ChemWiki - User contributions [en]

MRD:MolecularReactionDynamics01523148

2020-05-08T22:21:19Z

Yyl18: /* 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. */

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

The transition state is mathematically defined by ∂V(ri)/∂ri=0, which r value represents the distance between atoms.The transition state can be identified at the position between maximum and minimum on the Potential Energy Surface and also by finding out the energy value around the trajectory from the contour plot.
[[File:Inkedtransition state surface plot1234 LI.jpg]]

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

The best estimate of the transition state position is around 90.775pm as it has no F and oscillation between each atoms as there is no oscillation occurs from the internuclear distance vs time plot. This means the picture captured is the transition state.
[[File:Question 1.2.png]]

===Comment on how the mep and the trajectory you just calculated differ.===
From the images below, using r1 = rts+δ, r2 = rts, with δ=1. The MEP and dynamics calculation types are compared.
[[File:MES 01523148.png]]
Dynamics contour plot
[[File:Trajectory1.3 01523148.png]]
MEP contour plot

From the pictures above, it shows that dynamics include the calculation of oscillation of bonds as from the Dynamics contour plot we can see there is oscillating curve around B-C distances. This means dynamics calculation involve calculation of vibrational energy, while no kinetic energy will be accounted in MEP calculation.

[[File:1.3.2 0152314888 InterNuclear Dynamics.png]]
Internulcear Dynamics plot
[[File:1.3.2 0152314888 InterNuclear MEP.png]]
Internuclear MEP plot
[[File:1.3.2 01523148 MomentaVsTime Dynamics.png]]
MomentaVsTime Dynamics plot
[[File:1.3.2 01523148 MomentaVsTime MEP.png]]
MomentaVsTime MEP plot

===Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable"
|-
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Since BC distance is starting to diminish, it shows that H(b)-H(c) bond is forming while H(a)-H(b) bond is breaking. The particles go through the transition state when H(c) collides with H(b).
|[[File:Surface Plot 1.3.1 01523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|Since no AB distance can be found from the image, that means no bond formation or bond breaking between H(a) and H(b). Therefore no collision found during the interactions of three particles and cannot overcome the energy barrier, reason of no collision might be due to repulsion.
|[[File:Surface Plot 1.3.2 1523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The reaction is similar to the scenario one. Only difference is H(c) is moving at a higher speed toward H(b) but same outcome has reached because energy barrier is overcame.
|[[File:Surface Plot 1.3.3 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|The H(b)-H(c) bond is formed first by having H(c) colliding into H(a)-H(b). This collision crosses the transition energy barrier at first. However the diagram shows the recrossing of transition state region which breaks the H(b)-H(c) bond and reform H(a)-H(b) back to original, with greater vibrational energy.
|[[File:Surface Plot 1.3.4 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|This reaction is similar to that of scenario 4. However instead of H(c) leaving the system, the H(c) particle goes back to the system and form H(b)-H(c) bond which breaks the oscillating H(a)-H(b) bond. The system crosses and recrosses the transition state 3 times in total, leaving with greater oscillation of H(c)-H(b).
|[[File:Surface Plot 1.3.5 1523148.png|250px]]}

|}

===Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?===

Transition state theory treats reaction classically such that quantum tunnelling cannot occur. Also energy must obey Boltzmann distribution rule. There is always equilibrium between reactants and transition state. All systems that have already crossed the TS region cannot recross the areas from product back to reactants. Therefore the reaction rate values from TST prediction is expected to be greater than experimental values (like scenario 5 above) if any quantum effect like barrier-recrossing. Although sometimes tunnelling might occur that give underestimate value compared to experimental one because the reaction don't need to overcome the energy barrier then. However, the recrossing barrier effect usually dominates over that of tunnelling effect, therefore mostly overestimate of reaction rate values happens.

==Exercise 2: F-H-H system==
===PES inspection===
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions According to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===
F+H-H reaction is exothermic while H+H-F is endothermic. Since enthalpy of HH is +436 kJ/mol while HF is +562 kJ/mol, the formation of HF is exothermic as the energy released from H-F formation is greater than H-H breaking, vice versa.
From PES, it proves the classification as from the (F+H2) PES, the potential energy along reaction is actually lower than that of (H+HF) inspection.

[[File:Surface Plot 01523148 2.png]]
PES plot of H+HF
[[File:Surface Plot 01523148 1.png]]
PES plot of F+HH

===Locate the approximate position of the transition state.===
Endothermic reaction means the late transition state resembles the products meanwhile exothermic reaction resembles early transition state which is the reactants. Hence from (F+H-H) reaction, the exothermic reaction allows us to predict the transition state is around 74pm, which is the bond length of H-H, in one of the internuclear distances.
Through trial and error, AB=181.1pm and BC=74.4875pm allow us to obtain the transition state, with potential energy of -433.981.The internuclear distance plot by MEP below shows that no oscillation occurs around and hence confirm the capture of transition state.
[[File:Fig2.2 01523148 InternuclearDistance.png]]

===Report the activation energy for both reactions.===
Energy of transition state - Energy of reactants = activation energy. Since the transition state energy was found from above, the energy of reactants can be found from the following image, using 185pm AB and 50pm BC, with 1000number of steps and 0.15fs of size. The picture below depicts how energy transforms throughout the (F+H2) reaction. There are two energy states captures in the graph because there are two flat lines in the reaction observed. The first flat line shows the energy of -434.751 kJ/mol, which is energy of these two reactants combined (F+H2), meanwhile the second flat line shows the energy of -559 kJ/mol, which resembles the energy of the products (H+HF).
Therefore, the activation energy of H+HF endothermic reaction is -433.981-(-434.751)=+0.77kJ/mol
The activation energy of F+HF exothermic reaction is -433.981-(-559)=+125 kJ/mol
[[File:Fig2.3 01523148.png]]
The energy state of reactants(H+HF)
[[File:Fig2.3.2 01523148.png]]
The energy state of reactants(F+H2)

===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.===
Momentum of F-H is dependent on the rate of reaction, as supported from the plots below. This explains that the reaction energy is released as kinetic energy, more specifically, translational and vibrational form of kinetic energy. Since kinetic energy can convert into form of heat, heat can be measured by calorimetry, using E=mc(dT).
IR spectroscopy and vibrational spectroscopy can also measure the vibrational kinetic energy absorbed from the bond oscillation of HF, with IR measuring the transition from ground state to excited state while vibrational spectroscopy measuring the overtones of peaks in the spectra(e.g.0-->1,1-->2,1-->0).
[[File:2.4 01523148 1.png]]

AB=184pm,-1g.pm/fs.mol, BC=74pm,1g.pm/fs.mol

[[File:2.4 01523148 2.png]]

AB=184pm and -5g.pm/fs.mol, BC=74pm,5g.pm/fs.mol

[[File:2.4 01523148 3 .png]]

AB=184pm and -10g.pm/fs.mol, BC=74pm, 10g.pm/fs.mol

===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 rule states that the translational energy is more effective than vibration energy in overcoming an early transition state barrier, and vibrational energy is more effective than translational energy when overcoming a late transition state region.

<nowiki> </nowiki>From exothermic (F+H2) reaction, it resembles early transition state barrier. Therefore, from the images below, with increasing momentum and decreased energy of H-F vibration, the greater proportion of translational energy is more likely to overcome the transition energy barrier.

From endothermic(H+HF) reaction, the late transition state barrier is resembled. The smaller proportion of p1 to p2 ratio will be more likely to form reactive trajectory because transition state barrier is more likely to be overcame by increased proportion of vibration energy.
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
! p2/ g.mol-1.pm.fs-1
! Etot
! Reactive?
! Description of the dynamics
! Illustration of the trajectory

|-
| -1.3
| 9
|KE=+70.189kJ/mol
Total energy=-363.767kJ/mol
|no
|exothermic
|[[File:2.5 01523148 1.png|250px]]

|-
| -2
| 0.005
|KE=+2.115kJ/mol
Total energy=431.841kJ/mol
|yes
|exothermic
|[[File:2.5 01523148 2.png|250px]]

|-
| -1
| -5.1
|KE= +1.105kJ/mol
Total energy=-413.709kJ/mol
|no
|endothermic
|[[File:2.4 01523148 3.png|250px]]

|-
| -5.1
| -10.1
|KE= 43.632kJ/mol

Total energy=-371.183kJ/mol
|yes
|endothermic
|[[File:2.5 01523148 4.png|250px]]
|}
all scenarios with r1=184pm, r2=74pm, scenario 1&2 with atom A=F and atom B&C=H, while scenario 3&4 with atom A&B=H and atom C=F

MRD:MolecularReactionDynamics01523148

2020-05-08T22:17:58Z

Yyl18: /* Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? */

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

The transition state is mathematically defined by ∂V(ri)/∂ri=0, which r value represents the distance between atoms.The transition state can be identified at the position between maximum and minimum on the Potential Energy Surface and also by finding out the energy value around the trajectory from the contour plot.
[[File:Inkedtransition state surface plot1234 LI.jpg]]

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

The best estimate of the transition state position is around 90.775pm as it has no F and oscillation between each atoms as there is no oscillation occurs from the internuclear distance vs time plot. This means the picture captured is the transition state.
[[File:Question 1.2.png]]

===Comment on how the mep and the trajectory you just calculated differ.===
From the images below, using r1 = rts+δ, r2 = rts, with δ=1. The MEP and dynamics calculation types are compared.
[[File:MES 01523148.png]]
Dynamics contour plot
[[File:Trajectory1.3 01523148.png]]
MEP contour plot

From the pictures above, it shows that dynamics include the calculation of oscillation of bonds as from the Dynamics contour plot we can see there is oscillating curve around B-C distances. This means dynamics calculation involve calculation of vibrational energy, while no kinetic energy will be accounted in MEP calculation.

[[File:1.3.2 0152314888 InterNuclear Dynamics.png]]
Internulcear Dynamics plot
[[File:1.3.2 0152314888 InterNuclear MEP.png]]
Internuclear MEP plot
[[File:1.3.2 01523148 MomentaVsTime Dynamics.png]]
MomentaVsTime Dynamics plot
[[File:1.3.2 01523148 MomentaVsTime MEP.png]]
MomentaVsTime MEP plot

===Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable"
|-
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Since BC distance is starting to diminish, it shows that H(b)-H(c) bond is forming while H(a)-H(b) bond is breaking. The particles go through the transition state when H(c) collides with H(b).
|[[File:Surface Plot 1.3.1 01523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|Since no AB distance can be found from the image, that means no bond formation or bond breaking between H(a) and H(b). Therefore no collision found during the interactions of three particles and cannot overcome the energy barrier, reason of no collision might be due to repulsion.
|[[File:Surface Plot 1.3.2 1523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The reaction is similar to the scenario one. Only difference is H(c) is moving at a higher speed toward H(b) but same outcome has reached because energy barrier is overcame.
|[[File:Surface Plot 1.3.3 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|The H(b)-H(c) bond is formed first by having H(c) colliding into H(a)-H(b). This collision crosses the transition energy barrier at first. However the diagram shows the recrossing of transition state region which breaks the H(b)-H(c) bond and reform H(a)-H(b) back to original, with greater vibrational energy.
|[[File:Surface Plot 1.3.4 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|This reaction is similar to that of scenario 4. However instead of H(c) leaving the system, the H(c) particle goes back to the system and form H(b)-H(c) bond which breaks the oscillating H(a)-H(b) bond. The system crosses and recrosses the transition state 3 times in total, leaving with greater oscillation of H(c)-H(b).
|[[File:Surface Plot 1.3.5 1523148.png|250px]]}

|}

===Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?===

Transition state theory treats reaction classically such that quantum tunnelling cannot occur. Also energy must obey Boltzmann distribution rule. There is always equilibrium between reactants and transition state. All systems that have already crossed the TS region cannot recross the areas from product back to reactants. Therefore the reaction rate values from TST prediction is expected to be greater than experimental values (like scenario 5 above) if any quantum effect like barrier-recrossing. Although sometimes tunnelling might occur that give underestimate value compared to experimental one because the reaction don't need to overcome the energy barrier then. However, the recrossing barrier effect usually dominates over that of tunnelling effect, therefore mostly overestimate of reaction rate values happens.

==Exercise 2: F-H-H system==
===PES inspection===
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions According to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===
F+H-H reaction is exothermic while H+H-F is endothermic. Since enthalpy of HH is +436 kJ/mol while HF is +562 kJ/mol, the formation of HF is exothermic as the energy released from H-F formation is greater than H-H breaking, vice versa.
From PES, it proves the classification as from the (F+H2) PES, the potential energy along reaction is actually lower than that of (H+HF) inspection.

[[File:Surface Plot 01523148 2.png]]
PES plot of H+HF
[[File:Surface Plot 01523148 1.png]]
PES plot of F+HH

===Locate the approximate position of the transition state.===
Endothermic reaction means the late transition state resembles the products meanwhile exothermic reaction resembles early transition state which is the reactants. Hence from (F+H-H) reaction, the exothermic reaction allows us to predict the transition state is around 74pm, which is the bond length of H-H, in one of the internuclear distances.
Through trial and error, AB=181.1pm and BC=74.4875pm allow us to obtain the transition state, with potential energy of -433.981.The internuclear distance plot by MEP below shows that no oscillation occurs around and hence confirm the capture of transition state.
[[File:Fig2.2 01523148 InternuclearDistance.png]]

===Report the activation energy for both reactions.===
Energy of transition state - Energy of reactants = activation energy. Since the transition state energy was found from above, the energy of reactants can be found from the following image, using 185pm AB and 50pm BC, with 1000number of steps and 0.15fs of size. The picture below depicts how energy transforms throughout the (F+H2) reaction. There are two energy states captures in the graph because there are two flat lines in the reaction observed. The first flat line shows the energy of -434.751 kJ/mol, which is energy of these two reactants combined (F+H2), meanwhile the second flat line shows the energy of -559 kJ/mol, which resembles the energy of the products (H+HF).
Therefore, the activation energy of H+HF endothermic reaction is -433.981-(-434.751)=+0.77kJ/mol
The activation energy of F+HF exothermic reaction is -433.981-(-559)=+125 kJ/mol
[[File:Fig2.3 01523148.png]]
The energy state of reactants(H+HF)
[[File:Fig2.3.2 01523148.png]]
The energy state of reactants(F+H2)

===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.===
Momentum of F-H is dependent on the rate of reaction, as supported from the plots below. This explains that the reaction energy is released as kinetic energy, more specifically, translational and vibrational form of kinetic energy. Since kinetic energy can convert into form of heat, heat can be measured by calorimetry, using E=mc(dT).
IR spectroscopy and vibrational spectroscopy can also measure the vibrational kinetic energy absorbed from the bond oscillation of HF, with IR measuring the transition from ground state to excited state while vibrational spectroscopy measuring the overtones of peaks in the spectra.
[[File:2.4 01523148 1.png]]

AB=184pm,-1g.pm/fs.mol, BC=74pm,1g.pm/fs.mol

[[File:2.4 01523148 2.png]]

AB=184pm and -5g.pm/fs.mol, BC=74pm,5g.pm/fs.mol

[[File:2.4 01523148 3 .png]]

AB=184pm and -10g.pm/fs.mol, BC=74pm, 10g.pm/fs.mol

===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 rule states that the translational energy is more effective than vibration energy in overcoming an early transition state barrier, and vibrational energy is more effective than translational energy when overcoming a late transition state region.

