ChemWiki - User contributions [en]

MRD:01512138

2020-06-26T10:19:57Z

Mys18: /* EXERCISE 1: H + H2 system */

{{fontcolor1|green|Overall this is a very well put together report in terms of presentation and also your answers. You have provided clear explanations and in depth discussions throughout. Great job in including references. Keep up the great work! [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 11:19, 26 June 2020 (BST)}}

=='''EXERCISE 1: H + H2 system'''==
===Dynamics From the Transition State Region===

The transition state on a potential energy surface (PES) is mathematically defined as the (stationary) point at which the gradient of the potential is zero: ∂V(ri)/∂ri=0. The transition-state structure is a saddle point that lies on the lowest energy path between two minima: physically it is the highest energy point on a reaction coordinate and the minima correspond to stable chemical species (reactants and products).

If a trajectory with zero initial momentum is launched at the transition state it will remain at that exact point forever. When the geometry is perturbed from this point in the direction of the reactants or products, the system will 'roll' towards the reactant or product states respectively. This makes the location of the transition state discernible by initiating one of the trajectories close to the transition state. The reactants, products and intermediates share the mathematical property of zero gradient and are all stationary points of the PES. The transition state can be distinguished from local minima by considering the Hessian matrix features: the eigenvalues of the computed matrix will disclose the mathematical nature of any stationary point. For the local minima (of non-linear molecules)the 3N-6 eigenvalues are all positive as movement in any direction will increase the energy. For the transition state, 3N-7 of these eigenvalues will be positive but one will be negative. This saddle point has minimum energy paths in the directions of local minima (reactants, products and intermediates) which represent the reaction coordinates, but the also represents a point of maximum energy. The eigenvector associated with he negative eigenvalue will reveal the direction of the reaction coordinate path.

{{fontcolor1|green|Nice! To distinguish between TS and LM I would look to taking the second derivative. If it is a local minimum then I would expect a positive value in all directions. If it is a TS, as you mentioned is a saddle point on a PES, so it will give a negative value but in an orthogonal direction to that the value will then be positive - that will identify the TS. [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 11:04, 26 June 2020 (BST)}}

====Trajectories from r1 = r2: locating the transition state====

For the symmetric H + H2 surface the transition state was located considering r1 = r2 and p1= p2 = 0.0 g.mol-1.pm.fs-1. An estimate of the transition state position (rts) was found at r1 = r2 = 90.77 ppm. This is verified by looking at a plot of the internuclear distance as a function of time: the gradient of the the distance A-B = B-C = A-C = 0 showing that there is no oscillation of the atoms along the ridge r1 = r2.

[[File:Surface_Plot_H_+_H2.png|300px|thumb|centre|Figure 1. A plot showing the internuclear distance as a function of time at the transition state position for the H+H2 system.]]

{{fontcolor1|green|Good![[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 11:05, 26 June 2020 (BST)}}

====Trajectories from r1 = rts+δ, r2 = rts====

Figure 2. shows the minimum energy path (''mep'') trajectory run at the transition state; the trajectory simply follows the valley floor to H1+ H2-H3. Any point on the path is at an energy minimum in all directions perpendicular to the path and there is "infinite friction" acting on the system. This frictional force means that the system can gain no momentum and "rolls" along the lowest energy path. Contrary to this, the dynamics trajectory (Figure 3.) shows the system has some vibrational energy as the atoms oscillate. This is because of the momentum is increasing and the force on the atoms builds up and leads to vibration. The dynamics calculation therefore represents the system more realistically than the ''mep'' trajectory, since the atoms will have some vibrational energy.

[[File:Surface_Plot_MEP_displaced.png|250px|thumb|left|Figure 2. A plot showing the ''mep'' trajectory of the H + H2 system, where the initial position of H was displaced by 1 pm from the transition state. ]]

[[File:Surface_Plot_Dynamics_displaced.png|250px|thumb|right|Figure 3. A plot showing the dynamics trajectory of the H + H2 system, where the initial position of H was displaced by 1 pm from the transition state.]]

{{fontcolor1|green|Exactly! and we visually see a more wavy line for dynamic. [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 11:06, 26 June 2020 (BST)}}

====Trajectories from r2 = rts+δ, r1 = rts====

Looking at the “Internuclear Distances vs Time” dynamics trajectories for the displacement described above, r1 = rts+δ, r2 = rts, against r2 = rts+δ, r1 = rts it is apparent that the products of each reactive pathway are different. In the dynamics trajectory described above, the products are HB-HC + HA where as with new the initial conditions, r2 = rts+δ, r1 = rts, the products are HA-HC + HB. The trajectories are identical however the change in the displacement of specific atoms leads to this variation in the products.

[[File:internuclear_r1_dynamics_heg18.png|250px|thumb|left|Figure 4. An “Internuclear Distances vs Time” plot showing the dynamics trajectory of the H + H2 system with initial conditions: r1 = rts+δ, r2 = rts. ]]

[[File:internuclear_r2_dynamics_heg18.png|250px|thumb|right|Figure 5. An “Internuclear Distances vs Time” plot showing the dynamics trajectory of the H + H2 system with initial conditions: r2 = rts+δ, r1 = rts. ]]

Likewise in the "Momenta vs Time" plots for each set of initial conditions, the plots illustrate how in each trajectory leads to the system 'rolling' into different sets of products. The products are represented in each case by the oscillating momentum as a function of time.

[[File:momenta_r1_dynamics_heg18.png|250px|thumb|left|Figure 6. A "Momenta vs Time" plot showing the dynamics trajectory of the H + H2 system with initial conditions: r1 = rts+δ, r2 = rts. ]]

[[File:momenta_r2_dynamics_heg18.png|250px|thumb|right|Figure 7. A "Momenta vs Time" plot showing the dynamics trajectory of the H + H2 system with initial conditions: r2 = rts+δ, r1 = rts. ]]

===Reactive and Unreactive Trajectories===

====Determining the conditions for a reactive trajectory starting in the region of the reactants and passing near the transition state====

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes || The energy of the system is greater than the activation energy of the reaction, so the system passes over the transition state and forms the products. The atoms have some vibrational energy after the reaction has occurred and oscillate as they follow the reaction path. || [[File: Table_1_heg.png|200px|none]]
|-
| -3.1 || -4.1 || -420.077 || No || The energy of the system is less than the activation energy of the reaction, so the system does not reach the saddle point and the reaction does not occur (ie. an unreactive trajectory) and the system falls back to the reactants. || [[File: Table_2_heg.png|200px|none]]
|-
| -3.1 || -5.1 || -413.977 || Yes || For this reactive pathway, the system has sufficient energy to pass the saddle point and hence form the products. However, since the initial momentum of one of the atoms was greater, the resulting products have more vibrational energy than the product in the first trajectory. || [[File: Table_3_heg.png|200px|none]]
|-
| -5.1 || -10.1 || -357.277 || No || The energy of the system is sufficient to overcome the saddle point of the reaction pathway, however the products formed have a considerable amount of vibrational energy and the bond formed is broken as the system passes over the saddle point again and reverts back to the reactants. This trajectory is called barrier recrossing and since it ends with the reactants being formed, it is not a reactive pathway. || [[File: Table_4_heg.png|200px|none]]
|-
| -5.1 || -10.6 || -349.477 || Yes || Here the trajectory is a reactive pathway, since it ends with the product formation. The system has the largest amount of energy and so fluctuates about the transition state until it eventually favours the products. The system has a considerable amount of vibrational energy and the atoms oscillate as they follow the reaction pathway. || [[File: Table_5_heg.png|200px|none]]
|}

The table of reaction trajectories above reveals the conditions necessary for a relative pathway: not only must the system have sufficient kinetic energy to overcome the activation barrier of the reaction pathway, but they must also have an appropriate balance of this kinetic energy with the vibrational energy of the products obtained. If the vibrational energy of the products is too large, and sufficient to overcome the activation barrier to reform the products (e.g. barrier recrossing), the reactants may reform and the pathway is therefore unreactive.

{{fontcolor1|green|Precisely! Really good finds. [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 11:07, 26 June 2020 (BST)}}

====Transition State Theory====

Transition State Theory allows for the average transmission rate with which the reactant trajectories reach the divide between the reactants and products on a potential energy surface. This is a functional way for calculating the rate of chemical reactions however, it makes some assumptions that limit its accuracy. One key assumption is the presumption that all trajectories with sufficient kinetic energy to surpass the activation barrier along a reaction coordinate will be reactive: the system cannot recross the barrier in the direction of the reactants.<ref> M. J. Pilling, P. W. Seakins Reaction Kinetics, 2nd ed., OUP, 1995 </ref><ref>K. J. Laidler Chemical Kinetics 3rd ed., Harper-Collins, 1987</ref> In making this assumption the potential for barrier recrossing is ignored which, as demonstrated above, can occur practically. This means that in reality, a proportion of the trajectories with sufficient energy to surpass the activation barrier will end up reverting to the reactants, hence reducing the amount of reactive trajectories. Therefore, the actual rate of the reaction is lower than that predicted by Transition State Theory and is somewhat overestimated by the theory when compared to experimental values.

{{fontcolor1|green|This a great discussion. What are the other TST assumptions. What is one assumption that may not have as big of an impact but still must be considered (hint - quantum tunnelling...)[[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 11:09, 26 June 2020 (BST)}}

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

===PES inspection===

Inspection of the potential energy surfaces for the F + H2 reaction the reaction type can be assigned according to the energetics. From the potential energy surfaces it can be seen that the activation barrier occurs early in the reaction pathway. This tells us, with reference to Hammond's Postulate, that the reaction is exothermic. Hammond's postulate states that the transition state for a reaction resembles either the reactants or products depending on which it is closest to in energy. In an exothermic reaction the transition state is closer to the reactants: it is early or 'reactant like'. Where as in an endothermic reaction the transition state is located closer to the products and is said to be late or 'product-like'. In a similar sense, the potential energy surfaces for the H +HF system reveal that the reaction is endothermic in nature; the activation barrier occurs late in the reaction pathway.2 The energetic properties are related to the bond strength of the species involved. In the F + H2 reaction a strong H-F bond is formed which will release energy and is therefore exothermic in nature. In the H +HF reaction this strong H-F bond is being broken. This is energetically demanding and therefore the products will be higher in energy than the reactants- the reaction is endothermic.

{{fontcolor1|green|Good. We can take this further and explain why HF has a stronger bond than H2 (your bonding knowledge) and then you can search literature for bond energy values for HF and H2 to further support your explanation. [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 11:11, 26 June 2020 (BST)}}

===Locating the Transition State===

With the above knowledge of the reaction energetics, the transition state of the F-H-H system was located as H-F=181.3 pm and H-H=74.5 pm as demonstrated in Figure. 8. This is verified by looking at a plot of the internuclear distance as a function of time: the gradient of the the distance F-H = H-H = F-H = 0 demonstrating that there is very little oscillation of the atoms and hence very little force acting on the system.

[[File:internuclear_F_H_H_heg18.png|250px|thumb|centre|Figure 8. An "Internuclear Distances vs Time" plot showing the ''mep'' trajectory of the F-H-H system close to the transition state. ]]

A contour plot showing the Hessian matrix eigenvectors can be used to exemplify this also: one vector shows the path of steepest accent, while the other shows the lowest energy decent towards the products. Since this is a property of the transition state it suggests this is a good estimate for the transition state location.

[[File:orthogonal_eigenvectors_ts_F_H_H_heg18.png|250px|thumb|centre|Figure 8. A contour plot zoomed in around the transition state location, showing the orthogonal eigenvectors of the F-H-H system close to the transition state. ]]

==== The Activation Energy ====

The activation energy for the reaction generating H-H and H-F were determined by running a trajectory displaced slight from the transition state, and looking at the "Energy vs Time" plots. In order to calculate the activation energy, the difference between the reactants energy and the transition state energy was determined in each case (as demonstrated in the figures below). This gave the H-H and H-F activation energy as 0.910 KJmol-1 and 126.447 KJmol-1 respectively.

[[File:energy_time_H_H_heg18.png|250px|thumb|left|Figure 9. An Energy vs Time plot used to determine the activation energy barrier for the formation of H-H. ]]

[[File:energy_time_H_F_heg18.png|300px|thumb|right|Figure 10. An Energy vs Time plot used to determine the activation energy barrier for the formation of H-F. ]]

{{fontcolor1|green|Spot on! [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 11:12, 26 June 2020 (BST)}}

===Reaction dynamics===

For the F + H2, since the attacking atom A (F) is much heavier than atoms B-C (H-H), it approaches atom B while the B-C bond breaks and the repulsion is reduced. This repulsion leads to atom B recoiling, but A-B does not recoil since the bond is still extended and hasn't yet fully formed.[2] This leads to a large vibrational energy in the product in comparison to the reactant molecule H2, as seen in figure 11. This is because the energy released due to the B-C repulsion pushes B closer to atom A and there is a resulting vibration in the product A-B (F-H). This is known as ''mixed energy release''. This increase in kinetic energy in the products compared to the reactants can be measured by calorimetry: the increased vibrational energy of the system causes a temperature change that can be measured experimentally. Since the reaction is exothermic, heat energy will be released and the temperature of the surroundings will increase.

{{fontcolor1|green|That is true. We can even look into IR and something called Infrared chemiluminescence (consider molecules being excited to excited states and then back to ground state). [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 11:15, 26 June 2020 (BST)}}

[[File:momenta_time_F_H_H_heg18.png|250px|thumb|centre|Figure 11. A "Momenta vs Time" plot for the F +H2 reactive trajectory. ]]

Calculations starting on the reactants side of F + H2 were run, initiated at the bottom of the well rHH = 74 pm with a momentum pFH = -1.0 g.mol-1.pm.fs-1- exploring a variety values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1. The animations and “Momenta vs Time” plots showed that negative momenta resulted in a system with a lower kinetic energy than the corresponding positive momenta of the same magnitude. When the system initiated with greater vibrational energy, barrier recrossing occurred where there was an imbalance of translational and vibrational energy of the system. When this occurred the animation showed the product H2 molecule had less kinetic energy than in the initial reactant form (ie a loss of vibrational and translational kinetic energy occurred on barrier recrossing). For the reactive pathways, where HF was generated, the opposite occurred. Here, the HF product molecule had greater energy in the form of translational kinetic energy (seen in the animation) and also greater vibrational energy (seen in the contour plot) than in the reactant H2 molecule.

When the momentum pFH was increased slightly to pFH = -1.6 g.mol-1.pm.fs-1 yet the overall energy of the system was considerably reduced by decreasing pHH to pHH = 0.2 g.mol-1.pm.fs-1 the trajectory was still reactive (and no barrier recrossing occurred). During the reaction it is clear that potential energy was converted into kinetic energy since the products move away and oscillate faster.

==== The Reverse Reaction: H + HF====

After attempting to find the reverse reaction trajectory by testing various initial conditions (low vibration motion of the H-F bond and high values of pHH), it was unsuccessful in generating a reactive reverse pathway. Likewise, use of the inversion of momentum procedure did not yield a reactive pathway: in no case did the reactants pass the activation barrier necessary to form the products. However, according to Polanyi's rules and the principle of microscopic reversibility (ie the rates of the forwards and reverse reactions are equal), for an exothermic reaction the energy provided for the reaction to occur is mostly in the form of translational energy.[2] Translational energy in the reactants is there most effective way to cross the activation barrier and therefore increases the rate of an exothermic process. Likewise for an endothermic reaction state, the converse is true, where the transition state is "product-like" or late, the energy provided to overcome the saddle point is mostly in the form of vibrational energy. Since this reverse reaction is endothermic, it is expected that vibrational energy in the reactants will be more efficient for overcoming the saddle point- this was evident when trying to obtain a reactive pathway.

{{fontcolor1|green|Good, it would have been nice to see the trajectories you tested. Additionally, you can include 'late/early' energy barriers to make your answers fuller. But good understanding of Polanyi's rules shown! [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 11:18, 26 June 2020 (BST)}}

=== References ===

MRD:01512138

2020-06-26T10:18:34Z

Mys18: /* The Reverse Reaction: H + HF */

=='''EXERCISE 1: H + H2 system'''==
===Dynamics From the Transition State Region===

The transition state on a potential energy surface (PES) is mathematically defined as the (stationary) point at which the gradient of the potential is zero: ∂V(ri)/∂ri=0. The transition-state structure is a saddle point that lies on the lowest energy path between two minima: physically it is the highest energy point on a reaction coordinate and the minima correspond to stable chemical species (reactants and products).

If a trajectory with zero initial momentum is launched at the transition state it will remain at that exact point forever. When the geometry is perturbed from this point in the direction of the reactants or products, the system will 'roll' towards the reactant or product states respectively. This makes the location of the transition state discernible by initiating one of the trajectories close to the transition state. The reactants, products and intermediates share the mathematical property of zero gradient and are all stationary points of the PES. The transition state can be distinguished from local minima by considering the Hessian matrix features: the eigenvalues of the computed matrix will disclose the mathematical nature of any stationary point. For the local minima (of non-linear molecules)the 3N-6 eigenvalues are all positive as movement in any direction will increase the energy. For the transition state, 3N-7 of these eigenvalues will be positive but one will be negative. This saddle point has minimum energy paths in the directions of local minima (reactants, products and intermediates) which represent the reaction coordinates, but the also represents a point of maximum energy. The eigenvector associated with he negative eigenvalue will reveal the direction of the reaction coordinate path.

{{fontcolor1|green|Nice! To distinguish between TS and LM I would look to taking the second derivative. If it is a local minimum then I would expect a positive value in all directions. If it is a TS, as you mentioned is a saddle point on a PES, so it will give a negative value but in an orthogonal direction to that the value will then be positive - that will identify the TS. [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 11:04, 26 June 2020 (BST)}}

====Trajectories from r1 = r2: locating the transition state====

For the symmetric H + H2 surface the transition state was located considering r1 = r2 and p1= p2 = 0.0 g.mol-1.pm.fs-1. An estimate of the transition state position (rts) was found at r1 = r2 = 90.77 ppm. This is verified by looking at a plot of the internuclear distance as a function of time: the gradient of the the distance A-B = B-C = A-C = 0 showing that there is no oscillation of the atoms along the ridge r1 = r2.

[[File:Surface_Plot_H_+_H2.png|300px|thumb|centre|Figure 1. A plot showing the internuclear distance as a function of time at the transition state position for the H+H2 system.]]

{{fontcolor1|green|Good![[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 11:05, 26 June 2020 (BST)}}

====Trajectories from r1 = rts+δ, r2 = rts====

Figure 2. shows the minimum energy path (''mep'') trajectory run at the transition state; the trajectory simply follows the valley floor to H1+ H2-H3. Any point on the path is at an energy minimum in all directions perpendicular to the path and there is "infinite friction" acting on the system. This frictional force means that the system can gain no momentum and "rolls" along the lowest energy path. Contrary to this, the dynamics trajectory (Figure 3.) shows the system has some vibrational energy as the atoms oscillate. This is because of the momentum is increasing and the force on the atoms builds up and leads to vibration. The dynamics calculation therefore represents the system more realistically than the ''mep'' trajectory, since the atoms will have some vibrational energy.

[[File:Surface_Plot_MEP_displaced.png|250px|thumb|left|Figure 2. A plot showing the ''mep'' trajectory of the H + H2 system, where the initial position of H was displaced by 1 pm from the transition state. ]]

[[File:Surface_Plot_Dynamics_displaced.png|250px|thumb|right|Figure 3. A plot showing the dynamics trajectory of the H + H2 system, where the initial position of H was displaced by 1 pm from the transition state.]]

{{fontcolor1|green|Exactly! and we visually see a more wavy line for dynamic. [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 11:06, 26 June 2020 (BST)}}

====Trajectories from r2 = rts+δ, r1 = rts====

Looking at the “Internuclear Distances vs Time” dynamics trajectories for the displacement described above, r1 = rts+δ, r2 = rts, against r2 = rts+δ, r1 = rts it is apparent that the products of each reactive pathway are different. In the dynamics trajectory described above, the products are HB-HC + HA where as with new the initial conditions, r2 = rts+δ, r1 = rts, the products are HA-HC + HB. The trajectories are identical however the change in the displacement of specific atoms leads to this variation in the products.

