+ system]]

Using tabulated bond enthalpies, the enthalpy of reaction (ΔHR) can be calculated simply by considering the enthalpy of bonds broken (ΔHbroken) and the bonds formed (ΔHformed):

ΔHR = ΔHbroken - ΔHformed

H-F Bond = 569 kJ/mol and H-H Bond = 436 kJ/mol.<ref>T. L. Cottrell, ''The Strengths of Chemical Bonds'', 2nd ed., Butterworth, London, 1958</ref>

For the F+ + H2 → HF + H+ system ΔHR = '''-133 kJ/mol''' (exothermic) and for the H+ + HF → H2 + F+ system ΔHR = '''+133 kJ/mol''' which confirms the prediction of the initial potential surface calculations.

==Question 7 - Approximate Transition State Position==
The approximate position for the transition state is rHaHb = 0.74 Å and rHbF = 1.81 Å (3 s.f.). This was achieved using a similar trial and error method used in ''Question 2'' with the help of the data cursor.

From ''Figure 11'' it can be seen that when this separation is given zero momentum there is virtually no resulting movement and therefore supports that these values are a good approximation for the position of the transition state.

For the opposite reaction, the transition state will also be in the same position for the opposite reaction since they pass through the same transition state.

[[File:Cap160084.PNG|thumb|centre|''Figure 11'': A plot showing Internuclear distance vs Time for the for H2 + F+ → H+ + HF system.]]

==Question 8 - Activation Energies==
The activation energy is defined as the difference in potential energy from the reactants to the products. Using one system only, it is possible to calculate the activation energy for both reactions as they are essentially the same but in reverse, the difference between reactants and the transition state tells you the activation energy of the reverse reaction.

[[File:Cap170084.PNG|thumb|centre|''Figure 12'': A surface plot H2 + F+ → H+ + HF system showing the potential energy for the '''reactants''' kcal/mol (Z axis).]]

[[File:Cap180084.PNG|thumb|centre|''Figure 13'': A surface plot H2 + F+ → H+ + HF system showing the potential energy for the '''transition state''' kcal/mol (Z axis).]]

[[File:Cap190084.PNG|thumb|centre|''Figure 14'': A surface plot H2 + F+ → H+ + HF system showing the potential energy for the '''products''' in kcal/mol (Z axis).]]

The activation energy for H2 + F+ → H+ + HF = (-103.7) - (-103.9) = 0.2 kcal/mol = 0.84 kj/mol.

The activation energy for H+ + HF → H2 + F+ = (-103.7) - (-133.9) = 30.2 kcal/mol = 126 kj/mol.

==Question 9 - Reaction Dynamics - Reaction Energy==
Energy can only be transformed from one form to another. Potential energy can be transformed into vibrational energy and the opposite is also true. for the exothermic reaction of F+ + H2 → HF + H+ system ΔHR, the excess potential energy can be converted to kinetic energy (vibrational energy or translational energy). Some of the energy is also lost as heat to the surroundings.

The extent of reaction can be confirmed experimentally by a calorimetry to measure the transfer of heat to and from the system. The difference in vibrational energy of the products and reactants can be probed using IR spectroscopy methods to give a more detailed distribution of energy conversions between the reactants and products.

==Question 10 - Reaction Dynamics - 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.

According to Poliyani's Rules, a reaction that proceeds via an early transition state (exothermic), relative translational energy between the colliding species is the most important factor in overcoming the activation energy. For a reaction with a late transition state (endothermic), relative vibrational energies is more of a governing factor in the efficacy of the reaction.<ref>Guo, Hua, ''Control of chemical reactivity by transition-state and beyond'', Chem. Sci., 7,pp 3992-4003, 2016</ref>

[[File:Cap200084.PNG|thumb|centre|''Figure 15'': A surface plot for the '''exothermic''' H2 + F+ → H+ + HF system showing a '''successful''' reaction.]]

[[File:Cap210084.PNG|thumb|centre|''Figure 16'': A surface plot for the '''exothermic''' H2 + F+ → H+ + HF system showing a '''unsuccessful''' reaction.]]

Comparing ''Figure 15'' and ''Figure 16'' it is clear that there is a larger vibrational contribution (greater amplitude of oscillation in ''Figure 16'') has a significant effect on the reaction. This reaction is exothermic and according to Polanyi's rules, a larger translational contribution would yield a more efficient reaction.

Although ''Figure 15'' shows a successful reaction, a significant amount of vibrational energy is observed in the formation of the products suggesting that there is still significant amount of vibrational energy in the system making the reaction inefficient. This can be seen in the recrossing of the transition state which is quite wasteful as energy is used to arrange the atoms several times over, illustrating that Polanyi's Rules have a profound effect on the efficiency of this reaction.

