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MRD:ny517

2019-06-06T23:21:26Z

Cq3417:

== Molecular Reaction Dynamics ==
=== Exercise 1: H + H2 ===

''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 defined as the maximum energy point on the minimum energy path and the point on the potential energy surface when the gradient (partial first derivative) of the potential is equal to zero (∂V(ri)/∂ri=0). The transition state can be identified as a saddle point which means it has a partial second derivative greater than and less than zero (dependant on the variables and direction of axis). A local minimum is the point where it is solely a minimum and so the second derivative is always a positive value.

{{fontcolor|red|(Good.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 22:40, 6 June 2019 (BST)}}

''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 to 3 decimal places was:
rts=r1=r2= 0.908 Å

This was estimated through looking at the contour plot of the initial conditions set for a H + H2 reaction and following the reactive trajectory to the point where r1 was approximate to r2 - which was in the region of 0.9 Å (as seen on the contour plot below). This is because for a system with identical atoms, the equidistant point is equal to the transition state position. After trialling values in this region, 0.908 Å produced a trajectory with a constant internuclear distance between both atoms (as seen on the plot below) indicating that the trajectory will not fall off the ridge. This is a transition state since the atoms are stationary over time when they have no initial momenta.

[[File:Ny517_q2.JPG|300px|thumb|left|Contour plot from initial conditions]] [[File:Ny517_q2_Internuclear_distance.JPG|600px|thumb|center|Internuclear Distance vs Time plot where rts = 0.908 Å]]

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

The mep produces a trajectory that follows the valley floor and is also known as the reaction path. It shows a straight-line path along this trajectory because the velocity is always reset to zero after each time step. This does not provide a realistic account of the motion of atoms during the reaction since it assumes the atoms have no mass. The “Dynamics” calculation provides a more realistic trajectory since it accounts for the mass of the system and so the motion of the atoms in the gas phase will be inertial. In a MEP, kinetic energy is assumed to be 0 which explains why there are no oscillations along the path. As the potential energy decreases towards its minimum from the transition state, there is no kinetic energy producing a momentum and causing the molecule to travel up the potential energy curve at small bond distances so it remains at the potential energy minimum resulting in no oscillations.

{{fontcolor|red|(Good explaination.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 22:40, 6 June 2019 (BST)}}

[[File:Ny517_q3_mep.JPG|500px|thumb|left|mep surface plot]] [[File:Ny517_q3_dynamics.JPG|500px|thumb|center|Dynamics surface plot]]

If the initial conditions were reversed, the A-B and B-C plots would switch for the “Distances vs Time” and “Momenta vs Time” graphs. This is because the magnitude of the forces in the reaction do not change but the direction does. As the time increases, the A-B distance increases whilst the B-C bond distance stays constant (bar slight oscillations) at 0.75 because atom A is travelling further away from the BC diatomic molecule. At 5 seconds the A-B bond distance is 18 Å. As the time increases, the A-B momentum plateaus at 2.5 kg ms-1 whilst the B-C momentum oscillates about an average of 1.25 kg ms-1. If the values were reversed, the B-C bond distance would increase whilst the A-B bond distance would remain constant.

[[File:Ny517_q3_larget_bond.JPG|500px|thumb|left|Distance vs Time]] [[File:Ny517_q3_larget_momentum.JPG|500px|thumb|center|Momenta vs Time]]

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

Initial positions '''r1''' = 0.74 Å and '''r2''' = 2.0 Å

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.0 ||Yes || Reactants travel through transition state then start to vibrate ||Plot 1
|-
| -1.5 || -2.0 ||-100.4 ||No || Reactants do not overcome activation energy and rebound off the transition state. Atoms vibrate on approach and after reaching transition state ||Plot 2
|-
| -1.5 || -2.5 ||-98.9 ||Yes || Reactants vibrate whilst travelling through transition state then vibrate more vigorously ||Plot 3
|-
| -2.5 || -5.0 ||-85.0 ||No || Reactants pass through transition state then pass back through the transition state to reform the reactants ||Plot 4
|-
| -2.5 || -5.2 ||-83.4 ||Yes || Reactants travel through transition state, then reform reactants and pass through the transition state again. This occurs with strong vibrations throughout reaction ||Plot 5
|}

[[File:Ny517_plot1.JPG|300px]]

[[File:Ny517_plot2.JPG|300px]]

[[File:Ny517_plot3.JPG|300px]]

[[File:Ny517_plot4.JPG|300px]]

[[File:Ny517_plot5.JPG|300px]]

{{fontcolor|red|(Maybe next time try to label under the plot.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 22:40, 6 June 2019 (BST)}}

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

Transition state theory (TST) is used to qualitatively understand the reaction rates of elementary chemical reactions with the idea that all reactions occur through a transition state. One assumption is that once the reagents gain an energy sufficient enough to overcome the activation energy, they will react to form the product. In reality this is not the case as is seen in the calculated examples above where some reactants reached the transition state but reformed the reagents and not the products. This is because of another assumption of TST which is that the energy is looked at classically rather than quantum mechanically so fails to account for the quantization of molecular vibrations. Failing to factor this is why TST predictions for reaction rates will typically be higher than experimental values.

{{fontcolor|red|(Good, but any reference?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 22:40, 6 June 2019 (BST)}}

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

''By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic).''

 F + H2 

[[File:Ny517_h2_plus_f.JPG|300px|]]

H-H distance: 0.74 Å

F-H distance: 2.00 Å

H-H momentum: -1.5 kg ms-1

F-H momentum: -2.5 kg ms-1

The reactants are higher in energy than the products therefore the reaction is exothermic.

 H + HF 

[[File:Ny517_f_h_plus_h.JPG|300px|]]

F-H distance: 0.74 Å

H-H distance: 2.00 Å

F-H momentum: -1.5 kg ms-1

H-H momentum: -2.5 kg ms-1

The reactants are lower in energy than the products therefore the reaction is endothermic.

''How does this relate to the bond strength of the chemical species involved?''

Through inspection of the potential energy surfaces, it is clear that the bond strength of H-F is greater than the bond strength of H-H since more energy is needed to break the H-F bond in the second reaction.

''Locate the approximate position of the transition state.''

[[File:Ny517_hf_transition_state_distance.JPG|400px|thumb|left|Internuclear Distance vs Time]][[File:Ny517_hf_transition_state_contour.JPG|400px|thumb|center|Contour plot]]

H-H distance: 0.745 Å

H-F distance: 1.815 Å

To find the transition state, both momenta were set to equal zero. Hammond's postulate states that the transition state of a reaction resembles either the reactants or the products, to whichever it is closer in energy. The transition state will be closer in energy to the reactants of the exothermic (first) reaction and the products of the endothermic (second) reaction. The contour plot shows the atoms do not fall off the ridge and the Internuclear Distances plot shows the distances remained constant (bar slight oscillations).

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

The activation energy is defined as the difference in energy between the reactants and transition state. The energy of the transition state is known, which is -103.752 kcal mol-1. 0.3 A was added to and subtracted from the HF bond length to calculate the minimum energy difference between the reactants and transition state for both reactions. The reactant energies were extracted using an mep calculation. The calculated activation energies are stated below:

[[File:Ny517_activation_exo.JPG|300px|]] [[File:Ny517_activation_endo.JPG|300px|]]

*For the F + H2 system: 0.236 kcal mol-1
*For the H + HF system: 30.256 kcal mol-1

=== Reaction Dynamics ===

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

For a F + H2 reaction trajectory:

H-H distance: 0.744832 Å

H-F distance: 2.00 Å

H-H momentum: -0.7 kg ms-1

H-F momentum: -1.6 kg ms-1

[[File:Ny517_reaction_dynamics_momenta.JPG|400px|thumb|left|Momenta vs Time]][[File:Ny517_reaction_dynamics_energy.JPG|400px|thumb|center|Energy vs Time]]

It is seen that the B-C momentum plateaus over time due to the H2 bond being split and the H atoms becoming further apart so the bond vibrations begin to stop. The A-B momentum oscillates as the B-C momentum stops oscillating which is due to the H-F atom being formed. The H-F bond vibrations are larger than the H-H bond vibrations due to the strength of the bond being greater and hence a have a higher enthalpy of formation. If the H-F bond is stronger than the H-H bond then this will result in an exothermic reaction. This could be confirmed experimentally through a reaction calorimeter. The plot displaying energy over time indicates that the total energy is constant and hence conserved. It also shows that the the potential and kinetic energy convert between each other as the atoms vibrate between their bonds. Notice that the energy oscillations become larger after the transition state- similar to that seen in the momentum plot.

{{fontcolor|red|(Do you mean 'hence have a'?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 22:40, 6 June 2019 (BST)}}

Changing momenta:

*Keeping the same initial conditions but changing the H-H momentum:

[[File:Ny517_momentum_-3.JPG|500px|thumb|left|pHH=-3 kg ms-1]]
[[File:Ny517_momentum_-2.JPG|500px|thumb|center|pHH=-2 kg ms-1]]
[[File:Ny517_momentum_-1.JPG|500px|thumb|left|pHH=-1 kg ms-1]]
[[File:Ny517_momentum_0.JPG|500px|thumb|center|pHH=0 kg ms-1]]
[[File:Ny517_momentum_1.JPG|500px|thumb|left|pHH=1 kg ms-1]]
[[File:Ny517_momentum_2.JPG|500px|thumb|center|pHH=2 kg ms-1]]
[[File:Ny517_momentum_1.JPG|500px|thumb|left|pHH=3 kg ms-1]]

Only when pHH was equal to -1 and -2 kg ms-1 was there a reaction trajectory. When the time step was doubled for pHH=0 a reaction trajectory was produced indicating that the H2 bond was broken and a HF bond was formed. For the other reactions, the reactants passed the transition state but went back and reformed the reactants.

When pFH = -0.8 kg ms-1 and pHH = 0.1 kg ms-1, a reaction trajectory occurs:

[[File:Ny517_momentum_change.JPG||300px|]]

For the reverse reaction, H + HF:

The initial conditions are set with :

H-H distance: 2.00 Å

H-F distance: 0.74 Å

H-H momentum: -0.59 kg ms-1

H-F momentum: -6.55 kg ms-1

[[File:Ny517_reverse_1.JPG||300px|]]

Keeping the atom positions constant, the momenta were varied to have a low vibrational motion on on the H-F bond and an arbitrarily high value of pHH above the activation energy and obtain a reactive trajectory:

[[File:Ny517_reverse_2.JPG||300px|]]

H-H distance: 2.00 Å

F-H distance: 0.74 Å

H-H momentum: -7.55 kg ms-1

F-H momentum: -0.35 kg ms-1

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

The efficiency of a reaction is dependant on the position of the transition state. If there is a late transition state, the vibrational energy is the most influential on the efficiency of the reaction and hence its ability to overcome the reaction energy. If there is an early transition state, the translational energy is the most influencial on the efficiency. Using this, it can be deduced that for the endothermic H + HF system, the reaction is affected more by the vibrational energy and that for the exothermic F + H2 system, the reaction is affected more by the translational energy.

{{fontcolor|red|(Good, but any reference?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 22:40, 6 June 2019 (BST)}}

MRD:lz3717

2019-06-06T23:14:20Z

Cq3417:

'''===Molecular Reaction Dynamics: Applications to Triatomic systems===
'''

==Exercise 1: H + H2 system==
===Q1: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 the saddle point in this diagram, which is defined as the maximum on the minimum energy path linking reactants and the products.

Based on the mathematical expression, suppose there is a function '''f(x,y)''' with two different variables '''x''' and '''y'''.For a stationary point'''(x0,y0)''',the first derivatives of '''x''' and '''y''' must equal to zero.('''fx=fy=0''').

The second derivatives is illustrated as '''fxx''' and '''fyy'''.If a stationary point that '''fxxfyy-fxy2>0''' and '''fxx>0''',it is a local minimum point of this function.

If a stationary point that only satisfies '''fxxfyy-fxy2>0''',it is stated as the saddle point. [[File:Zhangleisherryqwer.png|350px|thumb|A potential energy surface diagram.]]

{{fontcolor|red|(Good.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 00:13, 7 June 2019 (BST)}}

===Q2: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.===

{| class="wikitable"
![[File:SHERRYISCIWHEDLIFHWE.png|500px|left|thumb|A contour plot to provide rts.]]
![[File:Sherryqw19981234.png|500px|right|thumb|Intermolecular distance vs time.]]
|}

The best estimate of rts is 0.9077.As seen on the contour map, it is the saddle point of PES which illustrates the transition state. For the internuclear distance vs time graph,at rts,internuclear

distance does not change with time,which shows that the atoms stop moving when it is transition state,the potential energy reaches maximum and kinetic energy should be 0.

{{fontcolor|red|(Good illustration.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 00:13, 7 June 2019 (BST)}}

===Q3:Comment on how the mep and the trajectory you just calculated differ.===

Set the parameter '''r1=0.9177,r2=0.9077,p1=p2=0'''.

The difference to calculate reaction path by MEP and dynamic types are shown below:

{| class="wikitable"
![[File:Lz3717 MEP.png|500px|left|thumb|The contour plot calculated by MEP.]]
![[File:Lz3717 dynamics.png|500px|right|thumb|The contour plot calculated by dynamic type.]]
|}

It can be seen that in the MEP calculation,the molecule doesn't show any vibrations(velocity is set to be zero),while the molecule is vibrating in the dynamic calculation.

{| class="wikitable"
![[File:Lz3717 energy diagrammep.png|500px|left|thumb|The energy vs steps diagram by MEP calculation.]]
![[File:Lz3717 energydiagramdynamic.png|500px|right|thumb|The energy vs time diagram by dynamic calculation.]]
|}

The MEP calculation shows the kinetic energy is always zero and the total energy is only potential energy,which means that the total energy is not conserved.However,the dynamic calculation shows conservation of total energy(kinetic energy increases while potential energy decreases).

===Q4: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" border="1"

! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || YES || [[File:Lz3717 Surface Plot1.png]] || AB and C are approaching to each other because C has enough initial momentum(kinetic energy),so C collides with AB successfully and B breaks bond with A and forms bond with C.
|-
| -1.5 || -2.0 || -100.456 || NO || [[File:Lz3717 Surface Plot2.png]] || AB and C are approaching to each other because C doesn't gain sufficient kinetic energy so C cannot collide with AB and finally C leaves without forming bond with AB.
|-
| -1.5 || -2.5 || -98.956 || YES || [[File:Lz3717 Surface Plot3.png]] || AB and C are approaching to each other but this time C has sufficient kinetic energy so C has successful collision with AB and forms new molecule BC.
|-
| -2.5 || -5.0 || -84.956 || NO || [[File:Lz3717 Surface Plot4.png]] || AB and C are approaching to each other.With high initial momentum, C collide with AB successfully and forms bonds with B at first.However,the newly formed bond BC proceeds high kinetic energy so that finally the new bond breaks and B leaves back towards A,reforming the reactant AB.The AB starts to vibrate.
|-
| -2.5 || -5.2 || -83.416 || YES || [[File:Lz3717 Surface Plot5.png]] || AB and C are approaching to each other.With high initial momentum, C collide with AB successfully and forms bonds with B at first.However,the newly formed bond BC proceeds high kinetic energy so that finally the new bond breaks and B leaves back towards A to reform the reactant.This time,the reactant AB still has the high kinetic energy,so the vibrating product BC still forms.
|}

Conclusion:In order to make the reaction happen,reactants have to gain enough kinetic energy to lead sufficient collision.Higher the kinetic energy,the more possible that the reaction will happen.However,high kinetic energy

cannot guarantee the successful reaction even if it passes the transition state,reactants may reform at this situation.

{{fontcolor|red|(Good discussion.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 00:13, 7 June 2019 (BST)}}

===Q5:State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?===

There are following three main assumptions of Transition State Theory:
<ref name="TST" />
<references>
<ref name="TST">Steinfeld J I, Francisco J S, Hase W L.''Chemical Kinetics and Dynamics.'' 2nd ed. New Jersey: Prentice-Hall, Inc; 1998</ref>
</references>

1.Reactants are in constant equilibrium with the transition state structure.

2.The energy of the particles follow a Boltzmann distribution.

3.Once reactants become the transition state, the transition state structure does not collapse back to the reactants.

From the results obtained above,the product does will come back to reform the reactant even if it already crosses the transition state,which has a conflict with one of the assumptions.Therefore,the rate of the reaction in

reality is much slower than that theoretically because the product goes back to reform the reactant.

==Exercise 2: F - H - H system==
===Q6:By the inspecting the potential energy surfaces,classify the F + H2 and H+HF reactions according to their energies(exothermic or endothermic).How does this relate to the bond strength of the chemical species involved?Locate the approximate position of the transition state.===

{| class="wikitable"
![[File:Lz3717 Surface Plot F+H2.png|500px|left|thumb|The surface plot for F + H2 reaction.]]
![[File:Lz3717 surface plot H+HF.png|500px|right|thumb|The surface plot for H + HF reaction.]]
|}

For the F+H2→HF+H reaction,the position of transition state is located at '''HF=1.812''' and '''HH=0.745'''.The reaction is '''exothermic''' because the energy needed for HF bond forming(-565KJ/mol) is larger than that for HH bond

breaking(+436KJ/mol).For the H+HF→F+H2 reaction,the position of transition state is located at HH=1.812 and HF=0.745.The reaction is '''endothermic''' because the energy needed for HH bond forming(-436KJ/mol) is lower than

that for HF bond breaking(+565KJ/mol).The F+H2→HF+H exothermic reaction shows that the H-F bond is stronger than H-H bond.

{| class="wikitable"
![[File:Lz3717 contour FH2.png|500px|left|thumb|The contour plot for locating the transition state F+H2.]]
![[File:Lz3717 distance FH2.png|500px|right|thumb|The internuclear distance vs steps plot for locating the transition state F+H2.The gradient of the graph is zero which illustrates that it reaches transition state.]]
|}

{| class="wikitable"
![[File:Lz3717 contour HHF.png|500px|left|thumb|The contour plot for locating the transition state H+HF.]]
![[File:Lz3717 distance HHF.png|500px|right|thumb|The internuclear distance vs steps plot for locating the transition state H+HF.The gradient of the graph is zero which illustrates that it reaches transition state.]]
|}

{{fontcolor|red|(Nice illustration.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 00:13, 7 June 2019 (BST)}}

===Q7:Report the activation energy for both reactions.===

When the distance HF is changed to be 0.745,a structure neighbouring transition state is obtained.Because the distance HF is smaller than the real distance of HF in transition state,so the reaction tends to move towards reactants F and H2.As seen on the graph,the difference between highest energy and lowest energy is calculated to be the '''activation energy'''.

{| class="wikitable"
![[File:Lz3717 energy FH2.png|500px|left|thumb|The total energy change plot of F+H2 reaction.]]
![[File:Lz3717 energy HHF.png|500px|right|thumb|The total energy change plot of H+HF reaction.]]
|}

Overall:

The '''activation energy''' of '''F+H2''' reaction is calculated to be +0.131kcal/mol.

The '''activation energy''' of '''H+HF''' reaction is calculated to be +31.21kcal/mol.

===Q8: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.===

Because now the total energy is conserved,so the calculation type needs to be changed to dynamics.The initial conditions of F+H2 is set to be r1=1.90,r2=0.74,p1=-0.8,p2=-0.2.

{| class="wikitable"
![[File:Lz3717 contour123 FH2.png|500px|left|thumb|The reaction pathway of F+H2 reaction on contour plot.]]
![[File:Lz3717 energy123 FH2.png|500px|right|thumb|The energy vs time graph of F+H2 reaction.]]
![[File:Lz3717 momentum FH2.png|500px|right|thumb|The momentum vs time graph of the F+H2 reaction.]]
|}

As it mentioned before,this F+H2 reaction is an exothermic reaction.After the reaction,there is an increase in kinetic energy and decrease in potential energy,which can be observed by the increase in temperature of the reaction.The total energy is still conserved because the excess kinetic energy transfers to the vibration energy of H-F bond,which can be proved by the fluctuated changing momentum of A-B bond in the momentum vs time plot.It means that the H-F bond is vibrating.This phenomenon can be observed from IR spectrum because H-F bond is IR active.

===Q9:Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.===

{| class="wikitable"
![[File:Lz3717 lowvibration FH2.png|500px|left|thumb|The reaction pathway of F+H2,in which the parameters are set to have high translation energy and low vibration energy.The reaction happens successfully.The reactant gains enough kinetic energy to overcome the energy barrier.]]
![[File:Lz3717 highvibration FH2.png|500px|right|thumb|The reaction pathway of F+H2,in which the parameters are set to have low translation energy and low vibration energy.The reaction doesn't occur.The reactant spends more time on oscillating itself.]]
|}

For F+H2 reaction,because it is a exothermic reaction,according to the Hammond postulate,the energy of transition state resembles the energy of reactants,so it has an early energy barrier.Thus,the translation energy

plays a more important role than vibration energy when crossing the energy barrier.

{| class="wikitable"
![[File:Lz3717 highvibration HHF.png|500px|left|thumb|The reaction pathway of H+HF,in which the parameters are set to have low translation energy and high vibration energy.The reaction happens successfully.The molecule gains enough vibration energy to become excited to reach energy barrier.]]
![[File:Lz3717 lowvibration HHF.png|500px|right|thumb|The reaction pathway of H+HF,in which the parameters are set to have high translation energy and low vibration energy.The reaction doesn't occur.The reactant doesn't vibrate effectively and bounces back.]]
|}

For H+HF reaction,because it is a endothermic reaction,according to the Hammond postulate,the energy of transition state resembles the energy of products,so it has an late energy barrier.Thus,the vibration energy

plays a more important role than translation energy when crossing the energy barrier.

{{fontcolor|red|(Good, but any reference?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 00:13, 7 June 2019 (BST)}}

MRD:lz3717

2019-06-06T23:13:52Z

Cq3417:

'''===Molecular Reaction Dynamics: Applications to Triatomic systems===
'''

==Exercise 1: H + H2 system==
===Q1: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 the saddle point in this diagram, which is defined as the maximum on the minimum energy path linking reactants and the products.

Based on the mathematical expression, suppose there is a function '''f(x,y)''' with two different variables '''x''' and '''y'''.For a stationary point'''(x0,y0)''',the first derivatives of '''x''' and '''y''' must equal to zero.('''fx=fy=0''').

The second derivatives is illustrated as '''fxx''' and '''fyy'''.If a stationary point that '''fxxfyy-fxy2>0''' and '''fxx>0''',it is a local minimum point of this function.

If a stationary point that only satisfies '''fxxfyy-fxy2>0''',it is stated as the saddle point. [[File:Zhangleisherryqwer.png|350px|thumb|A potential energy surface diagram.]]

{{fontcolor|red|(Good.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 00:13, 7 June 2019 (BST)}}

===Q2: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.===

{| class="wikitable"
![[File:SHERRYISCIWHEDLIFHWE.png|500px|left|thumb|A contour plot to provide rts.]]
![[File:Sherryqw19981234.png|500px|right|thumb|Intermolecular distance vs time.]]
|}

The best estimate of rts is 0.9077.As seen on the contour map, it is the saddle point of PES which illustrates the transition state. For the internuclear distance vs time graph,at rts,internuclear

distance does not change with time,which shows that the atoms stop moving when it is transition state,the potential energy reaches maximum and kinetic energy should be 0.

{{fontcolor|red|(Good illustration.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 00:13, 7 June 2019 (BST)}}

===Q3:Comment on how the mep and the trajectory you just calculated differ.===

Set the parameter '''r1=0.9177,r2=0.9077,p1=p2=0'''.

The difference to calculate reaction path by MEP and dynamic types are shown below:

{| class="wikitable"
![[File:Lz3717 MEP.png|500px|left|thumb|The contour plot calculated by MEP.]]
![[File:Lz3717 dynamics.png|500px|right|thumb|The contour plot calculated by dynamic type.]]
|}

It can be seen that in the MEP calculation,the molecule doesn't show any vibrations(velocity is set to be zero),while the molecule is vibrating in the dynamic calculation.

{| class="wikitable"
![[File:Lz3717 energy diagrammep.png|500px|left|thumb|The energy vs steps diagram by MEP calculation.]]
![[File:Lz3717 energydiagramdynamic.png|500px|right|thumb|The energy vs time diagram by dynamic calculation.]]
|}

The MEP calculation shows the kinetic energy is always zero and the total energy is only potential energy,which means that the total energy is not conserved.However,the dynamic calculation shows conservation of total energy(kinetic energy increases while potential energy decreases).

===Q4: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" border="1"

! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || YES || [[File:Lz3717 Surface Plot1.png]] || AB and C are approaching to each other because C has enough initial momentum(kinetic energy),so C collides with AB successfully and B breaks bond with A and forms bond with C.
|-
| -1.5 || -2.0 || -100.456 || NO || [[File:Lz3717 Surface Plot2.png]] || AB and C are approaching to each other because C doesn't gain sufficient kinetic energy so C cannot collide with AB and finally C leaves without forming bond with AB.
|-
| -1.5 || -2.5 || -98.956 || YES || [[File:Lz3717 Surface Plot3.png]] || AB and C are approaching to each other but this time C has sufficient kinetic energy so C has successful collision with AB and forms new molecule BC.
|-
| -2.5 || -5.0 || -84.956 || NO || [[File:Lz3717 Surface Plot4.png]] || AB and C are approaching to each other.With high initial momentum, C collide with AB successfully and forms bonds with B at first.However,the newly formed bond BC proceeds high kinetic energy so that finally the new bond breaks and B leaves back towards A,reforming the reactant AB.The AB starts to vibrate.
|-
| -2.5 || -5.2 || -83.416 || YES || [[File:Lz3717 Surface Plot5.png]] || AB and C are approaching to each other.With high initial momentum, C collide with AB successfully and forms bonds with B at first.However,the newly formed bond BC proceeds high kinetic energy so that finally the new bond breaks and B leaves back towards A to reform the reactant.This time,the reactant AB still has the high kinetic energy,so the vibrating product BC still forms.
|}

Conclusion:In order to make the reaction happen,reactants have to gain enough kinetic energy to lead sufficient collision.Higher the kinetic energy,the more possible that the reaction will happen.However,high kinetic energy

cannot guarantee the successful reaction even if it passes the transition state,reactants may reform at this situation.

{{fontcolor|red|(Good discussion.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 00:13, 7 June 2019 (BST)}}

===Q5:State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?===

There are following three main assumptions of Transition State Theory:
<ref name="TST" />
<references>
<ref name="TST">Steinfeld J I, Francisco J S, Hase W L.''Chemical Kinetics and Dynamics.'' 2nd ed. New Jersey: Prentice-Hall, Inc; 1998</ref>
</references>

1.Reactants are in constant equilibrium with the transition state structure.

2.The energy of the particles follow a Boltzmann distribution.

3.Once reactants become the transition state, the transition state structure does not collapse back to the reactants.

From the results obtained above,the product does will come back to reform the reactant even if it already crosses the transition state,which has a conflict with one of the assumptions.Therefore,the rate of the reaction in

reality is much slower than that theoretically because the product goes back to reform the reactant.

==Exercise 2: F - H - H system==
===Q6:By the inspecting the potential energy surfaces,classify the F + H2 and H+HF reactions according to their energies(exothermic or endothermic).How does this relate to the bond strength of the chemical species involved?Locate the approximate position of the transition state.===

{| class="wikitable"
![[File:Lz3717 Surface Plot F+H2.png|500px|left|thumb|The surface plot for F + H2 reaction.]]
![[File:Lz3717 surface plot H+HF.png|500px|right|thumb|The surface plot for H + HF reaction.]]
|}

For the F+H2→HF+H reaction,the position of transition state is located at '''HF=1.812''' and '''HH=0.745'''.The reaction is '''exothermic''' because the energy needed for HF bond forming(-565KJ/mol) is larger than that for HH bond

breaking(+436KJ/mol).For the H+HF→F+H2 reaction,the position of transition state is located at HH=1.812 and HF=0.745.The reaction is '''endothermic''' because the energy needed for HH bond forming(-436KJ/mol) is lower than

that for HF bond breaking(+565KJ/mol).The F+H2→HF+H exothermic reaction shows that the H-F bond is stronger than H-H bond.

{| class="wikitable"
![[File:Lz3717 contour FH2.png|500px|left|thumb|The contour plot for locating the transition state F+H2.]]
![[File:Lz3717 distance FH2.png|500px|right|thumb|The internuclear distance vs steps plot for locating the transition state F+H2.The gradient of the graph is zero which illustrates that it reaches transition state.]]
|}

{| class="wikitable"
![[File:Lz3717 contour HHF.png|500px|left|thumb|The contour plot for locating the transition state H+HF.]]
![[File:Lz3717 distance HHF.png|500px|right|thumb|The internuclear distance vs steps plot for locating the transition state H+HF.The gradient of the graph is zero which illustrates that it reaches transition state.]]
|}

{{fontcolor|red|(Nice illustration.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 00:13, 7 June 2019 (BST)}}

===Q7:Report the activation energy for both reactions.===

When the distance HF is changed to be 0.745,a structure neighbouring transition state is obtained.Because the distance HF is smaller than the real distance of HF in transition state,so the reaction tends to move towards reactants F and H2.As seen on the graph,the difference between highest energy and lowest energy is calculated to be the '''activation energy'''.

{| class="wikitable"
![[File:Lz3717 energy FH2.png|500px|left|thumb|The total energy change plot of F+H2 reaction.]]
![[File:Lz3717 energy HHF.png|500px|right|thumb|The total energy change plot of H+HF reaction.]]
|}

Overall:

The '''activation energy''' of '''F+H2''' reaction is calculated to be +0.131kcal/mol.

The '''activation energy''' of '''H+HF''' reaction is calculated to be +31.21kcal/mol.

===Q8: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.===

Because now the total energy is conserved,so the calculation type needs to be changed to dynamics.The initial conditions of F+H2 is set to be r1=1.90,r2=0.74,p1=-0.8,p2=-0.2.

{| class="wikitable"
![[File:Lz3717 contour123 FH2.png|500px|left|thumb|The reaction pathway of F+H2 reaction on contour plot.]]
![[File:Lz3717 energy123 FH2.png|500px|right|thumb|The energy vs time graph of F+H2 reaction.]]
![[File:Lz3717 momentum FH2.png|500px|right|thumb|The momentum vs time graph of the F+H2 reaction.]]
|}

As it mentioned before,this F+H2 reaction is an exothermic reaction.After the reaction,there is an increase in kinetic energy and decrease in potential energy,which can be observed by the increase in temperature of the reaction.The total energy is still conserved because the excess kinetic energy transfers to the vibration energy of H-F bond,which can be proved by the fluctuated changing momentum of A-B bond in the momentum vs time plot.It means that the H-F bond is vibrating.This phenomenon can be observed from IR spectrum because H-F bond is IR active.

===Q9:Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.===

{| class="wikitable"
![[File:Lz3717 lowvibration FH2.png|500px|left|thumb|The reaction pathway of F+H2,in which the parameters are set to have high translation energy and low vibration energy.The reaction happens successfully.The reactant gains enough kinetic energy to overcome the energy barrier.]]
![[File:Lz3717 highvibration FH2.png|500px|right|thumb|The reaction pathway of F+H2,in which the parameters are set to have low translation energy and low vibration energy.The reaction doesn't occur.The reactant spends more time on oscillating itself.]]
|}

For F+H2 reaction,because it is a exothermic reaction,according to the Hammond postulate,the energy of transition state resembles the energy of reactants,so it has an early energy barrier.Thus,the translation energy

plays a more important role than vibration energy when crossing the energy barrier.

{| class="wikitable"
![[File:Lz3717 highvibration HHF.png|500px|left|thumb|The reaction pathway of H+HF,in which the parameters are set to have low translation energy and high vibration energy.The reaction happens successfully.The molecule gains enough vibration energy to become excited to reach energy barrier.]]
![[File:Lz3717 lowvibration HHF.png|500px|right|thumb|The reaction pathway of H+HF,in which the parameters are set to have high translation energy and low vibration energy.The reaction doesn't occur.The reactant doesn't vibrate effectively and bounces back.]]
|}

For H+HF reaction,because it is a endothermic reaction,according to the Hammond postulate,the energy of transition state resembles the energy of products,so it has an late energy barrier.Thus,the vibration energy

plays a more important role than translation energy when crossing the energy barrier.

{{fontcolor|red|(Good, but nay reference?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 00:13, 7 June 2019 (BST)}}

MRD:mx4417

2019-06-06T22:33:03Z

Cq3417:

===Molecular reaction Dynamics===
The objectives of this wiki report is to provide a study on triatomic system reactivities, which involves the collision between one atom and a diatomic molecule.
====EXERCISE 1: H + H2 system====
=====Q1: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 of the potential energy surface diagram is the Saddle Point of the diagram, which is local maximum on the lowest energy pathway.

[[file:Q1.mx4417.PNG]]

The Mathmetical definition of saddle point: <math>\operatorname{grad}f = \nabla f</math>=0

[[file:Q1.3mx4417.PNG]]
<ref>https://bb.imperial.ac.uk/bbcswebdav/pid-1216108-dt-content-rid-3932941_1/courses/DSS-CH1_MPC1-17_18/MPC1-CourseNotes%281%29.pdf</ref>

[[file:Q1.24417.PNG|thumb|700px|none| The potential energy surface diagram.]]

{{fontcolor|red|(I think you can state more for this.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 22:58, 6 June 2019 (BST)}}

=====Q2: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.=====
With initial condition r1 = r2, and p1 = p2 = 0.0.
It was found that: r(ts)= 0.9077 Å.
[[file:Q2.1mx4417.png|thumb|700px|none| The internuclear distance vs time diagram with MEP calculation type.]]
As the gradient of both lines in the graph are zero and the distance was kept constant, so it is a saddle point and thus the transition state of the model.

{{fontcolor|red|(Ok, but why?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 22:58, 6 June 2019 (BST)}}

=====Q3:Comment on how the mep and the trajectory you just calculated differ.=====
Trajectories from r1 = 0.9077 Å + 0.01, r2 = 0.9077 Å, p1= p2 = 0

{| class="wikitable"
![[file:Q3.3mx4417.png|thumb|700px|Figure 1:The MEP calculation type. ]]
![[file:Q3.4mx4417.png|thumb|700px|Figure 2:The dynamics calculation type]]
|}
{| class="wikitable"
![[file:Q3.5.mx4417mep.png|thumb|700px| Figure 3:The MEP calculation type(momenta vs time). ]]
![[file:Q3.6mx4417.png|thumb|700px|Figure 4:The dynamics calculation type(momenta vs time)]]
|}
For dynamics method, the vibration motion can be seen as the trajectory is oscillating. For MEP methods, the kinetic energy is set to be zero (see figure 3), thus no oscillation.

=====Q4: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?=====
For the initial positions r1 = 0.74 and r2 = 2.0
{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Contour plot
|-
| -1.25 || -2.5 ||-99.018 || YES||The atom C approached the molecule AB, and as atom C had enough momentum to overcome the energy barrier at the Transition state,thus, it reacted with AB to form molecule BC. ||[[file:Q4.1mx4417.png]]
|-
| -1.5 || -2.0 || -100.456||NO ||The atom C approached the molecule AB, but atom C didnot have enough momentum to overcome the energy barrier at the Transition state,thus, it did not react with AB to form molecule BC. ||
[[file:Q4.2mx4417.png]]
|-
| -1.5 || -2.5 ||-98.956 ||YES || The atom C approached the molecule AB, and as atom C had enough momentum to overcome the energy barrier at the Transition state,thus, it reacted with AB to form molecule BC.||
[[file:Q4.3mx4417.png]]
|-
| -2.5 || -5.0 ||-84.956 ||NO ||The atom C approached the molecule AB, the trajectory passed the transition point, but the product reformed the vibrating excited reactant. ||
[[file:Q4.4mx4417.png]]
|-
| -2.5 || -5.2 ||-83.416 || YES||The atom C approached the molecule AB, the atom C had enough momenta and the trajectory passed the transition point, it formed the vibrating excited product. ||
[[file:Q4.5mx4417.png]]
|}
Conclusion: For the reactants start with same potential energy, but with different momentum (kinetic energy), increases kinetic energy not always make the reaction reactive. As high kinetic energy may cause the re-formation of the vibrating excited reactant.

{{fontcolor|red|(Good illustration.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 23:28, 6 June 2019 (BST)}}

=====Q5:State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?=====
Transition state theory (TST):
1. A molecular system that has crossed the transition state in the direction of a product cannot turn around and reform reactant.

2. In the transition state, motion along the reaction coordinate may be separated from the other motion and treated classically as translation.

3. Even in the absence of equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws.

4. Electronic and nuclear motions are separated due to the large difference in mass.(using the Born-Oppenheimer approximation)

<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase, Chemical Kinetic and Dynamics, Prentice-Hall, New Jersey, 1998</ref>

The first theory is not correct as the table shows the reactant could be reformed under certain circumstances, and the reformation of the reactants decreases the rate of the reaction.
The second one is also not as correct as it predicted, in the calculations (contour plot), there is some vibration close to the transition state.
The other cannot be verified due to insufficient data.

====Exercise 2: F-H-H system====
=====Q6: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?Locate the approximate position of the transition state.=====
Type of reaction: F + H2 → HF + H ( r1=1.81, r2=0.74), which is an exothermic reaction, as formation of the HF Bond is -565 KJ/mol and bond breaking H2 = +436KJ/mol.
{| class="wikitable"
![[File:Q5.1mx4417.png|500px | thumb |The Contour plot of the transition point of F + H2 → HF + H ]]
![[File:Q5.2mx4417.PNG|500px | thumb |The internuclear distance vs time plot, horizontal line repersents the transition state is reached.]]
![[
|}

Type of reaction: H + HF→ H2 + F ( r2=1.82, r1=0.74 ), which is an exothermic reaction, as breaking of the HF Bond is +565 KJ/mol and bond formation H2 = -436KJ/mol.

