ChemWiki - User contributions [en]

MRD:syk2017

2019-05-10T16:45:22Z

Syk2017: /* Reaction dynamics */

== H+H2 system ==

=== Dynamics from the transition state region ===

==== Transition state ====
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

[[File:Q0 Surface Plot.png|center|400px|thumb|[Figure 1] Black line is located within the MEP]]

The transition state is a short-lived state of molecules or atoms at a local maximum in the reaction coordinate. Therefore, on a potential energy surface diagram, it can be defined as a local maximum of the minimum energy path (MEP). Mathematically it is a point where the gradient of potential is zero ( ∂V('''ri''')/∂'''ri'''=0). Although it is located in relatively low potential energy level, it actually is a local maximum point. It can be distinguished from other local minimum or local maximum points, as it should be located within the region of MEP.

'''

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.

'''<nowiki/>'''

[[File:Q2 Surface Plot.png|left|400px|thumb|[Figure 2] Position of transition state represented as 'x' on the contour plot]]
[[File:Q2 Surface Plot2.png|center|400px|thumb|[Figure 3] Internuclear distance VS time graph at transition state]]

The best estimate is achieved at rts=0.907. 

Since H+H2 is symmetric, the rAB and rBC should be equal. The figure 2 has only two lines, representing rBC and rAC . The rAB is not shown because rBC is overlapping it.

The definition of transition state suggests that the internuclear distance should stay constant throughout the given time, as ∂V('''ri''')/∂'''ri'''=0. The figure 2 shows that there is minimum vibration among the H atoms and no translation, suggesting the transition state.

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

Following graphs are generated by calculation on r1=0.917 and r2=0.907 when p1=p2=0.

[[File:Q3 contour Plot MEP.png|left|400px|thumb|[Figure 4] Contour graph of MEP]]
[[File:Q3 contour Plot DY.png|center|400px|thumb|[Figure 5] Contour graph of Dynamic]]
[[File:Q3 internu Plot MEP.png|left|400px|thumb|[Figure 6] Internuclear distance VS time graph of MEP]]
[[File:Q3 internu Plot DY.png|center|400px|thumb|[Figure 7] Internuclear distance VS time graph of Dynamic]]

Comment on how the ''mep'' and the trajectory you just calculated differ.

In the MEP trajectory, the motion of molecule is infinitely slow that the momenta is reset to zero in every time step. This effectively takes the vibrational energy out of the system. Therefore, smooth line is shown MEP trajectories, when wavy line is shown in dynamic trajectory. Comparing figure 6 and figure 7, the Dynamic trajectories shows linearly increasing internuclear distance, and the MEP trajectories show decrease in the gradient. This is because MET plot is taken as each point being stationary (zero momenta), when the dynamic plot includes the momentum.

For the calculation r1=rts and r2=rts+0.01, the reaction would be in the other way around.

=== Reactive and unreactive trajectories ===
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?

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!P1
!P2
!Etot
!Contour plot
!Reactive?
!Description of the dynamics
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|[[File:Q5 contour1.png|frameless]]
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|[[File:Q5 contour2.png|frameless]]
|No
|Atom C has not enough momentum to collide with AB.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|[[File:Q5 contour3.png|frameless]]
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|[[File:Q5 contour4.png|frameless]]
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|[[File:Q5 contour5.png|frameless]]
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The transition stat theory (TST) is used to calculate reaction constant k.

The main assumptions are:

1. The electronic motion and nuclear motion is separated in the way similar to Born-Oppenheimer approximation

2. Molecules which have passed the transition state in the direction of product cannot reform the reactant.

3. In the transition state, motion along the reaction coordinate is separated from the other motions and treated as translation.

4. Transition state molecules are distributed among their states according to the Maxwell-Bpltzman law, even there is no equalibrium set.

5. The system is in quasi-equalibrium, where the reaction process is slow enough for internal equalibrium.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River </ref>

Therefore, the experimental value is likely to be slower than the theoretical value, because the disrupting factors, such as backward reaction, are ignored in assumptions.

However, the TST also ignores tunnelling effect of quantum mechanics, which reduces energy barrier. This would make the experimental value faster than the theoretical value.

== F-H-H system ==

=== PES inspection ===

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

[[File:Q6 surface F+H2.png|left|400px|thumb|[Figure 8]F+H2 surface plot]]

[[File:Q6 surface H+HF.png|center|400px|thumb|[Figure 9] H+HF surface plot]]

For the potential energy surface diagram of F+H2, the energy level goes down as the F atom and H2 molecule gets closer. This means that the energy level of product is lower than that of reactant, indicating it is a exothermic reaction. The reaction of H+HF shows completely opposite situation. The energy level increases as H-H gets closer, indicating that the reaction is endothermic. This is directly related to the bond energy. An exothermic reaction is when the bond energy of new bond of product is greater than broken bonds of reactant. Therefore, H-H bond is weaker than H-F bond.

Locate the approximate position of the transition state.
[[File:Q6 contour1.png|centre|thumb|400x400px|[Figure 9] Position of the transition state of F+H2 shown in contour plot]]
It is achieved at rFH=1.813 and rHH=0.74, when momenta=0.

Report the activation energy for both reactions.

'''''F-H + H reaction:'''''

The reaction is achieved at rFH=1.65 and rHH=0.74

Activation energy =104.125-131.053=+26.928 kcalmol-1

[[File:Ea F+HH.png|centre|thumb|400x400px]]

'''''H-H + F reaction:'''''

The reaction is achieved at rFH=1.89 and rHH=0.74

Activation energy = 103.708-103.762= +0.054kcalmol-1

[[File:Ea H+HF.png|centre|thumb|400x400px|[Figure 11] Energy VS time graph]]

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

The reaction is achieved at rFH=1.65 and rHH=0.74.

[[File:Inlight1.png|centre|thumb|400x400px|[Figure 12] Momentum VS time graph. ]]
Energy is released due to the formation of H-F bond. And this causes the increase in oscillation of the bonds, as shown in [Figure 12]. This means the kinetic energy of the molecules has increased. Since temperature is defined as average kinetic energy, this can be experimentally identified 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

Polanyi's empirical rules states vibration is more effective to promote a late-barrier reaction than the translation.

[[File:HH2.png|center|thumb|400x400px|[Figure 13] Contour plot at PHH=-2]]
[[File:HH1.png|center|thumb|400x400px|[Figure 14] Contour plot at PHH=-1]]
[[File:HH0.5.png|center|thumb|400x400px|[Figure 15] Contour plot at PHH=-0.5]]

The illustrations above shows change in the collision path due to change in momentum of H. The momentum on F-H bond was set to 0.02. These plots are to investigate the effect of translation. All three plots have not shown reactive systems, but it gets worse if the translation is greater.

[[File:HH1HF1.png|center|thumb|400x400px|[Figure 13] Contour plot at PFH=1.0]]
[[File:HH1HF6.0.png|center|thumb|400x400px|[Figure 14] Contour plot at PFH=6.0]]
[[File:HH1HF6.5.png|center|thumb|400x400px|[Figure 15] Contour plot at PFH=6.5]]

The illustration above shows the change in the collision path as the vibration of F-H bond increases. The momentum on H-H was set to -1.09. At pFH=6.0, the collision was almost reactive but it eventually returns to the reactants. At pFH=6.5, the system is finally found to be reactive. This clearly agrees with Polanyi's empirical rules, as increase in vibration is effective in promoting the reaction compared to the increase in translation.

MRD:syk2017

2019-05-10T16:37:34Z

Syk2017: /* Reaction dynamics */

== H+H2 system ==

=== Dynamics from the transition state region ===

==== Transition state ====
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

[[File:Q0 Surface Plot.png|center|400px|thumb|[Figure 1] Black line is located within the MEP]]

The transition state is a short-lived state of molecules or atoms at a local maximum in the reaction coordinate. Therefore, on a potential energy surface diagram, it can be defined as a local maximum of the minimum energy path (MEP). Mathematically it is a point where the gradient of potential is zero ( ∂V('''ri''')/∂'''ri'''=0). Although it is located in relatively low potential energy level, it actually is a local maximum point. It can be distinguished from other local minimum or local maximum points, as it should be located within the region of MEP.

'''

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.

'''<nowiki/>'''

[[File:Q2 Surface Plot.png|left|400px|thumb|[Figure 2] Position of transition state represented as 'x' on the contour plot]]
[[File:Q2 Surface Plot2.png|center|400px|thumb|[Figure 3] Internuclear distance VS time graph at transition state]]

The best estimate is achieved at rts=0.907. 

Since H+H2 is symmetric, the rAB and rBC should be equal. The figure 2 has only two lines, representing rBC and rAC . The rAB is not shown because rBC is overlapping it.

The definition of transition state suggests that the internuclear distance should stay constant throughout the given time, as ∂V('''ri''')/∂'''ri'''=0. The figure 2 shows that there is minimum vibration among the H atoms and no translation, suggesting the transition state.

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

Following graphs are generated by calculation on r1=0.917 and r2=0.907 when p1=p2=0.

[[File:Q3 contour Plot MEP.png|left|400px|thumb|[Figure 4] Contour graph of MEP]]
[[File:Q3 contour Plot DY.png|center|400px|thumb|[Figure 5] Contour graph of Dynamic]]
[[File:Q3 internu Plot MEP.png|left|400px|thumb|[Figure 6] Internuclear distance VS time graph of MEP]]
[[File:Q3 internu Plot DY.png|center|400px|thumb|[Figure 7] Internuclear distance VS time graph of Dynamic]]

Comment on how the ''mep'' and the trajectory you just calculated differ.

In the MEP trajectory, the motion of molecule is infinitely slow that the momenta is reset to zero in every time step. This effectively takes the vibrational energy out of the system. Therefore, smooth line is shown MEP trajectories, when wavy line is shown in dynamic trajectory. Comparing figure 6 and figure 7, the Dynamic trajectories shows linearly increasing internuclear distance, and the MEP trajectories show decrease in the gradient. This is because MET plot is taken as each point being stationary (zero momenta), when the dynamic plot includes the momentum.

For the calculation r1=rts and r2=rts+0.01, the reaction would be in the other way around.

=== Reactive and unreactive trajectories ===
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?

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!P1
!P2
!Etot
!Contour plot
!Reactive?
!Description of the dynamics
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|[[File:Q5 contour1.png|frameless]]
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|[[File:Q5 contour2.png|frameless]]
|No
|Atom C has not enough momentum to collide with AB.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|[[File:Q5 contour3.png|frameless]]
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|[[File:Q5 contour4.png|frameless]]
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|[[File:Q5 contour5.png|frameless]]
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The transition stat theory (TST) is used to calculate reaction constant k.

The main assumptions are:

1. The electronic motion and nuclear motion is separated in the way similar to Born-Oppenheimer approximation

2. Molecules which have passed the transition state in the direction of product cannot reform the reactant.

3. In the transition state, motion along the reaction coordinate is separated from the other motions and treated as translation.

4. Transition state molecules are distributed among their states according to the Maxwell-Bpltzman law, even there is no equalibrium set.

5. The system is in quasi-equalibrium, where the reaction process is slow enough for internal equalibrium.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River </ref>

Therefore, the experimental value is likely to be slower than the theoretical value, because the disrupting factors, such as backward reaction, are ignored in assumptions.

However, the TST also ignores tunnelling effect of quantum mechanics, which reduces energy barrier. This would make the experimental value faster than the theoretical value.

== F-H-H system ==

=== PES inspection ===

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

[[File:Q6 surface F+H2.png|left|400px|thumb|[Figure 8]F+H2 surface plot]]

[[File:Q6 surface H+HF.png|center|400px|thumb|[Figure 9] H+HF surface plot]]

For the potential energy surface diagram of F+H2, the energy level goes down as the F atom and H2 molecule gets closer. This means that the energy level of product is lower than that of reactant, indicating it is a exothermic reaction. The reaction of H+HF shows completely opposite situation. The energy level increases as H-H gets closer, indicating that the reaction is endothermic. This is directly related to the bond energy. An exothermic reaction is when the bond energy of new bond of product is greater than broken bonds of reactant. Therefore, H-H bond is weaker than H-F bond.

Locate the approximate position of the transition state.
[[File:Q6 contour1.png|centre|thumb|400x400px|[Figure 9] Position of the transition state of F+H2 shown in contour plot]]
It is achieved at rFH=1.813 and rHH=0.74, when momenta=0.

Report the activation energy for both reactions.

'''''F-H + H reaction:'''''

The reaction is achieved at rFH=1.65 and rHH=0.74

Activation energy =104.125-131.053=+26.928 kcalmol-1

[[File:Ea F+HH.png|centre|thumb|400x400px]]

'''''H-H + F reaction:'''''

The reaction is achieved at rFH=1.89 and rHH=0.74

Activation energy = 103.708-103.762= +0.054kcalmol-1

[[File:Ea H+HF.png|centre|thumb|400x400px|[Figure 11] Energy VS time graph]]

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

The reaction is achieved at rFH=1.65 and rHH=0.74.

[[File:Inlight1.png|centre|thumb|400x400px|[Figure 12] Momentum VS time graph. ]]
Energy is released due to the formation of H-F bond. And this causes the increase in oscillation of the bonds, as shown in [Figure 12]. This means the kinetic energy of the molecules has increased. Since temperature is defined as average kinetic energy, this can be experimentally identified 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

Polanyi's empirical rules states vibration is more effective to promote a late-barrier reaction than the translation.

[[File:HH2.png|center|thumb|400x400px|[Figure 13] Contour plot at PHH=-2]]
[[File:HH1.png|center|thumb|400x400px|[Figure 14] Contour plot at PHH=-1]]
[[File:HH0.5.png|center|thumb|400x400px|[Figure 15] Contour plot at PHH=-0.5]]

The illustrations above shows change in the collision path due to change in momentum of H. The momentum on F-H bond was set to 0.02. These plots are to investigate the effect of translation. All three plots have not shown reactive systems, but it gets worse if the translation is greater.

[[File:HH1HF1.png|center|thumb|400x400px|[Figure 13] Contour plot at PFH=1.0]]
[[File:HH1HF6.0.png|center|thumb|400x400px|[Figure 14] Contour plot at PFH=6.0]]
[[File:HH1HF6.5.png|center|thumb|400x400px|[Figure 15] Contour plot at PFH=6.5]]

The illustration above shows the change in the collision path as the vibration of F-H bond increases. The momentum on H-H was set to -1.09. At pFH=6.0, the collision was almost reactive but it eventually returns to the reactants. At pFH=6.5, the system is finally found to be reactive. This clearly agrees with Polanyi's empirical rules, as increase in vibration is effective in promoting the reaction compared to the increase in translation.

MRD:syk2017

2019-05-10T16:34:22Z

Syk2017: /* Reaction dynamics */

== H+H2 system ==

=== Dynamics from the transition state region ===

==== Transition state ====
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

[[File:Q0 Surface Plot.png|center|400px|thumb|[Figure 1] Black line is located within the MEP]]

The transition state is a short-lived state of molecules or atoms at a local maximum in the reaction coordinate. Therefore, on a potential energy surface diagram, it can be defined as a local maximum of the minimum energy path (MEP). Mathematically it is a point where the gradient of potential is zero ( ∂V('''ri''')/∂'''ri'''=0). Although it is located in relatively low potential energy level, it actually is a local maximum point. It can be distinguished from other local minimum or local maximum points, as it should be located within the region of MEP.

'''

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.

'''<nowiki/>'''

[[File:Q2 Surface Plot.png|left|400px|thumb|[Figure 2] Position of transition state represented as 'x' on the contour plot]]
[[File:Q2 Surface Plot2.png|center|400px|thumb|[Figure 3] Internuclear distance VS time graph at transition state]]

The best estimate is achieved at rts=0.907. 

Since H+H2 is symmetric, the rAB and rBC should be equal. The figure 2 has only two lines, representing rBC and rAC . The rAB is not shown because rBC is overlapping it.

The definition of transition state suggests that the internuclear distance should stay constant throughout the given time, as ∂V('''ri''')/∂'''ri'''=0. The figure 2 shows that there is minimum vibration among the H atoms and no translation, suggesting the transition state.

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

Following graphs are generated by calculation on r1=0.917 and r2=0.907 when p1=p2=0.

[[File:Q3 contour Plot MEP.png|left|400px|thumb|[Figure 4] Contour graph of MEP]]
[[File:Q3 contour Plot DY.png|center|400px|thumb|[Figure 5] Contour graph of Dynamic]]
[[File:Q3 internu Plot MEP.png|left|400px|thumb|[Figure 6] Internuclear distance VS time graph of MEP]]
[[File:Q3 internu Plot DY.png|center|400px|thumb|[Figure 7] Internuclear distance VS time graph of Dynamic]]

Comment on how the ''mep'' and the trajectory you just calculated differ.

In the MEP trajectory, the motion of molecule is infinitely slow that the momenta is reset to zero in every time step. This effectively takes the vibrational energy out of the system. Therefore, smooth line is shown MEP trajectories, when wavy line is shown in dynamic trajectory. Comparing figure 6 and figure 7, the Dynamic trajectories shows linearly increasing internuclear distance, and the MEP trajectories show decrease in the gradient. This is because MET plot is taken as each point being stationary (zero momenta), when the dynamic plot includes the momentum.

For the calculation r1=rts and r2=rts+0.01, the reaction would be in the other way around.

=== Reactive and unreactive trajectories ===
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?

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!P1
!P2
!Etot
!Contour plot
!Reactive?
!Description of the dynamics
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|[[File:Q5 contour1.png|frameless]]
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|[[File:Q5 contour2.png|frameless]]
|No
|Atom C has not enough momentum to collide with AB.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|[[File:Q5 contour3.png|frameless]]
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|[[File:Q5 contour4.png|frameless]]
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|[[File:Q5 contour5.png|frameless]]
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The transition stat theory (TST) is used to calculate reaction constant k.

The main assumptions are:

1. The electronic motion and nuclear motion is separated in the way similar to Born-Oppenheimer approximation

2. Molecules which have passed the transition state in the direction of product cannot reform the reactant.

3. In the transition state, motion along the reaction coordinate is separated from the other motions and treated as translation.

4. Transition state molecules are distributed among their states according to the Maxwell-Bpltzman law, even there is no equalibrium set.

5. The system is in quasi-equalibrium, where the reaction process is slow enough for internal equalibrium.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River </ref>

Therefore, the experimental value is likely to be slower than the theoretical value, because the disrupting factors, such as backward reaction, are ignored in assumptions.

However, the TST also ignores tunnelling effect of quantum mechanics, which reduces energy barrier. This would make the experimental value faster than the theoretical value.

== F-H-H system ==

=== PES inspection ===

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

[[File:Q6 surface F+H2.png|left|400px|thumb|[Figure 8]F+H2 surface plot]]

[[File:Q6 surface H+HF.png|center|400px|thumb|[Figure 9] H+HF surface plot]]

For the potential energy surface diagram of F+H2, the energy level goes down as the F atom and H2 molecule gets closer. This means that the energy level of product is lower than that of reactant, indicating it is a exothermic reaction. The reaction of H+HF shows completely opposite situation. The energy level increases as H-H gets closer, indicating that the reaction is endothermic. This is directly related to the bond energy. An exothermic reaction is when the bond energy of new bond of product is greater than broken bonds of reactant. Therefore, H-H bond is weaker than H-F bond.

Locate the approximate position of the transition state.
[[File:Q6 contour1.png|centre|thumb|400x400px|[Figure 9] Position of the transition state of F+H2 shown in contour plot]]
It is achieved at rFH=1.813 and rHH=0.74, when momenta=0.

Report the activation energy for both reactions.

'''''F-H + H reaction:'''''

The reaction is achieved at rFH=1.65 and rHH=0.74

Activation energy =104.125-131.053=+26.928 kcalmol-1

[[File:Ea F+HH.png|centre|thumb|400x400px]]

'''''H-H + F reaction:'''''

The reaction is achieved at rFH=1.89 and rHH=0.74

Activation energy = 103.708-103.762= +0.054kcalmol-1

[[File:Ea H+HF.png|centre|thumb|400x400px|[Figure 11] Energy VS time graph]]

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

The reaction is achieved at rFH=1.65 and rHH=0.74.

[[File:Inlight1.png|centre|thumb|400x400px|[Figure 12] Momentum VS time graph. ]]
Energy is released due to the formation of H-F bond. And this causes the increase in oscillation of the bonds, as shown in [Figure 12]. This means the kinetic energy of the molecules has increased. Since temperature is defined as average kinetic energy, this can be experimentally identified 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

Polanyi's empirical rules states vibration is more effective to promote a late-barrier reaction than the translation.

[[File:HH2.png|center|thumb|400x400px|[Figure 13] Contour plot at PHH=-2]]
[[File:HH1.png|center|thumb|400x400px|[Figure 14] Contour plot at PHH=-1]]
[[File:HH0.5.png|center|thumb|400x400px|[Figure 15] Contour plot at PHH=-0.5]]

The illustrations above shows change in the collision path due to change in momentum of H. The momentum on F-H bond was set to 0.02. These plots are to investigate the effect of translation. All three plots have not shown reactive systems, but it gets worse if the translation is greater.

[[File:HH1HF1.png|center|thumb|400x400px|[Figure 13] Contour plot at PFH=1.0]]
[[File:HH1HF6.0.png|center|thumb|400x400px|[Figure 14] Contour plot at PFH=6.0]]
[[File:HH1HF6.5.png|center|thumb|400x400px|[Figure 15] Contour plot at PFH=6.5]]

The illustration above shows the change in the collision path as the vibration of F-H bond increases. The momentum on H-H was set to -1.09. At pFH=6.5, the system is found to be reactive. This clearly agrees with Polanyi's empirical rules, as increase in vibration is effective in promoting the reaction.

MRD:syk2017

2019-05-10T16:33:09Z

Syk2017: /* Reaction dynamics */

== H+H2 system ==

=== Dynamics from the transition state region ===

==== Transition state ====
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

[[File:Q0 Surface Plot.png|center|400px|thumb|[Figure 1] Black line is located within the MEP]]

The transition state is a short-lived state of molecules or atoms at a local maximum in the reaction coordinate. Therefore, on a potential energy surface diagram, it can be defined as a local maximum of the minimum energy path (MEP). Mathematically it is a point where the gradient of potential is zero ( ∂V('''ri''')/∂'''ri'''=0). Although it is located in relatively low potential energy level, it actually is a local maximum point. It can be distinguished from other local minimum or local maximum points, as it should be located within the region of MEP.

'''

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.

'''<nowiki/>'''

[[File:Q2 Surface Plot.png|left|400px|thumb|[Figure 2] Position of transition state represented as 'x' on the contour plot]]
[[File:Q2 Surface Plot2.png|center|400px|thumb|[Figure 3] Internuclear distance VS time graph at transition state]]

The best estimate is achieved at rts=0.907. 

Since H+H2 is symmetric, the rAB and rBC should be equal. The figure 2 has only two lines, representing rBC and rAC . The rAB is not shown because rBC is overlapping it.

The definition of transition state suggests that the internuclear distance should stay constant throughout the given time, as ∂V('''ri''')/∂'''ri'''=0. The figure 2 shows that there is minimum vibration among the H atoms and no translation, suggesting the transition state.

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

Following graphs are generated by calculation on r1=0.917 and r2=0.907 when p1=p2=0.

[[File:Q3 contour Plot MEP.png|left|400px|thumb|[Figure 4] Contour graph of MEP]]
[[File:Q3 contour Plot DY.png|center|400px|thumb|[Figure 5] Contour graph of Dynamic]]
[[File:Q3 internu Plot MEP.png|left|400px|thumb|[Figure 6] Internuclear distance VS time graph of MEP]]
[[File:Q3 internu Plot DY.png|center|400px|thumb|[Figure 7] Internuclear distance VS time graph of Dynamic]]

Comment on how the ''mep'' and the trajectory you just calculated differ.

In the MEP trajectory, the motion of molecule is infinitely slow that the momenta is reset to zero in every time step. This effectively takes the vibrational energy out of the system. Therefore, smooth line is shown MEP trajectories, when wavy line is shown in dynamic trajectory. Comparing figure 6 and figure 7, the Dynamic trajectories shows linearly increasing internuclear distance, and the MEP trajectories show decrease in the gradient. This is because MET plot is taken as each point being stationary (zero momenta), when the dynamic plot includes the momentum.

For the calculation r1=rts and r2=rts+0.01, the reaction would be in the other way around.

=== Reactive and unreactive trajectories ===
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?

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!P1
!P2
!Etot
!Contour plot
!Reactive?
!Description of the dynamics
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|[[File:Q5 contour1.png|frameless]]
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|[[File:Q5 contour2.png|frameless]]
|No
|Atom C has not enough momentum to collide with AB.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|[[File:Q5 contour3.png|frameless]]
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|[[File:Q5 contour4.png|frameless]]
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|[[File:Q5 contour5.png|frameless]]
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The transition stat theory (TST) is used to calculate reaction constant k.

The main assumptions are:

1. The electronic motion and nuclear motion is separated in the way similar to Born-Oppenheimer approximation

2. Molecules which have passed the transition state in the direction of product cannot reform the reactant.

3. In the transition state, motion along the reaction coordinate is separated from the other motions and treated as translation.

4. Transition state molecules are distributed among their states according to the Maxwell-Bpltzman law, even there is no equalibrium set.

5. The system is in quasi-equalibrium, where the reaction process is slow enough for internal equalibrium.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River </ref>

Therefore, the experimental value is likely to be slower than the theoretical value, because the disrupting factors, such as backward reaction, are ignored in assumptions.

However, the TST also ignores tunnelling effect of quantum mechanics, which reduces energy barrier. This would make the experimental value faster than the theoretical value.

== F-H-H system ==

=== PES inspection ===

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

[[File:Q6 surface F+H2.png|left|400px|thumb|[Figure 8]F+H2 surface plot]]

[[File:Q6 surface H+HF.png|center|400px|thumb|[Figure 9] H+HF surface plot]]

For the potential energy surface diagram of F+H2, the energy level goes down as the F atom and H2 molecule gets closer. This means that the energy level of product is lower than that of reactant, indicating it is a exothermic reaction. The reaction of H+HF shows completely opposite situation. The energy level increases as H-H gets closer, indicating that the reaction is endothermic. This is directly related to the bond energy. An exothermic reaction is when the bond energy of new bond of product is greater than broken bonds of reactant. Therefore, H-H bond is weaker than H-F bond.

Locate the approximate position of the transition state.
[[File:Q6 contour1.png|centre|thumb|400x400px|[Figure 9] Position of the transition state of F+H2 shown in contour plot]]
It is achieved at rFH=1.813 and rHH=0.74, when momenta=0.

Report the activation energy for both reactions.

'''''F-H + H reaction:'''''

The reaction is achieved at rFH=1.65 and rHH=0.74

Activation energy =104.125-131.053=+26.928 kcalmol-1

[[File:Ea F+HH.png|centre|thumb|400x400px]]

'''''H-H + F reaction:'''''

The reaction is achieved at rFH=1.89 and rHH=0.74

Activation energy = 103.708-103.762= +0.054kcalmol-1

[[File:Ea H+HF.png|centre|thumb|400x400px|[Figure 11] Energy VS time graph]]

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

The reaction is achieved at rFH=1.65 and rHH=0.74.

[[File:Inlight1.png|centre|thumb|400x400px|[Figure 12] Momentum VS time graph. ]]
Energy is released due to the formation of H-F bond. And this causes the increase in oscillation of the bonds, as shown in [Figure 12]. This means the kinetic energy of the molecules has increased. Since temperature is defined as average kinetic energy, this can be experimentally identified 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

Polanyi's empirical rules states vibration is more effective to promote a late-barrier reaction than the translation.

[[File:HH2.png|center|thumb|400x400px|[Figure 13] Contour plot at PHH=-2]]
[[File:HH1.png|center|thumb|400x400px|[Figure 14] Contour plot at PHH=-1]]
[[File:HH0.5.png|center|thumb|400x400px|[Figure 15] Contour plot at PHH=-0.5]]

The illustrations above shows change in the collision path due to change in momentum of H. The momentum on F-H bond was set to 0.02. These plots are to investigate the effect of translation. All three plots have not shown reactive systems, but it gets worse if the translation is greater.

[[File:HH1HF1.png|center|thumb|400x400px|[Figure 13] Contour plot at PFH=1.0]]
[[File:HH1HF6.0.png|center|thumb|400x400px|[Figure 14] Contour plot at PFH=6.0]]
[[File:HH1HF6.5.png|center|thumb|400x400px|[Figure 15] Contour plot at PFH=6.5]]

The illustration above shows the change in the collision path as the vibration of F-H bond increases. The momentum on H-H was set to -1.09. At pFH=6.5, the system is found to be reactive. This clearly shows Polanyi's empirical rules.

MRD:syk2017

2019-05-10T16:32:15Z

Syk2017: /* Reaction dynamics */

== H+H2 system ==

=== Dynamics from the transition state region ===

==== Transition state ====
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

[[File:Q0 Surface Plot.png|center|400px|thumb|[Figure 1] Black line is located within the MEP]]

The transition state is a short-lived state of molecules or atoms at a local maximum in the reaction coordinate. Therefore, on a potential energy surface diagram, it can be defined as a local maximum of the minimum energy path (MEP). Mathematically it is a point where the gradient of potential is zero ( ∂V('''ri''')/∂'''ri'''=0). Although it is located in relatively low potential energy level, it actually is a local maximum point. It can be distinguished from other local minimum or local maximum points, as it should be located within the region of MEP.

'''

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.

'''<nowiki/>'''

[[File:Q2 Surface Plot.png|left|400px|thumb|[Figure 2] Position of transition state represented as 'x' on the contour plot]]
[[File:Q2 Surface Plot2.png|center|400px|thumb|[Figure 3] Internuclear distance VS time graph at transition state]]

The best estimate is achieved at rts=0.907. 

Since H+H2 is symmetric, the rAB and rBC should be equal. The figure 2 has only two lines, representing rBC and rAC . The rAB is not shown because rBC is overlapping it.

The definition of transition state suggests that the internuclear distance should stay constant throughout the given time, as ∂V('''ri''')/∂'''ri'''=0. The figure 2 shows that there is minimum vibration among the H atoms and no translation, suggesting the transition state.

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

Following graphs are generated by calculation on r1=0.917 and r2=0.907 when p1=p2=0.

[[File:Q3 contour Plot MEP.png|left|400px|thumb|[Figure 4] Contour graph of MEP]]
[[File:Q3 contour Plot DY.png|center|400px|thumb|[Figure 5] Contour graph of Dynamic]]
[[File:Q3 internu Plot MEP.png|left|400px|thumb|[Figure 6] Internuclear distance VS time graph of MEP]]
[[File:Q3 internu Plot DY.png|center|400px|thumb|[Figure 7] Internuclear distance VS time graph of Dynamic]]

Comment on how the ''mep'' and the trajectory you just calculated differ.

In the MEP trajectory, the motion of molecule is infinitely slow that the momenta is reset to zero in every time step. This effectively takes the vibrational energy out of the system. Therefore, smooth line is shown MEP trajectories, when wavy line is shown in dynamic trajectory. Comparing figure 6 and figure 7, the Dynamic trajectories shows linearly increasing internuclear distance, and the MEP trajectories show decrease in the gradient. This is because MET plot is taken as each point being stationary (zero momenta), when the dynamic plot includes the momentum.

For the calculation r1=rts and r2=rts+0.01, the reaction would be in the other way around.

=== Reactive and unreactive trajectories ===
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?

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!P1
!P2
!Etot
!Contour plot
!Reactive?
!Description of the dynamics
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|[[File:Q5 contour1.png|frameless]]
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|[[File:Q5 contour2.png|frameless]]
|No
|Atom C has not enough momentum to collide with AB.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|[[File:Q5 contour3.png|frameless]]
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|[[File:Q5 contour4.png|frameless]]
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|[[File:Q5 contour5.png|frameless]]
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The transition stat theory (TST) is used to calculate reaction constant k.

The main assumptions are:

1. The electronic motion and nuclear motion is separated in the way similar to Born-Oppenheimer approximation

2. Molecules which have passed the transition state in the direction of product cannot reform the reactant.

3. In the transition state, motion along the reaction coordinate is separated from the other motions and treated as translation.

4. Transition state molecules are distributed among their states according to the Maxwell-Bpltzman law, even there is no equalibrium set.

5. The system is in quasi-equalibrium, where the reaction process is slow enough for internal equalibrium.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River </ref>

Therefore, the experimental value is likely to be slower than the theoretical value, because the disrupting factors, such as backward reaction, are ignored in assumptions.

However, the TST also ignores tunnelling effect of quantum mechanics, which reduces energy barrier. This would make the experimental value faster than the theoretical value.

== F-H-H system ==

=== PES inspection ===

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

[[File:Q6 surface F+H2.png|left|400px|thumb|[Figure 8]F+H2 surface plot]]

[[File:Q6 surface H+HF.png|center|400px|thumb|[Figure 9] H+HF surface plot]]

For the potential energy surface diagram of F+H2, the energy level goes down as the F atom and H2 molecule gets closer. This means that the energy level of product is lower than that of reactant, indicating it is a exothermic reaction. The reaction of H+HF shows completely opposite situation. The energy level increases as H-H gets closer, indicating that the reaction is endothermic. This is directly related to the bond energy. An exothermic reaction is when the bond energy of new bond of product is greater than broken bonds of reactant. Therefore, H-H bond is weaker than H-F bond.

Locate the approximate position of the transition state.
[[File:Q6 contour1.png|centre|thumb|400x400px|[Figure 9] Position of the transition state of F+H2 shown in contour plot]]
It is achieved at rFH=1.813 and rHH=0.74, when momenta=0.

Report the activation energy for both reactions.

'''''F-H + H reaction:'''''

The reaction is achieved at rFH=1.65 and rHH=0.74

Activation energy =104.125-131.053=+26.928 kcalmol-1

[[File:Ea F+HH.png|centre|thumb|400x400px]]

'''''H-H + F reaction:'''''

The reaction is achieved at rFH=1.89 and rHH=0.74

Activation energy = 103.708-103.762= +0.054kcalmol-1

[[File:Ea H+HF.png|centre|thumb|400x400px|[Figure 11] Energy VS time graph]]

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

The reaction is achieved at rFH=1.65 and rHH=0.74.

[[File:Inlight1.png|centre|thumb|400x400px|[Figure 12] Momentum VS time graph. ]]
Energy is released due to the formation of H-F bond. And this causes the increase in oscillation of the bonds, as shown in [Figure 12]. This means the kinetic energy of the molecules has increased. Since temperature is defined as average kinetic energy, this can be experimentally identified 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

Polanyi's empirical rules states vibration is more effective to promote a late-barrier reaction than the translation.

[[File:HH2.png|thumb|400x400px|[Figure 13] Contour plot at PHH=-2]]
[[File:HH1.png|thumb|400x400px|[Figure 14] Contour plot at PHH=-1]]
[[File:HH0.5.png|thumb|400x400px|[Figure 15] Contour plot at PHH=-0.5]]

The illustrations above shows change in the collision path due to change in momentum of H. The momentum on F-H bond was set to 0.02. These plots are to investigate the effect of translation. All three plots have not shown reactive systems, but it gets worse if the translation is greater.

[[File:HH1HF1.png|left|thumb|400x400px|[Figure 13] Contour plot at PFH=1.0]]
[[File:HH1HF6.0.png|center|thumb|400x400px|[Figure 14] Contour plot at PFH=6.0]]
[[File:HH1HF6.5.png|right|thumb|400x400px|[Figure 15] Contour plot at PFH=6.5]]

The illustration above shows the change in the collision path as the vibration of F-H bond increases. The momentum on H-H was set to -1.09. At pFH=6.5, the system is found to be reactive. This clearly shows Polanyi's empirical rules.

MRD:syk2017

2019-05-10T16:31:29Z

Syk2017: /* Reaction dynamics */

== H+H2 system ==

=== Dynamics from the transition state region ===

==== Transition state ====
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

[[File:Q0 Surface Plot.png|center|400px|thumb|[Figure 1] Black line is located within the MEP]]

The transition state is a short-lived state of molecules or atoms at a local maximum in the reaction coordinate. Therefore, on a potential energy surface diagram, it can be defined as a local maximum of the minimum energy path (MEP). Mathematically it is a point where the gradient of potential is zero ( ∂V('''ri''')/∂'''ri'''=0). Although it is located in relatively low potential energy level, it actually is a local maximum point. It can be distinguished from other local minimum or local maximum points, as it should be located within the region of MEP.

'''

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.

'''<nowiki/>'''

[[File:Q2 Surface Plot.png|left|400px|thumb|[Figure 2] Position of transition state represented as 'x' on the contour plot]]
[[File:Q2 Surface Plot2.png|center|400px|thumb|[Figure 3] Internuclear distance VS time graph at transition state]]

The best estimate is achieved at rts=0.907. 

Since H+H2 is symmetric, the rAB and rBC should be equal. The figure 2 has only two lines, representing rBC and rAC . The rAB is not shown because rBC is overlapping it.

The definition of transition state suggests that the internuclear distance should stay constant throughout the given time, as ∂V('''ri''')/∂'''ri'''=0. The figure 2 shows that there is minimum vibration among the H atoms and no translation, suggesting the transition state.

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

Following graphs are generated by calculation on r1=0.917 and r2=0.907 when p1=p2=0.

[[File:Q3 contour Plot MEP.png|left|400px|thumb|[Figure 4] Contour graph of MEP]]
[[File:Q3 contour Plot DY.png|center|400px|thumb|[Figure 5] Contour graph of Dynamic]]
[[File:Q3 internu Plot MEP.png|left|400px|thumb|[Figure 6] Internuclear distance VS time graph of MEP]]
[[File:Q3 internu Plot DY.png|center|400px|thumb|[Figure 7] Internuclear distance VS time graph of Dynamic]]

Comment on how the ''mep'' and the trajectory you just calculated differ.

In the MEP trajectory, the motion of molecule is infinitely slow that the momenta is reset to zero in every time step. This effectively takes the vibrational energy out of the system. Therefore, smooth line is shown MEP trajectories, when wavy line is shown in dynamic trajectory. Comparing figure 6 and figure 7, the Dynamic trajectories shows linearly increasing internuclear distance, and the MEP trajectories show decrease in the gradient. This is because MET plot is taken as each point being stationary (zero momenta), when the dynamic plot includes the momentum.

For the calculation r1=rts and r2=rts+0.01, the reaction would be in the other way around.

=== Reactive and unreactive trajectories ===
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?

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!P1
!P2
!Etot
!Contour plot
!Reactive?
!Description of the dynamics
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|[[File:Q5 contour1.png|frameless]]
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|[[File:Q5 contour2.png|frameless]]
|No
|Atom C has not enough momentum to collide with AB.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|[[File:Q5 contour3.png|frameless]]
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|[[File:Q5 contour4.png|frameless]]
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|[[File:Q5 contour5.png|frameless]]
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The transition stat theory (TST) is used to calculate reaction constant k.

The main assumptions are:

1. The electronic motion and nuclear motion is separated in the way similar to Born-Oppenheimer approximation

2. Molecules which have passed the transition state in the direction of product cannot reform the reactant.

3. In the transition state, motion along the reaction coordinate is separated from the other motions and treated as translation.

4. Transition state molecules are distributed among their states according to the Maxwell-Bpltzman law, even there is no equalibrium set.

5. The system is in quasi-equalibrium, where the reaction process is slow enough for internal equalibrium.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River </ref>

Therefore, the experimental value is likely to be slower than the theoretical value, because the disrupting factors, such as backward reaction, are ignored in assumptions.

However, the TST also ignores tunnelling effect of quantum mechanics, which reduces energy barrier. This would make the experimental value faster than the theoretical value.

== F-H-H system ==

=== PES inspection ===

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

[[File:Q6 surface F+H2.png|left|400px|thumb|[Figure 8]F+H2 surface plot]]

[[File:Q6 surface H+HF.png|center|400px|thumb|[Figure 9] H+HF surface plot]]

For the potential energy surface diagram of F+H2, the energy level goes down as the F atom and H2 molecule gets closer. This means that the energy level of product is lower than that of reactant, indicating it is a exothermic reaction. The reaction of H+HF shows completely opposite situation. The energy level increases as H-H gets closer, indicating that the reaction is endothermic. This is directly related to the bond energy. An exothermic reaction is when the bond energy of new bond of product is greater than broken bonds of reactant. Therefore, H-H bond is weaker than H-F bond.

Locate the approximate position of the transition state.
[[File:Q6 contour1.png|centre|thumb|400x400px|[Figure 9] Position of the transition state of F+H2 shown in contour plot]]
It is achieved at rFH=1.813 and rHH=0.74, when momenta=0.

Report the activation energy for both reactions.

'''''F-H + H reaction:'''''

The reaction is achieved at rFH=1.65 and rHH=0.74

Activation energy =104.125-131.053=+26.928 kcalmol-1

[[File:Ea F+HH.png|centre|thumb|400x400px]]

'''''H-H + F reaction:'''''

The reaction is achieved at rFH=1.89 and rHH=0.74

Activation energy = 103.708-103.762= +0.054kcalmol-1

[[File:Ea H+HF.png|centre|thumb|400x400px|[Figure 11] Energy VS time graph]]

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

The reaction is achieved at rFH=1.65 and rHH=0.74.

[[File:Inlight1.png|centre|thumb|400x400px|[Figure 12] Momentum VS time graph. ]]
Energy is released due to the formation of H-F bond. And this causes the increase in oscillation of the bonds, as shown in [Figure 12]. This means the kinetic energy of the molecules has increased. Since temperature is defined as average kinetic energy, this can be experimentally identified 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

Polanyi's empirical rules states vibration is more effective to promote a late-barrier reaction than the translation.

[[File:HH2.png|left|thumb|400x400px|[Figure 13] Contour plot at PHH=-2]]
[[File:HH1.png|center|thumb|400x400px|[Figure 14] Contour plot at PHH=-1]]
[[File:HH0.5.png|right|thumb|400x400px|[Figure 15] Contour plot at PHH=-0.5]]

The illustrations above shows change in the collision path due to change in momentum of H. The momentum on F-H bond was set to 0.02. These plots are to investigate the effect of translation. All three plots have not shown reactive systems, but it gets worse if the translation is greater.

[[File:HH1HF1.png|left|thumb|400x400px|[Figure 13] Contour plot at PFH=1.0]]
[[File:HH1HF6.0.png|center|thumb|400x400px|[Figure 14] Contour plot at PFH=6.0]]
[[File:HH1HF6.5.png|right|thumb|400x400px|[Figure 15] Contour plot at PFH=6.5]]

The illustration above shows the change in the collision path as the vibration of F-H bond increases. The momentum on H-H was set to -1.09. At pFH=6.5, the system is found to be reactive. This clearly shows Polanyi's empirical rules.

File:HH1.png

2019-05-10T16:30:58Z

Syk2017:

MRD:syk2017

2019-05-10T16:30:13Z

Syk2017: /* Reaction dynamics */

== H+H2 system ==

=== Dynamics from the transition state region ===

==== Transition state ====
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

[[File:Q0 Surface Plot.png|center|400px|thumb|[Figure 1] Black line is located within the MEP]]

The transition state is a short-lived state of molecules or atoms at a local maximum in the reaction coordinate. Therefore, on a potential energy surface diagram, it can be defined as a local maximum of the minimum energy path (MEP). Mathematically it is a point where the gradient of potential is zero ( ∂V('''ri''')/∂'''ri'''=0). Although it is located in relatively low potential energy level, it actually is a local maximum point. It can be distinguished from other local minimum or local maximum points, as it should be located within the region of MEP.

'''

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.

'''<nowiki/>'''

[[File:Q2 Surface Plot.png|left|400px|thumb|[Figure 2] Position of transition state represented as 'x' on the contour plot]]
[[File:Q2 Surface Plot2.png|center|400px|thumb|[Figure 3] Internuclear distance VS time graph at transition state]]

The best estimate is achieved at rts=0.907. 

Since H+H2 is symmetric, the rAB and rBC should be equal. The figure 2 has only two lines, representing rBC and rAC . The rAB is not shown because rBC is overlapping it.

The definition of transition state suggests that the internuclear distance should stay constant throughout the given time, as ∂V('''ri''')/∂'''ri'''=0. The figure 2 shows that there is minimum vibration among the H atoms and no translation, suggesting the transition state.

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

Following graphs are generated by calculation on r1=0.917 and r2=0.907 when p1=p2=0.

[[File:Q3 contour Plot MEP.png|left|400px|thumb|[Figure 4] Contour graph of MEP]]
[[File:Q3 contour Plot DY.png|center|400px|thumb|[Figure 5] Contour graph of Dynamic]]
[[File:Q3 internu Plot MEP.png|left|400px|thumb|[Figure 6] Internuclear distance VS time graph of MEP]]
[[File:Q3 internu Plot DY.png|center|400px|thumb|[Figure 7] Internuclear distance VS time graph of Dynamic]]

Comment on how the ''mep'' and the trajectory you just calculated differ.

In the MEP trajectory, the motion of molecule is infinitely slow that the momenta is reset to zero in every time step. This effectively takes the vibrational energy out of the system. Therefore, smooth line is shown MEP trajectories, when wavy line is shown in dynamic trajectory. Comparing figure 6 and figure 7, the Dynamic trajectories shows linearly increasing internuclear distance, and the MEP trajectories show decrease in the gradient. This is because MET plot is taken as each point being stationary (zero momenta), when the dynamic plot includes the momentum.

For the calculation r1=rts and r2=rts+0.01, the reaction would be in the other way around.

=== Reactive and unreactive trajectories ===
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?

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!P1
!P2
!Etot
!Contour plot
!Reactive?
!Description of the dynamics
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|[[File:Q5 contour1.png|frameless]]
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|[[File:Q5 contour2.png|frameless]]
|No
|Atom C has not enough momentum to collide with AB.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|[[File:Q5 contour3.png|frameless]]
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|[[File:Q5 contour4.png|frameless]]
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|[[File:Q5 contour5.png|frameless]]
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The transition stat theory (TST) is used to calculate reaction constant k.

The main assumptions are:

1. The electronic motion and nuclear motion is separated in the way similar to Born-Oppenheimer approximation

2. Molecules which have passed the transition state in the direction of product cannot reform the reactant.

3. In the transition state, motion along the reaction coordinate is separated from the other motions and treated as translation.

4. Transition state molecules are distributed among their states according to the Maxwell-Bpltzman law, even there is no equalibrium set.

5. The system is in quasi-equalibrium, where the reaction process is slow enough for internal equalibrium.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River </ref>

Therefore, the experimental value is likely to be slower than the theoretical value, because the disrupting factors, such as backward reaction, are ignored in assumptions.

However, the TST also ignores tunnelling effect of quantum mechanics, which reduces energy barrier. This would make the experimental value faster than the theoretical value.

== F-H-H system ==

=== PES inspection ===

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

[[File:Q6 surface F+H2.png|left|400px|thumb|[Figure 8]F+H2 surface plot]]

[[File:Q6 surface H+HF.png|center|400px|thumb|[Figure 9] H+HF surface plot]]

For the potential energy surface diagram of F+H2, the energy level goes down as the F atom and H2 molecule gets closer. This means that the energy level of product is lower than that of reactant, indicating it is a exothermic reaction. The reaction of H+HF shows completely opposite situation. The energy level increases as H-H gets closer, indicating that the reaction is endothermic. This is directly related to the bond energy. An exothermic reaction is when the bond energy of new bond of product is greater than broken bonds of reactant. Therefore, H-H bond is weaker than H-F bond.

Locate the approximate position of the transition state.
[[File:Q6 contour1.png|centre|thumb|400x400px|[Figure 9] Position of the transition state of F+H2 shown in contour plot]]
It is achieved at rFH=1.813 and rHH=0.74, when momenta=0.

Report the activation energy for both reactions.

'''''F-H + H reaction:'''''

The reaction is achieved at rFH=1.65 and rHH=0.74

Activation energy =104.125-131.053=+26.928 kcalmol-1

[[File:Ea F+HH.png|centre|thumb|400x400px]]

'''''H-H + F reaction:'''''

The reaction is achieved at rFH=1.89 and rHH=0.74

Activation energy = 103.708-103.762= +0.054kcalmol-1

[[File:Ea H+HF.png|centre|thumb|400x400px|[Figure 11] Energy VS time graph]]

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

The reaction is achieved at rFH=1.65 and rHH=0.74.

[[File:Inlight1.png|centre|thumb|400x400px|[Figure 12] Momentum VS time graph. ]]
Energy is released due to the formation of H-F bond. And this causes the increase in oscillation of the bonds, as shown in [Figure 12]. This means the kinetic energy of the molecules has increased. Since temperature is defined as average kinetic energy, this can be experimentally identified 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

Polanyi's empirical rules states vibration is more effective to promote a late-barrier reaction than the translation.

[[File:HH2.png|left|thumb|400x400px|[Figure 13] Contour plot at PHH=-2]]
[[File:HH1.png|center|thumb|400x400px|[Figure 14] Contour plot at PHH=-1]]
[[File:HH0.5.png|right|thumb|400x400px|[Figure 15] Contour plot at PHH=-0.5]]

The illustrations above shows change in the collision path due to change in momentum of H. The momentum on F-H bond was set to 0.02. These plots are to investigate the effect of translation. All three plots have not shown reactive systems, but it gets worse if the translation is greater.

[[File:HH1HF1.png|left|thumb|400x400px|[Figure 13] Contour plot at PFH=1.0]]
[[File:HH1HF6.0.png|center|thumb|400x400px|[Figure 14] Contour plot at PFH=6.0]]
[[File:HH1HF6.5.png|right|thumb|400x400px|[Figure 15] Contour plot at PFH=6.5]]

The illustration above shows the change in the collision path as the vibration of F-H bond increases. The momentum on H-H was set to -1.09. At pFH=6.5, the system is found to be reactive. This clearly shows Polanyi's empirical rules.

File:HH1HF6.5.png

2019-05-10T16:29:55Z

Syk2017:

File:HH1HF6.0.png

2019-05-10T16:29:45Z

Syk2017:

File:HH1HF1.png

2019-05-10T16:29:35Z

Syk2017:

File:HH0.5.png

2019-05-10T16:29:19Z

Syk2017:

MRD:syk2017

2019-05-10T16:16:49Z

Syk2017: /* Reaction dynamics */

File:HH2.png

2019-05-10T16:15:59Z

Syk2017:

File:HH -1 HF 1.png

2019-05-10T16:11:15Z

Syk2017:

MRD:syk2017

2019-05-10T15:17:02Z

Syk2017: /* Reaction dynamics */

MRD:syk2017

2019-05-10T14:38:26Z

Syk2017: /* Reaction dynamics */

== H+H2 system ==

=== Dynamics from the transition state region ===

==== Transition state ====
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

[[File:Q0 Surface Plot.png|center|400px|thumb|[Figure 1] Black line is located within the MEP]]

The transition state is a short-lived state of molecules or atoms at a local maximum in the reaction coordinate. Therefore, on a potential energy surface diagram, it can be defined as a local maximum of the minimum energy path (MEP). Mathematically it is a point where the gradient of potential is zero ( ∂V('''ri''')/∂'''ri'''=0). Although it is located in relatively low potential energy level, it actually is a local maximum point. It can be distinguished from other local minimum or local maximum points, as it should be located within the region of MEP.

'''

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.

'''<nowiki/>'''

[[File:Q2 Surface Plot.png|left|400px|thumb|[Figure 2] Position of transition state represented as 'x' on the contour plot]]
[[File:Q2 Surface Plot2.png|center|400px|thumb|[Figure 3] Internuclear distance VS time graph at transition state]]

The best estimate is achieved at rts=0.907. 

Since H+H2 is symmetric, the rAB and rBC should be equal. The figure 2 has only two lines, representing rBC and rAC . The rAB is not shown because rBC is overlapping it.

The definition of transition state suggests that the internuclear distance should stay constant throughout the given time, as ∂V('''ri''')/∂'''ri'''=0. The figure 2 shows that there is minimum vibration among the H atoms and no translation, suggesting the transition state.

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

Following graphs are generated by calculation on r1=0.917 and r2=0.907 when p1=p2=0.

[[File:Q3 contour Plot MEP.png|left|400px|thumb|[Figure 4] Contour graph of MEP]]
[[File:Q3 contour Plot DY.png|center|400px|thumb|[Figure 5] Contour graph of Dynamic]]
[[File:Q3 internu Plot MEP.png|left|400px|thumb|[Figure 6] Internuclear distance VS time graph of MEP]]
[[File:Q3 internu Plot DY.png|center|400px|thumb|[Figure 7] Internuclear distance VS time graph of Dynamic]]

Comment on how the ''mep'' and the trajectory you just calculated differ.

In the MEP trajectory, the motion of molecule is infinitely slow that the momenta is reset to zero in every time step. This effectively takes the vibrational energy out of the system. Therefore, smooth line is shown MEP trajectories, when wavy line is shown in dynamic trajectory. Comparing figure 6 and figure 7, the Dynamic trajectories shows linearly increasing internuclear distance, and the MEP trajectories show decrease in the gradient. This is because MET plot is taken as each point being stationary (zero momenta), when the dynamic plot includes the momentum.

For the calculation r1=rts and r2=rts+0.01, the reaction would be in the other way around.

=== Reactive and unreactive trajectories ===
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?

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!P1
!P2
!Etot
!Contour plot
!Reactive?
!Description of the dynamics
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|[[File:Q5 contour1.png|frameless]]
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|[[File:Q5 contour2.png|frameless]]
|No
|Atom C has not enough momentum to collide with AB.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|[[File:Q5 contour3.png|frameless]]
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|[[File:Q5 contour4.png|frameless]]
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|[[File:Q5 contour5.png|frameless]]
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The transition stat theory (TST) is used to calculate reaction constant k.

The main assumptions are:

1. The electronic motion and nuclear motion is separated in the way similar to Born-Oppenheimer approximation

2. Molecules which have passed the transition state in the direction of product cannot reform the reactant.

3. In the transition state, motion along the reaction coordinate is separated from the other motions and treated as translation.

4. Transition state molecules are distributed among their states according to the Maxwell-Bpltzman law, even there is no equalibrium set.

5. The system is in quasi-equalibrium, where the reaction process is slow enough for internal equalibrium.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River </ref>

Therefore, the experimental value is likely to be slower than the theoretical value, because the disrupting factors, such as backward reaction, are ignored in assumptions.

However, the TST also ignores tunnelling effect of quantum mechanics, which reduces energy barrier. This would make the experimental value faster than the theoretical value.

== F-H-H system ==

=== PES inspection ===

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

[[File:Q6 surface F+H2.png|left|400px|thumb|[Figure 8]F+H2 surface plot]]

[[File:Q6 surface H+HF.png|center|400px|thumb|[Figure 9] H+HF surface plot]]

For the potential energy surface diagram of F+H2, the energy level goes down as the F atom and H2 molecule gets closer. This means that the energy level of product is lower than that of reactant, indicating it is a exothermic reaction. The reaction of H+HF shows completely opposite situation. The energy level increases as H-H gets closer, indicating that the reaction is endothermic. This is directly related to the bond energy. An exothermic reaction is when the bond energy of new bond of product is greater than broken bonds of reactant. Therefore, H-H bond is weaker than H-F bond.

Locate the approximate position of the transition state.
[[File:Q6 contour1.png|centre|thumb|400x400px|[Figure 9] Position of the transition state of F+H2 shown in contour plot]]
It is achieved at rFH=1.813 and rHH=0.74, when momenta=0.

Report the activation energy for both reactions.

'''''F-H + H reaction:'''''

The reaction is achieved at rFH=1.65 and rHH=0.74

Activation energy =104.125-131.053=+26.928 kcalmol-1

[[File:Ea F+HH.png|centre|thumb|400x400px]]

'''''H-H + F reaction:'''''

The reaction is achieved at rFH=1.89 and rHH=0.74

Activation energy = 103.708-103.762= +0.054kcalmol-1

[[File:Ea H+HF.png|centre|thumb|400x400px|[Figure 11] Energy VS time graph]]

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

The reaction is achieved at rFH=1.65 and rHH=0.74.

[[File:Inlight1.png|centre|thumb|400x400px|[Figure 12] Momentum VS time graph. ]]
Energy is released due to the formation of H-F bond. And this causes the increase in oscillation of the bonds, as shown in [Figure 12]. This means the kinetic energy of the molecules has increased. Since temperature is defined as average kinetic energy, this can be experimentally identified 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

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!P1
!P2
!Etot
!Contour plot
!Reactive?
!Description of the dynamics
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|[[File:Q5 contour1.png|frameless]]
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|[[File:Q5 contour2.png|frameless]]
|No
|Atom C has not enough momentum to collide with AB.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|[[File:Q5 contour3.png|frameless]]
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|[[File:Q5 contour4.png|frameless]]
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|[[File:Q5 contour5.png|frameless]]
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

MRD:syk2017

2019-05-10T14:36:21Z

Syk2017: /* Reaction dynamics */

File:Inlight1.png

2019-05-10T14:29:39Z

Syk2017:

MRD:syk2017

2019-05-10T14:26:58Z

Syk2017: /* PES inspection */

== H+H2 system ==

=== Dynamics from the transition state region ===

==== Transition state ====
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

[[File:Q0 Surface Plot.png|center|400px|thumb|[Figure 1] Black line is located within the MEP]]

The transition state is a short-lived state of molecules or atoms at a local maximum in the reaction coordinate. Therefore, on a potential energy surface diagram, it can be defined as a local maximum of the minimum energy path (MEP). Mathematically it is a point where the gradient of potential is zero ( ∂V('''ri''')/∂'''ri'''=0). Although it is located in relatively low potential energy level, it actually is a local maximum point. It can be distinguished from other local minimum or local maximum points, as it should be located within the region of MEP.

'''

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.

'''<nowiki/>'''

[[File:Q2 Surface Plot.png|left|400px|thumb|[Figure 2] Position of transition state represented as 'x' on the contour plot]]
[[File:Q2 Surface Plot2.png|center|400px|thumb|[Figure 3] Internuclear distance VS time graph at transition state]]

The best estimate is achieved at rts=0.907. 

Since H+H2 is symmetric, the rAB and rBC should be equal. The figure 2 has only two lines, representing rBC and rAC . The rAB is not shown because rBC is overlapping it.

The definition of transition state suggests that the internuclear distance should stay constant throughout the given time, as ∂V('''ri''')/∂'''ri'''=0. The figure 2 shows that there is minimum vibration among the H atoms and no translation, suggesting the transition state.

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

Following graphs are generated by calculation on r1=0.917 and r2=0.907 when p1=p2=0.

[[File:Q3 contour Plot MEP.png|left|400px|thumb|[Figure 4] Contour graph of MEP]]
[[File:Q3 contour Plot DY.png|center|400px|thumb|[Figure 5] Contour graph of Dynamic]]
[[File:Q3 internu Plot MEP.png|left|400px|thumb|[Figure 6] Internuclear distance VS time graph of MEP]]
[[File:Q3 internu Plot DY.png|center|400px|thumb|[Figure 7] Internuclear distance VS time graph of Dynamic]]

Comment on how the ''mep'' and the trajectory you just calculated differ.

In the MEP trajectory, the motion of molecule is infinitely slow that the momenta is reset to zero in every time step. This effectively takes the vibrational energy out of the system. Therefore, smooth line is shown MEP trajectories, when wavy line is shown in dynamic trajectory. Comparing figure 6 and figure 7, the Dynamic trajectories shows linearly increasing internuclear distance, and the MEP trajectories show decrease in the gradient. This is because MET plot is taken as each point being stationary (zero momenta), when the dynamic plot includes the momentum.

For the calculation r1=rts and r2=rts+0.01, the reaction would be in the other way around.

=== Reactive and unreactive trajectories ===
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?

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!P1
!P2
!Etot
!Contour plot
!Reactive?
!Description of the dynamics
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|[[File:Q5 contour1.png|frameless]]
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|[[File:Q5 contour2.png|frameless]]
|No
|Atom C has not enough momentum to collide with AB.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|[[File:Q5 contour3.png|frameless]]
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|[[File:Q5 contour4.png|frameless]]
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|[[File:Q5 contour5.png|frameless]]
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The transition stat theory (TST) is used to calculate reaction constant k.

The main assumptions are:

1. The electronic motion and nuclear motion is separated in the way similar to Born-Oppenheimer approximation

2. Molecules which have passed the transition state in the direction of product cannot reform the reactant.

3. In the transition state, motion along the reaction coordinate is separated from the other motions and treated as translation.

4. Transition state molecules are distributed among their states according to the Maxwell-Bpltzman law, even there is no equalibrium set.

5. The system is in quasi-equalibrium, where the reaction process is slow enough for internal equalibrium.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River </ref>

Therefore, the experimental value is likely to be slower than the theoretical value, because the disrupting factors, such as backward reaction, are ignored in assumptions.

However, the TST also ignores tunnelling effect of quantum mechanics, which reduces energy barrier. This would make the experimental value faster than the theoretical value.

== F-H-H system ==

=== PES inspection ===

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

[[File:Q6 surface F+H2.png|left|400px|thumb|[Figure 8]F+H2 surface plot]]

[[File:Q6 surface H+HF.png|center|400px|thumb|[Figure 9] H+HF surface plot]]

For the potential energy surface diagram of F+H2, the energy level goes down as the F atom and H2 molecule gets closer. This means that the energy level of product is lower than that of reactant, indicating it is a exothermic reaction. The reaction of H+HF shows completely opposite situation. The energy level increases as H-H gets closer, indicating that the reaction is endothermic. This is directly related to the bond energy. An exothermic reaction is when the bond energy of new bond of product is greater than broken bonds of reactant. Therefore, H-H bond is weaker than H-F bond.

Locate the approximate position of the transition state.
[[File:Q6 contour1.png|centre|thumb|400x400px|[Figure 9] Position of the transition state of F+H2 shown in contour plot]]
It is achieved at rFH=1.813 and rHH=0.74, when momenta=0.

Report the activation energy for both reactions.

'''''F-H + H reaction:'''''

The reaction is achieved at rFH=1.65 and rHH=0.74

Activation energy =104.125-131.053=+26.928 kcalmol-1

[[File:Ea F+HH.png|centre|thumb|400x400px]]

'''''H-H + F reaction:'''''

The reaction is achieved at rFH=1.89 and rHH=0.74

Activation energy = 103.708-103.762= +0.054kcalmol-1

[[File:Ea H+HF.png|centre|thumb|400x400px|[Figure 11] Energy VS time graph]]

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

Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state

File:Ea F+HH.png

2019-05-10T14:26:06Z

Syk2017:

File:Ea H+HF.png

2019-05-10T14:16:43Z

Syk2017: Syk2017 uploaded a new version of File:Ea H+HF.png

MRD:syk2017

2019-05-10T13:23:54Z

Syk2017: /* PES inspection */

== H+H2 system ==

=== Dynamics from the transition state region ===

==== Transition state ====
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

[[File:Q0 Surface Plot.png|center|400px|thumb|[Figure 1] Black line is located within the MEP]]

The transition state is a short-lived state of molecules or atoms at a local maximum in the reaction coordinate. Therefore, on a potential energy surface diagram, it can be defined as a local maximum of the minimum energy path (MEP). Mathematically it is a point where the gradient of potential is zero ( ∂V('''ri''')/∂'''ri'''=0). Although it is located in relatively low potential energy level, it actually is a local maximum point. It can be distinguished from other local minimum or local maximum points, as it should be located within the region of MEP.

'''

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.

'''<nowiki/>'''

[[File:Q2 Surface Plot.png|left|400px|thumb|[Figure 2] Position of transition state represented as 'x' on the contour plot]]
[[File:Q2 Surface Plot2.png|center|400px|thumb|[Figure 3] Internuclear distance VS time graph at transition state]]

The best estimate is achieved at rts=0.907. 

Since H+H2 is symmetric, the rAB and rBC should be equal. The figure 2 has only two lines, representing rBC and rAC . The rAB is not shown because rBC is overlapping it.

The definition of transition state suggests that the internuclear distance should stay constant throughout the given time, as ∂V('''ri''')/∂'''ri'''=0. The figure 2 shows that there is minimum vibration among the H atoms and no translation, suggesting the transition state.

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

Following graphs are generated by calculation on r1=0.917 and r2=0.907 when p1=p2=0.

[[File:Q3 contour Plot MEP.png|left|400px|thumb|[Figure 4] Contour graph of MEP]]
[[File:Q3 contour Plot DY.png|center|400px|thumb|[Figure 5] Contour graph of Dynamic]]
[[File:Q3 internu Plot MEP.png|left|400px|thumb|[Figure 6] Internuclear distance VS time graph of MEP]]
[[File:Q3 internu Plot DY.png|center|400px|thumb|[Figure 7] Internuclear distance VS time graph of Dynamic]]

Comment on how the ''mep'' and the trajectory you just calculated differ.

In the MEP trajectory, the motion of molecule is infinitely slow that the momenta is reset to zero in every time step. This effectively takes the vibrational energy out of the system. Therefore, smooth line is shown MEP trajectories, when wavy line is shown in dynamic trajectory. Comparing figure 6 and figure 7, the Dynamic trajectories shows linearly increasing internuclear distance, and the MEP trajectories show decrease in the gradient. This is because MET plot is taken as each point being stationary (zero momenta), when the dynamic plot includes the momentum.

For the calculation r1=rts and r2=rts+0.01, the reaction would be in the other way around.

=== Reactive and unreactive trajectories ===
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?

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!P1
!P2
!Etot
!Contour plot
!Reactive?
!Description of the dynamics
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|[[File:Q5 contour1.png|frameless]]
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|[[File:Q5 contour2.png|frameless]]
|No
|Atom C has not enough momentum to collide with AB.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|[[File:Q5 contour3.png|frameless]]
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|[[File:Q5 contour4.png|frameless]]
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|[[File:Q5 contour5.png|frameless]]
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The transition stat theory (TST) is used to calculate reaction constant k.

The main assumptions are:

1. The electronic motion and nuclear motion is separated in the way similar to Born-Oppenheimer approximation

2. Molecules which have passed the transition state in the direction of product cannot reform the reactant.

3. In the transition state, motion along the reaction coordinate is separated from the other motions and treated as translation.

4. Transition state molecules are distributed among their states according to the Maxwell-Bpltzman law, even there is no equalibrium set.

5. The system is in quasi-equalibrium, where the reaction process is slow enough for internal equalibrium.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River </ref>

Therefore, the experimental value is likely to be slower than the theoretical value, because the disrupting factors, such as backward reaction, are ignored in assumptions.

However, the TST also ignores tunnelling effect of quantum mechanics, which reduces energy barrier. This would make the experimental value faster than the theoretical value.

== F-H-H system ==

=== PES inspection ===

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

[[File:Q6 surface F+H2.png|left|400px|thumb|[Figure 8]F+H2 surface plot]]

[[File:Q6 surface H+HF.png|center|400px|thumb|[Figure 9] H+HF surface plot]]

For the potential energy surface diagram of F+H2, the energy level goes down as the F atom and H2 molecule gets closer. This means that the energy level of product is lower than that of reactant, indicating it is a exothermic reaction. The reaction of H+HF shows completely opposite situation. The energy level increases as H-H gets closer, indicating that the reaction is endothermic. This is directly related to the bond energy. An exothermic reaction is when the bond energy of new bond of product is greater than broken bonds of reactant. Therefore, H-H bond is weaker than H-F bond.

Locate the approximate position of the transition state.

[[File:Q6 surface H+HF.png|center|400px|thumb|[Figure 9] Position of the transition state of F+H2 shown in contour plot]]

It is achieved at rFH=1.813 and rHH=0.74, when momenta=0.

Report the activation energy for both reactions.

H-H + F reaction = +0.250 kcalmol-1
Total energy=
Potential energy=
Activation energy =

F-H + H reaction = +29.943 kcalmol-1
*The transition state is achieved at rFH=0.92 and rHH=2.0

Total energy=103.702
Potential energy=103.772
Activation energy = 0.070

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

Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state

File:Q6 contour1.png

2019-05-10T12:40:16Z

Syk2017:

MRD:syk2017

2019-05-10T12:11:49Z

Syk2017: /* PES inspection */

== H+H2 system ==

=== Dynamics from the transition state region ===

==== Transition state ====
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

[[File:Q0 Surface Plot.png|center|400px|thumb|[Figure 1] Black line is located within the MEP]]

The transition state is a short-lived state of molecules or atoms at a local maximum in the reaction coordinate. Therefore, on a potential energy surface diagram, it can be defined as a local maximum of the minimum energy path (MEP). Mathematically it is a point where the gradient of potential is zero ( ∂V('''ri''')/∂'''ri'''=0). Although it is located in relatively low potential energy level, it actually is a local maximum point. It can be distinguished from other local minimum or local maximum points, as it should be located within the region of MEP.

'''

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.

'''<nowiki/>'''

[[File:Q2 Surface Plot.png|left|400px|thumb|[Figure 2] Position of transition state represented as 'x' on the contour plot]]
[[File:Q2 Surface Plot2.png|center|400px|thumb|[Figure 3] Internuclear distance VS time graph at transition state]]

The best estimate is achieved at rts=0.907. 

Since H+H2 is symmetric, the rAB and rBC should be equal. The figure 2 has only two lines, representing rBC and rAC . The rAB is not shown because rBC is overlapping it.

The definition of transition state suggests that the internuclear distance should stay constant throughout the given time, as ∂V('''ri''')/∂'''ri'''=0. The figure 2 shows that there is minimum vibration among the H atoms and no translation, suggesting the transition state.

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

Following graphs are generated by calculation on r1=0.917 and r2=0.907 when p1=p2=0.

[[File:Q3 contour Plot MEP.png|left|400px|thumb|[Figure 4] Contour graph of MEP]]
[[File:Q3 contour Plot DY.png|center|400px|thumb|[Figure 5] Contour graph of Dynamic]]
[[File:Q3 internu Plot MEP.png|left|400px|thumb|[Figure 6] Internuclear distance VS time graph of MEP]]
[[File:Q3 internu Plot DY.png|center|400px|thumb|[Figure 7] Internuclear distance VS time graph of Dynamic]]

Comment on how the ''mep'' and the trajectory you just calculated differ.

In the MEP trajectory, the motion of molecule is infinitely slow that the momenta is reset to zero in every time step. This effectively takes the vibrational energy out of the system. Therefore, smooth line is shown MEP trajectories, when wavy line is shown in dynamic trajectory. Comparing figure 6 and figure 7, the Dynamic trajectories shows linearly increasing internuclear distance, and the MEP trajectories show decrease in the gradient. This is because MET plot is taken as each point being stationary (zero momenta), when the dynamic plot includes the momentum.

For the calculation r1=rts and r2=rts+0.01, the reaction would be in the other way around.

=== Reactive and unreactive trajectories ===
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?

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!P1
!P2
!Etot
!Contour plot
!Reactive?
!Description of the dynamics
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|[[File:Q5 contour1.png|frameless]]
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|[[File:Q5 contour2.png|frameless]]
|No
|Atom C has not enough momentum to collide with AB.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|[[File:Q5 contour3.png|frameless]]
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|[[File:Q5 contour4.png|frameless]]
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|[[File:Q5 contour5.png|frameless]]
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The transition stat theory (TST) is used to calculate reaction constant k.

The main assumptions are:

1. The electronic motion and nuclear motion is separated in the way similar to Born-Oppenheimer approximation

2. Molecules which have passed the transition state in the direction of product cannot reform the reactant.

3. In the transition state, motion along the reaction coordinate is separated from the other motions and treated as translation.

4. Transition state molecules are distributed among their states according to the Maxwell-Bpltzman law, even there is no equalibrium set.

5. The system is in quasi-equalibrium, where the reaction process is slow enough for internal equalibrium.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River </ref>

Therefore, the experimental value is likely to be slower than the theoretical value, because the disrupting factors, such as backward reaction, are ignored in assumptions.

However, the TST also ignores tunnelling effect of quantum mechanics, which reduces energy barrier. This would make the experimental value faster than the theoretical value.

== F-H-H system ==

=== PES inspection ===

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

[[File:Q6 surface F+H2.png|left|400px|thumb|[Figure 8]F+H2 surface plot]]

[[File:Q6 surface H+HF.png|center|400px|thumb|[Figure 9] H+HF surface plot]]

For the potential energy surface diagram of F+H2, the energy level goes down as the F atom and H2 molecule gets closer. This

The potential energy surface diagram of F + H2 shows the path of the reaction going from a higher energy minimum to a lower energy minimum, indicating that the products are at a lower energy than the reactants, which therefore corresponds to an exothermic reaction.

The potential energy surface diagram of H + HF shows the path of the reaction going from a lower energy minimum reaction to a higher energy minimum, indicating that the reactants are at a lower energy than the products, which therefore corresponds to an endothermic reaction.

Energy is taken in to break bonds, and energy is given out to make bonds. When the energy taken in is greater than the energy given out, this results in an endothermic reaction, and the opposite leads to an exothermic one. With F + H2, it's exothermic, which means that the bond formed (i.e. H-F) has a greater bond enthalpy than H2. With H + HF, it's endothermic, which suggests that the bond formed (i.e. H-H) has a lower bond enthalpy than the bond broken (i.e. H-F).

Locate the approximate position of the transition state.

Report the activation energy for both reactions.

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

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

MRD:syk2017

2019-05-10T12:10:05Z

Syk2017: /* PES inspection */

== H+H2 system ==

=== Dynamics from the transition state region ===

==== Transition state ====
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

[[File:Q0 Surface Plot.png|center|400px|thumb|[Figure 1] Black line is located within the MEP]]

The transition state is a short-lived state of molecules or atoms at a local maximum in the reaction coordinate. Therefore, on a potential energy surface diagram, it can be defined as a local maximum of the minimum energy path (MEP). Mathematically it is a point where the gradient of potential is zero ( ∂V('''ri''')/∂'''ri'''=0). Although it is located in relatively low potential energy level, it actually is a local maximum point. It can be distinguished from other local minimum or local maximum points, as it should be located within the region of MEP.

'''

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.

'''<nowiki/>'''

[[File:Q2 Surface Plot.png|left|400px|thumb|[Figure 2] Position of transition state represented as 'x' on the contour plot]]
[[File:Q2 Surface Plot2.png|center|400px|thumb|[Figure 3] Internuclear distance VS time graph at transition state]]

The best estimate is achieved at rts=0.907. 

Since H+H2 is symmetric, the rAB and rBC should be equal. The figure 2 has only two lines, representing rBC and rAC . The rAB is not shown because rBC is overlapping it.

The definition of transition state suggests that the internuclear distance should stay constant throughout the given time, as ∂V('''ri''')/∂'''ri'''=0. The figure 2 shows that there is minimum vibration among the H atoms and no translation, suggesting the transition state.

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

Following graphs are generated by calculation on r1=0.917 and r2=0.907 when p1=p2=0.

[[File:Q3 contour Plot MEP.png|left|400px|thumb|[Figure 4] Contour graph of MEP]]
[[File:Q3 contour Plot DY.png|center|400px|thumb|[Figure 5] Contour graph of Dynamic]]
[[File:Q3 internu Plot MEP.png|left|400px|thumb|[Figure 6] Internuclear distance VS time graph of MEP]]
[[File:Q3 internu Plot DY.png|center|400px|thumb|[Figure 7] Internuclear distance VS time graph of Dynamic]]

Comment on how the ''mep'' and the trajectory you just calculated differ.