= References =

MRD:10969OOR

2017-06-02T02:53:15Z

Lt912: /* mechanism of release of the reaction energy */

=''' Molecular Reaction Dynamics: Applications to Triatomic systems'''=
== Introduction ==
The experiment is to analyze reactivity and to locate transition states of a triatomic system mainly by looking at the potential surface plots and internuclear distance vs time plots drawn by Matlab.
A H-H-H and a H-H-F/F-H-H systems are studied in this experiment. A potential energy surface represents the energy of a system as a function of atoms' relative position. In the surface plot, the trajectories represent a path cross the potential surface.Moreover, the trajectories help to visualize the reaction pathway and reactivity. Based on Hammond postulate, locations of transition states and activation energies are found for various systems. Transition State theory is also discussed in this experiment for example, once the system reaches the transition states with sufficient energy then the products can form. The experiment also investigates about Polanyi's empirical rules as different transition states positions will have different translational and vibrational modes in the energy distribution, which directly affect the efficiency of the reaction.

== H + H2 system ==

=== Dynamics from the transition state region ===

: The total gradient of the potential energy surface at a minimum and at a transition structure is '''0'''. The transition state is the maximum position on the minimum energy path, the minimum is at the lowest position on the potential energy curve therefore gradients at the transition state and at the minimum are both zero.(first derivative=0) However, minima and transition states can be distinguished by working out second derivatives at each point as transition states have second derivatives less than zero and the minima have second derivatives greater than zero.

: Because at the transition state of the reaction, r1 = r2, therefore distance can be first roughly estimated by finding the intersection(rts = 0.9176 Å, t = 0.38 s) in the Figure 1 and then by testing conditions at around rts=0.91 Å, p=0 N/s, the best estimate('''rts = 0.908 Å''') is obtained until the internuclear distance vs time graph shows straight lines as molecules at the transition state have almost no vibrational energy.

{| class="wikitable" border="1"

|-
| [[File:Roughtest for ts.PNG|thumb|450px|center|Figure 1 Internuclear Distance vs Time at r1=0.74 Å, r2=2.30 Å.]]|| [[File:best estimate.PNG|450px|thumb|center|Figure 2 Internuclear Distance vs Time at transition state.]]
|-
|}

=== Comparing MEP and Dynamics calculation types ===

* Initial conditions are set to be ''' r1 = rts + 0.01 = 0.918 Å, r2 = rts''', initial momenta p1= p2

{| class="wikitable" border="10"

|-
| [[File:Mepr1 surface.PNG|450px|thumb|center|Figure 3 Surface plot (MEP) at r1 = 0.918.]] || [[File:Dynr1 surface.PNG|450px|thumb|center|Figure 4 Surface plot (Dynamics) at r1 = 0.918 Å.]]
|-
|}

: Graphs obtained by MEP calculation type generally have smooth curves, which is because MEP calculation sets the velocity to zero in each time step therefore atoms moves in an infinitely slow motion. Dynamics calculation type takes the realistic motions of atoms into account thus the fluctuated curves and trajectories imply the consideration of vibrational motions.

{| class="wikitable" border="1"
|-
| [[File:Mepr1 interdistance.PNG|450px|thumb|center|Figure 5 internuclear distance vs time (MEP) at r1=0.918.]] || [[File:Dynr1 interdistance.PNG|450px|thumb|center|Figure 6 internuclear distance vs time(Dynamics) at r1 = 0.918 Å.]]
|-
|}

{| class="wikitable" border="1"

|-
| [[File:Mepr1 intermomenta.PNG|450px|thumb|center|Figure 7 Internuclear momenta vs time (MEP) at r1=0.918.]] || [[File:Dynr1 intermomenta.PNG|450px|thumb|center|Figure 8 internuclear momenta vs time (Dynamics) at r1 = 0.918.]]
|-
|}

* Initial conditions are set to be '''r2 = rts + 0.01 = 0.918, r1 = rts''', initial momenta p1= p2.

{| class="wikitable" border="1"

|-
| [[File:Mepr2 surface.PNG|450px|thumb|center|Figure 9 Surface Plot (MEP) at r2 = 0.918.]] || [[File:Dynr2 surface.PNG|450px|thumb|center|Figure 10 Surface Plot (Dynamics) at r2 = 0.918.]]
|-
|}

Because now r2 is greater than r1 therefore the trajectory in the surface plot passes through the entrance channel(reactant side).

=== Reactive and unreactive trajectories ===

: Five trajectories with the same initial positions (r1=0.74 and r2= 2.0) but with different combinations of momenta were tested to see if they have enough kinetic energy to overcome the activation barrier. In the potential energy surface plot of a reactive trajectory, wavy lines should appear in exit channels.