{| class="wikitable"
![[File:Q5.3mx4417.png |500px | thumb |The Contour plot of the transition point of H + HF→ H2 + F ]]
![[File:Q5.4mx4417.PNG |500px | thumb |The internuclear distance vs time plot, horizontal line repersents the transition state is reached ]]
|}

=====Q7:Report the activation energy for both reactions.=====
{| class="wikitable"
![[File:Q5.6mx4417.png |500px | thumb |The energy vs time plot of H + HF→ H2 + F, the difference in energy is the energy difference of TS and products. As this reaction is exothermic, the activation energy is higher. 30.174 kcal/mol]]
![[File:Q5.7mx4417.png |500px | thumb |The energy vs time plot of F + H2 → HF + H, the difference in energy is the energy difference of TS and products. As this reaction is endothermic, the activation energy is lower (0.137 kcal/mol). ]]
|}

=====Q8: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.=====

{| class="wikitable"
![[File:Q8.1mx4417.png|500px | thumb |The Contour plot with the initial condition shown. It can be seen clearly the reaction occurs. The black line shows the reaction pathway of F+ H2. ]]
![[File:Q8.2mx4417.png |500px | thumb |The Energy vs time plot.The reaction energy is generated by the increase in kinetic energy and the decrease in potential energy ( with the total energy is conserved ). Because of that, the system is 'heat up',there is an increase in temperature.There is also energy trapped in vibration, so maybe the IR emission is possible. ]]
![[File:Q8.3mx4417.png |500px | thumb |The Momenta vs time plot. The B-C (H-H)bond is initially vibrating and then flattened after the reaction. Also, the A-B (H-F) bond forms, and the energy is transferred to the vibration energy of A-B bond. ]]
|}
=====Q9: 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.=====
For reaction pathway of F + H2 → HF + H, it has an early transition state. As the reaction is exothermic, according to Hammond's postulate: the transition state resembles the reactant.
{| class="wikitable"
![[File:Q9.1.mx4417.PNG |500px | thumb |The parameter is set to have high vibration and low translation energy. The reaction does not occur.]]
![[File:Q9.2mx4417.PNG |500px | thumb | The parameter is set to have low vibration and high translation energy. The reaction occurs.]]
|}
In conculsion, for early transiton state, the translation energy is more dominant for passing over the barrier and let the reaction to occur.

For reaction pathway of H + HF→ H2 + F, it has a late transition state. As the reaction is endothermic, according to Hammond's postulate: the transition state resembles the products.
{| class="wikitable"
![[File:Q9.3mx4417.PNG |500px | thumb |The parameter is set to have low vibration and high translation energy. The reaction does not occur.]]
![[File:Q9.4mx4417.PNG |500px | thumb | The parameter is set to have high vibration and low translation energy. The reaction occurs.]]
|}
In conculsion, for late transiton state, the vibration energy is more dominant for passing over the barrier and let the reaction to occur.

{{fontcolor|red|(Good illustration and explanation.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 23:28, 6 June 2019 (BST)}}

MRD:mx4417

2019-06-06T22:31:47Z

Cq3417:

===Molecular reaction Dynamics===
The objectives of this wiki report is to provide a study on triatomic system reactivities, which involves the collision between one atom and a diatomic molecule.
====EXERCISE 1: H + H2 system====
=====Q1: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 of the potential energy surface diagram is the Saddle Point of the diagram, which is local maximum on the lowest energy pathway.

[[file:Q1.mx4417.PNG]]

The Mathmetical definition of saddle point: <math>\operatorname{grad}f = \nabla f</math>=0

[[file:Q1.3mx4417.PNG]]
<ref>https://bb.imperial.ac.uk/bbcswebdav/pid-1216108-dt-content-rid-3932941_1/courses/DSS-CH1_MPC1-17_18/MPC1-CourseNotes%281%29.pdf</ref>

[[file:Q1.24417.PNG|thumb|700px|none| The potential energy surface diagram.]]

{{fontcolor|red|(I think you can state more for this.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 22:58, 6 June 2019 (BST)}}

=====Q2: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.=====
With initial condition r1 = r2, and p1 = p2 = 0.0.
It was found that: r(ts)= 0.9077 Å.
[[file:Q2.1mx4417.png|thumb|700px|none| The internuclear distance vs time diagram with MEP calculation type.]]
As the gradient of both lines in the graph are zero and the distance was kept constant, so it is a saddle point and thus the transition state of the model.

{{fontcolor|red|(Ok, but why?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 22:58, 6 June 2019 (BST)}}

=====Q3:Comment on how the mep and the trajectory you just calculated differ.=====
Trajectories from r1 = 0.9077 Å + 0.01, r2 = 0.9077 Å, p1= p2 = 0

{| class="wikitable"
![[file:Q3.3mx4417.png|thumb|700px|Figure 1:The MEP calculation type. ]]
![[file:Q3.4mx4417.png|thumb|700px|Figure 2:The dynamics calculation type]]
|}
{| class="wikitable"
![[file:Q3.5.mx4417mep.png|thumb|700px| Figure 3:The MEP calculation type(momenta vs time). ]]
![[file:Q3.6mx4417.png|thumb|700px|Figure 4:The dynamics calculation type(momenta vs time)]]
|}
For dynamics method, the vibration motion can be seen as the trajectory is oscillating. For MEP methods, the kinetic energy is set to be zero (see figure 3), thus no oscillation.

=====Q4: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?=====
For the initial positions r1 = 0.74 and r2 = 2.0
{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Contour plot
|-
| -1.25 || -2.5 ||-99.018 || YES||The atom C approached the molecule AB, and as atom C had enough momentum to overcome the energy barrier at the Transition state,thus, it reacted with AB to form molecule BC. ||[[file:Q4.1mx4417.png]]
|-
| -1.5 || -2.0 || -100.456||NO ||The atom C approached the molecule AB, but atom C didnot have enough momentum to overcome the energy barrier at the Transition state,thus, it did not react with AB to form molecule BC. ||
[[file:Q4.2mx4417.png]]
|-
| -1.5 || -2.5 ||-98.956 ||YES || The atom C approached the molecule AB, and as atom C had enough momentum to overcome the energy barrier at the Transition state,thus, it reacted with AB to form molecule BC.||
[[file:Q4.3mx4417.png]]
|-
| -2.5 || -5.0 ||-84.956 ||NO ||The atom C approached the molecule AB, the trajectory passed the transition point, but the product reformed the vibrating excited reactant. ||
[[file:Q4.4mx4417.png]]
|-
| -2.5 || -5.2 ||-83.416 || YES||The atom C approached the molecule AB, the atom C had enough momenta and the trajectory passed the transition point, it formed the vibrating excited product. ||
[[file:Q4.5mx4417.png]]
|}
Conclusion: For the reactants start with same potential energy, but with different momentum (kinetic energy), increases kinetic energy not always make the reaction reactive. As high kinetic energy may cause the re-formation of the vibrating excited reactant.

{{fontcolor|red|(Good illustration.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 23:28, 6 June 2019 (BST)}}

=====Q5:State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?=====
Transition state theory (TST):
1. A molecular system that has crossed the transition state in the direction of a product cannot turn around and reform reactant.

2. In the transition state, motion along the reaction coordinate may be separated from the other motion and treated classically as translation.

3. Even in the absence of equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws.

4. Electronic and nuclear motions are separated due to the large difference in mass.(using the Born-Oppenheimer approximation)

<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase, Chemical Kinetic and Dynamics, Prentice-Hall, New Jersey, 1998</ref>

The first theory is not correct as the table shows the reactant could be reformed under certain circumstances, and the reformation of the reactants decreases the rate of the reaction.
The second one is also not as correct as it predicted, in the calculations (contour plot), there is some vibration close to the transition state.
The other cannot be verified due to insufficient data.

====Exercise 2: F-H-H system====
=====Q6: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?Locate the approximate position of the transition state.=====
Type of reaction: F + H2 → HF + H ( r1=1.81, r2=0.74), which is an exothermic reaction, as formation of the HF Bond is -565 KJ/mol and bond breaking H2 = +436KJ/mol.
{| class="wikitable"
![[File:Q5.1mx4417.png|500px | thumb |The Contour plot of the transition point of F + H2 → HF + H ]]
![[File:Q5.2mx4417.PNG|500px | thumb |The internuclear distance vs time plot, horizontal line repersents the transition state is reached.]]
![[
|}

Type of reaction: H + HF→ H2 + F ( r2=1.82, r1=0.74 ), which is an exothermic reaction, as breaking of the HF Bond is +565 KJ/mol and bond formation H2 = -436KJ/mol.

{| class="wikitable"
![[File:Q5.3mx4417.png |500px | thumb |The Contour plot of the transition point of H + HF→ H2 + F ]]
![[File:Q5.4mx4417.PNG |500px | thumb |The internuclear distance vs time plot, horizontal line repersents the transition state is reached ]]
|}

=====Q7:Report the activation energy for both reactions.=====
{| class="wikitable"
![[File:Q5.6mx4417.png |500px | thumb |The energy vs time plot of H + HF→ H2 + F, the difference in energy is the energy difference of TS and products. As this reaction is exothermic, the activation energy is higher. 30.174 kcal/mol]]
![[File:Q5.7mx4417.png |500px | thumb |The energy vs time plot of F + H2 → HF + H, the difference in energy is the energy difference of TS and products. As this reaction is endothermic, the activation energy is lower (0.137 kcal/mol). ]]
|}

=====Q8: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.=====

{| class="wikitable"
![[File:Q8.1mx4417.png|500px | thumb |The Contour plot with the initial condition shown. It can be seen clearly the reaction occurs. The black line shows the reaction pathway of F+ H2. ]]
![[File:Q8.2mx4417.png |500px | thumb |The Energy vs time plot.The reaction energy is generated by the increase in kinetic energy and the decrease in potential energy ( with the total energy is conserved ). Because of that, the system is 'heat up',there is an increase in temperature.There is also energy trapped in vibration, so maybe the IR emission is possible. ]]
![[File:Q8.3mx4417.png |500px | thumb |The Momenta vs time plot. The B-C (H-H)bond is initially vibrating and then flattened after the reaction. Also, the A-B (H-F) bond forms, and the energy is transferred to the vibration energy of A-B bond. ]]
|}
=====Q9: 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.=====
For reaction pathway of F + H2 → HF + H, it has an early transition state. As the reaction is exothermic, according to Hammond's postulate: the transition state resembles the reactant.
{| class="wikitable"
![[File:Q9.1.mx4417.PNG |500px | thumb |The parameter is set to have high vibration and low translation energy. The reaction does not occur.]]
![[File:Q9.2mx4417.PNG |500px | thumb | The parameter is set to have low vibration and high translation energy. The reaction occurs.]]
|}
In conculsion, for early transiton state, the translation energy is more dominant for passing over the barrier and let the reaction to occur.

For reaction pathway of H + HF→ H2 + F, it has a late transition state. As the reaction is endothermic, according to Hammond's postulate: the transition state resembles the products.
{| class="wikitable"
![[File:Q9.3mx4417.PNG |500px | thumb |The parameter is set to have low vibration and high translation energy. The reaction does not occur.]]
![[File:Q9.4mx4417.PNG |500px | thumb | The parameter is set to have high vibration and low translation energy. The reaction occurs.]]
|}
In conculsion, for late transiton state, the vibration energy is more dominant for passing over the barrier and let the reaction to occur.

{{fontcolor|red|(Good illustration and explaination.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 23:28, 6 June 2019 (BST)}}

MRD:mx4417

2019-06-06T22:28:49Z

Cq3417:

===Molecular reaction Dynamics===
The objectives of this wiki report is to provide a study on triatomic system reactivities, which involves the collision between one atom and a diatomic molecule.
====EXERCISE 1: H + H2 system====
=====Q1: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 of the potential energy surface diagram is the Saddle Point of the diagram, which is local maximum on the lowest energy pathway.

[[file:Q1.mx4417.PNG]]

The Mathmetical definition of saddle point: <math>\operatorname{grad}f = \nabla f</math>=0

[[file:Q1.3mx4417.PNG]]
<ref>https://bb.imperial.ac.uk/bbcswebdav/pid-1216108-dt-content-rid-3932941_1/courses/DSS-CH1_MPC1-17_18/MPC1-CourseNotes%281%29.pdf</ref>

[[file:Q1.24417.PNG|thumb|700px|none| The potential energy surface diagram.]]

{{fontcolor|red|(I think you can state more for this.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 22:58, 6 June 2019 (BST)}}

=====Q2: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.=====
With initial condition r1 = r2, and p1 = p2 = 0.0.
It was found that: r(ts)= 0.9077 Å.
[[file:Q2.1mx4417.png|thumb|700px|none| The internuclear distance vs time diagram with MEP calculation type.]]
As the gradient of both lines in the graph are zero and the distance was kept constant, so it is a saddle point and thus the transition state of the model.

{{fontcolor|red|(Ok, but why?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 22:58, 6 June 2019 (BST)}}

=====Q3:Comment on how the mep and the trajectory you just calculated differ.=====
Trajectories from r1 = 0.9077 Å + 0.01, r2 = 0.9077 Å, p1= p2 = 0

{| class="wikitable"
![[file:Q3.3mx4417.png|thumb|700px|Figure 1:The MEP calculation type. ]]
![[file:Q3.4mx4417.png|thumb|700px|Figure 2:The dynamics calculation type]]
|}
{| class="wikitable"
![[file:Q3.5.mx4417mep.png|thumb|700px| Figure 3:The MEP calculation type(momenta vs time). ]]
![[file:Q3.6mx4417.png|thumb|700px|Figure 4:The dynamics calculation type(momenta vs time)]]
|}
For dynamics method, the vibration motion can be seen as the trajectory is oscillating. For MEP methods, the kinetic energy is set to be zero (see figure 3), thus no oscillation.

=====Q4: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?=====
For the initial positions r1 = 0.74 and r2 = 2.0
{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Contour plot
|-
| -1.25 || -2.5 ||-99.018 || YES||The atom C approached the molecule AB, and as atom C had enough momentum to overcome the energy barrier at the Transition state,thus, it reacted with AB to form molecule BC. ||[[file:Q4.1mx4417.png]]
|-
| -1.5 || -2.0 || -100.456||NO ||The atom C approached the molecule AB, but atom C didnot have enough momentum to overcome the energy barrier at the Transition state,thus, it did not react with AB to form molecule BC. ||
[[file:Q4.2mx4417.png]]
|-
| -1.5 || -2.5 ||-98.956 ||YES || The atom C approached the molecule AB, and as atom C had enough momentum to overcome the energy barrier at the Transition state,thus, it reacted with AB to form molecule BC.||
[[file:Q4.3mx4417.png]]
|-
| -2.5 || -5.0 ||-84.956 ||NO ||The atom C approached the molecule AB, the trajectory passed the transition point, but the product reformed the vibrating excited reactant. ||
[[file:Q4.4mx4417.png]]
|-
| -2.5 || -5.2 ||-83.416 || YES||The atom C approached the molecule AB, the atom C had enough momenta and the trajectory passed the transition point, it formed the vibrating excited product. ||
[[file:Q4.5mx4417.png]]
|}
Conclusion: For the reactants start with same potential energy, but with different momentum (kinetic energy), increases kinetic energy not always make the reaction reactive. As high kinetic energy may cause the re-formation of the vibrating excited reactant.

{{fontcolor|red|(Good illustration.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 23:28, 6 June 2019 (BST)}}

=====Q5:State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?=====
Transition state theory (TST):
1. A molecular system that has crossed the transition state in the direction of a product cannot turn around and reform reactant.

2. In the transition state, motion along the reaction coordinate may be separated from the other motion and treated classically as translation.

3. Even in the absence of equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws.

4. Electronic and nuclear motions are separated due to the large difference in mass.(using the Born-Oppenheimer approximation)

<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase, Chemical Kinetic and Dynamics, Prentice-Hall, New Jersey, 1998</ref>

The first theory is not correct as the table shows the reactant could be reformed under certain circumstances, and the reformation of the reactants decreases the rate of the reaction.
The second one is also not as correct as it predicted, in the calculations (contour plot), there is some vibration close to the transition state.
The other cannot be verified due to insufficient data.

====Exercise 2: F-H-H system====
=====Q6: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?Locate the approximate position of the transition state.=====
Type of reaction: F + H2 → HF + H ( r1=1.81, r2=0.74), which is an exothermic reaction, as formation of the HF Bond is -565 KJ/mol and bond breaking H2 = +436KJ/mol.
{| class="wikitable"
![[File:Q5.1mx4417.png|500px | thumb |The Contour plot of the transition point of F + H2 → HF + H ]]
![[File:Q5.2mx4417.PNG|500px | thumb |The internuclear distance vs time plot, horizontal line repersents the transition state is reached.]]
![[
|}

Type of reaction: H + HF→ H2 + F ( r2=1.82, r1=0.74 ), which is an exothermic reaction, as breaking of the HF Bond is +565 KJ/mol and bond formation H2 = -436KJ/mol.

{| class="wikitable"
![[File:Q5.3mx4417.png |500px | thumb |The Contour plot of the transition point of H + HF→ H2 + F ]]
![[File:Q5.4mx4417.PNG |500px | thumb |The internuclear distance vs time plot, horizontal line repersents the transition state is reached ]]
|}

=====Q7:Report the activation energy for both reactions.=====
{| class="wikitable"
![[File:Q5.6mx4417.png |500px | thumb |The energy vs time plot of H + HF→ H2 + F, the difference in energy is the energy difference of TS and products. As this reaction is exothermic, the activation energy is higher. 30.174 kcal/mol]]
![[File:Q5.7mx4417.png |500px | thumb |The energy vs time plot of F + H2 → HF + H, the difference in energy is the energy difference of TS and products. As this reaction is endothermic, the activation energy is lower (0.137 kcal/mol). ]]
|}

=====Q8: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.=====

{| class="wikitable"
![[File:Q8.1mx4417.png|500px | thumb |The Contour plot with the initial condition shown. It can be seen clearly the reaction occurs. The black line shows the reaction pathway of F+ H2. ]]
![[File:Q8.2mx4417.png |500px | thumb |The Energy vs time plot.The reaction energy is generated by the increase in kinetic energy and the decrease in potential energy ( with the total energy is conserved ). Because of that, the system is 'heat up',there is an increase in temperature.There is also energy trapped in vibration, so maybe the IR emission is possible. ]]
![[File:Q8.3mx4417.png |500px | thumb |The Momenta vs time plot. The B-C (H-H)bond is initially vibrating and then flattened after the reaction. Also, the A-B (H-F) bond forms, and the energy is transferred to the vibration energy of A-B bond. ]]
|}
=====Q9: 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.=====
For reaction pathway of F + H2 → HF + H, it has an early transition state. As the reaction is exothermic, according to Hammond's postulate: the transition state resembles the reactant.
{| class="wikitable"
![[File:Q9.1.mx4417.PNG |500px | thumb |The parameter is set to have high vibration and low translation energy. The reaction does not occur.]]
![[File:Q9.2mx4417.PNG |500px | thumb | The parameter is set to have low vibration and high translation energy. The reaction occurs.]]
|}
In conculsion, for early transiton state, the translation energy is more dominant for passing over the barrier and let the reaction to occur.

For reaction pathway of H + HF→ H2 + F, it has a late transition state. As the reaction is endothermic, according to Hammond's postulate: the transition state resembles the products.
{| class="wikitable"
![[File:Q9.3mx4417.PNG |500px | thumb |The parameter is set to have low vibration and high translation energy. The reaction does not occur.]]
![[File:Q9.4mx4417.PNG |500px | thumb | The parameter is set to have high vibration and low translation energy. The reaction occurs.]]
|}
In conculsion, for late transiton state, the vibration energy is more dominant for passing over the barrier and let the reaction to occur.

{{fontcolor|red|(Good illustration but I think you can make more discussion about this concept.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 23:28, 6 June 2019 (BST)}}

MRD:mx4417

2019-06-06T21:58:03Z

Cq3417:

===Molecular reaction Dynamics===
The objectives of this wiki report is to provide a study on triatomic system reactivities, which involves the collision between one atom and a diatomic molecule.
====EXERCISE 1: H + H2 system====
=====Q1: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 of the potential energy surface diagram is the Saddle Point of the diagram, which is local maximum on the lowest energy pathway.

[[file:Q1.mx4417.PNG]]

The Mathmetical definition of saddle point: <math>\operatorname{grad}f = \nabla f</math>=0

[[file:Q1.3mx4417.PNG]]
<ref>https://bb.imperial.ac.uk/bbcswebdav/pid-1216108-dt-content-rid-3932941_1/courses/DSS-CH1_MPC1-17_18/MPC1-CourseNotes%281%29.pdf</ref>

[[file:Q1.24417.PNG|thumb|700px|none| The potential energy surface diagram.]]

{{fontcolor|red|(I think you can state more for this.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 22:58, 6 June 2019 (BST)}}

=====Q2: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.=====
With initial condition r1 = r2, and p1 = p2 = 0.0.
It was found that: r(ts)= 0.9077 Å.
[[file:Q2.1mx4417.png|thumb|700px|none| The internuclear distance vs time diagram with MEP calculation type.]]
As the gradient of both lines in the graph are zero and the distance was kept constant, so it is a saddle point and thus the transition state of the model.

{{fontcolor|red|(Ok, but why?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 22:58, 6 June 2019 (BST)}}

=====Q3:Comment on how the mep and the trajectory you just calculated differ.=====
Trajectories from r1 = 0.9077 Å + 0.01, r2 = 0.9077 Å, p1= p2 = 0

{| class="wikitable"
![[file:Q3.3mx4417.png|thumb|700px|Figure 1:The MEP calculation type. ]]
![[file:Q3.4mx4417.png|thumb|700px|Figure 2:The dynamics calculation type]]
|}
{| class="wikitable"
![[file:Q3.5.mx4417mep.png|thumb|700px| Figure 3:The MEP calculation type(momenta vs time). ]]
![[file:Q3.6mx4417.png|thumb|700px|Figure 4:The dynamics calculation type(momenta vs time)]]
|}
For dynamics method, the vibration motion can be seen as the trajectory is oscillating. For MEP methods, the kinetic energy is set to be zero (see figure 3), thus no oscillation.

=====Q4: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?=====
For the initial positions r1 = 0.74 and r2 = 2.0
{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Contour plot
|-
| -1.25 || -2.5 ||-99.018 || YES||The atom C approached the molecule AB, and as atom C had enough momentum to overcome the energy barrier at the Transition state,thus, it reacted with AB to form molecule BC. ||[[file:Q4.1mx4417.png]]
|-
| -1.5 || -2.0 || -100.456||NO ||The atom C approached the molecule AB, but atom C didnot have enough momentum to overcome the energy barrier at the Transition state,thus, it did not react with AB to form molecule BC. ||
[[file:Q4.2mx4417.png]]
|-
| -1.5 || -2.5 ||-98.956 ||YES || The atom C approached the molecule AB, and as atom C had enough momentum to overcome the energy barrier at the Transition state,thus, it reacted with AB to form molecule BC.||
[[file:Q4.3mx4417.png]]
|-
| -2.5 || -5.0 ||-84.956 ||NO ||The atom C approached the molecule AB, the trajectory passed the transition point, but the product reformed the vibrating excited reactant. ||
[[file:Q4.4mx4417.png]]
|-
| -2.5 || -5.2 ||-83.416 || YES||The atom C approached the molecule AB, the atom C had enough momenta and the trajectory passed the transition point, it formed the vibrating excited product. ||
[[file:Q4.5mx4417.png]]
|}
Conclusion: For the reactants start with same potential energy, but with different momentum (kinetic energy), increases kinetic energy not always make the reaction reactive. As high kinetic energy may cause the re-formation of the vibrating excited reactant.

=====Q5:State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?=====
Transition state theory (TST):
1. A molecular system that has crossed the transition state in the direction of a product cannot turn around and reform reactant.

2. In the transition state, motion along the reaction coordinate may be separated from the other motion and treated classically as translation.

3. Even in the absence of equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws.

4. Electronic and nuclear motions are separated due to the large difference in mass.(using the Born-Oppenheimer approximation)

<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase, Chemical Kinetic and Dynamics, Prentice-Hall, New Jersey, 1998</ref>

The first theory is not correct as the table shows the reactant could be reformed under certain circumstances, and the reformation of the reactants decreases the rate of the reaction.
The second one is also not as correct as it predicted, in the calculations (contour plot), there is some vibration close to the transition state.
The other cannot be verified due to insufficient data.

====Exercise 2: F-H-H system====
=====Q6: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?Locate the approximate position of the transition state.=====
Type of reaction: F + H2 → HF + H ( r1=1.81, r2=0.74), which is an exothermic reaction, as formation of the HF Bond is -565 KJ/mol and bond breaking H2 = +436KJ/mol.
{| class="wikitable"
![[File:Q5.1mx4417.png|500px | thumb |The Contour plot of the transition point of F + H2 → HF + H ]]
![[File:Q5.2mx4417.PNG|500px | thumb |The internuclear distance vs time plot, horizontal line repersents the transition state is reached.]]
![[
|}

Type of reaction: H + HF→ H2 + F ( r2=1.82, r1=0.74 ), which is an exothermic reaction, as breaking of the HF Bond is +565 KJ/mol and bond formation H2 = -436KJ/mol.

{| class="wikitable"
![[File:Q5.3mx4417.png |500px | thumb |The Contour plot of the transition point of H + HF→ H2 + F ]]
![[File:Q5.4mx4417.PNG |500px | thumb |The internuclear distance vs time plot, horizontal line repersents the transition state is reached ]]
|}

=====Q7:Report the activation energy for both reactions.=====
{| class="wikitable"
![[File:Q5.6mx4417.png |500px | thumb |The energy vs time plot of H + HF→ H2 + F, the difference in energy is the energy difference of TS and products. As this reaction is exothermic, the activation energy is higher. 30.174 kcal/mol]]
![[File:Q5.7mx4417.png |500px | thumb |The energy vs time plot of F + H2 → HF + H, the difference in energy is the energy difference of TS and products. As this reaction is endothermic, the activation energy is lower (0.137 kcal/mol). ]]
|}

=====Q8: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.=====

{| class="wikitable"
![[File:Q8.1mx4417.png|500px | thumb |The Contour plot with the initial condition shown. It can be seen clearly the reaction occurs. The black line shows the reaction pathway of F+ H2. ]]
![[File:Q8.2mx4417.png |500px | thumb |The Energy vs time plot.The reaction energy is generated by the increase in kinetic energy and the decrease in potential energy ( with the total energy is conserved ). Because of that, the system is 'heat up',there is an increase in temperature.There is also energy trapped in vibration, so maybe the IR emission is possible. ]]
![[File:Q8.3mx4417.png |500px | thumb |The Momenta vs time plot. The B-C (H-H)bond is initially vibrating and then flattened after the reaction. Also, the A-B (H-F) bond forms, and the energy is transferred to the vibration energy of A-B bond. ]]
|}
=====Q9: 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.=====
For reaction pathway of F + H2 → HF + H, it has an early transition state. As the reaction is exothermic, according to Hammond's postulate: the transition state resembles the reactant.
{| class="wikitable"
![[File:Q9.1.mx4417.PNG |500px | thumb |The parameter is set to have high vibration and low translation energy. The reaction does not occur.]]
![[File:Q9.2mx4417.PNG |500px | thumb | The parameter is set to have low vibration and high translation energy. The reaction occurs.]]
|}
In conculsion, for early transiton state, the translation energy is more dominant for passing over the barrier and let the reaction to occur.

For reaction pathway of H + HF→ H2 + F, it has a late transition state. As the reaction is endothermic, according to Hammond's postulate: the transition state resembles the products.
{| class="wikitable"
![[File:Q9.3mx4417.PNG |500px | thumb |The parameter is set to have low vibration and high translation energy. The reaction does not occur.]]
![[File:Q9.4mx4417.PNG |500px | thumb | The parameter is set to have high vibration and low translation energy. The reaction occurs.]]
|}
In conculsion, for late transiton state, the vibration energy is more dominant for passing over the barrier and let the reaction to occur.

MRD:ny517

2019-06-06T21:40:55Z

Cq3417:

== Molecular Reaction Dynamics ==
=== Exercise 1: H + H2 ===

''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 defined as the maximum energy point on the minimum energy path and the point on the potential energy surface when the gradient (partial first derivative) of the potential is equal to zero (∂V(ri)/∂ri=0). The transition state can be identified as a saddle point which means it has a partial second derivative greater than and less than zero (dependant on the variables and direction of axis). A local minimum is the point where it is solely a minimum and so the second derivative is always a positive value.

{{fontcolor|red|(Good.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 22:40, 6 June 2019 (BST)}}

''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 to 3 decimal places was:
rts=r1=r2= 0.908 Å

This was estimated through looking at the contour plot of the initial conditions set for a H + H2 reaction and following the reactive trajectory to the point where r1 was approximate to r2 - which was in the region of 0.9 Å (as seen on the contour plot below). This is because for a system with identical atoms, the equidistant point is equal to the transition state position. After trialling values in this region, 0.908 Å produced a trajectory with a constant internuclear distance between both atoms (as seen on the plot below) indicating that the trajectory will not fall off the ridge. This is a transition state since the atoms are stationary over time when they have no initial momenta.

[[File:Ny517_q2.JPG|300px|thumb|left|Contour plot from initial conditions]] [[File:Ny517_q2_Internuclear_distance.JPG|600px|thumb|center|Internuclear Distance vs Time plot where rts = 0.908 Å]]

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

The mep produces a trajectory that follows the valley floor and is also known as the reaction path. It shows a straight-line path along this trajectory because the velocity is always reset to zero after each time step. This does not provide a realistic account of the motion of atoms during the reaction since it assumes the atoms have no mass. The “Dynamics” calculation provides a more realistic trajectory since it accounts for the mass of the system and so the motion of the atoms in the gas phase will be inertial. In a MEP, kinetic energy is assumed to be 0 which explains why there are no oscillations along the path. As the potential energy decreases towards its minimum from the transition state, there is no kinetic energy producing a momentum and causing the molecule to travel up the potential energy curve at small bond distances so it remains at the potential energy minimum resulting in no oscillations.

{{fontcolor|red|(Good explaination.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 22:40, 6 June 2019 (BST)}}

[[File:Ny517_q3_mep.JPG|500px|thumb|left|mep surface plot]] [[File:Ny517_q3_dynamics.JPG|500px|thumb|center|Dynamics surface plot]]

If the initial conditions were reversed, the A-B and B-C plots would switch for the “Distances vs Time” and “Momenta vs Time” graphs. This is because the magnitude of the forces in the reaction do not change but the direction does. As the time increases, the A-B distance increases whilst the B-C bond distance stays constant (bar slight oscillations) at 0.75 because atom A is travelling further away from the BC diatomic molecule. At 5 seconds the A-B bond distance is 18 Å. As the time increases, the A-B momentum plateaus at 2.5 kg ms-1 whilst the B-C momentum oscillates about an average of 1.25 kg ms-1. If the values were reversed, the B-C bond distance would increase whilst the A-B bond distance would remain constant.

[[File:Ny517_q3_larget_bond.JPG|500px|thumb|left|Distance vs Time]] [[File:Ny517_q3_larget_momentum.JPG|500px|thumb|center|Momenta vs Time]]

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

Initial positions '''r1''' = 0.74 Å and '''r2''' = 2.0 Å

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 ||-99.0 ||Yes || Reactants travel through transition state then start to vibrate ||Plot 1
|-
| -1.5 || -2.0 ||-100.4 ||No || Reactants do not overcome activation energy and rebound off the transition state. Atoms vibrate on approach and after reaching transition state ||Plot 2
|-
| -1.5 || -2.5 ||-98.9 ||Yes || Reactants vibrate whilst travelling through transition state then vibrate more vigorously ||Plot 3
|-
| -2.5 || -5.0 ||-85.0 ||No || Reactants pass through transition state then pass back through the transition state to reform the reactants ||Plot 4
|-
| -2.5 || -5.2 ||-83.4 ||Yes || Reactants travel through transition state, then reform reactants and pass through the transition state again. This occurs with strong vibrations throughout reaction ||Plot 5
|}

[[File:Ny517_plot1.JPG|300px]]

[[File:Ny517_plot2.JPG|300px]]

[[File:Ny517_plot3.JPG|300px]]

[[File:Ny517_plot4.JPG|300px]]

[[File:Ny517_plot5.JPG|300px]]

{{fontcolor|red|(Maybe next time try to label under the plot.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 22:40, 6 June 2019 (BST)}}

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

Transition state theory (TST) is used to qualitatively understand the reaction rates of elementary chemical reactions with the idea that all reactions occur through a transition state. One assumption is that once the reagents gain an energy sufficient enough to overcome the activation energy, they will react to form the product. In reality this is not the case as is seen in the calculated examples above where some reactants reached the transition state but reformed the reagents and not the products. This is because of another assumption of TST which is that the energy is looked at classically rather than quantum mechanically so fails to account for the quantization of molecular vibrations. Failing to factor this is why TST predictions for reaction rates will typically be higher than experimental values.

{{fontcolor|red|(Good, but any reference?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 22:40, 6 June 2019 (BST)}}

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

''By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic).''

 F + H2 

[[File:Ny517_h2_plus_f.JPG|300px|]]

H-H distance: 0.74 Å

F-H distance: 2.00 Å

H-H momentum: -1.5 kg ms-1

F-H momentum: -2.5 kg ms-1

The reactants are higher in energy than the products therefore the reaction is exothermic.

 H + HF 

[[File:Ny517_f_h_plus_h.JPG|300px|]]

F-H distance: 0.74 Å

H-H distance: 2.00 Å

F-H momentum: -1.5 kg ms-1

H-H momentum: -2.5 kg ms-1

The reactants are lower in energy than the products therefore the reaction is endothermic.

''How does this relate to the bond strength of the chemical species involved?''

Through inspection of the potential energy surfaces, it is clear that the bond strength of H-F is greater than the bond strength of H-H since more energy is needed to break the H-F bond in the second reaction.

''Locate the approximate position of the transition state.''

[[File:Ny517_hf_transition_state_distance.JPG|400px|thumb|left|Internuclear Distance vs Time]][[File:Ny517_hf_transition_state_contour.JPG|400px|thumb|center|Contour plot]]

H-H distance: 0.745 Å

H-F distance: 1.815 Å

To find the transition state, both momenta were set to equal zero. Hammond's postulate states that the transition state of a reaction resembles either the reactants or the products, to whichever it is closer in energy. The transition state will be closer in energy to the reactants of the exothermic (first) reaction and the products of the endothermic (second) reaction. The contour plot shows the atoms do not fall off the ridge and the Internuclear Distances plot shows the distances remained constant (bar slight oscillations).

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

The activation energy is defined as the difference in energy between the reactants and transition state. The energy of the transition state is known, which is -103.752 kcal mol-1. 0.3 A was added to and subtracted from the HF bond length to calculate the minimum energy difference between the reactants and transition state for both reactions. The reactant energies were extracted using an mep calculation. The calculated activation energies are stated below:

[[File:Ny517_activation_exo.JPG|300px|]] [[File:Ny517_activation_endo.JPG|300px|]]

*For the F + H2 system: 0.236 kcal mol-1
*For the H + HF system: 30.256 kcal mol-1

{{fontcolor|red|(Good.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 22:40, 6 June 2019 (BST)}}

=== Reaction Dynamics ===

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

For a F + H2 reaction trajectory:

H-H distance: 0.744832 Å

H-F distance: 2.00 Å

H-H momentum: -0.7 kg ms-1

H-F momentum: -1.6 kg ms-1

[[File:Ny517_reaction_dynamics_momenta.JPG|400px|thumb|left|Momenta vs Time]][[File:Ny517_reaction_dynamics_energy.JPG|400px|thumb|center|Energy vs Time]]

It is seen that the B-C momentum plateaus over time due to the H2 bond being split and the H atoms becoming further apart so the bond vibrations begin to stop. The A-B momentum oscillates as the B-C momentum stops oscillating which is due to the H-F atom being formed. The H-F bond vibrations are larger than the H-H bond vibrations due to the strength of the bond being greater and hence a have a higher enthalpy of formation. If the H-F bond is stronger than the H-H bond then this will result in an exothermic reaction. This could be confirmed experimentally through a reaction calorimeter. The plot displaying energy over time indicates that the total energy is constant and hence conserved. It also shows that the the potential and kinetic energy convert between each other as the atoms vibrate between their bonds. Notice that the energy oscillations become larger after the transition state- similar to that seen in the momentum plot.

{{fontcolor|red|(Do you mean 'hence have a'?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 22:40, 6 June 2019 (BST)}}

Changing momenta:

*Keeping the same initial conditions but changing the H-H momentum:

[[File:Ny517_momentum_-3.JPG|500px|thumb|left|pHH=-3 kg ms-1]]
[[File:Ny517_momentum_-2.JPG|500px|thumb|center|pHH=-2 kg ms-1]]
[[File:Ny517_momentum_-1.JPG|500px|thumb|left|pHH=-1 kg ms-1]]
[[File:Ny517_momentum_0.JPG|500px|thumb|center|pHH=0 kg ms-1]]
[[File:Ny517_momentum_1.JPG|500px|thumb|left|pHH=1 kg ms-1]]
[[File:Ny517_momentum_2.JPG|500px|thumb|center|pHH=2 kg ms-1]]
[[File:Ny517_momentum_1.JPG|500px|thumb|left|pHH=3 kg ms-1]]

Only when pHH was equal to -1 and -2 kg ms-1 was there a reaction trajectory. When the time step was doubled for pHH=0 a reaction trajectory was produced indicating that the H2 bond was broken and a HF bond was formed. For the other reactions, the reactants passed the transition state but went back and reformed the reactants.

When pFH = -0.8 kg ms-1 and pHH = 0.1 kg ms-1, a reaction trajectory occurs:

[[File:Ny517_momentum_change.JPG||300px|]]

For the reverse reaction, H + HF:

The initial conditions are set with :

H-H distance: 2.00 Å

H-F distance: 0.74 Å

H-H momentum: -0.59 kg ms-1

H-F momentum: -6.55 kg ms-1

[[File:Ny517_reverse_1.JPG||300px|]]

Keeping the atom positions constant, the momenta were varied to have a low vibrational motion on on the H-F bond and an arbitrarily high value of pHH above the activation energy and obtain a reactive trajectory:

[[File:Ny517_reverse_2.JPG||300px|]]

H-H distance: 2.00 Å

F-H distance: 0.74 Å

H-H momentum: -7.55 kg ms-1

F-H momentum: -0.35 kg ms-1

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

The efficiency of a reaction is dependant on the position of the transition state. If there is a late transition state, the vibrational energy is the most influential on the efficiency of the reaction and hence its ability to overcome the reaction energy. If there is an early transition state, the translational energy is the most influencial on the efficiency. Using this, it can be deduced that for the endothermic H + HF system, the reaction is affected more by the vibrational energy and that for the exothermic F + H2 system, the reaction is affected more by the translational energy.

{{fontcolor|red|(Good, but any reference?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 22:40, 6 June 2019 (BST)}}

MRD:pp4717

2019-05-31T11:30:34Z

Cq3417:

===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 generated from making a skew plot to form coordinates q1 and q2. The coordinates q1 and q2 are then differentiated and the minimum of one curve is the maximum of the other. The point at which the minimum and maximum converge is the transition state.

{{fontcolor|red|(Some more details of this definition are required here.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 01:03, 30 May 2019 (BST)}}

===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 transition state position was 0.9075 Ǎ

[[File:TSpp4717.PNG]]

The Internuclear Distances vs Time plot shows two straight lines corresponding to the bonds A-C and B-C which helped to identify the transition state. There are no fluctuations of the distance with time of the bonds in the transition state because they have no potential energy.

{{fontcolor|red|(Any comment with the aid of contour plot?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 01:03, 30 May 2019 (BST)}}

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

MEP does not consider the momentum that the molecule already has, only the momentum that was defined in the initial conditions in the first step. In the dynamic trajectory, the residual momentum is considered which is why there are vibrations present and the point at which the potential energy is zero occurs at a larger distance.
{| class="wikitable"
!p1
!p2
!E(total)
!Reactive?
!Description of the dynamics
!Illustration of trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.119</nowiki>
|yes
|Central proton is transferred from A to B.
|[[File:pp47171.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.619</nowiki>
|no
|Central proton is not transferred.
|[[File:pp47172.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.119</nowiki>
|yes
|Central proton is transferred from A to B.
|[[File:pp47173.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-85.119</nowiki>
|no
|Central proton continuously moves from A to B.
|[[File:pp47174.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.579</nowiki>
|yes
|Central proton is initially attached to A, then B to A.
|[[File:pp47175.png]]
|}

{{fontcolor|red|(Again, more details are required here.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 01:03, 30 May 2019 (BST)}}

{{fontcolor|red|(You missed a question here.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 01:03, 30 May 2019 (BST)}}

=== 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 + H2 is an exothermic reaction; this shows that the bond strength is strong.

H + HF is endothermic; this shows that the bond strength is weak.

===Locate the approximate position of the transition state.===

 For the F + H2 reaction: 

[[File:Ts01333428.PNG]]

The transition state is located at 1.83 Ǎ.

 For the H + HF reaction: 

[[File:TSIND01333428.png]]

The transition state is located at 1.802 Â

{{fontcolor|red|(My question is how did you get these results?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 01:03, 30 May 2019 (BST)}}

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

 For the F + H2 reaction:

[[File:ActivationenergyFHH pp01333428.png]]

The activation energy for the F + H2 reaction = 29.474 KCal/mol

 For the H + HF reaction: 

[[File:TS201333428.png]]

Activation energy = 0.009 KCal/mol

{{fontcolor|red|(Any more comments instead of just presenting the values?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 01:03, 30 May 2019 (BST)}}

=== 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.===

[[File:Initialconditions01333428.PNG]]

These set of initial conditions give the reactive trajectory for the F + H2 reaction.

{{fontcolor|red|(This answer is completely insufficient for this question.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 01:03, 30 May 2019 (BST)}}

{{fontcolor|red|(A question is missing here.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 01:03, 30 May 2019 (BST)}}

{{fontcolor|red|(Overall, this is an uncompleted report with two missed questions and insufficient discussion.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 01:03, 30 May 2019 (BST)}}

MRD:pp4717

2019-05-30T00:03:16Z

Cq3417:

===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 generated from making a skew plot to form coordinates q1 and q2. The coordinates q1 and q2 are then differentiated and the minimum of one curve is the maximum of the other. The point at which the minimum and maximum converge is the transition state.

{{fontcolor|red|(Some more details of this definition are required here.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 01:03, 30 May 2019 (BST)}}

===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 transition state position was 0.9075 Ǎ

[[File:TSpp4717.PNG]]

The Internuclear Distances vs Time plot shows two straight lines corresponding to the bonds A-C and B-C which helped to identify the transition state. There are no fluctuations of the distance with time of the bonds in the transition state because they have no potential energy.

{{fontcolor|red|(Any comment with the aid of contour plot?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 01:03, 30 May 2019 (BST)}}

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

MEP does not consider the momentum that the molecule already has, only the momentum that was defined in the initial conditions in the first step. In the dynamic trajectory, the residual momentum is considered which is why there are vibrations present and the point at which the potential energy is zero occurs at a larger distance.
{| class="wikitable"
!p1
!p2
!E(total)
!Reactive?
!Description of the dynamics
!Illustration of trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.119</nowiki>
|yes
|Central proton is transferred from A to B.
|[[File:pp47171.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.619</nowiki>
|no
|Central proton is not transferred.
|[[File:pp47172.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.119</nowiki>
|yes
|Central proton is transferred from A to B.
|[[File:pp47173.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-85.119</nowiki>
|no
|Central proton continuously moves from A to B.
|[[File:pp47174.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.579</nowiki>
|yes
|Central proton is initially attached to A, then B to A.
|[[File:pp47175.png]]
|}

{{fontcolor|red|(Again, more details are required here.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 01:03, 30 May 2019 (BST)}}

{{fontcolor|red|(You missed a question here.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 01:03, 30 May 2019 (BST)}}

=== 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 + H2 is an exothermic reaction; this shows that the bond strength is strong.

H + HF is endothermic; this shows that the bond strength is weak.

===Locate the approximate position of the transition state.===

 For the F + H2 reaction: 

[[File:Ts01333428.PNG]]

The transition state is located at 1.83 Ǎ.

 For the H + HF reaction: 

[[File:TSIND01333428.png]]

The transition state is located at 1.802 Â

{{fontcolor|red|(My question is how did you get these results?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 01:03, 30 May 2019 (BST)}}

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

 For the F + H2 reaction:

[[File:ActivationenergyFHH pp01333428.png]]

The activation energy for the F + H2 reaction = 29.474 KCal/mol

 For the H + HF reaction: 

[[File:TS201333428.png]]

Activation energy = 0.009 KCal/mol

{{fontcolor|red|(Any more comments instead of just presenting the values?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 01:03, 30 May 2019 (BST)}}

=== 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.===

[[File:Initialconditions01333428.PNG]]

These set of initial conditions give the reactive trajectory for the F + H2 reaction.

{{fontcolor|red|(This answer is completely insufficient for this question.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 01:03, 30 May 2019 (BST)}}

{{fontcolor|red|(A question is missing here.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 01:03, 30 May 2019 (BST)}}

{{fontcolor|red|(Overall, this is a uncompleted report with two missed questions and insufficient discussion.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 01:03, 30 May 2019 (BST)}}

MRD:CP3MD mm1517

2019-05-29T22:17:00Z

Cq3417:

== EXERCISE 1: H + H2 system ==

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

==== 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 can be defined as the maximum on a minimum energy pathway. An example of a minimum energy pathway is shown on Figure 1 as a coloured wavy line, where the maximum corresponds to the transition state of this reaction between the chosen reactants and products.

[[File:Capture 1 2.PNG|frame|centre|Figure 1 - Graph of rAB and rBC displaying the minimum energy pathway, the saddle point, and the orthogonal lines r1 and r2|271x271px]]

The transition state has the unique property of ∂V('''ri''')/∂'''ri'''=0. This point on the pathway can be labelled as a saddle point, meaning it is stable yet does not correspond to a local maxima or minima.

To conform that the chosen point is a saddle point, the second derivatives of the lines r1 and r2 must be taken, which are diagonal vectors relative to rAB and rBC and orthogonal relative to each other. Tangent to the saddle point on the minimum energy pathway is r2, whereas r1 cuts orthogonally through saddle point.

In order to identify the local maximum and minima, the second derivative must be taken.

∂2V('''r1''')/∂'''r1'''2''' '''> 0

∂2V('''r2''')/∂'''r2'''2''' '''< 0

Hence r1 is the local minima and r2 is the local maxima.

{{fontcolor|red|(Good definition.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 23:17, 29 May 2019 (BST)}}

==== ''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.'' ====
Due to the symmetrical surface of H + H2 , its transition state must have the property r1 = r2. As established in the previous question, the transition state has a gradient of zero. If they are unequal, a trajectory line is observed. Such a line results in oscillations about the saddle point, meaning it is non longer a transition state.

[[File:Capture 2.PNG|frame|centre|Figure 2 - Graph of distance against time for when tts = 0.9097 Å and momenta is 0]] In order to find rts, various values of r1 = r2 were tested, keeping the momenta as zero. An optimum value of 0.9097 Å''' '''was found to give the plot with minimum oscillations (Figure 2).

{{fontcolor|red|(Good illustration.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 23:17, 29 May 2019 (BST)}}

==== ''Comment on how the MEP and the trajectory you just calculated differ.'' ====
The minimum energy path (MEP) is a trajectory that corresponds to infinitely slow motion. The conditions were changed to where the system is slightly displaced from the transition state by δ. '''Figure's 3''' and '''4''' below correspond to when δ=0.01;

'''r1''' = '''rts'''+0.01, '''r2''' = '''rts '''and '''p1''' = '''p2'''.

As the MEP corresponds to infinitely slow motion, the plot is a straight line, therefore has no vibrations, whereas the dynamic plot is wavy. Therefore when compared to the dynamic graph, the energy path has fewer oscillations.

[[File:Capture 3.PNG|frame|centre|Figure 3 - MEP plot of rAB and rBC where rAB = 0.9197 Å and rBC = 0.9097 Å]] [[File:Capture 4.PNG|frame|centre|Figure 4 - Dynamic plot of rAB and rBC where rAB = 0.9197 Å and rBC = 0.9097 Å]]

=== Reactive and unreactive trajectories ===

==== ''Complete the table below 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" border=1
! p1 !! p2 !! Ek !! Ep !! Etot !! Illustration of the trajectory
!Reactive?!! Description of the dynamics
|-
| -1.25 || -2.5 || +4.687
|| -103.706
|| -99.018
|| [[File:Capture 5.PNG]]
|Yes
|| A collision of A-B with C occurs, with enough energy to overcome the Ea barrier and thus form BC. It is seen that AB has little/no vibrational energy, whereas BC has more vibrational energy than the reactants.
|-
| -1.5 || -2.0 || +3.250
|| -103.706
|| -100.456
|| [[File:Capture 6.PNG]]
|No
|| A collision occurs, however it does not have the energy to overcome Ea. As a result the formation of BC does not occur and AB "bounces" back and stays intact. Yet its oscillations have increased, so vibrational energy was gained.
|-
| -1.5 || -2.5 ||+4.750|| -103.706
|| -98.956
|| [[File:Capture 7.PNG]]
|Yes
|| AB and C collide, and successfully overcome the Ea barrier to form BC. The increase in magnitude of p1 results in higher vibrations in BC than previously. BC has larger oscillations than AB and so has a larger vibrational energy.
|-
| -2.5 || -5.0 ||+18.750|| -103.706
|| -85.956
|| [[File:Capture 8.PNG]]
|No
|| Molecules have a significantly higher energy than previously, thus easily surpass the Ea. However, it is observed that the minimum energy pathway passes back through the transition state, reforming the reactants with a higher vibrational energy than before, shown by larger oscillations.
|-
| -2.5 || -5.2 || +20.290
|| -103.706
|| -83.416
|| [[File:Capture 912.png]]
|Yes
|| A similar case to the previous example, however in this case they collide with even higher energy. BC's high energy seems to result in it reforming into AB, yet despite this, the minimum energy pathway still passes back through the transition state, so ultimately the reaction went to completion.
|}
From the table it can be concluded that higher momenta and higher energy of reactants do not always result in a successful reaction. It may cross the barrier, which is what theory predicts, yet there is the possibility of the products crossing back through the transition state and reforming the reactants.

==== State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ====
The assumptions made in the Transition State Theory (TST) are as follows:
* Distribution of the reactants follows the Maxwell-Boltzmann distribution.
* Quasi-equilibium is established, where A + B ⇌ [AB]‡ → P. This shows that the reactants and products are not in equilibrium with each other, the reactants are in equilibrium with the transition state, and that the products cannot reversibly reform the reactants.
* Obey the Born-Oppenheimer approximation, where nuclear and electronic motions are independent of one another.
The Transition State Theory allows prediction of the rate constant, however the predicted value will deviate from the experimental value. This is because its assumptions also leads to limitations in the values it obtains.

Its assumption that once the quasi-equilibrium has been established, the products cannot reform the reactants. However, this is not the case in reality, where barrier recrossing can occur if the energy of products is high enough to break apart and re form the reactants. Therefore the predicted rate constant using TST will be higher than the experimental value.

{{fontcolor|red|(Any reference here?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 23:17, 29 May 2019 (BST)}}

== 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?''''' ====

Figure 5 shows that the initial reaction, F + H2, begins at the higher end of the energy surface, thus it is exothermic. As a result, it can be deduced that the H-F bond is stronger than the H-H bond in the reactants. Comparatively to this, Figure 6<nowiki/>'s reaction, H + HF, begins at the lower end of the energy surface, so is endothermic. This correlates with theory, as due to the electronegativity of F, the covalent bond between H and F will be heavily polarised, resulting in a stronger bond.

[[File:Capture 11.png|frame|centre|Figure 5 - Graph of F + H2 with 0 momentum on each particle]] [[File:Capture 10.png|frame|centre|Figure 6 - Graph of H + HF with 0 momentum on each particle]]

==== '''''Locate the approximate position of the transition state.''''' ====

The AB distance and the momenta of both molecules were kept constant, and the BC distance was changed till the trajectory of the molecules appeared to be as motionless as possible. Thus through trial and error the transition state was found and is shown in Figure 6. Additionally, the steps were increased to 2000.

[[File:Capture 12.png |frame|centre|Figure 7 - Graph of F + H2 with the transition state shown]] [[File:Capture 16.png |frame|centre|Figure 8 - Inter nuclear Distance - Time graph with the transition state shown as straight lines]]

{{fontcolor|red|(Good, but how about the value of rHH and rHF?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 23:17, 29 May 2019 (BST)}}

==== ''Report the activation energy for both reactions.'' ====
The inter nuclear distances were displaced about the transition state by values between +0.01/-0.01, and the energy-step graphs observed to identify the activation barrier for H2 + F and HF + H.

Figure 9 HF + H reaction Ea = + 30.102 kcalmol -1

Figure 10 H2 + F reaction Ea = + 0.220 kcalmol -1

[[File:Capture 13.png|frame|centre|Figure 9 - Energy - Step graph of HF + H showing the activation energy Ea]] [[File:Capture 14.png|frame|centre|Figure 10 - Energy - Step graph of H2 + F showing the activation energy Ea]]

{{fontcolor|red|(I think you can discuss this in detail.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 23:17, 29 May 2019 (BST)}}

=== Reaction dynamics ===

==== ''In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'' ====
A set of initial conditions were identified that result in a reactive trajectory of H2 + F. The values chosen were rHH = 0.730 Å and rHH-F = 1.688 Å, and produced '''Figure 10'''. From this dynamic Momentum-Time graph, it can be seen that the A-B value tends to fixed value as the distance between them increases. Additionally as shown from '''Figure 5''' we know that this reaction is exothermic, so the formation of the H-F bond releases energy. This is represented in the large increase of B-C oscillation. Due to the release in energy, we know that the chemical energy has been converted into kinetic energy, which supports the rise in temperature seen when an exothermic reaction takes place.

{{fontcolor|red|(I think you can increase the steps and achieve better plot.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 23:17, 29 May 2019 (BST)}}

[[File:Capture 15.png|frame|centre|Figure 11 - Momentum-Time graph of H2 + F showing a reactive trajectory, A, B = H, C= F Ea]]

==== ''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 (TS).'' ====
Polanyi's empirical rules state that vibration energy is more efficient at promoting a the activation of a late transition-state than translational energy, and vise-versa applies for an early transition state.

It was determined from '''Figure 5 '''that the reaction H2 + F is exothermic, and from '''Figure 6 '''that the reaction HF + H is endothermic.

From Hammond's Postulate, it can be concluded that the reaction H2 + F has an early TS which resembles the reactants, so is more efficiently activated using translational energy. For HF + H, the TS is late and resembles the reactants, thus is more efficiently activated by vibrational energy.

The chosen parameters were rAB = 0.74 Å and rBC = 1.81 Å, and the momenta varied.

To begin with, H2 + F was focused on. Hammond's postulate concluded that it is more efficiently activated by translational energy, so this was tested. Conditions for a greater vibrational energy were set, pAB = -3 and pBC = -0.8, in '''Figure 12'''. From this, you can observe that the reaction was unsuccessful. However when the momenta were changed so that there was higher translational energy, pAB = -0.5 and pBC = -2.5, the reaction was successful, '''Figure 13'''.

[[File:H2+F bad.png|frame|centre|Figure 12 - H2 + F unsuccessful reaction with greater vibrational energy, pAB = -3 and pBC = -0.8]]

[[File:H2+F good.png|frame|centre|Figure 13 - H2 + F successful reaction with greater translational energy, pAB = -0.5 and pBC = -2.5]]

To conclude, HF + H was observed. Hammond's postulate concluded that it is more efficiently activated by vibrational energy.. Conditions for a greater translational energy were set, pAB = -0.5 and pBC = -2.5, in '''Figure 14'''. From this, you can observe that the reaction was unsuccessful. However when the momenta were changed so that there was higher vibrational energy, pAB = -2 and pBC = -1, the reaction was successful, '''Figure 15'''.

[[File:HF+H bad.png|frame|centre|Figure 14 - HF + H unsuccessful reaction with greater translational energy, pAB = -0.5 and pBC = -2.5]]

[[File:HF+H good.png|frame|centre|Figure 15 - HF + H successful reaction with greater vibrational energy, pAB = -2 and pBC = -1]]

{{fontcolor|red|(I think you should put a reference here.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 23:17, 29 May 2019 (BST)}}

{{fontcolor|red|(Overall, this is a report with good illustration and can be improved by more discussion and listing reference.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 23:17, 29 May 2019 (BST)}}

MRD:frances

2019-05-24T13:04:51Z

Cq3417:

==Exercise 1==

===Dynamics on a Potential Energy Surface Diagram===

====Definition of parameters====

AB distance = r1

BC distance = r2

On a potential energy surface diagram, the transition state is mathematically defined as maximum point along a minimum energy pathway. The maximum along the minimum energy trajectory is shown in the first image below from a side on view. It can mathematically defined as a saddle point, which is a local minimum in one direction but a local maximum in a direction orthogonal to this direction. From this definition, taking a first derivative of the inflection point of the reaction path, ∂V('<nowiki/>r1)/∂r1 and ∂V(r2)/∂r2 will give 0 for both derivatives, indicating a turning point. This can be distinguished from a local minimum point by taking the second derivative to the inflection point or the derivative of orthogonal vectors, q1 and q2. We can define two additional important vectors, q1 and q2. These are diagonal vectors relative to r1 and r2 and the second partial derivatives with respect to these values gives specific information about the nature of the transition state point. One derivative result will indicate a maximum point and the derivative in the orthogonal direction will indicate a minimum point i.e. ∂V('''q1''')/∂'''q1''' <0 and ∂V('''q2''')/∂'''q2''' >0. In one direction the second derivative will be positive indicating a minimum point, however in the other it will be negative indicating a maximum point. This is the definition of the saddle point.

[[File:SurfacePlot3817.png|thumb|center|Potential Energy Plot rotated to clearly show the saddle point in the middle as flat surface either side of which is a reduction in potential energy.]]

[[File: Animation23817.png|thumb|center|Contour plot annotated to show the two newly defined vectors q1 and q2.]]

===Trajectories from r1 = r2: locating the transition state===

The transition state position ('''rts''') can be estimated as 0.91 (2sf) Å. This value can be obtained through running a MEP type calculation with '''r1''' = '''r2''', and '''p1''' = '''p2''' = 0.0. This allows you to calculate the transition state point as at 0 momentum the trajectory oscillates on the ridge. Arbitrarily selecting '''r1''' = 0.6, the system increases to a distance of 0.9077 before plateauing, indicating the transition state distance has been reached. The system comes to rest at this point.

[[File:TransitionState3817.png|thumb|center|Internuclear Distance vs Time plot calculated using an MEP calculation to show the transition state inter-nuclear distance as 0.91 (2sf) Å.]]

[[File:zoomedtransitionstate3817.png|thumb|center|Internuclear Distance vs Time plot calculated using an MEP calculation to show the transition state inter-nuclear distance.]]

{{fontcolor1|red|(I think including the contour plot here is better.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 20:18, 22 May 2019 (BST)}}

=== Reaction Path Trajectories using MEP and Dynamic Calculations ===

Running a trajectory calculation with the parameters'''r1''' = 0.9077, i.e. transition state distance, '''r2''' = 0.9087 and '''p1''' = '''p2''' = 0, gives two slightly different trajectories depending on which type of calculation you use. In the case of the MEP calculation, the reaction trajectory essentially runs along the minimum with no oscillations, in contrast to the dynamic calculation which oscillates along the minimum energy trajectory. These are both shown below. The reason they differ in appearance can be explained using a sphere rolling on a potential energy surface analogy. The MEP takes recordings at infinitesimally small speeds which means the sphere does not take into account any previous momentum. As a result, it has no momentum to 'roll' up the other potential energy surface and as such, the trajectory follows a straight line. In contrast, the sphere representing the dynamic calculation is aware of its previous momentum it has gained from starting higher on the potential energy surface. As a result, when it reaches the minimum, it is able to slightly roll up the opposite surface. Hence, an oscillating trajectory is observed.

[[File:dynamicsurfaceplot.png|thumb|center|Dynamic Plot: A trajectory superimposed on a surface plot calculated using the dynamic calculation method.]]

[[File:mepsurfaceplot.png|thumb|center|MEP Plot: A trajectory superimposed on a surface plot calculated using the MEP calculation method.]]

{{fontcolor1|red|(I think here contour plot is better than surface plot.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 20:18, 22 May 2019 (BST)}}

The final values values of momentum and internuclear distance can be recorded at large t.

r1(2.5)= 8.97 Å

r2(2.5)= 0.68 Å

p1(2.5)= 2.48

p2(2.5)= 0.90

If we change the parameters so that r1 = rts and r2 = rts +0.01 instead, the only difference between the plots is that the lines switch over, the magnitude of the values remains the same. This is because we are dealing with a completely symmetrical system and so the momentum and internuclear distances will be the same irrespective of whether we are considering the exchange of a H atom in either case.

The graphs of Internuclear distance vs time and momentum vs time, from which the above values were calculated from, are shown below.

[[File:distance1.png|thumb|center|Dynamic Plot: Internuclear distance Vs Time graph with an extended number of steps to determine a value of internuclear distance at large t.]]

[[File:momentum2.png|thumb|center|Dynamic Plot: Momentum Vs Time graph with an extended number of steps to determine a value of momentum associated with the respective distances at large t.]]

===Reactive and unreactive trajectories ===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99.12 || Yes || [[File:contour1.png|thumb|center|]] || Molecule AB approaches Atom C with enough momentum and energy to overcome the activation barrier and so a reaction occurs.
|-
| -1.5 || -2.0 || -100.46 || No || [[File:contour2.png|thumb|center]] || Molecule AB approaches C however does not collide with enough energy to overcome the activation energy barrier therefore, no reaction occurs and the two particle entities separate in opposite directions.
|-
| -1.5 || -2.5 || -98.98 || Yes || [[File:contour3.png|thumb|center]] || Atom C and bonded atoms A and B come together and collide with enough momentum and energy for a reaction to occur, hence the bond between A and C breaks and a bond is formed between atoms C and D. Following the collision, the particles then part in opposite directions with the newly formed BC molecule oscillating.
|-
| -2.5 || -5.0 || -84.96 ||No|| [[File:contour4.png|thumb|center|]] || Molecule AB approaches atom C, highly oscillating. It appears that molecule AB dissociates at the transition state, as the internnuclear distance of BC appears to be less than that of AB, however molecule AB reforms and molecule AB separates from C. Again, molecule AB is highly oscillating.
|-
| -2.5 || -5.2 || -83.42 || Yes || [[File:contour7.png|thumb|center|]]|| Molecule AB approaches Atom C, highly oscillating. Upon collision, atom B appears to assiciate with atom C, followed by an oscillation back between atom A and C before finally bonding to atom C.
|}

{{fontcolor1|red|(The contour plot of your first example in this table is a little bit ambiguous.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 20:18, 22 May 2019 (BST)}}

This table is showing that total energy from the reaction is not the only factor that dictates whether or not a reaction occurs. In other words, not all trajectories starting with the same positions with values of momenta greater than those resulting in a previous successful collision, will result in a successful collision reaction. Having enough kinetic energy to overcome the activation barrier is not the only condition that needs to be satisfied in order to have a successful reaction. There are many other factors to take into account such as the ratio of contribution of vibrational and translational energy to the total energy that will impact whether or not a reaction occurs. These factors are discussed later. We can also see in the case of the penultimate row value, that barrier recrossing is possible. Therefore the average number of collisions that would be predicated to occurr successfully, based on energetics alone, would in fact be greater than that observed experimentally.

===Transition State Theory===

Transition State theory is the idea that rates of reaction can be studied by examining activated complexes at the saddle point. These complexes are in quasi-equilibrium with the reactant molecules.

The main assumptions of transition state theory are as follows:

1. The theory assumes that each reactive intermediate is long-lived enough that a Boltzmann distribution of energies is reached before the reaction then continues onto the next step. However, this can often lead to limitations of the TST, as if the intermediates are very short lived, the reaction does not have time to re-equilibriate its energy levels. In this case the transition state will have residual momentum from its reaction trajectory that is able to push it over to the products. In addition, the model also assumes that all reactant molecules are distributes according to the Boltzman distribution. As a result, the predictions made by the model may overestimate the number of successful reactions.

2. Another assumption is that the theory is based on atomic nuclei behaving according to classical mechanics. Unless atoms react with enough energy to overcome the activation energy, a reaction does not occur. However, the phenomenon of quantum mechanical tunnelling results in the possibility of particles still crossing the barrier even if the collision does not occur with enough energy to cross the activation barrier. However, this phenomenon will be relatively small.

3. The third assumption is that the reaction system will pass over the lowest energy saddle point on the potential energy surface. This is sufficient for low temperatures however at high temp, molecules populate higher vibrational modes therefore motions are more complex and so collisions may occur with more complex transition states that do not occur on the simple saddle point Transition State Theory predicts. As a result, the energetics and dynamics associated with the prediction may not be valid.

4. It also assumes that once the transition state has been crossed and products of the reaction formed, there is no way for the entities to then recross the transition state without altering the reaction conditions in a way that is then favourable for them to do so. In other words, it does not take into account barrier recrossing, hence it predicts a greater number of reactions occurr to completion of products than is observed in reality. Hence, the predicted rate constant for the reaction will be greater than the observed.

{{fontcolor1|red|(Good statement, any reference?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 20:18, 22 May 2019 (BST)}}

How will Transition State Theory predictions for reaction rate values compare with experimental values?

As can be seen from the table and as has been touched on above in point 4, the experimental values obtained do not follow transition state theory, as even in cases where the values predict a reaction to have enough energy to overcome the activation barrier, a reaction does not always occur. If the reactions did all obey the TST, all reactions would occur as they all have energy to overcome the activation barrier. As a result, predicted reaction rate values will be greater than those observed.

==Exercise 2==

The F + H2 reaction is exothermic as the reactant energies i.e. an atom of F and H2 molecule are higher than the product energies on the potential energy surface graph. However in the case of HF + H, the reaction is endothermic as the reactants are lower in energy than the products and hence the reaction proceeds in a downhill direction.

Endothermic reaction of HF + H:

[[File:endothermic3817.png]]

Exothermic reaction of F and H2:

[[File:exothermic3817.png]]

The blue dot represents the staring reactants on the potential energy surface.

The energetics of the reaction relate to the bond strength as the net energy taken in or released in the reaction is indicative of the relative strength of the reactant and product bonds. Bond formation is exothermic, whereas bond breaking is endothermic. Therefore, if the reactant bond strength is greater than the product bond strength, a greater amount of energy is required to break the bond than is released upon formation of the new bond. Hence, an endothermic reaction indicates that reactant bond strength is greater than product bond strength and an exothermic reaction indicates that reactant bond strength is weaker than product bond strength, as more energy has been released upon formation of the stronger product bond than was inputted to break the weaker reactant bond. This can be further illustrated by considering the initial H2 + H reaction where the energy of the reactants and products are equal as the bond being formed and created is identical, therefore taking in and releasing equal energy. As a result the potential energy surfaces for reactants and products are equal and hence no net energy is released. This is shown below.

[[File:hh3817.png]]

=== Locating the approximate position of the transition state ===

The transition state is a saddle point at which, if given no additional energy or momentum, the reaction trajectory will remain stationary.

a) For the reaction F + H2, the approximate position of the transition state can be located by finding the combination of AB and BC distances at which, given no momentum and no energy, the particle trajectory will remain stationary. This can be located when:

A = F

B = H

C = H

AB distance = 1.808 Å

BC distance = 0.705 Å

This is shown graphically plotted below.

[[File:ContourH2.png]]

The small variation of the trajectory position is shown below, a zoomed in version of the above graph. This graph shows the precision of the approximate position, there is very small deviation from the point.

[[File:contou2.png]]

b) For the reaction HF + H, the approximate position of the transition state will simply be the same coordinates associated with the opposite atoms as the transition state is the same regardless of the direction in which the reaction proceeds. Plotting these on a contour plot for the reverse reaction gives the same result as above, a stationary trajectory point, indicating that we area at a saddle point. This graph is shown below where,

A = H

B = H

C = F

AB distance = 0.705 Å

BC distance = 1.808 Å

[[File:contourhf.png]]

===Estimating the activation energy ===

To obtain an approximation for the activation energy of the reaction, an MEP calculation can be performed (with 5000 steps) on a structure with bond lengths adjusted slightly from the transition state values. By plotting an energy V time graph, shown below, it is possible to estimate the activation energy for the reaction as the difference between the total energy and the potential energy at high values of t.

a) For the H2 + F reaction, this can be approximated as +0.255 kcal/mol.

The graphical representation of this is shown below:

[[File:h23817.png]]

b) For the HF + H reaction, this can be approximated from taking away the potential energy of the HF molecule from the total (transition state) energy. Therefore, there are an additional energy values required for this calculation, the potential energy of the HF molecule.

Using the values below:

HF potential energy = -134.03
Total energy/ Transition State energy = -103.75
H2 energy = -104.02

The activation energy of the reaction HF + H can be approximated as + 30.01 kcal/mol.

{{fontcolor1|red|(Any diagrams here?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 20:18, 22 May 2019 (BST)}}

=== Reaction Dynamics ===

A set of conditions that results in a reactive trajectory for F + H2 are as follows:

A = F, B = H, C= H

AB distance = 1.4

BC distance = 1

AB momentum = BC momentum = 0

The mechanism of release of reaction energy is as follows. This reaction is an exothermic reaction, hence energy is released to the surroundings. The mechanism of release is the conversion of excess potential energy stored in the bonds of the reactants converted to kinetic energy. Molecules are constantly switching energy between potential and kinetic and hence this is a dynamic process. Upon reaction completion, energy is converted to translational energy of the H atom leaving the HF molecule as well as vibrational energy of the molecules. Therefore, even though the kinetic energy is constantly switching between vibrational and translational energy, the time average kinetic energy has increased following reaction completion.

This is shown in the graph below, where the time average kinetic energy increases, and the time average potential energy decreases:

[[File:Kineticenergy3817.png]]

In addition, an analysis of the momentum time graph shows that energy released from forming the stronger H-F bond is released and converted into a large increase in BC oscillation. The AB value however tends to a fixed value as the product molecule and particle move away from each other and so do not influence each others momentum. This means the translational energy represented by the orange line, remains constant. The HF bond will continue to oscillate at a steady rate however, with the increase in oscillation proportional to the energy released from the chemical bonds and hence the increase in temperature of the reaction.

[[File:Momenta23817.png]]

In summary, chemical Potential energy is being converted to Kinetic energy.

Experimentally this can be confirmed through measuring the temperature of the reaction. Kinetic energy is all energy an object possesses due to motion and so in general the kinetic energy of the molecules increases due to the release of energy, therefore the temperature of the surroundings of an exothermic reaction will increase. As a result, using calorimetry, you can measure the increase in kinetic energy in the surroundings.

=== 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.===

Contributions from the energy occur as vibrational and translational.

'''In the exothermic reaction of F + H2 '''

pFH = the translational contribution to the energy
pHH = the vibrational contribution to the energy

A number of simulations were performed in order to determine the empirical rules for the reaction.

First, initially setting the conditions of a reaction of F + H2 as rHH = 0.74, with a momentum pFH = -0.5. Testing a range of values of pFH between -3 and 3, it is observed that at values close to the limits, the reaction is successful and goes to completion. However, at values in between, it is observed that either the reaction is not successful or there is a significant amount of barrier recrossing. Therefore if the vibrational energy of the molecule approaching is too low, the reaction will not be successful. As we have been told, we are placing a much greater amount of energy in the system, much greater than that required to surpass the activation energy barrier. The fact that some of the reactions even with this magnitude of energy are unsuccessful is further proof against TST and indication that the type of energy (vibrational or transnational) supplied to the system is important.

Using the increasing the momentum of the FH bond to -0.8 and reducing the momentum of the HH vibration to 0.1 you observe the reaction does occur, however there appears to be barrier recrossing.

This reaction is exothermic and hence has an early transition state.

This shows that increasing the translational contribution even slightly for an early transition state has a greater impact on the success of the reaction than any changes to vibrational contributions.

'''In the endothermic reaction FH + H '''

This reaction is endothermic and hence has a late transition state. In this case, it was observed that adding vibrational energy to the reaction (ie in the form of momentum pFH) was of greatest importance. A reaction still occurred despite low values of pHH, if there was still a high enough magnitude of vibrational energy.

This can be illustrated in the below contour plot which was generated using parameters of rFH = 0.9, rHH= 1.3 pHH = 0.2 and pFH = 12. In this case, the translational contribution to the energy is very small, yet the vibrational contribution is very large and hence the reaction still occurs. This illustrates the rules.

[[File:endoplot.png |thumb|center| A contour plot showing a successful endothermic reaction with vibrational energy of 12 despite the small translational energy of 0.2. This is a clear illustration of Polanyi's rules where the vibrational energy is the most important factor for an endothermic reaction with a late transition state.]]

'''Polanyi's rules'''

Polanyi's rules state that a certain distribution of energy in reactants and products is more favoured to promote a successful reaction. The specifics of this distribution are dependant on whether the reaction has an early or late transition state. Using Hammond's postulate, it can be assumed that an exothermic reaction has an early transition state in which the transition state closely resembles the reactants and in an endothermic reaction the transition state energy most closely resembles the products.

Polanyi's rules relate concept of early or late transition state to the most favourable distribution of energy for the highest efficiency of reaction[1]. These rules state that vibrational energy is more efficient in promoting a late transition state reaction than translational energy, whereas the reverse is true for an early transition state reaction.

[[For an early transition state]] (exothermic reaction), translational energy is most important in determining the success of the reaction.

An example of this case is shown below. Starting with reaction parameters: rFH = 2 rHH = 0.74 and pHH = -1.5, pHH = -0.26, a reaction does not occur. However, only slightly increasing the FH momentum, related to the translational energy, from -0.26 to -0.28 results in a successful reaction. This proves that translational energy is most important in determining the success of a reaction. This process is illustrated below in the two graphs where just a small change in translational energy results in the reaction overcoming the transition state barrier and the reaction occurring successfully. As was indicated above, it took much greater changes to the vibrational motion in the range pHH = -3 to 3 in order to achieve a successful reaction. This further confirms the notion that translational energy is most important factor for the efficiency of the exothermic reaction.

[[File:Notworking3817.png|thumb|center| A contour plot showing an unsuccessful reaction with translational energy of -0.26.]]

[[File:Working3817.png |thumb|center| A contour plot showing a successful reaction with translational energy of -0.28.]]

[[For a late transition state]] (endothermic reaction), vibrational energy is most important in determining the success of the reaction.

These rules are in line with the observations made above about the importance of the translation and vibrational energy distribution for the exothermic reaction of F + H2 and the reverse endothermic reaction of FH + H.

These rules are successful to a certain degree, however, varying the conditions even slightly may result in an unsuccessful reaction even where the rules may predict that a reaction would be successful.

'''Conclusion'''

Polanyi's rules state that for a late transition state (endothermic reaction), vibrational energy is most important in determining the success of the reaction, yet for an early transition state (exothermic reaction), translational energy is most important in determining the success of the reaction. These rules are useful to a certain extent as they provide a quick guide, much like with TST, as to whether or not a reaction will be successful and the parameters needed in order to improve reactions efficiency. However, these rules are also a broad overview and should not be taken as the whole story as often small tweaks in conditions will act against the rules. Such an example includes

{{fontcolor1|red|(Overall, well done) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 20:18, 22 May 2019 (BST)}}

==Reference==

1. J. Phys. Chem. Lett., 2012, 3 (23), pp 3416–3419

MRD:jp2517

2019-05-23T20:24:13Z

Cq3417:

== Wiki Page J Powell ==
==Exercise 1==
===Question 1===
 1a) How is the transition state mathematically defined on a Potential Energy Surface diagram?

On a Potential Energy Surface (PES) diagram, the transition state is defined as a saddle point - that is to say, a point in 3D space which is a minimum in one direction, and a maximum orthogonal to it.

 1b) How can the transition state be identified?

To define the saddle point, two new functions, orthogonal to each other, need to be defined. Q1 and Q2 will be parallel with the line of the minimum and maximum at the saddle point (hence why they are orthogonal to each other). Q2 and Q1 represent 45 degree rotations (it just happens that it is 45 degrees in this case, it could be a different angle) of the AB and BC axes, namely, that Q2 is the line tangential to the maximum potential energy along the minimum energy pathway, while Q1 is the line perpendicular to this, also passing through the maximum potential energy along the minimum energy pathway. The fact that they are orthogonal to each other means they provide a good basis set for all points. This setup is shown in Figure 1

[[File:jp2517contourplot.PNG|right|thumb|400 px|Figure 1: A contour plot with the orthogonal coordinate system|400 px]]

To mathematically locate the saddle point, it's necessary to find the first-order derivatives along Q1 and Q2, to locate the stationary points. In a saddle point, these will both occur at the same value. Second-order differentiation can confirm that a maximum has been found on Q2 and a maximum on Q1

 1c) How can it be distinguished from the local energy minimum of the potential energy surface?

While a local energy minimum may be a minimum compared to some points around it, it may not be the global minimum. However, the local minimum can be distinguished from the global energy minimum by differentiating to find all the stationary points along a line, and comparing the values for V at which these occur. The global minimum will occur at the lowest value for V out of all the minima.

{{fontcolor|red|(Good.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 21:24, 23 May 2019 (BST)}}

===Question 2===

 2. Finding a best estimate of transition state position 

Hammond's postulate states that in a reaction which goes between products and reactants with a transition state, the transition state will be most structurally similar to either the reactants or products, based on which it is closer to in energy. However, this only really works if the reactants and products are very different in energy. In this example, the reactants and products are identical in energy, therefore the transition state will be exactly halfway between the two, making it the structural average of both of them.

We can therefore say that r1 = r2 at the transition state, meaning that on the contour plot, the transition state lies somewhere along the diagonal line going from bottom left to top right.

From visual inspection of the contour plot, it can be estimated that this occurs somewhere around r1 = r2 = 0.9 Å. Use of a plot of Internuclear Distance against Time can confirm that the point has been found, as when the molecules are put precisely at the transition state, they will sit there and nothing will happen, thus the plot of Internuclear Distance vs Time will show two flat lines. If the molecules are even slightly off the transition state point, they will oscillate back and forth, leading to a sinusoidal curve in the Internuclear Distance vs Time plot.

The less-efficient way to find the transition state point is to iteratively solve for r1 and r2 by trying different values for each and obtaining a progressively flatter line. The best estimate I could perform, to 3DP for the transition state, is r1 = r2 = 0.907 Å. The plot of Internuclear Distance vs Time at this value is pretty flat, but not quite perfect, as seen in Figure 2.

[[File:jp2517indistvstime.PNG|thumb|250 px|Figure 2: Internuclear Distance vs Time at a less accurate value for the transition state]]

[[File:jp2517indistvstime.PNG|thumb|250 px|Figure 3: Internuclear Distance vs Time at a more accurate value for the transition state]]

There is a better way to do this, which provides a more accurate answer, is to use the MEP (Minimum Energy Pathway). Normally, the molecule carries its momentum over when it passes the stationary point, resulting in oscillation. However, with MEP enabled, momenta and velocities are reset to zero after each infinitesimally small time step. Thus, if the molecule ends up on the transition state, a minima, it will subsequently have zero momentum and velocity in the next time step, given that the gradient at the point is zero. Thus, the molecule remains at the stationary point, and zooming in on the end of the molecule's trajectory can give us a value for r1 = r2. From this method, the value can be estimated, to 6DP, as 0.907742 Å and the corresponding plot of Internuclear Distance vs Time confirms this (Figure 3)

This value differs from the estimate value as it is more accurate, and the trajectory calculated is different in that it no longer oscillates, due to momentum and velocity being reset after each infinitesimally small time step. The trajectory instead stops when the molecule reaches the minimum point (Figure 4)

[[File:jp2517trajectory.PNG|thumb|200 px|Figure 4: MEP used to find the transition state position]]

MEP calculations show the lowest energy pathway for the reaction, whilst dynamics analysis shows the reaction pathway. The MEP can provide the result of the location for the transition state location in terms of distances between the three atoms.

{{fontcolor|red|(Good illutstration.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 21:24, 23 May 2019 (BST)}}

===Question 3===

 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" border=1
! p1 !! p2 !! Etot !! Reactive? !! Trajectory !! Description of the dynamics
|-
| -1.25 || -2.5 || -99 || Yes || [[File:jp2517Traj1.PNG|300 px]] || Collides smoothly and reacts
|-
| -1.5 || -2.0 || -100 || No || [[File:jp2517Traj2.PNG|300 px]] || Bounce off unreacted after moving together slowly
|-
| -1.5 || -2.5 || -99 || Yes || [[File:jp2517Traj3.PNG|300 px]] || Molecules collide and react, moving away from each other whilst oscillating
|-
| -2.5 || -5.0 || -85 || No || [[File:jp2517Traj4.PNG|300 px]] || Atoms collide and a few cycles of the central atom being exchanged take place, before the system ends up the same as the reactants were
|-
| -2.5 || -5.2 || -83 || Yes || [[File:jp2517Traj5.PNG|300 px]] || Molecules collide and exchange the central atom twice before moving away, having reacted
|}

The interesting conclusion from the table is that simply adding more energy to the reactants is not sufficient to guarantee a reaction. For example, the molecules react when their momenta are -1.5 and -2.5, but not when their momenta are increased to -2.5 and -5.0. This shows that the molecules also need to be in the correct vibrational state when they collide, otherwise the energy will be converted to forms not involved in bonding, such as kinetic or thermal.

Too much energy and they will simply bounce off each other, unreacted.

===Question 4===

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

Transition State Theory posits that there is a quasi-equilibrium between products and reactants at the transition state, in which molecules with sufficient energy to reach the activation barrier move between semi-reactants and semi-products before falling down one side of the transition state into reactants or products.

Its main assumptions are that:

1. Atomic nuclei behave according to classical mechanics, and that quantum tunnelling effects are negligible. Thus, that the reaction does not occur unless molecules collide with enough energy to form the transition structure.

2. Each intermediate is long-lived enough to reach a Boltzmann Distribution of energies before continuing. If they are too short-lived, the momentum of the reaction trajectory can carry over to influence the selectivity of the reaction.

3. That the reactants pass precisely over the saddle point, which is a decent approximation at low temperatures but falls down at higher temperatures as molecules will have more energy than is needed to surpass the activation barrier.

Given these assumptions, predicted reaction rate values will be greater than experimental ones as the reactions will be less efficient in reality.

{{fontcolor|red|(Good, any reference?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 13:22, 23 May 2019 (BST)}}

==Exercise 2==

===Question 1===

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

The reaction of F + H2 to HF + H is exothermic. This is shown on the figure in that the products have lower potential energy (in essence, they are more stable) than the reactants. Applying Hammond's Postulate, we can estimate that the transition state will be early

The reaction of HF + H to F + H2 is endothermic. As shown on the diagram, the reactants are at higher energy than the products, which makes sense according to the bond strengths of the chemical species involved. The HF bond is much stronger than the H2 bond. The activated complex is most likely to be closer in structure and energy to the products than the reactants, making it a late transition state, according to Hammond's Postulate.

{{fontcolor|red|(Nice and clear.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 21:24, 23 May 2019 (BST)}}

===Question 2===

 Locate the approximate position of the transition state 

F + H2 to HF + H

There are plenty of minima along the minimum energy pathway, however the one which corresponds to the transition state (saddle point) will be the one closest to the sharp drop in energy, which represents the energy pathway of the reaction after it has overcome the activation barrier. By using the MEP method, the transition state was estimated to lie at r1 = 1.808 Å and r2 = 0.745 Å, as confirmed by a plot of Internuclear distance vs time, which shows a flat line.

[[File:jp2517Exothermic.PNG|300 px|Figure 5: ]][[File:jp2517Contourplotfexo.PNG|300 px|Figure 6: ]][[File:jp2517Indistfexo.PNG|300 px|Figure 7: ]]

HF + H to F + H2

Given that this reaction is the same as the first one, just reversed in direction, the position of the transition state will be the same value but with r1 and r2 swapped, hence r1 = 0.745 Å and r2 = 1.808 Å

[[File:Endothermic.PNG|300 px|Figure 8: ]][[File:jp2517Contourplotfendo.PNG|300 px|Figure 9: ]][[File:jp2517Indistfendo.PNG|300 px|Figure 10: ]]

{{fontcolor|red|(Good plots.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 21:24, 23 May 2019 (BST)}}

===Question 3===

 Report the activation energy for both reactions 

Common sense says that the activation energy for the endothermic direction will be much larger than that of the exothermic direction, due to the lateness of the transition state in the endothermic direction requiring much more energy to reach. For the endothermic reaction of F + H2 to HF + H, the activation energy can be measured by using an MEP calculation starting at a structure neighbouring the transition state; just far enough away that it spontaneously falls down the energy well (to the exothermic products), given that the structure is no longer quite at the transition state. From this, a graph of energy vs time yielded the activation energy for the reaction.

Energy of transition state: -103.752 kJmol-1

Energy of HF + H: -133.827 kJmol-1

Activation Energy (Endothermic) = +30.1 kJmol-1 to 1DP

[[File:jp2517Actenergyendo.PNG|400 px]]

For the Exothermic Direction, the activation energy will be very small on account of the transition state being early.

Energy of H2 + F: -103.95 kJmol-1

Activation Energy (Exothermic): +0.2 kJmol-1 to 1DP

[[File:jp2517Actenergyexo.PNG|400 px]]

===Question 4===

A set of initial conditions that results in a reactive trajectory for F + H2 is as follows:

Atom A = F, Atom B = H, Atom C = H

r1 = 2.8 Å, r2 = 0.8 Å

AB Momentum = -2.5, BC Momentum = -2.5

The momentum vs time graph for the above conditions, which produce a reaction, is as follows:

[[File:Jp2517Momentumvstime.PNG|400 px]]

As can be interpreted from the graph, the oscillating H2 molecule approaches the F atom, appears to react before springing back and returning to the other H atom, then falling back to react to form HF. This is a good demonstration of a temporary equilibrium between reactants and products formed at the transition state. The orange line shows the vibrations of the middle atom, which become dampened as it approaches and reacts with the fluorine atom. Meanwhile, the Fluorine atom (blue line) did not previously show much oscillation but does once reacted, as energy has been transferred to it.

 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 

Given that this is an exothermic reaction, and that energy is conserved overall, some energy must be released into the reaction (to influence the products) and/or the surroundings. Potential energy has been exchanged for kinetic energy, which could be confirmed experimentally by calorimetry; observation of a rise in temperature as the reaction progresses. This is due to the energy released into the system/surroundings by the formation of the stronger H-F bond. These findings are shown in the image below, in which the total energy of the system remains constant while energy is converted between kinetic and potential energy.

[[File:Jp2517Energyq4.PNG|400 px]]

===Question 5===

 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 

Simulations were run with Atom A = F, Atom B = H and Atom C = H, starting at r1 = 1.4 Å and r2 = 0.74 Å, AB Momentum = -0.5 and changing BC Momentum between -3 and 3. It can be seen from this configuration that AB (pFH) Momentum corresponds to the translational energy of the system and BC Momentum (pHH) corresponds to the vibrational energy of the system.

It was observed that, for the exothermic reaction, low values of vibrational energy (BC Momentum) slowed the reaction or prevented it from happening.

When pHH was increased to -0.8, the reaction still occurs, however the system lingers at equilibrium longer, with plenty of recrossing of the transition barrier.

For the reverse, endothermic reaction of H + HF to H2 + F, it was observed that the reaction still happened despite very low values of p(HH), ie, translational energy, provided that there were adequate levels of p(FH), ie, vibrational energy.

Combining this with the findings of the simulations for the exothermic reaction, it can be seen that vibrational energy is much more influential in deciding the success/failure of an endothermic reaction, one which has a late transition state, than translational energy. Conversely, in the exothermic reaction, translational energy seemed to be the most important factor. It's also worth noting that the reactions seem to be more 'efficient', that is to say that there is less time spent in equilibrium at the transition state, oscillating between products and reactants before finally falling either side to one of them, when the correct form of energy for the type of transition state (early vs late) is provided.

These findings are summarised by a series of rules proposed by Hungarian Physicist Polanyi, which relate preference for an early or late transition state to the distribution of vibrational or translational energy available for the reaction of highest efficiency. Therefore, translational energy is better at promoting a reaction in which the transition state is early, and vibrational energy for if the transition state is late. [1]

The two figures above support Polanyi's Rules; from the left image to the right image, the AB momentum of the exothermic reaction (which is the translational energy) increases from -0.5 to -0.8, and this increase in translational energy causes the reaction to happen in the right image where it didn't previously.

[[File:Jp2517Support1.PNG|400 px]][[File:Jp2517Support2.PNG|400 px]]

The two figures underneath further support Polanyi's Rule for the endothermic case, where from left to right, vibrational energy increases, and the system goes from a non-reaction to a reaction.

[[File:Jp2517Support3.PNG|400 px]][[File:Jp2517Support4.PNG|400 px]]

Polanyi's Rules generally apply to most cases, but don't work all the time, as there are some exceptions. In the two images below, from left to right, increasing the translational energy from 0 to 3 in the exothermic case should cause a non-reaction to react, but it actually causes a reaction to not react.

[[File:Jp2517Against1.PNG|400 px]][[File:Jp2517Against2.PNG|400 px]]

Thus, the conclusion we can draw is that Polanyi's Rules are generally quite accurate but have a few exceptions, especially when going to higher levels of either form of energy. They can be treated more like guidelines.

{{fontcolor|red|(Well done, I think your report shows good understanding of this concept.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 13:23, 23 May 2019 (BST)}}

==References==

1. Zhang, Z., Zhou, Y., Zhang, D. H., Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction, The Journal of Physical Chemistry Letters, 2012, 3 (23), pp 3416–3419

MRD:01366863

2019-05-23T16:37:40Z

Cq3417:

== Molecular Reaction Dynamics ==

=== Introduction ===

By calculating and simulating the molecular reaction dynamic trajectory, the reaction of a diatomic molecule and an atom will be analysed. The information of the chemical kinetics, such as the transition state, the activation barrier and energetics of the reaction will be obtained and help analysing the reaction system.

=== EXERCISE 1 : H + H2 System ===
==== Q1. 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 structure is a saddle point in the potential energy surface.

Saddle point is one of the three kinds of stationary points, which include local minimum, local maximum and saddle point.

For function f(x,y):

Stationary point: the partial derivatives fx = 0 and fy = 0.

Minimum: the second derivatives (fxxfyy - f2xy) > 0 and fxx > 0.

Saddle Point: (fxxfyy - f2xy) < 0

{{fontcolor|red|(Good, but I think you can describe it in detail rather than simply put on the equations.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 17:37, 23 May 2019 (BST)}}

[[File:Saddle_point_ly617.png | thumb | left | 500px | The Contour page that shows a transition state (saddle point)]]

{{fontcolor|red|(I think a surface plot is better.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 17:37, 23 May 2019 (BST)}}

==== Q2. Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory. ====

[[File:Y2C10.png]]

Set r1 = r2 = 0.95, p1 = p2 = 0 :

[[File:0.95_ly617.png|thumb|500px|left|A Contour plot of H + H2 system]]

[[File:Distance_vs_time_ly617.png|thumb|centre|500px|A plot of internuclear distance vs. time of H + H2 system]]

rts = 0.908

The line A-B and B-C separates at r = 0.908, which means the internuclear distances between A-B and B-C are the same, indicating the existence of the transition state.

{{fontcolor|red|(Any more comment with the internuclear distance vs time plot?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 17:37, 23 May 2019 (BST)}}

==== Q3. Comment on how the mep and the trajectory you just calculated differ. ====

Set r1 = rts + 0.01 = 0.918, r2 = rts = 0.908, p1 = p2 = 0 :

[[File:q3_ly617.png|thumb|left|500px|A Contour plot of the system using MEP]] [[File:Q3_2_ly617.png|thumb|500px|centre|A Contour plot of the system using dynamics]]

The minimum energy path (MEP) is a special trajectory that shows a slow motion. The momenta and velocity are set to 0, hence the kinetic energy as well (Ek = p2 / 2m = mv2/2). Thereby, in the MEP plot the vibrational motion (vibrating wavy line) was not shown. The dynamic method describes a realistic motion of atoms where the energy is conserved, hence the vibrational motion can be seen in the dynamics plot.

{{fontcolor|red|(Your description is not clear here.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 17:37, 23 May 2019 (BST)}}

==== Q4. 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" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Plot
|-
| -1.25 || -2.5 || -99.018 || Reactive || Molecule AB reacts with atom C. Atom C gets close to molecule AB and passes through transition state, giving product molecule BC and atom A with BC vibrating. || [[File:Q4_1_ly617.png | 350px]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || Atom C gets closer but does not react with Molecule AB successfully since not enough energy to overcome the transition state energy barrier. || [[File:Q4_2_ly617.png | 350px]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || Molecule AB react with atom C. Atom C gets close to molecule AB and passes through transition state, giving product molecule BC and atom A with BC vibrating. || [[File:Q4_3_ly617.png | 350px]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || Atom C gets closer and reacts with Molecule AB to form molecule BC, which exists shortly and reacts with atom A to give atom C and molecule AB. The process finally ends up with the original molecule AB and atom C being reformed instead of the product. This behaviour violates the assumption of ' the molecular system that crosses the transition states in direction of product can not return and reform reactants' hence the reaction doesn't work. || [[File:Q4_4_ly617.png | 350px]]
|-
| -2.5 || -5.2 || -83.415 || Reactive || Atom C gets close to molecule AB and crosses the transition state, they react to give product molecule BC and atom A. Shortly after, molecule BC divides and molecule AB forms again. The process repeats several times, the transition state has been recrossed and molecule BC and atom A are formed at the end. || [[File:Q4_5_ly617.png | 350px]]
|}

Conclusion:
When the internuclear distances r1 and r2 stay constant, the momenta and the energy can decide whether a reaction can undergo or not. The higher the kinetic energy is, the more likelihood for the reaction to be reactive. Though -1.5 < r1 < -0.8 and p2 = -2.5 is stated to be the reactive range, the parameters outside the range are able to provide a reactive condition.
The formation of the product is not permanent and the reformation of the reactants is possible.

==== Q5. State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ====

Assumptions:

1. Molecular system that have crossed the transition state in the direction of products cannot return around and reform reactants.

2. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation.

3. Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzman laws.

According to the fourth and fifth row of the previous table, it can be seen that in these reactions the transition state has been crossed repeatedly, though the fourth reaction is unreactive while the fifth is reactive. This behaviour violates the first assumption listed above. The experimental rate of reaction could be lower than the theoretical rate since in some situations the recrossing of transition state may occur, and it will lower the rate of reacting because certain part of the reactants do not transform into product successfully.

{{fontcolor|red|(Good, any reference?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 17:37, 23 May 2019 (BST)}}

=== EXERCISE 2 : F - H - H System ===

==== Q6. 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? Locate the approximate position of the transition state. ====

Bond energy of H-H = +436 kJ/mol

Bond energy of H-F = +565 kJ/mol

{| class="wikitable" border=1
! Reaction Undergo !! Bond Breaking and Making !! Energy Change !! Energetics
|-
| F + H2 → H + HF || Break H-H bond, form new H-F bond || ΔE = +436 - 565 = -129 kJ/mol || Exothermic
|-
| H + HF → F + H2 || Break H-F bond, form new H-H bond || ΔE = +565 - 436 = +129 kJ/mol || Endothermic
|}

These two reactions are actually the same except for their opposite reaction direction (the second reaction is the reverse reaction of the first one), hence there should only be one transition state.

'''F + H2 :'''

Set A and B as H, C as F. Set r1 = 0.745, r2 = 1.811, p1 = p2 = 0.

[[File:Surface_Plot_FH2_ly617.png|thumb|left|500px|The surface plot of the F + H2 reaction]][[File:Q6distance1_ly617.png|centre|400px|The plot of internuclear distance vs. time of the F + H2 reaction]]
[[File:Contour_Plot_FH2_ly617.png|thumb|left|500px|The Contour plot of the F + H2 reaction]]

It can be seen that from the surface plot, from the reactant entry channel to product exit channel, the spot goes off a slope which means the reaction releases energy as heat in this process, hence it is exothermic. The distances between A-B and B-C are constant.
Transition state: r1 = 0.745, r2 = 1.811.

'''H + HF :'''

Set A as F, set B and C as H. Set r1 = 0.745 (now r1 = rH-F), r2 = 1.811, p1 = p2 = 0.

[[File:Surface_Plot_HHF_ly617.png|thumb|left|400px|The surface plot of the H + HF reaction]]
[[File:Q62.png|thumb|left|400px|The Contour plot of the H + HF reaction]]

{{fontcolor|red|(I can't see the saddle point on this plot.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 17:37, 23 May 2019 (BST)}}

{{fontcolor|red|(Like you said before, they are sharing same transition state and the reaction is reversible. Why you changed r1 here?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 17:37, 23 May 2019 (BST)}}

It can be seen that from the surface plot, from the reactant entry channel to product exit channel, the spot climbs up a slope which means the reaction need to absorb energy to continue this process, hence it is endothermic.

Transition state: r1 = 0.745, r2 = 1.811. The plot does not match the transition state perfectly and it may be due to the difference bond length between the reactant H-F (about 0.917) and the previous reactant H-H.

==== Q7. Report the activation energy for both reactions. ====

'''F + H2 → H + HF :'''

r1 = 0.745, r2 = 1.90, p1 = p2 = 0.

Activation energy = 104.0 - 103.8 = 0.22 kcal/mol

{{fontcolor|red|(Good, but how to get this plot?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 17:37, 23 May 2019 (BST)}}

[[File:AE1_ly617.png|frame|left|The plot of energy vs. time of F + H2 reaction by dynamics method]]

'''H + HF → F + H2 :'''

r1 = 0.745, r2 = 1.75, p1 = p2 = 0.

Activation energy = 133.76 - 103.52 = 30.24 kcal/mol

{{fontcolor|red|(Again, I think you need to describe more here.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 17:37, 23 May 2019 (BST)}}

[[File:Ea2_ly617.png|frame|left|The plot of energy vs. time of H + HF reaction by MEP]]

==== Q8. 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. ====

Set A and B as H, C as F, set r1 = 0.745, r2 = 1.9, p1 = -0.2, p2 = -0.8.

[[File:Contour_Plot_ly617.png|thumb|left|400px|The Contour plot of the F + H2 reaction]][[File:Momenta_Plot_ly617.png|centre|400px|The plot of momenta vs. time of the F + H2 reaction]]
[[File:Q8_Surface_Plot_ly617.png|thumb|left|500px|The surface plot of the F + H2 reaction]]

The reaction is exothermic. It can be seen from the surface plot that the potential energy decreases along the reaction coordination. Since the energy is conserved, the kinetic energy increases and is channeled into the vibration and rotation of HF molecule. The temperature of the reaction system goes up with the kinetic energy.

{{fontcolor|red|(channelled?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 17:37, 23 May 2019 (BST)}}

{{fontcolor|red|(Ok, but you have three plots here, any more comments on them?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 17:37, 23 May 2019 (BST)}}

==== Q9. 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. ====

Using IR chemiluminescence, the energy distribution of HF can be measured.

By the Hammond's postulate, for exothermic reactions, the transition state favourably resembles the reactants which is closer in energy, hence an early transition state observed. In this case, the reactants with larger distribution of transitional energy will be more likely to cross the barrier successfully, and generally resulting in a vibrational excitation of the product.

As for endothermic reactions, transition state resembles product, a late transition state would be observed and the vibrational energy contributes to cross the barrier more effectively than the transitional energy does, and it
leads to high transitional excitation of product.

p1 (momentum of molecule bond A-B) represents the vibrational energy, p2 represents the transitional energy.

{{fontcolor|red|(Any reference here?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 17:37, 23 May 2019 (BST)}}

'''F + H2 (early transition state) :'''

{| class="wikitable" border=1
! Dominant Mode !! r1 !! r2 !! p1 !! p2 !! Contour Plot
|-
| Transitional || 0.74 || 1.8 || -0.7 || -8 || [[File:Q9trans_ly617.png|frame|The Contour Plot of higher transitional energy of F + H2]]
|-
| Vibrational || 0.74 || 1.8 || -3 || -0.5 || [[File:Q9vib_ly617.png|frame|The Contour Plot of higher vibrational energy of F + H2]]
|}

'''H + HF (late transition state) :'''

{| class="wikitable" border=1
! Dominant Mode !! r1 !! r2 !! p1 !! p2 !! Contour Plot
|-
| Transitional || 0.74 || 1.8 || -1.5 || -0.5 || [[File:Q9trans2_ly617.png|frame|The Contour Plot of higher transitional energy of H + HF]]
|-
| Vibrational || 0.92 || 1.81 || -10 || -0.25 || [[File:Q9vib2_ly617.png|frame|The Contour Plot of higher vibrational energy of H + HF]]
|}

Hence, the reaction with early transition state favours more transitional energy, and the reaction with a late transition state favours more vibrational energy.

{{fontcolor|red|(It is better to give detailed explanation on your experimental data rather than put on them and left few or no comments.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 17:37, 23 May 2019 (BST)}}

{{fontcolor|red|(Overall, I think you need to pay more attention on your description of this concept to show us your understanding and effort.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 17:37, 23 May 2019 (BST)}}

=== Reference ===

1. J. I. Steinfeld, J. S. Francisco, W. L. Hase, Chemical Kinetic and Dynamics (1st Edition), Prentice-Hall, 1998, p293-311.

MRD:jp2517

2019-05-23T12:23:55Z

Cq3417:

== Wiki Page J Powell ==
==Exercise 1==
===Question 1===
 1a) How is the transition state mathematically defined on a Potential Energy Surface diagram?

On a Potential Energy Surface (PES) diagram, the transition state is defined as a saddle point - that is to say, a point in 3D space which is a minimum in one direction, and a maximum orthogonal to it.

 1b) How can the transition state be identified?

To define the saddle point, two new functions, orthogonal to each other, need to be defined. Q1 and Q2 will be parallel with the line of the minimum and maximum at the saddle point (hence why they are orthogonal to each other). Q2 and Q1 represent 45 degree rotations (it just happens that it is 45 degrees in this case, it could be a different angle) of the AB and BC axes, namely, that Q2 is the line tangential to the maximum potential energy along the minimum energy pathway, while Q1 is the line perpendicular to this, also passing through the maximum potential energy along the minimum energy pathway. The fact that they are orthogonal to each other means they provide a good basis set for all points. This setup is shown in Figure 1

[[File:jp2517contourplot.PNG|right|thumb|400 px|Figure 1: A contour plot with the orthogonal coordinate system|400 px]]

To mathematically locate the saddle point, it's necessary to find the first-order derivatives along Q1 and Q2, to locate the stationary points. In a saddle point, these will both occur at the same value. Second-order differentiation can confirm that a maximum has been found on Q2 and a maximum on Q1

 1c) How can it be distinguished from the local energy minimum of the potential energy surface?

While a local energy minimum may be a minimum compared to some points around it, it may not be the global minimum. However, the local minimum can be distinguished from the global energy minimum by differentiating to find all the stationary points along a line, and comparing the values for V at which these occur. The global minimum will occur at the lowest value for V out of all the minima.

===Question 2===

 2. Finding a best estimate of transition state position 

Hammond's postulate states that in a reaction which goes between products and reactants with a transition state, the transition state will be most structurally similar to either the reactants or products, based on which it is closer to in energy. However, this only really works if the reactants and products are very different in energy. In this example, the reactants and products are identical in energy, therefore the transition state will be exactly halfway between the two, making it the structural average of both of them.

We can therefore say that r1 = r2 at the transition state, meaning that on the contour plot, the transition state lies somewhere along the diagonal line going from bottom left to top right.

From visual inspection of the contour plot, it can be estimated that this occurs somewhere around r1 = r2 = 0.9 Å. Use of a plot of Internuclear Distance against Time can confirm that the point has been found, as when the molecules are put precisely at the transition state, they will sit there and nothing will happen, thus the plot of Internuclear Distance vs Time will show two flat lines. If the molecules are even slightly off the transition state point, they will oscillate back and forth, leading to a sinusoidal curve in the Internuclear Distance vs Time plot.

The less-efficient way to find the transition state point is to iteratively solve for r1 and r2 by trying different values for each and obtaining a progressively flatter line. The best estimate I could perform, to 3DP for the transition state, is r1 = r2 = 0.907 Å. The plot of Internuclear Distance vs Time at this value is pretty flat, but not quite perfect, as seen in Figure 2.

[[File:jp2517indistvstime.PNG|thumb|250 px|Figure 2: Internuclear Distance vs Time at a less accurate value for the transition state]]

[[File:jp2517indistvstime.PNG|thumb|250 px|Figure 3: Internuclear Distance vs Time at a more accurate value for the transition state]]

There is a better way to do this, which provides a more accurate answer, is to use the MEP (Minimum Energy Pathway). Normally, the molecule carries its momentum over when it passes the stationary point, resulting in oscillation. However, with MEP enabled, momenta and velocities are reset to zero after each infinitesimally small time step. Thus, if the molecule ends up on the transition state, a minima, it will subsequently have zero momentum and velocity in the next time step, given that the gradient at the point is zero. Thus, the molecule remains at the stationary point, and zooming in on the end of the molecule's trajectory can give us a value for r1 = r2. From this method, the value can be estimated, to 6DP, as 0.907742 Å and the corresponding plot of Internuclear Distance vs Time confirms this (Figure 3)

This value differs from the estimate value as it is more accurate, and the trajectory calculated is different in that it no longer oscillates, due to momentum and velocity being reset after each infinitesimally small time step. The trajectory instead stops when the molecule reaches the minimum point (Figure 4)

[[File:jp2517trajectory.PNG|thumb|200 px|Figure 4: MEP used to find the transition state position]]

MEP calculations show the lowest energy pathway for the reaction, whilst dynamics analysis shows the reaction pathway. The MEP can provide the result of the location for the transition state location in terms of distances between the three atoms.

===Question 3===

 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" border=1
! p1 !! p2 !! Etot !! Reactive? !! Trajectory !! Description of the dynamics
|-
| -1.25 || -2.5 || -99 || Yes || [[File:jp2517Traj1.PNG|300 px]] || Collides smoothly and reacts
|-
| -1.5 || -2.0 || -100 || No || [[File:jp2517Traj2.PNG|300 px]] || Bounce off unreacted after moving together slowly
|-
| -1.5 || -2.5 || -99 || Yes || [[File:jp2517Traj3.PNG|300 px]] || Molecules collide and react, moving away from each other whilst oscillating
|-
| -2.5 || -5.0 || -85 || No || [[File:jp2517Traj4.PNG|300 px]] || Atoms collide and a few cycles of the central atom being exchanged take place, before the system ends up the same as the reactants were
|-
| -2.5 || -5.2 || -83 || Yes || [[File:jp2517Traj5.PNG|300 px]] || Molecules collide and exchange the central atom twice before moving away, having reacted
|}

The interesting conclusion from the table is that simply adding more energy to the reactants is not sufficient to guarantee a reaction. For example, the molecules react when their momenta are -1.5 and -2.5, but not when their momenta are increased to -2.5 and -5.0. This shows that the molecules also need to be in the correct vibrational state when they collide, otherwise the energy will be converted to forms not involved in bonding, such as kinetic or thermal.

Too much energy and they will simply bounce off each other, unreacted.

===Question 4===

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

Transition State Theory posits that there is a quasi-equilibrium between products and reactants at the transition state, in which molecules with sufficient energy to reach the activation barrier move between semi-reactants and semi-products before falling down one side of the transition state into reactants or products.

Its main assumptions are that:

1. Atomic nuclei behave according to classical mechanics, and that quantum tunnelling effects are negligible. Thus, that the reaction does not occur unless molecules collide with enough energy to form the transition structure.

2. Each intermediate is long-lived enough to reach a Boltzmann Distribution of energies before continuing. If they are too short-lived, the momentum of the reaction trajectory can carry over to influence the selectivity of the reaction.

3. That the reactants pass precisely over the saddle point, which is a decent approximation at low temperatures but falls down at higher temperatures as molecules will have more energy than is needed to surpass the activation barrier.

Given these assumptions, predicted reaction rate values will be greater than experimental ones as the reactions will be less efficient in reality.

{{fontcolor|red|(Good, any reference?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 13:22, 23 May 2019 (BST)}}

==Exercise 2==

===Question 1===

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

The reaction of F + H2 to HF + H is exothermic. This is shown on the figure in that the products have lower potential energy (in essence, they are more stable) than the reactants. Applying Hammond's Postulate, we can estimate that the transition state will be early

The reaction of HF + H to F + H2 is endothermic. As shown on the diagram, the reactants are at higher energy than the products, which makes sense according to the bond strengths of the chemical species involved. The HF bond is much stronger than the H2 bond. The activated complex is most likely to be closer in structure and energy to the products than the reactants, making it a late transition state, according to Hammond's Postulate.

===Question 2===

 Locate the approximate position of the transition state 

F + H2 to HF + H

There are plenty of minima along the minimum energy pathway, however the one which corresponds to the transition state (saddle point) will be the one closest to the sharp drop in energy, which represents the energy pathway of the reaction after it has overcome the activation barrier. By using the MEP method, the transition state was estimated to lie at r1 = 1.808 Å and r2 = 0.745 Å, as confirmed by a plot of Internuclear distance vs time, which shows a flat line.

[[File:jp2517Exothermic.PNG|300 px|Figure 5: ]][[File:jp2517Contourplotfexo.PNG|300 px|Figure 6: ]][[File:jp2517Indistfexo.PNG|300 px|Figure 7: ]]

HF + H to F + H2

Given that this reaction is the same as the first one, just reversed in direction, the position of the transition state will be the same value but with r1 and r2 swapped, hence r1 = 0.745 Å and r2 = 1.808 Å

[[File:Endothermic.PNG|300 px|Figure 8: ]][[File:jp2517Contourplotfendo.PNG|300 px|Figure 9: ]][[File:jp2517Indistfendo.PNG|300 px|Figure 10: ]]

===Question 3===

 Report the activation energy for both reactions 

Common sense says that the activation energy for the endothermic direction will be much larger than that of the exothermic direction, due to the lateness of the transition state in the endothermic direction requiring much more energy to reach. For the endothermic reaction of F + H2 to HF + H, the activation energy can be measured by using an MEP calculation starting at a structure neighbouring the transition state; just far enough away that it spontaneously falls down the energy well (to the exothermic products), given that the structure is no longer quite at the transition state. From this, a graph of energy vs time yielded the activation energy for the reaction.

Energy of transition state: -103.752 kJmol-1

Energy of HF + H: -133.827 kJmol-1

Activation Energy (Endothermic) = +30.1 kJmol-1 to 1DP

[[File:jp2517Actenergyendo.PNG|400 px]]

For the Exothermic Direction, the activation energy will be very small on account of the transition state being early.

Energy of H2 + F: -103.95 kJmol-1

Activation Energy (Exothermic): +0.2 kJmol-1 to 1DP

[[File:jp2517Actenergyexo.PNG|400 px]]

===Question 4===

A set of initial conditions that results in a reactive trajectory for F + H2 is as follows:

Atom A = F, Atom B = H, Atom C = H

r1 = 2.8 Å, r2 = 0.8 Å

AB Momentum = -2.5, BC Momentum = -2.5

The momentum vs time graph for the above conditions, which produce a reaction, is as follows:

[[File:Jp2517Momentumvstime.PNG|400 px]]

As can be interpreted from the graph, the oscillating H2 molecule approaches the F atom, appears to react before springing back and returning to the other H atom, then falling back to react to form HF. This is a good demonstration of a temporary equilibrium between reactants and products formed at the transition state. The orange line shows the vibrations of the middle atom, which become dampened as it approaches and reacts with the fluorine atom. Meanwhile, the Fluorine atom (blue line) did not previously show much oscillation but does once reacted, as energy has been transferred to it.

 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 

Given that this is an exothermic reaction, and that energy is conserved overall, some energy must be released into the reaction (to influence the products) and/or the surroundings. Potential energy has been exchanged for kinetic energy, which could be confirmed experimentally by calorimetry; observation of a rise in temperature as the reaction progresses. This is due to the energy released into the system/surroundings by the formation of the stronger H-F bond. These findings are shown in the image below, in which the total energy of the system remains constant while energy is converted between kinetic and potential energy.

[[File:Jp2517Energyq4.PNG|400 px]]

===Question 5===

 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 

Simulations were run with Atom A = F, Atom B = H and Atom C = H, starting at r1 = 1.4 Å and r2 = 0.74 Å, AB Momentum = -0.5 and changing BC Momentum between -3 and 3. It can be seen from this configuration that AB (pFH) Momentum corresponds to the translational energy of the system and BC Momentum (pHH) corresponds to the vibrational energy of the system.

It was observed that, for the exothermic reaction, low values of vibrational energy (BC Momentum) slowed the reaction or prevented it from happening.

When pHH was increased to -0.8, the reaction still occurs, however the system lingers at equilibrium longer, with plenty of recrossing of the transition barrier.

For the reverse, endothermic reaction of H + HF to H2 + F, it was observed that the reaction still happened despite very low values of p(HH), ie, translational energy, provided that there were adequate levels of p(FH), ie, vibrational energy.

Combining this with the findings of the simulations for the exothermic reaction, it can be seen that vibrational energy is much more influential in deciding the success/failure of an endothermic reaction, one which has a late transition state, than translational energy. Conversely, in the exothermic reaction, translational energy seemed to be the most important factor. It's also worth noting that the reactions seem to be more 'efficient', that is to say that there is less time spent in equilibrium at the transition state, oscillating between products and reactants before finally falling either side to one of them, when the correct form of energy for the type of transition state (early vs late) is provided.

These findings are summarised by a series of rules proposed by Hungarian Physicist Polanyi, which relate preference for an early or late transition state to the distribution of vibrational or translational energy available for the reaction of highest efficiency. Therefore, translational energy is better at promoting a reaction in which the transition state is early, and vibrational energy for if the transition state is late. [1]

The two figures above support Polanyi's Rules; from the left image to the right image, the AB momentum of the exothermic reaction (which is the translational energy) increases from -0.5 to -0.8, and this increase in translational energy causes the reaction to happen in the right image where it didn't previously.

[[File:Jp2517Support1.PNG|400 px]][[File:Jp2517Support2.PNG|400 px]]

The two figures underneath further support Polanyi's Rule for the endothermic case, where from left to right, vibrational energy increases, and the system goes from a non-reaction to a reaction.

[[File:Jp2517Support3.PNG|400 px]][[File:Jp2517Support4.PNG|400 px]]

Polanyi's Rules generally apply to most cases, but don't work all the time, as there are some exceptions. In the two images below, from left to right, increasing the translational energy from 0 to 3 in the exothermic case should cause a non-reaction to react, but it actually causes a reaction to not react.

[[File:Jp2517Against1.PNG|400 px]][[File:Jp2517Against2.PNG|400 px]]

Thus, the conclusion we can draw is that Polanyi's Rules are generally quite accurate but have a few exceptions, especially when going to higher levels of either form of energy. They can be treated more like guidelines.

{{fontcolor|red|(Well done, I think your report shows good understanding of this concept.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 13:23, 23 May 2019 (BST)}}

==References==

1. Zhang, Z., Zhou, Y., Zhang, D. H., Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction, The Journal of Physical Chemistry Letters, 2012, 3 (23), pp 3416–3419

MRD:jp2517

2019-05-23T12:22:43Z

Cq3417:

== Wiki Page J Powell ==
==Exercise 1==
===Question 1===
 1a) How is the transition state mathematically defined on a Potential Energy Surface diagram?

On a Potential Energy Surface (PES) diagram, the transition state is defined as a saddle point - that is to say, a point in 3D space which is a minimum in one direction, and a maximum orthogonal to it.

 1b) How can the transition state be identified?

To define the saddle point, two new functions, orthogonal to each other, need to be defined. Q1 and Q2 will be parallel with the line of the minimum and maximum at the saddle point (hence why they are orthogonal to each other). Q2 and Q1 represent 45 degree rotations (it just happens that it is 45 degrees in this case, it could be a different angle) of the AB and BC axes, namely, that Q2 is the line tangential to the maximum potential energy along the minimum energy pathway, while Q1 is the line perpendicular to this, also passing through the maximum potential energy along the minimum energy pathway. The fact that they are orthogonal to each other means they provide a good basis set for all points. This setup is shown in Figure 1

[[File:jp2517contourplot.PNG|right|thumb|400 px|Figure 1: A contour plot with the orthogonal coordinate system|400 px]]

To mathematically locate the saddle point, it's necessary to find the first-order derivatives along Q1 and Q2, to locate the stationary points. In a saddle point, these will both occur at the same value. Second-order differentiation can confirm that a maximum has been found on Q2 and a maximum on Q1

 1c) How can it be distinguished from the local energy minimum of the potential energy surface?

While a local energy minimum may be a minimum compared to some points around it, it may not be the global minimum. However, the local minimum can be distinguished from the global energy minimum by differentiating to find all the stationary points along a line, and comparing the values for V at which these occur. The global minimum will occur at the lowest value for V out of all the minima.

===Question 2===

 2. Finding a best estimate of transition state position 

Hammond's postulate states that in a reaction which goes between products and reactants with a transition state, the transition state will be most structurally similar to either the reactants or products, based on which it is closer to in energy. However, this only really works if the reactants and products are very different in energy. In this example, the reactants and products are identical in energy, therefore the transition state will be exactly halfway between the two, making it the structural average of both of them.

We can therefore say that r1 = r2 at the transition state, meaning that on the contour plot, the transition state lies somewhere along the diagonal line going from bottom left to top right.

From visual inspection of the contour plot, it can be estimated that this occurs somewhere around r1 = r2 = 0.9 Å. Use of a plot of Internuclear Distance against Time can confirm that the point has been found, as when the molecules are put precisely at the transition state, they will sit there and nothing will happen, thus the plot of Internuclear Distance vs Time will show two flat lines. If the molecules are even slightly off the transition state point, they will oscillate back and forth, leading to a sinusoidal curve in the Internuclear Distance vs Time plot.

The less-efficient way to find the transition state point is to iteratively solve for r1 and r2 by trying different values for each and obtaining a progressively flatter line. The best estimate I could perform, to 3DP for the transition state, is r1 = r2 = 0.907 Å. The plot of Internuclear Distance vs Time at this value is pretty flat, but not quite perfect, as seen in Figure 2.

[[File:jp2517indistvstime.PNG|thumb|250 px|Figure 2: Internuclear Distance vs Time at a less accurate value for the transition state]]

[[File:jp2517indistvstime.PNG|thumb|250 px|Figure 3: Internuclear Distance vs Time at a more accurate value for the transition state]]

There is a better way to do this, which provides a more accurate answer, is to use the MEP (Minimum Energy Pathway). Normally, the molecule carries its momentum over when it passes the stationary point, resulting in oscillation. However, with MEP enabled, momenta and velocities are reset to zero after each infinitesimally small time step. Thus, if the molecule ends up on the transition state, a minima, it will subsequently have zero momentum and velocity in the next time step, given that the gradient at the point is zero. Thus, the molecule remains at the stationary point, and zooming in on the end of the molecule's trajectory can give us a value for r1 = r2. From this method, the value can be estimated, to 6DP, as 0.907742 Å and the corresponding plot of Internuclear Distance vs Time confirms this (Figure 3)

This value differs from the estimate value as it is more accurate, and the trajectory calculated is different in that it no longer oscillates, due to momentum and velocity being reset after each infinitesimally small time step. The trajectory instead stops when the molecule reaches the minimum point (Figure 4)

[[File:jp2517trajectory.PNG|thumb|200 px|Figure 4: MEP used to find the transition state position]]

MEP calculations show the lowest energy pathway for the reaction, whilst dynamics analysis shows the reaction pathway. The MEP can provide the result of the location for the transition state location in terms of distances between the three atoms.

===Question 3===

 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" border=1
! p1 !! p2 !! Etot !! Reactive? !! Trajectory !! Description of the dynamics
|-
| -1.25 || -2.5 || -99 || Yes || [[File:jp2517Traj1.PNG|300 px]] || Collides smoothly and reacts
|-
| -1.5 || -2.0 || -100 || No || [[File:jp2517Traj2.PNG|300 px]] || Bounce off unreacted after moving together slowly
|-
| -1.5 || -2.5 || -99 || Yes || [[File:jp2517Traj3.PNG|300 px]] || Molecules collide and react, moving away from each other whilst oscillating
|-
| -2.5 || -5.0 || -85 || No || [[File:jp2517Traj4.PNG|300 px]] || Atoms collide and a few cycles of the central atom being exchanged take place, before the system ends up the same as the reactants were
|-
| -2.5 || -5.2 || -83 || Yes || [[File:jp2517Traj5.PNG|300 px]] || Molecules collide and exchange the central atom twice before moving away, having reacted
|}

The interesting conclusion from the table is that simply adding more energy to the reactants is not sufficient to guarantee a reaction. For example, the molecules react when their momenta are -1.5 and -2.5, but not when their momenta are increased to -2.5 and -5.0. This shows that the molecules also need to be in the correct vibrational state when they collide, otherwise the energy will be converted to forms not involved in bonding, such as kinetic or thermal.

Too much energy and they will simply bounce off each other, unreacted.

===Question 4===

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

Transition State Theory posits that there is a quasi-equilibrium between products and reactants at the transition state, in which molecules with sufficient energy to reach the activation barrier move between semi-reactants and semi-products before falling down one side of the transition state into reactants or products.

Its main assumptions are that:

1. Atomic nuclei behave according to classical mechanics, and that quantum tunnelling effects are negligible. Thus, that the reaction does not occur unless molecules collide with enough energy to form the transition structure.

2. Each intermediate is long-lived enough to reach a Boltzmann Distribution of energies before continuing. If they are too short-lived, the momentum of the reaction trajectory can carry over to influence the selectivity of the reaction.

3. That the reactants pass precisely over the saddle point, which is a decent approximation at low temperatures but falls down at higher temperatures as molecules will have more energy than is needed to surpass the activation barrier.

Given these assumptions, predicted reaction rate values will be greater than experimental ones as the reactions will be less efficient in reality.

{{fontcolor|red|(Good, any reference?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 13:22, 23 May 2019 (BST)}}

==Exercise 2==

===Question 1===

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

The reaction of F + H2 to HF + H is exothermic. This is shown on the figure in that the products have lower potential energy (in essence, they are more stable) than the reactants. Applying Hammond's Postulate, we can estimate that the transition state will be early

The reaction of HF + H to F + H2 is endothermic. As shown on the diagram, the reactants are at higher energy than the products, which makes sense according to the bond strengths of the chemical species involved. The HF bond is much stronger than the H2 bond. The activated complex is most likely to be closer in structure and energy to the products than the reactants, making it a late transition state, according to Hammond's Postulate.

===Question 2===

 Locate the approximate position of the transition state 

F + H2 to HF + H

There are plenty of minima along the minimum energy pathway, however the one which corresponds to the transition state (saddle point) will be the one closest to the sharp drop in energy, which represents the energy pathway of the reaction after it has overcome the activation barrier. By using the MEP method, the transition state was estimated to lie at r1 = 1.808 Å and r2 = 0.745 Å, as confirmed by a plot of Internuclear distance vs time, which shows a flat line.

[[File:jp2517Exothermic.PNG|300 px|Figure 5: ]][[File:jp2517Contourplotfexo.PNG|300 px|Figure 6: ]][[File:jp2517Indistfexo.PNG|300 px|Figure 7: ]]

HF + H to F + H2

Given that this reaction is the same as the first one, just reversed in direction, the position of the transition state will be the same value but with r1 and r2 swapped, hence r1 = 0.745 Å and r2 = 1.808 Å

[[File:Endothermic.PNG|300 px|Figure 8: ]][[File:jp2517Contourplotfendo.PNG|300 px|Figure 9: ]][[File:jp2517Indistfendo.PNG|300 px|Figure 10: ]]

===Question 3===

 Report the activation energy for both reactions 

Common sense says that the activation energy for the endothermic direction will be much larger than that of the exothermic direction, due to the lateness of the transition state in the endothermic direction requiring much more energy to reach. For the endothermic reaction of F + H2 to HF + H, the activation energy can be measured by using an MEP calculation starting at a structure neighbouring the transition state; just far enough away that it spontaneously falls down the energy well (to the exothermic products), given that the structure is no longer quite at the transition state. From this, a graph of energy vs time yielded the activation energy for the reaction.

Energy of transition state: -103.752 kJmol-1

Energy of HF + H: -133.827 kJmol-1

Activation Energy (Endothermic) = +30.1 kJmol-1 to 1DP

[[File:jp2517Actenergyendo.PNG|400 px]]

For the Exothermic Direction, the activation energy will be very small on account of the transition state being early.

Energy of H2 + F: -103.95 kJmol-1

Activation Energy (Exothermic): +0.2 kJmol-1 to 1DP

[[File:jp2517Actenergyexo.PNG|400 px]]

===Question 4===

A set of initial conditions that results in a reactive trajectory for F + H2 is as follows:

Atom A = F, Atom B = H, Atom C = H

r1 = 2.8 Å, r2 = 0.8 Å

AB Momentum = -2.5, BC Momentum = -2.5

The momentum vs time graph for the above conditions, which produce a reaction, is as follows:

[[File:Jp2517Momentumvstime.PNG|400 px]]

As can be interpreted from the graph, the oscillating H2 molecule approaches the F atom, appears to react before springing back and returning to the other H atom, then falling back to react to form HF. This is a good demonstration of a temporary equilibrium between reactants and products formed at the transition state. The orange line shows the vibrations of the middle atom, which become dampened as it approaches and reacts with the fluorine atom. Meanwhile, the Fluorine atom (blue line) did not previously show much oscillation but does once reacted, as energy has been transferred to it.

 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 

Given that this is an exothermic reaction, and that energy is conserved overall, some energy must be released into the reaction (to influence the products) and/or the surroundings. Potential energy has been exchanged for kinetic energy, which could be confirmed experimentally by calorimetry; observation of a rise in temperature as the reaction progresses. This is due to the energy released into the system/surroundings by the formation of the stronger H-F bond. These findings are shown in the image below, in which the total energy of the system remains constant while energy is converted between kinetic and potential energy.

[[File:Jp2517Energyq4.PNG|400 px]]

===Question 5===

 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 

Simulations were run with Atom A = F, Atom B = H and Atom C = H, starting at r1 = 1.4 Å and r2 = 0.74 Å, AB Momentum = -0.5 and changing BC Momentum between -3 and 3. It can be seen from this configuration that AB (pFH) Momentum corresponds to the translational energy of the system and BC Momentum (pHH) corresponds to the vibrational energy of the system.

It was observed that, for the exothermic reaction, low values of vibrational energy (BC Momentum) slowed the reaction or prevented it from happening.

When pHH was increased to -0.8, the reaction still occurs, however the system lingers at equilibrium longer, with plenty of recrossing of the transition barrier.

For the reverse, endothermic reaction of H + HF to H2 + F, it was observed that the reaction still happened despite very low values of p(HH), ie, translational energy, provided that there were adequate levels of p(FH), ie, vibrational energy.

Combining this with the findings of the simulations for the exothermic reaction, it can be seen that vibrational energy is much more influential in deciding the success/failure of an endothermic reaction, one which has a late transition state, than translational energy. Conversely, in the exothermic reaction, translational energy seemed to be the most important factor. It's also worth noting that the reactions seem to be more 'efficient', that is to say that there is less time spent in equilibrium at the transition state, oscillating between products and reactants before finally falling either side to one of them, when the correct form of energy for the type of transition state (early vs late) is provided.

These findings are summarised by a series of rules proposed by Hungarian Physicist Polanyi, which relate preference for an early or late transition state to the distribution of vibrational or translational energy available for the reaction of highest efficiency. Therefore, translational energy is better at promoting a reaction in which the transition state is early, and vibrational energy for if the transition state is late. [1]

The two figures above support Polanyi's Rules; from the left image to the right image, the AB momentum of the exothermic reaction (which is the translational energy) increases from -0.5 to -0.8, and this increase in translational energy causes the reaction to happen in the right image where it didn't previously.

[[File:Jp2517Support1.PNG|400 px]][[File:Jp2517Support2.PNG|400 px]]

The two figures underneath further support Polanyi's Rule for the endothermic case, where from left to right, vibrational energy increases, and the system goes from a non-reaction to a reaction.

[[File:Jp2517Support3.PNG|400 px]][[File:Jp2517Support4.PNG|400 px]]

Polanyi's Rules generally apply to most cases, but don't work all the time, as there are some exceptions. In the two images below, from left to right, increasing the translational energy from 0 to 3 in the exothermic case should cause a non-reaction to react, but it actually causes a reaction to not react.

[[File:Jp2517Against1.PNG|400 px]][[File:Jp2517Against2.PNG|400 px]]

Thus, the conclusion we can draw is that Polanyi's Rules are generally quite accurate but have a few exceptions, especially when going to higher levels of either form of energy. They can be treated more like guidelines.

{{fontcolor|red|(Well done, I think your report shows good understanding of this concept.)}}

==References==

1. Zhang, Z., Zhou, Y., Zhang, D. H., Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction, The Journal of Physical Chemistry Letters, 2012, 3 (23), pp 3416–3419

MRD:frances

2019-05-22T19:20:22Z

Cq3417:

==Exercise 1==

===Dynamics on a Potential Energy Surface Diagram===

====Definition of parameters====

AB distance = r1

BC distance = r2

On a potential energy surface diagram, the transition state is mathematically defined as maximum point along a minimum energy pathway. The maximum along the minimum energy trajectory is shown in the first image below from a side on view. It can mathematically defined as a saddle point, which is a local minimum in one direction but a local maximum in a direction orthogonal to this direction. From this definition, taking a first derivative of the inflection point of the reaction path, ∂V('<nowiki/>r1)/∂r1 and ∂V(r2)/∂r2 will give 0 for both derivatives, indicating a turning point. This can be distinguished from a local minimum point by taking the second derivative to the inflection point or the derivative of orthogonal vectors, q1 and q2. We can define two additional important vectors, q1 and q2. These are diagonal vectors relative to r1 and r2 and the second partial derivatives with respect to these values gives specific information about the nature of the transition state point. One derivative result will indicate a maximum point and the derivative in the orthogonal direction will indicate a minimum point i.e. ∂V('''q1''')/∂'''q1''' <0 and ∂V('''q2''')/∂'''q2''' >0. In one direction the second derivative will be positive indicating a minimum point, however in the other it will be negative indicating a maximum point. This is the definition of the saddle point.

[[File:SurfacePlot3817.png|thumb|center|Potential Energy Plot rotated to clearly show the saddle point in the middle as flat surface either side of which is a reduction in potential energy.]]

[[File: Animation23817.png|thumb|center|Contour plot annotated to show the two newly defined vectors q1 and q2.]]

===Trajectories from r1 = r2: locating the transition state===

The transition state position ('''rts''') can be estimated as 0.91 (2sf) Å. This value can be obtained through running a MEP type calculation with '''r1''' = '''r2''', and '''p1''' = '''p2''' = 0.0. This allows you to calculate the transition state point as at 0 momentum the trajectory oscillates on the ridge. Arbitrarily selecting '''r1''' = 0.6, the system increases to a distance of 0.9077 before plateauing, indicating the transition state distance has been reached. The system comes to rest at this point.

[[File:TransitionState3817.png|thumb|center|Internuclear Distance vs Time plot calculated using an MEP calculation to show the transition state inter-nuclear distance as 0.91 (2sf) Å.]]

[[File:zoomedtransitionstate3817.png|thumb|center|Internuclear Distance vs Time plot calculated using an MEP calculation to show the transition state inter-nuclear distance.]]

{{fontcolor1|red|(I think including the counter plot here is better.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 20:18, 22 May 2019 (BST)}}

=== Reaction Path Trajectories using MEP and Dynamic Calculations ===

Running a trajectory calculation with the parameters'''r1''' = 0.9077, i.e. transition state distance, '''r2''' = 0.9087 and '''p1''' = '''p2''' = 0, gives two slightly different trajectories depending on which type of calculation you use. In the case of the MEP calculation, the reaction trajectory essentially runs along the minimum with no oscillations, in contrast to the dynamic calculation which oscillates along the minimum energy trajectory. These are both shown below. The reason they differ in appearance can be explained using a sphere rolling on a potential energy surface analogy. The MEP takes recordings at infinitesimally small speeds which means the sphere does not take into account any previous momentum. As a result, it has no momentum to 'roll' up the other potential energy surface and as such, the trajectory follows a straight line. In contrast, the sphere representing the dynamic calculation is aware of its previous momentum it has gained from starting higher on the potential energy surface. As a result, when it reaches the minimum, it is able to slightly roll up the opposite surface. Hence, an oscillating trajectory is observed.

[[File:dynamicsurfaceplot.png|thumb|center|Dynamic Plot: A trajectory superimposed on a surface plot calculated using the dynamic calculation method.]]

[[File:mepsurfaceplot.png|thumb|center|MEP Plot: A trajectory superimposed on a surface plot calculated using the MEP calculation method.]]

{{fontcolor1|red|(I think here counter plot is better than surface plot.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 20:18, 22 May 2019 (BST)}}

The final values values of momentum and internuclear distance can be recorded at large t.

r1(2.5)= 8.97 Å

r2(2.5)= 0.68 Å

p1(2.5)= 2.48

p2(2.5)= 0.90

If we change the parameters so that r1 = rts and r2 = rts +0.01 instead, the only difference between the plots is that the lines switch over, the magnitude of the values remains the same. This is because we are dealing with a completely symmetrical system and so the momentum and internuclear distances will be the same irrespective of whether we are considering the exchange of a H atom in either case.

The graphs of Internuclear distance vs time and momentum vs time, from which the above values were calculated from, are shown below.

[[File:distance1.png|thumb|center|Dynamic Plot: Internuclear distance Vs Time graph with an extended number of steps to determine a value of internuclear distance at large t.]]

[[File:momentum2.png|thumb|center|Dynamic Plot: Momentum Vs Time graph with an extended number of steps to determine a value of momentum associated with the respective distances at large t.]]

===Reactive and unreactive trajectories ===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99.12 || Yes || [[File:contour1.png|thumb|center|]] || Molecule AB approaches Atom C with enough momentum and energy to overcome the activation barrier and so a reaction occurs.
|-
| -1.5 || -2.0 || -100.46 || No || [[File:contour2.png|thumb|center]] || Molecule AB approaches C however does not collide with enough energy to overcome the activation energy barrier therefore, no reaction occurs and the two particle entities separate in opposite directions.
|-
| -1.5 || -2.5 || -98.98 || Yes || [[File:contour3.png|thumb|center]] || Atom C and bonded atoms A and B come together and collide with enough momentum and energy for a reaction to occur, hence the bond between A and C breaks and a bond is formed between atoms C and D. Following the collision, the particles then part in opposite directions with the newly formed BC molecule oscillating.
|-
| -2.5 || -5.0 || -84.96 ||No|| [[File:contour4.png|thumb|center|]] || Molecule AB approaches atom C, highly oscillating. It appears that molecule AB dissociates at the transition state, as the internnuclear distance of BC appears to be less than that of AB, however molecule AB reforms and molecule AB separates from C. Again, molecule AB is highly oscillating.
|-
| -2.5 || -5.2 || -83.42 || Yes || [[File:contour7.png|thumb|center|]]|| Molecule AB approaches Atom C, highly oscillating. Upon collision, atom B appears to assiciate with atom C, followed by an oscillation back between atom A and C before finally bonding to atom C.
|}

{{fontcolor1|red|(The counter plot of your first example in this table is a little bit ambiguous.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 20:18, 22 May 2019 (BST)}}

This table is showing that total energy from the reaction is not the only factor that dictates whether or not a reaction occurs. In other words, not all trajectories starting with the same positions with values of momenta greater than those resulting in a previous successful collision, will result in a successful collision reaction. Having enough kinetic energy to overcome the activation barrier is not the only condition that needs to be satisfied in order to have a successful reaction. There are many other factors to take into account such as the ratio of contribution of vibrational and translational energy to the total energy that will impact whether or not a reaction occurs. These factors are discussed later. We can also see in the case of the penultimate row value, that barrier recrossing is possible. Therefore the average number of collisions that would be predicated to occurr successfully, based on energetics alone, would in fact be greater than that observed experimentally.

===Transition State Theory===

Transition State theory is the idea that rates of reaction can be studied by examining activated complexes at the saddle point. These complexes are in quasi-equilibrium with the reactant molecules.

The main assumptions of transition state theory are as follows:

1. The theory assumes that each reactive intermediate is long-lived enough that a Boltzmann distribution of energies is reached before the reaction then continues onto the next step. However, this can often lead to limitations of the TST, as if the intermediates are very short lived, the reaction does not have time to re-equilibriate its energy levels. In this case the transition state will have residual momentum from its reaction trajectory that is able to push it over to the products. In addition, the model also assumes that all reactant molecules are distributes according to the Boltzman distribution. As a result, the predictions made by the model may overestimate the number of successful reactions.

2. Another assumption is that the theory is based on atomic nuclei behaving according to classical mechanics. Unless atoms react with enough energy to overcome the activation energy, a reaction does not occur. However, the phenomenon of quantum mechanical tunnelling results in the possibility of particles still crossing the barrier even if the collision does not occur with enough energy to cross the activation barrier. However, this phenomenon will be relatively small.

3. The third assumption is that the reaction system will pass over the lowest energy saddle point on the potential energy surface. This is sufficient for low temperatures however at high temp, molecules populate higher vibrational modes therefore motions are more complex and so collisions may occur with more complex transition states that do not occur on the simple saddle point Transition State Theory predicts. As a result, the energetics and dynamics associated with the prediction may not be valid.

4. It also assumes that once the transition state has been crossed and products of the reaction formed, there is no way for the entities to then recross the transition state without altering the reaction conditions in a way that is then favourable for them to do so. In other words, it does not take into account barrier recrossing, hence it predicts a greater number of reactions occurr to completion of products than is observed in reality. Hence, the predicted rate constant for the reaction will be greater than the observed.

{{fontcolor1|red|(Good statement, any reference?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 20:18, 22 May 2019 (BST)}}

How will Transition State Theory predictions for reaction rate values compare with experimental values?

As can be seen from the table and as has been touched on above in point 4, the experimental values obtained do not follow transition state theory, as even in cases where the values predict a reaction to have enough energy to overcome the activation barrier, a reaction does not always occur. If the reactions did all obey the TST, all reactions would occur as they all have energy to overcome the activation barrier. As a result, predicted reaction rate values will be greater than those observed.

==Exercise 2==

The F + H2 reaction is exothermic as the reactant energies i.e. an atom of F and H2 molecule are higher than the product energies on the potential energy surface graph. However in the case of HF + H, the reaction is endothermic as the reactants are lower in energy than the products and hence the reaction proceeds in a downhill direction.

Endothermic reaction of HF + H:

[[File:endothermic3817.png]]

Exothermic reaction of F and H2:

[[File:exothermic3817.png]]

The blue dot represents the staring reactants on the potential energy surface.

The energetics of the reaction relate to the bond strength as the net energy taken in or released in the reaction is indicative of the relative strength of the reactant and product bonds. Bond formation is exothermic, whereas bond breaking is endothermic. Therefore, if the reactant bond strength is greater than the product bond strength, a greater amount of energy is required to break the bond than is released upon formation of the new bond. Hence, an endothermic reaction indicates that reactant bond strength is greater than product bond strength and an exothermic reaction indicates that reactant bond strength is weaker than product bond strength, as more energy has been released upon formation of the stronger product bond than was inputted to break the weaker reactant bond. This can be further illustrated by considering the initial H2 + H reaction where the energy of the reactants and products are equal as the bond being formed and created is identical, therefore taking in and releasing equal energy. As a result the potential energy surfaces for reactants and products are equal and hence no net energy is released. This is shown below.

[[File:hh3817.png]]

=== Locating the approximate position of the transition state ===

The transition state is a saddle point at which, if given no additional energy or momentum, the reaction trajectory will remain stationary.

a) For the reaction F + H2, the approximate position of the transition state can be located by finding the combination of AB and BC distances at which, given no momentum and no energy, the particle trajectory will remain stationary. This can be located when:

A = F

B = H

C = H

AB distance = 1.808 Å

BC distance = 0.705 Å

This is shown graphically plotted below.

[[File:ContourH2.png]]

The small variation of the trajectory position is shown below, a zoomed in version of the above graph. This graph shows the precision of the approximate position, there is very small deviation from the point.

[[File:contou2.png]]

b) For the reaction HF + H, the approximate position of the transition state will simply be the same coordinates associated with the opposite atoms as the transition state is the same regardless of the direction in which the reaction proceeds. Plotting these on a contour plot for the reverse reaction gives the same result as above, a stationary trajectory point, indicating that we area at a saddle point. This graph is shown below where,

A = H

B = H

C = F

AB distance = 0.705 Å

BC distance = 1.808 Å

[[File:contourhf.png]]

===Estimating the activation energy ===

To obtain an approximation for the activation energy of the reaction, an MEP calculation can be performed (with 5000 steps) on a structure with bond lengths adjusted slightly from the transition state values. By plotting an energy V time graph, shown below, it is possible to estimate the activation energy for the reaction as the difference between the total energy and the potential energy at high values of t.

a) For the H2 + F reaction, this can be approximated as +0.255 kcal/mol.

The graphical representation of this is shown below:

[[File:h23817.png]]

b) For the HF + H reaction, this can be approximated from taking away the potential energy of the HF molecule from the total (transition state) energy. Therefore, there are an additional energy values required for this calculation, the potential energy of the HF molecule.

Using the values below:

HF potential energy = -134.03
Total energy/ Transition State energy = -103.75
H2 energy = -104.02

The activation energy of the reaction HF + H can be approximated as + 30.01 kcal/mol.

{{fontcolor1|red|(Any diagrams here?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 20:18, 22 May 2019 (BST)}}

=== Reaction Dynamics ===

A set of conditions that results in a reactive trajectory for F + H2 are as follows:

A = F, B = H, C= H

AB distance = 1.4

BC distance = 1

AB momentum = BC momentum = 0

The mechanism of release of reaction energy is as follows. This reaction is an exothermic reaction, hence energy is released to the surroundings. The mechanism of release is the conversion of excess potential energy stored in the bonds of the reactants converted to kinetic energy. Molecules are constantly switching energy between potential and kinetic and hence this is a dynamic process. Upon reaction completion, energy is converted to translational energy of the H atom leaving the HF molecule as well as vibrational energy of the molecules. Therefore, even though the kinetic energy is constantly switching between vibrational and translational energy, the time average kinetic energy has increased following reaction completion.

This is shown in the graph below, where the time average kinetic energy increases, and the time average potential energy decreases:

[[File:Kineticenergy3817.png]]

In addition, an analysis of the momentum time graph shows that energy released from forming the stronger H-F bond is released and converted into a large increase in BC oscillation. The AB value however tends to a fixed value as the product molecule and particle move away from each other and so do not influence each others momentum. This means the translational energy represented by the orange line, remains constant. The HF bond will continue to oscillate at a steady rate however, with the increase in oscillation proportional to the energy released from the chemical bonds and hence the increase in temperature of the reaction.

[[File:Momenta23817.png]]

In summary, chemical Potential energy is being converted to Kinetic energy.

Experimentally this can be confirmed through measuring the temperature of the reaction. Kinetic energy is all energy an object possesses due to motion and so in general the kinetic energy of the molecules increases due to the release of energy, therefore the temperature of the surroundings of an exothermic reaction will increase. As a result, using calorimetry, you can measure the increase in kinetic energy in the surroundings.

=== 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.===

Contributions from the energy occur as vibrational and translational.

'''In the exothermic reaction of F + H2 '''

pFH = the translational contribution to the energy
pHH = the vibrational contribution to the energy

A number of simulations were performed in order to determine the empirical rules for the reaction.

First, initially setting the conditions of a reaction of F + H2 as rHH = 0.74, with a momentum pFH = -0.5. Testing a range of values of pFH between -3 and 3, it is observed that at values close to the limits, the reaction is successful and goes to completion. However, at values in between, it is observed that either the reaction is not successful or there is a significant amount of barrier recrossing. Therefore if the vibrational energy of the molecule approaching is too low, the reaction will not be successful. As we have been told, we are placing a much greater amount of energy in the system, much greater than that required to surpass the activation energy barrier. The fact that some of the reactions even with this magnitude of energy are unsuccessful is further proof against TST and indication that the type of energy (vibrational or transnational) supplied to the system is important.

Using the increasing the momentum of the FH bond to -0.8 and reducing the momentum of the HH vibration to 0.1 you observe the reaction does occur, however there appears to be barrier recrossing.

This reaction is exothermic and hence has an early transition state.

This shows that increasing the translational contribution even slightly for an early transition state has a greater impact on the success of the reaction than any changes to vibrational contributions.

'''In the endothermic reaction FH + H '''

This reaction is endothermic and hence has a late transition state. In this case, it was observed that adding vibrational energy to the reaction (ie in the form of momentum pFH) was of greatest importance. A reaction still occurred despite low values of pHH, if there was still a high enough magnitude of vibrational energy.

This can be illustrated in the below contour plot which was generated using parameters of rFH = 0.9, rHH= 1.3 pHH = 0.2 and pFH = 12. In this case, the translational contribution to the energy is very small, yet the vibrational contribution is very large and hence the reaction still occurs. This illustrates the rules.

[[File:endoplot.png |thumb|center| A contour plot showing a successful endothermic reaction with vibrational energy of 12 despite the small translational energy of 0.2. This is a clear illustration of Polanyi's rules where the vibrational energy is the most important factor for an endothermic reaction with a late transition state.]]

'''Polanyi's rules'''

Polanyi's rules state that a certain distribution of energy in reactants and products is more favoured to promote a successful reaction. The specifics of this distribution are dependant on whether the reaction has an early or late transition state. Using Hammond's postulate, it can be assumed that an exothermic reaction has an early transition state in which the transition state closely resembles the reactants and in an endothermic reaction the transition state energy most closely resembles the products.

Polanyi's rules relate concept of early or late transition state to the most favourable distribution of energy for the highest efficiency of reaction[1]. These rules state that vibrational energy is more efficient in promoting a late transition state reaction than translational energy, whereas the reverse is true for an early transition state reaction.

[[For an early transition state]] (exothermic reaction), translational energy is most important in determining the success of the reaction.

An example of this case is shown below. Starting with reaction parameters: rFH = 2 rHH = 0.74 and pHH = -1.5, pHH = -0.26, a reaction does not occur. However, only slightly increasing the FH momentum, related to the translational energy, from -0.26 to -0.28 results in a successful reaction. This proves that translational energy is most important in determining the success of a reaction. This process is illustrated below in the two graphs where just a small change in translational energy results in the reaction overcoming the transition state barrier and the reaction occurring successfully. As was indicated above, it took much greater changes to the vibrational motion in the range pHH = -3 to 3 in order to achieve a successful reaction. This further confirms the notion that translational energy is most important factor for the efficiency of the exothermic reaction.

[[File:Notworking3817.png|thumb|center| A contour plot showing an unsuccessful reaction with translational energy of -0.26.]]

[[File:Working3817.png |thumb|center| A contour plot showing a successful reaction with translational energy of -0.28.]]

[[For a late transition state]] (endothermic reaction), vibrational energy is most important in determining the success of the reaction.

These rules are in line with the observations made above about the importance of the translation and vibrational energy distribution for the exothermic reaction of F + H2 and the reverse endothermic reaction of FH + H.

These rules are successful to a certain degree, however, varying the conditions even slightly may result in an unsuccessful reaction even where the rules may predict that a reaction would be successful.

'''Conclusion'''

Polanyi's rules state that for a late transition state (endothermic reaction), vibrational energy is most important in determining the success of the reaction, yet for an early transition state (exothermic reaction), translational energy is most important in determining the success of the reaction. These rules are useful to a certain extent as they provide a quick guide, much like with TST, as to whether or not a reaction will be successful and the parameters needed in order to improve reactions efficiency. However, these rules are also a broad overview and should not be taken as the whole story as often small tweaks in conditions will act against the rules. Such an example includes

{{fontcolor1|red|(Overall, well done) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 20:18, 22 May 2019 (BST)}}

==Reference==

1. J. Phys. Chem. Lett., 2012, 3 (23), pp 3416–3419

MRD:frances

2019-05-22T19:19:18Z

Cq3417:

==Exercise 1==

===Dynamics on a Potential Energy Surface Diagram===

====Definition of parameters====

AB distance = r1

BC distance = r2

On a potential energy surface diagram, the transition state is mathematically defined as maximum point along a minimum energy pathway. The maximum along the minimum energy trajectory is shown in the first image below from a side on view. It can mathematically defined as a saddle point, which is a local minimum in one direction but a local maximum in a direction orthogonal to this direction. From this definition, taking a first derivative of the inflection point of the reaction path, ∂V('<nowiki/>r1)/∂r1 and ∂V(r2)/∂r2 will give 0 for both derivatives, indicating a turning point. This can be distinguished from a local minimum point by taking the second derivative to the inflection point or the derivative of orthogonal vectors, q1 and q2. We can define two additional important vectors, q1 and q2. These are diagonal vectors relative to r1 and r2 and the second partial derivatives with respect to these values gives specific information about the nature of the transition state point. One derivative result will indicate a maximum point and the derivative in the orthogonal direction will indicate a minimum point i.e. ∂V('''q1''')/∂'''q1''' <0 and ∂V('''q2''')/∂'''q2''' >0. In one direction the second derivative will be positive indicating a minimum point, however in the other it will be negative indicating a maximum point. This is the definition of the saddle point.

[[File:SurfacePlot3817.png|thumb|center|Potential Energy Plot rotated to clearly show the saddle point in the middle as flat surface either side of which is a reduction in potential energy.]]

[[File: Animation23817.png|thumb|center|Contour plot annotated to show the two newly defined vectors q1 and q2.]]

===Trajectories from r1 = r2: locating the transition state===

The transition state position ('''rts''') can be estimated as 0.91 (2sf) Å. This value can be obtained through running a MEP type calculation with '''r1''' = '''r2''', and '''p1''' = '''p2''' = 0.0. This allows you to calculate the transition state point as at 0 momentum the trajectory oscillates on the ridge. Arbitrarily selecting '''r1''' = 0.6, the system increases to a distance of 0.9077 before plateauing, indicating the transition state distance has been reached. The system comes to rest at this point.

[[File:TransitionState3817.png|thumb|center|Internuclear Distance vs Time plot calculated using an MEP calculation to show the transition state inter-nuclear distance as 0.91 (2sf) Å.]]

[[File:zoomedtransitionstate3817.png|thumb|center|Internuclear Distance vs Time plot calculated using an MEP calculation to show the transition state inter-nuclear distance.]]

{{fontcolor1|red|(I think additionally including the counter plot here is better.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 20:18, 22 May 2019 (BST)}}

=== Reaction Path Trajectories using MEP and Dynamic Calculations ===

Running a trajectory calculation with the parameters'''r1''' = 0.9077, i.e. transition state distance, '''r2''' = 0.9087 and '''p1''' = '''p2''' = 0, gives two slightly different trajectories depending on which type of calculation you use. In the case of the MEP calculation, the reaction trajectory essentially runs along the minimum with no oscillations, in contrast to the dynamic calculation which oscillates along the minimum energy trajectory. These are both shown below. The reason they differ in appearance can be explained using a sphere rolling on a potential energy surface analogy. The MEP takes recordings at infinitesimally small speeds which means the sphere does not take into account any previous momentum. As a result, it has no momentum to 'roll' up the other potential energy surface and as such, the trajectory follows a straight line. In contrast, the sphere representing the dynamic calculation is aware of its previous momentum it has gained from starting higher on the potential energy surface. As a result, when it reaches the minimum, it is able to slightly roll up the opposite surface. Hence, an oscillating trajectory is observed.

[[File:dynamicsurfaceplot.png|thumb|center|Dynamic Plot: A trajectory superimposed on a surface plot calculated using the dynamic calculation method.]]

[[File:mepsurfaceplot.png|thumb|center|MEP Plot: A trajectory superimposed on a surface plot calculated using the MEP calculation method.]]

{{fontcolor1|red|(I think here counter plot is better than surface plot.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 20:18, 22 May 2019 (BST)}}

The final values values of momentum and internuclear distance can be recorded at large t.

r1(2.5)= 8.97 Å

r2(2.5)= 0.68 Å

p1(2.5)= 2.48

p2(2.5)= 0.90

If we change the parameters so that r1 = rts and r2 = rts +0.01 instead, the only difference between the plots is that the lines switch over, the magnitude of the values remains the same. This is because we are dealing with a completely symmetrical system and so the momentum and internuclear distances will be the same irrespective of whether we are considering the exchange of a H atom in either case.

The graphs of Internuclear distance vs time and momentum vs time, from which the above values were calculated from, are shown below.

[[File:distance1.png|thumb|center|Dynamic Plot: Internuclear distance Vs Time graph with an extended number of steps to determine a value of internuclear distance at large t.]]

[[File:momentum2.png|thumb|center|Dynamic Plot: Momentum Vs Time graph with an extended number of steps to determine a value of momentum associated with the respective distances at large t.]]

===Reactive and unreactive trajectories ===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99.12 || Yes || [[File:contour1.png|thumb|center|]] || Molecule AB approaches Atom C with enough momentum and energy to overcome the activation barrier and so a reaction occurs.
|-
| -1.5 || -2.0 || -100.46 || No || [[File:contour2.png|thumb|center]] || Molecule AB approaches C however does not collide with enough energy to overcome the activation energy barrier therefore, no reaction occurs and the two particle entities separate in opposite directions.
|-
| -1.5 || -2.5 || -98.98 || Yes || [[File:contour3.png|thumb|center]] || Atom C and bonded atoms A and B come together and collide with enough momentum and energy for a reaction to occur, hence the bond between A and C breaks and a bond is formed between atoms C and D. Following the collision, the particles then part in opposite directions with the newly formed BC molecule oscillating.
|-
| -2.5 || -5.0 || -84.96 ||No|| [[File:contour4.png|thumb|center|]] || Molecule AB approaches atom C, highly oscillating. It appears that molecule AB dissociates at the transition state, as the internnuclear distance of BC appears to be less than that of AB, however molecule AB reforms and molecule AB separates from C. Again, molecule AB is highly oscillating.
|-
| -2.5 || -5.2 || -83.42 || Yes || [[File:contour7.png|thumb|center|]]|| Molecule AB approaches Atom C, highly oscillating. Upon collision, atom B appears to assiciate with atom C, followed by an oscillation back between atom A and C before finally bonding to atom C.
|}

{{fontcolor1|red|(The counter plot of your first example in this table is a little bit ambiguous.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 20:18, 22 May 2019 (BST)}}

This table is showing that total energy from the reaction is not the only factor that dictates whether or not a reaction occurs. In other words, not all trajectories starting with the same positions with values of momenta greater than those resulting in a previous successful collision, will result in a successful collision reaction. Having enough kinetic energy to overcome the activation barrier is not the only condition that needs to be satisfied in order to have a successful reaction. There are many other factors to take into account such as the ratio of contribution of vibrational and translational energy to the total energy that will impact whether or not a reaction occurs. These factors are discussed later. We can also see in the case of the penultimate row value, that barrier recrossing is possible. Therefore the average number of collisions that would be predicated to occurr successfully, based on energetics alone, would in fact be greater than that observed experimentally.

===Transition State Theory===

Transition State theory is the idea that rates of reaction can be studied by examining activated complexes at the saddle point. These complexes are in quasi-equilibrium with the reactant molecules.

The main assumptions of transition state theory are as follows:

1. The theory assumes that each reactive intermediate is long-lived enough that a Boltzmann distribution of energies is reached before the reaction then continues onto the next step. However, this can often lead to limitations of the TST, as if the intermediates are very short lived, the reaction does not have time to re-equilibriate its energy levels. In this case the transition state will have residual momentum from its reaction trajectory that is able to push it over to the products. In addition, the model also assumes that all reactant molecules are distributes according to the Boltzman distribution. As a result, the predictions made by the model may overestimate the number of successful reactions.

2. Another assumption is that the theory is based on atomic nuclei behaving according to classical mechanics. Unless atoms react with enough energy to overcome the activation energy, a reaction does not occur. However, the phenomenon of quantum mechanical tunnelling results in the possibility of particles still crossing the barrier even if the collision does not occur with enough energy to cross the activation barrier. However, this phenomenon will be relatively small.

3. The third assumption is that the reaction system will pass over the lowest energy saddle point on the potential energy surface. This is sufficient for low temperatures however at high temp, molecules populate higher vibrational modes therefore motions are more complex and so collisions may occur with more complex transition states that do not occur on the simple saddle point Transition State Theory predicts. As a result, the energetics and dynamics associated with the prediction may not be valid.

4. It also assumes that once the transition state has been crossed and products of the reaction formed, there is no way for the entities to then recross the transition state without altering the reaction conditions in a way that is then favourable for them to do so. In other words, it does not take into account barrier recrossing, hence it predicts a greater number of reactions occurr to completion of products than is observed in reality. Hence, the predicted rate constant for the reaction will be greater than the observed.

{{fontcolor1|red|(Good statement, any reference?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 20:18, 22 May 2019 (BST)}}

How will Transition State Theory predictions for reaction rate values compare with experimental values?

As can be seen from the table and as has been touched on above in point 4, the experimental values obtained do not follow transition state theory, as even in cases where the values predict a reaction to have enough energy to overcome the activation barrier, a reaction does not always occur. If the reactions did all obey the TST, all reactions would occur as they all have energy to overcome the activation barrier. As a result, predicted reaction rate values will be greater than those observed.

==Exercise 2==

The F + H2 reaction is exothermic as the reactant energies i.e. an atom of F and H2 molecule are higher than the product energies on the potential energy surface graph. However in the case of HF + H, the reaction is endothermic as the reactants are lower in energy than the products and hence the reaction proceeds in a downhill direction.

Endothermic reaction of HF + H:

[[File:endothermic3817.png]]

Exothermic reaction of F and H2:

[[File:exothermic3817.png]]

The blue dot represents the staring reactants on the potential energy surface.

The energetics of the reaction relate to the bond strength as the net energy taken in or released in the reaction is indicative of the relative strength of the reactant and product bonds. Bond formation is exothermic, whereas bond breaking is endothermic. Therefore, if the reactant bond strength is greater than the product bond strength, a greater amount of energy is required to break the bond than is released upon formation of the new bond. Hence, an endothermic reaction indicates that reactant bond strength is greater than product bond strength and an exothermic reaction indicates that reactant bond strength is weaker than product bond strength, as more energy has been released upon formation of the stronger product bond than was inputted to break the weaker reactant bond. This can be further illustrated by considering the initial H2 + H reaction where the energy of the reactants and products are equal as the bond being formed and created is identical, therefore taking in and releasing equal energy. As a result the potential energy surfaces for reactants and products are equal and hence no net energy is released. This is shown below.

[[File:hh3817.png]]

=== Locating the approximate position of the transition state ===

The transition state is a saddle point at which, if given no additional energy or momentum, the reaction trajectory will remain stationary.

a) For the reaction F + H2, the approximate position of the transition state can be located by finding the combination of AB and BC distances at which, given no momentum and no energy, the particle trajectory will remain stationary. This can be located when:

A = F

B = H

C = H

AB distance = 1.808 Å

BC distance = 0.705 Å

This is shown graphically plotted below.

[[File:ContourH2.png]]

The small variation of the trajectory position is shown below, a zoomed in version of the above graph. This graph shows the precision of the approximate position, there is very small deviation from the point.

[[File:contou2.png]]

b) For the reaction HF + H, the approximate position of the transition state will simply be the same coordinates associated with the opposite atoms as the transition state is the same regardless of the direction in which the reaction proceeds. Plotting these on a contour plot for the reverse reaction gives the same result as above, a stationary trajectory point, indicating that we area at a saddle point. This graph is shown below where,

A = H

B = H

C = F

AB distance = 0.705 Å

BC distance = 1.808 Å

[[File:contourhf.png]]

===Estimating the activation energy ===

To obtain an approximation for the activation energy of the reaction, an MEP calculation can be performed (with 5000 steps) on a structure with bond lengths adjusted slightly from the transition state values. By plotting an energy V time graph, shown below, it is possible to estimate the activation energy for the reaction as the difference between the total energy and the potential energy at high values of t.

a) For the H2 + F reaction, this can be approximated as +0.255 kcal/mol.

The graphical representation of this is shown below:

[[File:h23817.png]]

b) For the HF + H reaction, this can be approximated from taking away the potential energy of the HF molecule from the total (transition state) energy. Therefore, there are an additional energy values required for this calculation, the potential energy of the HF molecule.

Using the values below:

HF potential energy = -134.03
Total energy/ Transition State energy = -103.75
H2 energy = -104.02

The activation energy of the reaction HF + H can be approximated as + 30.01 kcal/mol.

{{fontcolor1|red|(Any diagrams here?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 20:18, 22 May 2019 (BST)}}

=== Reaction Dynamics ===

A set of conditions that results in a reactive trajectory for F + H2 are as follows:

A = F, B = H, C= H

AB distance = 1.4

BC distance = 1

AB momentum = BC momentum = 0

The mechanism of release of reaction energy is as follows. This reaction is an exothermic reaction, hence energy is released to the surroundings. The mechanism of release is the conversion of excess potential energy stored in the bonds of the reactants converted to kinetic energy. Molecules are constantly switching energy between potential and kinetic and hence this is a dynamic process. Upon reaction completion, energy is converted to translational energy of the H atom leaving the HF molecule as well as vibrational energy of the molecules. Therefore, even though the kinetic energy is constantly switching between vibrational and translational energy, the time average kinetic energy has increased following reaction completion.

This is shown in the graph below, where the time average kinetic energy increases, and the time average potential energy decreases:

[[File:Kineticenergy3817.png]]

In addition, an analysis of the momentum time graph shows that energy released from forming the stronger H-F bond is released and converted into a large increase in BC oscillation. The AB value however tends to a fixed value as the product molecule and particle move away from each other and so do not influence each others momentum. This means the translational energy represented by the orange line, remains constant. The HF bond will continue to oscillate at a steady rate however, with the increase in oscillation proportional to the energy released from the chemical bonds and hence the increase in temperature of the reaction.

[[File:Momenta23817.png]]

In summary, chemical Potential energy is being converted to Kinetic energy.

Experimentally this can be confirmed through measuring the temperature of the reaction. Kinetic energy is all energy an object possesses due to motion and so in general the kinetic energy of the molecules increases due to the release of energy, therefore the temperature of the surroundings of an exothermic reaction will increase. As a result, using calorimetry, you can measure the increase in kinetic energy in the surroundings.

=== 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.===

Contributions from the energy occur as vibrational and translational.

'''In the exothermic reaction of F + H2 '''

pFH = the translational contribution to the energy
pHH = the vibrational contribution to the energy

A number of simulations were performed in order to determine the empirical rules for the reaction.

First, initially setting the conditions of a reaction of F + H2 as rHH = 0.74, with a momentum pFH = -0.5. Testing a range of values of pFH between -3 and 3, it is observed that at values close to the limits, the reaction is successful and goes to completion. However, at values in between, it is observed that either the reaction is not successful or there is a significant amount of barrier recrossing. Therefore if the vibrational energy of the molecule approaching is too low, the reaction will not be successful. As we have been told, we are placing a much greater amount of energy in the system, much greater than that required to surpass the activation energy barrier. The fact that some of the reactions even with this magnitude of energy are unsuccessful is further proof against TST and indication that the type of energy (vibrational or transnational) supplied to the system is important.

Using the increasing the momentum of the FH bond to -0.8 and reducing the momentum of the HH vibration to 0.1 you observe the reaction does occur, however there appears to be barrier recrossing.

This reaction is exothermic and hence has an early transition state.

This shows that increasing the translational contribution even slightly for an early transition state has a greater impact on the success of the reaction than any changes to vibrational contributions.

'''In the endothermic reaction FH + H '''

This reaction is endothermic and hence has a late transition state. In this case, it was observed that adding vibrational energy to the reaction (ie in the form of momentum pFH) was of greatest importance. A reaction still occurred despite low values of pHH, if there was still a high enough magnitude of vibrational energy.

This can be illustrated in the below contour plot which was generated using parameters of rFH = 0.9, rHH= 1.3 pHH = 0.2 and pFH = 12. In this case, the translational contribution to the energy is very small, yet the vibrational contribution is very large and hence the reaction still occurs. This illustrates the rules.

[[File:endoplot.png |thumb|center| A contour plot showing a successful endothermic reaction with vibrational energy of 12 despite the small translational energy of 0.2. This is a clear illustration of Polanyi's rules where the vibrational energy is the most important factor for an endothermic reaction with a late transition state.]]

'''Polanyi's rules'''

Polanyi's rules state that a certain distribution of energy in reactants and products is more favoured to promote a successful reaction. The specifics of this distribution are dependant on whether the reaction has an early or late transition state. Using Hammond's postulate, it can be assumed that an exothermic reaction has an early transition state in which the transition state closely resembles the reactants and in an endothermic reaction the transition state energy most closely resembles the products.

Polanyi's rules relate concept of early or late transition state to the most favourable distribution of energy for the highest efficiency of reaction[1]. These rules state that vibrational energy is more efficient in promoting a late transition state reaction than translational energy, whereas the reverse is true for an early transition state reaction.

[[For an early transition state]] (exothermic reaction), translational energy is most important in determining the success of the reaction.

An example of this case is shown below. Starting with reaction parameters: rFH = 2 rHH = 0.74 and pHH = -1.5, pHH = -0.26, a reaction does not occur. However, only slightly increasing the FH momentum, related to the translational energy, from -0.26 to -0.28 results in a successful reaction. This proves that translational energy is most important in determining the success of a reaction. This process is illustrated below in the two graphs where just a small change in translational energy results in the reaction overcoming the transition state barrier and the reaction occurring successfully. As was indicated above, it took much greater changes to the vibrational motion in the range pHH = -3 to 3 in order to achieve a successful reaction. This further confirms the notion that translational energy is most important factor for the efficiency of the exothermic reaction.

[[File:Notworking3817.png|thumb|center| A contour plot showing an unsuccessful reaction with translational energy of -0.26.]]

[[File:Working3817.png |thumb|center| A contour plot showing a successful reaction with translational energy of -0.28.]]

[[For a late transition state]] (endothermic reaction), vibrational energy is most important in determining the success of the reaction.

These rules are in line with the observations made above about the importance of the translation and vibrational energy distribution for the exothermic reaction of F + H2 and the reverse endothermic reaction of FH + H.

These rules are successful to a certain degree, however, varying the conditions even slightly may result in an unsuccessful reaction even where the rules may predict that a reaction would be successful.

'''Conclusion'''

Polanyi's rules state that for a late transition state (endothermic reaction), vibrational energy is most important in determining the success of the reaction, yet for an early transition state (exothermic reaction), translational energy is most important in determining the success of the reaction. These rules are useful to a certain extent as they provide a quick guide, much like with TST, as to whether or not a reaction will be successful and the parameters needed in order to improve reactions efficiency. However, these rules are also a broad overview and should not be taken as the whole story as often small tweaks in conditions will act against the rules. Such an example includes

{{fontcolor1|red|(Overall, well done) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 20:18, 22 May 2019 (BST)}}

==Reference==

1. J. Phys. Chem. Lett., 2012, 3 (23), pp 3416–3419

MRD:frances

2019-05-22T19:18:48Z

Cq3417:

==Exercise 1==

===Dynamics on a Potential Energy Surface Diagram===

====Definition of parameters====

AB distance = r1

BC distance = r2

On a potential energy surface diagram, the transition state is mathematically defined as maximum point along a minimum energy pathway. The maximum along the minimum energy trajectory is shown in the first image below from a side on view. It can mathematically defined as a saddle point, which is a local minimum in one direction but a local maximum in a direction orthogonal to this direction. From this definition, taking a first derivative of the inflection point of the reaction path, ∂V('<nowiki/>r1)/∂r1 and ∂V(r2)/∂r2 will give 0 for both derivatives, indicating a turning point. This can be distinguished from a local minimum point by taking the second derivative to the inflection point or the derivative of orthogonal vectors, q1 and q2. We can define two additional important vectors, q1 and q2. These are diagonal vectors relative to r1 and r2 and the second partial derivatives with respect to these values gives specific information about the nature of the transition state point. One derivative result will indicate a maximum point and the derivative in the orthogonal direction will indicate a minimum point i.e. ∂V('''q1''')/∂'''q1''' <0 and ∂V('''q2''')/∂'''q2''' >0. In one direction the second derivative will be positive indicating a minimum point, however in the other it will be negative indicating a maximum point. This is the definition of the saddle point.

[[File:SurfacePlot3817.png|thumb|center|Potential Energy Plot rotated to clearly show the saddle point in the middle as flat surface either side of which is a reduction in potential energy.]]

[[File: Animation23817.png|thumb|center|Contour plot annotated to show the two newly defined vectors q1 and q2.]]

===Trajectories from r1 = r2: locating the transition state===

The transition state position ('''rts''') can be estimated as 0.91 (2sf) Å. This value can be obtained through running a MEP type calculation with '''r1''' = '''r2''', and '''p1''' = '''p2''' = 0.0. This allows you to calculate the transition state point as at 0 momentum the trajectory oscillates on the ridge. Arbitrarily selecting '''r1''' = 0.6, the system increases to a distance of 0.9077 before plateauing, indicating the transition state distance has been reached. The system comes to rest at this point.

[[File:TransitionState3817.png|thumb|center|Internuclear Distance vs Time plot calculated using an MEP calculation to show the transition state inter-nuclear distance as 0.91 (2sf) Å.]]

[[File:zoomedtransitionstate3817.png|thumb|center|Internuclear Distance vs Time plot calculated using an MEP calculation to show the transition state inter-nuclear distance.]]

{{fontcolor1|red|(I think additionally include the counter plot here is better.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 20:18, 22 May 2019 (BST)}}

=== Reaction Path Trajectories using MEP and Dynamic Calculations ===

Running a trajectory calculation with the parameters'''r1''' = 0.9077, i.e. transition state distance, '''r2''' = 0.9087 and '''p1''' = '''p2''' = 0, gives two slightly different trajectories depending on which type of calculation you use. In the case of the MEP calculation, the reaction trajectory essentially runs along the minimum with no oscillations, in contrast to the dynamic calculation which oscillates along the minimum energy trajectory. These are both shown below. The reason they differ in appearance can be explained using a sphere rolling on a potential energy surface analogy. The MEP takes recordings at infinitesimally small speeds which means the sphere does not take into account any previous momentum. As a result, it has no momentum to 'roll' up the other potential energy surface and as such, the trajectory follows a straight line. In contrast, the sphere representing the dynamic calculation is aware of its previous momentum it has gained from starting higher on the potential energy surface. As a result, when it reaches the minimum, it is able to slightly roll up the opposite surface. Hence, an oscillating trajectory is observed.

[[File:dynamicsurfaceplot.png|thumb|center|Dynamic Plot: A trajectory superimposed on a surface plot calculated using the dynamic calculation method.]]

[[File:mepsurfaceplot.png|thumb|center|MEP Plot: A trajectory superimposed on a surface plot calculated using the MEP calculation method.]]

{{fontcolor1|red|(I think here counter plot is better than surface plot.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 20:18, 22 May 2019 (BST)}}

The final values values of momentum and internuclear distance can be recorded at large t.

r1(2.5)= 8.97 Å

r2(2.5)= 0.68 Å

p1(2.5)= 2.48

p2(2.5)= 0.90

If we change the parameters so that r1 = rts and r2 = rts +0.01 instead, the only difference between the plots is that the lines switch over, the magnitude of the values remains the same. This is because we are dealing with a completely symmetrical system and so the momentum and internuclear distances will be the same irrespective of whether we are considering the exchange of a H atom in either case.

The graphs of Internuclear distance vs time and momentum vs time, from which the above values were calculated from, are shown below.

[[File:distance1.png|thumb|center|Dynamic Plot: Internuclear distance Vs Time graph with an extended number of steps to determine a value of internuclear distance at large t.]]

[[File:momentum2.png|thumb|center|Dynamic Plot: Momentum Vs Time graph with an extended number of steps to determine a value of momentum associated with the respective distances at large t.]]

===Reactive and unreactive trajectories ===

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99.12 || Yes || [[File:contour1.png|thumb|center|]] || Molecule AB approaches Atom C with enough momentum and energy to overcome the activation barrier and so a reaction occurs.
|-
| -1.5 || -2.0 || -100.46 || No || [[File:contour2.png|thumb|center]] || Molecule AB approaches C however does not collide with enough energy to overcome the activation energy barrier therefore, no reaction occurs and the two particle entities separate in opposite directions.
|-
| -1.5 || -2.5 || -98.98 || Yes || [[File:contour3.png|thumb|center]] || Atom C and bonded atoms A and B come together and collide with enough momentum and energy for a reaction to occur, hence the bond between A and C breaks and a bond is formed between atoms C and D. Following the collision, the particles then part in opposite directions with the newly formed BC molecule oscillating.
|-
| -2.5 || -5.0 || -84.96 ||No|| [[File:contour4.png|thumb|center|]] || Molecule AB approaches atom C, highly oscillating. It appears that molecule AB dissociates at the transition state, as the internnuclear distance of BC appears to be less than that of AB, however molecule AB reforms and molecule AB separates from C. Again, molecule AB is highly oscillating.
|-
| -2.5 || -5.2 || -83.42 || Yes || [[File:contour7.png|thumb|center|]]|| Molecule AB approaches Atom C, highly oscillating. Upon collision, atom B appears to assiciate with atom C, followed by an oscillation back between atom A and C before finally bonding to atom C.
|}

{{fontcolor1|red|(The counter plot of your first example in this table is a little bit ambiguous.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 20:18, 22 May 2019 (BST)}}

This table is showing that total energy from the reaction is not the only factor that dictates whether or not a reaction occurs. In other words, not all trajectories starting with the same positions with values of momenta greater than those resulting in a previous successful collision, will result in a successful collision reaction. Having enough kinetic energy to overcome the activation barrier is not the only condition that needs to be satisfied in order to have a successful reaction. There are many other factors to take into account such as the ratio of contribution of vibrational and translational energy to the total energy that will impact whether or not a reaction occurs. These factors are discussed later. We can also see in the case of the penultimate row value, that barrier recrossing is possible. Therefore the average number of collisions that would be predicated to occurr successfully, based on energetics alone, would in fact be greater than that observed experimentally.

===Transition State Theory===

Transition State theory is the idea that rates of reaction can be studied by examining activated complexes at the saddle point. These complexes are in quasi-equilibrium with the reactant molecules.

The main assumptions of transition state theory are as follows:

1. The theory assumes that each reactive intermediate is long-lived enough that a Boltzmann distribution of energies is reached before the reaction then continues onto the next step. However, this can often lead to limitations of the TST, as if the intermediates are very short lived, the reaction does not have time to re-equilibriate its energy levels. In this case the transition state will have residual momentum from its reaction trajectory that is able to push it over to the products. In addition, the model also assumes that all reactant molecules are distributes according to the Boltzman distribution. As a result, the predictions made by the model may overestimate the number of successful reactions.

2. Another assumption is that the theory is based on atomic nuclei behaving according to classical mechanics. Unless atoms react with enough energy to overcome the activation energy, a reaction does not occur. However, the phenomenon of quantum mechanical tunnelling results in the possibility of particles still crossing the barrier even if the collision does not occur with enough energy to cross the activation barrier. However, this phenomenon will be relatively small.

3. The third assumption is that the reaction system will pass over the lowest energy saddle point on the potential energy surface. This is sufficient for low temperatures however at high temp, molecules populate higher vibrational modes therefore motions are more complex and so collisions may occur with more complex transition states that do not occur on the simple saddle point Transition State Theory predicts. As a result, the energetics and dynamics associated with the prediction may not be valid.

4. It also assumes that once the transition state has been crossed and products of the reaction formed, there is no way for the entities to then recross the transition state without altering the reaction conditions in a way that is then favourable for them to do so. In other words, it does not take into account barrier recrossing, hence it predicts a greater number of reactions occurr to completion of products than is observed in reality. Hence, the predicted rate constant for the reaction will be greater than the observed.

{{fontcolor1|red|(Good statement, any reference?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 20:18, 22 May 2019 (BST)}}

How will Transition State Theory predictions for reaction rate values compare with experimental values?

As can be seen from the table and as has been touched on above in point 4, the experimental values obtained do not follow transition state theory, as even in cases where the values predict a reaction to have enough energy to overcome the activation barrier, a reaction does not always occur. If the reactions did all obey the TST, all reactions would occur as they all have energy to overcome the activation barrier. As a result, predicted reaction rate values will be greater than those observed.

==Exercise 2==

The F + H2 reaction is exothermic as the reactant energies i.e. an atom of F and H2 molecule are higher than the product energies on the potential energy surface graph. However in the case of HF + H, the reaction is endothermic as the reactants are lower in energy than the products and hence the reaction proceeds in a downhill direction.

Endothermic reaction of HF + H:

[[File:endothermic3817.png]]

Exothermic reaction of F and H2:

[[File:exothermic3817.png]]

The blue dot represents the staring reactants on the potential energy surface.

The energetics of the reaction relate to the bond strength as the net energy taken in or released in the reaction is indicative of the relative strength of the reactant and product bonds. Bond formation is exothermic, whereas bond breaking is endothermic. Therefore, if the reactant bond strength is greater than the product bond strength, a greater amount of energy is required to break the bond than is released upon formation of the new bond. Hence, an endothermic reaction indicates that reactant bond strength is greater than product bond strength and an exothermic reaction indicates that reactant bond strength is weaker than product bond strength, as more energy has been released upon formation of the stronger product bond than was inputted to break the weaker reactant bond. This can be further illustrated by considering the initial H2 + H reaction where the energy of the reactants and products are equal as the bond being formed and created is identical, therefore taking in and releasing equal energy. As a result the potential energy surfaces for reactants and products are equal and hence no net energy is released. This is shown below.

[[File:hh3817.png]]

=== Locating the approximate position of the transition state ===

The transition state is a saddle point at which, if given no additional energy or momentum, the reaction trajectory will remain stationary.

a) For the reaction F + H2, the approximate position of the transition state can be located by finding the combination of AB and BC distances at which, given no momentum and no energy, the particle trajectory will remain stationary. This can be located when:

A = F

B = H

C = H

AB distance = 1.808 Å

BC distance = 0.705 Å

This is shown graphically plotted below.

[[File:ContourH2.png]]

The small variation of the trajectory position is shown below, a zoomed in version of the above graph. This graph shows the precision of the approximate position, there is very small deviation from the point.

[[File:contou2.png]]

b) For the reaction HF + H, the approximate position of the transition state will simply be the same coordinates associated with the opposite atoms as the transition state is the same regardless of the direction in which the reaction proceeds. Plotting these on a contour plot for the reverse reaction gives the same result as above, a stationary trajectory point, indicating that we area at a saddle point. This graph is shown below where,

A = H

B = H

C = F

AB distance = 0.705 Å

BC distance = 1.808 Å

[[File:contourhf.png]]

===Estimating the activation energy ===

To obtain an approximation for the activation energy of the reaction, an MEP calculation can be performed (with 5000 steps) on a structure with bond lengths adjusted slightly from the transition state values. By plotting an energy V time graph, shown below, it is possible to estimate the activation energy for the reaction as the difference between the total energy and the potential energy at high values of t.

a) For the H2 + F reaction, this can be approximated as +0.255 kcal/mol.

The graphical representation of this is shown below:

[[File:h23817.png]]

b) For the HF + H reaction, this can be approximated from taking away the potential energy of the HF molecule from the total (transition state) energy. Therefore, there are an additional energy values required for this calculation, the potential energy of the HF molecule.

Using the values below:

HF potential energy = -134.03
Total energy/ Transition State energy = -103.75
H2 energy = -104.02

The activation energy of the reaction HF + H can be approximated as + 30.01 kcal/mol.

{{fontcolor1|red|(Any diagrams here?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 20:18, 22 May 2019 (BST)}}

=== Reaction Dynamics ===

A set of conditions that results in a reactive trajectory for F + H2 are as follows:

A = F, B = H, C= H

AB distance = 1.4

BC distance = 1

AB momentum = BC momentum = 0

The mechanism of release of reaction energy is as follows. This reaction is an exothermic reaction, hence energy is released to the surroundings. The mechanism of release is the conversion of excess potential energy stored in the bonds of the reactants converted to kinetic energy. Molecules are constantly switching energy between potential and kinetic and hence this is a dynamic process. Upon reaction completion, energy is converted to translational energy of the H atom leaving the HF molecule as well as vibrational energy of the molecules. Therefore, even though the kinetic energy is constantly switching between vibrational and translational energy, the time average kinetic energy has increased following reaction completion.

This is shown in the graph below, where the time average kinetic energy increases, and the time average potential energy decreases:

[[File:Kineticenergy3817.png]]

In addition, an analysis of the momentum time graph shows that energy released from forming the stronger H-F bond is released and converted into a large increase in BC oscillation. The AB value however tends to a fixed value as the product molecule and particle move away from each other and so do not influence each others momentum. This means the translational energy represented by the orange line, remains constant. The HF bond will continue to oscillate at a steady rate however, with the increase in oscillation proportional to the energy released from the chemical bonds and hence the increase in temperature of the reaction.

[[File:Momenta23817.png]]

In summary, chemical Potential energy is being converted to Kinetic energy.

Experimentally this can be confirmed through measuring the temperature of the reaction. Kinetic energy is all energy an object possesses due to motion and so in general the kinetic energy of the molecules increases due to the release of energy, therefore the temperature of the surroundings of an exothermic reaction will increase. As a result, using calorimetry, you can measure the increase in kinetic energy in the surroundings.

=== 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.===

Contributions from the energy occur as vibrational and translational.

'''In the exothermic reaction of F + H2 '''

pFH = the translational contribution to the energy
pHH = the vibrational contribution to the energy

A number of simulations were performed in order to determine the empirical rules for the reaction.

First, initially setting the conditions of a reaction of F + H2 as rHH = 0.74, with a momentum pFH = -0.5. Testing a range of values of pFH between -3 and 3, it is observed that at values close to the limits, the reaction is successful and goes to completion. However, at values in between, it is observed that either the reaction is not successful or there is a significant amount of barrier recrossing. Therefore if the vibrational energy of the molecule approaching is too low, the reaction will not be successful. As we have been told, we are placing a much greater amount of energy in the system, much greater than that required to surpass the activation energy barrier. The fact that some of the reactions even with this magnitude of energy are unsuccessful is further proof against TST and indication that the type of energy (vibrational or transnational) supplied to the system is important.

Using the increasing the momentum of the FH bond to -0.8 and reducing the momentum of the HH vibration to 0.1 you observe the reaction does occur, however there appears to be barrier recrossing.

This reaction is exothermic and hence has an early transition state.

This shows that increasing the translational contribution even slightly for an early transition state has a greater impact on the success of the reaction than any changes to vibrational contributions.

'''In the endothermic reaction FH + H '''

This reaction is endothermic and hence has a late transition state. In this case, it was observed that adding vibrational energy to the reaction (ie in the form of momentum pFH) was of greatest importance. A reaction still occurred despite low values of pHH, if there was still a high enough magnitude of vibrational energy.

This can be illustrated in the below contour plot which was generated using parameters of rFH = 0.9, rHH= 1.3 pHH = 0.2 and pFH = 12. In this case, the translational contribution to the energy is very small, yet the vibrational contribution is very large and hence the reaction still occurs. This illustrates the rules.

[[File:endoplot.png |thumb|center| A contour plot showing a successful endothermic reaction with vibrational energy of 12 despite the small translational energy of 0.2. This is a clear illustration of Polanyi's rules where the vibrational energy is the most important factor for an endothermic reaction with a late transition state.]]

'''Polanyi's rules'''

Polanyi's rules state that a certain distribution of energy in reactants and products is more favoured to promote a successful reaction. The specifics of this distribution are dependant on whether the reaction has an early or late transition state. Using Hammond's postulate, it can be assumed that an exothermic reaction has an early transition state in which the transition state closely resembles the reactants and in an endothermic reaction the transition state energy most closely resembles the products.

Polanyi's rules relate concept of early or late transition state to the most favourable distribution of energy for the highest efficiency of reaction[1]. These rules state that vibrational energy is more efficient in promoting a late transition state reaction than translational energy, whereas the reverse is true for an early transition state reaction.

[[For an early transition state]] (exothermic reaction), translational energy is most important in determining the success of the reaction.

An example of this case is shown below. Starting with reaction parameters: rFH = 2 rHH = 0.74 and pHH = -1.5, pHH = -0.26, a reaction does not occur. However, only slightly increasing the FH momentum, related to the translational energy, from -0.26 to -0.28 results in a successful reaction. This proves that translational energy is most important in determining the success of a reaction. This process is illustrated below in the two graphs where just a small change in translational energy results in the reaction overcoming the transition state barrier and the reaction occurring successfully. As was indicated above, it took much greater changes to the vibrational motion in the range pHH = -3 to 3 in order to achieve a successful reaction. This further confirms the notion that translational energy is most important factor for the efficiency of the exothermic reaction.

[[File:Notworking3817.png|thumb|center| A contour plot showing an unsuccessful reaction with translational energy of -0.26.]]

[[File:Working3817.png |thumb|center| A contour plot showing a successful reaction with translational energy of -0.28.]]

[[For a late transition state]] (endothermic reaction), vibrational energy is most important in determining the success of the reaction.

These rules are in line with the observations made above about the importance of the translation and vibrational energy distribution for the exothermic reaction of F + H2 and the reverse endothermic reaction of FH + H.

These rules are successful to a certain degree, however, varying the conditions even slightly may result in an unsuccessful reaction even where the rules may predict that a reaction would be successful.

'''Conclusion'''

Polanyi's rules state that for a late transition state (endothermic reaction), vibrational energy is most important in determining the success of the reaction, yet for an early transition state (exothermic reaction), translational energy is most important in determining the success of the reaction. These rules are useful to a certain extent as they provide a quick guide, much like with TST, as to whether or not a reaction will be successful and the parameters needed in order to improve reactions efficiency. However, these rules are also a broad overview and should not be taken as the whole story as often small tweaks in conditions will act against the rules. Such an example includes

{{fontcolor1|red|(Overall, well done) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 20:18, 22 May 2019 (BST)}}

==Reference==

1. J. Phys. Chem. Lett., 2012, 3 (23), pp 3416–3419

MRD:zz161701327485

2019-05-17T15:59:24Z

Cq3417:

== '''Exercise 1''' ==

'''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 as the point that the gradient of the potential equals to zero. It can be expressed as '''r'''i where ∂ V(ri)/∂ ri=0. On a potential energy surface diagram, the transition state is identified as the saddle part of the minimum energy path. In the diagram below, the transition state is where on the minimum energy path (the black line) AB = 0.75-1.00 and BC = 1.00-1.25. The transition state is not only a local minimum of the potential energy surface but where the energy goes down most steeply along the minimum energy path linking reactants and products.
 (The definition is ok but you need to say '''second partial''' derivative. Besides, pay more attention on how you describe this concept.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:35, 15 May 2019 (BST)

[[File:Surface_Plot-1.png]]

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

In the plot below, it is set that r1=r1=1, p1=p2=0. The best estimate of the transition state position is where AC is minimum, where T = 0.156 or 0.473 or 0.797, AC = 1.654, rts= AB = BC = 0.825. This is the transition point because the gradient of the potential is zero at the point (local minimum of the potential energy surface), and AC reaches the minimum which means the greatest probability of interactions among A, B and C.

 (Seems that your system didn't reach the transition state and I think you can report better result.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:35, 15 May 2019 (BST)

[[File:Animation-11.png]]

'''Comment on how the mep and the trajectory you just calculated differ. '''
[[File:Plot-323.png|thumb|none|350px|"dynamics"]]
[[File:Plot-324.png|thumb|none|350px|"MEP"]]
For the mep, atom C almost always keeps a same distance (around 0.75) from atom B, while atom deviates from atom B quickly with an increasing speed. This shows that BC becomes a molecule after the transition state.
Foe (for) the trajectory, A-B and B-C keep almost the same distance until T=0.6, and later B-C vibrates more greatly than A-B. In this case, it is not clear to see which two atoms become a molecule. The three atoms are reasonably close to each other at all times.
 (I think you can describe more about how vibration affects the difference.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:35, 15 May 2019 (BST)

'''Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. Take note of the final values of the positions r1(t) r2(t) and the average momenta p1(t) p2(t) at large t. What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. What do you observe? '''

[[File:Plot-325.png|thumb|none|350px|"p1(t) and p2(t)"]]

In the trajectory mode, the final value of r1 = 1.00, r2 = 0.88. Both of the average momenta at large t equals to zero. If we used the initial conditions r'1 = rts and r'2 = rts+0.01 instead, the graph of r'1(t) and r'2(t) would interconvert, where r'1(t) = r2(t), r'2(t) = r1(t), p'1(t) = p2(t) and p'2(t) = p1(t).

 (You confused me.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:35, 15 May 2019 (BST)

'''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" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99.018 || Reactive || Molecule AB (with weak vibration) and atom C moves close to each other, BC becomes a molecule and moves away from C, molecule BC vibrates weakly
|-
| -1.5 || -2.0 || -100.456 || Unreactive || Molecule AB (with weak vibration) and atom C moves close to each other, and moves away; molecule AB vibrates weakly
|-
| -1.5 || -2.5 || -98.956 || Reactive || Molecule AB and atom C moves close to each other, BC becomes a molecule and moves away from C, where atom A has a greater speed than BC, molecule BC vibrates weakly
|-
| -2.5 || -5.0 || -84.956 || Unreactive || Molecule AB and atom C moves close to each other, BC gets closer to each other (with great vibration) shortly, then AB becomes a molecule (with great vibration) again and moves away from C
|-
| -2.5 || -5.2 || -83.416 || Reactive|| Molecule AB and atom C moves close to each other, then BC gets closer (with great vibration) shortly, then AB gets closer (with great vibration) again shortly. After a long transition state, BC finally becomes a molecule (with great vibration)
|}

From the table above, we can conclude that the reaction is reactive when there is a relatively large difference between p1 and p2, otherwise, the reaction is unreactive. When p1 and p2 are relatively low (more negative), they will experience a longer transition state, and molecules would vibrate more strongly.

 (I think you need to emphasise the role of transition state here.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:35, 15 May 2019 (BST)

[[File:mk-1.png|thumb|none|350px|"p1 = -1.25, p2 = -2.5"]]
[[File:mk-2.png|thumb|none|350px|"p1 = -1.5, p2 = -2.0"]]
[[File:mk-3.png|thumb|none|350px|"p1 = -1.5, p2 = -2.5"]]
[[File:mk-4.png|thumb|none|350px|"p1 = -2.5, p2 = -5.0"]]
[[File:mk-5.png|thumb|none|350px|"p1 = -2.5, p2 = -5.2"]]

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

The main assumptions of Transition State Theory are: Transition state is defined as the saddle point of the potential energy surface. The activated complexes would reach quasi-equilibrium with reactants and could be converted to products. [1]

From the experimental results, we can confirm that all reactive reactions have a transition state. Unreactive reactions may have a transition state (4th example, p1=-2.5, p2=-5.0) and may not have a transition state (2nd example, p1=-1.5, p2=-2.0). In the 4th example, the activated complexes are in equilibrium with the reactant molecules but did not convert into products.

 (You need to discuss this in detail and show your understanding of this topic.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:35, 15 May 2019 (BST)

== '''Exercise 2''' ==

'''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? Locate the approximate position of the transition state. Report the activation energy for both reactions. '''

F + H2 is an exothermic reaction, and H + HF is an endothermic reaction. This is reasonable because H-H bond energy is 432 KJ/mol, and H-F bond energy is 565 KJ/mol [2]. Therefore, it requires more energy to break an H-F bond than forming an H-H bond, vice versa.
The approximate position of the transition state of H+HF is AB=3.45, BC=0.92. The approximate position of the transition state of F+H2 is AB=1.82, BC=0.74.
The activation energy of H+HF is 29, and the activation energy of F+H2 is 0.2.

 (Again, detail please.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:35, 15 May 2019 (BST)

[[File:Surface_Plot-z1.png|thumb|none|350px|"transition state position for H+HF"]]
[[File:Surface_Plot-z2.png|thumb|none|350px|"transition state position for F+H2"]]
[[File:Surface_Plot-z3.png|thumb|none|350px|"Activation Energy for H+HF"]]
[[File:Surface_Plot-z4.png|thumb|none|350px|"Activation Energy for 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.'''

When r(HF)=1.82, r(HH)=0.74, p(F)=-0.7, p(HH)=-0.1, the reaction could exactly happens.

[[File:Surface_Plot-z9.png]]

For exothermic reactions, some energy in the system will be transformed to heat and release to the surroundings. Experimentally, this can be observed by rise of temperature of the surroundings and decrease of temperature of the system.

For endothermic reactions, the system would absorb energy from the surroundings. Experimentally, this can be observed by decrease of temperature of the surroundings and rise of temperature of the system.

F+H2 is an exothermic reaction. So experimentally, decrease of temperature of the surroundings and rise of temperature of the system should be observed. From the graph above, we can see that the new molecule F-H vibrates greatly.

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

Polanyi's empirical rules state that vibrational energy is more efficient than translational energy in activating a late-barrier reaction, but the reverse is true for an early-barrier reaction [3].

Experimentally, for reaction F+H2, where r(HH)=0.74, p(FH)=-0.5, when p(HH) is less than 1.76, the reaction could happen, and otherwise it cannot happen. F+H2 is an exothermic reaction and is a early-barrier reaction. From the graph, we can see that the reactions happens when there is more translational energy than vibrational energy.

Experimentally, H+HF is an endothermic reaction and is a late-barrier reaction. From the graph, we can see that the reactions happens when there is more vibrational energy than translational energy.

Therefore, Polanyi's empirical rules can be proved by the experiments.

[[File:F+H2, yes.png|thumb|none|350px|"F+H2, the reaction happens"]]
[[File:F+H2, no.png|thumb|none|350px|"F+H2, the reaction does not happen"]]
[[File:H+HF, yes.png|thumb|none|350px|"H+HF, the reaction happens"]]
[[File:H+HF,_no2.png|thumb|none|350px|"H+HF, the reaction does not happen"]]

 (Overall, this report can be improved by describing more detials.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:35, 15 May 2019 (BST)

== '''References''' ==

[1]D. M. Hirst Potential Energy Surfaces: Molecular Structure and Reaction Dynamics Taylor and Francis, 1985

[2]Chemical Bond Energies: http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html

[3] Wiley-VCH Verlag GmbH & Co. KGaA, Weinheim: Dynamics of the Reaction of Methane with Chlorine Atom on an Accurate Potential Energy Surface, 2011

MRD:jwy17mrd

2019-05-17T15:33:20Z

Cq3417:

= Molecular Reaction Dynamics =

== 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?"'''''<nowiki/>''

To define a transition state, we shall consider 2 new 2-D basis vectors orthoganol to each other, q1 and q2, with q2 parallel to the trajectory of the reaction. These 2 basis vectors are diagonal to the existing basis vectors, which are r1 and r2. The transition state would be at the saddle point of the potential energy surface, where the criteria are as follows:

- The first order partial derivative of the potential energy surface w.r.t q1 and q2 are zero.

- The second order partial derivative of the potential energy surface w.r.t q1 is positive and negative w.r.t to q2.

It is important that both criteria to be fulfilled so that the point on the potential energy surface corresponds to the transition state.

A local minimum, while fulfill the first criteria, falls short to fulfill the second because the second order partial derivative of the potential energy surface w.r.t q1 and q2 is positive.
[[File:internuclear_vs_time_jwy17.png|thumb|The internuclear distance between the H atoms at distance of 0.91 Angstrom.]]

== Trasition state of H2 + H System ==
'''"''Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory"'''''<nowiki/>'' ''

At the transition state, where hydrogens A and C are equidistant to the centre hydrogen B, would not oscillate if they have no momenta and do not posses excess potential energy that can be converted to kinetic energy. At 0.91 Angstrom, the Internuclear Distance Vs Time plot of the hydrogen atoms with 0 initial momenta showed a very minute oscillation. Therefore, by inspection (~0.8 Angstrom) and trial and error, the '''rts '''is 0.91 Angstrom.

{{fontcolor1|red| (I think you can report better result for transition state with the aid of contour plot.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:31, 15 May 2019 (BST)}}

== Trajectory of r1 = rts+δ, r2 = rts ==
'''"''Comment on how the mep and the trajectory you just calculated differ." (show the potential energy surface top view dynamics vs mep)'''''

The mep shows a non-oscilating trajectory, as it only displays the the path with the least potential energy. Realistically, the offset of the position of one of the hydrogens puts it on a negative gradient, moving the reaction off the ridge. The vibration arises the the difference between the equilibrium position of the product/reactant and the transition state position. As the equilibrium distance between the 2 hydrogens shift from the transition state position to that of hydrogen molecules, the hydrogen atoms find themselves at a position above the equilibrium. The potential energy can be converted to kinetic energy and vibrate, with inertia keeping it in motion.

== Reactivity ==
'''"''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?"'''''<nowiki/>'' ''
{| class="wikitable"
!P1 AB
!P2 BC
!Etot (kJ/mol)
!Reaction?
!Description of the dynamics
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|Yes
|The molecule does not vibrate much initally and vibrates after passing through the transition state
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|No
|The molecule vibrates but does not overcome the activation barrier and does not form a new molecule.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|Yes
|The molecule and new molecule both vibrates.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|No
|A new molecule was transient, as the molecules breaks to form a new molecule, but reverts back to the orginal molecule.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|Yes
|The new molecule reverts back to the orginal molecule once and then back to the new molecule.
|}
{| class="wikitable"
!P1 AB
!P2 BC
!Reaction?
!Reaction trajectories
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|Yes
|[[File:1p25 2p5 jwy17.png|frameless]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|No
|[[File:1p5 2p0 jwy17.png|frameless|294x294px]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|Yes
|[[File:1p5 2p5 jwy17.png|frameless]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|No
|[[File:-2.5,-5.0_jwy17.png|frameless|292x292px]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|Yes
|[[File:2p5 5p2 jwy17.png|frameless]]
|}
The amount of total energy the system does not indicate whether the reaction is reactive or not. Despite having the enough energy to overcome the activation energy barrier, the reactivity of the reaction is not guaranteed.

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

In Transition State Theory, it is assumed that:

'''1.The reactants and transition state follows a Boltzzman Distribution.'''

In the cases where for rapid reactions, the transition state may disappear rapidly, causing its concentration to be less than expected in the solution. This assumption overestimates the experimental rate value, as the transition state concentration is more than reality.

'''2. The reaction tranjectory only crosses the transition barrier once to form the product.'''

This is not true, as shown in the reactivity section, that can re-cross the barrier to form the reactant or re-cross in the direction of the products twice. This assumption also then overestimates the rate value because the actual value involves multiple re-crossing until it forms the reactants or products, hence less than the prediction.

{{fontcolor1|red| (I think a citation of assumption is required here.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:31, 15 May 2019 (BST)}}

= F-H-H System =

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

'''''Locate the approximate position of the transition state."'''''

By inspection, the reaction is exothermic, as the H + HF'' ''part of the potential energy surface miinium is lower than'' the ''F + H2. This means that HF has a much strong bond strength than the H2 speci

{{fontcolor1|red| (It seems that something is missing here.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:31, 15 May 2019 (BST)}}

[[File:F-H-H transition jwy17.png|thumb| The x marks the tansition state of the F-H-H system, where A is the Fluorine atom. The transition state has a BC distance of 0.74 Angstrom, similar to a H2 equilibrium distance.]]

To find the transition state. The equilibrium position of the H2 are first estimated (~0.74 Angstrom). then by trial and error, the F-H position was adjusted until the trajectory of the reaction is not obvious (~ 1.82 Angstrom). The transition state fits the Hammond postulate, as it is a exothermic reaction and the transition state resembles a H2 molecule.

== Activation energy ==
'''"''Report the activation energy for both reactions"'''''

H + HF to F + H2''''' ''''' : -103. 777 - (-133.269) = 30.0kcal/mol

F + H2 to H + HF: -103.738 - (-103.983) = 0.3 kcal/mol

''*Values of the total energy of both intial and final states are measured using the built in ruler''
[[File: Hf_to_hh_activation_energy_jwy17.png|thumb|The total energy of the system from transition state to HF. The drop in energy equals to the activation energy.]]
[[File: Hh_to_hf_activation_energy_jwy17.png|thumb|The total energy of the system from transition state to H2.]]

== Energy Dissipation ==
'''"''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."'''''

As the HF molecule is formed, the potential surface minimum drops dramamtically. As the total energy is conserved:
# The HF may find itself at a much higher vibrational state. The vibrational energy can be transfered to the H atom as kinetic energy.
# The reaction may directly result in a HF in a ground-state vibrational mode and the H atom realesed gains kinetic energy.
As the kinetic energy of the particles increase, so will the temperature, as they are proportional to each other. Therefore, as we perform the reaction, an increase in temperature or pressure would be observed.

{{fontcolor1|red| (I think you should include more experimental data to support your discussion.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:31, 15 May 2019 (BST)}}

== Polanyi's Rule<ref>L. Arnaut, S. Formosinho and H. Burrows, ''Chemical Kinetics From melecular Structure to Chemical Reactivity, Elsevier, Oxford, 2007''</ref> ==
'''''"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."'''''

'''Early Barrier'''

An early barrier is where the transition state would be on the PES when forming HF + H from F + H2 (Before turning the corner). A translational energy would be more efficient in carrying the trajectory over the transition state. The translational and vibrational energy requires to form the product is a balancing task, given too much of either can cause the tranjectory to 'bounce' back to the reactant.

'''Late Barrier'''

A late barrier is where the transition state would be on the PES when forming HF + H from F + H2 (After turning the corner). To access the transition state the HF molecule has to have a massive vibrational energy (large momentum). As with the early barrier, translational energy cannot be ignored and the right amount of translational energy is required still to reach the transition state.[[File:Late barrier jwy17.png|thumb|An example reactive trajectory of a late barrier (HF + H to H2 + F). The large vibrational energy of the HF will be required to reach the transition state.|left]][[File:Early barrier jwy17.png|An example trajectory of early barrier (H2 + F to HF + H). The H2 is very small vibrational energy, and has enough translational energy to move over the tansition state. |thumb|centre]]

{{fontcolor1|red| Overall, I think you can improve more if include more expeirmental data.[[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:31, 15 May 2019 (BST)}}

MRD:mz5717

2019-05-15T14:37:19Z

Cq3417:

== H + H2 SYSTEM ==

=== Transition State Determination ===

On a potential energy surface, the transition state is represented by a saddle point, which is mathematically defined as:

When the partial derivatives Vr1 and Vr2 both equal to 0 at a point, if the value of (Vr1r1Vr2r2 - V2r1r2) < 0 at that same point, then it is a saddle point.

Alternatively we can identify it by computing the second partial derivatives, Vr1r1 and Vr2r2. For a saddle point, one of them will be positive and the other one negative.

Additionally, when (Vr1r1Vr2r2 - V2r1r2) > 0, the point represents a local maximum/minimum. It is a local maximum if Vr1r1 < 0, or a local minimum if Vr1r1 > 0.

The transition state position is estimated to be 0.90743 Å. i.e. r1 = r2 = 0.90743 Å, p1 = p2 = 0. When subjected to this initial condition, the internuclear distances do not vary with time because the equilibrium established at the saddle point does not provide driving force for any forward or backward reaction.

This is illustrated with the plot below. The contour map shows the position of the TS with a cross.

{{fontcolor1|red|(Reasonable description of the transition state and good illustration of plots) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:37, 15 May 2019 (BST)}}

{| class="wikitable" border=1
|-
| [[File:TS_distances_mz5717.jpeg|thumb|400px|none|Figure 1: time-variation of internuclear distances.]] || [[File:TS_contour_mz5717.jpeg|thumb|400px|none|Figure 2: contour map of the potential energy surface showing the position of the TS.]]
|}

=== Reaction Trajectories ===

Dynamics and MEP are two methods of calculating the reaction trajectory. The table below compares the outcomes produced by these two methods.

{| class="wikitable" border=1
|-
| '''Dynamics''' (r1 = 0.90743, r2 = 0.91743, p1 = p2 = 0) || [[File:Dynamics_mz5717.jpeg|thumb|400px|none|Figure 3: reaction trajectory produced by Dynamics.]] || [[File:Dynamics_momenta_mz5717.jpeg|thumb|400px|none|Figure 4: time-variation of momenta produced by Dynamics. Blue curve shows the momentum oscillation of the new bond.]] ||[[File:Dynamics_distance_mz5717.jpeg|thumb|400px|none|Figure 5: time-variation of internuclear bond length produced by Dynamics. Blue curves shows the bond length oscillation of the new bond.]]|| The Dynamics calculation accounts for the conversion of kinetic energy to vibrational energy, as shown by the oscillating momentum and bond length of the new bond formed.
|-
| '''MEP''' (r1 = 0.90743, r2 = 0.91743, p1 = p2 = 0)|| [[File:MEP_mz5717.jpeg|thumb|400px|none|Figure 6: reaction trajectory produced by MEP.]] || [[File:MEP_momenta_mz5717.jpeg|thumb|400px|none|Figure 7: time-variation of momenta produced by MEP. There is zero motion.]] || [[File:MEP_distance_mz5717.jpeg|thumb|400px|none|Figure 8: time-variation of internuclear bond length produced by MEP. The length of the new bond stays constant.]] ||The MEP calculation does not account for the bond vibration, therefore the momentum of the newly formed molecule is a straight line, and there is no oscillation in the new bond.
|}

According to Figures 4 and 5, we set the initial conditions to be r1 = 0.74895, r2 = 1.0 (an arbitrary large distance), p1 = -1.2745, p2 = -2.4889, which corresponds with the final position of the trajectory only with the signs of the final momenta reversed. The resulting trajectory is almost the same as those we obtained previously.

{{fontcolor1|red|(Good explanation.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:37, 15 May 2019 (BST)}}

[[File:reverse_contour_mz5717.jpeg|thumb|400px|none|Figure 9: initial conditions corresponds to the final values of the above trajectory but with the signs of the momenta reversed. A very similar reaction path.]]

Next, we examine the effect of initial momenta and total energy on the reaction trajectory, i.e. if it is reactive or not, by giving the reactants varying amounts of initial momenta. The results are summarized by the table below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Reaction path !! Description of the dynamics
|-
| -1.25 || -2.5 || -99.018 || Y || [[File:Dynamics1_mz5717.jpeg|300px]] || More internal vibration after the transition state; energy sufficient to pass over Ea.
|-
| -1.5 || -2.0 || -100.456 || N || [[File:Dynamics2_mz5717.jpeg|300px]] || Reaction does not proceed; the molecules bounce back without reacting.
|-
| -1.5 || -2.5 || -98.956 || Y || [[File:Dynamics3_mz5717.jpeg|300px]] || Considerable internal vibration before & after reaching the TS; more than enough energy to overcome Ea.
|-
| -2.5 || -5.0 || -84.956 || N || [[File:Dynamics4_mz5717.jpeg|300px]] || The kinetic energy supplied is way above the activation energy so the reaction path goes past the saddle point. However it bounces back after a while, with most of the energy contributed to intramolecular vibration. So the starting materials ARE reactive and we do have products formed, but they eventually turn back to the reactants.
|-
| -2.5 || -5.2 || -83.416 || Y || [[File:Dynamics5_mz5717.jpeg|300px]] || Reaction proceeds to the formation of the products with much of the kinetic energy contributing to the internal vibration of the newly formed product.
|}

{{fontcolor1|red|(Excellent illustration though additionally showing the energy vs time plot can make it better.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:37, 15 May 2019 (BST)}}

=== Transition State Theory ===

The main assumptions of Transition State Theory are:

- The nuclei behave classically

- The reaction path passes through the saddle point representing the transition state

- Once the reaction passes through the TS, it will proceed to the formation of the product(s)

Under most circumstances, TST matches with experimental results to a good extent. However, from the simulation we can conclude that when the kinetic energy supplied is way above the activation energy, this is not necessarily the case. Here the experimental position of the transition state may largely deviate from the theoretical one represented by the saddle point on the potential energy surface. Because of the abundance of energy supplied, more molecules are able to occupy vibrational modes of higher energy, making their motion more complicated and harder to predict. This could be the underlying reason for the large deviation. It could be hard as well to predict the reaction rate under such circumstances; for example, when there is {{fontcolor1|red|a}} reverse reaction going on.

{{fontcolor1|red|(I think you may need to cite the description of this assumption.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:37, 15 May 2019 (BST)}}

== F - H - H SYSTEM ==

=== Transition State Determination ===

F + H2 --> H + HF is exothermic and has an early TS. The reaction releases energy because the new H-F bond formed is stronger than the H-H bond broken (F is much more electronegative than H). On the other hand, H + HF --> F + H2 is endothermic and has a late TS. The reaction requires input of energy because the H-F bond broken is much stronger than the H-H bond formed.

Setting initial momenta to be both zero, the position of the TS is estimated to be r1 = 0.745 Å and r2 = 1.80 Å.

{{fontcolor1|red|(Good, nice and clear.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:37, 15 May 2019 (BST)}}

{| class="wikitable" border=1
|-
| [[File:TS_HF_contour_mz5717.jpeg|thumb|400px|none|Figure 10: position of the TS marked by a cross.]] || [[File:TS_HF_distance_mz5717.jpeg|thumb|400px|none|Figure 11: no bond length variation when at equilibrium.]]
|}

=== Activation Energy ===

{| class="wikitable" border=1
|-
| [[File:Ea_energy_mz5717.jpeg|thumb|400px|none|Figure 12: reaction profile of H + HF --> F + H2.]] || [[File:Ea_energy_small_mz5717.jpeg|thumb|400px|none|Figure 13: reaction profile of F + H2 --> H + HF.]]
|}

The figures above show the energy profiles of H + HF and F + H2 respectively.

The activation energy of H + HF is determined as -103.867 - (-133.257) = 29.390 kcal/mol

The activation energy of F + H2 is determined as -103.752 - (-103.967) = 0.215 kcal/mol

{{fontcolor1|red|(How you get this plot is my question so I think you need to discuss this in detail.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:37, 15 May 2019 (BST)}}

=== Energy Distribution ===

First we focus on energy conversion. Below is a general case of a successful reaction.

{| class="wikitable" border=1
|-
| [[File:Energy_contour1_mz5717.jpeg|thumb|400px|none|Figure 14: a generic reactive path with r1 = 0.745, r2 = 1.40, p1 = 3, p2 = -0.5.]] || [[File:Energy_momenta1_mz5717.jpeg|thumb|400px|none|Figure 15: corresponding time-variation of momenta of this reaction path.]]
|}

Figure 15 shows that the vibration of bond A-B gradually ceases, which means that bond A-B corresponds to the H-H bond broken. In this case bond B-C must be the new H-F bond formed. The H-F bond continues its vibration and eventually becomes the only source of momentum, which means that the excess kinetic energy from the F atom was converted to vibrational energy after the reaction has occurred. The extent and modes of the H-F bond vibration can be measured by Raman spectroscopy.

{{fontcolor1|red|(It can be more clear if you include the energy vs time plot here.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:37, 15 May 2019 (BST)}}

Next, for the reaction F + H2 --> H + HF, we set the initial conditions to rHH = 0.74, rHF = 1.80 and pFH = -0.5. Different values of pHH, ranging from -3 to 3, are tested and the results are summarized in the table below.

{| class="wikitable" border=1
|-
| [[File:--3_mz5717.jpeg|thumb|200px|none|pHH = -3]] || [[File:--2.4_mz5717.jpeg|thumb|200px|none|pHH = -2.4]] || [[File:--1.8_mz5717.jpeg|thumb|200px|none|pHH = -1.8]] || [[File:--1.2_mz5717.jpeg|thumb|200px|none|pHH = -1.2]] || [[File:--0.6_mz5717.jpeg|thumb|200px|none|pHH = -0.6]]
|-
| [[File:-0_mz5717.jpeg|thumb|200px|none|pHH = 0]] || [[File:---0.6_mz5717.jpeg|thumb|200px|none|pHH = 0.6]] || [[File:---1.2_mz5717.jpeg|thumb|200px|none|pHH = 1.2]] || [[File:---1.8_mz5717.jpeg|thumb|200px|none|pHH = 1.8]] || [[File:---2.4_mz5717.jpeg|thumb|200px|none|pHH = 2.4]] || [[File:---3_mz5717.jpeg|thumb|200px|none|pHH = 3]]
|}

The table shows that when pHH has a value between -1.2 and 0, the reaction proceeds to the product side. In all other cases the reaction trajectory more or less bounces back to the reactant side, and the extent of this backward conversion increases as the absolute values of pHH increases. The only outlier is pHH = 2.4, where the trajectory generally stays on the product side.

Now increase pHF slightly to -0.8, and largely reduce pHH to 0.1. The trajectory below shows that despite having considerably lower total energy, the reaction still proceeds to the product side.

[[File:reduced_energy_mz5717.jpeg|thumb|400px|none|Figure 16: pHF = -0.8, pHH = 0.1, rHH = 0.74, rHF = 1.80. The total energy of the system has been significantly reduced.]]

Finally, for the reaction H + HF --> F + H2, we assume that there is very little vibration within the H-F bond (i.e. pHF = 0.2), and that the H atom is approaching with extremely high kinetic energy (pHH = 5). With the initial conditions rHH = 0.745, rHF = 1.80, the trajectory below is obtained. It is evident that the high kinetic energy possessed by the H atom is initially able to break the H-F bond and push the system to the product side; however, HF is reformed afterwards and then the entire process repeats. This is an oscillating reaction.

[[File:oscillate_mz5717.jpeg|thumb|400px|none|Figure 17: rHH = 0.745, rHF = 1.80, pHF = 0.2, pHH = 5. This set of initial conditions results in an oscillating reaction.]]

By decreasing pHH and increasing pHF, it is possible to obtain a reactive trajectory. For example, when we let other variables stay constant and change the rest to pHF = 0.3, pHH = 2.6, the reaction can indeed proceed to the product side as shown in the figure below.

[[File:balance_mz5717.jpeg|thumb|400px|none|Figure 18: when rHH = 0.745, rHF = 1.80, pHF = 0.3, pHH = 2.6, we obtain a reactive trajectory.]]

{{fontcolor1|red|(Where is your discussion that respects to the Polanyi's empirical rules?) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:37, 15 May 2019 (BST)}}

In conclusion, a proper distribution of kinetic and vibrational energy is crucial for the trajectory to be reactive. Generally in most cases, a high kinetic energy nearly guarantees a reactive path. The reaction F + H2 --> H + HF has an early TS, and does not need much energy to reach it; the other reaction H + HF --> F + H2 has a late TS, and requires much higher vibrational energy to reach it. The second reaction also shows us that just having high kinetic energy is not enough to trigger a successful reaction. By lowering the kinetic energy and increasing the vibrational energy we see that there is a delicate balance between the two that eventually leads to a reactive trajectory.

{{fontcolor1|red|(Overall this is a sufficient report that shows a good understanding of this concept. However, the citation of reference and, the combination of theories and experimental data are required to be noticed.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:37, 15 May 2019 (BST)}}

MRD:zz161701327485

2019-05-15T14:35:13Z

Cq3417:

== '''Exercise 1''' ==

'''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 as the point that the gradient of the potential equals to zero. It can be expressed as '''r'''i where ∂ V(ri)/∂ ri=0. On a potential energy surface diagram, the transition state is identified as the saddle part of the minimum energy path. In the diagram below, the transition state is where on the minimum energy path (the black line) AB = 0.75-1.00 and BC = 1.00-1.25. The transition state is not only a local minimum of the potential energy surface but where the energy goes down most steeply along the minimum energy path linking reactants and products.
 (The definition is ok but you need to say '''second partial''' derivative. Besides, pay more attention on how you describe this concept.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:35, 15 May 2019 (BST)

[[File:Surface_Plot-1.png]]

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

In the plot below, it is set that r1=r1=1, p1=p2=0. The best estimate of the transition state position is where AC is minimum, where T = 0.156 or 0.473 or 0.797, AC = 1.654, rts= AB = BC = 0.825. This is the transition point because the gradient of the potential is zero at the point (local minimum of the potential energy surface), and AC reaches the minimum which means the greatest probability of interactions among A, B and C.

 (Seems that your system didn't reach the transition state and I think you can report better result.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:35, 15 May 2019 (BST)

[[File:Animation-11.png]]

'''Comment on how the mep and the trajectory you just calculated differ. '''
[[File:Plot-323.png|thumb|none|350px|"dynamics"]]
[[File:Plot-324.png|thumb|none|350px|"MEP"]]
For the mep, atom C almost always keeps a same distance (around 0.75) from atom B, while atom deviates from atom B quickly with an increasing speed. This shows that BC becomes a molecule after the transition state.
Foe (for) the trajectory, A-B and B-C keep almost the same distance until T=0.6, and later B-C vibrates more greatly than A-B. In this case, it is not clear to see which two atoms become a molecule. The three atoms are reasonably close to each other at all times.
 (I didn't get your point.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:35, 15 May 2019 (BST)

'''Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. Take note of the final values of the positions r1(t) r2(t) and the average momenta p1(t) p2(t) at large t. What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. What do you observe? '''

[[File:Plot-325.png|thumb|none|350px|"p1(t) and p2(t)"]]

In the trajectory mode, the final value of r1 = 1.00, r2 = 0.88. Both of the average momenta at large t equals to zero. If we used the initial conditions r'1 = rts and r'2 = rts+0.01 instead, the graph of r'1(t) and r'2(t) would interconvert, where r'1(t) = r2(t), r'2(t) = r1(t), p'1(t) = p2(t) and p'2(t) = p1(t).

 (You confused me.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:35, 15 May 2019 (BST)

'''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" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99.018 || Reactive || Molecule AB (with weak vibration) and atom C moves close to each other, BC becomes a molecule and moves away from C, molecule BC vibrates weakly
|-
| -1.5 || -2.0 || -100.456 || Unreactive || Molecule AB (with weak vibration) and atom C moves close to each other, and moves away; molecule AB vibrates weakly
|-
| -1.5 || -2.5 || -98.956 || Reactive || Molecule AB and atom C moves close to each other, BC becomes a molecule and moves away from C, where atom A has a greater speed than BC, molecule BC vibrates weakly
|-
| -2.5 || -5.0 || -84.956 || Unreactive || Molecule AB and atom C moves close to each other, BC gets closer to each other (with great vibration) shortly, then AB becomes a molecule (with great vibration) again and moves away from C
|-
| -2.5 || -5.2 || -83.416 || Reactive|| Molecule AB and atom C moves close to each other, then BC gets closer (with great vibration) shortly, then AB gets closer (with great vibration) again shortly. After a long transition state, BC finally becomes a molecule (with great vibration)
|}

From the table above, we can conclude that the reaction is reactive when there is a relatively large difference between p1 and p2, otherwise, the reaction is unreactive. When p1 and p2 are relatively low (more negative), they will experience a longer transition state, and molecules would vibrate more strongly.

 (I think you need to emphasise the role of transition state here.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:35, 15 May 2019 (BST)

[[File:mk-1.png|thumb|none|350px|"p1 = -1.25, p2 = -2.5"]]
[[File:mk-2.png|thumb|none|350px|"p1 = -1.5, p2 = -2.0"]]
[[File:mk-3.png|thumb|none|350px|"p1 = -1.5, p2 = -2.5"]]
[[File:mk-4.png|thumb|none|350px|"p1 = -2.5, p2 = -5.0"]]
[[File:mk-5.png|thumb|none|350px|"p1 = -2.5, p2 = -5.2"]]

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

The main assumptions of Transition State Theory are: Transition state is defined as the saddle point of the potential energy surface. The activated complexes would reach quasi-equilibrium with reactants and could be converted to products. [1]

From the experimental results, we can confirm that all reactive reactions have a transition state. Unreactive reactions may have a transition state (4th example, p1=-2.5, p2=-5.0) and may not have a transition state (2nd example, p1=-1.5, p2=-2.0). In the 4th example, the activated complexes are in equilibrium with the reactant molecules but did not convert into products.

 (You need to discuss this in detail and show your understanding of this topic.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:35, 15 May 2019 (BST)

== '''Exercise 2''' ==

'''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? Locate the approximate position of the transition state. Report the activation energy for both reactions. '''

F + H2 is an exothermic reaction, and H + HF is an endothermic reaction. This is reasonable because H-H bond energy is 432 KJ/mol, and H-F bond energy is 565 KJ/mol [2]. Therefore, it requires more energy to break an H-F bond than forming an H-H bond, vice versa.
The approximate position of the transition state of H+HF is AB=3.45, BC=0.92. The approximate position of the transition state of F+H2 is AB=1.82, BC=0.74.
The activation energy of H+HF is 29, and the activation energy of F+H2 is 0.2.

 (Again, detail please.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:35, 15 May 2019 (BST)

[[File:Surface_Plot-z1.png|thumb|none|350px|"transition state position for H+HF"]]
[[File:Surface_Plot-z2.png|thumb|none|350px|"transition state position for F+H2"]]
[[File:Surface_Plot-z3.png|thumb|none|350px|"Activation Energy for H+HF"]]
[[File:Surface_Plot-z4.png|thumb|none|350px|"Activation Energy for 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.'''

When r(HF)=1.82, r(HH)=0.74, p(F)=-0.7, p(HH)=-0.1, the reaction could exactly happens.

[[File:Surface_Plot-z9.png]]

For exothermic reactions, some energy in the system will be transformed to heat and release to the surroundings. Experimentally, this can be observed by rise of temperature of the surroundings and decrease of temperature of the system.

For endothermic reactions, the system would absorb energy from the surroundings. Experimentally, this can be observed by decrease of temperature of the surroundings and rise of temperature of the system.

F+H2 is an exothermic reaction. So experimentally, decrease of temperature of the surroundings and rise of temperature of the system should be observed. From the graph above, we can see that the new molecule F-H vibrates greatly.

 (Seems that you are not fully understand what is happening.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:35, 15 May 2019 (BST)

 (I think you missed some questions.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:35, 15 May 2019 (BST)

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

Polanyi's empirical rules state that vibrational energy is more efficient than translational energy in activating a late-barrier reaction, but the reverse is true for an early-barrier reaction [3].

Experimentally, for reaction F+H2, where r(HH)=0.74, p(FH)=-0.5, when p(HH) is less than 1.76, the reaction could happen, and otherwise it cannot happen. F+H2 is an exothermic reaction and is a early-barrier reaction. From the graph, we can see that the reactions happens when there is more translational energy than vibrational energy.

Experimentally, H+HF is an endothermic reaction and is a late-barrier reaction. From the graph, we can see that the reactions happens when there is more vibrational energy than translational energy.

Therefore, Polanyi's empirical rules can be proved by the experiments.

[[File:F+H2, yes.png|thumb|none|350px|"F+H2, the reaction happens"]]
[[File:F+H2, no.png|thumb|none|350px|"F+H2, the reaction does not happen"]]
[[File:H+HF, yes.png|thumb|none|350px|"H+HF, the reaction happens"]]
[[File:H+HF,_no2.png|thumb|none|350px|"H+HF, the reaction does not happen"]]

 (Overall, this report is not sufficient since its not clear and elaborate enough.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:35, 15 May 2019 (BST)

== '''References''' ==

[1]D. M. Hirst Potential Energy Surfaces: Molecular Structure and Reaction Dynamics Taylor and Francis, 1985

[2]Chemical Bond Energies: http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html

[3] Wiley-VCH Verlag GmbH & Co. KGaA, Weinheim: Dynamics of the Reaction of Methane with Chlorine Atom on an Accurate Potential Energy Surface, 2011

MRD:jwy17mrd

2019-05-15T14:31:18Z

Cq3417:

= Molecular Reaction Dynamics =

== 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?"'''''<nowiki/>''

To define a transition state, we shall consider 2 new 2-D basis vectors orthoganol to each other, q1 and q2, with q2 parallel to the trajectory of the reaction. These 2 basis vectors are diagonal to the existing basis vectors, which are r1 and r2. The transition state would be at the saddle point of the potential energy surface, where the criteria are as follows:

- The first order partial derivative of the potential energy surface w.r.t q1 and q2 are zero.

- The second order partial derivative of the potential energy surface w.r.t q1 is positive and negative w.r.t to q2.

It is important that both criteria to be fulfilled so that the point on the potential energy surface corresponds to the transition state.

A local minimum, while fulfill the first criteria, falls short to fulfill the second because the second order partial derivative of the potential energy surface w.r.t q1 and q2 is positive.
[[File:internuclear_vs_time_jwy17.png|thumb|The internuclear distance between the H atoms at distance of 0.91 Angstrom.]]

== Trasition state of H2 + H System ==
'''"''Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory"'''''<nowiki/>'' ''

At the transition state, where hydrogens A and C are equidistant to the centre hydrogen B, would not oscillate if they have no momenta and do not posses excess potential energy that can be converted to kinetic energy. At 0.91 Angstrom, the Internuclear Distance Vs Time plot of the hydrogen atoms with 0 initial momenta showed a very minute oscillation. Therefore, by inspection (~0.8 Angstrom) and trial and error, the '''rts '''is 0.91 Angstrom.

{{fontcolor1|red| (I think you can report better result for transition state with the aid of contour plot.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:31, 15 May 2019 (BST)}}

== Trajectory of r1 = rts+δ, r2 = rts ==
'''"''Comment on how the mep and the trajectory you just calculated differ." (show the potential energy surface top view dynamics vs mep)'''''

The mep shows a non-oscilating trajectory, as it only displays the the path with the least potential energy. Realistically, the offset of the position of one of the hydrogens puts it on a negative gradient, moving the reaction off the ridge. The vibration arises the the difference between the equilibrium position of the product/reactant and the transition state position. As the equilibrium distance between the 2 hydrogens shift from the transition state position to that of hydrogen molecules, the hydrogen atoms find themselves at a position above the equilibrium. The potential energy can be converted to kinetic energy and vibrate, with inertia keeping it in motion.

== Reactivity ==
'''"''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?"'''''<nowiki/>'' ''
{| class="wikitable"
!P1 AB
!P2 BC
!Etot (kJ/mol)
!Reaction?
!Description of the dynamics
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|Yes
|The molecule does not vibrate much initally and vibrates after passing through the transition state
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|No
|The molecule vibrates but does not overcome the activation barrier and does not form a new molecule.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|Yes
|The molecule and new molecule both vibrates.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|No
|A new molecule was transient, as the molecules breaks to form a new molecule, but reverts back to the orginal molecule.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|Yes
|The new molecule reverts back to the orginal molecule once and then back to the new molecule.
|}
{| class="wikitable"
!P1 AB
!P2 BC
!Reaction?
!Reaction trajectories
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|Yes
|[[File:1p25 2p5 jwy17.png|frameless]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|No
|[[File:1p5 2p0 jwy17.png|frameless|294x294px]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|Yes
|[[File:1p5 2p5 jwy17.png|frameless]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|No
|[[File:-2.5,-5.0_jwy17.png|frameless|292x292px]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|Yes
|[[File:2p5 5p2 jwy17.png|frameless]]
|}
The amount of total energy the system does not indicate whether the reaction is reactive or not. Despite having the enough energy to overcome the activation energy barrier, the reactivity of the reaction is not guaranteed.

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

In Transition State Theory, it is assumed that:

'''1.The reactants and transition state follows a Boltzzman Distribution.'''

In the cases where for rapid reactions, the transition state may disappear rapidly, causing its concentration to be less than expected in the solution. This assumption overestimates the experimental rate value, as the transition state concentration is more than reality.

'''2. The reaction tranjectory only crosses the transition barrier once to form the product.'''

This is not true, as shown in the reactivity section, that can re-cross the barrier to form the reactant or re-cross in the direction of the products twice. This assumption also then overestimates the rate value because the actual value involves multiple re-crossing until it forms the reactants or products, hence less than the prediction.

{{fontcolor1|red| (I think a citation of assumption is required here.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:31, 15 May 2019 (BST)}}

= F-H-H System =

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

'''''Locate the approximate position of the transition state."'''''

By inspection, the reaction is exothermic, as the H + HF'' ''part of the potential energy surface miinium is lower than'' the ''F + H2. This means that HF has a much strong bond strength than the H2 speci

{{fontcolor1|red| (It seems that something is missing here.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:31, 15 May 2019 (BST)}}

[[File:F-H-H transition jwy17.png|thumb| The x marks the tansition state of the F-H-H system, where A is the Fluorine atom. The transition state has a BC distance of 0.74 Angstrom, similar to a H2 equilibrium distance.]]

To find the transition state. The equilibrium position of the H2 are first estimated (~0.74 Angstrom). then by trial and error, the F-H position was adjusted until the trajectory of the reaction is not obvious (~ 1.82 Angstrom). The transition state fits the Hammond postulate, as it is a exothermic reaction and the transition state resembles a H2 molecule.

== Activation energy ==
'''"''Report the activation energy for both reactions"'''''

H + HF to F + H2''''' ''''' : -103. 777 - (-133.269) = 30.0kcal/mol

F + H2 to H + HF: -103.738 - (-103.983) = 0.3 kcal/mol

''*Values of the total energy of both intial and final states are measured using the built in ruler''
[[File: Hf_to_hh_activation_energy_jwy17.png|thumb|The total energy of the system from transition state to HF. The drop in energy equals to the activation energy.]]
[[File: Hh_to_hf_activation_energy_jwy17.png|thumb|The total energy of the system from transition state to H2.]]

== Energy Dissipation ==
'''"''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."'''''

As the HF molecule is formed, the potential surface minimum drops dramamtically. As the total energy is conserved:
# The HF may find itself at a much higher vibrational state. The vibrational energy can be transfered to the H atom as kinetic energy.
# The reaction may directly result in a HF in a ground-state vibrational mode and the H atom realesed gains kinetic energy.
As the kinetic energy of the particles increase, so will the temperature, as they are proportional to each other. Therefore, as we perform the reaction, an increase in temperature or pressure would be observed.

{{fontcolor1|red| (I think you should include more experimental data to support your discussion.) [[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:31, 15 May 2019 (BST)}}

== Polanyi's Rule<ref>L. Arnaut, S. Formosinho and H. Burrows, ''Chemical Kinetics From melecular Structure to Chemical Reactivity, Elsevier, Oxford, 2007''</ref> ==
'''''"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."'''''

'''Early Barrier'''

An early barrier is where the transition state would be on the PES when forming HF + H from F + H2 (Before turning the corner). A translational energy would be more efficient in carrying the trajectory over the transition state. The translational and vibrational energy requires to form the product is a balancing task, given too much of either can cause the tranjectory to 'bounce' back to the reactant.

'''Late Barrier'''

A late barrier is where the transition state would be on the PES when forming HF + H from F + H2 (After turning the corner). To access the transition state the HF molecule has to have a massive vibrational energy (large momentum). As with the early barrier, translational energy cannot be ignored and the right amount of translational energy is required still to reach the transition state.[[File:Late barrier jwy17.png|thumb|An example reactive trajectory of a late barrier (HF + H to H2 + F). The large vibrational energy of the HF will be required to reach the transition state.|left]][[File:Early barrier jwy17.png|An example trajectory of early barrier (H2 + F to HF + H). The H2 is very small vibrational energy, and has enough translational energy to move over the tansition state. |thumb|centre]]

{{fontcolor1|red| Overall, this is a good report. However, more experimental data should be included to support the discussion.[[User:Cq3417|Cq3417]] ([[User talk:Cq3417|talk]]) 15:31, 15 May 2019 (BST)}}

MRD:jwy17mrd

2019-05-15T14:23:43Z

Cq3417:

= Molecular Reaction Dynamics =

== 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?"'''''<nowiki/>''

To define a transition state, we shall consider 2 new 2-D basis vectors orthoganol to each other, q1 and q2, with q2 parallel to the trajectory of the reaction. These 2 basis vectors are diagonal to the existing basis vectors, which are r1 and r2. The transition state would be at the saddle point of the potential energy surface, where the criteria are as follows:

- The first order partial derivative of the potential energy surface w.r.t q1 and q2 are zero.

- The second order partial derivative of the potential energy surface w.r.t q1 is positive and negative w.r.t to q2.

It is important that both criteria to be fulfilled so that the point on the potential energy surface corresponds to the transition state.

A local minimum, while fulfill the first criteria, falls short to fulfill the second because the second order partial derivative of the potential energy surface w.r.t q1 and q2 is positive.
[[File:internuclear_vs_time_jwy17.png|thumb|The internuclear distance between the H atoms at distance of 0.91 Angstrom.]]

== Trasition state of H2 + H System ==
'''"''Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory"'''''<nowiki/>'' ''

At the transition state, where hydrogens A and C are equidistant to the centre hydrogen B, would not oscillate if they have no momenta and do not posses excess potential energy that can be converted to kinetic energy. At 0.91 Angstrom, the Internuclear Distance Vs Time plot of the hydrogen atoms with 0 initial momenta showed a very minute oscillation. Therefore, by inspection (~0.8 Angstrom) and trial and error, the '''rts '''is 0.91 Angstrom.

{{fontcolor1|red| (I think you can report better result for transition state with the aid of contour plot.) }}

== Trajectory of r1 = rts+δ, r2 = rts ==
'''"''Comment on how the mep and the trajectory you just calculated differ." (show the potential energy surface top view dynamics vs mep)'''''

The mep shows a non-oscilating trajectory, as it only displays the the path with the least potential energy. Realistically, the offset of the position of one of the hydrogens puts it on a negative gradient, moving the reaction off the ridge. The vibration arises the the difference between the equilibrium position of the product/reactant and the transition state position. As the equilibrium distance between the 2 hydrogens shift from the transition state position to that of hydrogen molecules, the hydrogen atoms find themselves at a position above the equilibrium. The potential energy can be converted to kinetic energy and vibrate, with inertia keeping it in motion.

== Reactivity ==
'''"''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?"'''''<nowiki/>'' ''
{| class="wikitable"
!P1 AB
!P2 BC
!Etot (kJ/mol)
!Reaction?
!Description of the dynamics
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|Yes
|The molecule does not vibrate much initally and vibrates after passing through the transition state
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|No
|The molecule vibrates but does not overcome the activation barrier and does not form a new molecule.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|Yes
|The molecule and new molecule both vibrates.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|No
|A new molecule was transient, as the molecules breaks to form a new molecule, but reverts back to the orginal molecule.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|Yes
|The new molecule reverts back to the orginal molecule once and then back to the new molecule.
|}
{| class="wikitable"
!P1 AB
!P2 BC
!Reaction?
!Reaction trajectories
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|Yes
|[[File:1p25 2p5 jwy17.png|frameless]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|No
|[[File:1p5 2p0 jwy17.png|frameless|294x294px]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|Yes
|[[File:1p5 2p5 jwy17.png|frameless]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|No
|[[File:-2.5,-5.0_jwy17.png|frameless|292x292px]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|Yes
|[[File:2p5 5p2 jwy17.png|frameless]]
|}
The amount of total energy the system does not indicate whether the reaction is reactive or not. Despite having the enough energy to overcome the activation energy barrier, the reactivity of the reaction is not guaranteed.

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

In Transition State Theory, it is assumed that:

'''1.The reactants and transition state follows a Boltzzman Distribution.'''

In the cases where for rapid reactions, the transition state may disappear rapidly, causing its concentration to be less than expected in the solution. This assumption overestimates the experimental rate value, as the transition state concentration is more than reality.

'''2. The reaction tranjectory only crosses the transition barrier once to form the product.'''

This is not true, as shown in the reactivity section, that can re-cross the barrier to form the reactant or re-cross in the direction of the products twice. This assumption also then overestimates the rate value because the actual value involves multiple re-crossing until it forms the reactants or products, hence less than the prediction.

{{fontcolor1|red| (I think a citation of assumption is required here.) }}

= F-H-H System =

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

'''''Locate the approximate position of the transition state."'''''

By inspection, the reaction is exothermic, as the H + HF'' ''part of the potential energy surface miinium is lower than'' the ''F + H2. This means that HF has a much strong bond strength than the H2 speci

{{fontcolor1|red| (It seems that something is missing here.) }}

[[File:F-H-H transition jwy17.png|thumb| The x marks the tansition state of the F-H-H system, where A is the Fluorine atom. The transition state has a BC distance of 0.74 Angstrom, similar to a H2 equilibrium distance.]]

To find the transition state. The equilibrium position of the H2 are first estimated (~0.74 Angstrom). then by trial and error, the F-H position was adjusted until the trajectory of the reaction is not obvious (~ 1.82 Angstrom). The transition state fits the Hammond postulate, as it is a exothermic reaction and the transition state resembles a H2 molecule.

== Activation energy ==
'''"''Report the activation energy for both reactions"'''''

H + HF to F + H2''''' ''''' : -103. 777 - (-133.269) = 30.0kcal/mol

F + H2 to H + HF: -103.738 - (-103.983) = 0.3 kcal/mol

''*Values of the total energy of both intial and final states are measured using the built in ruler''
[[File: Hf_to_hh_activation_energy_jwy17.png|thumb|The total energy of the system from transition state to HF. The drop in energy equals to the activation energy.]]
[[File: Hh_to_hf_activation_energy_jwy17.png|thumb|The total energy of the system from transition state to H2.]]

== Energy Dissipation ==
'''"''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."'''''

As the HF molecule is formed, the potential surface minimum drops dramamtically. As the total energy is conserved:
# The HF may find itself at a much higher vibrational state. The vibrational energy can be transfered to the H atom as kinetic energy.
# The reaction may directly result in a HF in a ground-state vibrational mode and the H atom realesed gains kinetic energy.
As the kinetic energy of the particles increase, so will the temperature, as they are proportional to each other. Therefore, as we perform the reaction, an increase in temperature or pressure would be observed.

{{fontcolor1|red| (I think you should include more experimental data to support your discussion.) }}

== Polanyi's Rule<ref>L. Arnaut, S. Formosinho and H. Burrows, ''Chemical Kinetics From melecular Structure to Chemical Reactivity, Elsevier, Oxford, 2007''</ref> ==
'''''"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."'''''

'''Early Barrier'''

An early barrier is where the transition state would be on the PES when forming HF + H from F + H2 (Before turning the corner). A translational energy would be more efficient in carrying the trajectory over the transition state. The translational and vibrational energy requires to form the product is a balancing task, given too much of either can cause the tranjectory to 'bounce' back to the reactant.

'''Late Barrier'''

A late barrier is where the transition state would be on the PES when forming HF + H from F + H2 (After turning the corner). To access the transition state the HF molecule has to have a massive vibrational energy (large momentum). As with the early barrier, translational energy cannot be ignored and the right amount of translational energy is required still to reach the transition state.[[File:Late barrier jwy17.png|thumb|An example reactive trajectory of a late barrier (HF + H to H2 + F). The large vibrational energy of the HF will be required to reach the transition state.|left]][[File:Early barrier jwy17.png|An example trajectory of early barrier (H2 + F to HF + H). The H2 is very small vibrational energy, and has enough translational energy to move over the tansition state. |thumb|centre]]

{{fontcolor1|red| Overall, this is a good report. However, more experimental data should be included to support the discussion.}}

MRD:mz5717

2019-05-11T16:05:47Z

Cq3417:

== H + H2 SYSTEM ==

=== Transition State Determination ===

On a potential energy surface, the transition state is represented by a saddle point, which is mathematically defined as:

When the partial derivatives Vr1 and Vr2 both equal to 0 at a point, if the value of (Vr1r1Vr2r2 - V2r1r2) < 0 at that same point, then it is a saddle point.

Alternatively we can identify it by computing the second partial derivatives, Vr1r1 and Vr2r2. For a saddle point, one of them will be positive and the other one negative.

Additionally, when (Vr1r1Vr2r2 - V2r1r2) > 0, the point represents a local maximum/minimum. It is a local maximum if Vr1r1 < 0, or a local minimum if Vr1r1 > 0.

The transition state position is estimated to be 0.90743 Å. i.e. r1 = r2 = 0.90743 Å, p1 = p2 = 0. When subjected to this initial condition, the internuclear distances do not vary with time because the equilibrium established at the saddle point does not provide driving force for any forward or backward reaction.

This is illustrated with the plot below. The contour map shows the position of the TS with a cross.

{{fontcolor1|red|(Reasonable description of the transition state and good illustration of plots)}}

{| class="wikitable" border=1
|-
| [[File:TS_distances_mz5717.jpeg|thumb|400px|none|Figure 1: time-variation of internuclear distances.]] || [[File:TS_contour_mz5717.jpeg|thumb|400px|none|Figure 2: contour map of the potential energy surface showing the position of the TS.]]
|}

=== Reaction Trajectories ===

Dynamics and MEP are two methods of calculating the reaction trajectory. The table below compares the outcomes produced by these two methods.

{| class="wikitable" border=1
|-
| '''Dynamics''' (r1 = 0.90743, r2 = 0.91743, p1 = p2 = 0) || [[File:Dynamics_mz5717.jpeg|thumb|400px|none|Figure 3: reaction trajectory produced by Dynamics.]] || [[File:Dynamics_momenta_mz5717.jpeg|thumb|400px|none|Figure 4: time-variation of momenta produced by Dynamics. Blue curve shows the momentum oscillation of the new bond.]] ||[[File:Dynamics_distance_mz5717.jpeg|thumb|400px|none|Figure 5: time-variation of internuclear bond length produced by Dynamics. Blue curves shows the bond length oscillation of the new bond.]]|| The Dynamics calculation accounts for the conversion of kinetic energy to vibrational energy, as shown by the oscillating momentum and bond length of the new bond formed.
|-
| '''MEP''' (r1 = 0.90743, r2 = 0.91743, p1 = p2 = 0)|| [[File:MEP_mz5717.jpeg|thumb|400px|none|Figure 6: reaction trajectory produced by MEP.]] || [[File:MEP_momenta_mz5717.jpeg|thumb|400px|none|Figure 7: time-variation of momenta produced by MEP. There is zero motion.]] || [[File:MEP_distance_mz5717.jpeg|thumb|400px|none|Figure 8: time-variation of internuclear bond length produced by MEP. The length of the new bond stays constant.]] ||The MEP calculation does not account for the bond vibration, therefore the momentum of the newly formed molecule is a straight line, and there is no oscillation in the new bond.
|}

According to Figures 4 and 5, we set the initial conditions to be r1 = 0.74895, r2 = 1.0 (an arbitrary large distance), p1 = -1.2745, p2 = -2.4889, which corresponds with the final position of the trajectory only with the signs of the final momenta reversed. The resulting trajectory is almost the same as those we obtained previously.

{{fontcolor1|red|(Good explanation.)}}

[[File:reverse_contour_mz5717.jpeg|thumb|400px|none|Figure 9: initial conditions corresponds to the final values of the above trajectory but with the signs of the momenta reversed. A very similar reaction path.]]

Next, we examine the effect of initial momenta and total energy on the reaction trajectory, i.e. if it is reactive or not, by giving the reactants varying amounts of initial momenta. The results are summarized by the table below.

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Reaction path !! Description of the dynamics
|-
| -1.25 || -2.5 || -99.018 || Y || [[File:Dynamics1_mz5717.jpeg|300px]] || More internal vibration after the transition state; energy sufficient to pass over Ea.
|-
| -1.5 || -2.0 || -100.456 || N || [[File:Dynamics2_mz5717.jpeg|300px]] || Reaction does not proceed; the molecules bounce back without reacting.
|-
| -1.5 || -2.5 || -98.956 || Y || [[File:Dynamics3_mz5717.jpeg|300px]] || Considerable internal vibration before & after reaching the TS; more than enough energy to overcome Ea.
|-
| -2.5 || -5.0 || -84.956 || N || [[File:Dynamics4_mz5717.jpeg|300px]] || The kinetic energy supplied is way above the activation energy so the reaction path goes past the saddle point. However it bounces back after a while, with most of the energy contributed to intramolecular vibration. So the starting materials ARE reactive and we do have products formed, but they eventually turn back to the reactants.
|-
| -2.5 || -5.2 || -83.416 || Y || [[File:Dynamics5_mz5717.jpeg|300px]] || Reaction proceeds to the formation of the products with much of the kinetic energy contributing to the internal vibration of the newly formed product.
|}

{{fontcolor1|red|(Excellent illustration though additionally showing the energy vs time plot can make it better.)}}

=== Transition State Theory ===

The main assumptions of Transition State Theory are:

- The nuclei behave classically

- The reaction path passes through the saddle point representing the transition state

- Once the reaction passes through the TS, it will proceed to the formation of the product(s)

Under most circumstances, TST matches with experimental results to a good extent. However, from the simulation we can conclude that when the kinetic energy supplied is way above the activation energy, this is not necessarily the case. Here the experimental position of the transition state may largely deviate from the theoretical one represented by the saddle point on the potential energy surface. Because of the abundance of energy supplied, more molecules are able to occupy vibrational modes of higher energy, making their motion more complicated and harder to predict. This could be the underlying reason for the large deviation. It could be hard as well to predict the reaction rate under such circumstances; for example, when there is {{fontcolor1|red|a}} reverse reaction going on.

{{fontcolor1|red|(I think you may need to cite the description of this assumption.)}}

== F - H - H SYSTEM ==

=== Transition State Determination ===

F + H2 --> H + HF is exothermic and has an early TS. The reaction releases energy because the new H-F bond formed is stronger than the H-H bond broken (F is much more electronegative than H). On the other hand, H + HF --> F + H2 is endothermic and has a late TS. The reaction requires input of energy because the H-F bond broken is much stronger than the H-H bond formed.

Setting initial momenta to be both zero, the position of the TS is estimated to be r1 = 0.745 Å and r2 = 1.80 Å.

{{fontcolor1|red|(Good, nice and clear.)}}

{| class="wikitable" border=1
|-
| [[File:TS_HF_contour_mz5717.jpeg|thumb|400px|none|Figure 10: position of the TS marked by a cross.]] || [[File:TS_HF_distance_mz5717.jpeg|thumb|400px|none|Figure 11: no bond length variation when at equilibrium.]]
|}

=== Activation Energy ===

{| class="wikitable" border=1
|-
| [[File:Ea_energy_mz5717.jpeg|thumb|400px|none|Figure 12: reaction profile of H + HF --> F + H2.]] || [[File:Ea_energy_small_mz5717.jpeg|thumb|400px|none|Figure 13: reaction profile of F + H2 --> H + HF.]]
|}

The figures above show the energy profiles of H + HF and F + H2 respectively.

The activation energy of H + HF is determined as -103.867 - (-133.257) = 29.390 kcal/mol

The activation energy of F + H2 is determined as -103.752 - (-103.967) = 0.215 kcal/mol

{{fontcolor1|red|(How you get this plot is my question so I think you need to discuss this in detail.)}}

=== Energy Distribution ===

First we focus on energy conversion. Below is a general case of a successful reaction.

{| class="wikitable" border=1
|-
| [[File:Energy_contour1_mz5717.jpeg|thumb|400px|none|Figure 14: a generic reactive path with r1 = 0.745, r2 = 1.40, p1 = 3, p2 = -0.5.]] || [[File:Energy_momenta1_mz5717.jpeg|thumb|400px|none|Figure 15: corresponding time-variation of momenta of this reaction path.]]
|}

Figure 15 shows that the vibration of bond A-B gradually ceases, which means that bond A-B corresponds to the H-H bond broken. In this case bond B-C must be the new H-F bond formed. The H-F bond continues its vibration and eventually becomes the only source of momentum, which means that the excess kinetic energy from the F atom was converted to vibrational energy after the reaction has occurred. The extent and modes of the H-F bond vibration can be measured by Raman spectroscopy.

{{fontcolor1|red|(It can be more clear if you include the energy vs time plot here.)}}

Next, for the reaction F + H2 --> H + HF, we set the initial conditions to rHH = 0.74, rHF = 1.80 and pFH = -0.5. Different values of pHH, ranging from -3 to 3, are tested and the results are summarized in the table below.

{| class="wikitable" border=1
|-
| [[File:--3_mz5717.jpeg|thumb|200px|none|pHH = -3]] || [[File:--2.4_mz5717.jpeg|thumb|200px|none|pHH = -2.4]] || [[File:--1.8_mz5717.jpeg|thumb|200px|none|pHH = -1.8]] || [[File:--1.2_mz5717.jpeg|thumb|200px|none|pHH = -1.2]] || [[File:--0.6_mz5717.jpeg|thumb|200px|none|pHH = -0.6]]
|-
| [[File:-0_mz5717.jpeg|thumb|200px|none|pHH = 0]] || [[File:---0.6_mz5717.jpeg|thumb|200px|none|pHH = 0.6]] || [[File:---1.2_mz5717.jpeg|thumb|200px|none|pHH = 1.2]] || [[File:---1.8_mz5717.jpeg|thumb|200px|none|pHH = 1.8]] || [[File:---2.4_mz5717.jpeg|thumb|200px|none|pHH = 2.4]] || [[File:---3_mz5717.jpeg|thumb|200px|none|pHH = 3]]
|}

The table shows that when pHH has a value between -1.2 and 0, the reaction proceeds to the product side. In all other cases the reaction trajectory more or less bounces back to the reactant side, and the extent of this backward conversion increases as the absolute values of pHH increases. The only outlier is pHH = 2.4, where the trajectory generally stays on the product side.

Now increase pHF slightly to -0.8, and largely reduce pHH to 0.1. The trajectory below shows that despite having considerably lower total energy, the reaction still proceeds to the product side.

[[File:reduced_energy_mz5717.jpeg|thumb|400px|none|Figure 16: pHF = -0.8, pHH = 0.1, rHH = 0.74, rHF = 1.80. The total energy of the system has been significantly reduced.]]

Finally, for the reaction H + HF --> F + H2, we assume that there is very little vibration within the H-F bond (i.e. pHF = 0.2), and that the H atom is approaching with extremely high kinetic energy (pHH = 5). With the initial conditions rHH = 0.745, rHF = 1.80, the trajectory below is obtained. It is evident that the high kinetic energy possessed by the H atom is initially able to break the H-F bond and push the system to the product side; however, HF is reformed afterwards and then the entire process repeats. This is an oscillating reaction.

[[File:oscillate_mz5717.jpeg|thumb|400px|none|Figure 17: rHH = 0.745, rHF = 1.80, pHF = 0.2, pHH = 5. This set of initial conditions results in an oscillating reaction.]]

By decreasing pHH and increasing pHF, it is possible to obtain a reactive trajectory. For example, when we let other variables stay constant and change the rest to pHF = 0.3, pHH = 2.6, the reaction can indeed proceed to the product side as shown in the figure below.

[[File:balance_mz5717.jpeg|thumb|400px|none|Figure 18: when rHH = 0.745, rHF = 1.80, pHF = 0.3, pHH = 2.6, we obtain a reactive trajectory.]]

{{fontcolor1|red|(Where is your discussion that respects to the Polanyi's empirical rules?)}}

In conclusion, a proper distribution of kinetic and vibrational energy is crucial for the trajectory to be reactive. Generally in most cases, a high kinetic energy nearly guarantees a reactive path. The reaction F + H2 --> H + HF has an early TS, and does not need much energy to reach it; the other reaction H + HF --> F + H2 has a late TS, and requires much higher vibrational energy to reach it. The second reaction also shows us that just having high kinetic energy is not enough to trigger a successful reaction. By lowering the kinetic energy and increasing the vibrational energy we see that there is a delicate balance between the two that eventually leads to a reactive trajectory.

{{fontcolor1|red|(Overall this is a sufficient report that shows a good understanding of this concept. However, the citation of reference and, the combination of theories and experimental data are required to be noticed.)}}

MRD:zz161701327485

2019-05-03T20:03:01Z

Cq3417:

== '''Exercise 1''' ==

'''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 as the point that the gradient of the potential equals to zero. It can be expressed as '''r'''i where ∂ V(ri)/∂ ri=0. On a potential energy surface diagram, the transition state is identified as the saddle part of the minimum energy path. In the diagram below, the transition state is where on the minimum energy path (the black line) AB = 0.75-1.00 and BC = 1.00-1.25. The transition state is not only a local minimum of the potential energy surface but where the energy goes down most steeply along the minimum energy path linking reactants and products.
 (The definition is ok but you need to say '''second partial''' derivative. Besides, pay more attention on how you describe this concept.)

[[File:Surface_Plot-1.png]]

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

In the plot below, it is set that r1=r1=1, p1=p2=0. The best estimate of the transition state position is where AC is minimum, where T = 0.156 or 0.473 or 0.797, AC = 1.654, rts= AB = BC = 0.825. This is the transition point because the gradient of the potential is zero at the point (local minimum of the potential energy surface), and AC reaches the minimum which means the greatest probability of interactions among A, B and C.

 (Seems that your system didn't reach the transition state and I think you can report better result.)

[[File:Animation-11.png]]

'''Comment on how the mep and the trajectory you just calculated differ. '''
[[File:Plot-323.png|thumb|none|350px|"dynamics"]]
[[File:Plot-324.png|thumb|none|350px|"MEP"]]
For the mep, atom C almost always keeps a same distance (around 0.75) from atom B, while atom deviates from atom B quickly with an increasing speed. This shows that BC becomes a molecule after the transition state.
Foe (for) the trajectory, A-B and B-C keep almost the same distance until T=0.6, and later B-C vibrates more greatly than A-B. In this case, it is not clear to see which two atoms become a molecule. The three atoms are reasonably close to each other at all times.
 (I didn't get your point.)

'''Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. Take note of the final values of the positions r1(t) r2(t) and the average momenta p1(t) p2(t) at large t. What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. What do you observe? '''

[[File:Plot-325.png|thumb|none|350px|"p1(t) and p2(t)"]]

In the trajectory mode, the final value of r1 = 1.00, r2 = 0.88. Both of the average momenta at large t equals to zero. If we used the initial conditions r'1 = rts and r'2 = rts+0.01 instead, the graph of r'1(t) and r'2(t) would interconvert, where r'1(t) = r2(t), r'2(t) = r1(t), p'1(t) = p2(t) and p'2(t) = p1(t).

 (You confused me.)

'''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" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99.018 || Reactive || Molecule AB (with weak vibration) and atom C moves close to each other, BC becomes a molecule and moves away from C, molecule BC vibrates weakly
|-
| -1.5 || -2.0 || -100.456 || Unreactive || Molecule AB (with weak vibration) and atom C moves close to each other, and moves away; molecule AB vibrates weakly
|-
| -1.5 || -2.5 || -98.956 || Reactive || Molecule AB and atom C moves close to each other, BC becomes a molecule and moves away from C, where atom A has a greater speed than BC, molecule BC vibrates weakly
|-
| -2.5 || -5.0 || -84.956 || Unreactive || Molecule AB and atom C moves close to each other, BC gets closer to each other (with great vibration) shortly, then AB becomes a molecule (with great vibration) again and moves away from C
|-
| -2.5 || -5.2 || -83.416 || Reactive|| Molecule AB and atom C moves close to each other, then BC gets closer (with great vibration) shortly, then AB gets closer (with great vibration) again shortly. After a long transition state, BC finally becomes a molecule (with great vibration)
|}

From the table above, we can conclude that the reaction is reactive when there is a relatively large difference between p1 and p2, otherwise, the reaction is unreactive. When p1 and p2 are relatively low (more negative), they will experience a longer transition state, and molecules would vibrate more strongly.

 (I think you need to emphasise the role of transition state here.)

[[File:mk-1.png|thumb|none|350px|"p1 = -1.25, p2 = -2.5"]]
[[File:mk-2.png|thumb|none|350px|"p1 = -1.5, p2 = -2.0"]]
[[File:mk-3.png|thumb|none|350px|"p1 = -1.5, p2 = -2.5"]]
[[File:mk-4.png|thumb|none|350px|"p1 = -2.5, p2 = -5.0"]]
[[File:mk-5.png|thumb|none|350px|"p1 = -2.5, p2 = -5.2"]]

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

The main assumptions of Transition State Theory are: Transition state is defined as the saddle point of the potential energy surface. The activated complexes would reach quasi-equilibrium with reactants and could be converted to products. [1]

From the experimental results, we can confirm that all reactive reactions have a transition state. Unreactive reactions may have a transition state (4th example, p1=-2.5, p2=-5.0) and may not have a transition state (2nd example, p1=-1.5, p2=-2.0). In the 4th example, the activated complexes are in equilibrium with the reactant molecules but did not convert into products.

 (You need to discuss this in detail and show your understanding of this topic.)

== '''Exercise 2''' ==

'''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? Locate the approximate position of the transition state. Report the activation energy for both reactions. '''

F + H2 is an exothermic reaction, and H + HF is an endothermic reaction. This is reasonable because H-H bond energy is 432 KJ/mol, and H-F bond energy is 565 KJ/mol [2]. Therefore, it requires more energy to break an H-F bond than forming an H-H bond, vice versa.
The approximate position of the transition state of H+HF is AB=3.45, BC=0.92. The approximate position of the transition state of F+H2 is AB=1.82, BC=0.74.
The activation energy of H+HF is 29, and the activation energy of F+H2 is 0.2.

 (Again, detail please.)

[[File:Surface_Plot-z1.png|thumb|none|350px|"transition state position for H+HF"]]
[[File:Surface_Plot-z2.png|thumb|none|350px|"transition state position for F+H2"]]
[[File:Surface_Plot-z3.png|thumb|none|350px|"Activation Energy for H+HF"]]
[[File:Surface_Plot-z4.png|thumb|none|350px|"Activation Energy for 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.'''

When r(HF)=1.82, r(HH)=0.74, p(F)=-0.7, p(HH)=-0.1, the reaction could exactly happens.

[[File:Surface_Plot-z9.png]]

For exothermic reactions, some energy in the system will be transformed to heat and release to the surroundings. Experimentally, this can be observed by rise of temperature of the surroundings and decrease of temperature of the system.

For endothermic reactions, the system would absorb energy from the surroundings. Experimentally, this can be observed by decrease of temperature of the surroundings and rise of temperature of the system.

F+H2 is an exothermic reaction. So experimentally, decrease of temperature of the surroundings and rise of temperature of the system should be observed. From the graph above, we can see that the new molecule F-H vibrates greatly.

 (Seems that you are not fully understand what is happening.)

 (I think you missed some questions.)

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

Polanyi's empirical rules state that vibrational energy is more efficient than translational energy in activating a late-barrier reaction, but the reverse is true for an early-barrier reaction [3].

Experimentally, for reaction F+H2, where r(HH)=0.74, p(FH)=-0.5, when p(HH) is less than 1.76, the reaction could happen, and otherwise it cannot happen. F+H2 is an exothermic reaction and is a early-barrier reaction. From the graph, we can see that the reactions happens when there is more translational energy than vibrational energy.

Experimentally, H+HF is an endothermic reaction and is a late-barrier reaction. From the graph, we can see that the reactions happens when there is more vibrational energy than translational energy.

Therefore, Polanyi's empirical rules can be proved by the experiments.

[[File:F+H2, yes.png|thumb|none|350px|"F+H2, the reaction happens"]]
[[File:F+H2, no.png|thumb|none|350px|"F+H2, the reaction does not happen"]]
[[File:H+HF, yes.png|thumb|none|350px|"H+HF, the reaction happens"]]
[[File:H+HF,_no2.png|thumb|none|350px|"H+HF, the reaction does not happen"]]

 (Overall, your report is not good since its not clear and elaborate enough.)

== '''References''' ==

[1]D. M. Hirst Potential Energy Surfaces: Molecular Structure and Reaction Dynamics Taylor and Francis, 1985

[2]Chemical Bond Energies: http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html

[3] Wiley-VCH Verlag GmbH & Co. KGaA, Weinheim: Dynamics of the Reaction of Methane with Chlorine Atom on an Accurate Potential Energy Surface, 2011

MRD:zz161701327485

2019-05-03T19:04:30Z

Cq3417: /* Exercise 1 */

== '''Exercise 1''' ==

'''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 as the point that the gradient of the potential equals to zero. It can be expressed as '''r'''i where ∂ V(ri)/∂ ri=0. On a potential energy surface diagram, the transition state is identified as the saddle part of the minimum energy path. In the diagram below, the transition state is where on the minimum energy path (the black line) AB = 0.75-1.00 and BC = 1.00-1.25. The transition state is not only a local minimum of the potential energy surface but where the energy goes down most steeply along the minimum energy path linking reactants and products.
 (The definition is ok but you need to say '''second partial''' derivative. Besides, pay more attention on how you describe this concept.)

[[File:Surface_Plot-1.png]]

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

In the plot below, it is set that r1=r1=1, p1=p2=0. The best estimate of the transition state position is where AC is minimum, where T = 0.156 or 0.473 or 0.797, AC = 1.654, rts= AB = BC = 0.825. This is the transition point because the gradient of the potential is zero at the point (local minimum of the potential energy surface), and AC reaches the minimum which means the greatest probability of interactions among A, B and C.
 (Seems that your system didn't reach the transition state and I think you can report better result.)

[[File:Animation-11.png]]

'''Comment on how the mep and the trajectory you just calculated differ. '''
[[File:Plot-323.png|thumb|none|350px|"dynamics"]]
[[File:Plot-324.png|thumb|none|350px|"MEP"]]
For the mep, atom C almost always keeps a same distance (around 0.75) from atom B, while atom deviates from atom B quickly with an increasing speed. This shows that BC becomes a molecule after the transition state.
Foe the trajectory, A-B and B-C keep almost the same distance until T=0.6, and later B-C vibrates more greatly than A-B. In this case, it is not clear to see which two atoms become a molecule. The three atoms are reasonably close to each other at all times.

'''Look at the “Internuclear Distances vs Time” and “Momenta vs Time”. Take note of the final values of the positions r1(t) r2(t) and the average momenta p1(t) p2(t) at large t. What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead? Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. What do you observe? '''

[[File:Plot-325.png|thumb|none|350px|"p1(t) and p2(t)"]]

In the trajectory mode, the final value of r1 = 1.00, r2 = 0.88. Both of the average momenta at large t equals to zero. If we used the initial conditions r'1 = rts and r'2 = rts+0.01 instead, the graph of r'1(t) and r'2(t) would interconvert, where r'1(t) = r2(t), r'2(t) = r1(t), p'1(t) = p2(t) and p'2(t) = p1(t).

'''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" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics
|-
| -1.25 || -2.5 || -99.018 || Reactive || Molecule AB (with weak vibration) and atom C moves close to each other, BC becomes a molecule and moves away from C, molecule BC vibrates weakly
|-
| -1.5 || -2.0 || -100.456 || Unreactive || Molecule AB (with weak vibration) and atom C moves close to each other, and moves away; molecule AB vibrates weakly
|-
| -1.5 || -2.5 || -98.956 || Reactive || Molecule AB and atom C moves close to each other, BC becomes a molecule and moves away from C, where atom A has a greater speed than BC, molecule BC vibrates weakly
|-
| -2.5 || -5.0 || -84.956 || Unreactive || Molecule AB and atom C moves close to each other, BC gets closer to each other (with great vibration) shortly, then AB becomes a molecule (with great vibration) again and moves away from C
|-
| -2.5 || -5.2 || -83.416 || Reactive|| Molecule AB and atom C moves close to each other, then BC gets closer (with great vibration) shortly, then AB gets closer (with great vibration) again shortly. After a long transition state, BC finally becomes a molecule (with great vibration)
|}

From the table above, we can conclude that the reaction is reactive when there is a relatively large difference between p1 and p2, otherwise, the reaction is unreactive. When p1 and p2 are relatively low (more negative), they will experience a longer transition state, and molecules would vibrate more strongly.

[[File:mk-1.png|thumb|none|350px|"p1 = -1.25, p2 = -2.5"]]
[[File:mk-2.png|thumb|none|350px|"p1 = -1.5, p2 = -2.0"]]
[[File:mk-3.png|thumb|none|350px|"p1 = -1.5, p2 = -2.5"]]
[[File:mk-4.png|thumb|none|350px|"p1 = -2.5, p2 = -5.0"]]
[[File:mk-5.png|thumb|none|350px|"p1 = -2.5, p2 = -5.2"]]

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

The main assumptions of Transition State Theory are: Transition state is defined as the saddle point of the potential energy surface. The activated complexes would reach quasi-equilibrium with reactants and could be converted to products. [1]

From the experimental results, we can confirm that all reactive reactions have a transition state. Unreactive reactions may have a transition state (4th example, p1=-2.5, p2=-5.0) and may not have a transition state (2nd example, p1=-1.5, p2=-2.0). In the 4th example, the activated complexes are in equilibrium with the reactant molecules but did not convert into products.

== '''Exercise 2''' ==

'''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? Locate the approximate position of the transition state. Report the activation energy for both reactions. '''

F + H2 is an exothermic reaction, and H + HF is an endothermic reaction. This is reasonable because H-H bond energy is 432 KJ/mol, and H-F bond energy is 565 KJ/mol [2]. Therefore, it requires more energy to break an H-F bond than forming an H-H bond, vice versa.
The approximate position of the transition state of H+HF is AB=3.45, BC=0.92. The approximate position of the transition state of F+H2 is AB=1.82, BC=0.74.
The activation energy of H+HF is 29, and the activation energy of F+H2 is 0.2.

[[File:Surface_Plot-z1.png|thumb|none|350px|"transition state position for H+HF"]]
[[File:Surface_Plot-z2.png|thumb|none|350px|"transition state position for F+H2"]]
[[File:Surface_Plot-z3.png|thumb|none|350px|"Activation Energy for H+HF"]]
[[File:Surface_Plot-z4.png|thumb|none|350px|"Activation Energy for 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.'''

When r(HF)=1.82, r(HH)=0.74, p(F)=-0.7, p(HH)=-0.1, the reaction could exactly happens.

[[File:Surface_Plot-z9.png]]

For exothermic reactions, some energy in the system will be transformed to heat and release to the surroundings. Experimentally, this can be observed by rise of temperature of the surroundings and decrease of temperature of the system.

For endothermic reactions, the system would absorb energy from the surroundings. Experimentally, this can be observed by decrease of temperature of the surroundings and rise of temperature of the system.

F+H2 is an exothermic reaction. So experimentally, decrease of temperature of the surroundings and rise of temperature of the system should be observed. From the graph above, we can see that the new molecule F-H vibrates greatly.

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

Polanyi's empirical rules state that vibrational energy is more efficient than translational energy in activating a late-barrier reaction, but the reverse is true for an early-barrier reaction [3].

Experimentally, for reaction F+H2, where r(HH)=0.74, p(FH)=-0.5, when p(HH) is less than 1.76, the reaction could happen, and otherwise it cannot happen. F+H2 is an exothermic reaction and is a early-barrier reaction. From the graph, we can see that the reactions happens when there is more translational energy than vibrational energy.

Experimentally, H+HF is an endothermic reaction and is a late-barrier reaction. From the graph, we can see that the reactions happens when there is more vibrational energy than translational energy.

Therefore, Polanyi's empirical rules can be proved by the experiments.

[[File:F+H2, yes.png|thumb|none|350px|"F+H2, the reaction happens"]]
[[File:F+H2, no.png|thumb|none|350px|"F+H2, the reaction does not happen"]]
[[File:H+HF, yes.png|thumb|none|350px|"H+HF, the reaction happens"]]
[[File:H+HF,_no2.png|thumb|none|350px|"H+HF, the reaction does not happen"]]

== '''References''' ==

[1]D. M. Hirst Potential Energy Surfaces: Molecular Structure and Reaction Dynamics Taylor and Francis, 1985

[2]Chemical Bond Energies: http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html

[3] Wiley-VCH Verlag GmbH & Co. KGaA, Weinheim: Dynamics of the Reaction of Methane with Chlorine Atom on an Accurate Potential Energy Surface, 2011