In the MEP trajectory, the motion of molecule is infinitely slow that the momenta is reset to zero in every time step. This effectively takes the vibrational energy out of the system. Therefore, smooth line is shown MEP trajectories, when wavy line is shown in dynamic trajectory. Comparing figure 6 and figure 7, the Dynamic trajectories shows linearly increasing internuclear distance, and the MEP trajectories show decrease in the gradient. This is because MET plot is taken as each point being stationary (zero momenta), when the dynamic plot includes the momentum.

For the calculation r1=rts and r2=rts+0.01, the reaction would be in the other way around.

=== Reactive and unreactive trajectories ===
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?

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!P1
!P2
!Etot
!Contour plot
!Reactive?
!Description of the dynamics
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|[[File:Q5 contour1.png|frameless]]
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|[[File:Q5 contour2.png|frameless]]
|No
|Atom C has not enough momentum to collide with AB.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|[[File:Q5 contour3.png|frameless]]
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|[[File:Q5 contour4.png|frameless]]
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|[[File:Q5 contour5.png|frameless]]
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The transition stat theory (TST) is used to calculate reaction constant k.

The main assumptions are:

1. The electronic motion and nuclear motion is separated in the way similar to Born-Oppenheimer approximation

2. Molecules which have passed the transition state in the direction of product cannot reform the reactant.

3. In the transition state, motion along the reaction coordinate is separated from the other motions and treated as translation.

4. Transition state molecules are distributed among their states according to the Maxwell-Bpltzman law, even there is no equalibrium set.

5. The system is in quasi-equalibrium, where the reaction process is slow enough for internal equalibrium.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River </ref>

Therefore, the experimental value is likely to be slower than the theoretical value, because the disrupting factors, such as backward reaction, are ignored in assumptions.

However, the TST also ignores tunnelling effect of quantum mechanics, which reduces energy barrier. This would make the experimental value faster than the theoretical value.

== F-H-H system ==

=== PES inspection ===

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

[[File:Q6 surface F+H2.png|left|frame|[Figure 8] F+H2 surface plot]]

[[File:Q6 surface H+HF.png|thumb]]
For the potential energy surface diagram of F+H2, the energy level goes down as the F atom and H2 molecule gets closer. This

The potential energy surface diagram of F + H2 shows the path of the reaction going from a higher energy minimum to a lower energy minimum, indicating that the products are at a lower energy than the reactants, which therefore corresponds to an exothermic reaction.

The potential energy surface diagram of H + HF shows the path of the reaction going from a lower energy minimum reaction to a higher energy minimum, indicating that the reactants are at a lower energy than the products, which therefore corresponds to an endothermic reaction.

Energy is taken in to break bonds, and energy is given out to make bonds. When the energy taken in is greater than the energy given out, this results in an endothermic reaction, and the opposite leads to an exothermic one. With F + H2, it's exothermic, which means that the bond formed (i.e. H-F) has a greater bond enthalpy than H2. With H + HF, it's endothermic, which suggests that the bond formed (i.e. H-H) has a lower bond enthalpy than the bond broken (i.e. H-F).

Locate the approximate position of the transition state.

Report the activation energy for both reactions.

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

Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state

File:Q6 surface H+HF.png

2019-05-10T12:09:40Z

Syk2017:

File:Q6 surface F+H2.png

2019-05-10T12:07:42Z

Syk2017:

MRD:syk2017

2019-05-10T12:07:16Z

Syk2017: /* Trajectories from r1 = rts+δ, r2 = rts */

== H+H2 system ==

=== Dynamics from the transition state region ===

==== Transition state ====
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

[[File:Q0 Surface Plot.png|center|400px|thumb|[Figure 1] Black line is located within the MEP]]

The transition state is a short-lived state of molecules or atoms at a local maximum in the reaction coordinate. Therefore, on a potential energy surface diagram, it can be defined as a local maximum of the minimum energy path (MEP). Mathematically it is a point where the gradient of potential is zero ( ∂V('''ri''')/∂'''ri'''=0). Although it is located in relatively low potential energy level, it actually is a local maximum point. It can be distinguished from other local minimum or local maximum points, as it should be located within the region of MEP.

'''

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.

'''<nowiki/>'''

[[File:Q2 Surface Plot.png|left|400px|thumb|[Figure 2] Position of transition state represented as 'x' on the contour plot]]
[[File:Q2 Surface Plot2.png|center|400px|thumb|[Figure 3] Internuclear distance VS time graph at transition state]]

The best estimate is achieved at rts=0.907. 

Since H+H2 is symmetric, the rAB and rBC should be equal. The figure 2 has only two lines, representing rBC and rAC . The rAB is not shown because rBC is overlapping it.

The definition of transition state suggests that the internuclear distance should stay constant throughout the given time, as ∂V('''ri''')/∂'''ri'''=0. The figure 2 shows that there is minimum vibration among the H atoms and no translation, suggesting the transition state.

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

Following graphs are generated by calculation on r1=0.917 and r2=0.907 when p1=p2=0.

[[File:Q3 contour Plot MEP.png|left|400px|thumb|[Figure 4] Contour graph of MEP]]
[[File:Q3 contour Plot DY.png|center|400px|thumb|[Figure 5] Contour graph of Dynamic]]
[[File:Q3 internu Plot MEP.png|left|400px|thumb|[Figure 6] Internuclear distance VS time graph of MEP]]
[[File:Q3 internu Plot DY.png|center|400px|thumb|[Figure 7] Internuclear distance VS time graph of Dynamic]]

Comment on how the ''mep'' and the trajectory you just calculated differ.

In the MEP trajectory, the motion of molecule is infinitely slow that the momenta is reset to zero in every time step. This effectively takes the vibrational energy out of the system. Therefore, smooth line is shown MEP trajectories, when wavy line is shown in dynamic trajectory. Comparing figure 6 and figure 7, the Dynamic trajectories shows linearly increasing internuclear distance, and the MEP trajectories show decrease in the gradient. This is because MET plot is taken as each point being stationary (zero momenta), when the dynamic plot includes the momentum.

For the calculation r1=rts and r2=rts+0.01, the reaction would be in the other way around.

=== Reactive and unreactive trajectories ===
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?

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!P1
!P2
!Etot
!Contour plot
!Reactive?
!Description of the dynamics
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|[[File:Q5 contour1.png|frameless]]
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|[[File:Q5 contour2.png|frameless]]
|No
|Atom C has not enough momentum to collide with AB.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|[[File:Q5 contour3.png|frameless]]
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|[[File:Q5 contour4.png|frameless]]
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|[[File:Q5 contour5.png|frameless]]
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The transition stat theory (TST) is used to calculate reaction constant k.

The main assumptions are:

1. The electronic motion and nuclear motion is separated in the way similar to Born-Oppenheimer approximation

2. Molecules which have passed the transition state in the direction of product cannot reform the reactant.

3. In the transition state, motion along the reaction coordinate is separated from the other motions and treated as translation.

4. Transition state molecules are distributed among their states according to the Maxwell-Bpltzman law, even there is no equalibrium set.

5. The system is in quasi-equalibrium, where the reaction process is slow enough for internal equalibrium.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River </ref>

Therefore, the experimental value is likely to be slower than the theoretical value, because the disrupting factors, such as backward reaction, are ignored in assumptions.

However, the TST also ignores tunnelling effect of quantum mechanics, which reduces energy barrier. This would make the experimental value faster than the theoretical value.

== F-H-H system ==

=== PES inspection ===

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

For the potential energy surface diagram of F+H2, the energy level goes down as the F atom and H2 molecule gets closer. This

The potential energy surface diagram of F + H2 shows the path of the reaction going from a higher energy minimum to a lower energy minimum, indicating that the products are at a lower energy than the reactants, which therefore corresponds to an exothermic reaction.

The potential energy surface diagram of H + HF shows the path of the reaction going from a lower energy minimum reaction to a higher energy minimum, indicating that the reactants are at a lower energy than the products, which therefore corresponds to an endothermic reaction.

Energy is taken in to break bonds, and energy is given out to make bonds. When the energy taken in is greater than the energy given out, this results in an endothermic reaction, and the opposite leads to an exothermic one. With F + H2, it's exothermic, which means that the bond formed (i.e. H-F) has a greater bond enthalpy than H2. With H + HF, it's endothermic, which suggests that the bond formed (i.e. H-H) has a lower bond enthalpy than the bond broken (i.e. H-F).

Locate the approximate position of the transition state.

Report the activation energy for both reactions.

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

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

MRD:syk2017

2019-05-10T11:48:45Z

Syk2017: /* Reactive and unreactive trajectories */

== H+H2 system ==

=== Dynamics from the transition state region ===

==== Transition state ====
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

[[File:Q0 Surface Plot.png|center|400px|thumb|[Figure 1] Black line is located within the MEP]]

The transition state is a short-lived state of molecules or atoms at a local maximum in the reaction coordinate. Therefore, on a potential energy surface diagram, it can be defined as a local maximum of the minimum energy path (MEP). Mathematically it is a point where the gradient of potential is zero ( ∂V('''ri''')/∂'''ri'''=0). Although it is located in relatively low potential energy level, it actually is a local maximum point. It can be distinguished from other local minimum or local maximum points, as it should be located within the region of MEP.

'''

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.

'''<nowiki/>'''

[[File:Q2 Surface Plot.png|left|400px|thumb|[Figure 2] Position of transition state represented as 'x' on the contour plot]]
[[File:Q2 Surface Plot2.png|center|400px|thumb|[Figure 3] Internuclear distance VS time graph at transition state]]

The best estimate is achieved at rts=0.907. 

Since H+H2 is symmetric, the rAB and rBC should be equal. The figure 2 has only two lines, representing rBC and rAC . The rAB is not shown because rBC is overlapping it.

The definition of transition state suggests that the internuclear distance should stay constant throughout the given time, as ∂V('''ri''')/∂'''ri'''=0. The figure 2 shows that there is minimum vibration among the H atoms and no translation, suggesting the transition state.

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

Following graphs are generated by calculation on r1=0.917 and r2=0.907 when p1=p2=0.

[[File:Q3 contour Plot MEP.png|left|400px|thumb|[Figure 4] Contour graph of MEP]]
[[File:Q3 contour Plot DY.png|center|400px|thumb|[Figure 5] Contour graph of Dynamic]]
[[File:Q3 internu Plot MEP.png|left|400px|thumb|[Figure 6] Internuclear distance VS time graph of MEP]]
[[File:Q3 internu Plot DY.png|center|400px|thumb|[Figure 7] Internuclear distance VS time graph of Dynamic]]

Comment on how the ''mep'' and the trajectory you just calculated differ.

In the MEP trajectory, the motion of molecule is infinitely slow that the momenta is reset to zero in every time step. This effectively takes the vibrational energy out of the system. Therefore, smooth line is shown MEP trajectories, when wavy line is shown in dynamic trajectory. Comparing figure 6 and figure 7, the Dynamic trajectories shows linearly increasing internuclear distance, and the MEP trajectories show decrease in the gradient. This is because MET plot is taken as each point being stationary (zero momenta), when the dynamic plot includes the momentum.

For the calculation r1=rts and r2=rts+0.01, the reaction would be in the other way around.

=== Reactive and unreactive trajectories ===
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?

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!P1
!P2
!Etot
!Contour plot
!Reactive?
!Description of the dynamics
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|[[File:Q5 contour1.png|frameless]]
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|[[File:Q5 contour2.png|frameless]]
|No
|Atom C has not enough momentum to collide with AB.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|[[File:Q5 contour3.png|frameless]]
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|[[File:Q5 contour4.png|frameless]]
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|[[File:Q5 contour5.png|frameless]]
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The transition stat theory (TST) is used to calculate reaction constant k.

The main assumptions are:

1. The electronic motion and nuclear motion is separated in the way similar to Born-Oppenheimer approximation

2. Molecules which have passed the transition state in the direction of product cannot reform the reactant.

3. In the transition state, motion along the reaction coordinate is separated from the other motions and treated as translation.

4. Transition state molecules are distributed among their states according to the Maxwell-Bpltzman law, even there is no equalibrium set.

5. The system is in quasi-equalibrium, where the reaction process is slow enough for internal equalibrium.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River </ref>

Therefore, the experimental value is likely to be slower than the theoretical value, because the disrupting factors, such as backward reaction, are ignored in assumptions.

However, the TST also ignores tunnelling effect of quantum mechanics, which reduces energy barrier. This would make the experimental value faster than the theoretical value.

== F-H-H system ==

=== PES inspection ===

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

For the potential energy surface diagram of F+H2, the energy level goes down as the F atom and H2 molecule gets closer. This

The potential energy surface diagram of F + H2 shows the path of the reaction going from a higher energy minimum to a lower energy minimum, indicating that the products are at a lower energy than the reactants, which therefore corresponds to an exothermic reaction.

The potential energy surface diagram of H + HF shows the path of the reaction going from a lower energy minimum reaction to a higher energy minimum, indicating that the reactants are at a lower energy than the products, which therefore corresponds to an endothermic reaction.

Energy is taken in to break bonds, and energy is given out to make bonds. When the energy taken in is greater than the energy given out, this results in an endothermic reaction, and the opposite leads to an exothermic one. With F + H2, it's exothermic, which means that the bond formed (i.e. H-F) has a greater bond enthalpy than H2. With H + HF, it's endothermic, which suggests that the bond formed (i.e. H-H) has a lower bond enthalpy than the bond broken (i.e. H-F).

Locate the approximate position of the transition state.

Report the activation energy for both reactions.

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

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

MRD:syk2017

2019-05-10T11:42:22Z

Syk2017: /* Trajectories from r1 = rts+δ, r2 = rts */

== H+H2 system ==

=== Dynamics from the transition state region ===

==== Transition state ====
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

[[File:Q0 Surface Plot.png|center|400px|thumb|[Figure 1] Black line is located within the MEP]]

The transition state is a short-lived state of molecules or atoms at a local maximum in the reaction coordinate. Therefore, on a potential energy surface diagram, it can be defined as a local maximum of the minimum energy path (MEP). Mathematically it is a point where the gradient of potential is zero ( ∂V('''ri''')/∂'''ri'''=0). Although it is located in relatively low potential energy level, it actually is a local maximum point. It can be distinguished from other local minimum or local maximum points, as it should be located within the region of MEP.

'''

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.

'''<nowiki/>'''

[[File:Q2 Surface Plot.png|left|400px|thumb|[Figure 2] Position of transition state represented as 'x' on the contour plot]]
[[File:Q2 Surface Plot2.png|center|400px|thumb|[Figure 3] Internuclear distance VS time graph at transition state]]

The best estimate is achieved at rts=0.907. 

Since H+H2 is symmetric, the rAB and rBC should be equal. The figure 2 has only two lines, representing rBC and rAC . The rAB is not shown because rBC is overlapping it.

The definition of transition state suggests that the internuclear distance should stay constant throughout the given time, as ∂V('''ri''')/∂'''ri'''=0. The figure 2 shows that there is minimum vibration among the H atoms and no translation, suggesting the transition state.

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

Following graphs are generated by calculation on r1=0.917 and r2=0.907 when p1=p2=0.

[[File:Q3 contour Plot MEP.png|left|400px|thumb|[Figure 4] Contour graph of MEP]]
[[File:Q3 contour Plot DY.png|center|400px|thumb|[Figure 5] Contour graph of Dynamic]]
[[File:Q3 internu Plot MEP.png|left|400px|thumb|[Figure 6] Internuclear distance VS time graph of MEP]]
[[File:Q3 internu Plot DY.png|center|400px|thumb|[Figure 7] Internuclear distance VS time graph of Dynamic]]

Comment on how the ''mep'' and the trajectory you just calculated differ.

In the MEP trajectory, the motion of molecule is infinitely slow that the momenta is reset to zero in every time step. This effectively takes the vibrational energy out of the system. Therefore, smooth line is shown MEP trajectories, when wavy line is shown in dynamic trajectory. Comparing figure 6 and figure 7, the Dynamic trajectories shows linearly increasing internuclear distance, and the MEP trajectories show decrease in the gradient. This is because MET plot is taken as each point being stationary (zero momenta), when the dynamic plot includes the momentum.

For the calculation r1=rts and r2=rts+0.01, the reaction would be in the other way around.

=== Reactive and unreactive trajectories ===
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?

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!P1
!P2
!Etot
!Contour plot
!Reactive?
!Description of the dynamics
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|[[File:Q5 contour1.png|frameless]]
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|[[File:Q5 contour2.png|frameless]]
|No
|Atom C has not enough momentum to collide with AB.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|[[File:Q5 contour3.png|frameless]]
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|[[File:Q5 contour4.png|frameless]]
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|[[File:Q5 contour5.png|frameless]]
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The main assumptions are:
1.molecules which have passed the transition state in the direction of product cannot reform the reactant.
2. In the transition state, motion along the reaction coordinate is separated from the other motions and treated as translation.
3. Transition state molecules are distributed among their states according to the Maxwell-Bpltzman law.
4. The system is in quasi-equalibrium, where the reaction process is slow enough for internal equalibrium.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River </ref>

Therefore, the experimental value is likely to be slower than the theoretical value, because the disrupting factors are ignored as assumption.

== F-H-H system ==

=== PES inspection ===

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

For the potential energy surface diagram of F+H2, the energy level goes down as the F atom and H2 molecule gets closer. This

The potential energy surface diagram of F + H2 shows the path of the reaction going from a higher energy minimum to a lower energy minimum, indicating that the products are at a lower energy than the reactants, which therefore corresponds to an exothermic reaction.

The potential energy surface diagram of H + HF shows the path of the reaction going from a lower energy minimum reaction to a higher energy minimum, indicating that the reactants are at a lower energy than the products, which therefore corresponds to an endothermic reaction.

Energy is taken in to break bonds, and energy is given out to make bonds. When the energy taken in is greater than the energy given out, this results in an endothermic reaction, and the opposite leads to an exothermic one. With F + H2, it's exothermic, which means that the bond formed (i.e. H-F) has a greater bond enthalpy than H2. With H + HF, it's endothermic, which suggests that the bond formed (i.e. H-H) has a lower bond enthalpy than the bond broken (i.e. H-F).

Locate the approximate position of the transition state.

Report the activation energy for both reactions.

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

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

MRD:syk2017

2019-05-10T11:41:48Z

Syk2017: /* Trajectories from r1 = rts+δ, r2 = rts */

== H+H2 system ==

=== Dynamics from the transition state region ===

==== Transition state ====
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

[[File:Q0 Surface Plot.png|center|400px|thumb|[Figure 1] Black line is located within the MEP]]

The transition state is a short-lived state of molecules or atoms at a local maximum in the reaction coordinate. Therefore, on a potential energy surface diagram, it can be defined as a local maximum of the minimum energy path (MEP). Mathematically it is a point where the gradient of potential is zero ( ∂V('''ri''')/∂'''ri'''=0). Although it is located in relatively low potential energy level, it actually is a local maximum point. It can be distinguished from other local minimum or local maximum points, as it should be located within the region of MEP.

'''

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.

'''<nowiki/>'''

[[File:Q2 Surface Plot.png|left|400px|thumb|[Figure 2] Position of transition state represented as 'x' on the contour plot]]
[[File:Q2 Surface Plot2.png|center|400px|thumb|[Figure 3] Internuclear distance VS time graph at transition state]]

The best estimate is achieved at rts=0.907. 

Since H+H2 is symmetric, the rAB and rBC should be equal. The figure 2 has only two lines, representing rBC and rAC . The rAB is not shown because rBC is overlapping it.

The definition of transition state suggests that the internuclear distance should stay constant throughout the given time, as ∂V('''ri''')/∂'''ri'''=0. The figure 2 shows that there is minimum vibration among the H atoms and no translation, suggesting the transition state.

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

Following graphs are generated by calculation on r1=0.917 and r2=0.907 when p1=p2=0.[[File:Q3 contour Plot MEP.png|left|400px|thumb|[Figure 4] Contour graph of MEP]]
[[File:Q3 contour Plot DY.png|center|400px|thumb|[Figure 5] Contour graph of Dynamic]]
[[File:Q3 internu Plot MEP.png|left|400px|thumb|[Figure 6] Internuclear distance VS time graph of MEP]]
[[File:Q3 internu Plot DY.png|center|400px|thumb|[Figure 7] Internuclear distance VS time graph of Dynamic]]

Comment on how the ''mep'' and the trajectory you just calculated differ.

In the MEP trajectory, the motion of molecule is infinitely slow that the momenta is reset to zero in every time step. This effectively takes the vibrational energy out of the system. Therefore, smooth line is shown MEP trajectories, when wavy line is shown in dynamic trajectory. Comparing figure 6 and figure 7, the Dynamic trajectories shows linearly increasing internuclear distance, and the MEP trajectories show decrease in the gradient. This is because MET plot is taken as each point being stationary (zero momenta), when the dynamic plot includes the momentum.

For the calculation r1=rts and r2=rts+0.01, the reaction would be in the other way around.

=== Reactive and unreactive trajectories ===
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?

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!P1
!P2
!Etot
!Contour plot
!Reactive?
!Description of the dynamics
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|[[File:Q5 contour1.png|frameless]]
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|[[File:Q5 contour2.png|frameless]]
|No
|Atom C has not enough momentum to collide with AB.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|[[File:Q5 contour3.png|frameless]]
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|[[File:Q5 contour4.png|frameless]]
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|[[File:Q5 contour5.png|frameless]]
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The main assumptions are:
1.molecules which have passed the transition state in the direction of product cannot reform the reactant.
2. In the transition state, motion along the reaction coordinate is separated from the other motions and treated as translation.
3. Transition state molecules are distributed among their states according to the Maxwell-Bpltzman law.
4. The system is in quasi-equalibrium, where the reaction process is slow enough for internal equalibrium.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River </ref>

Therefore, the experimental value is likely to be slower than the theoretical value, because the disrupting factors are ignored as assumption.

== F-H-H system ==

=== PES inspection ===

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

For the potential energy surface diagram of F+H2, the energy level goes down as the F atom and H2 molecule gets closer. This

The potential energy surface diagram of F + H2 shows the path of the reaction going from a higher energy minimum to a lower energy minimum, indicating that the products are at a lower energy than the reactants, which therefore corresponds to an exothermic reaction.

The potential energy surface diagram of H + HF shows the path of the reaction going from a lower energy minimum reaction to a higher energy minimum, indicating that the reactants are at a lower energy than the products, which therefore corresponds to an endothermic reaction.

Energy is taken in to break bonds, and energy is given out to make bonds. When the energy taken in is greater than the energy given out, this results in an endothermic reaction, and the opposite leads to an exothermic one. With F + H2, it's exothermic, which means that the bond formed (i.e. H-F) has a greater bond enthalpy than H2. With H + HF, it's endothermic, which suggests that the bond formed (i.e. H-H) has a lower bond enthalpy than the bond broken (i.e. H-F).

Locate the approximate position of the transition state.

Report the activation energy for both reactions.

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

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

MRD:syk2017

2019-05-10T11:37:06Z

Syk2017: /* Trajectories from r1 = rts+δ, r2 = rts */

== H+H2 system ==

=== Dynamics from the transition state region ===

==== Transition state ====
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

[[File:Q0 Surface Plot.png|center|400px|thumb|[Figure 1] Black line is located within the MEP]]

The transition state is a short-lived state of molecules or atoms at a local maximum in the reaction coordinate. Therefore, on a potential energy surface diagram, it can be defined as a local maximum of the minimum energy path (MEP). Mathematically it is a point where the gradient of potential is zero ( ∂V('''ri''')/∂'''ri'''=0). Although it is located in relatively low potential energy level, it actually is a local maximum point. It can be distinguished from other local minimum or local maximum points, as it should be located within the region of MEP.

'''

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.

'''<nowiki/>'''

[[File:Q2 Surface Plot.png|left|400px|thumb|[Figure 2] Position of transition state represented as 'x' on the contour plot]]
[[File:Q2 Surface Plot2.png|center|400px|thumb|[Figure 3] Internuclear distance VS time graph at transition state]]

The best estimate is achieved at rts=0.907. 

Since H+H2 is symmetric, the rAB and rBC should be equal. The figure 2 has only two lines, representing rBC and rAC . The rAB is not shown because rBC is overlapping it.

The definition of transition state suggests that the internuclear distance should stay constant throughout the given time, as ∂V('''ri''')/∂'''ri'''=0. The figure 2 shows that there is minimum vibration among the H atoms and no translation, suggesting the transition state.

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

[[File:Q3 contour Plot MEP.png|left|400px|thumb|[Figure 4] Contour graph of MEP]]
[[File:Q3 contour Plot DY.png|center|400px|thumb|[Figure 5] Contour graph of Dynamic]]
[[File:Q3 internu Plot MEP.png|left|400px|thumb|[Figure 6] Internuclear distance VS time graph of MEP]]
[[File:Q3 internu Plot DY.png|center|400px|thumb|[Figure 7] Internuclear distance VS time graph of Dynamic]]

Calculation on r

Comment on how the ''mep'' and the trajectory you just calculated differ.

In the MEP trajectory, the motion of molecule is infinitely slow that the momenta is reset to zero in every time step. This effectively takes the vibrational energy out of the system. Therefore, smooth line is shown MEP trajectories, when wavy line is shown in dynamic trajectory. Comparing figure 6 and figure 7, the Dynamic trajectories shows linearly increasing internuclear distance, and the MEP trajectories show decrease in the gradient. This is because MET plot is taken as each point being stationary (zero momenta), when the dynamic plot includes the momentum.

For the calculation rAB=rts+0.01 and rBC=rts

=== Reactive and unreactive trajectories ===
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?

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!P1
!P2
!Etot
!Contour plot
!Reactive?
!Description of the dynamics
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|[[File:Q5 contour1.png|frameless]]
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|[[File:Q5 contour2.png|frameless]]
|No
|Atom C has not enough momentum to collide with AB.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|[[File:Q5 contour3.png|frameless]]
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|[[File:Q5 contour4.png|frameless]]
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|[[File:Q5 contour5.png|frameless]]
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The main assumptions are:
1.molecules which have passed the transition state in the direction of product cannot reform the reactant.
2. In the transition state, motion along the reaction coordinate is separated from the other motions and treated as translation.
3. Transition state molecules are distributed among their states according to the Maxwell-Bpltzman law.
4. The system is in quasi-equalibrium, where the reaction process is slow enough for internal equalibrium.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River </ref>

Therefore, the experimental value is likely to be slower than the theoretical value, because the disrupting factors are ignored as assumption.

== F-H-H system ==

=== PES inspection ===

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

For the potential energy surface diagram of F+H2, the energy level goes down as the F atom and H2 molecule gets closer. This

The potential energy surface diagram of F + H2 shows the path of the reaction going from a higher energy minimum to a lower energy minimum, indicating that the products are at a lower energy than the reactants, which therefore corresponds to an exothermic reaction.

The potential energy surface diagram of H + HF shows the path of the reaction going from a lower energy minimum reaction to a higher energy minimum, indicating that the reactants are at a lower energy than the products, which therefore corresponds to an endothermic reaction.

Energy is taken in to break bonds, and energy is given out to make bonds. When the energy taken in is greater than the energy given out, this results in an endothermic reaction, and the opposite leads to an exothermic one. With F + H2, it's exothermic, which means that the bond formed (i.e. H-F) has a greater bond enthalpy than H2. With H + HF, it's endothermic, which suggests that the bond formed (i.e. H-H) has a lower bond enthalpy than the bond broken (i.e. H-F).

Locate the approximate position of the transition state.

Report the activation energy for both reactions.

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

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

MRD:syk2017

2019-05-10T11:32:31Z

Syk2017: /* Transition state */

== H+H2 system ==

=== Dynamics from the transition state region ===

==== Transition state ====
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

[[File:Q0 Surface Plot.png|center|400px|thumb|[Figure 1] Black line is located within the MEP]]

The transition state is a short-lived state of molecules or atoms at a local maximum in the reaction coordinate. Therefore, on a potential energy surface diagram, it can be defined as a local maximum of the minimum energy path (MEP). Mathematically it is a point where the gradient of potential is zero ( ∂V('''ri''')/∂'''ri'''=0). Although it is located in relatively low potential energy level, it actually is a local maximum point. It can be distinguished from other local minimum or local maximum points, as it should be located within the region of MEP.

'''

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.

'''<nowiki/>'''

[[File:Q2 Surface Plot.png|left|400px|thumb|[Figure 2] Position of transition state represented as 'x' on the contour plot]]
[[File:Q2 Surface Plot2.png|center|400px|thumb|[Figure 3] Internuclear distance VS time graph at transition state]]

The best estimate is achieved at rts=0.907. 

Since H+H2 is symmetric, the rAB and rBC should be equal. The figure 2 has only two lines, representing rBC and rAC . The rAB is not shown because rBC is overlapping it.

The definition of transition state suggests that the internuclear distance should stay constant throughout the given time, as ∂V('''ri''')/∂'''ri'''=0. The figure 2 shows that there is minimum vibration among the H atoms and no translation, suggesting the transition state.

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

[[File:Q3 contour Plot MEP.png|left|400px|thumb|[Figure 4] Contour graph of MEP]]
[[File:Q3 contour Plot DY.png|center|400px|thumb|[Figure 5] Contour graph of Dynamic]]
[[File:Q3 internu Plot MEP.png|left|400px|thumb|[Figure 6] Internuclear distance VS time graph of MEP]]
[[File:Q3 internu Plot DY.png|center|400px|thumb|[Figure 7] Internuclear distance VS time graph of Dynamic]]

Comment on how the ''mep'' and the trajectory you just calculated differ.

In the MEP trajectory, the motion of molecule is infinitely slow that the momenta is reset to zero in every time step. This effectively takes the vibrational energy out of the system. Therefore, smooth line is shown MEP trajectories, when wavy line is shown in dynamic trajectory. Comparing figure 6 and figure 7, the Dynamic trajectories shows linearly increasing internuclear distance, and the MEP trajectories show decrease in the gradient. This would suggests some deceleration in the system as momenta is reset to zero in every time step.

=== Reactive and unreactive trajectories ===
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?

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!P1
!P2
!Etot
!Contour plot
!Reactive?
!Description of the dynamics
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|[[File:Q5 contour1.png|frameless]]
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|[[File:Q5 contour2.png|frameless]]
|No
|Atom C has not enough momentum to collide with AB.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|[[File:Q5 contour3.png|frameless]]
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|[[File:Q5 contour4.png|frameless]]
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|[[File:Q5 contour5.png|frameless]]
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The main assumptions are:
1.molecules which have passed the transition state in the direction of product cannot reform the reactant.
2. In the transition state, motion along the reaction coordinate is separated from the other motions and treated as translation.
3. Transition state molecules are distributed among their states according to the Maxwell-Bpltzman law.
4. The system is in quasi-equalibrium, where the reaction process is slow enough for internal equalibrium.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River </ref>

Therefore, the experimental value is likely to be slower than the theoretical value, because the disrupting factors are ignored as assumption.

== F-H-H system ==

=== PES inspection ===

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

For the potential energy surface diagram of F+H2, the energy level goes down as the F atom and H2 molecule gets closer. This

The potential energy surface diagram of F + H2 shows the path of the reaction going from a higher energy minimum to a lower energy minimum, indicating that the products are at a lower energy than the reactants, which therefore corresponds to an exothermic reaction.

The potential energy surface diagram of H + HF shows the path of the reaction going from a lower energy minimum reaction to a higher energy minimum, indicating that the reactants are at a lower energy than the products, which therefore corresponds to an endothermic reaction.

Energy is taken in to break bonds, and energy is given out to make bonds. When the energy taken in is greater than the energy given out, this results in an endothermic reaction, and the opposite leads to an exothermic one. With F + H2, it's exothermic, which means that the bond formed (i.e. H-F) has a greater bond enthalpy than H2. With H + HF, it's endothermic, which suggests that the bond formed (i.e. H-H) has a lower bond enthalpy than the bond broken (i.e. H-F).

Locate the approximate position of the transition state.

Report the activation energy for both reactions.

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

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

MRD:syk2017

2019-05-10T11:31:29Z

Syk2017: /* Reactive and unreactive trajectories */

== H+H2 system ==

=== Dynamics from the transition state region ===

==== Transition state ====
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

[[File:Q0 Surface Plot.png|center|400px|thumb|[Figure 1] Black line is located within the MEP]]

The transition state is a short-lived state of molecules or atoms at a local maximum in the reaction coordinate. Therefore, on a potential energy surface diagram, it can be defined as a local maximum of the minimum energy path (MEP). Mathematically it is a point where the gradient of potential is zero ( ∂V('''ri''')/∂'''ri'''=0). Although it is located in relatively low potential energy level, it actually is a local maximum point. It can be distinguished from other local minimum or local maximum points, as it should be located within the region of MEP.

'''

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.

'''<nowiki/>'''

[[File:Q2 Surface Plot.png|left|400px|thumb|[Figure 2] Position of transition state represented as 'x' on the contour plot]]
[[File:Q2 Surface Plot2.png|center|400px|thumb|[Figure 3] Internuclear distance VS time graph at transition state]]

The best estimate is achieved at rts=0.907. 

Since H+H2 is symmetric, the rAB and rBC should be equal. The figure 2 has only two lines, representing rBC and rAC . The rAB is not shown because rBC is overlapping it.

The definition of transition state suggests that the internuclear distance should stay constant throughout the given time, as ∂V('''ri''')/∂'''ri'''=0. The figure 2 shows that there is minimum vibration among the H atoms, suggesting the transition state.

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

[[File:Q3 contour Plot MEP.png|left|400px|thumb|[Figure 4] Contour graph of MEP]]
[[File:Q3 contour Plot DY.png|center|400px|thumb|[Figure 5] Contour graph of Dynamic]]
[[File:Q3 internu Plot MEP.png|left|400px|thumb|[Figure 6] Internuclear distance VS time graph of MEP]]
[[File:Q3 internu Plot DY.png|center|400px|thumb|[Figure 7] Internuclear distance VS time graph of Dynamic]]

Comment on how the ''mep'' and the trajectory you just calculated differ.

In the MEP trajectory, the motion of molecule is infinitely slow that the momenta is reset to zero in every time step. This effectively takes the vibrational energy out of the system. Therefore, smooth line is shown MEP trajectories, when wavy line is shown in dynamic trajectory. Comparing figure 6 and figure 7, the Dynamic trajectories shows linearly increasing internuclear distance, and the MEP trajectories show decrease in the gradient. This would suggests some deceleration in the system as momenta is reset to zero in every time step.

=== Reactive and unreactive trajectories ===
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?

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!P1
!P2
!Etot
!Contour plot
!Reactive?
!Description of the dynamics
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|[[File:Q5 contour1.png|frameless]]
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|[[File:Q5 contour2.png|frameless]]
|No
|Atom C has not enough momentum to collide with AB.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|[[File:Q5 contour3.png|frameless]]
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|[[File:Q5 contour4.png|frameless]]
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|[[File:Q5 contour5.png|frameless]]
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The main assumptions are:
1.molecules which have passed the transition state in the direction of product cannot reform the reactant.
2. In the transition state, motion along the reaction coordinate is separated from the other motions and treated as translation.
3. Transition state molecules are distributed among their states according to the Maxwell-Bpltzman law.
4. The system is in quasi-equalibrium, where the reaction process is slow enough for internal equalibrium.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River </ref>

Therefore, the experimental value is likely to be slower than the theoretical value, because the disrupting factors are ignored as assumption.

== F-H-H system ==

=== PES inspection ===

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

For the potential energy surface diagram of F+H2, the energy level goes down as the F atom and H2 molecule gets closer. This

The potential energy surface diagram of F + H2 shows the path of the reaction going from a higher energy minimum to a lower energy minimum, indicating that the products are at a lower energy than the reactants, which therefore corresponds to an exothermic reaction.

The potential energy surface diagram of H + HF shows the path of the reaction going from a lower energy minimum reaction to a higher energy minimum, indicating that the reactants are at a lower energy than the products, which therefore corresponds to an endothermic reaction.

Energy is taken in to break bonds, and energy is given out to make bonds. When the energy taken in is greater than the energy given out, this results in an endothermic reaction, and the opposite leads to an exothermic one. With F + H2, it's exothermic, which means that the bond formed (i.e. H-F) has a greater bond enthalpy than H2. With H + HF, it's endothermic, which suggests that the bond formed (i.e. H-H) has a lower bond enthalpy than the bond broken (i.e. H-F).

Locate the approximate position of the transition state.

Report the activation energy for both reactions.

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

Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state

File:Q5 contour5.png

2019-05-10T11:30:13Z

Syk2017:

File:Q5 contour4.png

2019-05-10T11:30:02Z

Syk2017:

File:Q5 contour3.png

2019-05-10T11:27:45Z

Syk2017:

File:Q5 contour2.png

2019-05-10T11:27:32Z

Syk2017:

File:Q5 contour1.png

2019-05-10T11:27:12Z

Syk2017:

MRD:syk2017

2019-05-10T11:23:58Z

Syk2017: /* Reactive and unreactive trajectories */

== H+H2 system ==

=== Dynamics from the transition state region ===

==== Transition state ====
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

[[File:Q0 Surface Plot.png|center|400px|thumb|[Figure 1] Black line is located within the MEP]]

The transition state is a short-lived state of molecules or atoms at a local maximum in the reaction coordinate. Therefore, on a potential energy surface diagram, it can be defined as a local maximum of the minimum energy path (MEP). Mathematically it is a point where the gradient of potential is zero ( ∂V('''ri''')/∂'''ri'''=0). Although it is located in relatively low potential energy level, it actually is a local maximum point. It can be distinguished from other local minimum or local maximum points, as it should be located within the region of MEP.

'''

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.

'''<nowiki/>'''

[[File:Q2 Surface Plot.png|left|400px|thumb|[Figure 2] Position of transition state represented as 'x' on the contour plot]]
[[File:Q2 Surface Plot2.png|center|400px|thumb|[Figure 3] Internuclear distance VS time graph at transition state]]

The best estimate is achieved at rts=0.907. 

Since H+H2 is symmetric, the rAB and rBC should be equal. The figure 2 has only two lines, representing rBC and rAC . The rAB is not shown because rBC is overlapping it.

The definition of transition state suggests that the internuclear distance should stay constant throughout the given time, as ∂V('''ri''')/∂'''ri'''=0. The figure 2 shows that there is minimum vibration among the H atoms, suggesting the transition state.

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

[[File:Q3 contour Plot MEP.png|left|400px|thumb|[Figure 4] Contour graph of MEP]]
[[File:Q3 contour Plot DY.png|center|400px|thumb|[Figure 5] Contour graph of Dynamic]]
[[File:Q3 internu Plot MEP.png|left|400px|thumb|[Figure 6] Internuclear distance VS time graph of MEP]]
[[File:Q3 internu Plot DY.png|center|400px|thumb|[Figure 7] Internuclear distance VS time graph of Dynamic]]

Comment on how the ''mep'' and the trajectory you just calculated differ.

In the MEP trajectory, the motion of molecule is infinitely slow that the momenta is reset to zero in every time step. This effectively takes the vibrational energy out of the system. Therefore, smooth line is shown MEP trajectories, when wavy line is shown in dynamic trajectory. Comparing figure 6 and figure 7, the Dynamic trajectories shows linearly increasing internuclear distance, and the MEP trajectories show decrease in the gradient. This would suggests some deceleration in the system as momenta is reset to zero in every time step.

=== Reactive and unreactive trajectories ===
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?

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!P1
!P2
!Etot
!Contour plot
!Reactive?
!Description of the dynamics
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|[[File:Q5 contour1.png|100px|thumb]
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|[[File:Q5 contour2.png|100px|thumb]
|No
|Atom C has not enough momentum to collide with AB.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|[[File:Q5 contour3.png|100px|thumb]
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|[[File:Q5 contour4.png|100px|thumb]
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|[[File:Q5 contour5.png|100px|thumb]
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The main assumptions are:
1.molecules which have passed the transition state in the direction of product cannot reform the reactant.
2. In the transition state, motion along the reaction coordinate is separated from the other motions and treated as translation.
3. Transition state molecules are distributed among their states according to the Maxwell-Bpltzman law.
4. The system is in quasi-equalibrium, where the reaction process is slow enough for internal equalibrium.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River </ref>

Therefore, the experimental value is likely to be slower than the theoretical value, because the disrupting factors are ignored as assumption.

== F-H-H system ==

=== PES inspection ===

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

For the potential energy surface diagram of F+H2, the energy level goes down as the F atom and H2 molecule gets closer. This

The potential energy surface diagram of F + H2 shows the path of the reaction going from a higher energy minimum to a lower energy minimum, indicating that the products are at a lower energy than the reactants, which therefore corresponds to an exothermic reaction.

The potential energy surface diagram of H + HF shows the path of the reaction going from a lower energy minimum reaction to a higher energy minimum, indicating that the reactants are at a lower energy than the products, which therefore corresponds to an endothermic reaction.

Energy is taken in to break bonds, and energy is given out to make bonds. When the energy taken in is greater than the energy given out, this results in an endothermic reaction, and the opposite leads to an exothermic one. With F + H2, it's exothermic, which means that the bond formed (i.e. H-F) has a greater bond enthalpy than H2. With H + HF, it's endothermic, which suggests that the bond formed (i.e. H-H) has a lower bond enthalpy than the bond broken (i.e. H-F).

Locate the approximate position of the transition state.

Report the activation energy for both reactions.

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

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

MRD:syk2017

2019-05-10T11:21:48Z

Syk2017: /* Reactive and unreactive trajectories */

== H+H2 system ==

=== Dynamics from the transition state region ===

==== Transition state ====
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

[[File:Q0 Surface Plot.png|center|400px|thumb|[Figure 1] Black line is located within the MEP]]

The transition state is a short-lived state of molecules or atoms at a local maximum in the reaction coordinate. Therefore, on a potential energy surface diagram, it can be defined as a local maximum of the minimum energy path (MEP). Mathematically it is a point where the gradient of potential is zero ( ∂V('''ri''')/∂'''ri'''=0). Although it is located in relatively low potential energy level, it actually is a local maximum point. It can be distinguished from other local minimum or local maximum points, as it should be located within the region of MEP.

'''

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.

'''<nowiki/>'''

[[File:Q2 Surface Plot.png|left|400px|thumb|[Figure 2] Position of transition state represented as 'x' on the contour plot]]
[[File:Q2 Surface Plot2.png|center|400px|thumb|[Figure 3] Internuclear distance VS time graph at transition state]]

The best estimate is achieved at rts=0.907. 

Since H+H2 is symmetric, the rAB and rBC should be equal. The figure 2 has only two lines, representing rBC and rAC . The rAB is not shown because rBC is overlapping it.

The definition of transition state suggests that the internuclear distance should stay constant throughout the given time, as ∂V('''ri''')/∂'''ri'''=0. The figure 2 shows that there is minimum vibration among the H atoms, suggesting the transition state.

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

[[File:Q3 contour Plot MEP.png|left|400px|thumb|[Figure 4] Contour graph of MEP]]
[[File:Q3 contour Plot DY.png|center|400px|thumb|[Figure 5] Contour graph of Dynamic]]
[[File:Q3 internu Plot MEP.png|left|400px|thumb|[Figure 6] Internuclear distance VS time graph of MEP]]
[[File:Q3 internu Plot DY.png|center|400px|thumb|[Figure 7] Internuclear distance VS time graph of Dynamic]]

Comment on how the ''mep'' and the trajectory you just calculated differ.

In the MEP trajectory, the motion of molecule is infinitely slow that the momenta is reset to zero in every time step. This effectively takes the vibrational energy out of the system. Therefore, smooth line is shown MEP trajectories, when wavy line is shown in dynamic trajectory. Comparing figure 6 and figure 7, the Dynamic trajectories shows linearly increasing internuclear distance, and the MEP trajectories show decrease in the gradient. This would suggests some deceleration in the system as momenta is reset to zero in every time step.

=== Reactive and unreactive trajectories ===
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?

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!P1
!P2
!Etot
!Contour plot
!Reactive?
!Description of the dynamics
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|
|No
|Atom C has not enough momentum to collide with AB.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

[[File:Q5 contour1.png|left|300px|thumb|[Figure 3] Contour graph of MEP]]
[[File:Q5 contour2.png|center|300px|thumb|[Figure 3] Contour graph of MEP]]
[[File:Q5 contour3.png|right|300px|thumb|[Figure 3] Contour graph of MEP]]

[[File:Q5 contour4.png|left|300px|thumb|[Figure 3] Contour graph of MEP]]
[[File:Q5 contour5.png|right|300px|thumb|[Figure 3] Contour graph of MEP]]

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The main assumptions are:
1.molecules which have passed the transition state in the direction of product cannot reform the reactant.
2. In the transition state, motion along the reaction coordinate is separated from the other motions and treated as translation.
3. Transition state molecules are distributed among their states according to the Maxwell-Bpltzman law.
4. The system is in quasi-equalibrium, where the reaction process is slow enough for internal equalibrium.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River </ref>

Therefore, the experimental value is likely to be slower than the theoretical value, because the disrupting factors are ignored as assumption.

== F-H-H system ==

=== PES inspection ===

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

For the potential energy surface diagram of F+H2, the energy level goes down as the F atom and H2 molecule gets closer. This

The potential energy surface diagram of F + H2 shows the path of the reaction going from a higher energy minimum to a lower energy minimum, indicating that the products are at a lower energy than the reactants, which therefore corresponds to an exothermic reaction.

The potential energy surface diagram of H + HF shows the path of the reaction going from a lower energy minimum reaction to a higher energy minimum, indicating that the reactants are at a lower energy than the products, which therefore corresponds to an endothermic reaction.

Energy is taken in to break bonds, and energy is given out to make bonds. When the energy taken in is greater than the energy given out, this results in an endothermic reaction, and the opposite leads to an exothermic one. With F + H2, it's exothermic, which means that the bond formed (i.e. H-F) has a greater bond enthalpy than H2. With H + HF, it's endothermic, which suggests that the bond formed (i.e. H-H) has a lower bond enthalpy than the bond broken (i.e. H-F).

Locate the approximate position of the transition state.

Report the activation energy for both reactions.

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

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

MRD:syk2017

2019-05-10T11:20:46Z

Syk2017: /* Reactive and unreactive trajectories */

== H+H2 system ==

=== Dynamics from the transition state region ===

==== Transition state ====
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

[[File:Q0 Surface Plot.png|center|400px|thumb|[Figure 1] Black line is located within the MEP]]

The transition state is a short-lived state of molecules or atoms at a local maximum in the reaction coordinate. Therefore, on a potential energy surface diagram, it can be defined as a local maximum of the minimum energy path (MEP). Mathematically it is a point where the gradient of potential is zero ( ∂V('''ri''')/∂'''ri'''=0). Although it is located in relatively low potential energy level, it actually is a local maximum point. It can be distinguished from other local minimum or local maximum points, as it should be located within the region of MEP.

'''

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.

'''<nowiki/>'''

[[File:Q2 Surface Plot.png|left|400px|thumb|[Figure 2] Position of transition state represented as 'x' on the contour plot]]
[[File:Q2 Surface Plot2.png|center|400px|thumb|[Figure 3] Internuclear distance VS time graph at transition state]]

The best estimate is achieved at rts=0.907. 

Since H+H2 is symmetric, the rAB and rBC should be equal. The figure 2 has only two lines, representing rBC and rAC . The rAB is not shown because rBC is overlapping it.

The definition of transition state suggests that the internuclear distance should stay constant throughout the given time, as ∂V('''ri''')/∂'''ri'''=0. The figure 2 shows that there is minimum vibration among the H atoms, suggesting the transition state.

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

[[File:Q3 contour Plot MEP.png|left|400px|thumb|[Figure 4] Contour graph of MEP]]
[[File:Q3 contour Plot DY.png|center|400px|thumb|[Figure 5] Contour graph of Dynamic]]
[[File:Q3 internu Plot MEP.png|left|400px|thumb|[Figure 6] Internuclear distance VS time graph of MEP]]
[[File:Q3 internu Plot DY.png|center|400px|thumb|[Figure 7] Internuclear distance VS time graph of Dynamic]]

Comment on how the ''mep'' and the trajectory you just calculated differ.

In the MEP trajectory, the motion of molecule is infinitely slow that the momenta is reset to zero in every time step. This effectively takes the vibrational energy out of the system. Therefore, smooth line is shown MEP trajectories, when wavy line is shown in dynamic trajectory. Comparing figure 6 and figure 7, the Dynamic trajectories shows linearly increasing internuclear distance, and the MEP trajectories show decrease in the gradient. This would suggests some deceleration in the system as momenta is reset to zero in every time step.

=== Reactive and unreactive trajectories ===
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?

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!Combination
!P1
!P2
!Etot
!Reactive?
!Description of the dynamics
|-
|1
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|2
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|No
|Atom C has not enough momentum to collide with AB.
|-
|3
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|4
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|5
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

[[File:Q5 contour1.png|left|300px|thumb|[Figure 3] Contour graph of MEP]]
[[File:Q5 contour2.png|center|300px|thumb|[Figure 3] Contour graph of MEP]]
[[File:Q5 contour3.png|right|300px|thumb|[Figure 3] Contour graph of MEP]]

[[File:Q5 contour4.png|left|300px|thumb|[Figure 3] Contour graph of MEP]]
[[File:Q5 contour5.png|right|300px|thumb|[Figure 3] Contour graph of MEP]]

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The main assumptions are:
1.molecules which have passed the transition state in the direction of product cannot reform the reactant.
2. In the transition state, motion along the reaction coordinate is separated from the other motions and treated as translation.
3. Transition state molecules are distributed among their states according to the Maxwell-Bpltzman law.
4. The system is in quasi-equalibrium, where the reaction process is slow enough for internal equalibrium.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River </ref>

Therefore, the experimental value is likely to be slower than the theoretical value, because the disrupting factors are ignored as assumption.

== F-H-H system ==

=== PES inspection ===

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

For the potential energy surface diagram of F+H2, the energy level goes down as the F atom and H2 molecule gets closer. This

The potential energy surface diagram of F + H2 shows the path of the reaction going from a higher energy minimum to a lower energy minimum, indicating that the products are at a lower energy than the reactants, which therefore corresponds to an exothermic reaction.

The potential energy surface diagram of H + HF shows the path of the reaction going from a lower energy minimum reaction to a higher energy minimum, indicating that the reactants are at a lower energy than the products, which therefore corresponds to an endothermic reaction.

Energy is taken in to break bonds, and energy is given out to make bonds. When the energy taken in is greater than the energy given out, this results in an endothermic reaction, and the opposite leads to an exothermic one. With F + H2, it's exothermic, which means that the bond formed (i.e. H-F) has a greater bond enthalpy than H2. With H + HF, it's endothermic, which suggests that the bond formed (i.e. H-H) has a lower bond enthalpy than the bond broken (i.e. H-F).

Locate the approximate position of the transition state.

Report the activation energy for both reactions.

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

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

MRD:syk2017

2019-05-10T11:20:01Z

Syk2017: /* Reactive and unreactive trajectories */

== H+H2 system ==

=== Dynamics from the transition state region ===

==== Transition state ====
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

[[File:Q0 Surface Plot.png|center|400px|thumb|[Figure 1] Black line is located within the MEP]]

The transition state is a short-lived state of molecules or atoms at a local maximum in the reaction coordinate. Therefore, on a potential energy surface diagram, it can be defined as a local maximum of the minimum energy path (MEP). Mathematically it is a point where the gradient of potential is zero ( ∂V('''ri''')/∂'''ri'''=0). Although it is located in relatively low potential energy level, it actually is a local maximum point. It can be distinguished from other local minimum or local maximum points, as it should be located within the region of MEP.

'''

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.

'''<nowiki/>'''

[[File:Q2 Surface Plot.png|left|400px|thumb|[Figure 2] Position of transition state represented as 'x' on the contour plot]]
[[File:Q2 Surface Plot2.png|center|400px|thumb|[Figure 3] Internuclear distance VS time graph at transition state]]

The best estimate is achieved at rts=0.907. 

Since H+H2 is symmetric, the rAB and rBC should be equal. The figure 2 has only two lines, representing rBC and rAC . The rAB is not shown because rBC is overlapping it.

The definition of transition state suggests that the internuclear distance should stay constant throughout the given time, as ∂V('''ri''')/∂'''ri'''=0. The figure 2 shows that there is minimum vibration among the H atoms, suggesting the transition state.

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

[[File:Q3 contour Plot MEP.png|left|400px|thumb|[Figure 4] Contour graph of MEP]]
[[File:Q3 contour Plot DY.png|center|400px|thumb|[Figure 5] Contour graph of Dynamic]]
[[File:Q3 internu Plot MEP.png|left|400px|thumb|[Figure 6] Internuclear distance VS time graph of MEP]]
[[File:Q3 internu Plot DY.png|center|400px|thumb|[Figure 7] Internuclear distance VS time graph of Dynamic]]

Comment on how the ''mep'' and the trajectory you just calculated differ.

In the MEP trajectory, the motion of molecule is infinitely slow that the momenta is reset to zero in every time step. This effectively takes the vibrational energy out of the system. Therefore, smooth line is shown MEP trajectories, when wavy line is shown in dynamic trajectory. Comparing figure 6 and figure 7, the Dynamic trajectories shows linearly increasing internuclear distance, and the MEP trajectories show decrease in the gradient. This would suggests some deceleration in the system as momenta is reset to zero in every time step.

=== Reactive and unreactive trajectories ===
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?

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!Combination
!P1
!P2
!Etot
!Reactive?
!Description of the dynamics
|-
|1
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|2
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|No
|Atom C has not enough momentum to collide with AB.
|-
|3
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|4
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|5
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

[[File:Q5 contour1.png|left|400px|thumb|[Figure 3] Contour graph of MEP]]
[[File:Q5 contour2.png|center|400px|thumb|[Figure 3] Contour graph of MEP]]
[[File:Q5 contour3.png|right|400px|thumb|[Figure 3] Contour graph of MEP]]

[[File:Q5 contour4.png|left|400px|thumb|[Figure 3] Contour graph of MEP]]
[[File:Q5 contour5.png|right|400px|thumb|[Figure 3] Contour graph of MEP]]

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The main assumptions are:
1.molecules which have passed the transition state in the direction of product cannot reform the reactant.
2. In the transition state, motion along the reaction coordinate is separated from the other motions and treated as translation.
3. Transition state molecules are distributed among their states according to the Maxwell-Bpltzman law.
4. The system is in quasi-equalibrium, where the reaction process is slow enough for internal equalibrium.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River </ref>

Therefore, the experimental value is likely to be slower than the theoretical value, because the disrupting factors are ignored as assumption.

== F-H-H system ==

=== PES inspection ===

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

For the potential energy surface diagram of F+H2, the energy level goes down as the F atom and H2 molecule gets closer. This

The potential energy surface diagram of F + H2 shows the path of the reaction going from a higher energy minimum to a lower energy minimum, indicating that the products are at a lower energy than the reactants, which therefore corresponds to an exothermic reaction.

The potential energy surface diagram of H + HF shows the path of the reaction going from a lower energy minimum reaction to a higher energy minimum, indicating that the reactants are at a lower energy than the products, which therefore corresponds to an endothermic reaction.

Energy is taken in to break bonds, and energy is given out to make bonds. When the energy taken in is greater than the energy given out, this results in an endothermic reaction, and the opposite leads to an exothermic one. With F + H2, it's exothermic, which means that the bond formed (i.e. H-F) has a greater bond enthalpy than H2. With H + HF, it's endothermic, which suggests that the bond formed (i.e. H-H) has a lower bond enthalpy than the bond broken (i.e. H-F).

Locate the approximate position of the transition state.

Report the activation energy for both reactions.

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

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

MRD:syk2017

2019-05-10T11:19:25Z

Syk2017: /* Reactive and unreactive trajectories */

== H+H2 system ==

=== Dynamics from the transition state region ===

==== Transition state ====
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

[[File:Q0 Surface Plot.png|center|400px|thumb|[Figure 1] Black line is located within the MEP]]

The transition state is a short-lived state of molecules or atoms at a local maximum in the reaction coordinate. Therefore, on a potential energy surface diagram, it can be defined as a local maximum of the minimum energy path (MEP). Mathematically it is a point where the gradient of potential is zero ( ∂V('''ri''')/∂'''ri'''=0). Although it is located in relatively low potential energy level, it actually is a local maximum point. It can be distinguished from other local minimum or local maximum points, as it should be located within the region of MEP.

'''

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.

'''<nowiki/>'''

[[File:Q2 Surface Plot.png|left|400px|thumb|[Figure 2] Position of transition state represented as 'x' on the contour plot]]
[[File:Q2 Surface Plot2.png|center|400px|thumb|[Figure 3] Internuclear distance VS time graph at transition state]]

The best estimate is achieved at rts=0.907. 

Since H+H2 is symmetric, the rAB and rBC should be equal. The figure 2 has only two lines, representing rBC and rAC . The rAB is not shown because rBC is overlapping it.

The definition of transition state suggests that the internuclear distance should stay constant throughout the given time, as ∂V('''ri''')/∂'''ri'''=0. The figure 2 shows that there is minimum vibration among the H atoms, suggesting the transition state.

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

[[File:Q3 contour Plot MEP.png|left|400px|thumb|[Figure 4] Contour graph of MEP]]
[[File:Q3 contour Plot DY.png|center|400px|thumb|[Figure 5] Contour graph of Dynamic]]
[[File:Q3 internu Plot MEP.png|left|400px|thumb|[Figure 6] Internuclear distance VS time graph of MEP]]
[[File:Q3 internu Plot DY.png|center|400px|thumb|[Figure 7] Internuclear distance VS time graph of Dynamic]]

Comment on how the ''mep'' and the trajectory you just calculated differ.

In the MEP trajectory, the motion of molecule is infinitely slow that the momenta is reset to zero in every time step. This effectively takes the vibrational energy out of the system. Therefore, smooth line is shown MEP trajectories, when wavy line is shown in dynamic trajectory. Comparing figure 6 and figure 7, the Dynamic trajectories shows linearly increasing internuclear distance, and the MEP trajectories show decrease in the gradient. This would suggests some deceleration in the system as momenta is reset to zero in every time step.

=== Reactive and unreactive trajectories ===
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?

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!Combination
!P1
!P2
!Etot
!Reactive?
!Description of the dynamics
|-
|1
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|2
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|No
|Atom C has not enough momentum to collide with AB.
|-
|3
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|4
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|5
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

[[File:Q5 contour1.png|left|400px|thumb|[Figure 3] Contour graph of MEP]]
[[File:Q5 contour2.png|center|400px|thumb|[Figure 3] Contour graph of MEP]]
[[File:Q5 contour3.png|right|400px|thumb|[Figure 3] Contour graph of MEP]]
[[File:Q5 contour4.png|left|400px|thumb|[Figure 3] Contour graph of MEP]]
[[File:Q5 contour5.png|right|400px|thumb|[Figure 3] Contour graph of MEP]]

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The main assumptions are:
1.molecules which have passed the transition state in the direction of product cannot reform the reactant.
2. In the transition state, motion along the reaction coordinate is separated from the other motions and treated as translation.
3. Transition state molecules are distributed among their states according to the Maxwell-Bpltzman law.
4. The system is in quasi-equalibrium, where the reaction process is slow enough for internal equalibrium.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River </ref>

Therefore, the experimental value is likely to be slower than the theoretical value, because the disrupting factors are ignored as assumption.

== F-H-H system ==

=== PES inspection ===

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

For the potential energy surface diagram of F+H2, the energy level goes down as the F atom and H2 molecule gets closer. This

The potential energy surface diagram of F + H2 shows the path of the reaction going from a higher energy minimum to a lower energy minimum, indicating that the products are at a lower energy than the reactants, which therefore corresponds to an exothermic reaction.

The potential energy surface diagram of H + HF shows the path of the reaction going from a lower energy minimum reaction to a higher energy minimum, indicating that the reactants are at a lower energy than the products, which therefore corresponds to an endothermic reaction.

Energy is taken in to break bonds, and energy is given out to make bonds. When the energy taken in is greater than the energy given out, this results in an endothermic reaction, and the opposite leads to an exothermic one. With F + H2, it's exothermic, which means that the bond formed (i.e. H-F) has a greater bond enthalpy than H2. With H + HF, it's endothermic, which suggests that the bond formed (i.e. H-H) has a lower bond enthalpy than the bond broken (i.e. H-F).

Locate the approximate position of the transition state.

Report the activation energy for both reactions.

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

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

MRD:syk2017

2019-05-10T11:18:50Z

Syk2017: /* Reactive and unreactive trajectories */

== H+H2 system ==

=== Dynamics from the transition state region ===

==== Transition state ====
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

[[File:Q0 Surface Plot.png|center|400px|thumb|[Figure 1] Black line is located within the MEP]]

The transition state is a short-lived state of molecules or atoms at a local maximum in the reaction coordinate. Therefore, on a potential energy surface diagram, it can be defined as a local maximum of the minimum energy path (MEP). Mathematically it is a point where the gradient of potential is zero ( ∂V('''ri''')/∂'''ri'''=0). Although it is located in relatively low potential energy level, it actually is a local maximum point. It can be distinguished from other local minimum or local maximum points, as it should be located within the region of MEP.

'''

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.

'''<nowiki/>'''

[[File:Q2 Surface Plot.png|left|400px|thumb|[Figure 2] Position of transition state represented as 'x' on the contour plot]]
[[File:Q2 Surface Plot2.png|center|400px|thumb|[Figure 3] Internuclear distance VS time graph at transition state]]

The best estimate is achieved at rts=0.907. 

Since H+H2 is symmetric, the rAB and rBC should be equal. The figure 2 has only two lines, representing rBC and rAC . The rAB is not shown because rBC is overlapping it.

The definition of transition state suggests that the internuclear distance should stay constant throughout the given time, as ∂V('''ri''')/∂'''ri'''=0. The figure 2 shows that there is minimum vibration among the H atoms, suggesting the transition state.

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

[[File:Q3 contour Plot MEP.png|left|400px|thumb|[Figure 4] Contour graph of MEP]]
[[File:Q3 contour Plot DY.png|center|400px|thumb|[Figure 5] Contour graph of Dynamic]]
[[File:Q3 internu Plot MEP.png|left|400px|thumb|[Figure 6] Internuclear distance VS time graph of MEP]]
[[File:Q3 internu Plot DY.png|center|400px|thumb|[Figure 7] Internuclear distance VS time graph of Dynamic]]

Comment on how the ''mep'' and the trajectory you just calculated differ.

In the MEP trajectory, the motion of molecule is infinitely slow that the momenta is reset to zero in every time step. This effectively takes the vibrational energy out of the system. Therefore, smooth line is shown MEP trajectories, when wavy line is shown in dynamic trajectory. Comparing figure 6 and figure 7, the Dynamic trajectories shows linearly increasing internuclear distance, and the MEP trajectories show decrease in the gradient. This would suggests some deceleration in the system as momenta is reset to zero in every time step.

=== Reactive and unreactive trajectories ===
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?

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!Combination
!P1
!P2
!Etot
!Reactive?
!Description of the dynamics
|-
|1
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|2
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|No
|Atom C has not enough momentum to collide with AB.
|-
|3
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|4
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|5
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

[[File:Q5 contour1.png|left|400px|thumb|[Figure 3] Contour graph of MEP]]
[[File:Q5 contour2.png|left|400px|thumb|[Figure 3] Contour graph of MEP]]
[[File:Q5 contour3.png|left|400px|thumb|[Figure 3] Contour graph of MEP]]
[[File:Q5 contour4.png|left|400px|thumb|[Figure 3] Contour graph of MEP]]
[[File:Q5 contour5.png|left|400px|thumb|[Figure 3] Contour graph of MEP]]

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The main assumptions are:
1.molecules which have passed the transition state in the direction of product cannot reform the reactant.
2. In the transition state, motion along the reaction coordinate is separated from the other motions and treated as translation.
3. Transition state molecules are distributed among their states according to the Maxwell-Bpltzman law.
4. The system is in quasi-equalibrium, where the reaction process is slow enough for internal equalibrium.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River </ref>

Therefore, the experimental value is likely to be slower than the theoretical value, because the disrupting factors are ignored as assumption.

== F-H-H system ==

=== PES inspection ===

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

For the potential energy surface diagram of F+H2, the energy level goes down as the F atom and H2 molecule gets closer. This

The potential energy surface diagram of F + H2 shows the path of the reaction going from a higher energy minimum to a lower energy minimum, indicating that the products are at a lower energy than the reactants, which therefore corresponds to an exothermic reaction.

The potential energy surface diagram of H + HF shows the path of the reaction going from a lower energy minimum reaction to a higher energy minimum, indicating that the reactants are at a lower energy than the products, which therefore corresponds to an endothermic reaction.

Energy is taken in to break bonds, and energy is given out to make bonds. When the energy taken in is greater than the energy given out, this results in an endothermic reaction, and the opposite leads to an exothermic one. With F + H2, it's exothermic, which means that the bond formed (i.e. H-F) has a greater bond enthalpy than H2. With H + HF, it's endothermic, which suggests that the bond formed (i.e. H-H) has a lower bond enthalpy than the bond broken (i.e. H-F).

Locate the approximate position of the transition state.

Report the activation energy for both reactions.

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

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

MRD:syk2017

2019-05-10T11:17:57Z

Syk2017: /* Reactive and unreactive trajectories */

== H+H2 system ==

=== Dynamics from the transition state region ===

==== Transition state ====
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

[[File:Q0 Surface Plot.png|center|400px|thumb|[Figure 1] Black line is located within the MEP]]

The transition state is a short-lived state of molecules or atoms at a local maximum in the reaction coordinate. Therefore, on a potential energy surface diagram, it can be defined as a local maximum of the minimum energy path (MEP). Mathematically it is a point where the gradient of potential is zero ( ∂V('''ri''')/∂'''ri'''=0). Although it is located in relatively low potential energy level, it actually is a local maximum point. It can be distinguished from other local minimum or local maximum points, as it should be located within the region of MEP.

'''

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.

'''<nowiki/>'''

[[File:Q2 Surface Plot.png|left|400px|thumb|[Figure 2] Position of transition state represented as 'x' on the contour plot]]
[[File:Q2 Surface Plot2.png|center|400px|thumb|[Figure 3] Internuclear distance VS time graph at transition state]]

The best estimate is achieved at rts=0.907. 

Since H+H2 is symmetric, the rAB and rBC should be equal. The figure 2 has only two lines, representing rBC and rAC . The rAB is not shown because rBC is overlapping it.

The definition of transition state suggests that the internuclear distance should stay constant throughout the given time, as ∂V('''ri''')/∂'''ri'''=0. The figure 2 shows that there is minimum vibration among the H atoms, suggesting the transition state.

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

[[File:Q3 contour Plot MEP.png|left|400px|thumb|[Figure 4] Contour graph of MEP]]
[[File:Q3 contour Plot DY.png|center|400px|thumb|[Figure 5] Contour graph of Dynamic]]
[[File:Q3 internu Plot MEP.png|left|400px|thumb|[Figure 6] Internuclear distance VS time graph of MEP]]
[[File:Q3 internu Plot DY.png|center|400px|thumb|[Figure 7] Internuclear distance VS time graph of Dynamic]]

Comment on how the ''mep'' and the trajectory you just calculated differ.

In the MEP trajectory, the motion of molecule is infinitely slow that the momenta is reset to zero in every time step. This effectively takes the vibrational energy out of the system. Therefore, smooth line is shown MEP trajectories, when wavy line is shown in dynamic trajectory. Comparing figure 6 and figure 7, the Dynamic trajectories shows linearly increasing internuclear distance, and the MEP trajectories show decrease in the gradient. This would suggests some deceleration in the system as momenta is reset to zero in every time step.

=== Reactive and unreactive trajectories ===
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?

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!P1
!P2
!Etot
!Reactive?
!Description of the dynamics
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|No
|Atom C has not enough momentum to collide with AB.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

[[File:Q5 contour1.png|left|400px|thumb|[Figure 3] Contour graph of MEP]]
[[File:Q5 contour2.png|left|400px|thumb|[Figure 3] Contour graph of MEP]]
[[File:Q5 contour3.png|left|400px|thumb|[Figure 3] Contour graph of MEP]]
[[File:Q5 contour4.png|left|400px|thumb|[Figure 3] Contour graph of MEP]]
[[File:Q5 contour5.png|left|400px|thumb|[Figure 3] Contour graph of MEP]]

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The main assumptions are:
1.molecules which have passed the transition state in the direction of product cannot reform the reactant.
2. In the transition state, motion along the reaction coordinate is separated from the other motions and treated as translation.
3. Transition state molecules are distributed among their states according to the Maxwell-Bpltzman law.
4. The system is in quasi-equalibrium, where the reaction process is slow enough for internal equalibrium.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River </ref>

Therefore, the experimental value is likely to be slower than the theoretical value, because the disrupting factors are ignored as assumption.

== F-H-H system ==

=== PES inspection ===

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

For the potential energy surface diagram of F+H2, the energy level goes down as the F atom and H2 molecule gets closer. This

The potential energy surface diagram of F + H2 shows the path of the reaction going from a higher energy minimum to a lower energy minimum, indicating that the products are at a lower energy than the reactants, which therefore corresponds to an exothermic reaction.

The potential energy surface diagram of H + HF shows the path of the reaction going from a lower energy minimum reaction to a higher energy minimum, indicating that the reactants are at a lower energy than the products, which therefore corresponds to an endothermic reaction.

Energy is taken in to break bonds, and energy is given out to make bonds. When the energy taken in is greater than the energy given out, this results in an endothermic reaction, and the opposite leads to an exothermic one. With F + H2, it's exothermic, which means that the bond formed (i.e. H-F) has a greater bond enthalpy than H2. With H + HF, it's endothermic, which suggests that the bond formed (i.e. H-H) has a lower bond enthalpy than the bond broken (i.e. H-F).

Locate the approximate position of the transition state.

Report the activation energy for both reactions.

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

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

MRD:syk2017

2019-05-10T11:14:55Z

Syk2017: /* Dynamics from the transition state region */

== H+H2 system ==

=== Dynamics from the transition state region ===

==== Transition state ====
On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

[[File:Q0 Surface Plot.png|center|400px|thumb|[Figure 1] Black line is located within the MEP]]

The transition state is a short-lived state of molecules or atoms at a local maximum in the reaction coordinate. Therefore, on a potential energy surface diagram, it can be defined as a local maximum of the minimum energy path (MEP). Mathematically it is a point where the gradient of potential is zero ( ∂V('''ri''')/∂'''ri'''=0). Although it is located in relatively low potential energy level, it actually is a local maximum point. It can be distinguished from other local minimum or local maximum points, as it should be located within the region of MEP.

'''

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.

'''<nowiki/>'''

[[File:Q2 Surface Plot.png|left|400px|thumb|[Figure 2] Position of transition state represented as 'x' on the contour plot]]
[[File:Q2 Surface Plot2.png|center|400px|thumb|[Figure 3] Internuclear distance VS time graph at transition state]]

The best estimate is achieved at rts=0.907. 

Since H+H2 is symmetric, the rAB and rBC should be equal. The figure 2 has only two lines, representing rBC and rAC . The rAB is not shown because rBC is overlapping it.

The definition of transition state suggests that the internuclear distance should stay constant throughout the given time, as ∂V('''ri''')/∂'''ri'''=0. The figure 2 shows that there is minimum vibration among the H atoms, suggesting the transition state.

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

[[File:Q3 contour Plot MEP.png|left|400px|thumb|[Figure 4] Contour graph of MEP]]
[[File:Q3 contour Plot DY.png|center|400px|thumb|[Figure 5] Contour graph of Dynamic]]
[[File:Q3 internu Plot MEP.png|left|400px|thumb|[Figure 6] Internuclear distance VS time graph of MEP]]
[[File:Q3 internu Plot DY.png|center|400px|thumb|[Figure 7] Internuclear distance VS time graph of Dynamic]]

Comment on how the ''mep'' and the trajectory you just calculated differ.

In the MEP trajectory, the motion of molecule is infinitely slow that the momenta is reset to zero in every time step. This effectively takes the vibrational energy out of the system. Therefore, smooth line is shown MEP trajectories, when wavy line is shown in dynamic trajectory. Comparing figure 6 and figure 7, the Dynamic trajectories shows linearly increasing internuclear distance, and the MEP trajectories show decrease in the gradient. This would suggests some deceleration in the system as momenta is reset to zero in every time step.

=== Reactive and unreactive trajectories ===
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?

initial positions : '''r1''' = 0.74 and '''r2''' = 2.0
{| class="wikitable"
!P1
!P2
!Etot
!Reactive?
!Description of the dynamics
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|Yes
|Atom C collides with AB, then the product A+BC formed. The system had no vibrational energy initially, but the product does.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|No
|Atom C has not enough momentum to collide with AB.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|Yes
|Atom C collides with AB and product A+BC formed. The vibrational energy increases throughout the reaction.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|No
|Atom C slowly collides with AB to form product A+BC, but reforms the reaction after second collision by atom A.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|Yes
|Atom C slowly collides with AB to form product A+BC. After the secondary collisions between A-B and B-C, the product A+BC retained.
|}

[[File:Q6 contour Plot MEP.png|left|400px|thumb|[Figure 3] Contour graph of MEP]]
[[File:Q6 contour Plot MEP.png|left|400px|thumb|[Figure 3] Contour graph of MEP]]
[[File:Q6 contour Plot MEP.png|left|400px|thumb|[Figure 3] Contour graph of MEP]]
[[File:Q6 contour Plot MEP.png|left|400px|thumb|[Figure 3] Contour graph of MEP]]
[[File:Q6 contour Plot MEP.png|left|400px|thumb|[Figure 3] Contour graph of MEP]]

State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?

The main assumptions are:
1.molecules which have passed the transition state in the direction of product cannot reform the reactant.
2. In the transition state, motion along the reaction coordinate is separated from the other motions and treated as translation.
3. Transition state molecules are distributed among their states according to the Maxwell-Bpltzman law.
4. The system is in quasi-equalibrium, where the reaction process is slow enough for internal equalibrium.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, 1999, Upper Saddle River </ref>

Therefore, the experimental value is likely to be slower than the theoretical value, because the disrupting factors are ignored as assumption.

== F-H-H system ==

=== PES inspection ===

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

For the potential energy surface diagram of F+H2, the energy level goes down as the F atom and H2 molecule gets closer. This

The potential energy surface diagram of F + H2 shows the path of the reaction going from a higher energy minimum to a lower energy minimum, indicating that the products are at a lower energy than the reactants, which therefore corresponds to an exothermic reaction.

The potential energy surface diagram of H + HF shows the path of the reaction going from a lower energy minimum reaction to a higher energy minimum, indicating that the reactants are at a lower energy than the products, which therefore corresponds to an endothermic reaction.

Energy is taken in to break bonds, and energy is given out to make bonds. When the energy taken in is greater than the energy given out, this results in an endothermic reaction, and the opposite leads to an exothermic one. With F + H2, it's exothermic, which means that the bond formed (i.e. H-F) has a greater bond enthalpy than H2. With H + HF, it's endothermic, which suggests that the bond formed (i.e. H-H) has a lower bond enthalpy than the bond broken (i.e. H-F).

Locate the approximate position of the transition state.

Report the activation energy for both reactions.

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

Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state

File:Q3 internu Plot MEP.png

2019-05-10T11:14:48Z

Syk2017:

File:Q3 internu Plot DY.png

2019-05-10T11:14:24Z

Syk2017: