File:Yu ea 1.PNG

2018-05-25T16:19:31Z

Yr316:

Rep:YR316DYN

2018-05-25T16:19:14Z

Yr316: /* Question 8: Activation energy */

=Molecular Reaction Dynamics=

== H + H2 system==

===Question 1: minimum and saddle===

''What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.''

In this triatomic system where the HA is approaching HBHC with an initial momentum. r1 is the distance between HA and HB and r2 is the distance between HB and HC. The total potential energy of the system (<math>V(r_1,r_2)</math>) is changing with respect to both r1 and r2. Hence the gradient of potential energy consists of two components:<math> { \partial V(r_1,r_2)\over \partial r_1}</math> and <math> { \partial V(r_1,r_2)\over \partial r_1}</math>. The minima correspond to the minimum potential energy of the oscillating system HBHC (before collision)or HAHB (after collision). Mathematically, the minimum can be expressed as <math> { \partial V(r_1,r_2)\over \partial r_2}</math>=0 and <math> { \partial V(r_1,r_2)\over \partial r_1}</math>=0 respectively. <math> { \partial V(r_1,r_2)\over \partial r_2}</math> is partial differentiation, it differentiates V with respect to r2 while treating r1 as constant. The curvatures of the minima are described by the second partial derivative: Vr1r1 and Vr1r1 and smaller 0. They are the minimum point viewing from cross sections along r1 and r2 respectively. It is the same when Hc is leaving after the collision.

However, as said at the beginning while HBHC are oscillating, HA is approaching and this increases the potential energy of the system. The different minima at different r1 values increases to a saddle point. This is the maximum point in the trough created by joining the minima when viewing from potential surface. At this point both <math> { \partial V(r_1,r_2)\over \partial r_2}</math> and <math> { \partial V(r_1,r_2)\over \partial r_1}</math> are 0. The curvature at this point can be described by the determinant: Vr1r1•Vr1r1-V2r1r2. This point corresponds to the transition state of the system.

===Question 2: estimation of transition state===

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

The transition state is estimated to be at r1=r2=0.90773 as shown in figure 1.

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

figure 1. “Inter-nuclear Distances vs Time” plot at stationary point

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

figure2. “Inter-nuclear Distances vs Time” plot under one initial condition

Figure 1 shows under one initial condition when HA collides with HB HC the intercept of the A-B and B-C line gives a rough estimation of where the transition would be. A more accurate estimation can be done by setting the momentum to zero while changing values for distances (keep r1=r2 as the potential surface of H + H2 is symmetric. The trajectory would be stationary i.e. distance does not vary with time for a transition state as suggested by the definition of a stationary point(saddle as an example). while at other point it would produce a oscillation pattern over time as shown in figure 3.

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

figure 3. “Inter-nuclear Distances vs Time” plot at stationary point at other r1=r2 points other then the stationary points

===Question 3: comparing dynamics and MEP===

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

Figure 4 and 5 shows the reaction path calculation under dynamic method and MEP method respectively. Both of them have initial condition r1=0.91773, r2=0.90773, p1=p2=0. The pathway calculated with dynamic method is wavy, while the pathway calculated with MEP method is a smooth curve. MEP stands for minimum energy pathway, in each step of MEP the momentum was reset to zero, hence the inter-conversion between kinetic and the oscillation are ignored. HBH1 of the molecule was ignored and was set at minimum potential energy along the pathway. The step number chosen for dynamics was 150 steps while that of the MEP calculation was chosen to be 10000. The reason for MEP has a much larger step number is also because that in each step of MEP the momentum was reset to zero and the inertia of the particles are ignored, while the momentum in the dynamic method was cumulative. Hence, if the same step number was chosen for both method, MEP result would give a much shorter of path length due to this reset of momentum.

[[File:Yu_dynamic_150steps.PNG|400px]]

Figure 4. Pathway calculation with dynamic method

[[File:Yu_MEP_10000_steps.PNG|400px]]

Figure 5. Pathway calculation with MEP method

If the initial conditions was set to r1 = rts and r2 = rts+0.01 instead, the reaction pathways would be a mirror image of r1=rts+0.01 and r2=rts condition with orthogonal plane r1=r2 as the mirror plane as shown in figure 6. This is due to the fact that the initial position of the trajectory was reversed.
If the initial positions was set to be the same as the final positions of the previous trajectory, the new resultant trajectory would represent the reverse reaction of the previous trajectory.

[[File:yu_mirror1.PNG|400px]]
[[File:yu_mirror2.PNG|400px]]

Figure 6. mirror image to each other: r1=rts+0.01 and r2=rts(left) r1 = rts and r2 = rts+0.01 (right) with steps extended to 500

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

Figure 7. pathway initial positions correspond to the final positions with opposite momenta

=== Question 4: reactive and unreactive trajectory ===

''Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.''

{| class="wikitable" border=1
! p1 !! p2!!Energy
|-
| -1.25 || -2.5 || -99.018
|-
| -1.5 || -2.0 || -100.456
|-
| -1.5 || -2.5 || -98.956
|-
| -2.5 || -5.0 || -84.956
|-
| -2.5 || -5.2 || -83.416
|}

table 1. Energy tabel with various initial monenta when r1=0.74 and r2= 2.0

[[File:yu_q3_1.PNG|400px]]

Figure 7. trajectory with p1=-1.25 and p2=-2.5

'''Comment:''' Reactive

[[File:yu_q3_2.PNG|400px]]

Figure 8. trajectory with p1=-1.5 and p2=-2.0

'''Comment:''' Unreactive

[[File:yu_q3_3.PNG|400px]]

Figure 9. trajectory with p1=-1.5 and p2=-2.5

'''Comment:''' Reactive

[[File:yu_q3_4.PNG|400px]]

Figure 10. trajectory with p1=-2.5 and p2=-5.0

'''Comment:''' Unreactive

[[File:yu_q3_5.PNG|400px]]

Figure 11. trajectory with p1=-2.5 and p2=-5.2

'''Comment:''' Reactive

=== Question 5: 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?''

My answer:

The main assumption of transition state theory are give as:

1. The reactants (or products) are in equilibrium with the activated complex.

2. Quantum-tunneling effects are negligible and the Born-Oppenheimer approximation is invoked.

3. Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.

According to the assumptions, the transition state prediction for rate of reaction should be slower than the experimental value. It can be rationalized by the following two reasons:

1. As stated in the second assumption above. Transition State Theory predictions ignores quantum tunneling effects and treat them as noreactive, where in reality particles with energies lower than the activation energy could react by tunneling through the activation energy barrier.

2. The third assumption can be interpreted that the reaction trajectory are only allowed to cross the transition state once. However, the calculated reaction trajectory for last initial condition in question 4 shows that the reaction trajectory could cross the transition states multiple times and eventually reach the product side. This reactive condition is also ignored by the transition state prediction.

==The study F-H-H system==

===Question 6: energetic of the reactions===

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?

My answer:

Figure 12 and figure 13 shows the potential surface of a F + H2 reaction and a H + HF reaction respectively.

[[file: yu_FHH_condition.PNG|400px]]

Figure 12. Potential surface of Yu_HHF_condition.PNG reaction

In this condition the F atom is approaching the H2 molecule with the momentum showing in the figure 12. The F + H2 has a higher potential energy then HF+H. This shows that the reaction: F + H2 ---> HF + H is exothermic. This result indicates that H-F which is formed is bond is stronger than the H-H which is broken.

[[file: Yu_HHF_condition.PNG|400px]]

Figure 13. Potential surface of H + HF reaction

In this condition the H atom is approaching the HF molecule with the momentum showing in the figure 13. The H + HF has a lower potential energy then H2 + F. This shows that the reaction: H + HF ---> H2 + F is endothermic. This result agrees with the previous result that H-H which is formed is bond is weaker than the H-F which is broken.

===Question 7: Transition states===

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

My answer:
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 first reaction: F + H2 = HF + H is exothermic. Hence, the transition state is closer in energy to the reactants: F + H2 and it adopts the following asymmetric geometry:

F------H--H

•H-H bond is only slightly elongated

•F atom is much further away

[[file:yu_ts_FHH.PNG|400px]]
[[file:yu_stationary_1.PNG|400px]]

This is known as an early transition state. The method used to find this transition state is similar to the one used for H-H-H transition state. The H-H (BC) distance can be interpreted to be around 0.74 Å which is a typical H-H bond length. As shown in figure the transition state is at F-H=1.8085 Å and H-H=0.7509. Figure shows that the point is a stationary point and it lies closely to the reactants.

The second reaction: H + HF= H2 + F is endothermic. Hence, the transition state is closer in energy to the products: F + H2 and it adopts the following asymmetric geometry:

H--H-------F

•H-F bond is largely elongated

•H-H bond is almost formed

[[file:yu_ts_HHF.PNG|400px]]
[[file:yu_stationary_2.PNG|400px]]

This is known as a late transition state. The H-H (AB) distance can be interpreted to be around 0.74 Å. As shown in figure the transition state is at H-H=0.7495 Å and H-H=1.8080. Figure shows that the point is a stationary point and it lies closely to the products.

