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01564233wikireport

2020-05-15T22:23:07Z

Am12618: /* Transition State */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should prove that the point investigated is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Saddle Point.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be rAB=rBC= 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct. This is seconded by the fact that in the contour plot no trajectory is shown, just a point.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|center|thumb|Internuclear distances vs Time plot for the best transition state estimated]]

[[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|center|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum. The dynamic trajectory is the actual trajectory

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive or not using rAB= 74 pm, rBC= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep moving down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants.We can now see that not all the reactions that have same initial positions of atoms have the same results.

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course one of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory predictions overestimate the experimental values.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's larger and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distance rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

One reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment, the increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation. When energy is inserted to the system electrons get excited and move from the lowest vibrational state to a higher one. These electrons fall back to the ground state emitting photons that are monitored using IR technique. The process of emission of photons after electrons drop from higher vibrational energy levels to the ground state is called Fluorescence (Radiative decay)

====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 reactions according to their nature can become more efficient either by investing more in translational or vibrational energy. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using the Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction where the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

By looking at the other plot of the unreactive trajectory we can tell that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T21:39:11Z

Am12618: /* 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. */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Saddle Point.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be rAB=rBC= 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct. This is seconded by the fact that in the contour plot no trajectory is shown, just a point.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|center|thumb|Internuclear distances vs Time plot for the best transition state estimated]]

[[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|center|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum. The dynamic trajectory is the actual trajectory

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive or not using rAB= 74 pm, rBC= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep moving down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants.We can now see that not all the reactions that have same initial positions of atoms have the same results.

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course one of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory predictions overestimate the experimental values.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's larger and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distance rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

One reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment, the increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation. When energy is inserted to the system electrons get excited and move from the lowest vibrational state to a higher one. These electrons fall back to the ground state emitting photons that are monitored using IR technique. The process of emission of photons after electrons drop from higher vibrational energy levels to the ground state is called Fluorescence (Radiative decay)

====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 reactions according to their nature can become more efficient either by investing more in translational or vibrational energy. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using the Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction where the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

By looking at the other plot of the unreactive trajectory we can tell that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T21:33:37Z

Am12618: /* PES inspection */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Saddle Point.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be rAB=rBC= 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct. This is seconded by the fact that in the contour plot no trajectory is shown, just a point.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|center|thumb|Internuclear distances vs Time plot for the best transition state estimated]]

[[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|center|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum. The dynamic trajectory is the actual trajectory

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive or not using rAB= 74 pm, rBC= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep moving down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants.We can now see that not all the reactions that have same initial positions of atoms have the same results.

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course one of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory predictions overestimate the experimental values.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's larger and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distance rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

One reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment, the increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation. When energy is inserted to the system electrons get excited and move from the lowest vibrational state to a higher one. These electrons fall back to the ground state emitting photons that are monitored using IR technique. The process of emission of photons after electrons drop from higher vibrational energy levels to the ground state is called Fluorescence (Radiative decay)

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction where the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

By looking at the other plot of the unreactive trajectory we can tell that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T21:31:54Z

Am12618: /* 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? */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Saddle Point.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be rAB=rBC= 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct. This is seconded by the fact that in the contour plot no trajectory is shown, just a point.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|center|thumb|Internuclear distances vs Time plot for the best transition state estimated]]

[[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|center|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum. The dynamic trajectory is the actual trajectory

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive or not using rAB= 74 pm, rBC= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep moving down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants.We can now see that not all the reactions that have same initial positions of atoms have the same results.

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course one of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory predictions overestimate the experimental values.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's larger and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distance rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

One reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment, the increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation. When energy is inserted to the system electrons get excited and move from the lowest vibrational state to a higher one. These electrons fall back to the ground state emitting photons that are monitored using IR technique. The process of emission of photons after electrons drop from higher vibrational energy levels to the ground state is called Fluorescence (Radiative decay)

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction where the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

By looking at the other plot of the unreactive trajectory we can tell that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T21:29:53Z

Am12618: /* Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Inter-nuclear Distances vs Time” plot for a relevant trajectory. */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Saddle Point.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be rAB=rBC= 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct. This is seconded by the fact that in the contour plot no trajectory is shown, just a point.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|center|thumb|Internuclear distances vs Time plot for the best transition state estimated]]

[[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|center|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum. The dynamic trajectory is the actual trajectory

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive or not using rAB= 74 pm, rBC= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep moving down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants.We can now see that not all the reactions that have same initial positions of atoms have the same results.

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course one of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory predictions overestimate the experimental values.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is larger and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distance rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

One reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment, the increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation. When energy is inserted to the system electrons get excited and move from the lowest vibrational state to a higher one. These electrons fall back to the ground state emitting photons that are monitored using IR technique. The process of emission of photons after electrons drop from higher vibrational energy levels to the ground state is called Fluorescence (Radiative decay)

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction where the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

By looking at the other plot of the unreactive trajectory we can tell that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T21:29:31Z

Am12618: /* Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Inter-nuclear Distances vs Time” plot for a relevant trajectory. */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Saddle Point.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be rAB=rBC= 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct. This is seconded by the fact that in the contour plot no trajectory is shown. Just a point.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|center|thumb|Internuclear distances vs Time plot for the best transition state estimated]]

[[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|center|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum. The dynamic trajectory is the actual trajectory

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive or not using rAB= 74 pm, rBC= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep moving down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants.We can now see that not all the reactions that have same initial positions of atoms have the same results.

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course one of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory predictions overestimate the experimental values.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is larger and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distance rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

One reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment, the increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation. When energy is inserted to the system electrons get excited and move from the lowest vibrational state to a higher one. These electrons fall back to the ground state emitting photons that are monitored using IR technique. The process of emission of photons after electrons drop from higher vibrational energy levels to the ground state is called Fluorescence (Radiative decay)

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction where the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

By looking at the other plot of the unreactive trajectory we can tell that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T21:27:43Z

Am12618: /* Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Inter-nuclear Distances vs Time” plot for a relevant trajectory. */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Saddle Point.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be rAB=rBC= 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|center|thumb|Internuclear distances vs Time plot for the best transition state estimated]]

[[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|center|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum. The dynamic trajectory is the actual trajectory

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive or not using rAB= 74 pm, rBC= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep moving down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants.We can now see that not all the reactions that have same initial positions of atoms have the same results.

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course one of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory predictions overestimate the experimental values.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is larger and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distance rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

One reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment, the increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation. When energy is inserted to the system electrons get excited and move from the lowest vibrational state to a higher one. These electrons fall back to the ground state emitting photons that are monitored using IR technique. The process of emission of photons after electrons drop from higher vibrational energy levels to the ground state is called Fluorescence (Radiative decay)

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction where the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

By looking at the other plot of the unreactive trajectory we can tell that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T21:26:48Z

Am12618: /* Transition State */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Saddle Point.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be rAB=rBC</sun>= 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|center|thumb|Internuclear distances vs Time plot for the best transition state estimated]]

[[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|center|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum. The dynamic trajectory is the actual trajectory

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive or not using rAB= 74 pm, rBC= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep moving down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants.We can now see that not all the reactions that have same initial positions of atoms have the same results.

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course one of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory predictions overestimate the experimental values.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is larger and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distance rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

One reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment, the increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation. When energy is inserted to the system electrons get excited and move from the lowest vibrational state to a higher one. These electrons fall back to the ground state emitting photons that are monitored using IR technique. The process of emission of photons after electrons drop from higher vibrational energy levels to the ground state is called Fluorescence (Radiative decay)

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction where the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

By looking at the other plot of the unreactive trajectory we can tell that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T21:26:22Z

Am12618: /* Dynamics from the transition state region */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that Saddle Point.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be rAB=rBC</sun>= 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|center|thumb|Internuclear distances vs Time plot for the best transition state estimated]]

[[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|center|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum. The dynamic trajectory is the actual trajectory

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive or not using rAB= 74 pm, rBC= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep moving down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants.We can now see that not all the reactions that have same initial positions of atoms have the same results.

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course one of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory predictions overestimate the experimental values.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is larger and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distance rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

One reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment, the increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation. When energy is inserted to the system electrons get excited and move from the lowest vibrational state to a higher one. These electrons fall back to the ground state emitting photons that are monitored using IR technique. The process of emission of photons after electrons drop from higher vibrational energy levels to the ground state is called Fluorescence (Radiative decay)

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction where the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

By looking at the other plot of the unreactive trajectory we can tell that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T21:10:41Z

Am12618: /* Reaction dynamics */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be rAB=rBC</sun>= 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|center|thumb|Internuclear distances vs Time plot for the best transition state estimated]]

[[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|center|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum. The dynamic trajectory is the actual trajectory

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive or not using rAB= 74 pm, rBC= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep moving down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants.We can now see that not all the reactions that have same initial positions of atoms have the same results.

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course one of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory predictions overestimate the experimental values.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is larger and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distance rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

One reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment, the increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation. When energy is inserted to the system electrons get excited and move from the lowest vibrational state to a higher one. These electrons fall back to the ground state emitting photons that are monitored using IR technique. The process of emission of photons after electrons drop from higher vibrational energy levels to the ground state is called Fluorescence (Radiative decay)

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction where the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

By looking at the other plot of the unreactive trajectory we can tell that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T21:07:29Z

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be rAB=rBC</sun>= 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|center|thumb|Internuclear distances vs Time plot for the best transition state estimated]]

[[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|center|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum. The dynamic trajectory is the actual trajectory

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive or not using rAB= 74 pm, rBC= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep moving down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants.We can now see that not all the reactions that have same initial positions of atoms have the same results.

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course one of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory predictions overestimate the experimental values.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is larger and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distance rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

One reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment, the increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation. When energy is inserted to the system electrons get excited and move from the lowest vibrational state to a higher one. These electrons fall back to the ground state emitting photons that are monitored using IR technique.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction where the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

By looking at the other plot of the unreactive trajectory we can tell that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T21:04:02Z

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be rAB=rBC</sun>= 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|center|thumb|Internuclear distances vs Time plot for the best transition state estimated]]

[[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|center|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum. The dynamic trajectory is the actual trajectory

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive or not using rAB= 74 pm, rBC= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep moving down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants.We can now see that not all the reactions that have same initial positions of atoms have the same results.

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course one of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory predictions overestimate the experimental values.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is larger and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distance rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

One reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment, the increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation. When energy is inserted to the system electrons get excited and move from the lowest vibrational state to a higher one. These electrons fall back to the ground state emitting photons that are monitored using IR technique.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction where the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T21:02:40Z

Am12618: /* In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be rAB=rBC</sun>= 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|center|thumb|Internuclear distances vs Time plot for the best transition state estimated]]

[[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|center|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum. The dynamic trajectory is the actual trajectory

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive or not using rAB= 74 pm, rBC= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep moving down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants.We can now see that not all the reactions that have same initial positions of atoms have the same results.

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course one of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory predictions overestimate the experimental values.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is larger and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distance rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

One reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment, the increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation. When energy is inserted to the system electrons get excited and move from the lowest vibrational state to a higher one. These electrons fall back to the ground state emitting photons that are monitored using IR technique.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T20:57:04Z

Am12618: /* Exercise 2:F - H - H system */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be rAB=rBC</sun>= 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|center|thumb|Internuclear distances vs Time plot for the best transition state estimated]]

[[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|center|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum. The dynamic trajectory is the actual trajectory

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive or not using rAB= 74 pm, rBC= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep moving down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants.We can now see that not all the reactions that have same initial positions of atoms have the same results.

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course one of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory predictions overestimate the experimental values.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is larger and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distance rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T20:56:34Z

Am12618:

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be rAB=rBC</sun>= 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|center|thumb|Internuclear distances vs Time plot for the best transition state estimated]]

[[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|center|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum. The dynamic trajectory is the actual trajectory

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive or not using rAB= 74 pm, rBC= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep moving down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants.We can now see that not all the reactions that have same initial positions of atoms have the same results.

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course one of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory predictions overestimate the experimental values.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is larger and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distance rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T20:44:37Z

Am12618: /* Transition State Theory */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be rAB=rBC</sun>= 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|center|thumb|Internuclear distances vs Time plot for the best transition state estimated]]

[[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|center|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum. The dynamic trajectory is the actual trajectory

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive or not using rAB= 74 pm, rBC= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep moving down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants.We can now see that not all the reactions that have same initial positions of atoms have the same results.

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course one of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory predictions overestimate the experimental values.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is large and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T20:41:21Z

Am12618: /* Reactive and unreactive trajectories */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be rAB=rBC</sun>= 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|center|thumb|Internuclear distances vs Time plot for the best transition state estimated]]

[[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|center|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum. The dynamic trajectory is the actual trajectory

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive or not using rAB= 74 pm, rBC= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep moving down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants.We can now see that not all the reactions that have same initial positions of atoms have the same results.

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course on of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is large and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T20:38:39Z

Am12618: /* Reactive and unreactive trajectories */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be rAB=rBC</sun>= 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|center|thumb|Internuclear distances vs Time plot for the best transition state estimated]]

[[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|center|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum. The dynamic trajectory is the actual trajectory

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive or not using rAB= 74 pm, rBC= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep moving down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants. Furthermore, we can see that not all the reactions that have same initial positions of atoms have the same ending

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course on of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is large and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T20:37:17Z

Am12618: /* Reactive and unreactive trajectories */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be rAB=rBC</sun>= 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|center|thumb|Internuclear distances vs Time plot for the best transition state estimated]]

[[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|center|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum. The dynamic trajectory is the actual trajectory

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive or not using rAB= 74 pm, rBC= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep movind down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants. Furthermore, we can see that not all the reactions that have same initial positions of atoms have the same ending

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course on of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is large and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T20:36:16Z

Am12618: /* Exercise 1: H + H2 system */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be rAB=rBC</sun>= 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|center|thumb|Internuclear distances vs Time plot for the best transition state estimated]]

[[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|center|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum. The dynamic trajectory is the actual trajectory

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep movind down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants. Furthermore, we can see that not all the reactions that have same initial positions of atoms have the same ending

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course on of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is large and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T20:35:53Z

Am12618: /* Reactive and unreactive trajectories */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be rAB=rBC</sun>= 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|center|thumb|Internuclear distances vs Time plot for the best transition state estimated]]

[[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|center|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum. The dynamic trajectory is the actual trajectory

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep movind down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants. Furthermore, we can see that not all the reactions that have same initial positions of atoms have the same ending

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course on of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is large and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T20:35:27Z

Am12618: /* How do the mep and the trajectory calculated differ */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be rAB=rBC</sun>= 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|center|thumb|Internuclear distances vs Time plot for the best transition state estimated]]

[[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|center|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum. The dynamic trajectory is the actual trajectory

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep movind down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants. Furthermore, we can see that not all the reactions that have same initial positions of atoms have the same ending

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course on of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is large and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T20:34:04Z

Am12618: /* Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Inter-nuclear Distances vs Time” plot for a relevant trajectory. */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be rAB=rBC</sun>= 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|center|thumb|Internuclear distances vs Time plot for the best transition state estimated]]

[[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|center|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum.

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep movind down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants. Furthermore, we can see that not all the reactions that have same initial positions of atoms have the same ending

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course on of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is large and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T20:31:50Z

Am12618: /* Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Inter-nuclear Distances vs Time” plot for a relevant trajectory. */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be r1=r2 is 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|center|thumb|Internuclear distances vs Time plot for the best transition state estimated]]

[[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|center|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum.

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep movind down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants. Furthermore, we can see that not all the reactions that have same initial positions of atoms have the same ending

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course on of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is large and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T20:30:41Z

Am12618: /* Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Inter-nuclear Distances vs Time” plot for a relevant trajectory. */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be r1=r2 is 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of rAB=rBC=90.775 pm and p1=p2=0.png|center|thumb|Internuclear distances vs Time plot for the best transition state estimated]]

[[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|center|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum.

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep movind down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants. Furthermore, we can see that not all the reactions that have same initial positions of atoms have the same ending

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course on of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is large and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T20:30:21Z

Am12618: /* Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Inter-nuclear Distances vs Time” plot for a relevant trajectory. */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be r1=r2 is 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of rAB=rBC=90.775 pm and p1=p2=0.png|center|thumb|Internuclear distances vs Time plot for the best transition state estimated]]

[[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|center|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum.

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep movind down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants. Furthermore, we can see that not all the reactions that have same initial positions of atoms have the same ending

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course on of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is large and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png

2020-05-15T20:29:00Z

Am12618: Am12618 uploaded a new version of File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png

File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png

2020-05-15T20:28:59Z

Am12618: Am12618 uploaded a new version of File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png

01564233wikireport

2020-05-15T20:27:16Z

Am12618: /* Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Inter-nuclear Distances vs Time” plot for a relevant trajectory. */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be r1=r2 is 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of rAB=rBC=90.775 pm and p1=p2=0.png|centre|thumb|Internuclear distances vs Time plot for the best transition state estimated]] [[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|centre|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum.

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep movind down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants. Furthermore, we can see that not all the reactions that have same initial positions of atoms have the same ending

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course on of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is large and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T20:25:32Z

Am12618: /* Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Inter-nuclear Distances vs Time” plot for a relevant trajectory. */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which it is closer in energy with. In this case it resembles both since the atoms that are used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be r1=r2 is 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of rAB=rBC=90.775 pm and p1=p2=0.png|centre|thumb|Internuclear distances vs Time plot for the best transition state estimated]] [[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|centre|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum.

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep movind down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants. Furthermore, we can see that not all the reactions that have same initial positions of atoms have the same ending

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course on of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is large and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T20:23:11Z

Am12618: /* Dynamics from the transition state region */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction it is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which are cosest in energy to it. Inthis case it resembles both since the atoms that are used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be r1=r2 is 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of rAB=rBC=90.775 pm and p1=p2=0.png|centre|thumb|Internuclear distances vs Time plot for the best transition state estimated]] [[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|centre|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum.

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep movind down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants. Furthermore, we can see that not all the reactions that have same initial positions of atoms have the same ending

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course on of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is large and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T20:21:28Z

Am12618: /* Transition State */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since that means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which are cosest in energy to it. Inthis case it resembles both since the atoms that are used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be r1=r2 is 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of rAB=rBC=90.775 pm and p1=p2=0.png|centre|thumb|Internuclear distances vs Time plot for the best transition state estimated]] [[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|centre|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum.

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep movind down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants. Furthermore, we can see that not all the reactions that have same initial positions of atoms have the same ending

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course on of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is large and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T20:20:16Z

Am12618: /* Dynamics from the transition state region */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

The Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms, it is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since this means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which are cosest in energy to it. Inthis case it resembles both since the atoms that are used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be r1=r2 is 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of rAB=rBC=90.775 pm and p1=p2=0.png|centre|thumb|Internuclear distances vs Time plot for the best transition state estimated]] [[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|centre|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum.

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep movind down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants. Furthermore, we can see that not all the reactions that have same initial positions of atoms have the same ending

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course on of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is large and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T19:33:40Z

Am12618:

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since this means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which are cosest in energy to it. Inthis case it resembles both since the atoms that are used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be r1=r2 is 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of rAB=rBC=90.775 pm and p1=p2=0.png|centre|thumb|Internuclear distances vs Time plot for the best transition state estimated]] [[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|centre|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum.

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep movind down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants. Furthermore, we can see that not all the reactions that have same initial positions of atoms have the same ending

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course on of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF distance that's is large and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H (reactants higher in energy than the products) and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which it is the closest in energy with. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with the number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 kjmol-1.pm.fs-1, prBC= -5 kjmol-1.pm.fs-1
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

In this exothermic reaction energy can be released in two forms, either translational or vibrational.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature could be calculated using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational energy. To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy must be promoted more, the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According to these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibrational. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is higher than the vibrational. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased (oscillations are observed).

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T18:48:37Z

Am12618: /* Report the activation energy for both reactions */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since this means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which are cosest in energy to it. Inthis case it resembles both since the atoms that are used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be r1=r2 is 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of rAB=rBC=90.775 pm and p1=p2=0.png|centre|thumb|Internuclear distances vs Time plot for the best transition state estimated]] [[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|centre|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum.

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep movind down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants. Furthermore, we can see that not all the reactions that have same initial positions of atoms have the same ending

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course on of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. Depending on how we see it, the reactants can be AB or BC. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF thats is large and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which are the closest in energy. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with number of steps increased, starting from a structure that is similar to the Transition state and varying it depending on the direction of the reaction by a very small amount

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature would be observed using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational . To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy mus be promoted more the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibration. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is more. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased.

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T18:46:42Z

Am12618: /* Report the activation energy for both reactions */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since this means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which are cosest in energy to it. Inthis case it resembles both since the atoms that are used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be r1=r2 is 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of rAB=rBC=90.775 pm and p1=p2=0.png|centre|thumb|Internuclear distances vs Time plot for the best transition state estimated]] [[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|centre|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum.

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep movind down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants. Furthermore, we can see that not all the reactions that have same initial positions of atoms have the same ending

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course on of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. Depending on how we see it, the reactants can be AB or BC. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF thats is large and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which are the closest in energy. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with number of steps ...., starting from a structure that is similar to the Transition state

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature would be observed using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational . To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy mus be promoted more the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibration. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is more. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased.

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T18:46:14Z

Am12618: /* Exercise 2:F - H - H system */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since this means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which are cosest in energy to it. Inthis case it resembles both since the atoms that are used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be r1=r2 is 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of rAB=rBC=90.775 pm and p1=p2=0.png|centre|thumb|Internuclear distances vs Time plot for the best transition state estimated]] [[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|centre|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum.

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep movind down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants. Furthermore, we can see that not all the reactions that have same initial positions of atoms have the same ending

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course on of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. Depending on how we see it, the reactants can be AB or BC. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF thats is large and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which are the closest in energy. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with number of steps ...., starting from a structure that is similar to the Transition state

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature would be observed using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational . To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy mus be promoted more the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibration. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is more. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased.

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T18:44:42Z

Am12618: /* Locate the approximate position of the transition state. */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since this means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which are cosest in energy to it. Inthis case it resembles both since the atoms that are used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be r1=r2 is 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of rAB=rBC=90.775 pm and p1=p2=0.png|centre|thumb|Internuclear distances vs Time plot for the best transition state estimated]] [[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|centre|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum.

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep movind down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants. Furthermore, we can see that not all the reactions that have same initial positions of atoms have the same ending

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course on of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. Depending on how we see it, the reactants can be AB or BC. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF thats is large and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which are the closest in energy. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 kjmol-1.pm.fs-1 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with number of steps ...., starting from a structure that is similar to the Transition state

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature would be observed using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational . To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy mus be promoted more the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibration. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is more. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased.

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T18:40:13Z

Am12618: /* Exercise 2:F - H - H system */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since this means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which are cosest in energy to it. Inthis case it resembles both since the atoms that are used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be r1=r2 is 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of rAB=rBC=90.775 pm and p1=p2=0.png|centre|thumb|Internuclear distances vs Time plot for the best transition state estimated]] [[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|centre|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum.

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep movind down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants. Furthermore, we can see that not all the reactions that have same initial positions of atoms have the same ending

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course on of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. Depending on how we see it, the reactants can be AB or BC. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . In this case AB distance is the HF thats is large and BC is the H2 that is shorter, therefore reactants are H2 and F while products are HF and H. This indicates the exothermic nature of the reaction H-H + F → H-F + H and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H since more energy is released in its formation.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which are the closest in energy. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with number of steps ...., starting from a structure that is similar to the Transition state

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature would be observed using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational . To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy mus be promoted more the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibration. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is more. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased.

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T18:33:29Z

Am12618: /* Transition State Theory */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since this means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which are cosest in energy to it. Inthis case it resembles both since the atoms that are used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be r1=r2 is 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of rAB=rBC=90.775 pm and p1=p2=0.png|centre|thumb|Internuclear distances vs Time plot for the best transition state estimated]] [[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|centre|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum.

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep movind down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants. Furthermore, we can see that not all the reactions that have same initial positions of atoms have the same ending

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products(e.g fourth row of the above table).Kinetic energy is of course on of the things to considerate while exploring a reaction but it is not the only one. Therefore, Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and leads in slight underestimation of the real value of the rate, however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus it becomes even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. Depending on how we see it, the reactants can be AB or BC. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . This indicate the exothermic nature of the reaction H-H + F → H-F + H and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which are the closest in energy. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with number of steps ...., starting from a structure that is similar to the Transition state

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature would be observed using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational . To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy mus be promoted more the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibration. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is more. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased.

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T18:25:49Z

Am12618: /* Reactive and unreactive trajectories */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since this means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which are cosest in energy to it. Inthis case it resembles both since the atoms that are used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be r1=r2 is 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of rAB=rBC=90.775 pm and p1=p2=0.png|centre|thumb|Internuclear distances vs Time plot for the best transition state estimated]] [[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|centre|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum.

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep movind down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed collides with A, forms AB and that again collides with C that leads in the final formation of BC.
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants have the right amount of kinetic energy to overcome the activation energy barrier and reach the transition state however due to their vibrational energy they can't form the products and reform the reactants. Furthermore, we can see that not all the reactions that have same initial positions of atoms have the same ending

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products. Therefore,Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and slightly underestimates the real value of the rate is underestimated however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus because even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. Depending on how we see it, the reactants can be AB or BC. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . This indicate the exothermic nature of the reaction H-H + F → H-F + H and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which are the closest in energy. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with number of steps ...., starting from a structure that is similar to the Transition state

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature would be observed using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational . To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy mus be promoted more the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibration. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is more. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased.

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T18:19:56Z

Am12618: /* MOLECULAR REACTION DYNAMICS */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 or ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition state is a saddle point since it is both a maximum for the reaction coordinate pathway as well as a minimum to the pathway that is orthogonal to the reaction coordinate. Thus In order to distinguish the transition state from a local minimum we should find that it is a saddle point since this means it must be a maximum in the direction orthogonal to which it is a minimum.

The force acting on the atoms in the transition state is zero and so is the Kinetic energy. All the energy is therefore potential. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction is going to (products or reactants).
[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. According to Hammond's postulate the transition state would resemble more either the reactants or the products depending on which are cosest in energy to it. Inthis case it resembles both since the atoms that are used are the same.In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be r1=r2 is 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

The plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates two straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of rAB=rBC=90.775 pm and p1=p2=0.png|centre|thumb|Internuclear distances vs Time plot for the best transition state estimated]] [[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|centre|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules and no oscillation at all but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum.

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates (more than AB) and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the oscillating hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating and keep movind down the pathway they were in before
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide forming the product BC, the reactants are reformed from the collision of BC with A
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). The reactant AB oscillates and collides with atomic hydrogen C , however this time the BC formed
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants can reach the transition state however due to their vibrational energy they can't form the products and reform the reactants

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products. Therefore,Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and slightly underestimates the real value of the rate is underestimated however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus because even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. Depending on how we see it, the reactants can be AB or BC. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . This indicate the exothermic nature of the reaction H-H + F → H-F + H and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which are the closest in energy. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with number of steps ...., starting from a structure that is similar to the Transition state

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature would be observed using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational . To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy mus be promoted more the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibration. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is more. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased.

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T17:55:06Z

Am12618: /* Report the activation energy for both reactions */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 and ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition stte is a saddle point since it is both a maximum for the reaction coordinate as well as a minimum to the pathway that is orthogonal to the reaction coordinate. The force acting on the atoms in the transition state is zero and the absence of the force results in zero Kinetic energy. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction is going to (products or reactants).

[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be r1=r2 is 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

Below the plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates 2 straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|centre|thumb|Internuclear distances vs Time plot for the best transition state estimated]] [[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|centre|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum.

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates more than AB and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating as they were keep moving how they were
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide, from the transition state the reactants are reformed instead of the products.
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). From the reactant pathwy, the transition state is reached and products are formed
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants can reach the transition state however due to their vibrational energy they can't form the products and reform the reactants

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products. Therefore,Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and slightly underestimates the real value of the rate is underestimated however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus because even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. Depending on how we see it, the reactants can be AB or BC. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . This indicate the exothermic nature of the reaction H-H + F → H-F + H and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which are the closest in energy. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with number of steps ...., starting from a structure that is similar to the Transition state

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be 110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature would be observed using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational . To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy mus be promoted more the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibration. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is more. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased.

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T17:49:55Z

Am12618: /* Report the activation energy for both reactions */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 and ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition stte is a saddle point since it is both a maximum for the reaction coordinate as well as a minimum to the pathway that is orthogonal to the reaction coordinate. The force acting on the atoms in the transition state is zero and the absence of the force results in zero Kinetic energy. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction is going to (products or reactants).

[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be r1=r2 is 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

Below the plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates 2 straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|centre|thumb|Internuclear distances vs Time plot for the best transition state estimated]] [[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|centre|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum.

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates more than AB and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating as they were keep moving how they were
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide, from the transition state the reactants are reformed instead of the products.
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). From the reactant pathwy, the transition state is reached and products are formed
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants can reach the transition state however due to their vibrational energy they can't form the products and reform the reactants

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products. Therefore,Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and slightly underestimates the real value of the rate is underestimated however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus because even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. Depending on how we see it, the reactants can be AB or BC. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . This indicate the exothermic nature of the reaction H-H + F → H-F + H and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which are the closest in energy. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with number of steps ...., starting from a structure that is similar to the Transition state

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be -110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature would be observed using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational . To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy mus be promoted more the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibration. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is more. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased.

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T17:49:34Z

Am12618: /* Report the activation energy for both reactions */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 and ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition stte is a saddle point since it is both a maximum for the reaction coordinate as well as a minimum to the pathway that is orthogonal to the reaction coordinate. The force acting on the atoms in the transition state is zero and the absence of the force results in zero Kinetic energy. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction is going to (products or reactants).

[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be r1=r2 is 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

Below the plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates 2 straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|centre|thumb|Internuclear distances vs Time plot for the best transition state estimated]] [[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|centre|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum.

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates more than AB and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating as they were keep moving how they were
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide, from the transition state the reactants are reformed instead of the products.
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). From the reactant pathwy, the transition state is reached and products are formed
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants can reach the transition state however due to their vibrational energy they can't form the products and reform the reactants

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products. Therefore,Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and slightly underestimates the real value of the rate is underestimated however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus because even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. Depending on how we see it, the reactants can be AB or BC. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . This indicate the exothermic nature of the reaction H-H + F → H-F + H and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which are the closest in energy. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with number of steps ...., starting from a structure that is similar to the Transition state

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be -110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature would be observed using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational . To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy mus be promoted more the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibration. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is more. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased.

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T17:48:38Z

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 and ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition stte is a saddle point since it is both a maximum for the reaction coordinate as well as a minimum to the pathway that is orthogonal to the reaction coordinate. The force acting on the atoms in the transition state is zero and the absence of the force results in zero Kinetic energy. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction is going to (products or reactants).

[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be r1=r2 is 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

Below the plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates 2 straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|centre|thumb|Internuclear distances vs Time plot for the best transition state estimated]] [[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|centre|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum.

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates more than AB and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating as they were keep moving how they were
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide, from the transition state the reactants are reformed instead of the products.
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). From the reactant pathwy, the transition state is reached and products are formed
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants can reach the transition state however due to their vibrational energy they can't form the products and reform the reactants

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products. Therefore,Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and slightly underestimates the real value of the rate is underestimated however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus because even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. Depending on how we see it, the reactants can be AB or BC. The starting materials go through the transition state and they form the products.By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . This indicate the exothermic nature of the reaction H-H + F → H-F + H and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which are the closest in energy. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with number of steps ...., starting from a structure that is similar to the Transition state

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be -110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature would be observed using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational . To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy mus be promoted more the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibration. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is more. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased.

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T17:33:41Z

Am12618: /* Bibiliography */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 and ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition stte is a saddle point since it is both a maximum for the reaction coordinate as well as a minimum to the pathway that is orthogonal to the reaction coordinate. The force acting on the atoms in the transition state is zero and the absence of the force results in zero Kinetic energy. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction is going to (products or reactants).

[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be r1=r2 is 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

Below the plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates 2 straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|centre|thumb|Internuclear distances vs Time plot for the best transition state estimated]] [[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|centre|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum.

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates more than AB and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating as they were keep moving how they were
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide, from the transition state the reactants are reformed instead of the products.
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). From the reactant pathwy, the transition state is reached and products are formed
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants can reach the transition state however due to their vibrational energy they can't form the products and reform the reactants

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products. Therefore,Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and slightly underestimates the real value of the rate is underestimated however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus because even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. As it can be seen from the contour plot the starting materials are H-F and H , that pass through the transition state to form the products H-H and F. By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . This indicate the exothermic nature of the reaction H-H + F → H-F + H and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which are the closest in energy. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with number of steps ...., starting from a structure that is similar to the Transition state

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be -110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature would be observed using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational . To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy mus be promoted more the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibration. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is more. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased.

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

==Bibiliography==
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T17:31:48Z

Am12618: /* Transition State Theory */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 and ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition stte is a saddle point since it is both a maximum for the reaction coordinate as well as a minimum to the pathway that is orthogonal to the reaction coordinate. The force acting on the atoms in the transition state is zero and the absence of the force results in zero Kinetic energy. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction is going to (products or reactants).

[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be r1=r2 is 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

Below the plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates 2 straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|centre|thumb|Internuclear distances vs Time plot for the best transition state estimated]] [[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|centre|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum.

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates more than AB and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating as they were keep moving how they were
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide, from the transition state the reactants are reformed instead of the products.
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). From the reactant pathwy, the transition state is reached and products are formed
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants can reach the transition state however due to their vibrational energy they can't form the products and reform the reactants

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products. Therefore,Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and slightly underestimates the real value of the rate is underestimated however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus because even less important with increasing size of the atoms

*An additional assumption of the theory is that the Energy distribution is a Maxwell Boltzmann distribution and equilibrium theory may be used to find the concentration of the activated complexes

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. As it can be seen from the contour plot the starting materials are H-F and H , that pass through the transition state to form the products H-H and F. By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . This indicate the exothermic nature of the reaction H-H + F → H-F + H and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which are the closest in energy. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with number of steps ...., starting from a structure that is similar to the Transition state

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be -110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature would be observed using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational . To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy mus be promoted more the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibration. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is more. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased.

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

===Bibiliography===
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T17:21:12Z

Am12618: /* Bibiliography */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 and ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition stte is a saddle point since it is both a maximum for the reaction coordinate as well as a minimum to the pathway that is orthogonal to the reaction coordinate. The force acting on the atoms in the transition state is zero and the absence of the force results in zero Kinetic energy. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction is going to (products or reactants).

[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be r1=r2 is 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

Below the plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates 2 straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|centre|thumb|Internuclear distances vs Time plot for the best transition state estimated]] [[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|centre|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum.

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates more than AB and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating as they were keep moving how they were
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide, from the transition state the reactants are reformed instead of the products.
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). From the reactant pathwy, the transition state is reached and products are formed
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants can reach the transition state however due to their vibrational energy they can't form the products and reform the reactants

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products. Therefore,Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and slightly underestimates the real value of the rate is underestimated however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus because even less important with increasing size of the atoms

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. As it can be seen from the contour plot the starting materials are H-F and H , that pass through the transition state to form the products H-H and F. By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . This indicate the exothermic nature of the reaction H-H + F → H-F + H and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which are the closest in energy. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with number of steps ...., starting from a structure that is similar to the Transition state

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be -110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature would be observed using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational . To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy mus be promoted more the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibration. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is more. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased.

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

===Bibiliography===
*Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press

*J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007

*Chemical kinetics by Laidler, Keith J

01564233wikireport

2020-05-15T17:13:18Z

Am12618: /* Bibiliography */

=MOLECULAR REACTION DYNAMICS =

==Exercise 1: H + H2 system ==

===Dynamics from the transition state region===
==== Transition State ====

''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?''

Transition state on a potential energy surface diagram is the point in the minimum energy pathway that appears to be the maximum. In mathematical terms is the point on the curve where the first derivative of the potential (the gradient) is zero and the second derivative is smaller or larger than zero, i.e. ∂V(ri)/∂ri=0 and ∂2V(ri)/∂ri2 < 0, ∂2V(ri)/∂ri2 > 0. Transition stte is a saddle point since it is both a maximum for the reaction coordinate as well as a minimum to the pathway that is orthogonal to the reaction coordinate. The force acting on the atoms in the transition state is zero and the absence of the force results in zero Kinetic energy. The transition state for the reaction between molecular hydrogen and atomic hydrogen is symmetrical thus the distance between the atoms is equal. In addition to that, since the momenta in transition state are zero if no change in the geometry is made, it remains still. However if we think of the transition state as the top of a cliff and with a ball, a small push of that ball could make it roll towards either the reactants or the products. Thus to locate transition states we could start the trajectories near TS and observe the direction is going to (products or reactants).

[[File:Saddle point min and max.png|center|thumb|Potential Energy Surface plot that illustrates the Transition state.]]

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

As mentioned previously the Transition state here is symmetrical and the atoms are equidistant. In order to easily spot the transition state, different initial conditions where used and the ones where the forces along AB and BC atoms where ~0 kj.mol-1.pm-1 were reported. The initial conditions were found to be r1=r2 is 90.775 pm while p1=p2=0 kjmol-1.pm.fs-1 .

Below the plot of “Inter-nuclear Distances vs Time” is illustrated as well as the Contour plot. The “Inter-nuclear Distances vs Time” plot illustrates 2 straight lines that prove that the bonds AB and BC do not change in the transitions state thus the values found were correct.

[[File:Internuclear distances vs Time plot of r1=r2=90.775 pm and p1=p2=0.png|centre|thumb|Internuclear distances vs Time plot for the best transition state estimated]] [[File:Contour Plot of the best estimate of the transition state position(r1=r2=90.775 pm).png|centre|thumb|Contour plot for the best transition state estimated]]

====How do the mep and the trajectory calculated differ====

In the MEP trajectory , the atoms are moving very slowly and they are on the minimum energy pathway possible for the reaction to take place. To calculate the MEP trajectory the system was displaced from the Initial transition state conditions by a very small amount (1 pm) . However, The trajectory calculated using calculation type Dynamics is different . The two, are illustrated below. Their difference is based on the fact that MEP is related to zero inertia in the molecules but the dynamic trajectory involves a small oscillation in the reaction pathway due to the presence of momentum.

[[File:AM. MEP calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|left|thumb|Trajectory using Dynamics calculation for H-H-H system ]] [[File:AM Dynamics Calculation with r1 = rts+1 pm, r2 = rts and p1 = p2 = 0.png|right|thumb|Trajectory using Dynamics Calculation for H-H-H system]]

====Reactive and unreactive trajectories====

* The table below shows which trajectories are reactive using r1= 74 pm, r2= 200 pm but each time with different momenta
{| 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
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|As the hydrogen molecule AB and atomic hydrogen C approach each other the momentum is sufficient for the two to collide, overcome the energy barrier and form a molecule BC that oscillates more than AB and atomic hydrogen A
|[[File:AM01564233 1.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|As the hydrogen molecule and atomic hydrogen approach each other the momentum is not sufficient for the two to collide, the energy barrier is not crossed and instead of forming the new molecule , the reactants keep vibrating as they were keep moving how they were
|[[File:AM12618 2.png]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|In contrast to the above, a small energy change in the momentum 2 results in formation of products and overcoming of the energetic barrier. The oscillation in the new molecule is more observed than the starting one
|[[File:AM12618 3.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-352.277</nowiki>
|No
|The momentum value of atomic hydrogen C is very high thus it is approaching molecular hydrogen with high speed, however the trajectory is not reactive because, although the energy barrier is being reached , and the atoms collide, from the transition state the reactants are reformed instead of the products.
|[[File:AM12618 4.png]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|As it can be observed from here, a small change in the momentum leads to different results (comparison with the above). From the reactant pathwy, the transition state is reached and products are formed
|[[File:AM12618 5.png]]
|}
'''<nowiki/>'''

'''Conclusion from the table :'''
Not all trajectories can be reactive. There are cases where the reactants can reach the transition state however due to their vibrational energy they can't form the products and reform the reactants

====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, is a theory that takes into account the structure of TS and the properties of the reactants to calculate the rate of a reaction. An assumption that transition state theory makes is that all reactions with kinetic energy larger than the activation energy lead to completion. As soon as the reactants reach the transition state they can't go back and be reformed. The only pathway from there is to go to the products. In reality however, even when the transition state is reached because there is sufficient energy, it is still possible that the reactants are reformed instead of the reaction going to products. This can be shown in the results obtained above where not all trajectories that reached transition state led to formation of products. Therefore,Transition state theory overestimates the experimental value.

*Another assumption that is made from this theory is that everything can be treated classically, ignoring the quantum mechanics such as quantum mechanical tunneling. Tunneling in reality exists, but its effect is really small, and slightly underestimates the real value of the rate is underestimated however since this effect is really small the transition state theory overall overestimates the true rate value. Furthermore the tunneling is inversely proportional to the mass of the atoms, thus because even less important with increasing size of the atoms

==Exercise 2:F - H - H system ==

===PES inspection===

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

In the plots below, atom F is represented with A while the two H atoms are represented with B and C. As it can be seen from the contour plot the starting materials are H-F and H , that pass through the transition state to form the products H-H and F. By inspecting the potential energy surface, we can see that there is a drop in energy going from reactants to products . This indicate the exothermic nature of the reaction H-H + F → H-F + H and at the same time the endothermic nature of the reverse reaction H-F + H → H-H + F. The bond strength of H-F is higher than that of H-H.

[[File:Surface Plot exercise 2 PES PART.png|centre|thumb|Potential Energy surface for F-H-H system]] [[File:Contour Plot exercise 2 PES PART.png|centre|thumb|Contour plot for F-H-H System]]

====Locate the approximate position of the transition state.====
Hammond's Postulate states that the Transition state structure resembles more the structure of either the products or the reactants depending on which are the closest in energy. This indicates that for exothermic reactions the transition state resembles the reactants while in endothermic reactions it's product like. In this case the exothermic reaction is H-H + F → H-F + H and thus the TS will resemble H-H +F. Thus using Hammonds postulate as a guide, the same method as in exercise one was used to locate the transition state, however in this case the distances rAB is not equal to rBC since the Transition state would not be symmetrical and as mentioned above, would resemble H-H + F. What has helped in the process was looking at the contour plot and animation plot. No kinetic energy in transition state implies p1=p2=0 while rAB is approximately 181.1 pm, rBC is approximately 75 pm

[[File:New Internuclear distances vs Time plot H-H-F system.png|left|thumb|Internuclear distances vs Time plot for transition state estimate in H-H-F system]] [[File:TS PES plot.png|right|thumb|Potential energy surface plot for transition state in F-H-H system]]

====Report the activation energy for both reactions====

In order to estimate the activation energy for the reactions, a MEP calculation was used, with number of steps ...., starting from a structure that is similar to the Transition state

'''For the reaction H-H + F → H-F + H:'''

The exothermic nature of the reaction suggests that the activation energy required to form the transition state is small. After all, the TS already resembles the reactants and to overcome the energy barrier , a small amount of energy was needed. This was calculated and found to be 0.06 kJmol-1.

[[File:Energy vs time plot for exothermic reaction.png|thumb|center|Energy vs Time plot for the exothermic reaction H-H + F → H-F + H]]

'''For the reaction H-F + F → H-H + H'''

The endothermic nature of the reaction suggests that the activation energy required to form the transition state is large and in order to form the products a lot of amount of heat/energy needs to be supplied to the system to overcome the barrier. This was calculated and found to be -110.219 kJmol-1.

[[File:Plot5 Energy forms (170, 75, 0, 0, -544.341, -434.112).png|center|thumb|Energy vs Time plot for the endothermic reaction H-F + H → H-H + F]]

===Reaction dynamics===

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally====

A reactive trajectory for the reaction for F + H2, was found to be at rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5
The above reaction is exothermic thus energy in the form of heat is released into the system upon completion of reaction. By inspection of the contour plot we can see that the molecule H2 is approaching the Fluorine atom with intense vibration. Upon collision of the molecule with the atom, a bond is formed between the fluorine atom and the hydrogen resulting in HF molecule and atomic Hydrogen. The HF molecule is now vibrating more intensily than before. By looking at the Momenta vs Time plot we can observe that the total momentum of the system remains constant thus the total energy of the system is constant, but the kinetic and potential are alternating.

[[File:Contour plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|left|thumb|Contour plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]] [[File:Momenta vs Time plot of H2 going to HF reactive trajectory with initial HH=75 HF=190, pHH=-5 and pHF=-1.png|right|thumb|Momenta vs time plot of the reactive trajectory using rAB= 190 pm, rBC= 75 pm, pAB = -1 , prBC= -5]]

Since the reaction is exothermic in an experiment we would observe increase of temperature would be observed using a calorimeter. However caloremetry calculates the total heat produced, and with that we cannot distinguish between Kinetic translational and Kinetic vibrational . To distinguish between the two we can use the IR technique in order to monitor the IR photons emitted from the vibrational relaxation.

====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 by investing more in translational or vibrational energy, reactions can become more efficient. To determine which energy mus be promoted more the nature of the chemical reaction must be investigated using Hammond's Postulate that was mentioned earlier.
According these, an exothermic reaction, i.e reaction when the transition state is early and resembles the reactants, is more successful by investing more translational energy rather than the vibration. At the same time, for an endothermic reaction where the TS is late and resembles the products the reaction is more successfully promoted from higher vibrational energy.

The above can be verified by looking at the following plots. These describe the exothermic reaction of H-H + F → H-F + H.

[[File:Reactive trajectory of an exothermic reaction.png|center|thumb| Reactive trajectory for an exothermic reaction, with rAB=225 pm, rBC=75 pm, p1=-4, p2=-1]]

The first plot shows a reactive trajectory. As it is observed , starting form the reactants and going through the channel, the TS is crossed and then we go to the products channel. From reactants to the transition state not many vibrations appear thus the translational energy is more. However after overcoming barrier , from the TS to the reactants the vibrational energy is increased.

[[File:Exothermic Reaction Polyani's empirical Rules unsuccessful Reaction.png|center|thumb|Unreactive trajectory for an exothermic reaction with rAB=225 pm, rBC=75 pm, p1=-4, p2=-8]]

If you look at the other plot of the unreactive trajectory we can see that if we start with a high vibrational energy, the reaction can't be completed. The opposite thing would be observed for the reverse reaction HF + H → H2 + F

===Bibiliography===
<ref name="Atkins" >Atkins’ Physical Chemistry, P.W Atkins and J. De Paula 9th Edition, Oxford University Press</ref>

<ref name="Polanyi's rules">Atkins’ J. Chem. Phys. 138, 234104 (2013); https://doi.org/10.1063/1.4810007 Press</ref>