{| class="wikitable" border=1
! Trajectory !! p1 !! p2 !! Reactivity
|-
| 1|| -1.25 || -2.5 || Reactive
|-
| 2|| -1.5 || -2.0 || Unreactive
|-
| 3|| -1.5 || -2.5 || Reactive
|-
| 4|| -2.5 || -5.0 || unreactive
|-
| 5|| -2.5 || -5.2 || reactive
|}

==== Trajectory 1---reactive ====

{| class="wikitable" border="1"
|-
| [[File:T1.PNG|450px|thumb|center|Figure 11 surface plot for Trajectory 1.]] || [[File:T1 imd.PNG|450px|thumb|center|Figure 12 Internuclear distance vs time plot for Trajectory 1.]]
|-
|}

* The trajectory is reactive.
* The decreasing A-B bond distance indicates that HA approaches HB-HC. The increasing B-C bond distance and the wavy line in the exit channel indicate the formation of HA-HB and this also proves that the reaction has enough kinetic energy to surmount the energy barrier.
* The intersection between A-B bond and B-C bond curves in the Figure 12 indicate the formation of transition state.

==== Trajectory 2---unreactive ====
{| class="wikitable" border="1"
|-
| [[File:T2.PNG|450px|thumb|center|Figure 13 surface plot for Trajectory 2.]] || [[File:T2 idm.PNG|450px|thumb|center|Figure 14 Internuclear distance vs time plot for Trajectory 2.]]
|-
|}
* The trajectory is unreactive.
* The decrease in the first half of the A-B bond distance curve in the Internuclear distance vs time plot illustrates the fact that HA approaches HB-HC but the increasing second half of the A-B bond distance curve shows that Atom A bounces back. The trajectory bounces back to the entrance channel in the surface plot indicates that the reaction does not have enough kinetic energy to surmount the energy barrier.

==== Trajectory 3---reactive ====
{| class="wikitable" border="1"
|-
| [[File:T3.PNG|450px|thumb|center|Figure 15 surface plot for Trajectory 3.]] || [[File:T3 idm.PNG|450px|thumb|center|Figure 16 Internuclear distance vs time plot for Trajectory 3.]]
|-
|}
* The trajectory is reactive.
* The surface plot and intermolecular distance vs time plot for this trajectory has a similar shape as the trajectory 1. The progressing wavy line in the exit channel proves that the reaction has enough kinetic energy to overcome the energy barrier.

==== Trajectory 4---unreactive ====
{| class="wikitable" border="1"
|-
| [[File:T4.PNG|450px|thumb|center|Figure 17 surface plot for Trajectory 4.]] || [[File:T4 Imd.PNG|450px|thumb|center|Figure 18 Internuclear distance vs time plot for Trajectory 4.]]
|-
|}
* The trajectory is unreactive.
* The two intersections in the exit and entrance channel indicates that the transition states formed twice during the reaction. Though P2 is given enough momentum to collide with B-C and the reaction has enough kinetic energy to overcome the barrier after forming the first transition state, the energy is not in the right vibrational modes at the right time thus the trajectory is bounced off to the entrance channel. The excess momenta of Atom A allows Atom A to attempt to form the second transition state but then the reaction has not enough energy to overcome the kinetic barrier.
* Some of the translational energy of Atom A converts to the vibrational energy of B-C thus the trajectory shows a wavy line after bouncing back to entrance channel.

==== Trajectory 5---reactive ====
{| class="wikitable" border="1"
|-
| [[File:T5.PNG|450px|thumb|center|Figure 19 surface plot for Trajectory 3.]] || [[File:T5 imd.PNG|450px|thumb|center|Figure 20 Internuclear distance vs time plot for Trajectory 5.]]
|-
|}
* The trajectory is reactive.
* The three intersections in the internucleara distance vs time graph indicates that the transition states formed three times in the reaction and eventually formed the product.
* As P2 is given the highest momenta among all 5 trajectories therefore HA-HB has the highest vibrational energy and the oscillating curve in the exit channel proves that the reaction has enough kinetic energy to surmount the barrier.

=== Transition State Theory ===

==== Assumptions ====
* Reactants and transition states have a Boltzmann distribution of energy in every degree of freedom.<ref>G. Mills, H. Jdnsson and G. K. Schenter,.</ref>
* Atoms follow newton's laws and molecular vibrations are quantized or quantum tunneling effects are neglected.<ref>P. Pechukas, Ann.Rev.Phys.Chem, 1981, 159–177. </ref>
* The system reaches the transition state only once in the reaction process.
* TST assumes that a trajectory that reaches the transition state in the direction from reactants to products is a reactive trajectory and once the trajectory passes the transition state, it is almost impossible to find the way back to reactants.
* Conventional TST locates the transition state at the saddle point and all vibrations and rotations about the saddle point are out consideration.

==== Compare with experimental values ====
* In reality, trajectories may reach the transition states for several times before ultimately forming the product and these are counted in TST. As a result, TST often overestimates reaction rates.<ref>G. Mills, H. Jdnsson and G. K. Schenter,.</ref>
* Rate constants calculated without tunneling usually underestimate the exact rate constants at low temperature.

== F-H-H System ==
=== PES inspection ===
==== F + H2 ====
* reaction profile
[[File: Fhhreaction.PNG|thumb|450px|center|Figure 21 Surface plot of F+ H2.]]
F + H2 is an exothermic reaction. Because in the surface plot,the entrance channel(reactants) has a higher surface potential energy comparing to the products as H-F bond is stronger than H-H bond therefore heat energy is released in the reaction.
* the transition state location
[[File: Fhhts.PNG|thumb|450px|center|Figure 22 internuclear distance vs time of F+ H2.]]
According to Hammond postulate, the transition state resembles the structure of species neighboring it along the reaction coordinate thus in an exothermic reaction, the transition state is close in energy to the reactants. The transition state location on the potential surface can be found close to the reactants.
The transition state is found when rAB = 1.81 Å, rBC = 0.745 Å, momentum = 0.
* calculating activation energy
[[File: Fhhpevst.PNG|thumb|450px|center|Figure 23 potential energy vs time of F+ H2.]]
The activation energy for this reaction is the energy difference between the reactants and transition states, which is 0.2 Kcal/mol.

==== H + HF ====
* reaction profile
[[File: Hhfreaction.PNG|thumb|450px|center|Figure 24 Surface plot of F+ H2.]]
H + HF is an endothermic reaction. Because the entrance channel(reactants) has a lower surface potential energy comparing to the products. As H-F bond is much stronger than H-H bond therefore breaking H-F bond intakes heat energy. In an endothermic reaction pathway, the transition state resembles products more than the reactants.
* the transition state location
[[File: Fhfts.PNG|thumb|450px|center|Figure 25 internuclear distance vs time of H + HF.]]
* calculating activation energy
[[File: Newfhfidvst.PNG|thumb|450px|center|Figure 26 internuclear distance vs time of H + HF.]]
The activation energy for this reaction is the energy difference between the reactants and transition states, which is 30.1 Kcal/mol.

=== Reaction dynamics ===
==== mechanism of release of the reaction energy ====
{| class="wikitable" border="1"
|-
| [[File: 2.6.PNG|thumb|350px|center|Figure 27 surface plot of H + HF.]] || [[File:2.6imvst.PNG|350px|thumb|center|Figure 28 Internuclear momenta vs Time plot for H + H-F.]]|| [[File:2.6id.PNG|350px|thumb|center|Figure 29 Internuclear distance vs Time plot for H + H-F.]]
|-
|}
* The initial conditions are set as: rAB= 2.3 Å, rBC= 0.74 Å, pAB= -2.6 N/s,pBC= 0 N/s.
* F + H2 is a two-step reaction. In the first step, distance between H and F decreases as F atom approaches H atom. After forming the first transition state, the reaction fails to overcome the kinetic energy barrier thus the distance between H atoms increases slightly. The second transition state is reached as the F atom attack the H atom again. The A-B curve in the internuclear momenta vs time oscillates at a very high amplitude indicates that the formation of H-F bond is exothermic and the released heat energy converts to the vibrational energy of H-F bond.
* This can be confirmed by infrared chemiluminescence and calorimetry. Because the reaction is exothermic therefore it can be followed by the raising of the temperature.(Q = mcΔt, where Q is the heat released, c is the specific heat capacity and Δt is the changes in temperature.)
* As the reaction is exothermic, the released heat converts to vibrational energy and excites products. Infrared chemiluminescence can be used to measure the emission of the infrared photons from vibrationally excited products. The intensity in the chemiluminescence spectroscopy can distinguish between attractive and repulsive potential surfaces. The reaction with attractive potential surface proceeds more efficiently when the energy is mostly translational energy and the reaction with repulsive potential surface proceeds more efficiently when the energy is as vibrations.<ref>Atkins & De Paula Physical Chemistry, 9 edn., 2010. </ref>

{{fontcolor1|red|(In what form of energy is the "heat" stored?[[User:Lt912|Lt912]] ([[User talk:Lt912|talk]]) 03:53, 2 June 2017 (BST))}}

==== Polanyi's empirical rules ====
{| class="wikitable" border="1"
|-
| [[File: Newreactive.PNG|thumb|450px|center|Figure 30 surface plot of H + HF.]] || [[File:Reactive.PNG|450px|thumb|center|Figure 31 surface plot of H + H-F.]]
|-
|}
initial conditions for unreactice H + HF: rAB= 2 Å, rBC= 0.9 Å, pAB= -2.2 N/s,pBC= 1.9 N/s.
initial conditions for reactice H + HF: rAB= 2 Å, rBC= 0.9 Å, pAB= -0.6 N/s,pBC= 7.9 N/s.

