ChemWiki - User contributions [en]

MRD:ch6518

2020-05-15T20:33:12Z

Ch6518: /* Exercise 1: H + H2 system */

= Molecular Reaction Dymanics =

== Exercise 1: H + H2 system ==
* ''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 appears on potential energy surfaces as a saddle point⁠—it is a critical point (derivatives are zero) but it is neither a local minimum nor a local maximum. Unlike a local minimum, at which an infinitesimal perturbation in any direction will result in a positive derivative, or a local maximum at which perturbation will result in a negative derivative, perturbation from a saddle point will result in a derivative whose sign depends on the direction of the perturbation.
* ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' The transition state in a H+H2 system has a position '''rts = 90.75 ± 0.05 pm''' This was determined by approximating the position of the transition state by visual inspection of the contour plot for internuclear distances—this was estimated as 91 pm—then modifying the initial conditions around this approximation in a binary search-type method until the amplitude of oscillations in the “Internuclear Distances vs Time” plot were as close to zero as possible.
[[File:Ch6518 90.75pm xt.png|none|thumb|520x520px|The plot of internuclear distances vs time for a H+H2 reaction at the transition state position of 90.75 pm, showing zero oscillation]][[File:Ch6518 90.75pm cont.png|none|thumb|524x524px|The contour plot for the H+H2 reaction initially at the transition state position of 90.75 pm, showing no oscillation on the ridge]]
* ''Comment on how the mep and the trajectory you just calculated differ.'' The ''MEP'' (minimum energy path) of the reaction is simply the route that is parallel to and in the opposite direction of of the gradient vector (orthogonal to the contour) at every point; that is to say, it moves towards and then follows the trough/valley of the energy surface—the bond length in the product molecule will decrease to a constant. This shows the energy profile of a reaction clearly, but it fails to take into account the vibrational energy in molecules. In reality, as shown by the dynamics plot of the same reaction, there is vibrational energy in the system, so the interatomic distance between the two atoms in the product molecule will oscillate. As a result, the kinetic energy of the system will have the same total value as in the mep plot, but it will be comprised of both translational and vibrational kinetic energy.
[[File:Ch6518 mep.png|none|thumb|393x393px|The minimum energy path of the system following minor perturbation from the transition state]][[File:Ch6518 dynamic.png|none|thumb|387x387px|The dynamic trajectory of the system under the same initial conditions, showing oscillation of the AB distance]]
* ''Complete the table by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table? '''''POTENTIAL ENERGY OF TRANSITION STATE = -415.378 kJ.mol-1'''
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot / kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|The system initially has little to no AB oscilation. It overcomes the activation energy, passing the transition state, and continues to form the A+BC products with oscillation of BC
|[[File:Ch6518 mrd table traj 1.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The reactants approach each other, but the translational kinetic energy is insufficient to overcome the activation energy so the reactants separate again.
|[[File:Ch6518 mrd table traj 2.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The reactants approach each other, overcome the activation energy, and the products separate.
|[[File:Ch6518 mrd table traj 3.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No*
|The system overcomes the activation energy and forms BC, but the products oscillate back together and re-cross the transition state, separating again as AB and C.
|[[File:Ch6518 mrd table traj 4.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes*
|The system overcomes the transition state, then re-crosses it, re-forming the reactants, then crosses the transition state once more to form the BC product.
|[[File:Ch6518 mrd table traj 5.png|thumb]]
|}
From this table, it is evident that total energy alone is insufficient to determine whether the system will cross the transition state or not. It appears that increasing translational energy increases the likelihood of successful reaction up to a certain point, after which it makes the system more likely to cross back over the transition state. Conversely, increasing vibrational energy appears to make the system less likely to cross the transition state. A combination of these two factors will therefore determine if the trajectory is reactive or not.
* ''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ''Transition state theory assumes that all trajectories are reactive if they have total kinetic energies greater than the activation energies of the reaction. However, the fourth entry in the table demonstrates that this is not always the case; though the kinetic energy of the system was significantly higher than the activation energy required, the trajectory was overall unreactive. As a result, for this case Transition State Theory will predict a rate for a reaction that in reality does not happen.

== Exercise 2: F - H - H system ==
* ''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?'' The F + H2 → H + HF reaction is '''exothermic''', therefore the H + HF → F + H2 reaction is '''endothermic'''.This is because the H-F bond is much stronger than the H-H bond, therefore less work is done on the H-H bond to break it than the amount of work done on the surroundings by the formation of the H-F bond. This results in an exothermic reaction. Conversely, more work must be done to break the H-F bond than is done by forming a H-H bond, so this reaction is endothermic.[[File:Ch6518 f h2 exothermic.png|none|thumb|318x318px|The forward direction corresponds to a reactive pathway that leads to the formation of H and HF from F and H2. Though strong vibration is occurring, it is clear that the potential energy is decreasing from a maximum and kinetic energy is increasing.]]
* ''Locate the approximate position of the transition state.'' The transition state for the F-H-H system is at '''rF-H = 180.71 pm, rH-H = 74.6 pm'''. This was determined using the same method as for the H-H-H system.[[File:Ch6518 fhh trans xt.png|none|thumb|350x350px|The energy profile for the F-H-H system when started on the transition state with no momentum]][[File:Ch6518 fhh trans contour.png|none|thumb|354x354px|The trajectory contour plot for the F-H-H system when started on the transition state with no momentum]]
* ''Report the activation energy for both reactions''. The transition state has a potential energy of -433.979 kJ.mol-1. As H-H are further separated from F along the minimum energy path, the potential energy of the system tends to -435.039 kJ.mol-1. Consequently, the activation energy for the F + H2 → H + HF reaction must be '''+1.060 kJ.mol-1'''. For a system in which the H-F product has formed and is separate from the H, the energy along the MEP tends to -560.700 kJ.mol-1. Thus, the activation energy for the H + HF → F + H2 reaction is '''+126.721 kJ.mol-1'''. We can therefore see the enormous 120x difference in activation energy for the reaction in each direction
* ''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. ''At any point in time, kinetic energy + potential energy = constant. Therefore, when reaction (potential) energy is released, it is converted to kinetic energy, which comprises both vibrational and translational modes. The internuclear velocity profile of an exothermic reaction such as the formation of HF and H from H2 and F shows a large increase in the velocity and oscillation amplitude of the products compared to the atoms. This demonstrates the conversion of chemical potential energy into translation and vibration, Experimentally, this could of course be determined by simple calorimetry; the temperature of a substance is related by the Boltzmann distribution to the population of its kinetic energy states, so the increase in temperature for an exothermic reaction can be attributed to this release of reaction energy.[[File:Ch6518 vel profile fhh.png|none|thumb|491x491px|The internuclear velocity profile for an exothermic reaction (H2+F→HF+H), showing a large increase in the amplitude of oscillations]]
* ''Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. ''The success of a reaction seemed to be determined by how closely the trajectory of the components followed the minimum energy path—systems which deviated from this path were more likely to re-cross the transition state and form the reactants again. In the symmetric H+H2→H2+H reaction, the most efficient distribution was for all energy to be in the translational mode, since this allowed the system to follow the MEP closely. In the F + H2 → H + HF reaction, where the transition state was very close to the reactants, low-ampliude oscillation resulted in the most reactive trajectories, specifically oscillations in which the atoms in the reactant molecule were separating as the transition state was crossed. This suggests that low-amplitude, high-frequency oscillations are optimum as they keep the reaction path close to the minimum energy path, whilst increasing the chance that the momentum will be in the correct direction for any given point. Conversely, in the H + HF → F + H2 reaction, in which the transition state was very similar to the products and far from the reactants, higher-amplitude oscillations seemed to be favoured for much the same reason- they ensured the reaction closely followed the MEP. It may therefore be said that for a trajectory to be efficient, more energy must be distributed to the vibrational modes the further the transition state is from the reactants.[[File:Ch6518 close trajectory.png|left|thumb|364x364px|An example of an exothermic trajectory in which there is little energy in the vibrational modes of the system prior to crossing the transition state- the reaction plot passes over the transition state at the same point as the MEP]][[File:Ch6518 far trajectory.png|none|thumb|368x368px|An example of an exothermic reaction with greater amplitude of oscillation and thus greater deviation from the MEP when crossing the transition state- this trajectory re-crosses the transition state.]]