<nowiki> </nowiki>From exothermic (F+H2) reaction, it resembles early transition state barrier. Therefore, from the images below, with increasing momentum and decreased energy of H-F vibration, the greater proportion of translational energy is more likely to overcome the transition energy barrier.

From endothermic(H+HF) reaction, the late transition state barrier is resembled. The smaller proportion of p1 to p2 ratio will be more likely to form reactive trajectory because transition state barrier is more likely to be overcame by increased proportion of vibration energy.
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
! p2/ g.mol-1.pm.fs-1
! Etot
! Reactive?
! Description of the dynamics
! Illustration of the trajectory

|-
| -1.3
| 9
|KE=+70.189kJ/mol
Total energy=-363.767kJ/mol
|no
|exothermic
|[[File:2.5 01523148 1.png|250px]]

|-
| -2
| 0.005
|KE=+2.115kJ/mol
Total energy=431.841kJ/mol
|yes
|exothermic
|[[File:2.5 01523148 2.png|250px]]

|-
| -1
| -5.1
|KE= +1.105kJ/mol
Total energy=-413.709kJ/mol
|no
|endothermic
|[[File:2.4 01523148 3.png|250px]]

|-
| -5.1
| -10.1
|KE= 43.632kJ/mol

Total energy=-371.183kJ/mol
|yes
|endothermic
|[[File:2.5 01523148 4.png|250px]]
|}
all scenarios with r1=184pm, r2=74pm, scenario 1&2 with atom A=F and atom B&C=H, while scenario 3&4 with atom A&B=H and atom C=F

MRD:MolecularReactionDynamics01523148

2020-05-08T22:11:08Z

Yyl18: /* 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. */

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

The transition state is mathematically defined by ∂V(ri)/∂ri=0, which r value represents the distance between atoms.The transition state can be identified at the position between maximum and minimum on the Potential Energy Surface and also by finding out the energy value around the trajectory from the contour plot.
[[File:Inkedtransition state surface plot1234 LI.jpg]]

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

The best estimate of the transition state position is around 90.775pm as it has no F and oscillation between each atoms as there is no oscillation occurs from the internuclear distance vs time plot. This means the picture captured is the transition state.
[[File:Question 1.2.png]]

===Comment on how the mep and the trajectory you just calculated differ.===
From the images below, using r1 = rts+δ, r2 = rts, with δ=1. The MEP and dynamics calculation types are compared.
[[File:MES 01523148.png]]
Dynamics contour plot
[[File:Trajectory1.3 01523148.png]]
MEP contour plot

From the pictures above, it shows that dynamics include the calculation of oscillation of bonds as from the Dynamics contour plot we can see there is oscillating curve around B-C distances. This means dynamics calculation involve calculation of vibrational energy, while no kinetic energy will be accounted in MEP calculation.

[[File:1.3.2 0152314888 InterNuclear Dynamics.png]]
Internulcear Dynamics plot
[[File:1.3.2 0152314888 InterNuclear MEP.png]]
Internuclear MEP plot
[[File:1.3.2 01523148 MomentaVsTime Dynamics.png]]
MomentaVsTime Dynamics plot
[[File:1.3.2 01523148 MomentaVsTime MEP.png]]
MomentaVsTime MEP plot

===Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable"
|-
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Since BC distance is starting to diminish, it shows that H(b)-H(c) bond is forming while H(a)-H(b) bond is breaking. The particles go through the transition state when H(c) collides with H(b).
|[[File:Surface Plot 1.3.1 01523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|Since no AB distance can be found from the image, that means no bond formation or bond breaking between H(a) and H(b). Therefore no collision found during the interactions of three particles and cannot overcome the energy barrier, reason of no collision might be due to repulsion.
|[[File:Surface Plot 1.3.2 1523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The reaction is similar to the scenario one. Only difference is H(c) is moving at a higher speed toward H(b) but same outcome has reached because energy barrier is overcame.
|[[File:Surface Plot 1.3.3 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|The H(b)-H(c) bond is formed first by having H(c) colliding into H(a)-H(b). This collision crosses the transition energy barrier at first. However the diagram shows the recrossing of transition state region which breaks the H(b)-H(c) bond and reform H(a)-H(b) back to original, with greater vibrational energy.
|[[File:Surface Plot 1.3.4 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|This reaction is similar to that of scenario 4. However instead of H(c) leaving the system, the H(c) particle goes back to the system and form H(b)-H(c) bond which breaks the oscillating H(a)-H(b) bond. The system crosses and recrosses the transition state 3 times in total, leaving with greater oscillation of H(c)-H(b).
|[[File:Surface Plot 1.3.5 1523148.png|250px]]}

|}

===Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?===

Transition state theory treats reaction classically such that quantum tunnelling cannot occur. Also energy must obey Boltzmann distribution rule and all systems that have already crossed the TS region cannot recross the areas from product back to reactants. Therefore the reaction rate values from TST prediction is expected to be greater than experimental values (like scenario 5 above) if any quantum effect like barrier-recrossing or tunnelling, especially the former one occurs.

==Exercise 2: F-H-H system==
===PES inspection===
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions According to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===
F+H-H reaction is exothermic while H+H-F is endothermic. Since enthalpy of HH is +436 kJ/mol while HF is +562 kJ/mol, the formation of HF is exothermic as the energy released from H-F formation is greater than H-H breaking, vice versa.
From PES, it proves the classification as from the (F+H2) PES, the potential energy along reaction is actually lower than that of (H+HF) inspection.

[[File:Surface Plot 01523148 2.png]]
PES plot of H+HF
[[File:Surface Plot 01523148 1.png]]
PES plot of F+HH

===Locate the approximate position of the transition state.===
Endothermic reaction means the late transition state resembles the products meanwhile exothermic reaction resembles early transition state which is the reactants. Hence from (F+H-H) reaction, the exothermic reaction allows us to predict the transition state is around 74pm, which is the bond length of H-H, in one of the internuclear distances.
Through trial and error, AB=181.1pm and BC=74.4875pm allow us to obtain the transition state, with potential energy of -433.981.The internuclear distance plot by MEP below shows that no oscillation occurs around and hence confirm the capture of transition state.
[[File:Fig2.2 01523148 InternuclearDistance.png]]

===Report the activation energy for both reactions.===
Energy of transition state - Energy of reactants = activation energy. Since the transition state energy was found from above, the energy of reactants can be found from the following image, using 185pm AB and 50pm BC, with 1000number of steps and 0.15fs of size. The picture below depicts how energy transforms throughout the (F+H2) reaction. There are two energy states captures in the graph because there are two flat lines in the reaction observed. The first flat line shows the energy of -434.751 kJ/mol, which is energy of these two reactants combined (F+H2), meanwhile the second flat line shows the energy of -559 kJ/mol, which resembles the energy of the products (H+HF).
Therefore, the activation energy of H+HF endothermic reaction is -433.981-(-434.751)=+0.77kJ/mol
The activation energy of F+HF exothermic reaction is -433.981-(-559)=+125 kJ/mol
[[File:Fig2.3 01523148.png]]
The energy state of reactants(H+HF)
[[File:Fig2.3.2 01523148.png]]
The energy state of reactants(F+H2)

===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.===
Momentum of F-H is dependent on the rate of reaction, as supported from the plots below. This explains that the reaction energy is released as kinetic energy, more specifically, translational and vibrational form of kinetic energy. Since kinetic energy can convert into form of heat, heat can be measured by calorimetry, using E=mc(dT).
IR spectroscopy and vibrational spectroscopy can also measure the vibrational kinetic energy absorbed from the bond oscillation of HF, with IR measuring the transition from ground state to excited state while vibrational spectroscopy measuring the overtones of peaks in the spectra.
[[File:2.4 01523148 1.png]]

AB=184pm,-1g.pm/fs.mol, BC=74pm,1g.pm/fs.mol

[[File:2.4 01523148 2.png]]

AB=184pm and -5g.pm/fs.mol, BC=74pm,5g.pm/fs.mol

[[File:2.4 01523148 3 .png]]

AB=184pm and -10g.pm/fs.mol, BC=74pm, 10g.pm/fs.mol

===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 rule states that the translational energy is more effective than vibration energy in overcoming an early transition state barrier, and vibrational energy is more effective than translational energy when overcoming a late transition state region.

<nowiki> </nowiki>From exothermic (F+H2) reaction, it resembles early transition state barrier. Therefore, from the images below, with increasing momentum and decreased energy of H-F vibration, the greater proportion of translational energy is more likely to overcome the transition energy barrier.

From endothermic(H+HF) reaction, the late transition state barrier is resembled. The smaller proportion of p1 to p2 ratio will be more likely to form reactive trajectory because transition state barrier is more likely to be overcame by increased proportion of vibration energy.
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
! p2/ g.mol-1.pm.fs-1
! Etot
! Reactive?
! Description of the dynamics
! Illustration of the trajectory

|-
| -1.3
| 9
|KE=+70.189kJ/mol
Total energy=-363.767kJ/mol
|no
|exothermic
|[[File:2.5 01523148 1.png|250px]]

|-
| -2
| 0.005
|KE=+2.115kJ/mol
Total energy=431.841kJ/mol
|yes
|exothermic
|[[File:2.5 01523148 2.png|250px]]

|-
| -1
| -5.1
|KE= +1.105kJ/mol
Total energy=-413.709kJ/mol
|no
|endothermic
|[[File:2.4 01523148 3.png|250px]]

|-
| -5.1
| -10.1
|KE= 43.632kJ/mol

Total energy=-371.183kJ/mol
|yes
|endothermic
|[[File:2.5 01523148 4.png|250px]]
|}
all scenarios with r1=184pm, r2=74pm, scenario 1&2 with atom A=F and atom B&C=H, while scenario 3&4 with atom A&B=H and atom C=F

MRD:MolecularReactionDynamics01523148

2020-05-08T22:07:56Z

Yyl18: /* 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? */

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

The transition state is mathematically defined by ∂V(ri)/∂ri=0, which r value represents the distance between atoms.The transition state can be identified at the position between maximum and minimum on the Potential Energy Surface and also by finding out the energy value around the trajectory from the contour plot.
[[File:Inkedtransition state surface plot1234 LI.jpg]]

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

The best estimate of the transition state position is around 90.775pm as it has no F and oscillation between each atoms as there is no oscillation occurs from the internuclear distance vs time plot. This means the picture captured is the transition state.
[[File:Question 1.2.png]]

===Comment on how the mep and the trajectory you just calculated differ.===
From the images below, using r1 = rts+δ, r2 = rts, with δ=1. The MEP and dynamics calculation types are compared.
[[File:MES 01523148.png]]
Dynamics contour plot
[[File:Trajectory1.3 01523148.png]]
MEP contour plot

From the pictures above, it shows that dynamics include the calculation of oscillation of bonds as from the Dynamics contour plot we can see there is oscillating curve around B-C distances. This means dynamics calculation involve calculation of vibrational energy, while no kinetic energy will be accounted in MEP calculation.

[[File:1.3.2 0152314888 InterNuclear Dynamics.png]]
Internulcear Dynamics plot
[[File:1.3.2 0152314888 InterNuclear MEP.png]]
Internuclear MEP plot
[[File:1.3.2 01523148 MomentaVsTime Dynamics.png]]
MomentaVsTime Dynamics plot
[[File:1.3.2 01523148 MomentaVsTime MEP.png]]
MomentaVsTime MEP plot

===Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable"
|-
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Since BC distance is starting to diminish, it shows that H(b)-H(c) bond is forming while H(a)-H(b) bond is breaking. The particles go through the transition state when H(c) collides with H(b).
|[[File:Surface Plot 1.3.1 01523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|Since no AB distance can be found from the image, that means no bond formation or bond breaking between H(a) and H(b). Therefore no collision found during the interactions of three particles and cannot overcome the energy barrier, reason of no collision might be due to repulsion.
|[[File:Surface Plot 1.3.2 1523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The reaction is similar to the scenario one. Only difference is H(c) is moving at a higher speed toward H(b) but same outcome has reached because energy barrier is overcame.
|[[File:Surface Plot 1.3.3 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|The H(b)-H(c) bond is formed first by having H(c) colliding into H(a)-H(b). This collision crosses the transition energy barrier at first. However the diagram shows the recrossing of transition state region which breaks the H(b)-H(c) bond and reform H(a)-H(b) back to original, with greater vibrational energy.
|[[File:Surface Plot 1.3.4 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|This reaction is similar to that of scenario 4. However instead of H(c) leaving the system, the H(c) particle goes back to the system and form H(b)-H(c) bond which breaks the oscillating H(a)-H(b) bond. The system crosses and recrosses the transition state 3 times in total, leaving with greater oscillation of H(c)-H(b).
|[[File:Surface Plot 1.3.5 1523148.png|250px]]}

|}

===Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?===

Transition state theory treats reaction classically such that quantum tunnelling cannot occur. Also energy must obey Boltzmann distribution rule and all systems that have already crossed the TS region cannot recross the areas from product back to reactants. Therefore the reaction rate values from TST prediction is expected to be greater than experimental values (like scenario 5 above) if any quantum effect like barrier-recrossing or tunnelling, especially the former one occurs.

