== EXERCISE 1: H + H2 system ==

=== Transition States ===
'''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 PE surface diagram defines the transition state as the saddle point where the gradient is 0 in orthogonal directions but is not in a minimum or maximum. ∂V('''ri''')/∂'''ri'''=0. The transition state is the point where the chemical structure is most unstable and has the highest free energy. The double derivative of the potential energy function can be used to distinguish the maximum to the minimum. If the value for the double derivative at the flat point is negative, it is the transition state.

'''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 best estimate of rts = 0.908Å. The transition state position is when rAB and rBC is identical. This value was estimated using trial and error, where there was no change in intermolecular distance for time. This can be seen in figure 2 with the plot of internuclear distance against time.

The plot displays two lines as the distance of AB and BC is equal and are overlapping.
<div><ul>
<li style="display: inline-block;">[[File:Mr21d21.PNG|frame|Figure 1: Contour Plot with Transition State Distance of 0.908Å in 0 Initial Momentum]]
<li style="display: inline-block;">[[File:Mrdi4125nt.PNG|frame|Figure 2: Plot of Internuclear Distances Against Time for a Distance of 0.908Å]]
</ul></div>

=== Trajectories ===
'''Comment on how the ''minimum energy path ''and the trajectory you just calculated differ.'''
<div><ul>
<li style="display: inline-block;">[[File:Mepcoutou412r.PNG|none|thumb|Figure 3: MEP - Contour Graph]]
<li style="display: inline-block;">[[File:Mepi241nt.PNG|none|thumb|Figure 4: MEP - Intermolecular Distance vs. Time]]
<li style="display: inline-block;">[[File:Notmep1.PNG|none|thumb|Figure 5: Dynamics - Contour Graph]]
<li style="display: inline-block;">[[File:Notmep2.PNG|none|thumb|Figure 6: Dynamics - Intermolecular Distances vs. Time]]
</ul></div>

MEP models the atoms in infinitely slow motion where there is no momentum. This is contrary to running the simulation in dynamics where momentum is present and there is vibrational motion. The lack of vibrational motion in MEP shows smooth lines in the graph of intermolecular distances vs. time, when the dynamics model shows a curly line depicting the vibrational motion of the atoms. The graph of intermolecular distances vs. time also shows a key difference in the two calculation types where the gradient of AB and AC is positive in the dynamics model and is slowly decreasing in the MEP model. This is due to the absence of momenta in the MEP calculation method, and thus is decelerating .

'''Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

{| class="wikitable"
|}
{| class="wikitable"
!p1
!p2
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.02</nowiki>
|Yes
|Atom C collides with molecule AB to form molecule BC with vibrational energy and atom A
|[[File:Mrd1asf.PNG|thumb]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.46</nowiki>
|No
|Atom C does not have sufficient momentum to complete the displacement
|[[File:Mr2412d2.PNG|thumb]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.96</nowiki>
|Yes
|Atom C collides with molecule AB to undergo displacement. BC has more vibrational energy than AB
|[[File:M211rd3.PNG|thumb]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.96</nowiki>
|No
|Atom C collides with molecule AB to temporarily displace A, however overwhelming vibrational energy
destroys the bond and recreates molecule AB
|[[File:Mrd613w4.PNG|thumb]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.42</nowiki>
|Yes
|Atom C collides with molecule AB to temporarily displace A, and overwhelming vibrational energy
destroys the bond and recreates molecule AB, however the large momentum carried towards molecule AB

allows atom C to displace the molecule again
|[[File:Mrd52154.PNG|thumb]]
|}
'''State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The main objective of the transition state theory is to calculate the rate constant and also be able to qualitatively view chemical reactions. This theory however has several assumptions<ref name="assu" />:
# Born Oppenheimer approximation where nuclei-electron interactions are treated separately
# Atoms as reactants and in the transition state have energies that are Boltzmann distributed
# Molecules past the transition state can not go back to forming reactants.
# Motion in the transition state can be treated classically
# Molecules only form the transition state if a collision occurs with enough energy

Experimental values will likely be smaller due to real world effects on the reaction. Real reactions are usually in equilibrium and time is taken to reach these equilibriums.

== 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?'''

[[File:D2412ei.PNG|centre|frame|Figure 7: Potential Energy Surface Where A = F, B = H, C = H]]

The potential energy surfaces (figure 7) show the H-F bond deeper in energy than the H-H bond. This indicates that the H-F bond is stronger and conversion of H-F bond to a H-H bond requires energy and the H-H to H-F reaction releases energy. Therefore:

''F+H2 '': This reaction is exothermic

H''+HF'': This reaction is endothermic

'''Locate the approximate position of the transition state.'''
[[File:F32124f.PNG|none|thumb|Figure 8: Plot of Intermolecular Distances vs. Time in Transition State of F-H-H]]

F-H: 1.810

H-H: 0.745

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

The Hammond's postulate approximates that the transition state resembles the reactants in exothermic reactions and the product in endothermic reactions.

Activation Energies

F + H2 -> HF + H

[[File:Fd4212s32.PNG]]

rHF = 1.7

rHH = 0.74

-103.78 to -133.6 = 29.862kcal/mol

[[HF + H -> H2 + F]]

[[File:1632241112432.PNG]]

rHF = 1.85

rHH = 0.74

0.0133kcal/mol

===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.'''

<nowiki>*</nowiki>add momentum vs time graph

With the completion of the displacement reaction, energy is released as the H-F bond is deeper in energy and thus exothermic. The consequences of this would be that energy may be released in the form of vibrational energy and heat. To confirm the previous statement, the experiment could be monitored by a change in temperature or an IR spectrum of the product could be taken to compare the newly formed vibrational frequencies.

'''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.'''

==References==

<references>

<ref name="assu"> J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River</ref>
</references>

File:1632241112432.PNG

2019-05-28T15:05:56Z

Mh4417:

MRD:anything4417

2019-05-28T15:04:27Z

Mh4417: /* EXERCISE 2: F - H - H system */

== EXERCISE 1: H + H2 system ==

=== Transition States ===
'''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 PE surface diagram defines the transition state as the saddle point where the gradient is 0 in orthogonal directions but is not in a minimum or maximum. ∂V('''ri''')/∂'''ri'''=0. The transition state is the point where the chemical structure is most unstable and has the highest free energy. The double derivative of the potential energy function can be used to distinguish the maximum to the minimum. If the value for the double derivative at the flat point is negative, it is the transition state.

'''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 best estimate of rts = 0.908Å. The transition state position is when rAB and rBC is identical. This value was estimated using trial and error, where there was no change in intermolecular distance for time. This can be seen in figure 2 with the plot of internuclear distance against time.

The plot displays two lines as the distance of AB and BC is equal and are overlapping.
<div><ul>
<li style="display: inline-block;">[[File:Mr21d21.PNG|frame|Figure 1: Contour Plot with Transition State Distance of 0.908Å in 0 Initial Momentum]]
<li style="display: inline-block;">[[File:Mrdi4125nt.PNG|frame|Figure 2: Plot of Internuclear Distances Against Time for a Distance of 0.908Å]]
</ul></div>

=== Trajectories ===
'''Comment on how the ''minimum energy path ''and the trajectory you just calculated differ.'''
<div><ul>
<li style="display: inline-block;">[[File:Mepcoutou412r.PNG|none|thumb|Figure 3: MEP - Contour Graph]]
<li style="display: inline-block;">[[File:Mepi241nt.PNG|none|thumb|Figure 4: MEP - Intermolecular Distance vs. Time]]
<li style="display: inline-block;">[[File:Notmep1.PNG|none|thumb|Figure 5: Dynamics - Contour Graph]]
<li style="display: inline-block;">[[File:Notmep2.PNG|none|thumb|Figure 6: Dynamics - Intermolecular Distances vs. Time]]
</ul></div>

MEP models the atoms in infinitely slow motion where there is no momentum. This is contrary to running the simulation in dynamics where momentum is present and there is vibrational motion. The lack of vibrational motion in MEP shows smooth lines in the graph of intermolecular distances vs. time, when the dynamics model shows a curly line depicting the vibrational motion of the atoms. The graph of intermolecular distances vs. time also shows a key difference in the two calculation types where the gradient of AB and AC is positive in the dynamics model and is slowly decreasing in the MEP model. This is due to the absence of momenta in the MEP calculation method, and thus is decelerating .

'''Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

{| class="wikitable"
|}
{| class="wikitable"
!p1
!p2
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.02</nowiki>
|Yes
|Atom C collides with molecule AB to form molecule BC with vibrational energy and atom A
|[[File:Mrd1asf.PNG|thumb]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.46</nowiki>
|No
|Atom C does not have sufficient momentum to complete the displacement
|[[File:Mr2412d2.PNG|thumb]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.96</nowiki>
|Yes
|Atom C collides with molecule AB to undergo displacement. BC has more vibrational energy than AB
|[[File:M211rd3.PNG|thumb]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.96</nowiki>
|No
|Atom C collides with molecule AB to temporarily displace A, however overwhelming vibrational energy
destroys the bond and recreates molecule AB
|[[File:Mrd613w4.PNG|thumb]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.42</nowiki>
|Yes
|Atom C collides with molecule AB to temporarily displace A, and overwhelming vibrational energy
destroys the bond and recreates molecule AB, however the large momentum carried towards molecule AB

allows atom C to displace the molecule again
|[[File:Mrd52154.PNG|thumb]]
|}
'''State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

The main objective of the transition state theory is to calculate the rate constant and also be able to qualitatively view chemical reactions. This theory however has several assumptions<ref name="assu" />:
# Born Oppenheimer approximation where nuclei-electron interactions are treated separately
# Atoms as reactants and in the transition state have energies that are Boltzmann distributed
# Molecules past the transition state can not go back to forming reactants.
# Motion in the transition state can be treated classically
# Molecules only form the transition state if a collision occurs with enough energy

Experimental values will likely be smaller due to real world effects on the reaction. Real reactions are usually in equilibrium and time is taken to reach these equilibriums.

== 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?'''

[[File:D2412ei.PNG|centre|frame|Figure 7: Potential Energy Surface Where A = F, B = H, C = H]]

The potential energy surfaces (figure 7) show the H-F bond deeper in energy than the H-H bond. This indicates that the H-F bond is stronger and conversion of H-F bond to a H-H bond requires energy and the H-H to H-F reaction releases energy. Therefore:

''F+H2 '': This reaction is exothermic

H''+HF'': This reaction is endothermic

'''Locate the approximate position of the transition state.'''
[[File:F32124f.PNG|none|thumb|Figure 8: Plot of Intermolecular Distances vs. Time in Transition State of F-H-H]]

F-H: 1.810

H-H: 0.745

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

The Hammond's postulate approximates that the transition state resembles the reactants in exothermic reactions and the product in endothermic reactions.

Activation Energies

F + H2 -> HF + H

[[File:Fd4212s32.PNG]]

rHF = 1.7

rHH = 0.74

-103.78 to -133.6 = 29.862kcal/mol

HF + H -> H2 + F

rHF = 1.85

rHH = 0.74

0.0133kcal/mol

===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.'''

<nowiki>*</nowiki>add momentum vs time graph

With the completion of the displacement reaction, energy is released as the H-F bond is deeper in energy and thus exothermic. The consequences of this would be that energy may be released in the form of vibrational energy and heat. To confirm the previous statement, the experiment could be monitored by a change in temperature or an IR spectrum of the product could be taken to compare the newly formed vibrational frequencies.

'''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.'''

==References==

<references>

<ref name="assu"> J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River</ref>
</references>

== 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 PE surface diagram defines the transition state as the maximum point in the minimum energy path where ∂V('''ri''')/∂'''ri'''=0, i.e. line is flat. The transition state is the point where the chemical structure is most unstable and has the highest free energy. The double derivative of the potential energy function can be used to distinguish the maximum to the minimum. If the value for the double derivative at the flat point is negative, it is the transition state.

'''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.'''

0.908Å. This value was estimated using trial and error, where there was no change in intermolecular distance. This can be seen in figure 2 with the plot of internuclear distance against time.
<div><ul>
<li style="display: inline-block;">[[File:Mrdi4125nt.PNG|frame|Figure 2: Plot of Internuclear Distances Against Time for a Distance of 0.908Å]]
</ul></div>

'''Comment on how the ''minimum energy path ''and the trajectory you just calculated differ.'''
[[File:Mepcoutou412r.PNG|none|thumb]]
[[File:Mepi241nt.PNG|none|thumb]]
[[File:Notmep1.PNG|none|thumb]]
[[File:Notmep2.PNG|none|thumb]]
'''Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''

{| class="wikitable"
|}
{| class="wikitable"
!p1
!p2
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.02</nowiki>
|Yes
|Atom C collides with molecule AB to form molecule BC with vibrational energy and atom A
|[[File:Mrd1asf.PNG|thumb]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.46</nowiki>
|No
|Atom C does not have sufficient momentum to complete the displacement
|[[File:Mr2412d2.PNG|thumb]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.96</nowiki>
|Yes
|Atom C collides with molecule AB to undergo displacement. BC has more vibrational energy than AB
|[[File:M211rd3.PNG|thumb]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.96</nowiki>
|No
|Atom C collides with molecule AB to temporarily displace A, however overwhelming vibrational energy
destroys the bond and recreates molecule AB
|[[File:Mrd613w4.PNG|thumb]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.42</nowiki>
|Yes
|Atom C collides with molecule AB to temporarily displace A, and overwhelming vibrational energy
destroys the bond and recreates molecule AB, however the large momentum carried towards molecule AB

allows atom C to displace the molecule again
|[[File:Mrd52154.PNG|thumb]]
|}
'''State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''

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

File:Notmep2.PNG

2019-05-24T14:11:58Z

Mh4417:

File:Notmep1.PNG

2019-05-24T14:11:27Z

Mh4417:

File:Mepi241nt.PNG

2019-05-24T14:10:24Z

Mh4417:

MRD:anything4417

2019-05-24T13:29:06Z

Mh4417:

File:M211rd3.PNG

2019-05-24T13:28:33Z

Mh4417:

File:Mr2412d2.PNG

2019-05-24T13:28:09Z

Mh4417: