ChemWiki - User contributions [en]

MRD:xc2017

2019-05-22T15:25:21Z

Xc2017: /* 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? */

== 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?''===

'''Answer:''' The transition state is a saddle point in the potential energy surface diagram where ∂V(ri)/∂ri=0 and G<0 where G = frr(r0,r'0)fr'r'(r0,r'0)−f2rr'(r0,r'0).

The transition point was defined as a maximum point in the minimum energy path linking reactants and products. For triatomic reaction, the transition state can be identified in reaction path in the contour plot where the AB and BC distance are equal.

The local minimum point can be distinguished by secondary partial derivation equal to zero and G > 0.

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

[[File:MRD_XC_1.png|thumb|Picture.1 A snapshot of 'Internuclear Distance vs Time' plot at r1 = r2 = 0.9075Å]]

'''Answer:''' The best estimate transition position is r1=r2=0.9075 Å where the “Internuclear Distances vs Time” plot shows two straight lines (Picture.1). The picture of distance vs time plot shows the AB and BC distance are constant which indicated the transition state.

=== '' Comment on how the mep and the trajectory you just calculated differ.''===
[[File:MRD_DYN_0521.png|thumb|Picture.2 A snapshot of dynamic surcafe plot]]
[[File:MRD_MEP_0521.png|thumb|Picture.3 A snapshot of MEP surface plot]]
'''Answer:'''

''' BC = r1 = rts+0.01'''

''' AB= '''r2 = rts

The initial condition was set to r1 = rts+0.01 and r2 = rts''' '''''',''' thus the AB distance ('''r2''') =0.9075 Å, and the BC distance ('''r1''') = 0.9175 Å
According to Picture 2 and 3, the reaction path calculated by MEP method was more smooth than the path calculated by dynamic method. For Picture.2 (using dynamic method), the reaction path vibrated at the lowest energy. For the MEP method, the reaction pathway was more smooth and more fit to the lowest energy barrier.
The difference between two ways of calculation may be because that MEP corresponds to infinitely slow motion while the dynamic method doesn't. For the dynamic calculation, the molecules move in a relative fast speed and vibrate. This makes the energy calculation harder. Thus, for small initial momenta, MEP calculation was more fit to the surface plot.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
!r1/Å
|0.9175
|7.15
|3.75
|-
!r2/Å
|0.9075
|0.75
|0.76
|-
!p1
|0
|2.46
|0
|-
!p2
|0
|1.33
|0
|}
For the table below, the initial distance was switched for r1 and r2,

BC distance =r1 = rts

AB distance = r2 = rts +0.01

According to the table below, it can be seen that the final value for both distance and momenta switched as the initial distance switched, and this change does not depend on their calculation method.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
!r1/ Å
|0.9075
|0.74
|0.76
|-
!r2/ Å
|0.9175
|7.15
|3.75
|-
!p1
|0
|1.33
|0
|-
!p2
|0
|2.46
|0
|}

=== 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
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reactants have enough kinetic energy to pass though the transition state and complete the reaction and it can be seen from the contour plot: the trajectory passes though the transition state and goes to the products. Thus, it is a productive reaction.[[File:MRD_R1_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.2</nowiki>
|No
|The reactants do not have enough energy to pass through the transition state, thus the trajectory moves back to the reactants.[[File:MRD_R2_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reagents have enough energy to pass through the transition state. Thus, the trajectory passes though the transition state and moves towards to the product. [[File:MRD_R3_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.6</nowiki>
|No
|The reagents have enough energy topass through the transition state, however, the energy is too high thus it drops back to the reactants.[[File:MRD_R4_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-82.6</nowiki>
|Yes
|The trajectory does not moves to product initially, the reagents have excess energy to pass through the transition state, thus it goes back to the reactants. After that, the reactants still have enough energy to pass through the transition state, thus the reaction occurs.[[File:MRD_R5_0521.png|thumb| Contour plot at this condition.]]
|}
In conclusion, to complete a productive reaction, the reactants should have enough energy to pass through the transition state but not too high to reverse the product back into the reactants.

===''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?''===
'''Assumptions:'''[https://www.sciencedirect.com/topics/chemistry/transition-state-theory]<ref name = 'TS'/>
# Quantum-tunnelling effects are assumed negligible.
# The Born-Oppenheimer approximation is invoked.
# The atoms in the reactant state have energies that are Boltzmann distributed.
# Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.

== 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?''===
{| class="wikitable"
!Reaction
!Initial condition
!Potential surface plot and discription
|-
!F + H2 (ABC)
|AB distance = 1.00 Å

BC distance = 0.74 Å

AB momentum = 0

BC momentum = -1.5
|[[File:MDR_FHH_0522.png|thumb|left|A surface plot for F+ H2 reaction. ]] The reaction is exothermic as the energy for reactants are higher than the energy for products. Thus, the energy is released during the reaction. This is caused by the formation of strong H-F bond.
|-
!H +HF (ABC)
|AB distance = 1.00 Å

BC distance = 0.917 Å

AB momentum = 0

BC momentum = -10
|[[File:MRD_HHF_0522.png|thumb|left|A surface plot for H + HF reaction. ]] This reaction is endothermic as the potential energy of reactants are lower than the potential energy of products. In this case, the strong H-F bond was break and a weaker H-H bond was formed. Thud, the overall reaction was endothermic.
|}

===''Locate the approximate position of the transition state.''===
{| class="wikitable"
!Reaction (ABC)
!Position of Transition Point
!Contour plot
|-
!F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_0522.png|thumb|Contour plot for for the transition point for F + H2.]]
|-
!H + HF
|AB distance = 0.745 Å
BC distance = 1.820 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_0522_1.png|thumb|Contour plot for for the transition point for H + HF.]]
|}

===''Report the activation energy for both reactions.''===
{| class="wikitable"
!Reaction (ABC)
!Initial condition
!Energy vs Time plot
!Activation energy
|-
!F + H2
|AB distance = 1.820 Å

BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_EA_0522.png|thumb|Energy vs Time plot for F + H2]]
|Ea = (-103.751) - (-104.002) = 0.251 kcal/mol
|-
!H + HF
|AB distance = 1.820 Å

BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_EA_1_0522.png|thumb|Energy vs Time plot for H + HF]]
|Ea = (-101.187) - (-133.852) = 32.665 kcal/mol
|}

===''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.''===
{| class="wikitable"
!Initial condition
!Graph
!Explanation
|-
!F + H2 (A+BC)

AB distance = 1.0 Å

BC distance = 0.74Å

AB momentum = -1.5

BC momentum = 0
|[[File:MRD_FHH_R1_0522.png |thumb|A contour plot for F + H2]]
[[File:MRD_FHH_E1_0522.png |thumb|An Energy vs Time plot for F + H2]]
|According to the graphs, when the kinetic energy increase, the potential energy decreases. Thus, the total energy in the system is conserved. It can also be proofed by the constant total energy at -105.721 kcal/mol.

As this reaction is exothermic, the increase in kinetic energy can be confirmed by increase in temperature.
|}

===''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.''===
{| class="wikitable"
!H + HF
(A +BC)
!Initial condition
!Contour plot
!Explanation
|-
!'''High translation energy'''
|rAB= 1.5Å

rBC= 0.917Å

pAB = -8

pBC = -1
|[[File:MRD_HHF_HTE_0522.png|thumb|Contour plot for high translation energy.]]
| rowspan="2" |Polanyi's empirical rules states that the vibration energy has a higher efficiency than translation energy for a late transition state reaction.

For the reaction H + HF, it is an endothermic reaction, thus it has a late transition state. According to the contour plot shown on the left, it can be seen that the vibration energy is dominant the reaction, which proofs the Polanyi's empirical rules.
|-
!'''High vibration energy'''
|rAB= 1.5Å

rBC= 0.917Å

pAB = -1

pBC =-10
|[[File:MRD_HHF_HVE_1_0522.png|thumb|Contour plot for high vibration energy.]]
|}

==Reference==
<references>
<ref name="TS">T.Bligaard, J.K.Nørskov, in Chemical Bonding at Surfaces and Interfaces, Elsevier, Amsterdam, 2008, pp 255-321.</ref>
</references>

MRD:xc2017

2019-05-22T15:23:03Z

Xc2017: /* 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. */

== 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?''===

'''Answer:''' The transition state is a saddle point in the potential energy surface diagram where ∂V(ri)/∂ri=0 and G<0 where G = frr(r0,r'0)fr'r'(r0,r'0)−f2rr'(r0,r'0).

The transition point was defined as a maximum point in the minimum energy path linking reactants and products. For triatomic reaction, the transition state can be identified in reaction path in the contour plot where the AB and BC distance are equal.

The local minimum point can be distinguished by secondary partial derivation equal to zero and G > 0.

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

[[File:MRD_XC_1.png|thumb|Picture.1 A snapshot of 'Internuclear Distance vs Time' plot at r1 = r2 = 0.9075Å]]

'''Answer:''' The best estimate transition position is r1=r2=0.9075 Å where the “Internuclear Distances vs Time” plot shows two straight lines (Picture.1). The picture of distance vs time plot shows the AB and BC distance are constant which indicated the transition state.

=== '' Comment on how the mep and the trajectory you just calculated differ.''===
[[File:MRD_DYN_0521.png|thumb|Picture.2 A snapshot of dynamic surcafe plot]]
[[File:MRD_MEP_0521.png|thumb|Picture.3 A snapshot of MEP surface plot]]
'''Answer:'''

''' BC = r1 = rts+0.01'''

''' AB= '''r2 = rts

The initial condition was set to r1 = rts+0.01 and r2 = rts''' '''''',''' thus the AB distance ('''r2''') =0.9075 Å, and the BC distance ('''r1''') = 0.9175 Å
According to Picture 2 and 3, the reaction path calculated by MEP method was more smooth than the path calculated by dynamic method. For Picture.2 (using dynamic method), the reaction path vibrated at the lowest energy. For the MEP method, the reaction pathway was more smooth and more fit to the lowest energy barrier.
The difference between two ways of calculation may be because that MEP corresponds to infinitely slow motion while the dynamic method doesn't. For the dynamic calculation, the molecules move in a relative fast speed and vibrate. This makes the energy calculation harder. Thus, for small initial momenta, MEP calculation was more fit to the surface plot.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
!r1/Å
|0.9175
|7.15
|3.75
|-
!r2/Å
|0.9075
|0.75
|0.76
|-
!p1
|0
|2.46
|0
|-
!p2
|0
|1.33
|0
|}
For the table below, the initial distance was switched for r1 and r2,

BC distance =r1 = rts

AB distance = r2 = rts +0.01

According to the table below, it can be seen that the final value for both distance and momenta switched as the initial distance switched, and this change does not depend on their calculation method.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
!r1/ Å
|0.9075
|0.74
|0.76
|-
!r2/ Å
|0.9175
|7.15
|3.75
|-
!p1
|0
|1.33
|0
|-
!p2
|0
|2.46
|0
|}

=== 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
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reactants have enough kinetic energy to pass though the transition state and complete the reaction and it can be seen from the contour plot: the trajectory passes though the transition state and goes to the products. Thus, it is a productive reaction.[[File:MRD_R1_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.2</nowiki>
|No
|The reactants do not have enough energy to pass through the transition state, thus the trajectory moves back to the reactants.[[File:MRD_R2_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reagents have enough energy to pass through the transition state. Thus, the trajectory passes though the transition state and moves towards to the product. [[File:MRD_R3_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.6</nowiki>
|No
|The reagents have enough energy topass through the transition state, however, the energy is too high thus it drops back to the reactants.[[File:MRD_R4_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-82.6</nowiki>
|Yes
|The trajectory does not moves to product initially, the reagents have excess energy to pass through the transition state, thus it goes back to the reactants. After that, the reactants still have enough energy to pass through the transition state, thus the reaction occurs.[[File:MRD_R5_0521.png|thumb| Contour plot at this condition.]]
|}
In conclusion, to complete a productive reaction, the reactants should have enough energy to pass through the transition state but not too high to reverse the product back into the reactants.

===''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?''===
'''Assumptions:'''[https://www.sciencedirect.com/topics/chemistry/transition-state-theory]<ref name = 'TS'/>
# Quantum-tunnelling effects are assumed negligible.
# The Born-Oppenheimer approximation is invoked.
# The atoms in the reactant state have energies that are Boltzmann distributed.
# Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.

== 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?''===
{| class="wikitable"
!Reaction
!Initial condition
!Potential surface plot and discription
|-
!F + H2 (ABC)
|AB distance = 1.00 Å

BC distance = 0.74 Å

AB momentum = 0

BC momentum = -1.5
|[[File:MDR_FHH_0522.png|thumb|left|A surface plot for F+ H2 reaction. ]] The reaction is exothermix as the energy for reactants are higher than the energy for products. Thus, the energy is released during the reaction. This is caused by the formation of strong H-F bond.
|-
!H +HF (ABC)
|AB distance = 1.00 Å

BC distance = 0.917 Å

AB momentum = 0

BC momentum = -10
|[[File:MRD_HHF_0522.png|thumb|left|A surface plot for H + HF reaction. ]] This reaction is endothermic as the potential energy of reactants are lower than the potential energy of products. In this case, the strong H-F bond was break and a weaker H-H bond was formed. Thud, the overall reaction was endothermic.
|}

===''Locate the approximate position of the transition state.''===
{| class="wikitable"
!Reaction (ABC)
!Position of Transition Point
!Contour plot
|-
!F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_0522.png|thumb|Contour plot for for the transition point for F + H2.]]
|-
!H + HF
|AB distance = 0.745 Å
BC distance = 1.820 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_0522_1.png|thumb|Contour plot for for the transition point for H + HF.]]
|}

===''Report the activation energy for both reactions.''===
{| class="wikitable"
!Reaction (ABC)
!Initial condition
!Energy vs Time plot
!Activation energy
|-
!F + H2
|AB distance = 1.820 Å

BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_EA_0522.png|thumb|Energy vs Time plot for F + H2]]
|Ea = (-103.751) - (-104.002) = 0.251 kcal/mol
|-
!H + HF
|AB distance = 1.820 Å

BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_EA_1_0522.png|thumb|Energy vs Time plot for H + HF]]
|Ea = (-101.187) - (-133.852) = 32.665 kcal/mol
|}

===''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.''===
{| class="wikitable"
!Initial condition
!Graph
!Explanation
|-
!F + H2 (A+BC)

AB distance = 1.0 Å

BC distance = 0.74Å

AB momentum = -1.5

BC momentum = 0
|[[File:MRD_FHH_R1_0522.png |thumb|A contour plot for F + H2]]
[[File:MRD_FHH_E1_0522.png |thumb|An Energy vs Time plot for F + H2]]
|According to the graphs, when the kinetic energy increase, the potential energy decreases. Thus, the total energy in the system is conserved. It can also be proofed by the constant total energy at -105.721 kcal/mol.

As this reaction is exothermic, the increase in kinetic energy can be confirmed by increase in temperature.
|}

===''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.''===
{| class="wikitable"
!H + HF
(A +BC)
!Initial condition
!Contour plot
!Explanation
|-
!'''High translation energy'''
|rAB= 1.5Å

rBC= 0.917Å

pAB = -8

pBC = -1
|[[File:MRD_HHF_HTE_0522.png|thumb|Contour plot for high translation energy.]]
| rowspan="2" |Polanyi's empirical rules states that the vibration energy has a higher efficiency than translation energy for a late transition state reaction.

For the reaction H + HF, it is an endothermic reaction, thus it has a late transition state. According to the contour plot shown on the left, it can be seen that the vibration energy is dominant the reaction, which proofs the Polanyi's empirical rules.
|-
!'''High vibration energy'''
|rAB= 1.5Å

rBC= 0.917Å

pAB = -1

pBC =-10
|[[File:MRD_HHF_HVE_1_0522.png|thumb|Contour plot for high vibration energy.]]
|}

==Reference==
<references>
<ref name="TS">T.Bligaard, J.K.Nørskov, in Chemical Bonding at Surfaces and Interfaces, Elsevier, Amsterdam, 2008, pp 255-321.</ref>
</references>

MRD:xc2017

2019-05-22T15:17:35Z

== 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?''===

'''Answer:''' The transition state is a saddle point in the potential energy surface diagram where ∂V(ri)/∂ri=0 and G<0 where G = frr(r0,r'0)fr'r'(r0,r'0)−f2rr'(r0,r'0).

The transition point was defined as a maximum point in the minimum energy path linking reactants and products. For triatomic reaction, the transition state can be identified in reaction path in the contour plot where the AB and BC distance are equal.

The local minimum point can be distinguished by secondary partial derivation equal to zero and G > 0.

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

[[File:MRD_XC_1.png|thumb|Picture.1 A snapshot of 'Internuclear Distance vs Time' plot at r1 = r2 = 0.9075Å]]

'''Answer:''' The best estimate transition position is r1=r2=0.9075 Å where the “Internuclear Distances vs Time” plot shows two straight lines (Picture.1). The picture of distance vs time plot shows the AB and BC distance are constant which indicated the transition state.

=== '' Comment on how the mep and the trajectory you just calculated differ.''===
[[File:MRD_DYN_0521.png|thumb|Picture.2 A snapshot of dynamic surcafe plot]]
[[File:MRD_MEP_0521.png|thumb|Picture.3 A snapshot of MEP surface plot]]
'''Answer:'''

''' BC = r1 = rts+0.01'''

''' AB= '''r2 = rts

The initial condition was set to r1 = rts+0.01 and r2 = rts''' '''''',''' thus the AB distance ('''r2''') =0.9075 Å, and the BC distance ('''r1''') = 0.9175 Å
According to Picture 2 and 3, the reaction path calculated by MEP method was more smooth than the path calculated by dynamic method. For Picture.2 (using dynamic method), the reaction path vibrated at the lowest energy. For the MEP method, the reaction pathway was more smooth and more fit to the lowest energy barrier.
The difference between two ways of calculation may be because that MEP corresponds to infinitely slow motion while the dynamic method doesn't. For the dynamic calculation, the molecules move in a relative fast speed and vibrate. This makes the energy calculation harder. Thus, for small initial momenta, MEP calculation was more fit to the surface plot.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/Å
|0.9175
|7.15
|3.75
|-
|r2/Å
|0.9075
|0.75
|0.76
|-
|p1
|0
|2.46
|0
|-
|p2
|0
|1.33
|0
|}
For the table below, the initial distance was switched for r1 and r2,

BC distance =r1 = rts

AB distance = r2 = rts +0.01

According to the table below, it can be seen that the final value for both distance and momenta switched as the initial distance switched, and this change does not depend on their calculation method.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/ Å
|0.9075
|0.74
|0.76
|-
|r2/ Å
|0.9175
|7.15
|3.75
|-
|p1
|0
|1.33
|0
|-
|p2
|0
|2.46
|0
|}

=== 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
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reactants have enough kinetic energy to pass though the transition state and complete the reaction and it can be seen from the contour plot: the trajectory passes though the transition state and goes to the products. Thus, it is a productive reaction.[[File:MRD_R1_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.2</nowiki>
|No
|The reactants do not have enough energy to pass through the transition state, thus the trajectory moves back to the reactants.[[File:MRD_R2_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reagents have enough energy to pass through the transition state. Thus, the trajectory passes though the transition state and moves towards to the product. [[File:MRD_R3_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.6</nowiki>
|No
|The reagents have enough energy topass through the transition state, however, the energy is too high thus it drops back to the reactants.[[File:MRD_R4_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-82.6</nowiki>
|Yes
|The trajectory does not moves to product initially, the reagents have excess energy to pass through the transition state, thus it goes back to the reactants. After that, the reactants still have enough energy to pass through the transition state, thus the reaction occurs.[[File:MRD_R5_0521.png|thumb| Contour plot at this condition.]]
|}
In conclusion, to complete a productive reaction, the reactants should have enough energy to pass through the transition state but not too high to reverse the product back into the reactants.