==Exercise 2: F-H-H system==
===PES inspection===
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions According to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===
F+H-H reaction is exothermic while H+H-F is endothermic. Since enthalpy of HH is +436 kJ/mol while HF is +562 kJ/mol, the formation of HF is exothermic as the energy released from H-F formation is greater than H-H breaking, vice versa.
From PES, it proves the classification as from the (F+H2) PES, the potential energy along reaction is actually lower than that of (H+HF) inspection.

[[File:Surface Plot 01523148 2.png]]
PES plot of H+HF
[[File:Surface Plot 01523148 1.png]]
PES plot of F+HH

===Locate the approximate position of the transition state.===
Endothermic reaction means the late transition state resembles the products meanwhile exothermic reaction resembles early transition state which is the reactants. Hence from (F+H-H) reaction, the exothermic reaction allows us to predict the transition state is around 74pm, which is the bond length of H-H, in one of the internuclear distances.
Through trial and error, AB=181.1pm and BC=74.4875pm allow us to obtain the transition state, with potential energy of -433.981.The internuclear distance plot by MEP below shows that no oscillation occurs around and hence confirm the capture of transition state.
[[File:Fig2.2 01523148 InternuclearDistance.png]]

===Report the activation energy for both reactions.===
Energy of transition state - Energy of reactants = activation energy. Since the transition state energy was found from above, the energy of reactants can be found from the following image, using 185pm AB and 50pm BC, with 1000number of steps and 0.15fs of size. The picture below depicts how energy transforms throughout the (F+H2) reaction. There are two energy states captures in the graph because there are two flat lines in the reaction observed. The first flat line shows the energy of -434.751 kJ/mol, which is energy of these two reactants combined (F+H2), meanwhile the second flat line shows the energy of -559 kJ/mol, which resembles the energy of the products (H+HF).
Therefore, the activation energy of H+HF endothermic reaction is -433.981-(-434.751)=+0.77kJ/mol
The activation energy of F+HF exothermic reaction is -433.981-(-559)=+125 kJ/mol
[[File:Fig2.3 01523148.png]]
The energy state of reactants(H+HF)
[[File:Fig2.3.2 01523148.png]]
The energy state of reactants(F+H2)

===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.===
Momentum of F-H is dependent on the rate of reaction, as supported from the plots below. This explains that the reaction energy is released as kinetic energy, more specifically, translational and vibrational form of kinetic energy. Since kinetic energy can convert into form of heat, heat can be measured by calorimetry, using E=mc(dT).
IR spectroscopy and vibrational spectroscopy can also measure the vibrational kinetic energy absorbed from the bond oscillation of HF, with IR measuring the transition from ground state to excited state while vibrational spectroscopy measuring the overtones of peaks in the spectra.
[[File:2.4 01523148 1.png]]

AB=184pm,-1g.pm/fs.mol, BC=74pm,1g.pm/fs.mol

[[File:2.4 01523148 2.png]]

AB=184pm and -5g.pm/fs.mol, BC=74pm,5g.pm/fs.mol

[[File:2.4 01523148 3 .png]]

AB=184pm and -10g.pm/fs.mol, BC=74pm, 10g.pm/fs.mol

===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 rule states that the translational energy is more effective than vibration energy in overcoming an early transition state barrier, and vibrational energy is more effective than translational energy when overcoming a late transition state region.

<nowiki> </nowiki>From exothermic (F+H2) reaction, it resembles early transition state barrier. Therefore, from the images below, with increasing momentum and decreased energy of H-F vibration, the greater proportion of translational energy is more likely to overcome the transition energy barrier.

From endothermic(H+HF) reaction, the late transition state barrier is resembled. The smaller proportion of p1 to p2 ratio will be more likely to form reactive trajectory because transition state barrier is more likely to be overcame by increased proportion of vibration energy.
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
! p2/ g.mol-1.pm.fs-1
! Etot
! Reactive?
! Description of the dynamics
! Illustration of the trajectory

|-
| -1.3
| 9
|KE=+70.189kJ/mol
Total energy=-363.767kJ/mol
|no
|exothermic
|[[File:2.5 01523148 1.png|250px]]

|-
| -2
| 0.005
|KE=+2.115kJ/mol
Total energy=431.841kJ/mol
|yes
|exothermic
|[[File:2.5 01523148 2.png|250px]]

|-
| -1
| -5.1
|KE= +1.105kJ/mol
Total energy=-413.709kJ/mol
|no
|endothermic
|[[File:2.4 01523148 3.png|250px]]

|-
| -5.1
| -10.1
|KE= 43.632kJ/mol

Total energy=-371.183kJ/mol
|yes
|endothermic
|[[File:2.5 01523148 4.png|250px]]
|}
r1=184pm, r2=74pm

MRD:MolecularReactionDynamics01523148

2020-05-08T22:06:47Z

Yyl18:

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

The transition state is mathematically defined by ∂V(ri)/∂ri=0, which r value represents the distance between atoms.The transition state can be identified at the position between maximum and minimum on the Potential Energy Surface and also by finding out the energy value around the trajectory from the contour plot.
[[File:Inkedtransition state surface plot1234 LI.jpg]]

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

The best estimate of the transition state position is around 90.775pm as it has no F and oscillation between each atoms as there is no oscillation occurs from the internuclear distance vs time plot. This means the picture captured is the transition state.
[[File:Question 1.2.png]]

===Comment on how the mep and the trajectory you just calculated differ.===
From the images below, using r1 = rts+δ, r2 = rts, with δ=1. The MEP and dynamics calculation types are compared.
[[File:MES 01523148.png]]
Dynamics contour plot
[[File:Trajectory1.3 01523148.png]]
MEP contour plot

From the pictures above, it shows that dynamics include the calculation of oscillation of bonds as from the Dynamics contour plot we can see there is oscillating curve around B-C distances. This means dynamics calculation involve calculation of vibrational energy, while no kinetic energy will be accounted in MEP calculation.

[[File:1.3.2 0152314888 InterNuclear Dynamics.png]]
Internulcear Dynamics plot
[[File:1.3.2 0152314888 InterNuclear MEP.png]]
Internuclear MEP plot
[[File:1.3.2 01523148 MomentaVsTime Dynamics.png]]
MomentaVsTime Dynamics plot
[[File:1.3.2 01523148 MomentaVsTime MEP.png]]
MomentaVsTime MEP plot

===Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable"
|-
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Since BC distance is starting to diminish, it shows that H(b)-H(c) bond is forming while H(a)-H(b) bond is breaking. The particles go through the transition state when H(c) collides with H(b).
|[[File:Surface Plot 1.3.1 01523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|Since no AB distance can be found from the image, that means no bond formation or bond breaking between H(a) and H(b). Therefore no collision found during the interactions of three particles and cannot overcome the energy barrier, reason of no collision might be due to repulsion.
|[[File:Surface Plot 1.3.2 1523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The reaction is similar to the scenario one. Only difference is H(c) is moving at a higher speed toward H(b) but same outcome has reached because energy barrier is overcame.
|[[File:Surface Plot 1.3.3 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|The H(b)-H(c) bond is formed first by having H(c) colliding into H(a)-H(b). This collision crosses the transition energy barrier at first. However the diagram shows the recrossing of transition state region which breaks the H(b)-H(c) bond and reform H(a)-H(b) back to original, with greater vibrational energy.
|[[File:Surface Plot 1.3.4 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|This reaction is similar to that of scenario 4. However instead of H(c) leaving the system, the H(c) particle goes back to the system and form H(b)-H(c) bond which breaks the oscillating H(a)-H(b) bond. The system crosses and recrosses the transition state 3 times in total, leaving with greater oscillation of H(c)-H(b).
|[[File:Surface Plot 1.3.5 1523148.png|250px]]}

|}

===Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?===

Transition state theory treats reaction classically such that quantum tunnelling cannot occur. Also energy must obey Boltzmann distribution rule and all systems that have already crossed the TS region cannot recross the areas from product back to reactants. Therefore the reaction rate values from TST prediction is expected to be greater than experimental values (like scenario 5 above) if any quantum effect like barrier-recrossing or tunnelling, especially the former one occurs.

==Exercise 2: F-H-H system==
===PES inspection===
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions According to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===
F+H-H reaction is exothermic while H+H-F is endothermic. Since enthalpy of HH is +436 kJ/mol while HF is +562 kJ/mol, the formation of HF is exothermic as the energy released from H-F formation is greater than H-H breaking, vice versa.
From PES, it proves the classification as from the (F+H2) PES, the potential energy along reaction is actually lower than that of (H+HF) inspection.
[[File:Surface Plot 01523148 2.png]]
PES plot of H+HF
[[File:Surface Plot 01523148 1.png]]
PES plot of F+HH

===Locate the approximate position of the transition state.===
Endothermic reaction means the late transition state resembles the products meanwhile exothermic reaction resembles early transition state which is the reactants. Hence from (F+H-H) reaction, the exothermic reaction allows us to predict the transition state is around 74pm, which is the bond length of H-H, in one of the internuclear distances.
Through trial and error, AB=181.1pm and BC=74.4875pm allow us to obtain the transition state, with potential energy of -433.981.The internuclear distance plot by MEP below shows that no oscillation occurs around and hence confirm the capture of transition state.
[[File:Fig2.2 01523148 InternuclearDistance.png]]

===Report the activation energy for both reactions.===
Energy of transition state - Energy of reactants = activation energy. Since the transition state energy was found from above, the energy of reactants can be found from the following image, using 185pm AB and 50pm BC, with 1000number of steps and 0.15fs of size. The picture below depicts how energy transforms throughout the (F+H2) reaction. There are two energy states captures in the graph because there are two flat lines in the reaction observed. The first flat line shows the energy of -434.751 kJ/mol, which is energy of these two reactants combined (F+H2), meanwhile the second flat line shows the energy of -559 kJ/mol, which resembles the energy of the products (H+HF).
Therefore, the activation energy of H+HF endothermic reaction is -433.981-(-434.751)=+0.77kJ/mol
The activation energy of F+HF exothermic reaction is -433.981-(-559)=+125 kJ/mol
[[File:Fig2.3 01523148.png]]
The energy state of reactants(H+HF)
[[File:Fig2.3.2 01523148.png]]
The energy state of reactants(F+H2)

===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.===
Momentum of F-H is dependent on the rate of reaction, as supported from the plots below. This explains that the reaction energy is released as kinetic energy, more specifically, translational and vibrational form of kinetic energy. Since kinetic energy can convert into form of heat, heat can be measured by calorimetry, using E=mc(dT).
IR spectroscopy and vibrational spectroscopy can also measure the vibrational kinetic energy absorbed from the bond oscillation of HF, with IR measuring the transition from ground state to excited state while vibrational spectroscopy measuring the overtones of peaks in the spectra.
[[File:2.4 01523148 1.png]]

AB=184pm,-1g.pm/fs.mol, BC=74pm,1g.pm/fs.mol

[[File:2.4 01523148 2.png]]

AB=184pm and -5g.pm/fs.mol, BC=74pm,5g.pm/fs.mol

[[File:2.4 01523148 3 .png]]

AB=184pm and -10g.pm/fs.mol, BC=74pm, 10g.pm/fs.mol

===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 rule states that the translational energy is more effective than vibration energy in overcoming an early transition state barrier, and vibrational energy is more effective than translational energy when overcoming a late transition state region.

<nowiki> </nowiki>From exothermic (F+H2) reaction, it resembles early transition state barrier. Therefore, from the images below, with increasing momentum and decreased energy of H-F vibration, the greater proportion of translational energy is more likely to overcome the transition energy barrier.

From endothermic(H+HF) reaction, the late transition state barrier is resembled. The smaller proportion of p1 to p2 ratio will be more likely to form reactive trajectory because transition state barrier is more likely to be overcame by increased proportion of vibration energy.
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
! p2/ g.mol-1.pm.fs-1
! Etot
! Reactive?
! Description of the dynamics
! Illustration of the trajectory

|-
| -1.3
| 9
|KE=+70.189kJ/mol
Total energy=-363.767kJ/mol
|no
|exothermic
|[[File:2.5 01523148 1.png|250px]]

|-
| -2
| 0.005
|KE=+2.115kJ/mol
Total energy=431.841kJ/mol
|yes
|exothermic
|[[File:2.5 01523148 2.png|250px]]

|-
| -1
| -5.1
|KE= +1.105kJ/mol
Total energy=-413.709kJ/mol
|no
|endothermic
|[[File:2.4 01523148 3.png|250px]]

|-
| -5.1
| -10.1
|KE= 43.632kJ/mol

Total energy=-371.183kJ/mol
|yes
|endothermic
|[[File:2.5 01523148 4.png|250px]]
|}
r1=184pm, r2=74pm

MRD:MolecularReactionDynamics01523148

2020-05-08T22:06:12Z

Yyl18:

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

The transition state is mathematically defined by ∂V(ri)/∂ri=0, which r value represents the distance between atoms.The transition state can be identified at the position between maximum and minimum on the Potential Energy Surface and also by finding out the energy value around the trajectory from the contour plot.
[[File:Inkedtransition state surface plot1234 LI.jpg]]

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

The best estimate of the transition state position is around 90.775pm as it has no F and oscillation between each atoms as there is no oscillation occurs from the internuclear distance vs time plot. This means the picture captured is the transition state.
[[File:Question 1.2.png]]

===Comment on how the mep and the trajectory you just calculated differ.===
From the images below, using r1 = rts+δ, r2 = rts, with δ=1. The MEP and dynamics calculation types are compared.
[[File:MES 01523148.png]]
Dynamics contour plot
[[File:Trajectory1.3 01523148.png]]
MEP contour plot

From the pictures above, it shows that dynamics include the calculation of oscillation of bonds as from the Dynamics contour plot we can see there is oscillating curve around B-C distances. This means dynamics calculation involve calculation of vibrational energy, while no kinetic energy will be accounted in MEP calculation.

[[File:1.3.2 0152314888 InterNuclear Dynamics.png]]
Internulcear Dynamics plot
[[File:1.3.2 0152314888 InterNuclear MEP.png]]
Internuclear MEP plot
[[File:1.3.2 01523148 MomentaVsTime Dynamics.png]]
MomentaVsTime Dynamics plot
[[File:1.3.2 01523148 MomentaVsTime MEP.png]]
MomentaVsTime MEP plot

===Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable"
|-
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Since BC distance is starting to diminish, it shows that H(b)-H(c) bond is forming while H(a)-H(b) bond is breaking. The particles go through the transition state when H(c) collides with H(b).
|[[File:Surface Plot 1.3.1 01523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|Since no AB distance can be found from the image, that means no bond formation or bond breaking between H(a) and H(b). Therefore no collision found during the interactions of three particles and cannot overcome the energy barrier, reason of no collision might be due to repulsion.
|[[File:Surface Plot 1.3.2 1523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The reaction is similar to the scenario one. Only difference is H(c) is moving at a higher speed toward H(b) but same outcome has reached because energy barrier is overcame.
|[[File:Surface Plot 1.3.3 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|The H(b)-H(c) bond is formed first by having H(c) colliding into H(a)-H(b). This collision crosses the transition energy barrier at first. However the diagram shows the recrossing of transition state region which breaks the H(b)-H(c) bond and reform H(a)-H(b) back to original, with greater vibrational energy.
|[[File:Surface Plot 1.3.4 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|This reaction is similar to that of scenario 4. However instead of H(c) leaving the system, the H(c) particle goes back to the system and form H(b)-H(c) bond which breaks the oscillating H(a)-H(b) bond. The system crosses and recrosses the transition state 3 times in total, leaving with greater oscillation of H(c)-H(b).
|[[File:Surface Plot 1.3.5 1523148.png|250px]]}

|}

===Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?===

Transition state theory treats reaction classically such that quantum tunnelling cannot occur. Also energy must obey Boltzmann distribution rule and all systems that have already crossed the TS region cannot recross the areas from product back to reactants. Therefore the reaction rate values from TST prediction is expected to be greater than experimental values (like scenario 5 above) if any quantum effect like barrier-recrossing or tunnelling, especially the former one occurs.

==Exercise 2: F-H-H system==
===PES inspection===
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions According to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===
F+H-H reaction is exothermic while H+H-F is endothermic. Since enthalpy of HH is +436 kJ/mol while HF is +562 kJ/mol, the formation of HF is exothermic as the energy released from H-F formation is greater than H-H breaking, vice versa.
From PES, it proves the classification as from the (F+H2) PES, the potential energy along reaction is actually lower than that of (H+HF) inspection.
[[File:Surface Plot 01523148 2.png]]
PES plot of H+HF
[[File:Surface Plot 01523148 1.png]]
PES plot of F+HH

===Locate the approximate position of the transition state.===
Endothermic reaction means the late transition state resembles the products meanwhile exothermic reaction resembles early transition state which is the reactants. Hence from (F+H-H) reaction, the exothermic reaction allows us to predict the transition state is around 74pm, which is the bond length of H-H, in one of the internuclear distances.
Through trial and error, AB=181.1pm and BC=74.4875pm allow us to obtain the transition state, with potential energy of -433.981.The internuclear distance plot by MEP below shows that no oscillation occurs around and hence confirm the capture of transition state.
[[File:Fig2.2 01523148 InternuclearDistance.png]]

===Report the activation energy for both reactions.===
Energy of transition state - Energy of reactants = activation energy. Since the transition state energy was found from above, the energy of reactants can be found from the following image, using 185pm AB and 50pm BC, with 1000number of steps and 0.15fs of size. The picture below depicts how energy transforms throughout the (F+H2) reaction. There are two energy states captures in the graph because there are two flat lines in the reaction observed. The first flat line shows the energy of -434.751 kJ/mol, which is energy of these two reactants combined (F+H2), meanwhile the second flat line shows the energy of -559 kJ/mol, which resembles the energy of the products (H+HF).
Therefore, the activation energy of H+HF endothermic reaction is -433.981-(-434.751)=+0.77kJ/mol
The activation energy of F+HF exothermic reaction is -433.981-(-559)=+125 kJ/mol
[[File:Fig2.3 01523148.png]]
The energy state of reactants(H+HF)
[[File:Fig2.3.2 01523148.png]]
The energy state of reactants(F+H2)

===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.===
Momentum of F-H is dependent on the rate of reaction, as supported from the plots below. This explains that the reaction energy is released as kinetic energy, more specifically, translational and vibrational form of kinetic energy. Since kinetic energy can convert into form of heat, heat can be measured by calorimetry, using E=mc(dT).
IR spectroscopy and vibrational spectroscopy can also measure the vibrational kinetic energy absorbed from the bond oscillation of HF, with IR measuring the transition from ground state to excited state while vibrational spectroscopy measuring the overtones of peaks in the spectra.
[[File:2.4 01523148 1.png]]

AB=184pm,-1g.pm/fs.mol, BC=74pm,1g.pm/fs.mol

[[File:2.4 01523148 2.png]]

AB=184pm and -5g.pm/fs.mol, BC=74pm,5g.pm/fs.mol

[[File:2.4 01523148 3 .png]]

AB=184pm and -10g.pm/fs.mol, BC=74pm, 10g.pm/fs.mol

===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 rule states that the translational energy is more effective than vibration energy in overcoming an early transition state barrier, and vibrational energy is more effective than translational energy when overcoming a late transition state region.

<nowiki> </nowiki>From exothermic (F+H2) reaction, it resembles early transition state barrier. Therefore, from the images below, with increasing momentum and decreased energy of H-F vibration, the greater proportion of translational energy is more likely to overcome the transition energy barrier.

From endothermic(H+HF) reaction, the late transition state barrier is resembled. The smaller proportion of p1 to p2 ratio will be more likely to form reactive trajectory because transition state barrier is more likely to be overcame by increased proportion of vibration energy.
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
! p2/ g.mol-1.pm.fs-1
! Etot
! Reactive?
! Description of the dynamics
! Illustration of the trajectory

|-
| -1.3
| 9
|KE=+70.189kJ/mol
Total energy=-363.767kJ/mol
|no
|exothermic
|[[File:2.5 01523148 1.png]]

|-
| -2
| 0.005
|KE=+2.115kJ/mol
Total energy=431.841kJ/mol
|yes
|exothermic
|[[File:2.5 01523148 2.png]]

|-
| -1
| -5.1
|KE= +1.105kJ/mol
Total energy=-413.709kJ/mol
|no
|endothermic
|[[File:2.4 01523148 3.png]]

|-
| -5.1
| -10.1
|KE= 43.632kJ/mol

Total energy=-371.183kJ/mol
|yes
|endothermic
|[[File:2.5 01523148 4.png]]
|}
r1=184pm, r2=74pm

File:2.5 01523148 2.png

2020-05-08T22:05:50Z

Yyl18:

File:2.5 01523148 1.png

2020-05-08T22:04:39Z

Yyl18:

File:2.4 01523148 3.png

2020-05-08T22:04:08Z

Yyl18:

File:2.5 01523148 4.png

2020-05-08T22:03:39Z

Yyl18:

MRD:MolecularReactionDynamics01523148

2020-05-08T22:02:52Z

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

The transition state is mathematically defined by ∂V(ri)/∂ri=0, which r value represents the distance between atoms.The transition state can be identified at the position between maximum and minimum on the Potential Energy Surface and also by finding out the energy value around the trajectory from the contour plot.
[[File:Inkedtransition state surface plot1234 LI.jpg]]

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

The best estimate of the transition state position is around 90.775pm as it has no F and oscillation between each atoms as there is no oscillation occurs from the internuclear distance vs time plot. This means the picture captured is the transition state.
[[File:Question 1.2.png]]

===Comment on how the mep and the trajectory you just calculated differ.===
From the images below, using r1 = rts+δ, r2 = rts, with δ=1. The MEP and dynamics calculation types are compared.
[[File:MES 01523148.png]]
Dynamics contour plot
[[File:Trajectory1.3 01523148.png]]
MEP contour plot

From the pictures above, it shows that dynamics include the calculation of oscillation of bonds as from the Dynamics contour plot we can see there is oscillating curve around B-C distances. This means dynamics calculation involve calculation of vibrational energy, while no kinetic energy will be accounted in MEP calculation.

[[File:1.3.2 0152314888 InterNuclear Dynamics.png]]
Internulcear Dynamics plot
[[File:1.3.2 0152314888 InterNuclear MEP.png]]
Internuclear MEP plot
[[File:1.3.2 01523148 MomentaVsTime Dynamics.png]]
MomentaVsTime Dynamics plot
[[File:1.3.2 01523148 MomentaVsTime MEP.png]]
MomentaVsTime MEP plot

===Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable"
|-
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Since BC distance is starting to diminish, it shows that H(b)-H(c) bond is forming while H(a)-H(b) bond is breaking. The particles go through the transition state when H(c) collides with H(b).
|[[File:Surface Plot 1.3.1 01523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|Since no AB distance can be found from the image, that means no bond formation or bond breaking between H(a) and H(b). Therefore no collision found during the interactions of three particles and cannot overcome the energy barrier, reason of no collision might be due to repulsion.
|[[File:Surface Plot 1.3.2 1523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The reaction is similar to the scenario one. Only difference is H(c) is moving at a higher speed toward H(b) but same outcome has reached because energy barrier is overcame.
|[[File:Surface Plot 1.3.3 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|The H(b)-H(c) bond is formed first by having H(c) colliding into H(a)-H(b). This collision crosses the transition energy barrier at first. However the diagram shows the recrossing of transition state region which breaks the H(b)-H(c) bond and reform H(a)-H(b) back to original, with greater vibrational energy.
|[[File:Surface Plot 1.3.4 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|This reaction is similar to that of scenario 4. However instead of H(c) leaving the system, the H(c) particle goes back to the system and form H(b)-H(c) bond which breaks the oscillating H(a)-H(b) bond. The system crosses and recrosses the transition state 3 times in total, leaving with greater oscillation of H(c)-H(b).
|[[File:Surface Plot 1.3.5 1523148.png|250px]]}

|}

===Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?===

Transition state theory treats reaction classically such that quantum tunnelling cannot occur. Also energy must obey Boltzmann distribution rule and all systems that have already crossed the TS region cannot recross the areas from product back to reactants. Therefore the reaction rate values from TST prediction is expected to be greater than experimental values (like scenario 5 above) if any quantum effect like barrier-recrossing or tunnelling, especially the former one occurs.

==Exercise 2: F-H-H system==
===PES inspection===
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions According to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===
F+H-H reaction is exothermic while H+H-F is endothermic. Since enthalpy of HH is +436 kJ/mol while HF is +562 kJ/mol, the formation of HF is exothermic as the energy released from H-F formation is greater than H-H breaking, vice versa.
From PES, it proves the classification as from the (F+H2) PES, the potential energy along reaction is actually lower than that of (H+HF) inspection.
[[File:Surface Plot 01523148 2.png]]
PES plot of H+HF
[[File:Surface Plot 01523148 1.png]]
PES plot of F+HH

===Locate the approximate position of the transition state.===
Endothermic reaction means the late transition state resembles the products meanwhile exothermic reaction resembles early transition state which is the reactants. Hence from (F+H-H) reaction, the exothermic reaction allows us to predict the transition state is around 74pm, which is the bond length of H-H, in one of the internuclear distances.
Through trial and error, AB=181.1pm and BC=74.4875pm allow us to obtain the transition state, with potential energy of -433.981.The internuclear distance plot by MEP below shows that no oscillation occurs around and hence confirm the capture of transition state.
[[File:Fig2.2 01523148 InternuclearDistance.png]]

===Report the activation energy for both reactions.===
Energy of transition state - Energy of reactants = activation energy. Since the transition state energy was found from above, the energy of reactants can be found from the following image, using 185pm AB and 50pm BC, with 1000number of steps and 0.15fs of size. The picture below depicts how energy transforms throughout the (F+H2) reaction. There are two energy states captures in the graph because there are two flat lines in the reaction observed. The first flat line shows the energy of -434.751 kJ/mol, which is energy of these two reactants combined (F+H2), meanwhile the second flat line shows the energy of -559 kJ/mol, which resembles the energy of the products (H+HF).
Therefore, the activation energy of H+HF endothermic reaction is -433.981-(-434.751)=+0.77kJ/mol
The activation energy of F+HF exothermic reaction is -433.981-(-559)=+125 kJ/mol
[[File:Fig2.3 01523148.png]]
The energy state of reactants(H+HF)
[[File:Fig2.3.2 01523148.png]]
The energy state of reactants(F+H2)

===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.===
Momentum of F-H is dependent on the rate of reaction, as supported from the plots below. This explains that the reaction energy is released as kinetic energy, more specifically, translational and vibrational form of kinetic energy. Since kinetic energy can convert into form of heat, heat can be measured by calorimetry, using E=mc(dT).
IR spectroscopy and vibrational spectroscopy can also measure the vibrational kinetic energy absorbed from the bond oscillation of HF, with IR measuring the transition from ground state to excited state while vibrational spectroscopy measuring the overtones of peaks in the spectra.
[[File:2.4 01523148 1.png]]

AB=184pm,-1g.pm/fs.mol, BC=74pm,1g.pm/fs.mol

[[File:2.4 01523148 2.png]]

AB=184pm and -5g.pm/fs.mol, BC=74pm,5g.pm/fs.mol

[[File:2.4 01523148 3 .png]]

AB=184pm and -10g.pm/fs.mol, BC=74pm, 10g.pm/fs.mol

===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 rule states that the translational energy is more effective than vibration energy in overcoming an early transition state barrier, and vibrational energy is more effective than translational energy when overcoming a late transition state region.

<nowiki> </nowiki>From exothermic (F+H2) reaction, it resembles early transition state barrier. Therefore, from the images below, with increasing momentum and decreased energy of H-F vibration, the greater proportion of translational energy is more likely to overcome the transition energy barrier.