MRD:ch6518

2020-05-15T20:17:21Z

Ch6518: /* Exercise 2: F - H - H system */

= Molecular Reaction Dymanics =

== Exercise 1: H + H2 system ==
* ''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 appears on potential energy surfaces as a saddle point⁠—it is a critical point (derivatives are zero) but it is neither a local minimum nor a local maximum. Unlike a local minimum, at which an infinitesimal perturbation in any direction will result in a positive derivative, or a local maximum at which perturbation will result in a negative derivative, perturbation from a saddle point will result in a derivative whose sign depends on the direction of the perturbation.
* ''Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.'' The transition state in a H+H2 system has a position '''rts = 90.75 ± 0.05 pm''' This was determined by approximating the position of the transition state by visual inspection of the contour plot for internuclear distances—this was estimated as 91 pm—then modifying the initial conditions around this approximation in a binary search-type method until the amplitude of oscillations in the “Internuclear Distances vs Time” plot were as close to zero as possible.
[[File:Ch6518 90.75pm xt.png|none|thumb|520x520px|The plot of internuclear distances vs time for a H+H2 reaction at the transition state position of 90.75 pm, showing zero oscillation]][[File:Ch6518 90.75pm cont.png|none|thumb|524x524px|The contour plot for the H+H2 reaction initially at the transition state position of 90.75 pm, showing no oscillation on the ridge]]
* ''Comment on how the mep and the trajectory you just calculated differ.'' The ''MEP'' (minimum energy path) of the reaction is simply the route that is parallel to and in the opposite direction of of the gradient vector (orthogonal to the contour) at every point; that is to say, it moves towards and then follows the trough/valley of the energy surface—the bond length in the product molecule will decrease to a constant. This shows the energy profile of a reaction clearly, but it fails to take into account the vibrational energy in molecules. In reality, as shown by the dynamics plot of the same reaction, there is vibrational energy in the system, so the interatomic distance between the two atoms in the product molecule will oscillate. As a result, the kinetic energy of the system will have the same total value as in the mep plot, but it will be comprised of both translational and vibrational kinetic energy.
[[File:Ch6518 mep.png|none|thumb|393x393px|The minimum energy path of the system following minor perturbation from the transition state]][[File:Ch6518 dynamic.png|none|thumb|387x387px|The dynamic trajectory of the system under the same initial conditions, showing oscillation of the AB distance]]
* ''Complete the table by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?''
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot / kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|The system initially has little to no AB oscilation. It overcomes the activation energy, passing the transition state, and continues to form the A+BC products with oscillation of BC
|[[File:Ch6518 mrd table traj 1.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The reactants approach each other, but the translational kinetic energy is insufficient to overcome the activation energy so the reactants separate again.
|[[File:Ch6518 mrd table traj 2.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The reactants approach each other, overcome the activation energy, and the products separate.
|[[File:Ch6518 mrd table traj 3.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No*
|The system overcomes the activation energy and forms BC, but the energy is so high that the products oscillate back together and re-cross the transition state, separating again as AB and C.
|[[File:Ch6518 mrd table traj 4.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes*
|The system overcomes the transition state, then re-crosses it, re-forming the reactants, then crosses the transition state once more to form the BC product.
|[[File:Ch6518 mrd table traj 5.png|thumb]]
|}
From this table, it is evident that total energy alone is insufficient to determine whether the system will cross the transition state or not. It appears that increasing translational energy increases the likelihood of successful reaction up to a certain point, after which it makes the system more likely to cross back over the transition state. Conversely, increasing vibrational energy appears to make the system less likely to cross the transition state. A combination of these two factors will therefore determine if the trajectory is reactive or not.
* ''Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?''

== Exercise 2: F - H - H system ==
* ''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?'' The F + H2 → H + HF reaction is '''exothermic''', therefore the H + HF → F + H2 reaction is '''endothermic'''.This is because the H-F bond is much stronger than the H-H bond, therefore less work is done on the H-H bond to break it than the amount of work done on the surroundings by the formation of the H-F bond. This results in an exothermic reaction. Conversely, more work must be done to break the H-F bond than is done by forming a H-H bond, so this reaction is endothermic.[[File:Ch6518 f h2 exothermic.png|none|thumb|318x318px|The forward direction corresponds to a reactive pathway that leads to the formation of H and HF from F and H2. Though strong vibration is occurring, it is clear that the potential energy is decreasing from a maximum and kinetic energy is increasing.]]
* ''Locate the approximate position of the transition state.'' The transition state for the F-H-H system is at '''rF-H = 180.71 pm, rH-H = 74.6 pm'''. This was determined using the same method as for the H-H-H system.[[File:Ch6518 fhh trans xt.png|none|thumb|350x350px|The energy profile for the F-H-H system when started on the transition state with no momentum]][[File:Ch6518 fhh trans contour.png|none|thumb|354x354px|The trajectory contour plot for the F-H-H system when started on the transition state with no momentum]]
* ''Report the activation energy for both reactions''. The transition state has a potential energy of -433.979 kJ.mol-1. As H-H are further separated from F along the minimum energy path, the potential energy of the system tends to -435.039 kJ.mol-1. Consequently, the activation energy for the F + H2 → H + HF reaction must be '''+1.060 kJ.mol-1'''. For a system in which the H-F product has formed and is separate from the H, the energy along the MEP tends to -560.700 kJ.mol-1. Thus, the activation energy for the H + HF → F + H2 reaction is '''+126.721 kJ.mol-1'''. We can therefore see the enormous 120x difference in activation energy for the reaction in each direction
* ''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. ''At any point in time, kinetic energy + potential energy = constant. Therefore, when reaction (potential) energy is released, it is converted to kinetic energy, which comprises both vibrational and translational modes. The internuclear velocity profile of an exothermic reaction such as the formation of HF and H from H2 and F shows a large increase in the velocity and oscillation amplitude of the products compared to the atoms. This demonstrates the conversion of chemical potential energy into translation and vibration, Experimentally, this could of course be determined by simple calorimetry; the temperature of a substance is related by the Boltzmann distribution to the population of its kinetic energy states, so the increase in temperature for an exothermic reaction can be attributed to this release of reaction energy.[[File:Ch6518 vel profile fhh.png|none|thumb|491x491px|The internuclear velocity profile for an exothermic reaction (H2+F→HF+H), showing a large increase in the amplitude of oscillations]]
* ''Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. ''The success of a reaction seemed to be determined by how closely the trajectory of the components followed the minimum energy path—systems which deviated from this path were more likely to re-cross the transition state and form the reactants again. In the symmetric H+H2→H2+H reaction, the most efficient distribution was for all energy to be in the translational mode, since this allowed the system to follow the MEP closely. In the F + H2 → H + HF reaction, where the transition state was very close to the reactants, low-ampliude oscillation resulted in the most reactive trajectories, specifically oscillations in which the atoms in the reactant molecule were separating as the transition state was crossed. This suggests that low-amplitude, high-frequency oscillations are optimum as they keep the reaction path close to the minimum energy path, whilst increasing the chance that the momentum will be in the correct direction for any given point. Conversely, in the H + HF → F + H2 reaction, in which the transition state was very similar to the products and far from the reactants, higher-amplitude oscillations seemed to be favoured for much the same reason- they ensured the reaction closely followed the MEP. It may therefore be said that for a trajectory to be efficient, more energy must be distributed to the vibrational modes the further the transition state is from the reactants.[[File:Ch6518 close trajectory.png|left|thumb|364x364px|An example of an exothermic trajectory in which there is little energy in the vibrational modes of the system prior to crossing the transition state- the reaction plot passes over the transition state at the same point as the MEP]][[File:Ch6518 far trajectory.png|none|thumb|368x368px|An example of an exothermic reaction with greater amplitude of oscillation and thus greater deviation from the MEP when crossing the transition state- this trajectory re-crosses the transition state.]]

File:Ch6518 far trajectory.png

2020-05-15T20:12:28Z

Ch6518:

File:Ch6518 close trajectory.png

2020-05-15T20:09:41Z

Ch6518:

2020-05-15T17:55:08Z

Ch6518: /* Exercise 2: F - H - H system */

MRD:ch6518

2020-05-15T17:47:47Z

Ch6518:

= Molecular Reaction Dymanics =

== Exercise 1: H + H2 system ==
* 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 appears on potential energy surfaces as a saddle point⁠—it is a critical point (derivatives are zero) but it is neither a local minimum nor a local maximum. Unlike a local minimum, at which an infinitesimal perturbation in any direction will result in a positive derivative, or a local maximum at which perturbation will result in a negative derivative, perturbation from a saddle point will result in a derivative whose sign depends on the direction of the perturbation.
* Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. The transition state in a H+H2 system has a position '''rts = 90.75 ± 0.05 pm''' This was determined by approximating the position of the transition state by visual inspection of the contour plot for internuclear distances—this was estimated as 91 pm—then modifying the initial conditions around this approximation in a binary search-type method until the amplitude of oscillations in the “Internuclear Distances vs Time” plot were as close to zero as possible.
**[[File:Ch6518 90.75pm xt.png|none|thumb|520x520px|The plot of internuclear distances vs time for a H+H2 reaction at the transition state position of 90.75 pm, showing zero oscillation]][[File:Ch6518 90.75pm cont.png|none|thumb|524x524px|The contour plot for the H+H2 reaction initially at the transition state position of 90.75 pm, showing no oscillation on the ridge]]

* Comment on how the ''mep'' and the trajectory you just calculated differ. The ''mep'' (minimum energy path) of the reaction is simply the route that is parallel to and in the opposite direction of of the gradient vector (orthogonal to the contour) at every point; that is to say, it moves towards and then follows the trough/valley of the energy surface—the bond length in the product molecule will decrease to a constant. This shows the energy profile of a reaction clearly, but it fails to take into account the vibrational energy in molecules. In reality, as shown by the dynamics plot of the same reaction, there is vibrational energy in the system, so the interatomic distance between the two atoms in the product molecule will oscillate. As a result, the kinetic energy of the system will have the same total value as in the mep plot, but it will be comprised of both translational and vibrational kinetic energy.
[[File:Ch6518 mep.png|none|thumb|393x393px|The minimum energy path of the system following minor perturbation from the transition state]][[File:Ch6518 dynamic.png|none|thumb|387x387px|The dynamic trajectory of the system under the same initial conditions, showing oscillation of the AB distance]]
* Complete the table by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot / kJ.mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|The system initially has little to no AB oscilation. It overcomes the activation energy, passing the transition state, and continues to form the A+BC products with oscillation of BC
|[[File:Ch6518 mrd table traj 1.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|The reactants approach each other, but the translational kinetic energy is insufficient to overcome the activation energy so the reactants separate again.
|[[File:Ch6518 mrd table traj 2.png|thumb]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The reactants approach each other, overcome the activation energy, and the products separate.
|[[File:Ch6518 mrd table traj 3.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No*
|The system overcomes the activation energy and forms BC, but the energy is so high that the products oscillate back together and re-cross the transition state, separating again as AB and C.
|[[File:Ch6518 mrd table traj 4.png|thumb]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes*
|The system overcomes the transition state, then re-crosses it, re-forming the reactants, then crosses the transition state once more to form the BC product.
|[[File:Ch6518 mrd table traj 5.png|thumb]]
|}
From this table, it is evident that total energy alone is insufficient to determine whether the system will cross the transition state or not. It appears that increasing translational energy increases the likelihood of successful reaction up to a certain point, after which it makes the system more likely to cross back over the transition state. Conversely, increasing vibrational energy appears to make the system less likely to cross the transition state. A combination of these two factors will therefore determine if the trajectory is reactive or not.
* Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

== Exercise 2: F - H - H system ==
* 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? The F + H2 → H + HF reaction is exothermic, therefore the H + HF → F + H2 reaction is endothermic.[[File:Ch6518 f h2 exothermic.png|none|thumb|318x318px|The forward direction corresponds to a reactive pathway that leads to the formation of H and HF from F and H2. Though strong vibration is occurring, it is clear that the potential energy is decreasing from a maximum and kinetic energy is increasing.]]
* Locate the approximate position of the transition state. The transition state for the F-H-H system is at '''rF-H''' = 180.71 pm, '''rH-H''' = 74.6 pm. This was determined using the same method as for the H-H-H system.[[File:Ch6518 fhh trans xt.png|none|thumb|350x350px|The energy profile for the F-H-H system when started on the transition state with no momentum]][[File:Ch6518 fhh trans contour.png|none|thumb|354x354px|The trajectory contour plot for the F-H-H system when started on the transition state with no momentum]]
* Report the activation energy for both reactions. The transition state has a potential energy of -433.979 kJ.mol-1.
* 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.
* 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.

File:Ch6518 fhh trans contour.png

2020-05-15T17:45:19Z

Ch6518:

File:Ch6518 fhh trans xt.png

2020-05-15T17:43:58Z

Ch6518:

File:Ch6518 f h2 exothermic.png

2020-05-15T17:37:36Z

Ch6518:

MRD:ch6518

2019-03-15T16:05:23Z

Ch6518: /* Molecular Hydrogen H2 */

==Ammonia NH3==
====Structure====
{| class="wikitable" border="1"
|+ Optimisation of of NH3
! Calculation Method !! Basis Set !! Final Energy E / a.u. !! Point Group
|-
| RB3LYP || 631G(d,p) || -56.55776873 || C3V
|}

<pre>
Item Value Threshold Converged?
Maximum Force 0.000004 0.000450 YES
RMS Force 0.000004 0.000300 YES
Maximum Displacement 0.000072 0.001800 YES
RMS Displacement 0.000035 0.001200 YES
</pre>



<jmol><jmolApplet>
<title>NH3</title>
<color>black</color>
<size>400</size>
<uploadedFileContents>CH6518_NH3_OPTF_POP.LOG</uploadedFileContents>
<script>frame 7</script>
</jmolApplet></jmol>

Optimised N-H bond length (Angstrom): <math>1.02 \pm 0.01</math>

Optimised H-N-H bond angle (Degrees): <math>106 \pm 1</math>

Charge on N: +0.375

Charge on H: -1.125

(Dipole Moment: 1.8466 Db)

Opt file [[Media:CH6518_NH3_OPTF_POP.LOG| here]]

====Vibrational Modes====

[[File:Ch6518_nh3_vibrations.PNG]]

{| class="wikitable" border="1"
|+ Vibrational Modes of NH3
! Wavenumber / cm-1 !! Symmetry !! Intensity !! Bend or Stretch !! Image
|-
| 1090 || A1 || 145.4 || rowspan="3" | Bend || [[File:Ch6518_nh3_1090.png|x150px]]
|-
| 1694 || E || 13.6 || [[File:Ch6518_nh3_1694_1.png|x150px]]
|-
| 1694 || E || 13.6 || [[File:Ch6518_nh3_1694_2.png|x150px]]
|-
| 3461 || A1 || 1.1 || rowspan="3" | Stretch || [[File:Ch6518_nh3_3461.png|x150px]]
|-
| 3590 || E || 0.3 || [[File:Ch6518_nh3_3590_1.png|x150px]]
|-
| 3590 || E || 0.3 || [[File:Ch6518_nh3_3590_2.png|x150px]]
|}

Number of expected vibrational modes: <math>3N-6=3*4-6=6</math>

Two pairs of degenerate vibrational modes, one pair at 1694 cm-1, the other at 3590 cm-1

The vibrations at 1090 cm-1 and 3461 cm-1 are highly symmetric

The 1090 cm-1 is known as the "umbrella" mode due to its resemblance to an opening and closing umbrella

In an IR spectrum of gaseous ammonia, four bands would be predicted since there are vibrational modes at four different frequencies

==Haber-Bosch Process==

===Molecular Nitrogen N2===
====Structure====
{| class="wikitable" border="1"
|+ Optimisation of of N2
! Calculation Method !! Basis Set !! Final Energy E / a.u. !! Point Group
|-
| RB3LYP || 631G(d,p) || -109.52412868 || D∞h
|}

<pre>
Item Value Threshold Converged?
Maximum Force 0.000001 0.000450 YES
RMS Force 0.000001 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000000 0.001200 YES
</pre>

<jmol><jmolApplet>
<title>N2</title>
<color>black</color>
<size>400</size>
<uploadedFileContents>CH6518_N2_OPTF_POP.LOG</uploadedFileContents>
<script>frame 4</script>
</jmolApplet></jmol>

Optimised N≡N bond length (Angstrom): <math>1.11 \pm 0.01</math>

Charge on each N: 0

(Dipole moment: 0 Db)

The molecule [https://onlinelibrary.wiley.com/doi/full/10.1002/chem.201702727 (dinitrogen)-(2,2',2''-(phosphanetriyl)tris(1-(diphenylphosphanyl)-3-methyl-1H-indole))-ruthenium tetrahydrofuran solvate] (refcode DEKFUX) has a reported N≡N bond length of 1.086(6) Angstrom

Opt file [[Media:CH6518_N2_OPTF_POP.LOG| here]]

====Vibrational Modes====

[[File:Ch6518_n2_vibrations.PNG]]

{| class="wikitable" border="1"
|+ Vibrational Modes of N2
! Wavenumber / cm-1 !! Symmetry !! Intensity !! Bend or Stretch !! Image
|-
| 2457 || SGG || 0.0 (IR inactive) || Stretch || [[File:Ch6518_n2.png|x150px]]
|}

Number of expected vibrational modes: <math>3N-5=3*2-5=1</math>

===Molecular Hydrogen H2===
====Structure====
{| class="wikitable" border="1"
|+ Optimisation of of H2
! Calculation Method !! Basis Set !! Final Energy E / a.u. !! Point Group
|-
| RB3LYP || 631G(d,p) || -1.17853936 || D∞h
|}

<pre>
Item Value Threshold Converged?
Maximum Force 0.000000 0.000450 YES
RMS Force 0.000000 0.000300 YES
Maximum Displacement 0.000000 0.001800 YES
RMS Displacement 0.000001 0.001200 YES
</pre>

<jmol><jmolApplet>
<title>H2</title>
<color>black</color>
<size>400</size>
<uploadedFileContents>CH6518_H2_OPTF_POP.LOG</uploadedFileContents>
<script>frame 5</script>
</jmolApplet></jmol>

Optimised H-H bond length (Angstrom): <math>0.74 \pm 0.01</math>

Charge on each H: 0

(Dipole moment: 0 Db)

Opt file [[Media:CH6518_H2_OPTF_POP.LOG| here]]

====Vibrational Modes====

[[File:Ch6518_h2_vibrations.PNG]]

{| class="wikitable" border="1"
|+ Vibrational Modes of H2
! Wavenumber / cm-1 !! Symmetry !! Intensity !! Bend or Stretch !! Image
|-
| 4466 || SGG || 0.0 (IR Inactive) || Stretch || [[File:Ch6518_h2.png|x150px]]
|}

Number of expected vibrational modes: <math>3N-5=3*2-5=1</math>

===Energy Calculations===
<math>E(NH_3)=-56.55777 </math>

<math>2*E(NH_3)=-113.11554 </math>

<math>E(N_2)=-109.52413 </math>

<math>E(H_2)=-1.17854 </math>

<math>3*E(H_2)=-3.53562 </math>

<math>\Delta E=2*E(NH_3)-[E(N_2)+3*E(H_2)]= -113.11554-(-109.52413-3.53562)=-0.0557905 a.u. =-146.5 kJ mol^{-1} </math>

Hence ΔE = -146.5 kJ mol-1 < 0

So the reaction is exothermic, and thus the ammonia product is more stable than the reactants

==Sulfur Tetrafluoride SF4==
{| class="wikitable" border="1"
|+ Optimisation of of SF4
! Calculation Method !! Basis Set !! Final Energy E / a.u. !! Point Group
|-
| RB3LYP || 631G(d,p) || -797.45952460 || C2v
|}

<pre>
Item Value Threshold Converged?
Maximum Force 0.000148 0.000450 YES
RMS Force 0.000065 0.000300 YES
Maximum Displacement 0.000570 0.001800 YES
RMS Displacement 0.000281 0.001200 YES
</pre>

<jmol><jmolApplet>
<title>SF4</title>
<color>black</color>
<size>400</size>
<uploadedFileContents>CH6518_SF4_OPTF_POP.LOG</uploadedFileContents>
<script>frame 7</script>
</jmolApplet></jmol>

Optimised S-Fax bond length (Angstrom): <math>1.67 \pm 0.01</math>

Optimised S-Feq bond length (Angstrom): <math>1.59 \pm 0.01</math>

Optimised Fax-S-Fax bond angle (Degrees): <math>171 \pm 1</math>

Optimised Feq-S-Feq bond angle (Degrees): <math>102 \pm 1</math>

Optimised Fax-S-Feq bond angle (Degrees): <math>87 \pm 1</math>

Opt file [[Media:CH6518_SF4_OPTF_POP.LOG| here]]

[[File:Ch6518_sf4_vibrations.PNG]]

{| class="wikitable" border="1"
|+ Vibrational Modes of SF4
! Wavenumber / cm-1 !! Symmetry !! Intensity !! Bend or Stretch !! Image
|-
| 189 || A1 || 0.5 || rowspan="5" | Bend || [[File:Ch6518_sf4_189.png|x150px]]
|-
| 330 || B1 || 9.9 || [[File:Ch6518_sf4_330.png|x150px]]
|-
| 435 || A2 || 0.0 (IR inactive) || [[File:Ch6518_sf4_435.png|x150px]]
|-
| 488 || A1 || 20.5 || [[File:Ch6518_sf4_488.png|x150px]]
|-
| 496 || B2 || 2.1 || [[File:Ch6518_sf4_496.png|x150px]]
|-
| 584 || A1 || 2.7 || rowspan="4" | Stretch || [[File:Ch6518_sf4_584.png|x150px]]
|-
| 807 || B2 || 508.8 || [[File:Ch6518_sf4_807.png|x150px]]
|-
| 852 || B1 || 149.9 || [[File:Ch6518_sf4_852.png|x150px]]
|-
| 867 || A1 || 100.8 || [[File:Ch6518_sf4_867.png|x150px]]
|}

Number of expected vibrational modes: <math>3N-6=3*5-6=9</math>

Charge on S: +1.983

Charge on Fax: -0.537

Charge on Feq: -0.455

(Dipole moment: 0.8849 Db)

{| class="wikitable" border="1"
|+ Molecular Orbitals of SF4
! Orbital Number !! Energy / eV !! Orbital type !! Image
|-
| 9 || -6.17183 || Non-bonding || [[File:Ch6518_sf4_mo9.PNG|x300px]]
|-
| 10 || -1.32222 || Bonding || [[File:Ch6518_sf4_mo10.png|x300px]]
|-
| 13 || -1.17960 || Primarily bonding || [[File:Ch6518_sf4_mo13.PNG|x300px]]
|-
| 14 || -0.78968 || Primarily antibonding || [[File:Ch6514_sf4_mo14.PNG|x300px]]
|-
| 26 (HOMO) || -0.34280 || Bonding || [[File:Ch6518_sf4_mo26.PNG|x300px]]
|-
| 27 (LUMO) || -0.07939 || Antibonding || [[File:Ch6518_sf4_mo27.PNG|x300px]]
|}

==Independence Task - other small molecules==

===Carbon Monoxide CO===
{| class="wikitable" border="1"
|+ Optimisation of of CO
! Calculation Method !! Basis Set !! Final Energy E / a.u. !! Point Group
|-
| RB3LYP || 631G(d,p) || -113.30945314 || C∞v
|}