===''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?''===
'''Assumptions:'''[https://www.sciencedirect.com/topics/chemistry/transition-state-theory]<ref name = 'TS'/>
# Quantum-tunnelling effects are assumed negligible.
# The Born-Oppenheimer approximation is invoked.
# The atoms in the reactant state have energies that are Boltzmann distributed.
# Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.

== 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?''===
{| class="wikitable"
!Reaction
!Initial condition
!Potential surface plot and discription
|-
|F + H2 (ABC)
|AB distance = 1.00 Å
BC distance = 0.74 Å

AB momentum = 0
BC momentum = -1.5
|[[File:MDR_FHH_0522.png|thumb|left|A surface plot for F+ H2 reaction. ]] The reaction is exothermix as the energy for reactants are higher than the energy for products. Thus, the energy is released during the reaction. This is caused by the formation of strong H-F bond.
|-
|H +HF (ABC)
|AB distance = 1.00 Å
BC distance = 0.917 Å

AB momentum = 0
BC momentum = -10
|[[File:MRD_HHF_0522.png|thumb|left|A surface plot for H + HF reaction. ]] This reaction is endothermic as the potential energy of reactants are lower than the potential energy of products. In this case, the strong H-F bond was break and a weaker H-H bond was formed. Thud, the overall reaction was endothermic.
|}

===''Locate the approximate position of the transition state.''===
{| class="wikitable"
!Reaction (ABC)
!Position of Transition Point
!Contour plot
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_0522.png|thumb|Contour plot for for the transition point for F + H2.]]
|-
|H + HF
|AB distance = 0.745 Å
BC distance = 1.820 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_0522_1.png|thumb|Contour plot for for the transition point for H + HF.]]
|}

===''Report the activation energy for both reactions.''===
{| class="wikitable"
!Reaction (ABC)
!Initial condition
!Energy vs Time plot
!Activation energy
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_EA_0522.png|thumb|Energy vs Time plot for F + H2]]
|Ea = (-103.751) - (-104.002) = 0.251 kcal/mol
|-
|H + HF
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_EA_1_0522.png|thumb|Energy vs Time plot for H + HF]]
|Ea = (-101.187) - (-133.852) = 32.665 kcal/mol
|}

===''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.''===
{| class="wikitable"
!Initial condition
!Graph
!Explanation
|-
|F + H2 (A+BC)
AB distance = 1.0 Å

BC distance = 0.74Å

AB momentum = -1.5

BC momentum = 0
|[[File:MRD_FHH_R1_0522.png |thumb|A contour plot for F + H2]]
[[File:MRD_FHH_E1_0522.png |thumb|An Energy vs Time plot for F + H2]]
|According to the graphs, when the kinetic energy increase, the potential energy decreases. Thus, the total energy in the system is conserved. It can also be proofed by the constant total energy at -105.721 kcal/mol.

As this reaction is exothermic, the increase in kinetic energy can be confirmed by increase in temperature.
|}

===''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.''===
{| class="wikitable"
!H + HF
(A +BC)
!Initial condition
!Contour plot
!Explanation
|-
!'''High translation energy'''
|AB distance = 1.5 Å
BC distance = 0.917Å

AB momentum = -8

BC momentum = -1
|[[File:MRD_HHF_HTE_0522.png|thumb|Contour plot for high translation energy.]]
| rowspan="2" |Polanyi's empirical rules states that the vibration energy has a higher efficiency than translation energy for a late transition state reaction.

For the reaction H + HF, it is an endothermic reaction, thus it has a late transition state. According to the contour plot shown on the left, it can be seen that the vibration energy is dominant the reaction, which proofs the Polanyi's empirical rules.
|-
!'''High vibration energy'''
|AB distance = 1.5 Å
BC distance = 0.9.7 Å

AB momentum = -1

BC momentum = -10
|[[File:MRD_HHF_HVE_1_0522.png|thumb|Contour plot for high vibration energy.]]
|}

==Reference==
<references>
<ref name="TS">T.Bligaard, J.K.Nørskov, in Chemical Bonding at Surfaces and Interfaces, Elsevier, Amsterdam, 2008, pp 255-321.</ref>
</references>

MRD:xc2017

2019-05-22T15:09:42Z

== 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?''===

'''Answer:''' The transition state is a saddle point in the potential energy surface diagram where ∂V(ri)/∂ri=0 and G<0 where G = frr(r0,r'0)fr'r'(r0,r'0)−f2rr'(r0,r'0).

The transition point was defined as a maximum point in the minimum energy path linking reactants and products. For triatomic reaction, the transition state can be identified in reaction path in the contour plot where the AB and BC distance are equal.

The local minimum point can be distinguished by secondary partial derivation equal to zero and G > 0.

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

[[File:MRD_XC_1.png|thumb|Picture.1 A snapshot of 'Internuclear Distance vs Time' plot at r1 = r2 = 0.9075Å]]

'''Answer:''' The best estimate transition position is r1=r2=0.9075 Å where the “Internuclear Distances vs Time” plot shows two straight lines (Picture.1). The picture of distance vs time plot shows the AB and BC distance are constant which indicated the transition state.

=== '' Comment on how the mep and the trajectory you just calculated differ.''===
[[File:MRD_DYN_0521.png|thumb|Picture.2 A snapshot of dynamic surcafe plot]]
[[File:MRD_MEP_0521.png|thumb|Picture.3 A snapshot of MEP surface plot]]
'''Answer:'''

''' BC = r1 = rts+0.01'''

''' AB= '''r2 = rts

The initial condition was set to r1 = rts+0.01 and r2 = rts''' '''''',''' thus the AB distance ('''r2''') =0.9075 Å, and the BC distance ('''r1''') = 0.9175 Å
According to Picture 2 and 3, the reaction path calculated by MEP method was more smooth than the path calculated by dynamic method. For Picture.2 (using dynamic method), the reaction path vibrated at the lowest energy. For the MEP method, the reaction pathway was more smooth and more fit to the lowest energy barrier.
The difference between two ways of calculation may be because that MEP corresponds to infinitely slow motion while the dynamic method doesn't. For the dynamic calculation, the molecules move in a relative fast speed and vibrate. This makes the energy calculation harder. Thus, for small initial momenta, MEP calculation was more fit to the surface plot.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/Å
|0.9175
|7.15
|3.75
|-
|r2/Å
|0.9075
|0.75
|0.76
|-
|p1
|0
|2.46
|0
|-
|p2
|0
|1.33
|0
|}
For the table below, the initial distance was switched for r1 and r2,

BC distance =r1 = rts

AB distance = r2 = rts +0.01

According to the table below, it can be seen that the final value for both distance and momenta switched as the initial distance switched, and this change does not depend on their calculation method.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/ Å
|0.9075
|0.74
|0.76
|-
|r2/ Å
|0.9175
|7.15
|3.75
|-
|p1
|0
|1.33
|0
|-
|p2
|0
|2.46
|0
|}

=== 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
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reactants have enough kinetic energy to pass though the transition state and complete the reaction and it can be seen from the contour plot: the trajectory passes though the transition state and goes to the products. Thus, it is a productive reaction.[[File:MRD_R1_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.2</nowiki>
|No
|The reactants do not have enough energy to pass through the transition state, thus the trajectory moves back to the reactants.[[File:MRD_R2_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reagents have enough energy to pass through the transition state. Thus, the trajectory passes though the transition state and moves towards to the product. [[File:MRD_R3_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.6</nowiki>
|No
|The reagents have enough energy topass through the transition state, however, the energy is too high thus it drops back to the reactants.[[File:MRD_R4_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-82.6</nowiki>
|Yes
|The trajectory does not moves to product initially, the reagents have excess energy to pass through the transition state, thus it goes back to the reactants. After that, the reactants still have enough energy to pass through the transition state, thus the reaction occurs.[[File:MRD_R5_0521.png|thumb| Contour plot at this condition.]]
|}
In conclusion, to complete a productive reaction, the reactants should have enough energy to pass through the transition state but not too high to reverse the product back into the reactants.

===''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?''===
'''Assumptions:'''[https://www.sciencedirect.com/topics/chemistry/transition-state-theory]<ref name = 'TS'/>
# Quantum-tunnelling effects are assumed negligible.
# The Born-Oppenheimer approximation is invoked.
# The atoms in the reactant state have energies that are Boltzmann distributed.
# Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.

== 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?''===
{| class="wikitable"
!Reaction
!Initial condition
!Potential surface plot and discription
|-
|F + H2 (ABC)
|AB distance = 1.00 Å
BC distance = 0.74 Å

AB momentum = 0
BC momentum = -1.5
|[[File:MDR_FHH_0522.png|thumb|left|A surface plot for F+ H2 reaction. ]] The reaction is exothermix as the energy for reactants are higher than the energy for products. Thus, the energy is released during the reaction. This is caused by the formation of strong H-F bond.
|-
|H +HF (ABC)
|AB distance = 1.00 Å
BC distance = 0.917 Å

AB momentum = 0
BC momentum = -10
|[[File:MRD_HHF_0522.png|thumb|left|A surface plot for H + HF reaction. ]] This reaction is endothermic as the potential energy of reactants are lower than the potential energy of products. In this case, the strong H-F bond was break and a weaker H-H bond was formed. Thud, the overall reaction was endothermic.
|}

===''Locate the approximate position of the transition state.''===
{| class="wikitable"
!Reaction (ABC)
!Position of Transition Point
!Contour plot
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_0522.png|thumb|Contour plot for for the transition point for F + H2.]]
|-
|H + HF
|AB distance = 0.745 Å
BC distance = 1.820 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_0522_1.png|thumb|Contour plot for for the transition point for H + HF.]]
|}

===''Report the activation energy for both reactions.''===
{| class="wikitable"
!Reaction (ABC)
!Initial condition
!Energy vs Time plot
!Activation energy
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_EA_0522.png|thumb|Energy vs Time plot for F + H2]]
|Ea = (-103.751) - (-104.002) = 0.251 kcal/mol
|-
|H + HF
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_EA_1_0522.png|thumb|Energy vs Time plot for H + HF]]
|Ea = (-101.187) - (-133.852) = 32.665 kcal/mol
|}

===''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.''===
{| class="wikitable"
!Initial condition
!Graph
!Explanation
|-
|F + H2 (A+BC)
AB distance = 1.0 Å

BC distance = 0.74Å

AB momentum = -1.5

BC momentum = 0
|[[File:MRD_FHH_R1_0522.png |thumb|A contour plot for F + H2]]
[[File:MRD_FHH_E1_0522.png |thumb|An Energy vs Time plot for F + H2]]
|According to the graphs, when the kinetic energy increase, the potential energy decreases. Thus, the total energy in the system is conserved. It can also be proofed by the constant total energy at -105.721 kcal/mol.

As this reaction is exothermic, the increase in kinetic energy can be confirmed by increase in temperature.
|}

===''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.''===
{| class="wikitable"
!H + HF
(A +BC)
!Initial condition
!Contour plot
!Explanation
|-
|High translation energy
|AB distance = 1.5 Å
BC distance = 0.917Å

AB momentum = -8

BC momentum = -1
|[[File:MRD_HHF_HTE_0522.png|thumb|Contour plot for high translation energy.]]
| rowspan="2" |
|-
|High vibration energy
|AB distance = 1.5 Å
BC distance = 0.9.7 Å

AB momentum = -1

BC momentum = -10
|[[File:MRD_HHF_HVE_1_0522.png|thumb|Contour plot for high vibration energy.]]
|}

==Reference==
<references>
<ref name="TS">T.Bligaard, J.K.Nørskov, in Chemical Bonding at Surfaces and Interfaces, Elsevier, Amsterdam, 2008, pp 255-321.</ref>
</references>

MRD:xc2017

2019-05-22T15:09:05Z

== 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?''===

'''Answer:''' The transition state is a saddle point in the potential energy surface diagram where ∂V(ri)/∂ri=0 and G<0 where G = frr(r0,r'0)fr'r'(r0,r'0)−f2rr'(r0,r'0).

The transition point was defined as a maximum point in the minimum energy path linking reactants and products. For triatomic reaction, the transition state can be identified in reaction path in the contour plot where the AB and BC distance are equal.

The local minimum point can be distinguished by secondary partial derivation equal to zero and G > 0.

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

[[File:MRD_XC_1.png|thumb|Picture.1 A snapshot of 'Internuclear Distance vs Time' plot at r1 = r2 = 0.9075Å]]

'''Answer:''' The best estimate transition position is r1=r2=0.9075 Å where the “Internuclear Distances vs Time” plot shows two straight lines (Picture.1). The picture of distance vs time plot shows the AB and BC distance are constant which indicated the transition state.

=== '' Comment on how the mep and the trajectory you just calculated differ.''===
[[File:MRD_DYN_0521.png|thumb|Picture.2 A snapshot of dynamic surcafe plot]]
[[File:MRD_MEP_0521.png|thumb|Picture.3 A snapshot of MEP surface plot]]
'''Answer:'''

''' BC = r1 = rts+0.01'''

''' AB= '''r2 = rts

The initial condition was set to r1 = rts+0.01 and r2 = rts''' '''''',''' thus the AB distance ('''r2''') =0.9075 Å, and the BC distance ('''r1''') = 0.9175 Å
According to Picture 2 and 3, the reaction path calculated by MEP method was more smooth than the path calculated by dynamic method. For Picture.2 (using dynamic method), the reaction path vibrated at the lowest energy. For the MEP method, the reaction pathway was more smooth and more fit to the lowest energy barrier.
The difference between two ways of calculation may be because that MEP corresponds to infinitely slow motion while the dynamic method doesn't. For the dynamic calculation, the molecules move in a relative fast speed and vibrate. This makes the energy calculation harder. Thus, for small initial momenta, MEP calculation was more fit to the surface plot.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/Å
|0.9175
|7.15
|3.75
|-
|r2/Å
|0.9075
|0.75
|0.76
|-
|p1
|0
|2.46
|0
|-
|p2
|0
|1.33
|0
|}
For the table below, the initial distance was switched for r1 and r2,

BC distance =r1 = rts

AB distance = r2 = rts +0.01

According to the table below, it can be seen that the final value for both distance and momenta switched as the initial distance switched, and this change does not depend on their calculation method.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/ Å
|0.9075
|0.74
|0.76
|-
|r2/ Å
|0.9175
|7.15
|3.75
|-
|p1
|0
|1.33
|0
|-
|p2
|0
|2.46
|0
|}

=== 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
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reactants have enough kinetic energy to pass though the transition state and complete the reaction and it can be seen from the contour plot: the trajectory passes though the transition state and goes to the products. Thus, it is a productive reaction.[[File:MRD_R1_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.2</nowiki>
|No
|The reactants do not have enough energy to pass through the transition state, thus the trajectory moves back to the reactants.[[File:MRD_R2_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reagents have enough energy to pass through the transition state. Thus, the trajectory passes though the transition state and moves towards to the product. [[File:MRD_R3_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.6</nowiki>
|No
|The reagents have enough energy topass through the transition state, however, the energy is too high thus it drops back to the reactants.[[File:MRD_R4_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-82.6</nowiki>
|Yes
|The trajectory does not moves to product initially, the reagents have excess energy to pass through the transition state, thus it goes back to the reactants. After that, the reactants still have enough energy to pass through the transition state, thus the reaction occurs.[[File:MRD_R5_0521.png|thumb| Contour plot at this condition.]]
|}
In conclusion, to complete a productive reaction, the reactants should have enough energy to pass through the transition state but not too high to reverse the product back into the reactants.

===''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?''===
'''Assumptions:'''[https://www.sciencedirect.com/topics/chemistry/transition-state-theory]<ref name = 'TS'/>
# Quantum-tunnelling effects are assumed negligible.
# The Born-Oppenheimer approximation is invoked.
# The atoms in the reactant state have energies that are Boltzmann distributed.
# Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.

== 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?''===
{| class="wikitable"
!Reaction
!Initial condition
!Potential surface plot and discription
|-
|F + H2 (ABC)
|AB distance = 1.00 Å
BC distance = 0.74 Å

AB momentum = 0
BC momentum = -1.5
|[[File:MDR_FHH_0522.png|thumb|left|A surface plot for F+ H2 reaction. ]] The reaction is exothermix as the energy for reactants are higher than the energy for products. Thus, the energy is released during the reaction. This is caused by the formation of strong H-F bond.
|-
|H +HF (ABC)
|AB distance = 1.00 Å
BC distance = 0.917 Å

AB momentum = 0
BC momentum = -10
|[[File:MRD_HHF_0522.png|thumb|left|A surface plot for H + HF reaction. ]] This reaction is endothermic as the potential energy of reactants are lower than the potential energy of products. In this case, the strong H-F bond was break and a weaker H-H bond was formed. Thud, the overall reaction was endothermic.
|}

===''Locate the approximate position of the transition state.''===
{| class="wikitable"
!Reaction (ABC)
!Position of Transition Point
!Contour plot
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_0522.png|thumb|Contour plot for for the transition point for F + H2.]]
|-
|H + HF
|AB distance = 0.745 Å
BC distance = 1.820 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_0522_1.png|thumb|Contour plot for for the transition point for H + HF.]]
|}

===''Report the activation energy for both reactions.''===
{| class="wikitable"
!Reaction (ABC)
!Initial condition
!Energy vs Time plot
!Activation energy
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_EA_0522.png|thumb|Energy vs Time plot for F + H2]]
|Ea = (-103.751) - (-104.002) = 0.251 kcal/mol
|-
|H + HF
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_EA_1_0522.png|thumb|Energy vs Time plot for H + HF]]
|Ea = (-101.187) - (-133.852) = 32.665 kcal/mol
|}

===''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.''===
{| class="wikitable"
!Initial condition
!Graph
!Explanation
|-
|F + H2 (A+BC)
AB distance = 1.0 Å

BC distance = 0.74Å

AB momentum = -1.5

BC momentum = 0
|[[File:MRD_FHH_R1_0522.png |thumb|A contour plot for F + H2]]
[[File:MRD_FHH_E1_0522.png |thumb|An Energy vs Time plot for F + H2]]
|According to the graphs, when the kinetic energy increase, the potential energy decreases. Thus, the total energy in the system is conserved. It can also be proofed by the constant total energy at -105.721 kcal/mol.

As this reaction is exothermic, the increase in kinetic energy can be confirmed by increase in temperature.
|}

===''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.''===
{| class="wikitable"
!H + HF (A +BC)
!Initial condition
!Contour plot
!Explanation
|-
|High translation energy
|AB distance = 1.5 Å
BC distance = 0.917Å

AB momentum = -8

BC momentum = -1
|[[File:MRD_HHF_HTE_0522.png|thumb|Contour plot for high translation energy.]]
| rowspan="2" |
|-
|High vibration energy
|AB distance = 1.5 Å
BC distance = 0.9.7 Å

AB momentum = -1

BC momentum = -10
|[[File:MRD_HHF_HVE_1_0522.png|thumb|Contour plot for high vibration energy.]]
|}

==Reference==
<references>
<ref name="TS">T.Bligaard, J.K.Nørskov, in Chemical Bonding at Surfaces and Interfaces, Elsevier, Amsterdam, 2008, pp 255-321.</ref>
</references>

File:MRD HHF HVE 1 0522.png

2019-05-22T15:08:39Z

Xc2017:

MRD:xc2017

2019-05-22T15:08:09Z

== 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?''===

'''Answer:''' The transition state is a saddle point in the potential energy surface diagram where ∂V(ri)/∂ri=0 and G<0 where G = frr(r0,r'0)fr'r'(r0,r'0)−f2rr'(r0,r'0).

The transition point was defined as a maximum point in the minimum energy path linking reactants and products. For triatomic reaction, the transition state can be identified in reaction path in the contour plot where the AB and BC distance are equal.

The local minimum point can be distinguished by secondary partial derivation equal to zero and G > 0.

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

[[File:MRD_XC_1.png|thumb|Picture.1 A snapshot of 'Internuclear Distance vs Time' plot at r1 = r2 = 0.9075Å]]

'''Answer:''' The best estimate transition position is r1=r2=0.9075 Å where the “Internuclear Distances vs Time” plot shows two straight lines (Picture.1). The picture of distance vs time plot shows the AB and BC distance are constant which indicated the transition state.

=== '' Comment on how the mep and the trajectory you just calculated differ.''===
[[File:MRD_DYN_0521.png|thumb|Picture.2 A snapshot of dynamic surcafe plot]]
[[File:MRD_MEP_0521.png|thumb|Picture.3 A snapshot of MEP surface plot]]
'''Answer:'''

''' BC = r1 = rts+0.01'''

''' AB= '''r2 = rts

The initial condition was set to r1 = rts+0.01 and r2 = rts''' '''''',''' thus the AB distance ('''r2''') =0.9075 Å, and the BC distance ('''r1''') = 0.9175 Å
According to Picture 2 and 3, the reaction path calculated by MEP method was more smooth than the path calculated by dynamic method. For Picture.2 (using dynamic method), the reaction path vibrated at the lowest energy. For the MEP method, the reaction pathway was more smooth and more fit to the lowest energy barrier.
The difference between two ways of calculation may be because that MEP corresponds to infinitely slow motion while the dynamic method doesn't. For the dynamic calculation, the molecules move in a relative fast speed and vibrate. This makes the energy calculation harder. Thus, for small initial momenta, MEP calculation was more fit to the surface plot.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/Å
|0.9175
|7.15
|3.75
|-
|r2/Å
|0.9075
|0.75
|0.76
|-
|p1
|0
|2.46
|0
|-
|p2
|0
|1.33
|0
|}
For the table below, the initial distance was switched for r1 and r2,

BC distance =r1 = rts

AB distance = r2 = rts +0.01

According to the table below, it can be seen that the final value for both distance and momenta switched as the initial distance switched, and this change does not depend on their calculation method.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/ Å
|0.9075
|0.74
|0.76
|-
|r2/ Å
|0.9175
|7.15
|3.75
|-
|p1
|0
|1.33
|0
|-
|p2
|0
|2.46
|0
|}

=== 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
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reactants have enough kinetic energy to pass though the transition state and complete the reaction and it can be seen from the contour plot: the trajectory passes though the transition state and goes to the products. Thus, it is a productive reaction.[[File:MRD_R1_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.2</nowiki>
|No
|The reactants do not have enough energy to pass through the transition state, thus the trajectory moves back to the reactants.[[File:MRD_R2_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reagents have enough energy to pass through the transition state. Thus, the trajectory passes though the transition state and moves towards to the product. [[File:MRD_R3_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.6</nowiki>
|No
|The reagents have enough energy topass through the transition state, however, the energy is too high thus it drops back to the reactants.[[File:MRD_R4_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-82.6</nowiki>
|Yes
|The trajectory does not moves to product initially, the reagents have excess energy to pass through the transition state, thus it goes back to the reactants. After that, the reactants still have enough energy to pass through the transition state, thus the reaction occurs.[[File:MRD_R5_0521.png|thumb| Contour plot at this condition.]]
|}
In conclusion, to complete a productive reaction, the reactants should have enough energy to pass through the transition state but not too high to reverse the product back into the reactants.

===''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?''===
'''Assumptions:'''[https://www.sciencedirect.com/topics/chemistry/transition-state-theory]<ref name = 'TS'/>
# Quantum-tunnelling effects are assumed negligible.
# The Born-Oppenheimer approximation is invoked.
# The atoms in the reactant state have energies that are Boltzmann distributed.
# Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.

== 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?''===
{| class="wikitable"
!Reaction
!Initial condition
!Potential surface plot and discription
|-
|F + H2 (ABC)
|AB distance = 1.00 Å
BC distance = 0.74 Å

AB momentum = 0
BC momentum = -1.5
|[[File:MDR_FHH_0522.png|thumb|left|A surface plot for F+ H2 reaction. ]] The reaction is exothermix as the energy for reactants are higher than the energy for products. Thus, the energy is released during the reaction. This is caused by the formation of strong H-F bond.
|-
|H +HF (ABC)
|AB distance = 1.00 Å
BC distance = 0.917 Å

AB momentum = 0
BC momentum = -10
|[[File:MRD_HHF_0522.png|thumb|left|A surface plot for H + HF reaction. ]] This reaction is endothermic as the potential energy of reactants are lower than the potential energy of products. In this case, the strong H-F bond was break and a weaker H-H bond was formed. Thud, the overall reaction was endothermic.
|}

===''Locate the approximate position of the transition state.''===
{| class="wikitable"
!Reaction (ABC)
!Position of Transition Point
!Contour plot
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_0522.png|thumb|Contour plot for for the transition point for F + H2.]]
|-
|H + HF
|AB distance = 0.745 Å
BC distance = 1.820 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_0522_1.png|thumb|Contour plot for for the transition point for H + HF.]]
|}

===''Report the activation energy for both reactions.''===
{| class="wikitable"
!Reaction (ABC)
!Initial condition
!Energy vs Time plot
!Activation energy
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_EA_0522.png|thumb|Energy vs Time plot for F + H2]]
|Ea = (-103.751) - (-104.002) = 0.251 kcal/mol
|-
|H + HF
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_EA_1_0522.png|thumb|Energy vs Time plot for H + HF]]
|Ea = (-101.187) - (-133.852) = 32.665 kcal/mol
|}

===''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.''===
{| class="wikitable"
!Initial condition
!Graph
!Explanation
|-
|F + H2 (A+BC)
AB distance = 1.0 Å

BC distance = 0.74Å

AB momentum = -1.5

BC momentum = 0
|[[File:MRD_FHH_R1_0522.png |thumb|A contour plot for F + H2]]
[[File:MRD_FHH_E1_0522.png |thumb|An Energy vs Time plot for F + H2]]
|According to the graphs, when the kinetic energy increase, the potential energy decreases. Thus, the total energy in the system is conserved. It can also be proofed by the constant total energy at -105.721 kcal/mol.

As this reaction is exothermic, the increase in kinetic energy can be confirmed by increase in temperature.
|}

===''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.''===
{| class="wikitable"
!H + HF (A +BC)
!Initial condition
!Contour plot
!Explanation
|-
|High translation energy
|AB distance = 1.5 Å
BC distance = 0.917Å

AB momentum = -8

BC momentum = -1
|[[File:MRD_HHF_HTE_0522.png|thumb|Contour plot for high translation energy.]]
| rowspan="2" |
|-
|High vibration energy
|AB distance = 1.5 Å
BC distance = 0.9.7 Å

AB momentum = -1

BC momentum = -10
|
|}

==Reference==
<references>
<ref name="TS">T.Bligaard, J.K.Nørskov, in Chemical Bonding at Surfaces and Interfaces, Elsevier, Amsterdam, 2008, pp 255-321.</ref>
</references>

File:MRD HHF HTE 0522.png

2019-05-22T15:07:30Z

Xc2017:

MRD:xc2017

2019-05-22T14:58:26Z

== 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?''===

'''Answer:''' The transition state is a saddle point in the potential energy surface diagram where ∂V(ri)/∂ri=0 and G<0 where G = frr(r0,r'0)fr'r'(r0,r'0)−f2rr'(r0,r'0).

The transition point was defined as a maximum point in the minimum energy path linking reactants and products. For triatomic reaction, the transition state can be identified in reaction path in the contour plot where the AB and BC distance are equal.

The local minimum point can be distinguished by secondary partial derivation equal to zero and G > 0.

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

[[File:MRD_XC_1.png|thumb|Picture.1 A snapshot of 'Internuclear Distance vs Time' plot at r1 = r2 = 0.9075Å]]

'''Answer:''' The best estimate transition position is r1=r2=0.9075 Å where the “Internuclear Distances vs Time” plot shows two straight lines (Picture.1). The picture of distance vs time plot shows the AB and BC distance are constant which indicated the transition state.

=== '' Comment on how the mep and the trajectory you just calculated differ.''===
[[File:MRD_DYN_0521.png|thumb|Picture.2 A snapshot of dynamic surcafe plot]]
[[File:MRD_MEP_0521.png|thumb|Picture.3 A snapshot of MEP surface plot]]
'''Answer:'''

''' BC = r1 = rts+0.01'''

''' AB= '''r2 = rts

The initial condition was set to r1 = rts+0.01 and r2 = rts''' '''''',''' thus the AB distance ('''r2''') =0.9075 Å, and the BC distance ('''r1''') = 0.9175 Å
According to Picture 2 and 3, the reaction path calculated by MEP method was more smooth than the path calculated by dynamic method. For Picture.2 (using dynamic method), the reaction path vibrated at the lowest energy. For the MEP method, the reaction pathway was more smooth and more fit to the lowest energy barrier.
The difference between two ways of calculation may be because that MEP corresponds to infinitely slow motion while the dynamic method doesn't. For the dynamic calculation, the molecules move in a relative fast speed and vibrate. This makes the energy calculation harder. Thus, for small initial momenta, MEP calculation was more fit to the surface plot.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/Å
|0.9175
|7.15
|3.75
|-
|r2/Å
|0.9075
|0.75
|0.76
|-
|p1
|0
|2.46
|0
|-
|p2
|0
|1.33
|0
|}
For the table below, the initial distance was switched for r1 and r2,

BC distance =r1 = rts

AB distance = r2 = rts +0.01

According to the table below, it can be seen that the final value for both distance and momenta switched as the initial distance switched, and this change does not depend on their calculation method.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/ Å
|0.9075
|0.74
|0.76
|-
|r2/ Å
|0.9175
|7.15
|3.75
|-
|p1
|0
|1.33
|0
|-
|p2
|0
|2.46
|0
|}

=== 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
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reactants have enough kinetic energy to pass though the transition state and complete the reaction and it can be seen from the contour plot: the trajectory passes though the transition state and goes to the products. Thus, it is a productive reaction.[[File:MRD_R1_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.2</nowiki>
|No
|The reactants do not have enough energy to pass through the transition state, thus the trajectory moves back to the reactants.[[File:MRD_R2_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reagents have enough energy to pass through the transition state. Thus, the trajectory passes though the transition state and moves towards to the product. [[File:MRD_R3_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.6</nowiki>
|No
|The reagents have enough energy topass through the transition state, however, the energy is too high thus it drops back to the reactants.[[File:MRD_R4_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-82.6</nowiki>
|Yes
|The trajectory does not moves to product initially, the reagents have excess energy to pass through the transition state, thus it goes back to the reactants. After that, the reactants still have enough energy to pass through the transition state, thus the reaction occurs.[[File:MRD_R5_0521.png|thumb| Contour plot at this condition.]]
|}
In conclusion, to complete a productive reaction, the reactants should have enough energy to pass through the transition state but not too high to reverse the product back into the reactants.

===''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?''===
'''Assumptions:'''[https://www.sciencedirect.com/topics/chemistry/transition-state-theory]<ref name = 'TS'/>
# Quantum-tunnelling effects are assumed negligible.
# The Born-Oppenheimer approximation is invoked.
# The atoms in the reactant state have energies that are Boltzmann distributed.
# Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.

== 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?''===
{| class="wikitable"
!Reaction
!Initial condition
!Potential surface plot and discription
|-
|F + H2 (ABC)
|AB distance = 1.00 Å
BC distance = 0.74 Å

AB momentum = 0
BC momentum = -1.5
|[[File:MDR_FHH_0522.png|thumb|left|A surface plot for F+ H2 reaction. ]] The reaction is exothermix as the energy for reactants are higher than the energy for products. Thus, the energy is released during the reaction. This is caused by the formation of strong H-F bond.
|-
|H +HF (ABC)
|AB distance = 1.00 Å
BC distance = 0.917 Å

AB momentum = 0
BC momentum = -10
|[[File:MRD_HHF_0522.png|thumb|left|A surface plot for H + HF reaction. ]] This reaction is endothermic as the potential energy of reactants are lower than the potential energy of products. In this case, the strong H-F bond was break and a weaker H-H bond was formed. Thud, the overall reaction was endothermic.
|}

===''Locate the approximate position of the transition state.''===
{| class="wikitable"
!Reaction (ABC)
!Position of Transition Point
!Contour plot
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_0522.png|thumb|Contour plot for for the transition point for F + H2.]]
|-
|H + HF
|AB distance = 0.745 Å
BC distance = 1.820 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_0522_1.png|thumb|Contour plot for for the transition point for H + HF.]]
|}

===''Report the activation energy for both reactions.''===
{| class="wikitable"
!Reaction (ABC)
!Initial condition
!Energy vs Time plot
!Activation energy
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_EA_0522.png|thumb|Energy vs Time plot for F + H2]]
|Ea = (-103.751) - (-104.002) = 0.251 kcal/mol
|-
|H + HF
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_EA_1_0522.png|thumb|Energy vs Time plot for H + HF]]
|Ea = (-101.187) - (-133.852) = 32.665 kcal/mol
|}

===''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.''===
{| class="wikitable"
!Initial condition
!Graph
!Explanation
|-
|F + H2 (A+BC)
AB distance = 1.0 Å

BC distance = 0.74Å

AB momentum = -1.5

BC momentum = 0
|[[File:MRD_FHH_R1_0522.png |thumb|A contour plot for F + H2]]
[[File:MRD_FHH_E1_0522.png |thumb|An Energy vs Time plot for F + H2]]
|According to the graphs, when the kinetic energy increase, the potential energy decreases. Thus, the total energy in the system is conserved. It can also be proofed by the constant total energy at -105.721 kcal/mol.

As this reaction is exothermic, the increase in kinetic energy can be confirmed by increase in temperature.
|}

===''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.''===
{| class="wikitable"
!H + HF (A +BC)
!Initial condition
!Contour plot
!Explanation
|-
|High translation energy
|AB distance = 1.5 Å
BC distance = 0.917Å

AB momentum = -8

BC momentum = -1
|
| rowspan="2" |
|-
|High vibration energy
|AB distance = 1.5 Å
BC distance = 0.9.7 Å

AB momentum = -1

BC momentum = -10
|
|}

==Reference==
<references>
<ref name="TS">T.Bligaard, J.K.Nørskov, in Chemical Bonding at Surfaces and Interfaces, Elsevier, Amsterdam, 2008, pp 255-321.</ref>
</references>

MRD:xc2017

2019-05-22T13:56:12Z

Xc2017: /* 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. */

== 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?''===

'''Answer:''' The transition state is a saddle point in the potential energy surface diagram where ∂V(ri)/∂ri=0 and G<0 where G = frr(r0,r'0)fr'r'(r0,r'0)−f2rr'(r0,r'0).

The transition point was defined as a maximum point in the minimum energy path linking reactants and products. For triatomic reaction, the transition state can be identified in reaction path in the contour plot where the AB and BC distance are equal.

The local minimum point can be distinguished by secondary partial derivation equal to zero and G > 0.

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

[[File:MRD_XC_1.png|thumb|Picture.1 A snapshot of 'Internuclear Distance vs Time' plot at r1 = r2 = 0.9075Å]]

'''Answer:''' The best estimate transition position is r1=r2=0.9075 Å where the “Internuclear Distances vs Time” plot shows two straight lines (Picture.1). The picture of distance vs time plot shows the AB and BC distance are constant which indicated the transition state.

=== '' Comment on how the mep and the trajectory you just calculated differ.''===
[[File:MRD_DYN_0521.png|thumb|Picture.2 A snapshot of dynamic surcafe plot]]
[[File:MRD_MEP_0521.png|thumb|Picture.3 A snapshot of MEP surface plot]]
'''Answer:'''

''' BC = r1 = rts+0.01'''

''' AB= '''r2 = rts

The initial condition was set to r1 = rts+0.01 and r2 = rts''' '''''',''' thus the AB distance ('''r2''') =0.9075 Å, and the BC distance ('''r1''') = 0.9175 Å
According to Picture 2 and 3, the reaction path calculated by MEP method was more smooth than the path calculated by dynamic method. For Picture.2 (using dynamic method), the reaction path vibrated at the lowest energy. For the MEP method, the reaction pathway was more smooth and more fit to the lowest energy barrier.
The difference between two ways of calculation may be because that MEP corresponds to infinitely slow motion while the dynamic method doesn't. For the dynamic calculation, the molecules move in a relative fast speed and vibrate. This makes the energy calculation harder. Thus, for small initial momenta, MEP calculation was more fit to the surface plot.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/Å
|0.9175
|7.15
|3.75
|-
|r2/Å
|0.9075
|0.75
|0.76
|-
|p1
|0
|2.46
|0
|-
|p2
|0
|1.33
|0
|}
For the table below, the initial distance was switched for r1 and r2,

BC distance =r1 = rts

AB distance = r2 = rts +0.01

According to the table below, it can be seen that the final value for both distance and momenta switched as the initial distance switched, and this change does not depend on their calculation method.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/ Å
|0.9075
|0.74
|0.76
|-
|r2/ Å
|0.9175
|7.15
|3.75
|-
|p1
|0
|1.33
|0
|-
|p2
|0
|2.46
|0
|}

=== 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
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reactants have enough kinetic energy to pass though the transition state and complete the reaction and it can be seen from the contour plot: the trajectory passes though the transition state and goes to the products. Thus, it is a productive reaction.[[File:MRD_R1_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.2</nowiki>
|No
|The reactants do not have enough energy to pass through the transition state, thus the trajectory moves back to the reactants.[[File:MRD_R2_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reagents have enough energy to pass through the transition state. Thus, the trajectory passes though the transition state and moves towards to the product. [[File:MRD_R3_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.6</nowiki>
|No
|The reagents have enough energy topass through the transition state, however, the energy is too high thus it drops back to the reactants.[[File:MRD_R4_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-82.6</nowiki>
|Yes
|The trajectory does not moves to product initially, the reagents have excess energy to pass through the transition state, thus it goes back to the reactants. After that, the reactants still have enough energy to pass through the transition state, thus the reaction occurs.[[File:MRD_R5_0521.png|thumb| Contour plot at this condition.]]
|}
In conclusion, to complete a productive reaction, the reactants should have enough energy to pass through the transition state but not too high to reverse the product back into the reactants.

===''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?''===
'''Assumptions:'''[https://www.sciencedirect.com/topics/chemistry/transition-state-theory]<ref name = 'TS'/>
# Quantum-tunnelling effects are assumed negligible.
# The Born-Oppenheimer approximation is invoked.
# The atoms in the reactant state have energies that are Boltzmann distributed.
# Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.

== 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?''===
{| class="wikitable"
!Reaction
!Initial condition
!Potential surface plot and discription
|-
|F + H2 (ABC)
|AB distance = 1.00 Å
BC distance = 0.74 Å

AB momentum = 0
BC momentum = -1.5
|[[File:MDR_FHH_0522.png|thumb|left|A surface plot for F+ H2 reaction. ]] The reaction is exothermix as the energy for reactants are higher than the energy for products. Thus, the energy is released during the reaction. This is caused by the formation of strong H-F bond.
|-
|H +HF (ABC)
|AB distance = 1.00 Å
BC distance = 0.917 Å

AB momentum = 0
BC momentum = -10
|[[File:MRD_HHF_0522.png|thumb|left|A surface plot for H + HF reaction. ]] This reaction is endothermic as the potential energy of reactants are lower than the potential energy of products. In this case, the strong H-F bond was break and a weaker H-H bond was formed. Thud, the overall reaction was endothermic.
|}

===''Locate the approximate position of the transition state.''===
{| class="wikitable"
!Reaction (ABC)
!Position of Transition Point
!Contour plot
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_0522.png|thumb|Contour plot for for the transition point for F + H2.]]
|-
|H + HF
|AB distance = 0.745 Å
BC distance = 1.820 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_0522_1.png|thumb|Contour plot for for the transition point for H + HF.]]
|}

===''Report the activation energy for both reactions.''===
{| class="wikitable"
!Reaction (ABC)
!Initial condition
!Energy vs Time plot
!Activation energy
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_EA_0522.png|thumb|Energy vs Time plot for F + H2]]
|Ea = (-103.751) - (-104.002) = 0.251 kcal/mol
|-
|H + HF
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_EA_1_0522.png|thumb|Energy vs Time plot for H + HF]]
|Ea = (-101.187) - (-133.852) = 32.665 kcal/mol
|}

===''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.''===
{| class="wikitable"
!Initial condition
!Graph
!Explanation
|-
|F + H2 (A+BC)
AB distance = 1.0 Å

BC distance = 0.74Å

AB momentum = -1.5

BC momentum = 0
|[[File:MRD_FHH_R1_0522.png |thumb|A contour plot for F + H2]]
[[File:MRD_FHH_E1_0522.png |thumb|An Energy vs Time plot for F + H2]]
|According to the graphs, when the kinetic energy increase, the potential energy decreases. Thus, the total energy in the system is conserved. It can also be proofed by the constant total energy at -105.721 kcal/mol.

As this reaction is exothermic, the increase in kinetic energy can be confirmed by increase in temperature.
|}

===''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.''===

==Reference==
<references>
<ref name="TS">T.Bligaard, J.K.Nørskov, in Chemical Bonding at Surfaces and Interfaces, Elsevier, Amsterdam, 2008, pp 255-321.</ref>
</references>

MRD:xc2017

2019-05-22T13:55:45Z

Xc2017: /* 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. */

== 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?''===

'''Answer:''' The transition state is a saddle point in the potential energy surface diagram where ∂V(ri)/∂ri=0 and G<0 where G = frr(r0,r'0)fr'r'(r0,r'0)−f2rr'(r0,r'0).

The transition point was defined as a maximum point in the minimum energy path linking reactants and products. For triatomic reaction, the transition state can be identified in reaction path in the contour plot where the AB and BC distance are equal.

The local minimum point can be distinguished by secondary partial derivation equal to zero and G > 0.

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

[[File:MRD_XC_1.png|thumb|Picture.1 A snapshot of 'Internuclear Distance vs Time' plot at r1 = r2 = 0.9075Å]]

'''Answer:''' The best estimate transition position is r1=r2=0.9075 Å where the “Internuclear Distances vs Time” plot shows two straight lines (Picture.1). The picture of distance vs time plot shows the AB and BC distance are constant which indicated the transition state.

=== '' Comment on how the mep and the trajectory you just calculated differ.''===
[[File:MRD_DYN_0521.png|thumb|Picture.2 A snapshot of dynamic surcafe plot]]
[[File:MRD_MEP_0521.png|thumb|Picture.3 A snapshot of MEP surface plot]]
'''Answer:'''

''' BC = r1 = rts+0.01'''

''' AB= '''r2 = rts

The initial condition was set to r1 = rts+0.01 and r2 = rts''' '''''',''' thus the AB distance ('''r2''') =0.9075 Å, and the BC distance ('''r1''') = 0.9175 Å
According to Picture 2 and 3, the reaction path calculated by MEP method was more smooth than the path calculated by dynamic method. For Picture.2 (using dynamic method), the reaction path vibrated at the lowest energy. For the MEP method, the reaction pathway was more smooth and more fit to the lowest energy barrier.
The difference between two ways of calculation may be because that MEP corresponds to infinitely slow motion while the dynamic method doesn't. For the dynamic calculation, the molecules move in a relative fast speed and vibrate. This makes the energy calculation harder. Thus, for small initial momenta, MEP calculation was more fit to the surface plot.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/Å
|0.9175
|7.15
|3.75
|-
|r2/Å
|0.9075
|0.75
|0.76
|-
|p1
|0
|2.46
|0
|-
|p2
|0
|1.33
|0
|}
For the table below, the initial distance was switched for r1 and r2,

BC distance =r1 = rts

AB distance = r2 = rts +0.01

According to the table below, it can be seen that the final value for both distance and momenta switched as the initial distance switched, and this change does not depend on their calculation method.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/ Å
|0.9075
|0.74
|0.76
|-
|r2/ Å
|0.9175
|7.15
|3.75
|-
|p1
|0
|1.33
|0
|-
|p2
|0
|2.46
|0
|}

=== 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
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reactants have enough kinetic energy to pass though the transition state and complete the reaction and it can be seen from the contour plot: the trajectory passes though the transition state and goes to the products. Thus, it is a productive reaction.[[File:MRD_R1_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.2</nowiki>
|No
|The reactants do not have enough energy to pass through the transition state, thus the trajectory moves back to the reactants.[[File:MRD_R2_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reagents have enough energy to pass through the transition state. Thus, the trajectory passes though the transition state and moves towards to the product. [[File:MRD_R3_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.6</nowiki>
|No
|The reagents have enough energy topass through the transition state, however, the energy is too high thus it drops back to the reactants.[[File:MRD_R4_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-82.6</nowiki>
|Yes
|The trajectory does not moves to product initially, the reagents have excess energy to pass through the transition state, thus it goes back to the reactants. After that, the reactants still have enough energy to pass through the transition state, thus the reaction occurs.[[File:MRD_R5_0521.png|thumb| Contour plot at this condition.]]
|}
In conclusion, to complete a productive reaction, the reactants should have enough energy to pass through the transition state but not too high to reverse the product back into the reactants.

===''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?''===
'''Assumptions:'''[https://www.sciencedirect.com/topics/chemistry/transition-state-theory]<ref name = 'TS'/>
# Quantum-tunnelling effects are assumed negligible.
# The Born-Oppenheimer approximation is invoked.
# The atoms in the reactant state have energies that are Boltzmann distributed.
# Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.

== 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?''===
{| class="wikitable"
!Reaction
!Initial condition
!Potential surface plot and discription
|-
|F + H2 (ABC)
|AB distance = 1.00 Å
BC distance = 0.74 Å

AB momentum = 0
BC momentum = -1.5
|[[File:MDR_FHH_0522.png|thumb|left|A surface plot for F+ H2 reaction. ]] The reaction is exothermix as the energy for reactants are higher than the energy for products. Thus, the energy is released during the reaction. This is caused by the formation of strong H-F bond.
|-
|H +HF (ABC)
|AB distance = 1.00 Å
BC distance = 0.917 Å

AB momentum = 0
BC momentum = -10
|[[File:MRD_HHF_0522.png|thumb|left|A surface plot for H + HF reaction. ]] This reaction is endothermic as the potential energy of reactants are lower than the potential energy of products. In this case, the strong H-F bond was break and a weaker H-H bond was formed. Thud, the overall reaction was endothermic.
|}

===''Locate the approximate position of the transition state.''===
{| class="wikitable"
!Reaction (ABC)
!Position of Transition Point
!Contour plot
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_0522.png|thumb|Contour plot for for the transition point for F + H2.]]
|-
|H + HF
|AB distance = 0.745 Å
BC distance = 1.820 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_0522_1.png|thumb|Contour plot for for the transition point for H + HF.]]
|}

===''Report the activation energy for both reactions.''===
{| class="wikitable"
!Reaction (ABC)
!Initial condition
!Energy vs Time plot
!Activation energy
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_EA_0522.png|thumb|Energy vs Time plot for F + H2]]
|Ea = (-103.751) - (-104.002) = 0.251 kcal/mol
|-
|H + HF
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_EA_1_0522.png|thumb|Energy vs Time plot for H + HF]]
|Ea = (-101.187) - (-133.852) = 32.665 kcal/mol
|}

===''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.''===
{| class="wikitable"
!Initial condition
!Graph
!Explanation
|-
|F + H2 (A+BC)
AB distance = 1.0 Å

BC distance = 0.74Å

AB momentum = -1.5

BC momentum = 0
|[[File:MRD_FHH_R1_0522.png |thumb|A contour plot for F + H2]]
[[File:MRD_FHH_E1_0522.png |thumb|An Energy vs Time plot for F + H2]]
|According to the graphs, when the kinetic energy increase, the potential energy decreases. Thus, the total energy in the system is conserved. It can also be proofed by the constant total energy at -105.721 kcal/mol.

As this reaction is exothermic, the increase in kinetic energy can be confirmed by increase in temperature.
|}

===''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.''===

==Reference==
<references>
<ref name="TS">T.Bligaard, J.K.Nørskov, in Chemical Bonding at Surfaces and Interfaces, Elsevier, Amsterdam, 2008, pp 255-321.</ref>
</references>

MRD:xc2017

2019-05-22T13:54:14Z

Xc2017: /* 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. */

== 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?''===

'''Answer:''' The transition state is a saddle point in the potential energy surface diagram where ∂V(ri)/∂ri=0 and G<0 where G = frr(r0,r'0)fr'r'(r0,r'0)−f2rr'(r0,r'0).

The transition point was defined as a maximum point in the minimum energy path linking reactants and products. For triatomic reaction, the transition state can be identified in reaction path in the contour plot where the AB and BC distance are equal.

The local minimum point can be distinguished by secondary partial derivation equal to zero and G > 0.

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

[[File:MRD_XC_1.png|thumb|Picture.1 A snapshot of 'Internuclear Distance vs Time' plot at r1 = r2 = 0.9075Å]]

'''Answer:''' The best estimate transition position is r1=r2=0.9075 Å where the “Internuclear Distances vs Time” plot shows two straight lines (Picture.1). The picture of distance vs time plot shows the AB and BC distance are constant which indicated the transition state.

=== '' Comment on how the mep and the trajectory you just calculated differ.''===
[[File:MRD_DYN_0521.png|thumb|Picture.2 A snapshot of dynamic surcafe plot]]
[[File:MRD_MEP_0521.png|thumb|Picture.3 A snapshot of MEP surface plot]]
'''Answer:'''

''' BC = r1 = rts+0.01'''

''' AB= '''r2 = rts

The initial condition was set to r1 = rts+0.01 and r2 = rts''' '''''',''' thus the AB distance ('''r2''') =0.9075 Å, and the BC distance ('''r1''') = 0.9175 Å
According to Picture 2 and 3, the reaction path calculated by MEP method was more smooth than the path calculated by dynamic method. For Picture.2 (using dynamic method), the reaction path vibrated at the lowest energy. For the MEP method, the reaction pathway was more smooth and more fit to the lowest energy barrier.
The difference between two ways of calculation may be because that MEP corresponds to infinitely slow motion while the dynamic method doesn't. For the dynamic calculation, the molecules move in a relative fast speed and vibrate. This makes the energy calculation harder. Thus, for small initial momenta, MEP calculation was more fit to the surface plot.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/Å
|0.9175
|7.15
|3.75
|-
|r2/Å
|0.9075
|0.75
|0.76
|-
|p1
|0
|2.46
|0
|-
|p2
|0
|1.33
|0
|}
For the table below, the initial distance was switched for r1 and r2,

BC distance =r1 = rts

AB distance = r2 = rts +0.01

According to the table below, it can be seen that the final value for both distance and momenta switched as the initial distance switched, and this change does not depend on their calculation method.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/ Å
|0.9075
|0.74
|0.76
|-
|r2/ Å
|0.9175
|7.15
|3.75
|-
|p1
|0
|1.33
|0
|-
|p2
|0
|2.46
|0
|}

=== 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
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reactants have enough kinetic energy to pass though the transition state and complete the reaction and it can be seen from the contour plot: the trajectory passes though the transition state and goes to the products. Thus, it is a productive reaction.[[File:MRD_R1_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.2</nowiki>
|No
|The reactants do not have enough energy to pass through the transition state, thus the trajectory moves back to the reactants.[[File:MRD_R2_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reagents have enough energy to pass through the transition state. Thus, the trajectory passes though the transition state and moves towards to the product. [[File:MRD_R3_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.6</nowiki>
|No
|The reagents have enough energy topass through the transition state, however, the energy is too high thus it drops back to the reactants.[[File:MRD_R4_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-82.6</nowiki>
|Yes
|The trajectory does not moves to product initially, the reagents have excess energy to pass through the transition state, thus it goes back to the reactants. After that, the reactants still have enough energy to pass through the transition state, thus the reaction occurs.[[File:MRD_R5_0521.png|thumb| Contour plot at this condition.]]
|}
In conclusion, to complete a productive reaction, the reactants should have enough energy to pass through the transition state but not too high to reverse the product back into the reactants.

===''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?''===
'''Assumptions:'''[https://www.sciencedirect.com/topics/chemistry/transition-state-theory]<ref name = 'TS'/>
# Quantum-tunnelling effects are assumed negligible.
# The Born-Oppenheimer approximation is invoked.
# The atoms in the reactant state have energies that are Boltzmann distributed.
# Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.

== 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?''===
{| class="wikitable"
!Reaction
!Initial condition
!Potential surface plot and discription
|-
|F + H2 (ABC)
|AB distance = 1.00 Å
BC distance = 0.74 Å

AB momentum = 0
BC momentum = -1.5
|[[File:MDR_FHH_0522.png|thumb|left|A surface plot for F+ H2 reaction. ]] The reaction is exothermix as the energy for reactants are higher than the energy for products. Thus, the energy is released during the reaction. This is caused by the formation of strong H-F bond.
|-
|H +HF (ABC)
|AB distance = 1.00 Å
BC distance = 0.917 Å

AB momentum = 0
BC momentum = -10
|[[File:MRD_HHF_0522.png|thumb|left|A surface plot for H + HF reaction. ]] This reaction is endothermic as the potential energy of reactants are lower than the potential energy of products. In this case, the strong H-F bond was break and a weaker H-H bond was formed. Thud, the overall reaction was endothermic.
|}

===''Locate the approximate position of the transition state.''===
{| class="wikitable"
!Reaction (ABC)
!Position of Transition Point
!Contour plot
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_0522.png|thumb|Contour plot for for the transition point for F + H2.]]
|-
|H + HF
|AB distance = 0.745 Å
BC distance = 1.820 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_0522_1.png|thumb|Contour plot for for the transition point for H + HF.]]
|}

===''Report the activation energy for both reactions.''===
{| class="wikitable"
!Reaction (ABC)
!Initial condition
!Energy vs Time plot
!Activation energy
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_EA_0522.png|thumb|Energy vs Time plot for F + H2]]
|Ea = (-103.751) - (-104.002) = 0.251 kcal/mol
|-
|H + HF
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_EA_1_0522.png|thumb|Energy vs Time plot for H + HF]]
|Ea = (-101.187) - (-133.852) = 32.665 kcal/mol
|}

===''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.''===
{| class="wikitable"
!Initial condition
!Graph
!Explanation
|-
|F + H2 (A+BC)
AB distance = 1.0 Å

BC distance = 0.74Å

AB momentum = -1.5

BC momentum = 0
|[[File:MRD_FHH_R1_0522.png |thumb|A contour plot for F + H2]]
[[File:MRD_FHH_E1_0522.png |thumb|An Energy vs Time plot for F + H2]]
|According to the graphs, when the kinetic energy increase, the potential energy decreases. Thus, the total energy in the system is conserved. It can also be proofed by the constant total energy at -105.721 kcal/mol.

As this reaction is exothermic, the increase in kinetic energy can be confirmed by increase in temperature.
|}

==Reference==
<references>
<ref name="TS">T.Bligaard, J.K.Nørskov, in Chemical Bonding at Surfaces and Interfaces, Elsevier, Amsterdam, 2008, pp 255-321.</ref>
</references>

MRD:xc2017

2019-05-22T13:48:44Z

Xc2017: /* 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? */

== 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?''===

'''Answer:''' The transition state is a saddle point in the potential energy surface diagram where ∂V(ri)/∂ri=0 and G<0 where G = frr(r0,r'0)fr'r'(r0,r'0)−f2rr'(r0,r'0).

The transition point was defined as a maximum point in the minimum energy path linking reactants and products. For triatomic reaction, the transition state can be identified in reaction path in the contour plot where the AB and BC distance are equal.

The local minimum point can be distinguished by secondary partial derivation equal to zero and G > 0.

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

[[File:MRD_XC_1.png|thumb|Picture.1 A snapshot of 'Internuclear Distance vs Time' plot at r1 = r2 = 0.9075Å]]

'''Answer:''' The best estimate transition position is r1=r2=0.9075 Å where the “Internuclear Distances vs Time” plot shows two straight lines (Picture.1). The picture of distance vs time plot shows the AB and BC distance are constant which indicated the transition state.

=== '' Comment on how the mep and the trajectory you just calculated differ.''===
[[File:MRD_DYN_0521.png|thumb|Picture.2 A snapshot of dynamic surcafe plot]]
[[File:MRD_MEP_0521.png|thumb|Picture.3 A snapshot of MEP surface plot]]
'''Answer:'''

''' BC = r1 = rts+0.01'''

''' AB= '''r2 = rts

The initial condition was set to r1 = rts+0.01 and r2 = rts''' '''''',''' thus the AB distance ('''r2''') =0.9075 Å, and the BC distance ('''r1''') = 0.9175 Å
According to Picture 2 and 3, the reaction path calculated by MEP method was more smooth than the path calculated by dynamic method. For Picture.2 (using dynamic method), the reaction path vibrated at the lowest energy. For the MEP method, the reaction pathway was more smooth and more fit to the lowest energy barrier.
The difference between two ways of calculation may be because that MEP corresponds to infinitely slow motion while the dynamic method doesn't. For the dynamic calculation, the molecules move in a relative fast speed and vibrate. This makes the energy calculation harder. Thus, for small initial momenta, MEP calculation was more fit to the surface plot.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/Å
|0.9175
|7.15
|3.75
|-
|r2/Å
|0.9075
|0.75
|0.76
|-
|p1
|0
|2.46
|0
|-
|p2
|0
|1.33
|0
|}
For the table below, the initial distance was switched for r1 and r2,

BC distance =r1 = rts

AB distance = r2 = rts +0.01

According to the table below, it can be seen that the final value for both distance and momenta switched as the initial distance switched, and this change does not depend on their calculation method.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/ Å
|0.9075
|0.74
|0.76
|-
|r2/ Å
|0.9175
|7.15
|3.75
|-
|p1
|0
|1.33
|0
|-
|p2
|0
|2.46
|0
|}

=== 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
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reactants have enough kinetic energy to pass though the transition state and complete the reaction and it can be seen from the contour plot: the trajectory passes though the transition state and goes to the products. Thus, it is a productive reaction.[[File:MRD_R1_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.2</nowiki>
|No
|The reactants do not have enough energy to pass through the transition state, thus the trajectory moves back to the reactants.[[File:MRD_R2_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reagents have enough energy to pass through the transition state. Thus, the trajectory passes though the transition state and moves towards to the product. [[File:MRD_R3_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.6</nowiki>
|No
|The reagents have enough energy topass through the transition state, however, the energy is too high thus it drops back to the reactants.[[File:MRD_R4_0521.png|thumb| Contour plot at this condition.]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-82.6</nowiki>
|Yes
|The trajectory does not moves to product initially, the reagents have excess energy to pass through the transition state, thus it goes back to the reactants. After that, the reactants still have enough energy to pass through the transition state, thus the reaction occurs.[[File:MRD_R5_0521.png|thumb| Contour plot at this condition.]]
|}
In conclusion, to complete a productive reaction, the reactants should have enough energy to pass through the transition state but not too high to reverse the product back into the reactants.

===''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?''===
'''Assumptions:'''[https://www.sciencedirect.com/topics/chemistry/transition-state-theory]<ref name = 'TS'/>
# Quantum-tunnelling effects are assumed negligible.
# The Born-Oppenheimer approximation is invoked.
# The atoms in the reactant state have energies that are Boltzmann distributed.
# Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.

== 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?''===
{| class="wikitable"
!Reaction
!Initial condition
!Potential surface plot and discription
|-
|F + H2 (ABC)
|AB distance = 1.00 Å
BC distance = 0.74 Å

AB momentum = 0
BC momentum = -1.5
|[[File:MDR_FHH_0522.png|thumb|left|A surface plot for F+ H2 reaction. ]] The reaction is exothermix as the energy for reactants are higher than the energy for products. Thus, the energy is released during the reaction. This is caused by the formation of strong H-F bond.
|-
|H +HF (ABC)
|AB distance = 1.00 Å
BC distance = 0.917 Å

AB momentum = 0
BC momentum = -10
|[[File:MRD_HHF_0522.png|thumb|left|A surface plot for H + HF reaction. ]] This reaction is endothermic as the potential energy of reactants are lower than the potential energy of products. In this case, the strong H-F bond was break and a weaker H-H bond was formed. Thud, the overall reaction was endothermic.
|}

===''Locate the approximate position of the transition state.''===
{| class="wikitable"
!Reaction (ABC)
!Position of Transition Point
!Contour plot
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_0522.png|thumb|Contour plot for for the transition point for F + H2.]]
|-
|H + HF
|AB distance = 0.745 Å
BC distance = 1.820 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_0522_1.png|thumb|Contour plot for for the transition point for H + HF.]]
|}

===''Report the activation energy for both reactions.''===
{| class="wikitable"
!Reaction (ABC)
!Initial condition
!Energy vs Time plot
!Activation energy
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_EA_0522.png|thumb|Energy vs Time plot for F + H2]]
|Ea = (-103.751) - (-104.002) = 0.251 kcal/mol
|-
|H + HF
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_EA_1_0522.png|thumb|Energy vs Time plot for H + HF]]
|Ea = (-101.187) - (-133.852) = 32.665 kcal/mol
|}

===''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.''===
{| class="wikitable"
!Initial condition
!Graph
!Explanation
|-
|F + H2 (A+BC)
AB distance = 1.0 Å
BC distance = 0.74Å

AB momentum = -1.5

BC momentum = 0
|[[File:MRD_FHH_R1_0522.png |thumb|A contour plot for F + H2]]
[[File:MRD_FHH_E1_0522.png |thumb|An Energy vs Time plot for F + H2]]
|
|}

==Reference==
<references>
<ref name="TS">T.Bligaard, J.K.Nørskov, in Chemical Bonding at Surfaces and Interfaces, Elsevier, Amsterdam, 2008, pp 255-321.</ref>
</references>

MRD:xc2017

2019-05-22T13:46:42Z

Xc2017: /* 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. */

== 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?''===

'''Answer:''' The transition state is a saddle point in the potential energy surface diagram where ∂V(ri)/∂ri=0 and G<0 where G = frr(r0,r'0)fr'r'(r0,r'0)−f2rr'(r0,r'0).

The transition point was defined as a maximum point in the minimum energy path linking reactants and products. For triatomic reaction, the transition state can be identified in reaction path in the contour plot where the AB and BC distance are equal.

The local minimum point can be distinguished by secondary partial derivation equal to zero and G > 0.

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

[[File:MRD_XC_1.png|thumb|Picture.1 A snapshot of 'Internuclear Distance vs Time' plot at r1 = r2 = 0.9075Å]]

'''Answer:''' The best estimate transition position is r1=r2=0.9075 Å where the “Internuclear Distances vs Time” plot shows two straight lines (Picture.1). The picture of distance vs time plot shows the AB and BC distance are constant which indicated the transition state.

=== '' Comment on how the mep and the trajectory you just calculated differ.''===
[[File:MRD_DYN_0521.png|thumb|Picture.2 A snapshot of dynamic surcafe plot]]
[[File:MRD_MEP_0521.png|thumb|Picture.3 A snapshot of MEP surface plot]]
'''Answer:'''

''' BC = r1 = rts+0.01'''

''' AB= '''r2 = rts

The initial condition was set to r1 = rts+0.01 and r2 = rts''' '''''',''' thus the AB distance ('''r2''') =0.9075 Å, and the BC distance ('''r1''') = 0.9175 Å
According to Picture 2 and 3, the reaction path calculated by MEP method was more smooth than the path calculated by dynamic method. For Picture.2 (using dynamic method), the reaction path vibrated at the lowest energy. For the MEP method, the reaction pathway was more smooth and more fit to the lowest energy barrier.
The difference between two ways of calculation may be because that MEP corresponds to infinitely slow motion while the dynamic method doesn't. For the dynamic calculation, the molecules move in a relative fast speed and vibrate. This makes the energy calculation harder. Thus, for small initial momenta, MEP calculation was more fit to the surface plot.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/Å
|0.9175
|7.15
|3.75
|-
|r2/Å
|0.9075
|0.75
|0.76
|-
|p1
|0
|2.46
|0
|-
|p2
|0
|1.33
|0
|}
For the table below, the initial distance was switched for r1 and r2,

BC distance =r1 = rts

AB distance = r2 = rts +0.01

According to the table below, it can be seen that the final value for both distance and momenta switched as the initial distance switched, and this change does not depend on their calculation method.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/ Å
|0.9075
|0.74
|0.76
|-
|r2/ Å
|0.9175
|7.15
|3.75
|-
|p1
|0
|1.33
|0
|-
|p2
|0
|2.46
|0
|}

=== 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"
!p1
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reactants have enough kinetic energy to pass though the transition state and complete the reaction and it can be seen from the contour plot: the trajectory passes though the transition state and goes to the products. Thus, it is a productive reaction.[[File:MRD_R1_0521.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.2</nowiki>
|No
|The reactants do not have enough energy to pass through the transition state, thus the trajectory moves back to the reactants.[[File:MRD_R2_0521.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reagents have enough energy to pass through the transition state. Thus, the trajectory passes though the transition state and moves towards to the product. [[File:MRD_R3_0521.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.6</nowiki>
|No
|The reagents have enough energy topass through the transition state, however, the energy is too high thus it drops back to the reactants.[[File:MRD_R4_0521.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-82.6</nowiki>
|Yes
|The trajectory does not moves to product initially, the reagents have excess energy to pass through the transition state, thus it goes back to the reactants. After that, the reactants still have enough energy to pass through the transition state, thus the reaction occurs.[[File:MRD_R5_0521.png]]
|}
In conclusion, to complete a productive reaction, the reactants should have enough energy to pass through the transition state but not too high to reverse the product back into the reactants.

===''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?''===
'''Assumptions:'''[https://www.sciencedirect.com/topics/chemistry/transition-state-theory]<ref name = 'TS'/>
# Quantum-tunnelling effects are assumed negligible.
# The Born-Oppenheimer approximation is invoked.
# The atoms in the reactant state have energies that are Boltzmann distributed.
# Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.

== 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?''===
{| class="wikitable"
!Reaction
!Initial condition
!Potential surface plot and discription
|-
|F + H2 (ABC)
|AB distance = 1.00 Å
BC distance = 0.74 Å

AB momentum = 0
BC momentum = -1.5
|[[File:MDR_FHH_0522.png|thumb|left|A surface plot for F+ H2 reaction. ]] The reaction is exothermix as the energy for reactants are higher than the energy for products. Thus, the energy is released during the reaction. This is caused by the formation of strong H-F bond.
|-
|H +HF (ABC)
|AB distance = 1.00 Å
BC distance = 0.917 Å

AB momentum = 0
BC momentum = -10
|[[File:MRD_HHF_0522.png|thumb|left|A surface plot for H + HF reaction. ]] This reaction is endothermic as the potential energy of reactants are lower than the potential energy of products. In this case, the strong H-F bond was break and a weaker H-H bond was formed. Thud, the overall reaction was endothermic.
|}

===''Locate the approximate position of the transition state.''===
{| class="wikitable"
!Reaction (ABC)
!Position of Transition Point
!Contour plot
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_0522.png|thumb|Contour plot for for the transition point for F + H2.]]
|-
|H + HF
|AB distance = 0.745 Å
BC distance = 1.820 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_0522_1.png|thumb|Contour plot for for the transition point for H + HF.]]
|}

===''Report the activation energy for both reactions.''===
{| class="wikitable"
!Reaction (ABC)
!Initial condition
!Energy vs Time plot
!Activation energy
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_EA_0522.png|thumb|Energy vs Time plot for F + H2]]
|Ea = (-103.751) - (-104.002) = 0.251 kcal/mol
|-
|H + HF
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_EA_1_0522.png|thumb|Energy vs Time plot for H + HF]]
|Ea = (-101.187) - (-133.852) = 32.665 kcal/mol
|}

===''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.''===
{| class="wikitable"
!Initial condition
!Graph
!Explanation
|-
|F + H2 (A+BC)
AB distance = 1.0 Å
BC distance = 0.74Å

AB momentum = -1.5

BC momentum = 0
|[[File:MRD_FHH_R1_0522.png |thumb|A contour plot for F + H2]]
[[File:MRD_FHH_E1_0522.png |thumb|An Energy vs Time plot for F + H2]]
|
|}

==Reference==
<references>
<ref name="TS">T.Bligaard, J.K.Nørskov, in Chemical Bonding at Surfaces and Interfaces, Elsevier, Amsterdam, 2008, pp 255-321.</ref>
</references>

MRD:xc2017

2019-05-22T13:44:40Z

Xc2017: /* 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. */

== 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?''===

'''Answer:''' The transition state is a saddle point in the potential energy surface diagram where ∂V(ri)/∂ri=0 and G<0 where G = frr(r0,r'0)fr'r'(r0,r'0)−f2rr'(r0,r'0).

The transition point was defined as a maximum point in the minimum energy path linking reactants and products. For triatomic reaction, the transition state can be identified in reaction path in the contour plot where the AB and BC distance are equal.

The local minimum point can be distinguished by secondary partial derivation equal to zero and G > 0.

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

[[File:MRD_XC_1.png|thumb|Picture.1 A snapshot of 'Internuclear Distance vs Time' plot at r1 = r2 = 0.9075Å]]

'''Answer:''' The best estimate transition position is r1=r2=0.9075 Å where the “Internuclear Distances vs Time” plot shows two straight lines (Picture.1). The picture of distance vs time plot shows the AB and BC distance are constant which indicated the transition state.

=== '' Comment on how the mep and the trajectory you just calculated differ.''===
[[File:MRD_DYN_0521.png|thumb|Picture.2 A snapshot of dynamic surcafe plot]]
[[File:MRD_MEP_0521.png|thumb|Picture.3 A snapshot of MEP surface plot]]
'''Answer:'''

''' BC = r1 = rts+0.01'''

''' AB= '''r2 = rts

The initial condition was set to r1 = rts+0.01 and r2 = rts''' '''''',''' thus the AB distance ('''r2''') =0.9075 Å, and the BC distance ('''r1''') = 0.9175 Å
According to Picture 2 and 3, the reaction path calculated by MEP method was more smooth than the path calculated by dynamic method. For Picture.2 (using dynamic method), the reaction path vibrated at the lowest energy. For the MEP method, the reaction pathway was more smooth and more fit to the lowest energy barrier.
The difference between two ways of calculation may be because that MEP corresponds to infinitely slow motion while the dynamic method doesn't. For the dynamic calculation, the molecules move in a relative fast speed and vibrate. This makes the energy calculation harder. Thus, for small initial momenta, MEP calculation was more fit to the surface plot.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/Å
|0.9175
|7.15
|3.75
|-
|r2/Å
|0.9075
|0.75
|0.76
|-
|p1
|0
|2.46
|0
|-
|p2
|0
|1.33
|0
|}
For the table below, the initial distance was switched for r1 and r2,

BC distance =r1 = rts

AB distance = r2 = rts +0.01

According to the table below, it can be seen that the final value for both distance and momenta switched as the initial distance switched, and this change does not depend on their calculation method.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/ Å
|0.9075
|0.74
|0.76
|-
|r2/ Å
|0.9175
|7.15
|3.75
|-
|p1
|0
|1.33
|0
|-
|p2
|0
|2.46
|0
|}

=== 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"
!p1
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reactants have enough kinetic energy to pass though the transition state and complete the reaction and it can be seen from the contour plot: the trajectory passes though the transition state and goes to the products. Thus, it is a productive reaction.[[File:MRD_R1_0521.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.2</nowiki>
|No
|The reactants do not have enough energy to pass through the transition state, thus the trajectory moves back to the reactants.[[File:MRD_R2_0521.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reagents have enough energy to pass through the transition state. Thus, the trajectory passes though the transition state and moves towards to the product. [[File:MRD_R3_0521.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.6</nowiki>
|No
|The reagents have enough energy topass through the transition state, however, the energy is too high thus it drops back to the reactants.[[File:MRD_R4_0521.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-82.6</nowiki>
|Yes
|The trajectory does not moves to product initially, the reagents have excess energy to pass through the transition state, thus it goes back to the reactants. After that, the reactants still have enough energy to pass through the transition state, thus the reaction occurs.[[File:MRD_R5_0521.png]]
|}
In conclusion, to complete a productive reaction, the reactants should have enough energy to pass through the transition state but not too high to reverse the product back into the reactants.

===''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?''===
'''Assumptions:'''[https://www.sciencedirect.com/topics/chemistry/transition-state-theory]<ref name = 'TS'/>
# Quantum-tunnelling effects are assumed negligible.
# The Born-Oppenheimer approximation is invoked.
# The atoms in the reactant state have energies that are Boltzmann distributed.
# Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.

== 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?''===
{| class="wikitable"
!Reaction
!Initial condition
!Potential surface plot and discription
|-
|F + H2 (ABC)
|AB distance = 1.00 Å
BC distance = 0.74 Å

AB momentum = 0
BC momentum = -1.5
|[[File:MDR_FHH_0522.png|thumb|left|A surface plot for F+ H2 reaction. ]] The reaction is exothermix as the energy for reactants are higher than the energy for products. Thus, the energy is released during the reaction. This is caused by the formation of strong H-F bond.
|-
|H +HF (ABC)
|AB distance = 1.00 Å
BC distance = 0.917 Å

AB momentum = 0
BC momentum = -10
|[[File:MRD_HHF_0522.png|thumb|left|A surface plot for H + HF reaction. ]] This reaction is endothermic as the potential energy of reactants are lower than the potential energy of products. In this case, the strong H-F bond was break and a weaker H-H bond was formed. Thud, the overall reaction was endothermic.
|}

===''Locate the approximate position of the transition state.''===
{| class="wikitable"
!Reaction (ABC)
!Position of Transition Point
!Contour plot
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_0522.png|thumb|Contour plot for for the transition point for F + H2.]]
|-
|H + HF
|AB distance = 0.745 Å
BC distance = 1.820 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_0522_1.png|thumb|Contour plot for for the transition point for H + HF.]]
|}

===''Report the activation energy for both reactions.''===
{| class="wikitable"
!Reaction (ABC)
!Initial condition
!Energy vs Time plot
!Activation energy
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_EA_0522.png|thumb|Energy vs Time plot for F + H2]]
|Ea = (-103.751) - (-104.002) = 0.251 kcal/mol
|-
|H + HF
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_EA_1_0522.png|thumb|Energy vs Time plot for H + HF]]
|Ea = (-101.187) - (-133.852) = 32.665 kcal/mol
|}

===''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.''===
{| class="wikitable"
!Reaction (ABC)
!Initial condition
!Graph
!Explanation
|-
|F + H2
|AB distance = 1.0 Å
BC distance = 0.74Å

AB momentum = -1.5

BC momentum = 0
|[[File:MRD_FHH_R1_0522.png |thumb|A contour plot for F + H2]]
[[File:MRD_FHH_E1_0522.png |thumb|An Energy vs Time plot for F + H2]]
|
|-
|H + HF
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|
|
|}

==Reference==
<references>
<ref name="TS">T.Bligaard, J.K.Nørskov, in Chemical Bonding at Surfaces and Interfaces, Elsevier, Amsterdam, 2008, pp 255-321.</ref>
</references>

File:MRD FHH E1 0522.png

2019-05-22T13:44:07Z

Xc2017:

File:MRD FHH R1 0522.png

2019-05-22T13:42:05Z

Xc2017:

MRD:xc2017

2019-05-22T13:40:26Z

Xc2017: /* 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. */

== 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?''===

'''Answer:''' The transition state is a saddle point in the potential energy surface diagram where ∂V(ri)/∂ri=0 and G<0 where G = frr(r0,r'0)fr'r'(r0,r'0)−f2rr'(r0,r'0).

The transition point was defined as a maximum point in the minimum energy path linking reactants and products. For triatomic reaction, the transition state can be identified in reaction path in the contour plot where the AB and BC distance are equal.

The local minimum point can be distinguished by secondary partial derivation equal to zero and G > 0.

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

[[File:MRD_XC_1.png|thumb|Picture.1 A snapshot of 'Internuclear Distance vs Time' plot at r1 = r2 = 0.9075Å]]

'''Answer:''' The best estimate transition position is r1=r2=0.9075 Å where the “Internuclear Distances vs Time” plot shows two straight lines (Picture.1). The picture of distance vs time plot shows the AB and BC distance are constant which indicated the transition state.

=== '' Comment on how the mep and the trajectory you just calculated differ.''===
[[File:MRD_DYN_0521.png|thumb|Picture.2 A snapshot of dynamic surcafe plot]]
[[File:MRD_MEP_0521.png|thumb|Picture.3 A snapshot of MEP surface plot]]
'''Answer:'''

''' BC = r1 = rts+0.01'''

''' AB= '''r2 = rts

The initial condition was set to r1 = rts+0.01 and r2 = rts''' '''''',''' thus the AB distance ('''r2''') =0.9075 Å, and the BC distance ('''r1''') = 0.9175 Å
According to Picture 2 and 3, the reaction path calculated by MEP method was more smooth than the path calculated by dynamic method. For Picture.2 (using dynamic method), the reaction path vibrated at the lowest energy. For the MEP method, the reaction pathway was more smooth and more fit to the lowest energy barrier.
The difference between two ways of calculation may be because that MEP corresponds to infinitely slow motion while the dynamic method doesn't. For the dynamic calculation, the molecules move in a relative fast speed and vibrate. This makes the energy calculation harder. Thus, for small initial momenta, MEP calculation was more fit to the surface plot.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/Å
|0.9175
|7.15
|3.75
|-
|r2/Å
|0.9075
|0.75
|0.76
|-
|p1
|0
|2.46
|0
|-
|p2
|0
|1.33
|0
|}
For the table below, the initial distance was switched for r1 and r2,

BC distance =r1 = rts

AB distance = r2 = rts +0.01

According to the table below, it can be seen that the final value for both distance and momenta switched as the initial distance switched, and this change does not depend on their calculation method.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/ Å
|0.9075
|0.74
|0.76
|-
|r2/ Å
|0.9175
|7.15
|3.75
|-
|p1
|0
|1.33
|0
|-
|p2
|0
|2.46
|0
|}

=== 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"
!p1
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reactants have enough kinetic energy to pass though the transition state and complete the reaction and it can be seen from the contour plot: the trajectory passes though the transition state and goes to the products. Thus, it is a productive reaction.[[File:MRD_R1_0521.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.2</nowiki>
|No
|The reactants do not have enough energy to pass through the transition state, thus the trajectory moves back to the reactants.[[File:MRD_R2_0521.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reagents have enough energy to pass through the transition state. Thus, the trajectory passes though the transition state and moves towards to the product. [[File:MRD_R3_0521.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.6</nowiki>
|No
|The reagents have enough energy topass through the transition state, however, the energy is too high thus it drops back to the reactants.[[File:MRD_R4_0521.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-82.6</nowiki>
|Yes
|The trajectory does not moves to product initially, the reagents have excess energy to pass through the transition state, thus it goes back to the reactants. After that, the reactants still have enough energy to pass through the transition state, thus the reaction occurs.[[File:MRD_R5_0521.png]]
|}
In conclusion, to complete a productive reaction, the reactants should have enough energy to pass through the transition state but not too high to reverse the product back into the reactants.

===''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?''===
'''Assumptions:'''[https://www.sciencedirect.com/topics/chemistry/transition-state-theory]<ref name = 'TS'/>
# Quantum-tunnelling effects are assumed negligible.
# The Born-Oppenheimer approximation is invoked.
# The atoms in the reactant state have energies that are Boltzmann distributed.
# Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.

== 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?''===
{| class="wikitable"
!Reaction
!Initial condition
!Potential surface plot and discription
|-
|F + H2 (ABC)
|AB distance = 1.00 Å
BC distance = 0.74 Å

AB momentum = 0
BC momentum = -1.5
|[[File:MDR_FHH_0522.png|thumb|left|A surface plot for F+ H2 reaction. ]] The reaction is exothermix as the energy for reactants are higher than the energy for products. Thus, the energy is released during the reaction. This is caused by the formation of strong H-F bond.
|-
|H +HF (ABC)
|AB distance = 1.00 Å
BC distance = 0.917 Å

AB momentum = 0
BC momentum = -10
|[[File:MRD_HHF_0522.png|thumb|left|A surface plot for H + HF reaction. ]] This reaction is endothermic as the potential energy of reactants are lower than the potential energy of products. In this case, the strong H-F bond was break and a weaker H-H bond was formed. Thud, the overall reaction was endothermic.
|}

===''Locate the approximate position of the transition state.''===
{| class="wikitable"
!Reaction (ABC)
!Position of Transition Point
!Contour plot
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_0522.png|thumb|Contour plot for for the transition point for F + H2.]]
|-
|H + HF
|AB distance = 0.745 Å
BC distance = 1.820 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_0522_1.png|thumb|Contour plot for for the transition point for H + HF.]]
|}

===''Report the activation energy for both reactions.''===
{| class="wikitable"
!Reaction (ABC)
!Initial condition
!Energy vs Time plot
!Activation energy
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_EA_0522.png|thumb|Energy vs Time plot for F + H2]]
|Ea = (-103.751) - (-104.002) = 0.251 kcal/mol
|-
|H + HF
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_EA_1_0522.png|thumb|Energy vs Time plot for H + HF]]
|Ea = (-101.187) - (-133.852) = 32.665 kcal/mol
|}

===''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.''===
{| class="wikitable"
!Reaction (ABC)
!Initial condition
!Graph
!Explanation
|-
|F + H2
|AB distance = 1.0 Å
BC distance = 0.74Å

AB momentum = -1.5

BC momentum = 0
|
|
|-
|H + HF
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|
|
|}

==Reference==
<references>
<ref name="TS">T.Bligaard, J.K.Nørskov, in Chemical Bonding at Surfaces and Interfaces, Elsevier, Amsterdam, 2008, pp 255-321.</ref>
</references>

MRD:xc2017

2019-05-22T13:25:10Z

Xc2017: /* Report the activation energy for both reactions. */

== 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?''===

'''Answer:''' The transition state is a saddle point in the potential energy surface diagram where ∂V(ri)/∂ri=0 and G<0 where G = frr(r0,r'0)fr'r'(r0,r'0)−f2rr'(r0,r'0).

The transition point was defined as a maximum point in the minimum energy path linking reactants and products. For triatomic reaction, the transition state can be identified in reaction path in the contour plot where the AB and BC distance are equal.

The local minimum point can be distinguished by secondary partial derivation equal to zero and G > 0.

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

[[File:MRD_XC_1.png|thumb|Picture.1 A snapshot of 'Internuclear Distance vs Time' plot at r1 = r2 = 0.9075Å]]

'''Answer:''' The best estimate transition position is r1=r2=0.9075 Å where the “Internuclear Distances vs Time” plot shows two straight lines (Picture.1). The picture of distance vs time plot shows the AB and BC distance are constant which indicated the transition state.

=== '' Comment on how the mep and the trajectory you just calculated differ.''===
[[File:MRD_DYN_0521.png|thumb|Picture.2 A snapshot of dynamic surcafe plot]]
[[File:MRD_MEP_0521.png|thumb|Picture.3 A snapshot of MEP surface plot]]
'''Answer:'''

''' BC = r1 = rts+0.01'''

''' AB= '''r2 = rts

The initial condition was set to r1 = rts+0.01 and r2 = rts''' '''''',''' thus the AB distance ('''r2''') =0.9075 Å, and the BC distance ('''r1''') = 0.9175 Å
According to Picture 2 and 3, the reaction path calculated by MEP method was more smooth than the path calculated by dynamic method. For Picture.2 (using dynamic method), the reaction path vibrated at the lowest energy. For the MEP method, the reaction pathway was more smooth and more fit to the lowest energy barrier.
The difference between two ways of calculation may be because that MEP corresponds to infinitely slow motion while the dynamic method doesn't. For the dynamic calculation, the molecules move in a relative fast speed and vibrate. This makes the energy calculation harder. Thus, for small initial momenta, MEP calculation was more fit to the surface plot.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/Å
|0.9175
|7.15
|3.75
|-
|r2/Å
|0.9075
|0.75
|0.76
|-
|p1
|0
|2.46
|0
|-
|p2
|0
|1.33
|0
|}
For the table below, the initial distance was switched for r1 and r2,

BC distance =r1 = rts

AB distance = r2 = rts +0.01

According to the table below, it can be seen that the final value for both distance and momenta switched as the initial distance switched, and this change does not depend on their calculation method.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/ Å
|0.9075
|0.74
|0.76
|-
|r2/ Å
|0.9175
|7.15
|3.75
|-
|p1
|0
|1.33
|0
|-
|p2
|0
|2.46
|0
|}

=== 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"
!p1
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reactants have enough kinetic energy to pass though the transition state and complete the reaction and it can be seen from the contour plot: the trajectory passes though the transition state and goes to the products. Thus, it is a productive reaction.[[File:MRD_R1_0521.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.2</nowiki>
|No
|The reactants do not have enough energy to pass through the transition state, thus the trajectory moves back to the reactants.[[File:MRD_R2_0521.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reagents have enough energy to pass through the transition state. Thus, the trajectory passes though the transition state and moves towards to the product. [[File:MRD_R3_0521.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.6</nowiki>
|No
|The reagents have enough energy topass through the transition state, however, the energy is too high thus it drops back to the reactants.[[File:MRD_R4_0521.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-82.6</nowiki>
|Yes
|The trajectory does not moves to product initially, the reagents have excess energy to pass through the transition state, thus it goes back to the reactants. After that, the reactants still have enough energy to pass through the transition state, thus the reaction occurs.[[File:MRD_R5_0521.png]]
|}
In conclusion, to complete a productive reaction, the reactants should have enough energy to pass through the transition state but not too high to reverse the product back into the reactants.

===''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?''===
'''Assumptions:'''[https://www.sciencedirect.com/topics/chemistry/transition-state-theory]<ref name = 'TS'/>
# Quantum-tunnelling effects are assumed negligible.
# The Born-Oppenheimer approximation is invoked.
# The atoms in the reactant state have energies that are Boltzmann distributed.
# Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.

== 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?''===
{| class="wikitable"
!Reaction
!Initial condition
!Potential surface plot and discription
|-
|F + H2 (ABC)
|AB distance = 1.00 Å
BC distance = 0.74 Å

AB momentum = 0
BC momentum = -1.5
|[[File:MDR_FHH_0522.png|thumb|left|A surface plot for F+ H2 reaction. ]] The reaction is exothermix as the energy for reactants are higher than the energy for products. Thus, the energy is released during the reaction. This is caused by the formation of strong H-F bond.
|-
|H +HF (ABC)
|AB distance = 1.00 Å
BC distance = 0.017 Å

AB momentum = 0
BC momentum = -10
|[[File:MRD_HHF_0522.png|thumb|left|A surface plot for H + HF reaction. ]] This reaction is endothermic as the potential energy of reactants are lower than the potential energy of products. In this case, the strong H-F bond was break and a weaker H-H bond was formed. Thud, the overall reaction was endothermic.
|}

===''Locate the approximate position of the transition state.''===
{| class="wikitable"
!Reaction
!Position of Transition Point
!Contour plot
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_0522.png|thumb|Contour plot for for the transition point for F + H2.]]
|-
|H + HF
|AB distance = 0.745 Å
BC distance = 1.820 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_0522_1.png|thumb|Contour plot for for the transition point for H + HF.]]
|}

===''Report the activation energy for both reactions.''===
{| class="wikitable"
!Reaction
!Initial condition
!Energy vs Time plot
!Activation energy
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_EA_0522.png|thumb|Energy vs Time plot for F + H2]]
|Ea = (-103.751) - (-104.002) = 0.251 kcal/mol
|-
|H + HF
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_EA_1_0522.png|thumb|Energy vs Time plot for H + HF]]
|Ea = (-101.187) - (-133.852) = 32.665 kcal/mol
|}

===''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.''===

==Reference==
<references>
<ref name="TS">T.Bligaard, J.K.Nørskov, in Chemical Bonding at Surfaces and Interfaces, Elsevier, Amsterdam, 2008, pp 255-321.</ref>
</references>

MRD:xc2017

2019-05-22T13:24:01Z

Xc2017: /* Report the activation energy for both reactions. */

== 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?''===

'''Answer:''' The transition state is a saddle point in the potential energy surface diagram where ∂V(ri)/∂ri=0 and G<0 where G = frr(r0,r'0)fr'r'(r0,r'0)−f2rr'(r0,r'0).

The transition point was defined as a maximum point in the minimum energy path linking reactants and products. For triatomic reaction, the transition state can be identified in reaction path in the contour plot where the AB and BC distance are equal.

The local minimum point can be distinguished by secondary partial derivation equal to zero and G > 0.

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

[[File:MRD_XC_1.png|thumb|Picture.1 A snapshot of 'Internuclear Distance vs Time' plot at r1 = r2 = 0.9075Å]]

'''Answer:''' The best estimate transition position is r1=r2=0.9075 Å where the “Internuclear Distances vs Time” plot shows two straight lines (Picture.1). The picture of distance vs time plot shows the AB and BC distance are constant which indicated the transition state.

=== '' Comment on how the mep and the trajectory you just calculated differ.''===
[[File:MRD_DYN_0521.png|thumb|Picture.2 A snapshot of dynamic surcafe plot]]
[[File:MRD_MEP_0521.png|thumb|Picture.3 A snapshot of MEP surface plot]]
'''Answer:'''

''' BC = r1 = rts+0.01'''

''' AB= '''r2 = rts

The initial condition was set to r1 = rts+0.01 and r2 = rts''' '''''',''' thus the AB distance ('''r2''') =0.9075 Å, and the BC distance ('''r1''') = 0.9175 Å
According to Picture 2 and 3, the reaction path calculated by MEP method was more smooth than the path calculated by dynamic method. For Picture.2 (using dynamic method), the reaction path vibrated at the lowest energy. For the MEP method, the reaction pathway was more smooth and more fit to the lowest energy barrier.
The difference between two ways of calculation may be because that MEP corresponds to infinitely slow motion while the dynamic method doesn't. For the dynamic calculation, the molecules move in a relative fast speed and vibrate. This makes the energy calculation harder. Thus, for small initial momenta, MEP calculation was more fit to the surface plot.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/Å
|0.9175
|7.15
|3.75
|-
|r2/Å
|0.9075
|0.75
|0.76
|-
|p1
|0
|2.46
|0
|-
|p2
|0
|1.33
|0
|}
For the table below, the initial distance was switched for r1 and r2,

BC distance =r1 = rts

AB distance = r2 = rts +0.01

According to the table below, it can be seen that the final value for both distance and momenta switched as the initial distance switched, and this change does not depend on their calculation method.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/ Å
|0.9075
|0.74
|0.76
|-
|r2/ Å
|0.9175
|7.15
|3.75
|-
|p1
|0
|1.33
|0
|-
|p2
|0
|2.46
|0
|}

=== 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"
!p1
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reactants have enough kinetic energy to pass though the transition state and complete the reaction and it can be seen from the contour plot: the trajectory passes though the transition state and goes to the products. Thus, it is a productive reaction.[[File:MRD_R1_0521.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.2</nowiki>
|No
|The reactants do not have enough energy to pass through the transition state, thus the trajectory moves back to the reactants.[[File:MRD_R2_0521.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reagents have enough energy to pass through the transition state. Thus, the trajectory passes though the transition state and moves towards to the product. [[File:MRD_R3_0521.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.6</nowiki>
|No
|The reagents have enough energy topass through the transition state, however, the energy is too high thus it drops back to the reactants.[[File:MRD_R4_0521.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-82.6</nowiki>
|Yes
|The trajectory does not moves to product initially, the reagents have excess energy to pass through the transition state, thus it goes back to the reactants. After that, the reactants still have enough energy to pass through the transition state, thus the reaction occurs.[[File:MRD_R5_0521.png]]
|}
In conclusion, to complete a productive reaction, the reactants should have enough energy to pass through the transition state but not too high to reverse the product back into the reactants.

===''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?''===
'''Assumptions:'''[https://www.sciencedirect.com/topics/chemistry/transition-state-theory]<ref name = 'TS'/>
# Quantum-tunnelling effects are assumed negligible.
# The Born-Oppenheimer approximation is invoked.
# The atoms in the reactant state have energies that are Boltzmann distributed.
# Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.

== 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?''===
{| class="wikitable"
!Reaction
!Initial condition
!Potential surface plot and discription
|-
|F + H2 (ABC)
|AB distance = 1.00 Å
BC distance = 0.74 Å

AB momentum = 0
BC momentum = -1.5
|[[File:MDR_FHH_0522.png|thumb|left|A surface plot for F+ H2 reaction. ]] The reaction is exothermix as the energy for reactants are higher than the energy for products. Thus, the energy is released during the reaction. This is caused by the formation of strong H-F bond.
|-
|H +HF (ABC)
|AB distance = 1.00 Å
BC distance = 0.017 Å

AB momentum = 0
BC momentum = -10
|[[File:MRD_HHF_0522.png|thumb|left|A surface plot for H + HF reaction. ]] This reaction is endothermic as the potential energy of reactants are lower than the potential energy of products. In this case, the strong H-F bond was break and a weaker H-H bond was formed. Thud, the overall reaction was endothermic.
|}

===''Locate the approximate position of the transition state.''===
{| class="wikitable"
!Reaction
!Position of Transition Point
!Contour plot
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_0522.png|thumb|Contour plot for for the transition point for F + H2.]]
|-
|H + HF
|AB distance = 0.745 Å
BC distance = 1.820 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_0522_1.png|thumb|Contour plot for for the transition point for H + HF.]]
|}

===''Report the activation energy for both reactions.''===
{| class="wikitable"
!Reaction
!Initial condition
!Energy vs Time plot
!Activation energy
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_EA_0522.png|thumb|Energy vs Time plot for F + H2]]
|Ea = (-103.751) - (-104.002) = 0.251 kcal/mol
|-
|H + HF
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_EA_1_0522.png|thumb|Energy vs Time plot for H + HF]]
|Ea = (-101.187) - (-133.852) = 32.665 kcal/mol
|}

==Reference==
<references>
<ref name="TS">T.Bligaard, J.K.Nørskov, in Chemical Bonding at Surfaces and Interfaces, Elsevier, Amsterdam, 2008, pp 255-321.</ref>
</references>

File:MRD HHF EA 1 0522.png

2019-05-22T13:23:10Z

Xc2017:

MRD:xc2017

2019-05-22T13:22:58Z

Xc2017: /* Report the activation energy for both reactions. */

== 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?''===

'''Answer:''' The transition state is a saddle point in the potential energy surface diagram where ∂V(ri)/∂ri=0 and G<0 where G = frr(r0,r'0)fr'r'(r0,r'0)−f2rr'(r0,r'0).

The transition point was defined as a maximum point in the minimum energy path linking reactants and products. For triatomic reaction, the transition state can be identified in reaction path in the contour plot where the AB and BC distance are equal.

The local minimum point can be distinguished by secondary partial derivation equal to zero and G > 0.

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

[[File:MRD_XC_1.png|thumb|Picture.1 A snapshot of 'Internuclear Distance vs Time' plot at r1 = r2 = 0.9075Å]]

'''Answer:''' The best estimate transition position is r1=r2=0.9075 Å where the “Internuclear Distances vs Time” plot shows two straight lines (Picture.1). The picture of distance vs time plot shows the AB and BC distance are constant which indicated the transition state.

=== '' Comment on how the mep and the trajectory you just calculated differ.''===
[[File:MRD_DYN_0521.png|thumb|Picture.2 A snapshot of dynamic surcafe plot]]
[[File:MRD_MEP_0521.png|thumb|Picture.3 A snapshot of MEP surface plot]]
'''Answer:'''

''' BC = r1 = rts+0.01'''

''' AB= '''r2 = rts

The initial condition was set to r1 = rts+0.01 and r2 = rts''' '''''',''' thus the AB distance ('''r2''') =0.9075 Å, and the BC distance ('''r1''') = 0.9175 Å
According to Picture 2 and 3, the reaction path calculated by MEP method was more smooth than the path calculated by dynamic method. For Picture.2 (using dynamic method), the reaction path vibrated at the lowest energy. For the MEP method, the reaction pathway was more smooth and more fit to the lowest energy barrier.
The difference between two ways of calculation may be because that MEP corresponds to infinitely slow motion while the dynamic method doesn't. For the dynamic calculation, the molecules move in a relative fast speed and vibrate. This makes the energy calculation harder. Thus, for small initial momenta, MEP calculation was more fit to the surface plot.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/Å
|0.9175
|7.15
|3.75
|-
|r2/Å
|0.9075
|0.75
|0.76
|-
|p1
|0
|2.46
|0
|-
|p2
|0
|1.33
|0
|}
For the table below, the initial distance was switched for r1 and r2,

BC distance =r1 = rts

AB distance = r2 = rts +0.01

According to the table below, it can be seen that the final value for both distance and momenta switched as the initial distance switched, and this change does not depend on their calculation method.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/ Å
|0.9075
|0.74
|0.76
|-
|r2/ Å
|0.9175
|7.15
|3.75
|-
|p1
|0
|1.33
|0
|-
|p2
|0
|2.46
|0
|}

=== 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"
!p1
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reactants have enough kinetic energy to pass though the transition state and complete the reaction and it can be seen from the contour plot: the trajectory passes though the transition state and goes to the products. Thus, it is a productive reaction.[[File:MRD_R1_0521.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.2</nowiki>
|No
|The reactants do not have enough energy to pass through the transition state, thus the trajectory moves back to the reactants.[[File:MRD_R2_0521.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reagents have enough energy to pass through the transition state. Thus, the trajectory passes though the transition state and moves towards to the product. [[File:MRD_R3_0521.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.6</nowiki>
|No
|The reagents have enough energy topass through the transition state, however, the energy is too high thus it drops back to the reactants.[[File:MRD_R4_0521.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-82.6</nowiki>
|Yes
|The trajectory does not moves to product initially, the reagents have excess energy to pass through the transition state, thus it goes back to the reactants. After that, the reactants still have enough energy to pass through the transition state, thus the reaction occurs.[[File:MRD_R5_0521.png]]
|}
In conclusion, to complete a productive reaction, the reactants should have enough energy to pass through the transition state but not too high to reverse the product back into the reactants.

===''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?''===
'''Assumptions:'''[https://www.sciencedirect.com/topics/chemistry/transition-state-theory]<ref name = 'TS'/>
# Quantum-tunnelling effects are assumed negligible.
# The Born-Oppenheimer approximation is invoked.
# The atoms in the reactant state have energies that are Boltzmann distributed.
# Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.

== 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?''===
{| class="wikitable"
!Reaction
!Initial condition
!Potential surface plot and discription
|-
|F + H2 (ABC)
|AB distance = 1.00 Å
BC distance = 0.74 Å

AB momentum = 0
BC momentum = -1.5
|[[File:MDR_FHH_0522.png|thumb|left|A surface plot for F+ H2 reaction. ]] The reaction is exothermix as the energy for reactants are higher than the energy for products. Thus, the energy is released during the reaction. This is caused by the formation of strong H-F bond.
|-
|H +HF (ABC)
|AB distance = 1.00 Å
BC distance = 0.017 Å

AB momentum = 0
BC momentum = -10
|[[File:MRD_HHF_0522.png|thumb|left|A surface plot for H + HF reaction. ]] This reaction is endothermic as the potential energy of reactants are lower than the potential energy of products. In this case, the strong H-F bond was break and a weaker H-H bond was formed. Thud, the overall reaction was endothermic.
|}

===''Locate the approximate position of the transition state.''===
{| class="wikitable"
!Reaction
!Position of Transition Point
!Contour plot
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_0522.png|thumb|Contour plot for for the transition point for F + H2.]]
|-
|H + HF
|AB distance = 0.745 Å
BC distance = 1.820 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_0522_1.png|thumb|Contour plot for for the transition point for H + HF.]]
|}

===''Report the activation energy for both reactions.''===
{| class="wikitable"
!Reaction
!Initial condition
!Energy vs Time plot
!Activation energy
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_EA_0522.png|thumb|Energy vs Time plot for F + H2]]
|Ea = (-103.751) - (-104.002) = 0.251 kcal/mol
|-
|H + HF
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|
|Ea = (-101.187) - (-133.852) = 32.665 kcal/mol
|}

==Reference==
<references>
<ref name="TS">T.Bligaard, J.K.Nørskov, in Chemical Bonding at Surfaces and Interfaces, Elsevier, Amsterdam, 2008, pp 255-321.</ref>
</references>

MRD:xc2017

2019-05-22T13:20:04Z

Xc2017: /* Report the activation energy for both reactions. */

== 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?''===

'''Answer:''' The transition state is a saddle point in the potential energy surface diagram where ∂V(ri)/∂ri=0 and G<0 where G = frr(r0,r'0)fr'r'(r0,r'0)−f2rr'(r0,r'0).

The transition point was defined as a maximum point in the minimum energy path linking reactants and products. For triatomic reaction, the transition state can be identified in reaction path in the contour plot where the AB and BC distance are equal.

The local minimum point can be distinguished by secondary partial derivation equal to zero and G > 0.

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

[[File:MRD_XC_1.png|thumb|Picture.1 A snapshot of 'Internuclear Distance vs Time' plot at r1 = r2 = 0.9075Å]]

'''Answer:''' The best estimate transition position is r1=r2=0.9075 Å where the “Internuclear Distances vs Time” plot shows two straight lines (Picture.1). The picture of distance vs time plot shows the AB and BC distance are constant which indicated the transition state.

=== '' Comment on how the mep and the trajectory you just calculated differ.''===
[[File:MRD_DYN_0521.png|thumb|Picture.2 A snapshot of dynamic surcafe plot]]
[[File:MRD_MEP_0521.png|thumb|Picture.3 A snapshot of MEP surface plot]]
'''Answer:'''

''' BC = r1 = rts+0.01'''

''' AB= '''r2 = rts

The initial condition was set to r1 = rts+0.01 and r2 = rts''' '''''',''' thus the AB distance ('''r2''') =0.9075 Å, and the BC distance ('''r1''') = 0.9175 Å
According to Picture 2 and 3, the reaction path calculated by MEP method was more smooth than the path calculated by dynamic method. For Picture.2 (using dynamic method), the reaction path vibrated at the lowest energy. For the MEP method, the reaction pathway was more smooth and more fit to the lowest energy barrier.
The difference between two ways of calculation may be because that MEP corresponds to infinitely slow motion while the dynamic method doesn't. For the dynamic calculation, the molecules move in a relative fast speed and vibrate. This makes the energy calculation harder. Thus, for small initial momenta, MEP calculation was more fit to the surface plot.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/Å
|0.9175
|7.15
|3.75
|-
|r2/Å
|0.9075
|0.75
|0.76
|-
|p1
|0
|2.46
|0
|-
|p2
|0
|1.33
|0
|}
For the table below, the initial distance was switched for r1 and r2,

BC distance =r1 = rts

AB distance = r2 = rts +0.01

According to the table below, it can be seen that the final value for both distance and momenta switched as the initial distance switched, and this change does not depend on their calculation method.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/ Å
|0.9075
|0.74
|0.76
|-
|r2/ Å
|0.9175
|7.15
|3.75
|-
|p1
|0
|1.33
|0
|-
|p2
|0
|2.46
|0
|}

=== 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"
!p1
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reactants have enough kinetic energy to pass though the transition state and complete the reaction and it can be seen from the contour plot: the trajectory passes though the transition state and goes to the products. Thus, it is a productive reaction.[[File:MRD_R1_0521.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.2</nowiki>
|No
|The reactants do not have enough energy to pass through the transition state, thus the trajectory moves back to the reactants.[[File:MRD_R2_0521.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reagents have enough energy to pass through the transition state. Thus, the trajectory passes though the transition state and moves towards to the product. [[File:MRD_R3_0521.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.6</nowiki>
|No
|The reagents have enough energy topass through the transition state, however, the energy is too high thus it drops back to the reactants.[[File:MRD_R4_0521.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-82.6</nowiki>
|Yes
|The trajectory does not moves to product initially, the reagents have excess energy to pass through the transition state, thus it goes back to the reactants. After that, the reactants still have enough energy to pass through the transition state, thus the reaction occurs.[[File:MRD_R5_0521.png]]
|}
In conclusion, to complete a productive reaction, the reactants should have enough energy to pass through the transition state but not too high to reverse the product back into the reactants.

===''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?''===
'''Assumptions:'''[https://www.sciencedirect.com/topics/chemistry/transition-state-theory]<ref name = 'TS'/>
# Quantum-tunnelling effects are assumed negligible.
# The Born-Oppenheimer approximation is invoked.
# The atoms in the reactant state have energies that are Boltzmann distributed.
# Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.

== 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?''===
{| class="wikitable"
!Reaction
!Initial condition
!Potential surface plot and discription
|-
|F + H2 (ABC)
|AB distance = 1.00 Å
BC distance = 0.74 Å

AB momentum = 0
BC momentum = -1.5
|[[File:MDR_FHH_0522.png|thumb|left|A surface plot for F+ H2 reaction. ]] The reaction is exothermix as the energy for reactants are higher than the energy for products. Thus, the energy is released during the reaction. This is caused by the formation of strong H-F bond.
|-
|H +HF (ABC)
|AB distance = 1.00 Å
BC distance = 0.017 Å

AB momentum = 0
BC momentum = -10
|[[File:MRD_HHF_0522.png|thumb|left|A surface plot for H + HF reaction. ]] This reaction is endothermic as the potential energy of reactants are lower than the potential energy of products. In this case, the strong H-F bond was break and a weaker H-H bond was formed. Thud, the overall reaction was endothermic.
|}

===''Locate the approximate position of the transition state.''===
{| class="wikitable"
!Reaction
!Position of Transition Point
!Contour plot
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_0522.png|thumb|Contour plot for for the transition point for F + H2.]]
|-
|H + HF
|AB distance = 0.745 Å
BC distance = 1.820 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_0522_1.png|thumb|Contour plot for for the transition point for H + HF.]]
|}

===''Report the activation energy for both reactions.''===
{| class="wikitable"
!Reaction
!Initial condition
!Energy vs Time plot
!Activation energy
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|
|Ea = (-103.751) - (-104.002) = 0.251 kcal/mol
|-
|H + HF
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|
|Ea = (-101.187) - (-133.852) = 32.665 kcal/mol
|}

==Reference==
<references>
<ref name="TS">T.Bligaard, J.K.Nørskov, in Chemical Bonding at Surfaces and Interfaces, Elsevier, Amsterdam, 2008, pp 255-321.</ref>
</references>

File:MRD FHH EA 0522.png

2019-05-22T13:08:17Z

Xc2017:

MRD:xc2017

2019-05-22T12:45:26Z

Xc2017: /* Locate the approximate position of the transition state. */

== 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?''===

'''Answer:''' The transition state is a saddle point in the potential energy surface diagram where ∂V(ri)/∂ri=0 and G<0 where G = frr(r0,r'0)fr'r'(r0,r'0)−f2rr'(r0,r'0).

The transition point was defined as a maximum point in the minimum energy path linking reactants and products. For triatomic reaction, the transition state can be identified in reaction path in the contour plot where the AB and BC distance are equal.

The local minimum point can be distinguished by secondary partial derivation equal to zero and G > 0.

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

[[File:MRD_XC_1.png|thumb|Picture.1 A snapshot of 'Internuclear Distance vs Time' plot at r1 = r2 = 0.9075Å]]

'''Answer:''' The best estimate transition position is r1=r2=0.9075 Å where the “Internuclear Distances vs Time” plot shows two straight lines (Picture.1). The picture of distance vs time plot shows the AB and BC distance are constant which indicated the transition state.

=== '' Comment on how the mep and the trajectory you just calculated differ.''===
[[File:MRD_DYN_0521.png|thumb|Picture.2 A snapshot of dynamic surcafe plot]]
[[File:MRD_MEP_0521.png|thumb|Picture.3 A snapshot of MEP surface plot]]
'''Answer:'''

''' BC = r1 = rts+0.01'''

''' AB= '''r2 = rts

The initial condition was set to r1 = rts+0.01 and r2 = rts''' '''''',''' thus the AB distance ('''r2''') =0.9075 Å, and the BC distance ('''r1''') = 0.9175 Å
According to Picture 2 and 3, the reaction path calculated by MEP method was more smooth than the path calculated by dynamic method. For Picture.2 (using dynamic method), the reaction path vibrated at the lowest energy. For the MEP method, the reaction pathway was more smooth and more fit to the lowest energy barrier.
The difference between two ways of calculation may be because that MEP corresponds to infinitely slow motion while the dynamic method doesn't. For the dynamic calculation, the molecules move in a relative fast speed and vibrate. This makes the energy calculation harder. Thus, for small initial momenta, MEP calculation was more fit to the surface plot.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/Å
|0.9175
|7.15
|3.75
|-
|r2/Å
|0.9075
|0.75
|0.76
|-
|p1
|0
|2.46
|0
|-
|p2
|0
|1.33
|0
|}
For the table below, the initial distance was switched for r1 and r2,

BC distance =r1 = rts

AB distance = r2 = rts +0.01

According to the table below, it can be seen that the final value for both distance and momenta switched as the initial distance switched, and this change does not depend on their calculation method.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/ Å
|0.9075
|0.74
|0.76
|-
|r2/ Å
|0.9175
|7.15
|3.75
|-
|p1
|0
|1.33
|0
|-
|p2
|0
|2.46
|0
|}

=== 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"
!p1
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reactants have enough kinetic energy to pass though the transition state and complete the reaction and it can be seen from the contour plot: the trajectory passes though the transition state and goes to the products. Thus, it is a productive reaction.[[File:MRD_R1_0521.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.2</nowiki>
|No
|The reactants do not have enough energy to pass through the transition state, thus the trajectory moves back to the reactants.[[File:MRD_R2_0521.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reagents have enough energy to pass through the transition state. Thus, the trajectory passes though the transition state and moves towards to the product. [[File:MRD_R3_0521.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.6</nowiki>
|No
|The reagents have enough energy topass through the transition state, however, the energy is too high thus it drops back to the reactants.[[File:MRD_R4_0521.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-82.6</nowiki>
|Yes
|The trajectory does not moves to product initially, the reagents have excess energy to pass through the transition state, thus it goes back to the reactants. After that, the reactants still have enough energy to pass through the transition state, thus the reaction occurs.[[File:MRD_R5_0521.png]]
|}
In conclusion, to complete a productive reaction, the reactants should have enough energy to pass through the transition state but not too high to reverse the product back into the reactants.

===''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?''===
'''Assumptions:'''[https://www.sciencedirect.com/topics/chemistry/transition-state-theory]<ref name = 'TS'/>
# Quantum-tunnelling effects are assumed negligible.
# The Born-Oppenheimer approximation is invoked.
# The atoms in the reactant state have energies that are Boltzmann distributed.
# Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.

== 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?''===
{| class="wikitable"
!Reaction
!Initial condition
!Potential surface plot and discription
|-
|F + H2 (ABC)
|AB distance = 1.00 Å
BC distance = 0.74 Å

AB momentum = 0
BC momentum = -1.5
|[[File:MDR_FHH_0522.png|thumb|left|A surface plot for F+ H2 reaction. ]] The reaction is exothermix as the energy for reactants are higher than the energy for products. Thus, the energy is released during the reaction. This is caused by the formation of strong H-F bond.
|-
|H +HF (ABC)
|AB distance = 1.00 Å
BC distance = 0.017 Å

AB momentum = 0
BC momentum = -10
|[[File:MRD_HHF_0522.png|thumb|left|A surface plot for H + HF reaction. ]] This reaction is endothermic as the potential energy of reactants are lower than the potential energy of products. In this case, the strong H-F bond was break and a weaker H-H bond was formed. Thud, the overall reaction was endothermic.
|}

===''Locate the approximate position of the transition state.''===
{| class="wikitable"
!Reaction
!Position of Transition Point
!Contour plot
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_0522.png|thumb|Contour plot for for the transition point for F + H2.]]
|-
|H + HF
|AB distance = 0.745 Å
BC distance = 1.820 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_0522_1.png|thumb|Contour plot for for the transition point for H + HF.]]
|}

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

==Reference==
<references>
<ref name="TS">T.Bligaard, J.K.Nørskov, in Chemical Bonding at Surfaces and Interfaces, Elsevier, Amsterdam, 2008, pp 255-321.</ref>
</references>

MRD:xc2017

2019-05-22T12:44:45Z

Xc2017: /* Locate the approximate position of the transition state. */

== 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?''===

'''Answer:''' The transition state is a saddle point in the potential energy surface diagram where ∂V(ri)/∂ri=0 and G<0 where G = frr(r0,r'0)fr'r'(r0,r'0)−f2rr'(r0,r'0).

The transition point was defined as a maximum point in the minimum energy path linking reactants and products. For triatomic reaction, the transition state can be identified in reaction path in the contour plot where the AB and BC distance are equal.

The local minimum point can be distinguished by secondary partial derivation equal to zero and G > 0.

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

[[File:MRD_XC_1.png|thumb|Picture.1 A snapshot of 'Internuclear Distance vs Time' plot at r1 = r2 = 0.9075Å]]

'''Answer:''' The best estimate transition position is r1=r2=0.9075 Å where the “Internuclear Distances vs Time” plot shows two straight lines (Picture.1). The picture of distance vs time plot shows the AB and BC distance are constant which indicated the transition state.

=== '' Comment on how the mep and the trajectory you just calculated differ.''===
[[File:MRD_DYN_0521.png|thumb|Picture.2 A snapshot of dynamic surcafe plot]]
[[File:MRD_MEP_0521.png|thumb|Picture.3 A snapshot of MEP surface plot]]
'''Answer:'''

''' BC = r1 = rts+0.01'''

''' AB= '''r2 = rts

The initial condition was set to r1 = rts+0.01 and r2 = rts''' '''''',''' thus the AB distance ('''r2''') =0.9075 Å, and the BC distance ('''r1''') = 0.9175 Å
According to Picture 2 and 3, the reaction path calculated by MEP method was more smooth than the path calculated by dynamic method. For Picture.2 (using dynamic method), the reaction path vibrated at the lowest energy. For the MEP method, the reaction pathway was more smooth and more fit to the lowest energy barrier.
The difference between two ways of calculation may be because that MEP corresponds to infinitely slow motion while the dynamic method doesn't. For the dynamic calculation, the molecules move in a relative fast speed and vibrate. This makes the energy calculation harder. Thus, for small initial momenta, MEP calculation was more fit to the surface plot.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/Å
|0.9175
|7.15
|3.75
|-
|r2/Å
|0.9075
|0.75
|0.76
|-
|p1
|0
|2.46
|0
|-
|p2
|0
|1.33
|0
|}
For the table below, the initial distance was switched for r1 and r2,

BC distance =r1 = rts

AB distance = r2 = rts +0.01

According to the table below, it can be seen that the final value for both distance and momenta switched as the initial distance switched, and this change does not depend on their calculation method.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/ Å
|0.9075
|0.74
|0.76
|-
|r2/ Å
|0.9175
|7.15
|3.75
|-
|p1
|0
|1.33
|0
|-
|p2
|0
|2.46
|0
|}

=== 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"
!p1
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reactants have enough kinetic energy to pass though the transition state and complete the reaction and it can be seen from the contour plot: the trajectory passes though the transition state and goes to the products. Thus, it is a productive reaction.[[File:MRD_R1_0521.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.2</nowiki>
|No
|The reactants do not have enough energy to pass through the transition state, thus the trajectory moves back to the reactants.[[File:MRD_R2_0521.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reagents have enough energy to pass through the transition state. Thus, the trajectory passes though the transition state and moves towards to the product. [[File:MRD_R3_0521.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.6</nowiki>
|No
|The reagents have enough energy topass through the transition state, however, the energy is too high thus it drops back to the reactants.[[File:MRD_R4_0521.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-82.6</nowiki>
|Yes
|The trajectory does not moves to product initially, the reagents have excess energy to pass through the transition state, thus it goes back to the reactants. After that, the reactants still have enough energy to pass through the transition state, thus the reaction occurs.[[File:MRD_R5_0521.png]]
|}
In conclusion, to complete a productive reaction, the reactants should have enough energy to pass through the transition state but not too high to reverse the product back into the reactants.

===''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?''===
'''Assumptions:'''[https://www.sciencedirect.com/topics/chemistry/transition-state-theory]<ref name = 'TS'/>
# Quantum-tunnelling effects are assumed negligible.
# The Born-Oppenheimer approximation is invoked.
# The atoms in the reactant state have energies that are Boltzmann distributed.
# Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.

== 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?''===
{| class="wikitable"
!Reaction
!Initial condition
!Potential surface plot and discription
|-
|F + H2 (ABC)
|AB distance = 1.00 Å
BC distance = 0.74 Å

AB momentum = 0
BC momentum = -1.5
|[[File:MDR_FHH_0522.png|thumb|left|A surface plot for F+ H2 reaction. ]] The reaction is exothermix as the energy for reactants are higher than the energy for products. Thus, the energy is released during the reaction. This is caused by the formation of strong H-F bond.
|-
|H +HF (ABC)
|AB distance = 1.00 Å
BC distance = 0.017 Å

AB momentum = 0
BC momentum = -10
|[[File:MRD_HHF_0522.png|thumb|left|A surface plot for H + HF reaction. ]] This reaction is endothermic as the potential energy of reactants are lower than the potential energy of products. In this case, the strong H-F bond was break and a weaker H-H bond was formed. Thud, the overall reaction was endothermic.
|}

===''Locate the approximate position of the transition state.''===
{| class="wikitable"
!Reaction
!Position of Transition Point
!Contour plot
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_0522.png|thumb|Contour plot for for the transition point for F + H2.]]
|-
|H + HF
|AB distance = 0.745 Å
BC distance = 1.820 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_0522_1.png|thumb|Contour plot for for the transition point for H + HF.]]
|}

==Reference==
<references>
<ref name="TS">T.Bligaard, J.K.Nørskov, in Chemical Bonding at Surfaces and Interfaces, Elsevier, Amsterdam, 2008, pp 255-321.</ref>
</references>

MRD:xc2017

2019-05-22T12:44:13Z

Xc2017: /* Locate the approximate position of the transition state. */

== 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?''===

'''Answer:''' The transition state is a saddle point in the potential energy surface diagram where ∂V(ri)/∂ri=0 and G<0 where G = frr(r0,r'0)fr'r'(r0,r'0)−f2rr'(r0,r'0).

The transition point was defined as a maximum point in the minimum energy path linking reactants and products. For triatomic reaction, the transition state can be identified in reaction path in the contour plot where the AB and BC distance are equal.

The local minimum point can be distinguished by secondary partial derivation equal to zero and G > 0.

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

[[File:MRD_XC_1.png|thumb|Picture.1 A snapshot of 'Internuclear Distance vs Time' plot at r1 = r2 = 0.9075Å]]

'''Answer:''' The best estimate transition position is r1=r2=0.9075 Å where the “Internuclear Distances vs Time” plot shows two straight lines (Picture.1). The picture of distance vs time plot shows the AB and BC distance are constant which indicated the transition state.

=== '' Comment on how the mep and the trajectory you just calculated differ.''===
[[File:MRD_DYN_0521.png|thumb|Picture.2 A snapshot of dynamic surcafe plot]]
[[File:MRD_MEP_0521.png|thumb|Picture.3 A snapshot of MEP surface plot]]
'''Answer:'''

''' BC = r1 = rts+0.01'''

''' AB= '''r2 = rts

The initial condition was set to r1 = rts+0.01 and r2 = rts''' '''''',''' thus the AB distance ('''r2''') =0.9075 Å, and the BC distance ('''r1''') = 0.9175 Å
According to Picture 2 and 3, the reaction path calculated by MEP method was more smooth than the path calculated by dynamic method. For Picture.2 (using dynamic method), the reaction path vibrated at the lowest energy. For the MEP method, the reaction pathway was more smooth and more fit to the lowest energy barrier.
The difference between two ways of calculation may be because that MEP corresponds to infinitely slow motion while the dynamic method doesn't. For the dynamic calculation, the molecules move in a relative fast speed and vibrate. This makes the energy calculation harder. Thus, for small initial momenta, MEP calculation was more fit to the surface plot.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/Å
|0.9175
|7.15
|3.75
|-
|r2/Å
|0.9075
|0.75
|0.76
|-
|p1
|0
|2.46
|0
|-
|p2
|0
|1.33
|0
|}
For the table below, the initial distance was switched for r1 and r2,

BC distance =r1 = rts

AB distance = r2 = rts +0.01

According to the table below, it can be seen that the final value for both distance and momenta switched as the initial distance switched, and this change does not depend on their calculation method.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/ Å
|0.9075
|0.74
|0.76
|-
|r2/ Å
|0.9175
|7.15
|3.75
|-
|p1
|0
|1.33
|0
|-
|p2
|0
|2.46
|0
|}

=== 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"
!p1
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reactants have enough kinetic energy to pass though the transition state and complete the reaction and it can be seen from the contour plot: the trajectory passes though the transition state and goes to the products. Thus, it is a productive reaction.[[File:MRD_R1_0521.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.2</nowiki>
|No
|The reactants do not have enough energy to pass through the transition state, thus the trajectory moves back to the reactants.[[File:MRD_R2_0521.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reagents have enough energy to pass through the transition state. Thus, the trajectory passes though the transition state and moves towards to the product. [[File:MRD_R3_0521.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.6</nowiki>
|No
|The reagents have enough energy topass through the transition state, however, the energy is too high thus it drops back to the reactants.[[File:MRD_R4_0521.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-82.6</nowiki>
|Yes
|The trajectory does not moves to product initially, the reagents have excess energy to pass through the transition state, thus it goes back to the reactants. After that, the reactants still have enough energy to pass through the transition state, thus the reaction occurs.[[File:MRD_R5_0521.png]]
|}
In conclusion, to complete a productive reaction, the reactants should have enough energy to pass through the transition state but not too high to reverse the product back into the reactants.

===''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?''===
'''Assumptions:'''[https://www.sciencedirect.com/topics/chemistry/transition-state-theory]<ref name = 'TS'/>
# Quantum-tunnelling effects are assumed negligible.
# The Born-Oppenheimer approximation is invoked.
# The atoms in the reactant state have energies that are Boltzmann distributed.
# Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.

== 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?''===
{| class="wikitable"
!Reaction
!Initial condition
!Potential surface plot and discription
|-
|F + H2 (ABC)
|AB distance = 1.00 Å
BC distance = 0.74 Å

AB momentum = 0
BC momentum = -1.5
|[[File:MDR_FHH_0522.png|thumb|left|A surface plot for F+ H2 reaction. ]] The reaction is exothermix as the energy for reactants are higher than the energy for products. Thus, the energy is released during the reaction. This is caused by the formation of strong H-F bond.
|-
|H +HF (ABC)
|AB distance = 1.00 Å
BC distance = 0.017 Å

AB momentum = 0
BC momentum = -10
|[[File:MRD_HHF_0522.png|thumb|left|A surface plot for H + HF reaction. ]] This reaction is endothermic as the potential energy of reactants are lower than the potential energy of products. In this case, the strong H-F bond was break and a weaker H-H bond was formed. Thud, the overall reaction was endothermic.
|}

===''Locate the approximate position of the transition state.''===
{| class="wikitable"
!Reaction
!Position of Transition Point
!Contour plot
|-
|F + H2
|AB distance = 1.820 Å
BC distance = 0.745 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_FHH_0522.png|thumb|Contour plot for for the transition point.]]
|-
|H + HF
|AB distance = 0.745 Å
BC distance = 1.820 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_0522_1.png|thumb|Contour plot for for the transition point.]]
|}

==Reference==
<references>
<ref name="TS">T.Bligaard, J.K.Nørskov, in Chemical Bonding at Surfaces and Interfaces, Elsevier, Amsterdam, 2008, pp 255-321.</ref>
</references>

File:MRD FHH 0522.png

2019-05-22T12:43:22Z

Xc2017:

MRD:xc2017

2019-05-22T12:40:36Z

Xc2017: /* Locate the approximate position of the transition state. */

== 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?''===

'''Answer:''' The transition state is a saddle point in the potential energy surface diagram where ∂V(ri)/∂ri=0 and G<0 where G = frr(r0,r'0)fr'r'(r0,r'0)−f2rr'(r0,r'0).

The transition point was defined as a maximum point in the minimum energy path linking reactants and products. For triatomic reaction, the transition state can be identified in reaction path in the contour plot where the AB and BC distance are equal.

The local minimum point can be distinguished by secondary partial derivation equal to zero and G > 0.

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

[[File:MRD_XC_1.png|thumb|Picture.1 A snapshot of 'Internuclear Distance vs Time' plot at r1 = r2 = 0.9075Å]]

'''Answer:''' The best estimate transition position is r1=r2=0.9075 Å where the “Internuclear Distances vs Time” plot shows two straight lines (Picture.1). The picture of distance vs time plot shows the AB and BC distance are constant which indicated the transition state.

=== '' Comment on how the mep and the trajectory you just calculated differ.''===
[[File:MRD_DYN_0521.png|thumb|Picture.2 A snapshot of dynamic surcafe plot]]
[[File:MRD_MEP_0521.png|thumb|Picture.3 A snapshot of MEP surface plot]]
'''Answer:'''

''' BC = r1 = rts+0.01'''

''' AB= '''r2 = rts

The initial condition was set to r1 = rts+0.01 and r2 = rts''' '''''',''' thus the AB distance ('''r2''') =0.9075 Å, and the BC distance ('''r1''') = 0.9175 Å
According to Picture 2 and 3, the reaction path calculated by MEP method was more smooth than the path calculated by dynamic method. For Picture.2 (using dynamic method), the reaction path vibrated at the lowest energy. For the MEP method, the reaction pathway was more smooth and more fit to the lowest energy barrier.
The difference between two ways of calculation may be because that MEP corresponds to infinitely slow motion while the dynamic method doesn't. For the dynamic calculation, the molecules move in a relative fast speed and vibrate. This makes the energy calculation harder. Thus, for small initial momenta, MEP calculation was more fit to the surface plot.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/Å
|0.9175
|7.15
|3.75
|-
|r2/Å
|0.9075
|0.75
|0.76
|-
|p1
|0
|2.46
|0
|-
|p2
|0
|1.33
|0
|}
For the table below, the initial distance was switched for r1 and r2,

BC distance =r1 = rts

AB distance = r2 = rts +0.01

According to the table below, it can be seen that the final value for both distance and momenta switched as the initial distance switched, and this change does not depend on their calculation method.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/ Å
|0.9075
|0.74
|0.76
|-
|r2/ Å
|0.9175
|7.15
|3.75
|-
|p1
|0
|1.33
|0
|-
|p2
|0
|2.46
|0
|}

=== 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"
!p1
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reactants have enough kinetic energy to pass though the transition state and complete the reaction and it can be seen from the contour plot: the trajectory passes though the transition state and goes to the products. Thus, it is a productive reaction.[[File:MRD_R1_0521.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.2</nowiki>
|No
|The reactants do not have enough energy to pass through the transition state, thus the trajectory moves back to the reactants.[[File:MRD_R2_0521.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reagents have enough energy to pass through the transition state. Thus, the trajectory passes though the transition state and moves towards to the product. [[File:MRD_R3_0521.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.6</nowiki>
|No
|The reagents have enough energy topass through the transition state, however, the energy is too high thus it drops back to the reactants.[[File:MRD_R4_0521.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-82.6</nowiki>
|Yes
|The trajectory does not moves to product initially, the reagents have excess energy to pass through the transition state, thus it goes back to the reactants. After that, the reactants still have enough energy to pass through the transition state, thus the reaction occurs.[[File:MRD_R5_0521.png]]
|}
In conclusion, to complete a productive reaction, the reactants should have enough energy to pass through the transition state but not too high to reverse the product back into the reactants.

===''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?''===
'''Assumptions:'''[https://www.sciencedirect.com/topics/chemistry/transition-state-theory]<ref name = 'TS'/>
# Quantum-tunnelling effects are assumed negligible.
# The Born-Oppenheimer approximation is invoked.
# The atoms in the reactant state have energies that are Boltzmann distributed.
# Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.

== 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?''===
{| class="wikitable"
!Reaction
!Initial condition
!Potential surface plot and discription
|-
|F + H2 (ABC)
|AB distance = 1.00 Å
BC distance = 0.74 Å

AB momentum = 0
BC momentum = -1.5
|[[File:MDR_FHH_0522.png|thumb|left|A surface plot for F+ H2 reaction. ]] The reaction is exothermix as the energy for reactants are higher than the energy for products. Thus, the energy is released during the reaction. This is caused by the formation of strong H-F bond.
|-
|H +HF (ABC)
|AB distance = 1.00 Å
BC distance = 0.017 Å

AB momentum = 0
BC momentum = -10
|[[File:MRD_HHF_0522.png|thumb|left|A surface plot for H + HF reaction. ]] This reaction is endothermic as the potential energy of reactants are lower than the potential energy of products. In this case, the strong H-F bond was break and a weaker H-H bond was formed. Thud, the overall reaction was endothermic.
|}

===''Locate the approximate position of the transition state.''===
{| class="wikitable"
!Reaction
!Position of Transition Point
!Contour plot
|-
|F + H2
|AB distance = 1.00 Å
BC distance = 0.74 Å

AB momentum = 0

BC momentum = 0
|
|-
|H + HF
|AB distance = 0.745 Å
BC distance = 1.820 Å

AB momentum = 0

BC momentum = 0
|[[File:MRD_HHF_0522_1.png|thumb|Contour plot for for the transition point.]]
|}

==Reference==
<references>
<ref name="TS">T.Bligaard, J.K.Nørskov, in Chemical Bonding at Surfaces and Interfaces, Elsevier, Amsterdam, 2008, pp 255-321.</ref>
</references>

File:MRD HHF 0522 1.png

2019-05-22T12:36:28Z

Xc2017:

MRD:xc2017

2019-05-22T12:36:10Z

Xc2017: /* EXERCISE 2: F - H - H system */

== 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?''===

'''Answer:''' The transition state is a saddle point in the potential energy surface diagram where ∂V(ri)/∂ri=0 and G<0 where G = frr(r0,r'0)fr'r'(r0,r'0)−f2rr'(r0,r'0).

The transition point was defined as a maximum point in the minimum energy path linking reactants and products. For triatomic reaction, the transition state can be identified in reaction path in the contour plot where the AB and BC distance are equal.

The local minimum point can be distinguished by secondary partial derivation equal to zero and G > 0.

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

[[File:MRD_XC_1.png|thumb|Picture.1 A snapshot of 'Internuclear Distance vs Time' plot at r1 = r2 = 0.9075Å]]

'''Answer:''' The best estimate transition position is r1=r2=0.9075 Å where the “Internuclear Distances vs Time” plot shows two straight lines (Picture.1). The picture of distance vs time plot shows the AB and BC distance are constant which indicated the transition state.

=== '' Comment on how the mep and the trajectory you just calculated differ.''===
[[File:MRD_DYN_0521.png|thumb|Picture.2 A snapshot of dynamic surcafe plot]]
[[File:MRD_MEP_0521.png|thumb|Picture.3 A snapshot of MEP surface plot]]
'''Answer:'''

''' BC = r1 = rts+0.01'''

''' AB= '''r2 = rts

The initial condition was set to r1 = rts+0.01 and r2 = rts''' '''''',''' thus the AB distance ('''r2''') =0.9075 Å, and the BC distance ('''r1''') = 0.9175 Å
According to Picture 2 and 3, the reaction path calculated by MEP method was more smooth than the path calculated by dynamic method. For Picture.2 (using dynamic method), the reaction path vibrated at the lowest energy. For the MEP method, the reaction pathway was more smooth and more fit to the lowest energy barrier.
The difference between two ways of calculation may be because that MEP corresponds to infinitely slow motion while the dynamic method doesn't. For the dynamic calculation, the molecules move in a relative fast speed and vibrate. This makes the energy calculation harder. Thus, for small initial momenta, MEP calculation was more fit to the surface plot.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/Å
|0.9175
|7.15
|3.75
|-
|r2/Å
|0.9075
|0.75
|0.76
|-
|p1
|0
|2.46
|0
|-
|p2
|0
|1.33
|0
|}
For the table below, the initial distance was switched for r1 and r2,

BC distance =r1 = rts

AB distance = r2 = rts +0.01

According to the table below, it can be seen that the final value for both distance and momenta switched as the initial distance switched, and this change does not depend on their calculation method.
{| class="wikitable"
!
!Initial
!Final (dynamic)
!Final(MEP)
|-
|r1/ Å
|0.9075
|0.74
|0.76
|-
|r2/ Å
|0.9175
|7.15
|3.75
|-
|p1
|0
|1.33
|0
|-
|p2
|0
|2.46
|0
|}

=== 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"
!p1
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reactants have enough kinetic energy to pass though the transition state and complete the reaction and it can be seen from the contour plot: the trajectory passes though the transition state and goes to the products. Thus, it is a productive reaction.[[File:MRD_R1_0521.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.2</nowiki>
|No
|The reactants do not have enough energy to pass through the transition state, thus the trajectory moves back to the reactants.[[File:MRD_R2_0521.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|The reagents have enough energy to pass through the transition state. Thus, the trajectory passes though the transition state and moves towards to the product. [[File:MRD_R3_0521.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.6</nowiki>
|No
|The reagents have enough energy topass through the transition state, however, the energy is too high thus it drops back to the reactants.[[File:MRD_R4_0521.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-82.6</nowiki>
|Yes
|The trajectory does not moves to product initially, the reagents have excess energy to pass through the transition state, thus it goes back to the reactants. After that, the reactants still have enough energy to pass through the transition state, thus the reaction occurs.[[File:MRD_R5_0521.png]]
|}
In conclusion, to complete a productive reaction, the reactants should have enough energy to pass through the transition state but not too high to reverse the product back into the reactants.

===''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?''===
'''Assumptions:'''[https://www.sciencedirect.com/topics/chemistry/transition-state-theory]<ref name = 'TS'/>
# Quantum-tunnelling effects are assumed negligible.
# The Born-Oppenheimer approximation is invoked.
# The atoms in the reactant state have energies that are Boltzmann distributed.
# Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.

== 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?''===
{| class="wikitable"
!Reaction
!Initial condition
!Potential surface plot and discription
|-
|F + H2 (ABC)
|AB distance = 1.00 Å
BC distance = 0.74 Å

AB momentum = 0
BC momentum = -1.5
|[[File:MDR_FHH_0522.png|thumb|left|A surface plot for F+ H2 reaction. ]] The reaction is exothermix as the energy for reactants are higher than the energy for products. Thus, the energy is released during the reaction. This is caused by the formation of strong H-F bond.
|-
|H +HF (ABC)
|AB distance = 1.00 Å
BC distance = 0.017 Å

AB momentum = 0
BC momentum = -10
|[[File:MRD_HHF_0522.png|thumb|left|A surface plot for H + HF reaction. ]] This reaction is endothermic as the potential energy of reactants are lower than the potential energy of products. In this case, the strong H-F bond was break and a weaker H-H bond was formed. Thud, the overall reaction was endothermic.
|}

===''Locate the approximate position of the transition state.''===
{| class="wikitable"
!Reaction
!Position of Transition Point
!Contour plot
|-
|F + H2
|AB distance = 1.00 Å
BC distance = 0.74 Å

AB momentum = 0

BC momentum = 0
|
|-
|H + HF
|AB distance = 0.745 Å
BC distance = 1.820 Å

AB momentum = 0

BC momentum = 0
|
|}

==Reference==
<references>
<ref name="TS">T.Bligaard, J.K.Nørskov, in Chemical Bonding at Surfaces and Interfaces, Elsevier, Amsterdam, 2008, pp 255-321.</ref>
</references>

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Xc2017: /* 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? */