From endothermic(H+HF) reaction, the late transition state barrier is resembled. The smaller proportion of p1 to p2 ratio will be more likely to form reactive trajectory because transition state barrier is more likely to be overcame by increased proportion of vibration energy.
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
! p2/ g.mol-1.pm.fs-1
! Etot
! Reactive?
! Description of the dynamics
! Illustration of the trajectory

|-
| -1.3
| 9
|KE=+70.189kJ/mol
Total energy=-363.767kJ/mol
|no
|exothermic
|

|-
| -2
| 0.005
|KE=+2.115kJ/mol
Total energy=431.841kJ/mol
|yes
|exothermic
|

|-
| -1
| -5.1
|KE= +1.105kJ/mol
Total energy=-413.709kJ/mol
|no
|endothermic
|

|-
| -5.1
| -10.1
|KE= 43.632kJ/mol

Total energy=-371.183kJ/mol
|yes
|endothermic
|

|-
|
|
|}
r1=184pm, r2=74pm

MRD:MolecularReactionDynamics01523148

2020-05-08T21:35:07Z

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

The transition state is mathematically defined by ∂V(ri)/∂ri=0, which r value represents the distance between atoms.The transition state can be identified at the position between maximum and minimum on the Potential Energy Surface and also by finding out the energy value around the trajectory from the contour plot.
[[File:Inkedtransition state surface plot1234 LI.jpg]]

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

The best estimate of the transition state position is around 90.775pm as it has no F and oscillation between each atoms as there is no oscillation occurs from the internuclear distance vs time plot. This means the picture captured is the transition state.
[[File:Question 1.2.png]]

===Comment on how the mep and the trajectory you just calculated differ.===
From the images below, using r1 = rts+δ, r2 = rts, with δ=1. The MEP and dynamics calculation types are compared.
[[File:MES 01523148.png]]
Dynamics contour plot
[[File:Trajectory1.3 01523148.png]]
MEP contour plot

From the pictures above, it shows that dynamics include the calculation of oscillation of bonds as from the Dynamics contour plot we can see there is oscillating curve around B-C distances. This means dynamics calculation involve calculation of vibrational energy, while no kinetic energy will be accounted in MEP calculation.

[[File:1.3.2 0152314888 InterNuclear Dynamics.png]]
Internulcear Dynamics plot
[[File:1.3.2 0152314888 InterNuclear MEP.png]]
Internuclear MEP plot
[[File:1.3.2 01523148 MomentaVsTime Dynamics.png]]
MomentaVsTime Dynamics plot
[[File:1.3.2 01523148 MomentaVsTime MEP.png]]
MomentaVsTime MEP plot

===Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable"
|-
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Since BC distance is starting to diminish, it shows that H(b)-H(c) bond is forming while H(a)-H(b) bond is breaking. The particles go through the transition state when H(c) collides with H(b).
|[[File:Surface Plot 1.3.1 01523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|Since no AB distance can be found from the image, that means no bond formation or bond breaking between H(a) and H(b). Therefore no collision found during the interactions of three particles and cannot overcome the energy barrier, reason of no collision might be due to repulsion.
|[[File:Surface Plot 1.3.2 1523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The reaction is similar to the scenario one. Only difference is H(c) is moving at a higher speed toward H(b) but same outcome has reached because energy barrier is overcame.
|[[File:Surface Plot 1.3.3 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|The H(b)-H(c) bond is formed first by having H(c) colliding into H(a)-H(b). This collision crosses the transition energy barrier at first. However the diagram shows the recrossing of transition state region which breaks the H(b)-H(c) bond and reform H(a)-H(b) back to original, with greater vibrational energy.
|[[File:Surface Plot 1.3.4 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|This reaction is similar to that of scenario 4. However instead of H(c) leaving the system, the H(c) particle goes back to the system and form H(b)-H(c) bond which breaks the oscillating H(a)-H(b) bond. The system crosses and recrosses the transition state 3 times in total, leaving with greater oscillation of H(c)-H(b).
|[[File:Surface Plot 1.3.5 1523148.png|250px]]}

|}

===Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?===

Transition state theory treats reaction classically such that quantum tunnelling cannot occur. Also energy must obey Boltzmann distribution rule and all systems that have already crossed the TS region cannot recross the areas from product back to reactants. Therefore the reaction rate values from TST prediction is expected to be greater than experimental values (like scenario 5 above) if any quantum effect like barrier-recrossing or tunnelling, especially the former one occurs.

==Exercise 2: F-H-H system==
===PES inspection===
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions According to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===
F+H-H reaction is exothermic while H+H-F is endothermic. Since enthalpy of HH is +436 kJ/mol while HF is +562 kJ/mol, the formation of HF is exothermic as the energy released from H-F formation is greater than H-H breaking, vice versa.
From PES, it proves the classification as from the (F+H2) PES, the potential energy along reaction is actually lower than that of (H+HF) inspection.
[[File:Surface Plot 01523148 2.png]]
PES plot of H+HF
[[File:Surface Plot 01523148 1.png]]
PES plot of F+HH

===Locate the approximate position of the transition state.===
Endothermic reaction means the late transition state resembles the products meanwhile exothermic reaction resembles early transition state which is the reactants. Hence from (F+H-H) reaction, the exothermic reaction allows us to predict the transition state is around 74pm, which is the bond length of H-H, in one of the internuclear distances.
Through trial and error, AB=181.1pm and BC=74.4875pm allow us to obtain the transition state, with potential energy of -433.981.The internuclear distance plot by MEP below shows that no oscillation occurs around and hence confirm the capture of transition state.
[[File:Fig2.2 01523148 InternuclearDistance.png]]

===Report the activation energy for both reactions.===
Energy of transition state - Energy of reactants = activation energy. Since the transition state energy was found from above, the energy of reactants can be found from the following image, using 185pm AB and 50pm BC, with 1000number of steps and 0.15fs of size. The picture below depicts how energy transforms throughout the (F+H2) reaction. There are two energy states captures in the graph because there are two flat lines in the reaction observed. The first flat line shows the energy of -434.751 kJ/mol, which is energy of these two reactants combined (F+H2), meanwhile the second flat line shows the energy of -559 kJ/mol, which resembles the energy of the products (H+HF).
Therefore, the activation energy of H+HF endothermic reaction is -433.981-(-434.751)=+0.77kJ/mol
The activation energy of F+HF exothermic reaction is -433.981-(-559)=+125 kJ/mol
[[File:Fig2.3 01523148.png]]
The energy state of reactants(H+HF)
[[File:Fig2.3.2 01523148.png]]
The energy state of reactants(F+H2)

===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.===
Momentum of F-H is dependent on the rate of reaction, as supported from the plots below. This explains that the reaction energy is released as kinetic energy, more specifically, translational and vibrational form of kinetic energy. Since kinetic energy can convert into form of heat, heat can be measured by calorimetry, using E=mc(dT).
IR spectroscopy and vibrational spectroscopy can also measure the vibrational kinetic energy absorbed from the bond oscillation of HF, with IR measuring the transition from ground state to excited state while vibrational spectroscopy measuring the overtones of peaks in the spectra.
[[File:2.4 01523148 1.png]]
AB=184pm,-1g.pm/fs.mol, BC=74pm,1g.pm/fs.mol

[[File:2.4 01523148 2.png]]
AB=184pm and -5g.pm/fs.mol, BC=74pm,5g.pm/fs.mol

[[File:2.4 01523148 3 .png]]
AB=184pm and -10g.pm/fs.mol, BC=74pm, 10g.pm/fs.mol

===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 rule states that the translational energy is more effective than vibration energy in overcoming an early transition state barrier, and vibrational energy is more effective than translational energy when overcoming a late transition state region.
From exothermic (F+H2) reaction, it resembles early transition state barrier. Therefore, from the images below, with increasing momentum and decreased energy of H-F vibration, the greater proportion of translational energy is more likely to overcome the transition energy barrier.
From endothermic(H+HF) reaction, the late transition state barrier is resembled. The smaller proportion of p1 to p2 ratio will be more likely to form reactive trajectory because transition state barrier is more likely to be overcame by increased proportion of vibration energy.

MRD:MolecularReactionDynamics01523148

2020-05-08T21:34:50Z

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

The transition state is mathematically defined by ∂V(ri)/∂ri=0, which r value represents the distance between atoms.The transition state can be identified at the position between maximum and minimum on the Potential Energy Surface and also by finding out the energy value around the trajectory from the contour plot.
[[File:Inkedtransition state surface plot1234 LI.jpg]]

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

The best estimate of the transition state position is around 90.775pm as it has no F and oscillation between each atoms as there is no oscillation occurs from the internuclear distance vs time plot. This means the picture captured is the transition state.
[[File:Question 1.2.png]]

===Comment on how the mep and the trajectory you just calculated differ.===
From the images below, using r1 = rts+δ, r2 = rts, with δ=1. The MEP and dynamics calculation types are compared.
[[File:MES 01523148.png]]
Dynamics contour plot
[[File:Trajectory1.3 01523148.png]]
MEP contour plot

From the pictures above, it shows that dynamics include the calculation of oscillation of bonds as from the Dynamics contour plot we can see there is oscillating curve around B-C distances. This means dynamics calculation involve calculation of vibrational energy, while no kinetic energy will be accounted in MEP calculation.

[[File:1.3.2 0152314888 InterNuclear Dynamics.png]]
Internulcear Dynamics plot
[[File:1.3.2 0152314888 InterNuclear MEP.png]]
Internuclear MEP plot
[[File:1.3.2 01523148 MomentaVsTime Dynamics.png]]
MomentaVsTime Dynamics plot
[[File:1.3.2 01523148 MomentaVsTime MEP.png]]
MomentaVsTime MEP plot

===Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable"
|-
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Since BC distance is starting to diminish, it shows that H(b)-H(c) bond is forming while H(a)-H(b) bond is breaking. The particles go through the transition state when H(c) collides with H(b).
|[[File:Surface Plot 1.3.1 01523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|Since no AB distance can be found from the image, that means no bond formation or bond breaking between H(a) and H(b). Therefore no collision found during the interactions of three particles and cannot overcome the energy barrier, reason of no collision might be due to repulsion.
|[[File:Surface Plot 1.3.2 1523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The reaction is similar to the scenario one. Only difference is H(c) is moving at a higher speed toward H(b) but same outcome has reached because energy barrier is overcame.
|[[File:Surface Plot 1.3.3 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|The H(b)-H(c) bond is formed first by having H(c) colliding into H(a)-H(b). This collision crosses the transition energy barrier at first. However the diagram shows the recrossing of transition state region which breaks the H(b)-H(c) bond and reform H(a)-H(b) back to original, with greater vibrational energy.
|[[File:Surface Plot 1.3.4 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|This reaction is similar to that of scenario 4. However instead of H(c) leaving the system, the H(c) particle goes back to the system and form H(b)-H(c) bond which breaks the oscillating H(a)-H(b) bond. The system crosses and recrosses the transition state 3 times in total, leaving with greater oscillation of H(c)-H(b).
|[[File:Surface Plot 1.3.5 1523148.png|250px]]}

|}

===Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?===

Transition state theory treats reaction classically such that quantum tunnelling cannot occur. Also energy must obey Boltzmann distribution rule and all systems that have already crossed the TS region cannot recross the areas from product back to reactants. Therefore the reaction rate values from TST prediction is expected to be greater than experimental values (like scenario 5 above) if any quantum effect like barrier-recrossing or tunnelling, especially the former one occurs.

==Exercise 2: F-H-H system==
===PES inspection===
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions According to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===
F+H-H reaction is exothermic while H+H-F is endothermic. Since enthalpy of HH is +436 kJ/mol while HF is +562 kJ/mol, the formation of HF is exothermic as the energy released from H-F formation is greater than H-H breaking, vice versa.
From PES, it proves the classification as from the (F+H2) PES, the potential energy along reaction is actually lower than that of (H+HF) inspection.
[[File:Surface Plot 01523148 2.png]]
PES plot of H+HF
[[File:Surface Plot 01523148 1.png]]
PES plot of F+HH

===Locate the approximate position of the transition state.===
Endothermic reaction means the late transition state resembles the products meanwhile exothermic reaction resembles early transition state which is the reactants. Hence from (F+H-H) reaction, the exothermic reaction allows us to predict the transition state is around 74pm, which is the bond length of H-H, in one of the internuclear distances.
Through trial and error, AB=181.1pm and BC=74.4875pm allow us to obtain the transition state, with potential energy of -433.981.The internuclear distance plot by MEP below shows that no oscillation occurs around and hence confirm the capture of transition state.
[[File:Fig2.2 01523148 InternuclearDistance.png]]

===Report the activation energy for both reactions.===
Energy of transition state - Energy of reactants = activation energy. Since the transition state energy was found from above, the energy of reactants can be found from the following image, using 185pm AB and 50pm BC, with 1000number of steps and 0.15fs of size. The picture below depicts how energy transforms throughout the (F+H2) reaction. There are two energy states captures in the graph because there are two flat lines in the reaction observed. The first flat line shows the energy of -434.751 kJ/mol, which is energy of these two reactants combined (F+H2), meanwhile the second flat line shows the energy of -559 kJ/mol, which resembles the energy of the products (H+HF).
Therefore, the activation energy of H+HF endothermic reaction is -433.981-(-434.751)=+0.77kJ/mol
The activation energy of F+HF exothermic reaction is -433.981-(-559)=+125 kJ/mol
[[File:Fig2.3 01523148.png]]
The energy state of reactants(H+HF)
[[File:Fig2.3.2 01523148.png]]
The energy state of reactants(F+H2)

===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.===
Momentum of F-H is dependent on the rate of reaction, as supported from the plots below. This explains that the reaction energy is released as kinetic energy, more specifically, translational and vibrational form of kinetic energy. Since kinetic energy can convert into form of heat, heat can be measured by calorimetry, using E=mc(dT).
IR spectroscopy and vibrational spectroscopy can also measure the vibrational kinetic energy absorbed from the bond oscillation of HF, with IR measuring the transition from ground state to excited state while vibrational spectroscopy measuring the overtones of peaks in the spectra.
[[File:2.4 01523148 1.png]]
AB=184pm,-1g.pm/fs.mol, BC=74pm,1g.pm/fs.mol

[[File:2.4 01523148 2.png]]
AB=184pm and -5g.pm/fs.mol, BC=74pm,5g.pm/fs.mol

[[File:2.4 01523148 3 .png]]
AB=184pm and -10g.pm/fs.mol, BC=74pm, 10g.pm/fs.mol

===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 rule states that the translational energy is more effective than vibration energy in overcoming an early transition state barrier, and vibrational energy is more effective than translational energy when overcoming a late transition state region.

From exothermic (F+H2) reaction, it resembles early transition state barrier. Therefore, from the images below, with increasing momentum and decreased energy of H-F vibration, the greater proportion of translational energy is more likely to overcome the transition energy barrier.

From endothermic(H+HF) reaction, the late transition state barrier is resembled. The smaller proportion of p1 to p2 ratio will be more likely to form reactive trajectory because transition state barrier is more likely to be overcame by increased proportion of vibration energy.

MRD:MolecularReactionDynamics01523148

2020-05-08T20:31:38Z

Yyl18: /* 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? */

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

The transition state is mathematically defined by ∂V(ri)/∂ri=0, which r value represents the distance between atoms.The transition state can be identified at the position between maximum and minimum on the Potential Energy Surface and also by finding out the energy value around the trajectory from the contour plot.
[[File:Inkedtransition state surface plot1234 LI.jpg]]

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

The best estimate of the transition state position is around 90.775pm as it has no F and oscillation between each atoms as there is no oscillation occurs from the internuclear distance vs time plot. This means the picture captured is the transition state.
[[File:Question 1.2.png]]

===Comment on how the mep and the trajectory you just calculated differ.===
From the images below, using r1 = rts+δ, r2 = rts, with δ=1. The MEP and dynamics calculation types are compared.
[[File:MES 01523148.png]]
Dynamics contour plot
[[File:Trajectory1.3 01523148.png]]
MEP contour plot

From the pictures above, it shows that dynamics include the calculation of oscillation of bonds as from the Dynamics contour plot we can see there is oscillating curve around B-C distances. This means dynamics calculation involve calculation of vibrational energy, while no kinetic energy will be accounted in MEP calculation.

[[File:1.3.2 0152314888 InterNuclear Dynamics.png]]
Internulcear Dynamics plot
[[File:1.3.2 0152314888 InterNuclear MEP.png]]
Internuclear MEP plot
[[File:1.3.2 01523148 MomentaVsTime Dynamics.png]]
MomentaVsTime Dynamics plot
[[File:1.3.2 01523148 MomentaVsTime MEP.png]]
MomentaVsTime MEP plot

===Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable"
|-
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Since BC distance is starting to diminish, it shows that H(b)-H(c) bond is forming while H(a)-H(b) bond is breaking. The particles go through the transition state when H(c) collides with H(b).
|[[File:Surface Plot 1.3.1 01523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|Since no AB distance can be found from the image, that means no bond formation or bond breaking between H(a) and H(b). Therefore no collision found during the interactions of three particles and cannot overcome the energy barrier, reason of no collision might be due to repulsion.
|[[File:Surface Plot 1.3.2 1523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The reaction is similar to the scenario one. Only difference is H(c) is moving at a higher speed toward H(b) but same outcome has reached because energy barrier is overcame.
|[[File:Surface Plot 1.3.3 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|The H(b)-H(c) bond is formed first by having H(c) colliding into H(a)-H(b). This collision crosses the transition energy barrier at first. However the diagram shows the recrossing of transition state region which breaks the H(b)-H(c) bond and reform H(a)-H(b) back to original, with greater vibrational energy.
|[[File:Surface Plot 1.3.4 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|This reaction is similar to that of scenario 4. However instead of H(c) leaving the system, the H(c) particle goes back to the system and form H(b)-H(c) bond which breaks the oscillating H(a)-H(b) bond. The system crosses and recrosses the transition state 3 times in total, leaving with greater oscillation of H(c)-H(b).
|[[File:Surface Plot 1.3.5 1523148.png|250px]]}

|}

===Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?===

Transition state theory treats reaction classically such that quantum tunnelling cannot occur. Also energy must obey Boltzmann distribution rule and all systems that have already crossed the TS region cannot recross the areas from product back to reactants. Therefore the reaction rate values from TST prediction is expected to be greater than experimental values (like scenario 5 above) if any quantum effect like barrier-recrossing or tunnelling, especially the former one occurs.

==Exercise 2: F-H-H system==
===PES inspection===
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions According to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===
F+H-H reaction is exothermic while H+H-F is endothermic. Since enthalpy of HH is +436 kJ/mol while HF is +562 kJ/mol, the formation of HF is exothermic as the energy released from H-F formation is greater than H-H breaking, vice versa.
From PES, it proves the classification as from the (F+H2) PES, the potential energy along reaction is actually lower than that of (H+HF) inspection.
[[File:Surface Plot 01523148 2.png]]
PES plot of H+HF
[[File:Surface Plot 01523148 1.png]]
PES plot of F+HH

===Locate the approximate position of the transition state.===
Endothermic reaction means the late transition state resembles the products meanwhile exothermic reaction resembles early transition state which is the reactants. Hence from (F+H-H) reaction, the exothermic reaction allows us to predict the transition state is around 74pm, which is the bond length of H-H, in one of the internuclear distances.
Through trial and error, AB=181.1pm and BC=74.4875pm allow us to obtain the transition state, with potential energy of -433.981.The internuclear distance plot by MEP below shows that no oscillation occurs around and hence confirm the capture of transition state.
[[File:Fig2.2 01523148 InternuclearDistance.png]]

===Report the activation energy for both reactions.===
Energy of transition state - Energy of reactants = activation energy. Since the transition state energy was found from above, the energy of reactants can be found from the following image, using 185pm AB and 50pm BC, with 1000number of steps and 0.15fs of size. The picture below depicts how energy transforms throughout the (F+H2) reaction. There are two energy states captures in the graph because there are two flat lines in the reaction observed. The first flat line shows the energy of -434.751 kJ/mol, which is energy of these two reactants combined (F+H2), meanwhile the second flat line shows the energy of -559 kJ/mol, which resembles the energy of the products (H+HF).
Therefore, the activation energy of H+HF endothermic reaction is -433.981-(-434.751)=+0.77kJ/mol
The activation energy of F+HF exothermic reaction is -433.981-(-559)=+125 kJ/mol
[[File:Fig2.3 01523148.png]]
The energy state of reactants(H+HF)
[[File:Fig2.3.2 01523148.png]]
The energy state of reactants(F+H2)

===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.===
Momentum of F-H is dependent on the rate of reaction, as supported from the plots below. This explains that the reaction energy is released as kinetic energy, more specifically, translational and vibrational form of kinetic energy. Since kinetic energy can convert into form of heat, heat can be measured by calorimetry, using E=mc(dT).
IR spectroscopy and vibrational spectroscopy can also measure the vibrational kinetic energy absorbed from the bond oscillation of HF, with IR measuring the transition from ground state to excited state while vibrational spectroscopy measuring the overtones of peaks in the spectra.
[[File:2.4 01523148 1.png]]
AB=184pm,-1g.pm/fs.mol, BC=74pm,1g.pm/fs.mol

[[File:2.4 01523148 2.png]]
AB=184pm and -5g.pm/fs.mol, BC=74pm,5g.pm/fs.mol

[[File:2.4 01523148 3 .png]]
AB=184pm and -10g.pm/fs.mol, BC=74pm, 10g.pm/fs.mol

===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 rule states that the translational energy is more effective than vibration energy in overcoming an early transition state barrier.

MRD:MolecularReactionDynamics01523148

2020-05-08T20:30:16Z

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

The transition state is mathematically defined by ∂V(ri)/∂ri=0, which r value represents the distance between atoms.The transition state can be identified at the position between maximum and minimum on the Potential Energy Surface and also by finding out the energy value around the trajectory from the contour plot.
[[File:Inkedtransition state surface plot1234 LI.jpg]]

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

The best estimate of the transition state position is around 90.775pm as it has no F and oscillation between each atoms as there is no oscillation occurs from the internuclear distance vs time plot. This means the picture captured is the transition state.
[[File:Question 1.2.png]]

===Comment on how the mep and the trajectory you just calculated differ.===
From the images below, using r1 = rts+δ, r2 = rts, with δ=1. The MEP and dynamics calculation types are compared.
[[File:MES 01523148.png]]
Dynamics contour plot
[[File:Trajectory1.3 01523148.png]]
MEP contour plot

From the pictures above, it shows that dynamics include the calculation of oscillation of bonds as from the Dynamics contour plot we can see there is oscillating curve around B-C distances. This means dynamics calculation involve calculation of vibrational energy, while no kinetic energy will be accounted in MEP calculation.

[[File:1.3.2 0152314888 InterNuclear Dynamics.png]]
Internulcear Dynamics plot
[[File:1.3.2 0152314888 InterNuclear MEP.png]]
Internuclear MEP plot
[[File:1.3.2 01523148 MomentaVsTime Dynamics.png]]
MomentaVsTime Dynamics plot
[[File:1.3.2 01523148 MomentaVsTime MEP.png]]
MomentaVsTime MEP plot

===Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable"
|-
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Since BC distance is starting to diminish, it shows that H(b)-H(c) bond is forming while H(a)-H(b) bond is breaking. The particles go through the transition state when H(c) collides with H(b).
|[[File:Surface Plot 1.3.1 01523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|Since no AB distance can be found from the image, that means no bond formation or bond breaking between H(a) and H(b). Therefore no collision found during the interactions of three particles and cannot overcome the energy barrier, reason of no collision might be due to repulsion.
|[[File:Surface Plot 1.3.2 1523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The reaction is similar to the scenario one. Only difference is H(c) is moving at a higher speed toward H(b) but same outcome has reached because energy barrier is overcame.
|[[File:Surface Plot 1.3.3 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|The H(b)-H(c) bond is formed first by having H(c) colliding into H(a)-H(b). This collision crosses the transition energy barrier at first. However the diagram shows the recrossing of transition state region which breaks the H(b)-H(c) bond and reform H(a)-H(b) back to original.
|[[File:Surface Plot 1.3.4 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|This reaction is similar to that of scenario 4. However instead of H(c) leaving the system, the H(c) particle goes back to the system and form H(b)-H(c) bond which breaks the oscillating H(a)-H(b) bond. The system crosses and recrosses the transition state 3 times in total.
|[[File:Surface Plot 1.3.5 1523148.png|250px]]}

|}

===Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?===

Transition state theory treats reaction classically such that quantum tunnelling cannot occur. Also energy must obey Boltzmann distribution rule and all systems that have already crossed the TS region cannot recross the areas from product back to reactants. Therefore the reaction rate values from TST prediction is expected to be greater than experimental values (like scenario 5 above) if any quantum effect like barrier-recrossing or tunnelling, especially the former one occurs.

==Exercise 2: F-H-H system==
===PES inspection===
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions According to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===
F+H-H reaction is exothermic while H+H-F is endothermic. Since enthalpy of HH is +436 kJ/mol while HF is +562 kJ/mol, the formation of HF is exothermic as the energy released from H-F formation is greater than H-H breaking, vice versa.
From PES, it proves the classification as from the (F+H2) PES, the potential energy along reaction is actually lower than that of (H+HF) inspection.
[[File:Surface Plot 01523148 2.png]]
PES plot of H+HF
[[File:Surface Plot 01523148 1.png]]
PES plot of F+HH

===Locate the approximate position of the transition state.===
Endothermic reaction means the late transition state resembles the products meanwhile exothermic reaction resembles early transition state which is the reactants. Hence from (F+H-H) reaction, the exothermic reaction allows us to predict the transition state is around 74pm, which is the bond length of H-H, in one of the internuclear distances.
Through trial and error, AB=181.1pm and BC=74.4875pm allow us to obtain the transition state, with potential energy of -433.981.The internuclear distance plot by MEP below shows that no oscillation occurs around and hence confirm the capture of transition state.
[[File:Fig2.2 01523148 InternuclearDistance.png]]

===Report the activation energy for both reactions.===
Energy of transition state - Energy of reactants = activation energy. Since the transition state energy was found from above, the energy of reactants can be found from the following image, using 185pm AB and 50pm BC, with 1000number of steps and 0.15fs of size. The picture below depicts how energy transforms throughout the (F+H2) reaction. There are two energy states captures in the graph because there are two flat lines in the reaction observed. The first flat line shows the energy of -434.751 kJ/mol, which is energy of these two reactants combined (F+H2), meanwhile the second flat line shows the energy of -559 kJ/mol, which resembles the energy of the products (H+HF).
Therefore, the activation energy of H+HF endothermic reaction is -433.981-(-434.751)=+0.77kJ/mol
The activation energy of F+HF exothermic reaction is -433.981-(-559)=+125 kJ/mol
[[File:Fig2.3 01523148.png]]
The energy state of reactants(H+HF)
[[File:Fig2.3.2 01523148.png]]
The energy state of reactants(F+H2)

===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.===
Momentum of F-H is dependent on the rate of reaction, as supported from the plots below. This explains that the reaction energy is released as kinetic energy, more specifically, translational and vibrational form of kinetic energy. Since kinetic energy can convert into form of heat, heat can be measured by calorimetry, using E=mc(dT).
IR spectroscopy and vibrational spectroscopy can also measure the vibrational kinetic energy absorbed from the bond oscillation of HF, with IR measuring the transition from ground state to excited state while vibrational spectroscopy measuring the overtones of peaks in the spectra.
[[File:2.4 01523148 1.png]]
AB=184pm,-1g.pm/fs.mol, BC=74pm,1g.pm/fs.mol

[[File:2.4 01523148 2.png]]
AB=184pm and -5g.pm/fs.mol, BC=74pm,5g.pm/fs.mol

[[File:2.4 01523148 3 .png]]
AB=184pm and -10g.pm/fs.mol, BC=74pm, 10g.pm/fs.mol

===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 rule states that the translational energy is more effective than vibration energy in overcoming an early transition state barrier.

MRD:MolecularReactionDynamics01523148

2020-05-08T20:11:55Z

Yyl18: /* 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. */

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

The transition state is mathematically defined by ∂V(ri)/∂ri=0, which r value represents the distance between atoms.The transition state can be identified at the position between maximum and minimum on the Potential Energy Surface and also by finding out the energy value around the trajectory from the contour plot.
[[File:Inkedtransition state surface plot1234 LI.jpg]]

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

The best estimate of the transition state position is around 90.775pm as it has no F and oscillation between each atoms as there is no oscillation occurs from the internuclear distance vs time plot. This means the picture captured is the transition state.
[[File:Question 1.2.png]]

===Comment on how the mep and the trajectory you just calculated differ.===
From the images below, using r1 = rts+δ, r2 = rts, with δ=1. The MEP and dynamics calculation types are compared.
[[File:MES 01523148.png]]
Dynamics contour plot
[[File:Trajectory1.3 01523148.png]]
MEP contour plot

From the pictures above, it shows that dynamics include the calculation of oscillation of bonds as from the Dynamics contour plot we can see there is oscillating curve around B-C distances. This means dynamics calculation involve calculation of vibrational energy, while no kinetic energy will be accounted in MEP calculation.

[[File:1.3.2 0152314888 InterNuclear Dynamics.png]]
Internulcear Dynamics plot
[[File:1.3.2 0152314888 InterNuclear MEP.png]]
Internuclear MEP plot
[[File:1.3.2 01523148 MomentaVsTime Dynamics.png]]
MomentaVsTime Dynamics plot
[[File:1.3.2 01523148 MomentaVsTime MEP.png]]
MomentaVsTime MEP plot

===Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable"
|-
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Since BC distance is starting to diminish, it shows that H(b)-H(c) bond is forming while H(a)-H(b) bond is breaking. The particles go through the transition state when H(c) collides with H(b).
|[[File:Surface Plot 1.3.1 01523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|Since no AB distance can be found from the image, that means no bond formation or bond breaking between H(a) and H(b). Therefore no collision found during the interactions of three particles and cannot overcome the energy barrier, reason of no collision might be due to repulsion.
|[[File:Surface Plot 1.3.2 1523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The reaction is similar to the scenario one. Only difference is H(c) is moving at a higher speed toward H(b) but same outcome has reached because energy barrier is overcame.
|[[File:Surface Plot 1.3.3 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|The H(b)-H(c) bond is formed first by having H(c) colliding into H(a)-H(b). This collision crosses the transition energy barrier at first. However the diagram shows the recrossing of transition state region which breaks the H(b)-H(c) bond and reform H(a)-H(b) back to original.
|[[File:Surface Plot 1.3.4 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|This reaction is similar to that of scenario 4. However instead of H(c) leaving the system, the H(c) particle goes back to the system and form H(b)-H(c) bond which breaks the oscillating H(a)-H(b) bond. The system crosses and recrosses the transition state 3 times in total.
|[[File:Surface Plot 1.3.5 1523148.png|250px]]}

|}

===Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?===

Transition state theory treats reaction classically such that quantum tunnelling cannot occur. Also energy must obey Boltzmann distribution rule and all systems that have already crossed the TS region cannot recross the areas from product back to reactants. Therefore the reaction rate values from TST prediction is expected to be greater than experimental values (like scenario 5 above) if any quantum effect like barrier-recrossing or tunnelling, especially the former one occurs.

==Exercise 2: F-H-H system==
===PES inspection===
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions According to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===
F+H-H reaction is exothermic while H+H-F is endothermic. Since enthalpy of HH is +436 kJ/mol while HF is +562 kJ/mol, the formation of HF is exothermic as the energy released from H-F formation is greater than H-H breaking, vice versa.
From PES, it proves the classification as from the (F+H2) PES, the potential energy along reaction is actually lower than that of (H+HF) inspection.
[[File:Surface Plot 01523148 2.png]]
PES plot of H+HF
[[File:Surface Plot 01523148 1.png]]
PES plot of F+HH

===Locate the approximate position of the transition state.===
Endothermic reaction means the late transition state resembles the products meanwhile exothermic reaction resembles early transition state which is the reactants. Hence from (F+H-H) reaction, the exothermic reaction allows us to predict the transition state is around 74pm, which is the bond length of H-H, in one of the internuclear distances.
Through trial and error, AB=181.1pm and BC=74.4875pm allow us to obtain the transition state, with potential energy of -433.981.The internuclear distance plot by MEP below shows that no oscillation occurs around and hence confirm the capture of transition state.
[[File:Fig2.2 01523148 InternuclearDistance.png]]

===Report the activation energy for both reactions.===
Energy of transition state - Energy of reactants = activation energy. Since the transition state energy was found from above, the energy of reactants can be found from the following image, using 185pm AB and 50pm BC, with 1000number of steps and 0.15fs of size. The picture below depicts how energy transforms throughout the (F+H2) reaction. There are two energy states captures in the graph because there are two flat lines in the reaction observed. The first flat line shows the energy of -434.751 kJ/mol, which is energy of these two reactants combined (F+H2), meanwhile the second flat line shows the energy of -559 kJ/mol, which resembles the energy of the products (H+HF).
Therefore, the activation energy of H+HF endothermic reaction is -433.981-(-434.751)=+0.77kJ/mol
The activation energy of F+HF exothermic reaction is -433.981-(-559)=+125 kJ/mol
[[File:Fig2.3 01523148.png]]
The energy state of reactants(H+HF)
[[File:Fig2.3.2 01523148.png]]
The energy state of reactants(F+H2)

===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.===
Momentum of F-H is dependent on the rate of reaction, as supported from the plots below. This explains that the reaction energy is released as kinetic energy, more specifically, translational and vibrational form of kinetic energy. Since kinetic energy can convert into form of heat, heat can be measured by calorimetry, using E=mc(dT).
IR spectroscopy and vibrational spectroscopy can also measure the vibrational kinetic energy absorbed from the bond oscillation of HF, with IR measuring the transition from ground state to excited state while vibrational spectroscopy measuring the overtones of peaks in the spectra.
[[File:2.4 01523148 1.png]]
AB=184pm,-1g.pm/fs.mol, BC=74pm,1g.pm/fs.mol

[[File:2.4 01523148 2.png]]
AB=184pm and -5g.pm/fs.mol, BC=74pm,5g.pm/fs.mol

[[File:2.4 01523148 3 .png]]
AB=184pm and -10g.pm/fs.mol, BC=74pm, 10g.pm/fs.mol

===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:2.4 01523148 3 .png

2020-05-08T20:11:45Z

Yyl18:

File:2.4 01523148 2.png

2020-05-08T20:10:59Z

Yyl18:

File:2.4 01523148 1.png

2020-05-08T20:08:00Z

Yyl18:

MRD:MolecularReactionDynamics01523148

2020-05-08T20:05:12Z

Yyl18: /* Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? */

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

The transition state is mathematically defined by ∂V(ri)/∂ri=0, which r value represents the distance between atoms.The transition state can be identified at the position between maximum and minimum on the Potential Energy Surface and also by finding out the energy value around the trajectory from the contour plot.
[[File:Inkedtransition state surface plot1234 LI.jpg]]

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

The best estimate of the transition state position is around 90.775pm as it has no F and oscillation between each atoms as there is no oscillation occurs from the internuclear distance vs time plot. This means the picture captured is the transition state.
[[File:Question 1.2.png]]

===Comment on how the mep and the trajectory you just calculated differ.===
From the images below, using r1 = rts+δ, r2 = rts, with δ=1. The MEP and dynamics calculation types are compared.
[[File:MES 01523148.png]]
Dynamics contour plot
[[File:Trajectory1.3 01523148.png]]
MEP contour plot

From the pictures above, it shows that dynamics include the calculation of oscillation of bonds as from the Dynamics contour plot we can see there is oscillating curve around B-C distances. This means dynamics calculation involve calculation of vibrational energy, while no kinetic energy will be accounted in MEP calculation.

[[File:1.3.2 0152314888 InterNuclear Dynamics.png]]
Internulcear Dynamics plot
[[File:1.3.2 0152314888 InterNuclear MEP.png]]
Internuclear MEP plot
[[File:1.3.2 01523148 MomentaVsTime Dynamics.png]]
MomentaVsTime Dynamics plot
[[File:1.3.2 01523148 MomentaVsTime MEP.png]]
MomentaVsTime MEP plot

===Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable"
|-
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Since BC distance is starting to diminish, it shows that H(b)-H(c) bond is forming while H(a)-H(b) bond is breaking. The particles go through the transition state when H(c) collides with H(b).
|[[File:Surface Plot 1.3.1 01523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|Since no AB distance can be found from the image, that means no bond formation or bond breaking between H(a) and H(b). Therefore no collision found during the interactions of three particles and cannot overcome the energy barrier, reason of no collision might be due to repulsion.
|[[File:Surface Plot 1.3.2 1523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The reaction is similar to the scenario one. Only difference is H(c) is moving at a higher speed toward H(b) but same outcome has reached because energy barrier is overcame.
|[[File:Surface Plot 1.3.3 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|The H(b)-H(c) bond is formed first by having H(c) colliding into H(a)-H(b). This collision crosses the transition energy barrier at first. However the diagram shows the recrossing of transition state region which breaks the H(b)-H(c) bond and reform H(a)-H(b) back to original.
|[[File:Surface Plot 1.3.4 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|This reaction is similar to that of scenario 4. However instead of H(c) leaving the system, the H(c) particle goes back to the system and form H(b)-H(c) bond which breaks the oscillating H(a)-H(b) bond. The system crosses and recrosses the transition state 3 times in total.
|[[File:Surface Plot 1.3.5 1523148.png|250px]]}

|}

===Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?===

Transition state theory treats reaction classically such that quantum tunnelling cannot occur. Also energy must obey Boltzmann distribution rule and all systems that have already crossed the TS region cannot recross the areas from product back to reactants. Therefore the reaction rate values from TST prediction is expected to be greater than experimental values (like scenario 5 above) if any quantum effect like barrier-recrossing or tunnelling, especially the former one occurs.

==Exercise 2: F-H-H system==
===PES inspection===
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions According to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===
F+H-H reaction is exothermic while H+H-F is endothermic. Since enthalpy of HH is +436 kJ/mol while HF is +562 kJ/mol, the formation of HF is exothermic as the energy released from H-F formation is greater than H-H breaking, vice versa.
From PES, it proves the classification as from the (F+H2) PES, the potential energy along reaction is actually lower than that of (H+HF) inspection.
[[File:Surface Plot 01523148 2.png]]
PES plot of H+HF
[[File:Surface Plot 01523148 1.png]]
PES plot of F+HH

===Locate the approximate position of the transition state.===
Endothermic reaction means the late transition state resembles the products meanwhile exothermic reaction resembles early transition state which is the reactants. Hence from (F+H-H) reaction, the exothermic reaction allows us to predict the transition state is around 74pm, which is the bond length of H-H, in one of the internuclear distances.
Through trial and error, AB=181.1pm and BC=74.4875pm allow us to obtain the transition state, with potential energy of -433.981.The internuclear distance plot by MEP below shows that no oscillation occurs around and hence confirm the capture of transition state.
[[File:Fig2.2 01523148 InternuclearDistance.png]]

===Report the activation energy for both reactions.===
Energy of transition state - Energy of reactants = activation energy. Since the transition state energy was found from above, the energy of reactants can be found from the following image, using 185pm AB and 50pm BC, with 1000number of steps and 0.15fs of size. The picture below depicts how energy transforms throughout the (F+H2) reaction. There are two energy states captures in the graph because there are two flat lines in the reaction observed. The first flat line shows the energy of -434.751 kJ/mol, which is energy of these two reactants combined (F+H2), meanwhile the second flat line shows the energy of -559 kJ/mol, which resembles the energy of the products (H+HF).
Therefore, the activation energy of H+HF endothermic reaction is -433.981-(-434.751)=+0.77kJ/mol
The activation energy of F+HF exothermic reaction is -433.981-(-559)=+125 kJ/mol
[[File:Fig2.3 01523148.png]]
The energy state of reactants(H+HF)
[[File:Fig2.3.2 01523148.png]]
The energy state of reactants(F+H2)

===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.===
Momentum of F-H is dependent on the rate of reaction, as supported from the plots below. This explains that the reaction energy is released as kinetic energy, more specifically, translational and vibrational form of kinetic energy. Since kinetic energy can convert into form of heat, heat can be measured by calorimetry, using E=mc(dT).
IR spectroscopy and vibrational spectroscopy can also measure the vibrational kinetic energy absorbed from the bond oscillation of HF, with IR measuring the transition from ground state to excited state while vibrational spectroscopy measuring the overtones of peaks in the spectra.

===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:MolecularReactionDynamics01523148

2020-05-08T20:01:40Z

Yyl18: /* 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. */

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

The transition state is mathematically defined by ∂V(ri)/∂ri=0, which r value represents the distance between atoms.The transition state can be identified at the position between maximum and minimum on the Potential Energy Surface and also by finding out the energy value around the trajectory from the contour plot.
[[File:Inkedtransition state surface plot1234 LI.jpg]]

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

The best estimate of the transition state position is around 90.775pm as it has no F and oscillation between each atoms as there is no oscillation occurs from the internuclear distance vs time plot. This means the picture captured is the transition state.
[[File:Question 1.2.png]]

===Comment on how the mep and the trajectory you just calculated differ.===
From the images below, using r1 = rts+δ, r2 = rts, with δ=1. The MEP and dynamics calculation types are compared.
[[File:MES 01523148.png]]
Dynamics contour plot
[[File:Trajectory1.3 01523148.png]]
MEP contour plot

From the pictures above, it shows that dynamics include the calculation of oscillation of bonds as from the Dynamics contour plot we can see there is oscillating curve around B-C distances. This means dynamics calculation involve calculation of vibrational energy, while no kinetic energy will be accounted in MEP calculation.

[[File:1.3.2 0152314888 InterNuclear Dynamics.png]]
Internulcear Dynamics plot
[[File:1.3.2 0152314888 InterNuclear MEP.png]]
Internuclear MEP plot
[[File:1.3.2 01523148 MomentaVsTime Dynamics.png]]
MomentaVsTime Dynamics plot
[[File:1.3.2 01523148 MomentaVsTime MEP.png]]
MomentaVsTime MEP plot

===Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable"
|-
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Since BC distance is starting to diminish, it shows that H(b)-H(c) bond is forming while H(a)-H(b) bond is breaking. The particles go through the transition state when H(c) collides with H(b).
|[[File:Surface Plot 1.3.1 01523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|Since no AB distance can be found from the image, that means no bond formation or bond breaking between H(a) and H(b). Therefore no collision found during the interactions of three particles and cannot overcome the energy barrier, reason of no collision might be due to repulsion.
|[[File:Surface Plot 1.3.2 1523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The reaction is similar to the scenario one. Only difference is H(c) is moving at a higher speed toward H(b) but same outcome has reached because energy barrier is overcame.
|[[File:Surface Plot 1.3.3 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|The H(b)-H(c) bond is formed first by having H(c) colliding into H(a)-H(b). This collision crosses the transition energy barrier at first. However the diagram shows the recrossing of transition state region which breaks the H(b)-H(c) bond and reform H(a)-H(b) back to original.
|[[File:Surface Plot 1.3.4 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|This reaction is similar to that of scenario 4. However instead of H(c) leaving the system, the H(c) particle goes back to the system and form H(b)-H(c) bond which breaks the oscillating H(a)-H(b) bond. The system crosses and recrosses the transition state 3 times in total.
|[[File:Surface Plot 1.3.5 1523148.png|250px]]}

|}

===Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?===

Transition state theory treats reaction classically such that quantum tunnelling cannot occur. Also energy must obey Boltzmann distribution rule and all systems that have already crossed the TS region cannot recross the areas from product back to reactants. Therefore the reaction rate values from TST prediction greater than experimental values if any quantum effect like barrier-recrossing or tunnelling, especially the former one occurs.

==Exercise 2: F-H-H system==
===PES inspection===
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions According to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===
F+H-H reaction is exothermic while H+H-F is endothermic. Since enthalpy of HH is +436 kJ/mol while HF is +562 kJ/mol, the formation of HF is exothermic as the energy released from H-F formation is greater than H-H breaking, vice versa.
From PES, it proves the classification as from the (F+H2) PES, the potential energy along reaction is actually lower than that of (H+HF) inspection.
[[File:Surface Plot 01523148 2.png]]
PES plot of H+HF
[[File:Surface Plot 01523148 1.png]]
PES plot of F+HH

===Locate the approximate position of the transition state.===
Endothermic reaction means the late transition state resembles the products meanwhile exothermic reaction resembles early transition state which is the reactants. Hence from (F+H-H) reaction, the exothermic reaction allows us to predict the transition state is around 74pm, which is the bond length of H-H, in one of the internuclear distances.
Through trial and error, AB=181.1pm and BC=74.4875pm allow us to obtain the transition state, with potential energy of -433.981.The internuclear distance plot by MEP below shows that no oscillation occurs around and hence confirm the capture of transition state.
[[File:Fig2.2 01523148 InternuclearDistance.png]]

===Report the activation energy for both reactions.===
Energy of transition state - Energy of reactants = activation energy. Since the transition state energy was found from above, the energy of reactants can be found from the following image, using 185pm AB and 50pm BC, with 1000number of steps and 0.15fs of size. The picture below depicts how energy transforms throughout the (F+H2) reaction. There are two energy states captures in the graph because there are two flat lines in the reaction observed. The first flat line shows the energy of -434.751 kJ/mol, which is energy of these two reactants combined (F+H2), meanwhile the second flat line shows the energy of -559 kJ/mol, which resembles the energy of the products (H+HF).
Therefore, the activation energy of H+HF endothermic reaction is -433.981-(-434.751)=+0.77kJ/mol
The activation energy of F+HF exothermic reaction is -433.981-(-559)=+125 kJ/mol
[[File:Fig2.3 01523148.png]]
The energy state of reactants(H+HF)
[[File:Fig2.3.2 01523148.png]]
The energy state of reactants(F+H2)

===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.===
Momentum of F-H is dependent on the rate of reaction, as supported from the plots below. This explains that the reaction energy is released as kinetic energy, more specifically, translational and vibrational form of kinetic energy. Since kinetic energy can convert into form of heat, heat can be measured by calorimetry, using E=mc(dT).
IR spectroscopy and vibrational spectroscopy can also measure the vibrational kinetic energy absorbed from the bond oscillation of HF, with IR measuring the transition from ground state to excited state while vibrational spectroscopy measuring the overtones of peaks in the spectra.

===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:MolecularReactionDynamics01523148

2020-05-08T19:02:13Z

Yyl18: /* Comment on how the mep and the trajectory you just calculated differ. */

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

The transition state is mathematically defined by ∂V(ri)/∂ri=0, which r value represents the distance between atoms.The transition state can be identified at the position between maximum and minimum on the Potential Energy Surface and also by finding out the energy value around the trajectory from the contour plot.
[[File:Inkedtransition state surface plot1234 LI.jpg]]

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

The best estimate of the transition state position is around 90.775pm as it has no F and oscillation between each atoms as there is no oscillation occurs from the internuclear distance vs time plot. This means the picture captured is the transition state.
[[File:Question 1.2.png]]

===Comment on how the mep and the trajectory you just calculated differ.===
From the images below, using r1 = rts+δ, r2 = rts, with δ=1. The MEP and dynamics calculation types are compared.
[[File:MES 01523148.png]]
Dynamics contour plot
[[File:Trajectory1.3 01523148.png]]
MEP contour plot

From the pictures above, it shows that dynamics include the calculation of oscillation of bonds as from the Dynamics contour plot we can see there is oscillating curve around B-C distances. This means dynamics calculation involve calculation of vibrational energy, while no kinetic energy will be accounted in MEP calculation.

[[File:1.3.2 0152314888 InterNuclear Dynamics.png]]
Internulcear Dynamics plot
[[File:1.3.2 0152314888 InterNuclear MEP.png]]
Internuclear MEP plot
[[File:1.3.2 01523148 MomentaVsTime Dynamics.png]]
MomentaVsTime Dynamics plot
[[File:1.3.2 01523148 MomentaVsTime MEP.png]]
MomentaVsTime MEP plot

===Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable"
|-
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Since BC distance is starting to diminish, it shows that H(b)-H(c) bond is forming while H(a)-H(b) bond is breaking. The particles go through the transition state when H(c) collides with H(b).
|[[File:Surface Plot 1.3.1 01523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|Since no AB distance can be found from the image, that means no bond formation or bond breaking between H(a) and H(b). Therefore no collision found during the interactions of three particles and cannot overcome the energy barrier, reason of no collision might be due to repulsion.
|[[File:Surface Plot 1.3.2 1523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The reaction is similar to the scenario one. Only difference is H(c) is moving at a higher speed toward H(b) but same outcome has reached because energy barrier is overcame.
|[[File:Surface Plot 1.3.3 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|The H(b)-H(c) bond is formed first by having H(c) colliding into H(a)-H(b). This collision crosses the transition energy barrier at first. However the diagram shows the recrossing of transition state region which breaks the H(b)-H(c) bond and reform H(a)-H(b) back to original.
|[[File:Surface Plot 1.3.4 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|This reaction is similar to that of scenario 4. However instead of H(c) leaving the system, the H(c) particle goes back to the system and form H(b)-H(c) bond which breaks the oscillating H(a)-H(b) bond. The system crosses and recrosses the transition state 3 times in total.
|[[File:Surface Plot 1.3.5 1523148.png|250px]]}

|}

===Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?===

Transition state theory treats reaction classically such that quantum tunnelling cannot occur. Also energy must obey Boltzmann distribution rule and all systems that have already crossed the TS region cannot recross the areas from product back to reactants. Therefore the reaction rate values from TST prediction greater than experimental values if any quantum effect like barrier-recrossing or tunnelling, especially the former one occurs.

==Exercise 2: F-H-H system==
===PES inspection===
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions According to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===
F+H-H reaction is exothermic while H+H-F is endothermic. Since enthalpy of HH is +436 kJ/mol while HF is +562 kJ/mol, the formation of HF is exothermic as the energy released from H-F formation is greater than H-H breaking, vice versa.
From PES, it proves the classification as from the (F+H2) PES, the potential energy along reaction is actually lower than that of (H+HF) inspection.
[[File:Surface Plot 01523148 2.png]]
PES plot of H+HF
[[File:Surface Plot 01523148 1.png]]
PES plot of F+HH

===Locate the approximate position of the transition state.===
Endothermic reaction means the late transition state resembles the products meanwhile exothermic reaction resembles early transition state which is the reactants. Hence from (F+H-H) reaction, the exothermic reaction allows us to predict the transition state is around 74pm, which is the bond length of H-H, in one of the internuclear distances.
Through trial and error, AB=181.1pm and BC=74.4875pm allow us to obtain the transition state, with potential energy of -433.981.The internuclear distance plot by MEP below shows that no oscillation occurs around and hence confirm the capture of transition state.
[[File:Fig2.2 01523148 InternuclearDistance.png]]

===Report the activation energy for both reactions.===
Energy of transition state - Energy of reactants = activation energy. Since the transition state energy was found from above, the energy of reactants can be found from the following image, using 185pm AB and 50pm BC, with 1000number of steps and 0.15fs of size. The picture below depicts how energy transforms throughout the (F+H2) reaction. There are two energy states captures in the graph because there are two flat lines in the reaction observed. The first flat line shows the energy of -434.751 kJ/mol, which is energy of these two reactants combined (F+H2), meanwhile the second flat line shows the energy of -559 kJ/mol, which resembles the energy of the products (H+HF).
Therefore, the activation energy of H+HF endothermic reaction is -433.981-(-434.751)=+0.77kJ/mol
The activation energy of F+HF exothermic reaction is -433.981-(-559)=+125 kJ/mol
[[File:Fig2.3 01523148.png]]
The energy state of reactants(H+HF)
[[File:Fig2.3.2 01523148.png]]
The energy state of reactants(F+H2)

===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:MolecularReactionDynamics01523148

2020-05-08T18:40:30Z

Yyl18:

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

The transition state is mathematically defined by ∂V(ri)/∂ri=0, which r value represents the distance between atoms.The transition state can be identified at the position between maximum and minimum on the Potential Energy Surface and also by finding out the energy value around the trajectory from the contour plot.
[[File:Inkedtransition state surface plot1234 LI.jpg]]

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

The best estimate of the transition state position is around 90.775pm as it has no F and oscillation between each atoms as there is no oscillation occurs from the internuclear distance vs time plot. This means the picture captured is the transition state.
[[File:Question 1.2.png]]

===Comment on how the mep and the trajectory you just calculated differ.===
From the images below, using r1 = rts+δ, r2 = rts, with δ=1. The MEP and dynamics calculation types are compared.
[[File:MES 01523148.png]]
Dynamics contour plot
[[File:Trajectory1.3 01523148.png]]
MEP contour plot

From the pictures above, it shows that dynamics include the calculation of oscillation of bonds as from the Dynamics contour plot we can see there is oscillating curve around B-C distances. This means dynamics calculation involve calculation of vibrational energy.

[[File:1.3.2 0152314888 InterNuclear Dynamics.png]]
Internulcear Dynamics plot
[[File:1.3.2 0152314888 InterNuclear MEP.png]]
Internuclear MEP plot
[[File:1.3.2 01523148 MomentaVsTime Dynamics.png]]
MomentaVsTime Dynamics plot
[[File:1.3.2 01523148 MomentaVsTime MEP.png]]
MomentaVsTime MEP plot
===Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?===
{| class="wikitable"
|-
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|Since BC distance is starting to diminish, it shows that H(b)-H(c) bond is forming while H(a)-H(b) bond is breaking. The particles go through the transition state when H(c) collides with H(b).
|[[File:Surface Plot 1.3.1 01523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|Since no AB distance can be found from the image, that means no bond formation or bond breaking between H(a) and H(b). Therefore no collision found during the interactions of three particles and cannot overcome the energy barrier, reason of no collision might be due to repulsion.
|[[File:Surface Plot 1.3.2 1523148.png|250px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|The reaction is similar to the scenario one. Only difference is H(c) is moving at a higher speed toward H(b) but same outcome has reached because energy barrier is overcame.
|[[File:Surface Plot 1.3.3 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|The H(b)-H(c) bond is formed first by having H(c) colliding into H(a)-H(b). This collision crosses the transition energy barrier at first. However the diagram shows the recrossing of transition state region which breaks the H(b)-H(c) bond and reform H(a)-H(b) back to original.
|[[File:Surface Plot 1.3.4 1523148.png|250px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|This reaction is similar to that of scenario 4. However instead of H(c) leaving the system, the H(c) particle goes back to the system and form H(b)-H(c) bond which breaks the oscillating H(a)-H(b) bond. The system crosses and recrosses the transition state 3 times in total.
|[[File:Surface Plot 1.3.5 1523148.png|250px]]}

|}

===Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?===

Transition state theory treats reaction classically such that quantum tunnelling cannot occur. Also energy must obey Boltzmann distribution rule and all systems that have already crossed the TS region cannot recross the areas from product back to reactants. Therefore the reaction rate values from TST prediction greater than experimental values if any quantum effect like barrier-recrossing or tunnelling, especially the former one occurs.

==Exercise 2: F-H-H system==
===PES inspection===
===By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions According to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===
F+H-H reaction is exothermic while H+H-F is endothermic. Since enthalpy of HH is +436 kJ/mol while HF is +562 kJ/mol, the formation of HF is exothermic as the energy released from H-F formation is greater than H-H breaking, vice versa.
From PES, it proves the classification as from the (F+H2) PES, the potential energy along reaction is actually lower than that of (H+HF) inspection.
[[File:Surface Plot 01523148 2.png]]
PES plot of H+HF
[[File:Surface Plot 01523148 1.png]]
PES plot of F+HH

===Locate the approximate position of the transition state.===
Endothermic reaction means the late transition state resembles the products meanwhile exothermic reaction resembles early transition state which is the reactants. Hence from (F+H-H) reaction, the exothermic reaction allows us to predict the transition state is around 74pm, which is the bond length of H-H, in one of the internuclear distances.
Through trial and error, AB=181.1pm and BC=74.4875pm allow us to obtain the transition state, with potential energy of -433.981.The internuclear distance plot by MEP below shows that no oscillation occurs around and hence confirm the capture of transition state.
[[File:Fig2.2 01523148 InternuclearDistance.png]]

===Report the activation energy for both reactions.===
Energy of transition state - Energy of reactants = activation energy. Since the transition state energy was found from above, the energy of reactants can be found from the following image, using 185pm AB and 50pm BC, with 1000number of steps and 0.15fs of size. The picture below depicts how energy transforms throughout the (F+H2) reaction. There are two energy states captures in the graph because there are two flat lines in the reaction observed. The first flat line shows the energy of -434.751 kJ/mol, which is energy of these two reactants combined (F+H2), meanwhile the second flat line shows the energy of -559 kJ/mol, which resembles the energy of the products (H+HF).
Therefore, the activation energy of H+HF endothermic reaction is -433.981-(-434.751)=+0.77kJ/mol
The activation energy of F+HF exothermic reaction is -433.981-(-559)=+125 kJ/mol
[[File:Fig2.3 01523148.png]]
The energy state of reactants(H+HF)
[[File:Fig2.3.2 01523148.png]]
The energy state of reactants(F+H2)

===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:Surface Plot 01523148 2.png

2020-05-08T18:39:22Z

Yyl18: Yyl18 uploaded a new version of File:Surface Plot 01523148 2.png

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File:Surface Plot 01523148 1.png

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