[[File:internuclear_r1_dynamics_heg18.png|250px|thumb|left|Figure 4. An “Internuclear Distances vs Time” plot showing the dynamics trajectory of the H + H2 system with initial conditions: r1 = rts+δ, r2 = rts. ]]

[[File:internuclear_r2_dynamics_heg18.png|250px|thumb|right|Figure 5. An “Internuclear Distances vs Time” plot showing the dynamics trajectory of the H + H2 system with initial conditions: r2 = rts+δ, r1 = rts. ]]

Likewise in the "Momenta vs Time" plots for each set of initial conditions, the plots illustrate how in each trajectory leads to the system 'rolling' into different sets of products. The products are represented in each case by the oscillating momentum as a function of time.

[[File:momenta_r1_dynamics_heg18.png|250px|thumb|left|Figure 6. A "Momenta vs Time" plot showing the dynamics trajectory of the H + H2 system with initial conditions: r1 = rts+δ, r2 = rts. ]]

[[File:momenta_r2_dynamics_heg18.png|250px|thumb|right|Figure 7. A "Momenta vs Time" plot showing the dynamics trajectory of the H + H2 system with initial conditions: r2 = rts+δ, r1 = rts. ]]

===Reactive and Unreactive Trajectories===

====Determining the conditions for a reactive trajectory starting in the region of the reactants and passing near the transition state====

{| class="wikitable" border=1
! p1/ g.mol-1.pm.fs-1 !! p2/ g.mol-1.pm.fs-1 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -2.56 || -5.1 || -414.280 || Yes || The energy of the system is greater than the activation energy of the reaction, so the system passes over the transition state and forms the products. The atoms have some vibrational energy after the reaction has occurred and oscillate as they follow the reaction path. || [[File: Table_1_heg.png|200px|none]]
|-
| -3.1 || -4.1 || -420.077 || No || The energy of the system is less than the activation energy of the reaction, so the system does not reach the saddle point and the reaction does not occur (ie. an unreactive trajectory) and the system falls back to the reactants. || [[File: Table_2_heg.png|200px|none]]
|-
| -3.1 || -5.1 || -413.977 || Yes || For this reactive pathway, the system has sufficient energy to pass the saddle point and hence form the products. However, since the initial momentum of one of the atoms was greater, the resulting products have more vibrational energy than the product in the first trajectory. || [[File: Table_3_heg.png|200px|none]]
|-
| -5.1 || -10.1 || -357.277 || No || The energy of the system is sufficient to overcome the saddle point of the reaction pathway, however the products formed have a considerable amount of vibrational energy and the bond formed is broken as the system passes over the saddle point again and reverts back to the reactants. This trajectory is called barrier recrossing and since it ends with the reactants being formed, it is not a reactive pathway. || [[File: Table_4_heg.png|200px|none]]
|-
| -5.1 || -10.6 || -349.477 || Yes || Here the trajectory is a reactive pathway, since it ends with the product formation. The system has the largest amount of energy and so fluctuates about the transition state until it eventually favours the products. The system has a considerable amount of vibrational energy and the atoms oscillate as they follow the reaction pathway. || [[File: Table_5_heg.png|200px|none]]
|}

The table of reaction trajectories above reveals the conditions necessary for a relative pathway: not only must the system have sufficient kinetic energy to overcome the activation barrier of the reaction pathway, but they must also have an appropriate balance of this kinetic energy with the vibrational energy of the products obtained. If the vibrational energy of the products is too large, and sufficient to overcome the activation barrier to reform the products (e.g. barrier recrossing), the reactants may reform and the pathway is therefore unreactive.

{{fontcolor1|green|Precisely! Really good finds. [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 11:07, 26 June 2020 (BST)}}

====Transition State Theory====

Transition State Theory allows for the average transmission rate with which the reactant trajectories reach the divide between the reactants and products on a potential energy surface. This is a functional way for calculating the rate of chemical reactions however, it makes some assumptions that limit its accuracy. One key assumption is the presumption that all trajectories with sufficient kinetic energy to surpass the activation barrier along a reaction coordinate will be reactive: the system cannot recross the barrier in the direction of the reactants.<ref> M. J. Pilling, P. W. Seakins Reaction Kinetics, 2nd ed., OUP, 1995 </ref><ref>K. J. Laidler Chemical Kinetics 3rd ed., Harper-Collins, 1987</ref> In making this assumption the potential for barrier recrossing is ignored which, as demonstrated above, can occur practically. This means that in reality, a proportion of the trajectories with sufficient energy to surpass the activation barrier will end up reverting to the reactants, hence reducing the amount of reactive trajectories. Therefore, the actual rate of the reaction is lower than that predicted by Transition State Theory and is somewhat overestimated by the theory when compared to experimental values.

{{fontcolor1|green|This a great discussion. What are the other TST assumptions. What is one assumption that may not have as big of an impact but still must be considered (hint - quantum tunnelling...)[[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 11:09, 26 June 2020 (BST)}}

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

===PES inspection===

Inspection of the potential energy surfaces for the F + H2 reaction the reaction type can be assigned according to the energetics. From the potential energy surfaces it can be seen that the activation barrier occurs early in the reaction pathway. This tells us, with reference to Hammond's Postulate, that the reaction is exothermic. Hammond's postulate states that the transition state for a reaction resembles either the reactants or products depending on which it is closest to in energy. In an exothermic reaction the transition state is closer to the reactants: it is early or 'reactant like'. Where as in an endothermic reaction the transition state is located closer to the products and is said to be late or 'product-like'. In a similar sense, the potential energy surfaces for the H +HF system reveal that the reaction is endothermic in nature; the activation barrier occurs late in the reaction pathway.2 The energetic properties are related to the bond strength of the species involved. In the F + H2 reaction a strong H-F bond is formed which will release energy and is therefore exothermic in nature. In the H +HF reaction this strong H-F bond is being broken. This is energetically demanding and therefore the products will be higher in energy than the reactants- the reaction is endothermic.

{{fontcolor1|green|Good. We can take this further and explain why HF has a stronger bond than H2 (your bonding knowledge) and then you can search literature for bond energy values for HF and H2 to further support your explanation. [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 11:11, 26 June 2020 (BST)}}

===Locating the Transition State===

With the above knowledge of the reaction energetics, the transition state of the F-H-H system was located as H-F=181.3 pm and H-H=74.5 pm as demonstrated in Figure. 8. This is verified by looking at a plot of the internuclear distance as a function of time: the gradient of the the distance F-H = H-H = F-H = 0 demonstrating that there is very little oscillation of the atoms and hence very little force acting on the system.

[[File:internuclear_F_H_H_heg18.png|250px|thumb|centre|Figure 8. An "Internuclear Distances vs Time" plot showing the ''mep'' trajectory of the F-H-H system close to the transition state. ]]

A contour plot showing the Hessian matrix eigenvectors can be used to exemplify this also: one vector shows the path of steepest accent, while the other shows the lowest energy decent towards the products. Since this is a property of the transition state it suggests this is a good estimate for the transition state location.

[[File:orthogonal_eigenvectors_ts_F_H_H_heg18.png|250px|thumb|centre|Figure 8. A contour plot zoomed in around the transition state location, showing the orthogonal eigenvectors of the F-H-H system close to the transition state. ]]

==== The Activation Energy ====

The activation energy for the reaction generating H-H and H-F were determined by running a trajectory displaced slight from the transition state, and looking at the "Energy vs Time" plots. In order to calculate the activation energy, the difference between the reactants energy and the transition state energy was determined in each case (as demonstrated in the figures below). This gave the H-H and H-F activation energy as 0.910 KJmol-1 and 126.447 KJmol-1 respectively.

[[File:energy_time_H_H_heg18.png|250px|thumb|left|Figure 9. An Energy vs Time plot used to determine the activation energy barrier for the formation of H-H. ]]

[[File:energy_time_H_F_heg18.png|300px|thumb|right|Figure 10. An Energy vs Time plot used to determine the activation energy barrier for the formation of H-F. ]]

{{fontcolor1|green|Spot on! [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 11:12, 26 June 2020 (BST)}}

===Reaction dynamics===

For the F + H2, since the attacking atom A (F) is much heavier than atoms B-C (H-H), it approaches atom B while the B-C bond breaks and the repulsion is reduced. This repulsion leads to atom B recoiling, but A-B does not recoil since the bond is still extended and hasn't yet fully formed.[2] This leads to a large vibrational energy in the product in comparison to the reactant molecule H2, as seen in figure 11. This is because the energy released due to the B-C repulsion pushes B closer to atom A and there is a resulting vibration in the product A-B (F-H). This is known as ''mixed energy release''. This increase in kinetic energy in the products compared to the reactants can be measured by calorimetry: the increased vibrational energy of the system causes a temperature change that can be measured experimentally. Since the reaction is exothermic, heat energy will be released and the temperature of the surroundings will increase.

{{fontcolor1|green|That is true. We can even look into IR and something called Infrared chemiluminescence (consider molecules being excited to excited states and then back to ground state). [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 11:15, 26 June 2020 (BST)}}

[[File:momenta_time_F_H_H_heg18.png|250px|thumb|centre|Figure 11. A "Momenta vs Time" plot for the F +H2 reactive trajectory. ]]

Calculations starting on the reactants side of F + H2 were run, initiated at the bottom of the well rHH = 74 pm with a momentum pFH = -1.0 g.mol-1.pm.fs-1- exploring a variety values of pHH in the range -6.1 to 6.1 g.mol-1.pm.fs-1. The animations and “Momenta vs Time” plots showed that negative momenta resulted in a system with a lower kinetic energy than the corresponding positive momenta of the same magnitude. When the system initiated with greater vibrational energy, barrier recrossing occurred where there was an imbalance of translational and vibrational energy of the system. When this occurred the animation showed the product H2 molecule had less kinetic energy than in the initial reactant form (ie a loss of vibrational and translational kinetic energy occurred on barrier recrossing). For the reactive pathways, where HF was generated, the opposite occurred. Here, the HF product molecule had greater energy in the form of translational kinetic energy (seen in the animation) and also greater vibrational energy (seen in the contour plot) than in the reactant H2 molecule.

When the momentum pFH was increased slightly to pFH = -1.6 g.mol-1.pm.fs-1 yet the overall energy of the system was considerably reduced by decreasing pHH to pHH = 0.2 g.mol-1.pm.fs-1 the trajectory was still reactive (and no barrier recrossing occurred). During the reaction it is clear that potential energy was converted into kinetic energy since the products move away and oscillate faster.

==== The Reverse Reaction: H + HF====

After attempting to find the reverse reaction trajectory by testing various initial conditions (low vibration motion of the H-F bond and high values of pHH), it was unsuccessful in generating a reactive reverse pathway. Likewise, use of the inversion of momentum procedure did not yield a reactive pathway: in no case did the reactants pass the activation barrier necessary to form the products. However, according to Polanyi's rules and the principle of microscopic reversibility (ie the rates of the forwards and reverse reactions are equal), for an exothermic reaction the energy provided for the reaction to occur is mostly in the form of translational energy.[2] Translational energy in the reactants is there most effective way to cross the activation barrier and therefore increases the rate of an exothermic process. Likewise for an endothermic reaction state, the converse is true, where the transition state is "product-like" or late, the energy provided to overcome the saddle point is mostly in the form of vibrational energy. Since this reverse reaction is endothermic, it is expected that vibrational energy in the reactants will be more efficient for overcoming the saddle point- this was evident when trying to obtain a reactive pathway.

{{fontcolor1|green|Good, it would have been nice to see the trajectories you tested. Additionally, you can include 'late/early' energy barriers to make your answers fuller. But good understanding of Polanyi's rules shown! [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 11:18, 26 June 2020 (BST)}}

=== References ===

2020-06-24T22:27:54Z

Mys18:

'''''On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''''[[File:Gradient_123.png|400x400px|thumb|Figure 1|none]]The transition state is defined as the maximum point on the minimum energy path linking reactants and the products hence can be mathematically defined as ∂V(ri)/∂ri=0 which represents that the gradient of the potential is zero. It can be identified by starting a trajectory near the transition state and see whether they roll towards the reactants or products. This can be done several times with slightly varying trajectories in order to accurately identify the transition state. The transition state can be distinguished from the local minimum of the potential energy surface by taking the second derivative and checking whether ∂2V(ri)/∂r2i > 0 or ∂2V(ri)/∂r2i < 0 . ∂2V(ri)/∂r2i > 0 corresponds to the local minimum whereas ∂2V([[ri|ri]])/∂r2i < 0 corresponds to the maximum ( transition state).

{{fontcolor1|green| Fantastic. Your last point is not wrong wrt. local minimum. TS is a maximum along a minimum energy pathway. On a PES it can be seen a TS is a saddle point...see how the second derivative at this point differs from a local minimum in all directions. There you will have your answer in differentiating between the two. [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 23:13, 24 June 2020 (BST)}}

'''''Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''''[[File:Surface_Plot_01552413.png|495x495px|thumb|Figure 3]]
[[File:TS_distance_plot.png|480x480px|thumb|Figure 2|none]]Since the H + H2 surface is symmetric, the transition state corresponds to where r1=r2 and hence Figure 2 shows the intersection point at 20 fs with a distance of 90.8 pm. Hence AB=BC= 90.8 pm. Figure 3 shows this point (saddle point) is at the diagonal intersect further providing evidence the transition state is this distance and also the force of the bonds is 0 kJ/mol/pm.

{{fontcolor1|green| Nice job! [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 23:13, 24 June 2020 (BST)}}

'''''Comment on how the mep and the trajectory you just calculated differ'''''
[[File:Dynamic_01552413.png|443x443px|thumb|Figure 5]][[File:MEP_01552413.png|464x464px|thumb|Figure 4|none]]The trajectories, was found using r1 = rts+δ, r2 = rts where δ = 1 hence r1 = 91.8 pm and r2 = 90.8 pm and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1 . Figure 4 shows a mep plot which doesn't provide a realistic account of the motion of atoms during a reaction as the vibration of the atoms aren't taken into account. However, Figure 5 showing the dynamic plot does include the atomic vibrations hence provides a more realistic account of the motion of atoms.

{{fontcolor1|green| Exemplary! [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 23:20, 24 June 2020 (BST)}}

'''''Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''''
{| class="wikitable"
|-
|-
|-
|-
|-
|-
|}
{| class="wikitable"
|
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|-2.56
|-5.1
|-414.280
|Reactive
|Figure 5 shows a reactive trajectory as the kinetic energy is large enough to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. After the formation of AB and as rBC increases the molecule vibrates.
|[[File:Counter_01552413_1.png|400x400px|thumb|Figure 6]]

|-
|-3.1
|-4.1
|-420.081
|Unreactive
|Figure 6 shows an unreactive trajectory as the kinetic energy is not large enough to overcome the activation barrier and hence the collision doesn't have enough energy to break the BC bond and form the new AB bond. Compared to Figure 5 the momentum of the incoming A atom is reduced from -5.1 to -4.1 hence decreasing the kinetic energy and so the reactants are reformed
|[[File:Counter_01552413_2.png|400x400px|thumb|Figure 7]]
|-
|-3.1
|-5.1
|-413.978
|Reactive
|Figure 7 shows a reactive trajectory as the kinetic energy is large enough to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. After the formation of AB and as rBC increases the molecule vibrates. Compared to Figure 6 the momentum of the incoming A atom is increased from -4.1 to -5.1 hence increasing the kinetic energy and so the product forms.
|[[File:Counter_01552413_3.png|400x400px|thumb|Figure 8]]
|-
|-5.1
|-10.1
|-357.274
|Unreactive
|Figure 8 shows an unreactive trajectory. Even though there is enough kinetic energy to overcome the activation barrier the product AB starts to form but then the system reverts back to the reactants. This is barrier recrossing, where due to large kinetic energies there's a lot of extra vibrational energy which causes the trajectory to recross the barrier and therefore reform the reactants.
|[[File:Counter_01552413_4.png|400x400px|thumb|Figure 9]]
|-
|-5.1
|-10.6
|-349.476
|Reactive
|Figure 9 shows a reactive trajectory. There is enough kinetic energy to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. Similarly to Figure 7 barrier recrossing occurs where the system reverts back to the reactants but since atom A has higher energy -10.6 vs -10.1 (Figure 8), it's able to revert back to the product via a higher potential energy.
|[[File:Counter_01552413_5.png|400x400px|thumb|Figure 10]]
|}
|}
'''''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''''

The transition state theory (TST) states that a molecular collision that leads to reaction must pass through an intermediate state known as the transition state essentially where A-B is partially formed and B-C is partially broken as discussed previously. It assumes once a transition state has been achieved the transition state structure does not collapse back to the reactants. However this does not fit with experimental data as shown in Figure 9 the transition state collapses back to the reactants. Therefore TST overestimates the rate of this bimolecular reaction, since it assumes all reactions go to completion, but experimentally this violates the assumption. Furthermore, the TST is also based on the assumption that atomic nuclei behave according to the classic mechanics, hence it ignores Quantum Tunnelling which leads to the underestimation of the TST, but since this is a small effect as compared to barrier recrossing, the TST is mainly an overestimate of rate. The TST also fails for some reactions at high temperatures due to the more complex motions of molecules or at very low temperatures due to the quantum tunneling.

{{fontcolor1|green| Great discussion. I would recommend referencing these assumptions. [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 23:24, 24 June 2020 (BST)}}

'''''By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state.and the value of Activation energy'''''
[[File:ATS_01552413.png|left|452x452px|thumb|Figure 11]]
[[File:TS1_01552413.png|448x448px|centre|thumb|Figure 12]]

This reaction is exothermic as H-F is stronger than H-H. Even though H-H has stronger covalent character than H-F as the overlap of the 1s orbitals is greater than the overlap of 1s and 2p orbitals of H-F, the H-F molecule is polar due to the very electronegative fluorine atom, hence it has a greater ionic character so forms stronger bonds. H-H is non-polar and hence has no ionic character. The transition state was found by changing the distances of H-H and H-F till the force = 0 kJ/mol/pm. This occurred at HF= 181.1 pm and HH= 74.5 pm. This provides evidence of an early transition sate hence an exothermic process occurs. From this, the activation energy was obtained from the mep plot of energy vs time as shown in Figure 10 and so the activation energy = 121.76 kJ/mol

{{fontcolor1|green| Good. You could search the literature for experimental bond energies for H2 and HF to support your bonding discussion. [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 23:27, 24 June 2020 (BST)}}

{| class="wikitable"
!p= -6.1
!p=-4.1
!p=-2.1
!p=0
|-
|[[File:-6.1_01552413.png|left|250x250px]]
|[[File:-4.1_01552413.png|250x250px|centre]]
|[[File:-2.1_01552413.png|left|250x250px]]
|[[File:0_01552413.png|left|250x250px]]
|}

{| class="wikitable"
!p=2.1
!p=4.1
!p=6.1
|-
|[[File:2.1_01552413.png|left|250x250px]]
|[[File:4.1_01552413.png|left|250x250px]]
|[[File:6.1_01552413.png|left|250x250px]]
|}
The table above shows the contour diagrams at changing momentums of H-H. When p=-4.1, p=2.1, and p=4.1 the reaction proceeds to the products (reactive trajectories)whereas for p=-6.1, p= -2.1 p=0 and p=6.1 after reaching the transition state, it reverts back to the reactants hence these are unreactive trajectories. After reaching the TS for the reactive trajectories there is greater vibrational energy as the H-F bond is stronger than the H-H bond and hence the force constant will be higher therefore the vibrational energy will be larger. This further shows the difference between experimental results and the transition state theory as when the transition state has reached, the product (H-F) doesn't always form as it reverts back to the reactants.[[File:Momentaa_01552413.png|491x491px|right|thumb|Figure 13]][[File:Standard_01552413.png|498x498px|thumb|Figure 14|none]]

The mechanism of energy release can be interpreted, in terms of vibrational energy and translational energy as shown in Figure 13. At t = 0 fs H-F has no vibrational energy whereas H-H does, as shown by the oscillations on the graph. As the reaction proceeds H-F vibrates more and H-H vibrates less and eventually converges at p=3.12 g.mol-1.pm.fs-1 as no oscillations occur hence the energy is purely translational. H-F has greater oscillations than H-H as the H-F bond is stronger and so the force constant is greater so vibrational energy will be greater. For greater accuracy, IR spectroscopy can be used to confirm this as H-F is has a dipole moment so it is IR active, unlike H2 which is IR inactive hence transition between the energy levels is forbidden. Since H-F molecules can be excited, overtones are produced which suggests vibrational energy is present in the product.

'''''HF + H'''''

[[File:HF1_01552413.png|488x488px|thumb|Figure 15]][[File:HF2_01552413.png|500x500px|thumb|Figure 16|none]]

the bond distances found for the reactive trajectory was rHF = 74.49 pm and rHH= 181 pm with p = -3 and -5.1 repectively.

This reaction will be endothermic as the reactant H-F is lower in energy than the product H-H since H-F is a stronger bond than H-H as discussed previously. Via Hammond's postulate, the reactants are lower in energy compared to the products, hence a late transition state is observed. The transition state is located as previously discussed, was H-H = 74.12 pm and HF = 181.5 pm. The activation energy = 145.7 kJ/mol as shown in Figure 14.

'''''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.'''''

Te effect of translational and vibrational energies on the reaction efficiency and position of the TS can be explained by Polanyi's empirical rules. For an exothermic F + H2 reaction, the reaction barrier is usually located in the entrance valley of the reaction, that is, an early barrier. According to Polanyi's rules, reactant translational energy is then more effective than vibration to surmount the barrier to reaction, thus, accelerating the reaction rate. For an endothermic, H + HF, process a late transition barrier occurs, and hence the vibrational energy is more effective than translation to surmount the reaction barrier, thus, accelerating the reaction rate. '''''1'''''

'''''References'''''
# Yeston, J. (2012). Stretching the Polanyi Rules. ''Science'', 338(6113), pp.1397–1397.
# Schramm, V.L. (2007). Enzymatic Transition State Theory and Transition State Analogue Design. ''Journal of Biological Chemistry'', 282(39), pp.28297–28300.
# Levine, R.D. (1990). The steric factor in transition state theory and in collison theory. ''Chemical Physics Letters'', 175(4), pp.331–337. ‌ ‌
‌

MRD:01552413

2020-06-24T22:24:29Z

Mys18:

'''''On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''''[[File:Gradient_123.png|400x400px|thumb|Figure 1|none]]The transition state is defined as the maximum point on the minimum energy path linking reactants and the products hence can be mathematically defined as ∂V(ri)/∂ri=0 which represents that the gradient of the potential is zero. It can be identified by starting a trajectory near the transition state and see whether they roll towards the reactants or products. This can be done several times with slightly varying trajectories in order to accurately identify the transition state. The transition state can be distinguished from the local minimum of the potential energy surface by taking the second derivative and checking whether ∂2V(ri)/∂r2i > 0 or ∂2V(ri)/∂r2i < 0 . ∂2V(ri)/∂r2i > 0 corresponds to the local minimum whereas ∂2V([[ri|ri]])/∂r2i < 0 corresponds to the maximum ( transition state).

{{fontcolor1|green| Fantastic. Your last point is not wrong wrt. local minimum. TS is a maximum along a minimum energy pathway. On a PES it can be seen a TS is a saddle point...see how the second derivative at this point differs from a local minimum in all directions. There you will have your answer in differentiating between the two. [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 23:13, 24 June 2020 (BST)}}

'''''Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'''''[[File:Surface_Plot_01552413.png|495x495px|thumb|Figure 3]]
[[File:TS_distance_plot.png|480x480px|thumb|Figure 2|none]]Since the H + H2 surface is symmetric, the transition state corresponds to where r1=r2 and hence Figure 2 shows the intersection point at 20 fs with a distance of 90.8 pm. Hence AB=BC= 90.8 pm. Figure 3 shows this point (saddle point) is at the diagonal intersect further providing evidence the transition state is this distance and also the force of the bonds is 0 kJ/mol/pm.

{{fontcolor1|green| Nice job! [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 23:13, 24 June 2020 (BST)}}

'''''Comment on how the mep and the trajectory you just calculated differ'''''
[[File:Dynamic_01552413.png|443x443px|thumb|Figure 5]][[File:MEP_01552413.png|464x464px|thumb|Figure 4|none]]The trajectories, was found using r1 = rts+δ, r2 = rts where δ = 1 hence r1 = 91.8 pm and r2 = 90.8 pm and '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1 . Figure 4 shows a mep plot which doesn't provide a realistic account of the motion of atoms during a reaction as the vibration of the atoms aren't taken into account. However, Figure 5 showing the dynamic plot does include the atomic vibrations hence provides a more realistic account of the motion of atoms.

{{fontcolor1|green| Exemplary! [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 23:20, 24 June 2020 (BST)}}

'''''Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''''
{| class="wikitable"
|-
|-
|-
|-
|-
|-
|}
{| class="wikitable"
|
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|-2.56
|-5.1
|-414.280
|Reactive
|Figure 5 shows a reactive trajectory as the kinetic energy is large enough to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. After the formation of AB and as rBC increases the molecule vibrates.
|[[File:Counter_01552413_1.png|400x400px|thumb|Figure 6]]

|-
|-3.1
|-4.1
|-420.081
|Unreactive
|Figure 6 shows an unreactive trajectory as the kinetic energy is not large enough to overcome the activation barrier and hence the collision doesn't have enough energy to break the BC bond and form the new AB bond. Compared to Figure 5 the momentum of the incoming A atom is reduced from -5.1 to -4.1 hence decreasing the kinetic energy and so the reactants are reformed
|[[File:Counter_01552413_2.png|400x400px|thumb|Figure 7]]
|-
|-3.1
|-5.1
|-413.978
|Reactive
|Figure 7 shows a reactive trajectory as the kinetic energy is large enough to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. After the formation of AB and as rBC increases the molecule vibrates. Compared to Figure 6 the momentum of the incoming A atom is increased from -4.1 to -5.1 hence increasing the kinetic energy and so the product forms.
|[[File:Counter_01552413_3.png|400x400px|thumb|Figure 8]]
|-
|-5.1
|-10.1
|-357.274
|Unreactive
|Figure 8 shows an unreactive trajectory. Even though there is enough kinetic energy to overcome the activation barrier the product AB starts to form but then the system reverts back to the reactants. This is barrier recrossing, where due to large kinetic energies there's a lot of extra vibrational energy which causes the trajectory to recross the barrier and therefore reform the reactants.
|[[File:Counter_01552413_4.png|400x400px|thumb|Figure 9]]
|-
|-5.1
|-10.6
|-349.476
|Reactive
|Figure 9 shows a reactive trajectory. There is enough kinetic energy to overcome the activation barrier and hence the collision has enough energy to break the BC bond and form the new AB bond. Similarly to Figure 7 barrier recrossing occurs where the system reverts back to the reactants but since atom A has higher energy -10.6 vs -10.1 (Figure 8), it's able to revert back to the product via a higher potential energy.
|[[File:Counter_01552413_5.png|400x400px|thumb|Figure 10]]
|}
|}
'''''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''''

The transition state theory (TST) states that a molecular collision that leads to reaction must pass through an intermediate state known as the transition state essentially where A-B is partially formed and B-C is partially broken as discussed previously. It assumes once a transition state has been achieved the transition state structure does not collapse back to the reactants. However this does not fit with experimental data as shown in Figure 9 the transition state collapses back to the reactants. Therefore TST overestimates the rate of this bimolecular reaction, since it assumes all reactions go to completion, but experimentally this violates the assumption. Furthermore, the TST is also based on the assumption that atomic nuclei behave according to the classic mechanics, hence it ignores Quantum Tunnelling which leads to the underestimation of the TST, but since this is a small effect as compared to barrier recrossing, the TST is mainly an overestimate of rate. The TST also fails for some reactions at high temperatures due to the more complex motions of molecules or at very low temperatures due to the quantum tunneling.

{{fontcolor1|green| Great discussion. I would recommend referencing these assumptions. [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 23:24, 24 June 2020 (BST)}}

'''''By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state.and the value of Activation energy'''''
[[File:ATS_01552413.png|left|452x452px|thumb|Figure 11]]
[[File:TS1_01552413.png|448x448px|centre|thumb|Figure 12]]

This reaction is exothermic as H-F is stronger than H-H. Even though H-H has stronger covalent character than H-F as the overlap of the 1s orbitals is greater than the overlap of 1s and 2p orbitals of H-F, the H-F molecule is polar due to the very electronegative fluorine atom, hence it has a greater ionic character so forms stronger bonds. H-H is non-polar and hence has no ionic character. The transition state was found by changing the distances of H-H and H-F till the force = 0 kJ/mol/pm. This occurred at HF= 181.1 pm and HH= 74.5 pm. This provides evidence of an early transition sate hence an exothermic process occurs. From this, the activation energy was obtained from the mep plot of energy vs time as shown in Figure 10 and so the activation energy = 121.76 kJ/mol
{| class="wikitable"
!p= -6.1
!p=-4.1
!p=-2.1
!p=0
|-
|[[File:-6.1_01552413.png|left|250x250px]]
|[[File:-4.1_01552413.png|250x250px|centre]]
|[[File:-2.1_01552413.png|left|250x250px]]
|[[File:0_01552413.png|left|250x250px]]
|}

{| class="wikitable"
!p=2.1
!p=4.1
!p=6.1
|-
|[[File:2.1_01552413.png|left|250x250px]]
|[[File:4.1_01552413.png|left|250x250px]]
|[[File:6.1_01552413.png|left|250x250px]]
|}
The table above shows the contour diagrams at changing momentums of H-H. When p=-4.1, p=2.1, and p=4.1 the reaction proceeds to the products (reactive trajectories)whereas for p=-6.1, p= -2.1 p=0 and p=6.1 after reaching the transition state, it reverts back to the reactants hence these are unreactive trajectories. After reaching the TS for the reactive trajectories there is greater vibrational energy as the H-F bond is stronger than the H-H bond and hence the force constant will be higher therefore the vibrational energy will be larger. This further shows the difference between experimental results and the transition state theory as when the transition state has reached, the product (H-F) doesn't always form as it reverts back to the reactants.[[File:Momentaa_01552413.png|491x491px|right|thumb|Figure 13]][[File:Standard_01552413.png|498x498px|thumb|Figure 14|none]]

The mechanism of energy release can be interpreted, in terms of vibrational energy and translational energy as shown in Figure 13. At t = 0 fs H-F has no vibrational energy whereas H-H does, as shown by the oscillations on the graph. As the reaction proceeds H-F vibrates more and H-H vibrates less and eventually converges at p=3.12 g.mol-1.pm.fs-1 as no oscillations occur hence the energy is purely translational. H-F has greater oscillations than H-H as the H-F bond is stronger and so the force constant is greater so vibrational energy will be greater. For greater accuracy, IR spectroscopy can be used to confirm this as H-F is has a dipole moment so it is IR active, unlike H2 which is IR inactive hence transition between the energy levels is forbidden. Since H-F molecules can be excited, overtones are produced which suggests vibrational energy is present in the product.

'''''HF + H'''''

[[File:HF1_01552413.png|488x488px|thumb|Figure 15]][[File:HF2_01552413.png|500x500px|thumb|Figure 16|none]]

the bond distances found for the reactive trajectory was rHF = 74.49 pm and rHH= 181 pm with p = -3 and -5.1 repectively.

This reaction will be endothermic as the reactant H-F is lower in energy than the product H-H since H-F is a stronger bond than H-H as discussed previously. Via Hammond's postulate, the reactants are lower in energy compared to the products, hence a late transition state is observed. The transition state is located as previously discussed, was H-H = 74.12 pm and HF = 181.5 pm. The activation energy = 145.7 kJ/mol as shown in Figure 14.

'''''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.'''''

Te effect of translational and vibrational energies on the reaction efficiency and position of the TS can be explained by Polanyi's empirical rules. For an exothermic F + H2 reaction, the reaction barrier is usually located in the entrance valley of the reaction, that is, an early barrier. According to Polanyi's rules, reactant translational energy is then more effective than vibration to surmount the barrier to reaction, thus, accelerating the reaction rate. For an endothermic, H + HF, process a late transition barrier occurs, and hence the vibrational energy is more effective than translation to surmount the reaction barrier, thus, accelerating the reaction rate. '''''1'''''

'''''References'''''
# Yeston, J. (2012). Stretching the Polanyi Rules. ''Science'', 338(6113), pp.1397–1397.
# Schramm, V.L. (2007). Enzymatic Transition State Theory and Transition State Analogue Design. ''Journal of Biological Chemistry'', 282(39), pp.28297–28300.
# Levine, R.D. (1990). The steric factor in transition state theory and in collison theory. ''Chemical Physics Letters'', 175(4), pp.331–337. ‌ ‌
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