<pre>
Item Value Threshold Converged?
Maximum Force 0.000007 0.000450 YES
RMS Force 0.000007 0.000300 YES
Maximum Displacement 0.000003 0.001800 YES
RMS Displacement 0.000004 0.001200 YES
</pre>

<jmol><jmolApplet>
<title>CO</title>
<color>black</color>
<size>400</size>
<uploadedFileContents>CH6518_CO_OPTF_POP.LOG</uploadedFileContents>
<script>frame 7</script>
</jmolApplet></jmol>

Optimised C-O bond length (Angstrom): <math>1.14 \pm 0.01</math>

Opt file [[Media:CH6518_CO_OPTF_POP.LOG| here]]

[[File:Ch6518_co_vibrations.PNG]]

{| class="wikitable" border="1"
|+ Vibrational Modes of CO
! Wavenumber / cm-1 !! Symmetry !! Intensity !! Bend or Stretch !! Image
|-
| 2209 || SG || 68.0 || Stretch || [[File:Ch6518_co.png|x150px]]
|}

Number of expected vibrational modes: <math>3N-5=3*2-5=1</math>

Charge on C: +0.506

Charge on O: -0.506

(Dipole moment: 0.0599 Db)

===Cyanide CN-===
{| class="wikitable" border="1"
|+ Optimisation of of CN-
! Calculation Method !! Basis Set !! Final Energy E / a.u. !! Point Group
|-
| RB3LYP || 631G(d,p) || -92.82453153 || C∞v
|}

<pre>
Item Value Threshold Converged?
Maximum Force 0.000012 0.000450 YES
RMS Force 0.000012 0.000300 YES
Maximum Displacement 0.000005 0.001800 YES
RMS Displacement 0.000008 0.001200 YES
</pre>

<jmol><jmolApplet>
<title>CN-</title>
<color>black</color>
<size>400</size>
<uploadedFileContents>CH6518_CN-_OPTF_POP.LOG</uploadedFileContents>
<script>frame 7</script>
</jmolApplet></jmol>

Optimised C-N bond length (Angstrom): <math>1.18 \pm 0.01</math>

Opt file [[Media:CH6518_CN-_OPTF_POP.LOG| here]]

[[File:Ch6518_cn_vibrations.PNG]]

{| class="wikitable" border="1"
|+ Vibrational Modes of CN-
! Wavenumber / cm-1 !! Symmetry !! Intensity !! Bend or Stretch !! Image
|-
| 2139 || SG || 7.8 || Stretch || [[File:Ch6518_cn.png|x150px]]
|}

Number of expected vibrational modes: <math>3N-5=3*2-5=1</math>

Charge on C: -0.246

Charge on N:-0.754

(Dipole moment: 0.5236 Db)

===Hydrogen Cyanide HCN===
{| class="wikitable" border="1"
|+ Optimisation of of HCN
! Calculation Method !! Basis Set !! Final Energy E / a.u. !! Point Group
|-
| RB3LYP || 631G(d,p) || -93.42458132 || C∞v
|}

<pre>
Item Value Threshold Converged?
Maximum Force 0.000370 0.000450 YES
RMS Force 0.000255 0.000300 YES
Maximum Displacement 0.000676 0.001800 YES
RMS Displacement 0.000427 0.001200 YES
</pre>

<jmol><jmolApplet>
<title>HCN</title>
<color>black</color>
<size>400</size>
<uploadedFileContents>CH6518_HCN_OPTF_POP.LOG</uploadedFileContents>
<script>frame 7</script>
</jmolApplet></jmol>

Optimised C-N bond length (Angstrom): <math>1.07 \pm 0.01</math>

Optimised H-C bond length (Angstrom): <math>1.16 \pm 0.01</math>

Opt file [[Media:CH6518_HCN_OPTF_POP.LOG| here]]

[[File:Ch6518_hcn_vibrations.PNG]]

{| class="wikitable" border="1"
|+ Vibrational Modes of HCN
! Wavenumber / cm-1 !! Symmetry !! Intensity !! Bend or Stretch !! Image
|-
| 767 || PI || 35.3 || rowspan="2" | Bend || [[File:Ch6518_hcn_767_1.png|x150px]]
|-
| 767 || PI || 35.3 || [[File:Ch6518_hcn_767_2.png|x150px]]
|-
| 2215 || SG || 2.0 || rowspan="2" | Stretch || [[File:Ch6518_hcn_2215.png|x150px]]
|-
| 3480 || SG || 57.3 || [[File:Ch6518_hcn_3480.png|x150px]]
|}

Number of expected vibrational modes: <math>3N-5=3*3-5=4</math>

Charge on H: +0.234

Charge on C: +0.073

Charge on N: -0.308

(Dipole moment: 2.8933 Db)