The Polanyi's empirical rules, which state that the vibrational energy contributes more in a late barrier reaction than the translational energy. <ref>J. M. Bowman, 2012, 4, 4–7. </ref>
The endothermic reaction is a late barrier reaction as the transition state resemble more from the products than reactants thus a system has a higher vibrational energy will have a higher possibility to overcome the transition barrier. H + H-F is an endothermic reaction. It is clear that the system in the Figure 30 has a higher translational energy and lower vibrational energy than in the Figure 31 therefore higher the vibrational energy in the Figure 31, more contribution for the reaction to surmount the barrier.

== Conclusion ==
The surface plots can be used to identify the transition sates and reactivity. The trajectory travels from the entrance channel to the exit channel on the surface plot implies that the reaction is reactive. Otherwise, if the trajectory stops before travelling into the exit channel, the reaction is unreactive.
Transition State theory and Hammond Postulate help to identify transition states location and reaction pathway. In this experiment, it was proved that F + H-H reaction is exothermic and its reverse reaction (H + H-F) is endothermic, which has an higher activation energy. Polanyi's empirical rules can be illustrated by Figure 30 and 31, as vibrational energy is more efficient in promoting for a late barrier(endothermic) reaction than translational energy.

== Reference ==

Ct1515MRD

2017-06-02T02:38:49Z

Lt912: /* Exercise 1: H + H2 */

=='''Exercise 1: H + H2'''==
'''Question 1'''

'''What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.'''

At the minimum point the total gradient of potential energy is 0 and at the transition state (maxima) it will also be 0. They can be distinguished by taking the second derivative and the minima value will a postive value where as the maxima (transition state) value will give a negative second derivative.

{{fontcolor1|red|(This won't be true if you take the second derivative along the axis where the transition state sits at a minimum. [[User:Lt912|Lt912]] ([[User talk:Lt912|talk]]) 03:38, 2 June 2017 (BST))}}

'''Question 2''' - '''Location of Transition state'''

[[File:Transition state 2.png|500px|thumb|center|Figure 1]]
To find the transition state initially conditions with r1 = r2 = 1.4, and p1 = p2 = 0.0. An initial estimate of the transition state was found to be X= 0.9067 Å, Y= 0.9067 Å and Z= -98.8 Kcal/mol. This can be seen in Figure 1 and was found by roughly estimating the maxima value by using the curser.

[[File:Tranistion_State_internuclear_distance.png|500px|thumb|center|Figure 2]]
Upon adjustment of the internuclear distance when the bond length was 0.9079 Å the fluctuations of bond length became negligible. This can be seen in Figure 2 and shows that the trasition state bond lenth had been achieved.

'''Question 3''' - '''Calculating the reaction path'''

[[File:Reaction_Path.png|500px|thumb|center|Figure 3]]
[[File:Reaction_Path_Dynamics.png|500px|thumb|center|Figure 4]]

'''Comment on how the mep and the trajectory you just calculated differ.'''

Figure 3 and 4 show the reaction pathway with MEP analysis and Dynamics analysis respectively. A small change of 0.01 was made to the value of r2. As is can be seen the MEP analysis doesn't take into account the velocity and shows a very straight trajectory. Dynamics does take into account the velocity of the particles and therefore oscillates.
Final Value of A-B = 0.7454 Å, B-C = 5.281 Å.

'''Question 4''' - '''Reactive and unreactive trajectories'''

{| class="wikitable" border="1"
|+
! p1 !! p2 !! '''Unreactive or Reactive'''
|-
| -1.25 || -2.5 || Reactive
|-
| -1.5 || -2.0 || Unreactive
|-
| -1.5 || -2.5 || Reactive
|-
| -2.5 || -5.0 || Unreactive
|-
| -2.5 || -5.2 || Reactive
|-
|}

[[File:2,_0.74.png|500px|thumb|center|Figure 5: p1 = -1.25, p2 = -2.5]]
In this reaction trajectory the molecules overcome the activation barrier to react. The Ha approaches the Hb-Hc molecule decreasing its bond length and then upon reaction the Hb-Hc bond length increases as Hc moves away.

[[File:-1.5,_2.png|500px|thumb|center|Figure 6: p1 = -1.5, p2 = -2.0]]
In this reaction trajectory the molecules do not overcome the activation barrier and therefore do not react.The Ha approaches the Hb-Hc but without enough energy to over come activation energy.

[[File:-1.5,_-2.5.png|500px|thumb|center|Figure 7: p1 = -1.5, p2 = -2.5]]
In this reaction trajectory the molecules overcome the activation barrier to react.

[[File:-2.5, -5.png|500px|thumb|center|Figure 8: p1 = -2.5, p2 = -5.0]]
In this reaction trajectory the molecules overcome the activation barrier to react initially then return to be below activation energy and therefore do not end up reacting. This is due to the significant change in momentum.

[[File:-2.5,_-5.2_(ct).png|500px|thumb|center|Figure 9: p1 = -2.5, p2 = -5.2]]
In this reaction trajectory the molecules overcome the activation barrier to react initially then return to be un-reacted and then overcome the activation barrier once more to react. This is due to the momentum

'''Question 5''' - '''Transition State Theory'''

The transition state theory assumes equilibrium between the reactants and an activated transition state. Transition State Theory and experimental values will differ when transition states are very unstable and short lived as the momentum of the reaction trajectory of the reactants will affect that of the transition state and therefore the selectivity of the products.<ref name="[1]" /> Also this theory assumes that atomic nuclei behave according to classical systems and doesn't take into account the fact that nuclei can tunnel across the barrier without having the activation energy. As the transition state is a saddle point (a minimax point: in a 3D graph of internuclear distance of the reactants and products vs potential energy, it is a minima point between the internuclear distances but a maxima between internuclear distance and potential energy)<ref name="[2]"/> the reaction can also take alternative routes without having activation energy.

=='''Exercise 2'''==

'''F + H2 and FH + H'''

* '''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.'''

* '''Report the activation energy for both reactions.'''

[[File:HH-F.png|500px|thumb|center|Figure 10: HH and F reaction]]
This reaction appears to be exothermic as the energy of the products is lower than that of the reactants. This bares true because it will be in a lower energy state upon forming a very strong H-F bonds.

[[File:FH-H.png|500px|thumb|center|Figure 10: FH and H reaction]]
This reaction appears to be an endothermic reaction at the energy level of the products is a lot higher than that of the reactants. This bares true to the fact that he H-F bond is very strong and therefore will have a lower energy system.

When the transition state isn't obvious Hammonds postulate can be employed. It states that when the reaction is endothermic, the transition state represents that of the products as they are closer in energy to the transition state; if the reaction is exothermic the transition state represents that of the reactants as it is closer in energy to the maximum point on the energy diagram.<ref name="[3]" />

[[File:Tranistion_State_Zoom,_HF,_H.png|500px|thumb|centre|Figure 11: FH and H reaction]]
[[File:Transition_State,_HF,_H.png|500px|thumb|centre|Figure 12: FH and H reaction]]

In this endothermic reaction it can be seen that the transition state is when the bond length between H-F is 1.84 Å and the bond length between H-H is 0.74 Å. It can be seen that this is very similar to that of the products as the slight parabola is very small and can be seen in Figure 11 by the change in gradient either side of the point on the line selected. The activation energy is:
Energy of the transition state - Energy of reactants = -103.7-(-133.9 kcal/mol) = +30.2 Kcal/mol.

[[File:Transition_state_zoom,_HH,_F.png|500px|thumb|centre|Figure 13: HH and F reaction]]
[[File:Reactant_Energy_HH_-_F.png|500px|thumb|centre|Figure 14: Reactant Energy]]

The values for this reaction should be very similar in terms of the transition state value. The activation energy is very small for this reaction:
Energy of the transition state - Energy of reactants = -103.7-(-103.9)= + 0.2 kcal/mol
Therefore this has a lot smaller activation energy than that of the other reaction due to the fact it is an exothermic reaction.

'''In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?'''

[[File:Internuclear_momenta_vs_time.png|500px|thumb|centre|Figure 15: Internuclear_momenta_vs_time]]

As the potential energy of the products is lower than that in the reactants in an exothermic reaction, the kinetic energy of the reactants must be lower than that in the products due to the fact that overall energy must be conserved. Due to the H-F bond being a lot stronger than the H-H bond it will have a stronger vibrational energy and therefore higher kinetic energy. This vibrational mode will be seen in an IR spectra. There will be one specific to the H-H bond and the H-F bond with the H-F bond being at a higher energy (higher wavenumber). From Figure 15 it can be seen that the magnitude of momentum is much lager in the A-B bond (H-F) than that of the B-C (H-H) bond meaning that also the kinetic energy is a lot larger.

'''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 state that increased vibrational energy favours a late transition state, which occurs in an endothermic reaction where the products resemble the transition state. On the other hand, increased translational energy favours an early transition state, which occurs in an exothermic reaction where the reactants resemble the transition state.<ref name="[4]"/>

In the H-H + F reaction, as it is an exothermic reaction, high translational energy and low vibrational energy will provide a less efficient trajectory and therefore make the reaction less likely to go to completion. A increased momentum value is proportional to high translational energy and increased bond length is proportional to decreased vibrational energy. This relationship can be seen in Figure in 16 and 17, when the momentum was increased the trajectory was a lot less efficient and the reaction didn't go to completion.

[[File:Exothermic_Low_T_Energy.png|500px|thumb|centre|Figure 16: Exothermic - Low Translational Energy]]
[[File:Exothermic_High_T_energy.png|500px|thumb|centre|Figure 17: Exothermic - High Translational Energy]]

As the F-H + H reaction is endothermic it is therefore expected that a system with low translational energy but high vibrational energy will favour the transition state and give a less efficient trajectory and means the reaction is less likely to go to completion. A low momentum value is proportional to low translational energy and decreased bond length is proportional to increased vibrational energy. This relationship can be seen in Figure in 18 and 19, when the bond length was increased and therefore the vibrational energy decrease the trajectory was a lot more efficient and the reaction did go to completion.

[[File:Low_Vib_energy.png|500px|thumb|centre|Figure 18: Endothermic - Low Vibrational Energy]]
[[File:High_Vib_energy.png|500px|thumb|centre|Figure 19: Endothermic - High Vibrational Energy]]

=='''Bibliography'''==
<references>
<ref name="[1]">https://en.wikipedia.org/wiki/Transition_state_theory access date: 18.05.17</ref>
<ref name="[2]">https://en.wikipedia.org/wiki/Saddle_point access date: 18.05.17</ref>
<ref name="[3]">https://chem.libretexts.org/Reference/Organic_Chemistry_Glossary/Hammond%E2%80%99s_Postulate access date: 18.05.17</ref>
<ref name="[4]">"Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction"; Zhaojun Zhang, Yong Zhou,† and Dong H. Zhang*; dx.doi.org/10.1021/jz301649w; J. Phys. Chem. Lett. 2012, 3, 3416−3419 © 2012 American Chemical Society; Available at URL: http://pubs.acs.org/doi/pdf/10.1021/jz301649w access date: 18.05.17</ref>
</references>

MRD:JDN15

2017-06-02T02:34:33Z

Lt912: /* Reaction Dynamics */

== EXERCISE 1: H + H2 system ==

=== Value of gradient at minimum and at transition state. ===
'''''Q: What value does the total gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.'''''

The total gradient of the potential energy surface at minimum = '''0'''.

The total gradient of the potential energy surface at a transition structure = '''0'''.

While their total gradient are both zero, they can be differentiated by observing the curvature of the potential energy surfaces. At the minima, the curvature of the potential energy surface has a '''positive second derivate (i.e concave). '''However, at the transition structure, it is a saddle point as the '''second derivate will be negative'''.

{{fontcolor1|red|(As a saddle point it will the maximum along once axis and the minimum along another. [[User:Lt912|Lt912]] ([[User talk:Lt912|talk]]) 03:22, 2 June 2017 (BST))}}

=== Estimating Transition State Position (rts) ===
'''''Q:''''' '''''Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” screenshot for a relevant trajectory.'''''

rts = 0.9075 Å

[[File:IMG_1.png|600x600px|thumb|none]]

As see in the "Internuclear Distance vs Time" graph, we can observe that there is minimal oscillations. This represents that the atoms are near its equilibrium positions and it is near its minimum potential energy, which is its transition state.

=== '''Difference in Trajectory between MEP and Dynamic Calculation methods''' ===
'''''Q: Comment on how the mep and the trajectory you just calculated differ.'''''

Using MEP calculation type, the trajectory does not show any oscillations and hence produces a straight line. However, if dynamic calculation type is used, oscillations are observed as waves in the graph. This is because in MEP calculations, the velocity is reset to 0 after each step, therefore the trajectory will move in the direction of the steepest descent each step. This corresponds to the minima of the potential surface, which is the valley floor. In dynamic calculations, the atoms possess an initial velocity that affects its motion and hence trajectory, resulting in oscillations.

=== '''Reactive or Unreactive Trajectories''' ===
'''''Q: Complete the table by adding a column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a screenshot of the trajectory and a small description for what happens along the trajectory.'''''

{| class="wikitable"
!p1
!p2
!Reactivity
!Screenshot
!Description
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|Reactive
|[[File:IMG_2.png|400px|thumb|none]]
|At the start, reactant AB has low vibrational energy while C has translational energy, moving towards AB. Some translational energy is converted to vibrational energy during the reaction, where there is sufficient energy to form BC + A. At the end, the product BC has vibrational energy while A has translational energy, and moves away from BC.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|Unreactive
|[[File:JD_IMG_3.png|400px|thumb|none]]
|At the start, reactant AB has vibrational energy while C has translational energy, moving towards AB. However, C does not have sufficient kinetic energy for the reaction to occur, hence it bounces off the barrier and the reactant AB is regenerated with vibrational energy. C is reflected back and moves further away from B with almost the same amount of vibrational energy.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|Reactive
|[[File:JD_IMG_4.png|400px|thumb|none]]
|At the start, reactant AB has vibrational energy while C has translational energy, moving towards AB. There is sufficient energy to form BC + A, and hence A moves away from BC. The BC bond oscillates due to vibrational energy. The initial trajectory oscillates, which is unlike the first case. This is likely due to larger initial momentum in AB, resulting in higher vibrational energy. At the end, the product BC has vibrational energy while A has translational energy, and moves away from BC.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|Unreactive
|[[File:JD_IMG_5.png|400px|thumb|none]]
|At the start, both reactant AB and C mainly have translational energy. The system crosses the transition state and the product BC is formed temporarily. However, the system re-crosses the transition state and the reactant AB is regenerated. A large amount of translational energy from C is converted to vibrational energy in AB.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|Reactive
|[[File:JD_IMG_6.png|400px|thumb|none]]
|At the start, reactant AB has vibrational energy while C has translational energy. The system crosses the transition state region thrice and eventually the product BC is formed. At the end, product BC has large amount of vibrational energy, suggesting that some translational energy was converted to vibrational energy.
|}

=== Assumptions of Transition State Theory ===
'''''Q: 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?'''''

1. The Transition State Theory assumes that a quasi-equilibrium occurs between reactants and activated transition state complexes. This is when even though the reactant and products are not in equilibrium, the reactant is in equilibrium with the activated transition state complex.

2. The flux of activated complexes in the two directions are independent of each other.

3. Each intermediate is long-lived enough to reach a Boltzmann distribution of energies before continuing to the next step.

4. Unless atoms or molecules collide with enough energy to form the transition structure, the reaction does not occur.

5. The reaction system will pass over the lowest energy saddle point on the potential energy surface.

6. Barrier recrossing is not allowed.

{{fontcolor1|red|(Based on your dynamics calculations, are these assumptions valid? Which of these are unlikely to be true in a real system? [[User:Lt912|Lt912]] ([[User talk:Lt912|talk]]) 03:26, 2 June 2017 (BST))}}

== EXERCISE 2: F - H - H system ==

=== PES INSPECTION ===
'''''Q. Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). '''''

F+H2 → HF + H (1) is an exothermic process, while HF + H → F+H2 (2) is an endothermic process.

'''''Q.''''' '''''How does this relate to the bond strength of the chemical species involved?'''''

In (1), weaker H-H bonds (432 kJmol-1) are broken and stronger H-F bonds are formed (-565 kJmol-1), resulting in a overall stabilisation (exothermic) of the molecule.

In (2), stronger H-F bonds are broken and weaker H-H bonds are formed, resulting in a destabilisation (endothermic) of the molecule.

'''''Q.''''' '''''Locate the approximate position of the transition state.'''''
[[File:JDN15 Equilibriumpos.png|500px|thumb|none]]

The approximate positions are rH-F = 0.181 Å and rH-H = 0.745 Å.

'''''Q. Report the activation energy for both reactions.'''''
[[File:JD15 Reactant-Product.png|600px|thumb|none]]
Energy of Transition F-H-H = -103.3 kJmol-1

Energy of F+H2 = -103.8 kJmol-1

Activation of Energy for (1) = -103.8 - 103.3 = '''0.6 kJmol-1'''

Energy of HF + H = -133.9 kJmol-1

Activation Energy for (2) = -133.9 - 103.3 = '''30.5 kJmol-1'''

=== '''Reaction Dynamics''' ===
'''''Q. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. '''''
[[File:JDN15 Reaction surface.png|500px|thumb|none]]
[[File:JDN15 Movement 10.png|500px|thumb|none]]
At the start of the reaction, H2 has very little vibrational energy. As F moves towards the H2 , it is observed that the system recrosses into the transition state and back many times. The product HF and H eventually forms. HF possess a large amount of vibrational energy while H has little vibrational energy / translational energy only and moves away from HF.

{{fontcolor1|red|(It looks like you were starting your trajectory from the transition state rather than from the reactants. I am also not convinces that the transition satte is being recrossed may times. The cyclic motion seems to happening in the product channel.[[User:Lt912|Lt912]] ([[User talk:Lt912|talk]]) 03:28, 2 June 2017 (BST))}}

'''''Q.''''' '''''How could this be confirmed experimentally?'''''

This can be confirmed via Infrared (IR) chemiluminescence spectroscopy, which will detect the infrared vibration when the excited molecules return down to ground state.

The experiment is also exothermic and hence a calorimeter can be used to determine if the reaction has occurred.

'''''Q. 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.'''''

A reaction can only occur if the reactants possess sufficient energy that is distributed in the correct modes. If the energy modes are not in the correct level, the reaction will not proceed as the molecules will not approach in the right orientation or efficiently.

It is important to consider the energy modes at the transition state as it will affect the efficiency of the reaction. For example, in an exothermic reaction, according the Hammond's postulate, the energy modes of the transition state will resemble more like the reactant. Similarly, for an endothermic reaction, Hammond's postulate suggest that the energy mode of the transition state will resemble more like the product.

{{fontcolor1|red|(This is not a correct interpretation of Hammond's postulate - which does not address distribution of energy in the transition state. I think there is some confusion with Polanyi's rules here. [[User:Lt912|Lt912]] ([[User talk:Lt912|talk]]) 03:34, 2 June 2017 (BST))}}

MRD:JDN15

2017-06-02T02:30:12Z

Lt912: /* Reaction Dynamics */