===Question 8: Activation energy===

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

My answer:

Activation energy is the energy between the reactant and the transition state. If the initial condition is set to be slightly displaced from the transition state and closer to the reactant position, the trajectory will end in the reactant condition. The energy vs time graph would start as a plateau which resembles the energy of the transition state and drops and finally reaches a plateau with resembles the energy of the reactant. The energy difference between the two plateau would give the activation energy of the reaction.

For F + H2 ---> HF + H:

[[file: yu_ea_1.PNG|400px]]

Figure Activation energy for F + H2 ---> HF + H

{| class="wikitable"
|+
|-
| Energy of the transition state(kJ/mol) || Energy of the transition state(kJ/mol) || Activation energy(kJ/mol)
|-
| -103.752
| -103.987
| 0.235
|}

Table 2 Activation energy for F + H2 ---> HF + H

For H + HF= H2 + F:

[[file: yu_ea_2.PNG|400px]]

Figure Activation energy for H + HF= H2 + F

{| class="wikitable"
|+
|-
| Energy of the transition state(kJ/mol) || Energy of the transition state(kJ/mol) || Activation energy(kJ/mol)
|-
| -103.694
| -133.827
| 30.133
|}

Table 2 Activation energy for H + HF= H2 + F

===Question 9: the reaction trajectory===

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

My answer:

The following initial condition was set:

[[file: yu_initial_3.PNG]]

[[file: yu_fhh_reactive.PNG|400px]]

[[file: yu_traj_et.PNG|400px]]
[[file: yu_traj_pt.PNG|400px]]

The reaction is exothermic. The reaction energy are converted to kinetic energy and specifically vibrational energy of the newly formed H-F bond in this case. As shown on the contour plot, the oscillation along AB (HF in this case) distance axis is significantly larger than the oscillation along BC (HH in this case) distance. This can be confirmed by the energy versus time plot as it shows a constant conversion between kinetic and potential energy. The momentum versus time plot also show a vibration pattern for AB after the reaction and the amplitude is much larger than that of BC before the reaction.

The course of reaction can be studied by the flow method. The reaction mixture that traveled further down in reaction tube were left reacted for a longer time. Hence, the reaction mixture furthest away from the point of mixing resembles the product and the mixture at the point of mixing resemble the reactants, between these two points are the intermediates.

Carrying out IR spectroscopy measurement the points allows the study of the vibration of chemical bonds. As said above, the reaction is exothermic and allows higher vibrational energy levels to become populated. Therefore, besides the original Vn=0 to Vn=1 transition, hot band transitions such as transition from Vn=1 to Vn=2 become possible. This gives rise to overtone signals in the spectrum. Overtones peaks are lower in energy and intensity when compared to the absorption peak representing Vn=0 to Vn=1 transition. The more vibrational energy levels populated the more overtone bands can be observed. Hence the spectrum of the product would have more overtone signals compared to the reactant.

===Question 10: efficiency of the reaction===

''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.''

My answer:
As the higher translational and vibrational energy level occupied by the molecule the higher the efficiency of the reaction in general. However, that is not always the case. More detailed discussion can be achieved by quoting the Polanyi's rule:

The Polanyi's Empirical Rules states that translational energy is more efficient at promoting early transition state reactions, while vibrational energy is more efficient at promoting late transition state energy reactions.

We can look at specific examples of the reactions:

1.For a early transition state reaction: F + H2 ----> HF + H

[[file: yu_t_e.PNG|400px]]

As shown in the figure above the reaction trajectory is reactive. The H2 is vibrating with a small amplitude. The translation energy which can be roughly interpreted by PHF is large. The effect varying vibrational energy by varing PH2 value between -3 and 3 has a small impact. This agrees with the first part rules stated above.

2.For a late transition state reaction: H + HF ----> H2+ F

[[file: yu_v_l.PNG|400px]]

A reactive trajectory is show as the figure above. The H-F bond oscillates with a large amplitude this corresponds to the fact that the vibrational energy is more efficient in promoting this reaction. A much higher AB momentum is needed if BC vibration is set small, and will often cause the trajectory to bounce back resulting in a nonreactive condition.

Rep:YR316DYN

2018-05-25T16:14:36Z

Yr316: /* Question 8: Activation energy */

=Molecular Reaction Dynamics=

== H + H2 system==

===Question 1: minimum and saddle===

''What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.''

In this triatomic system where the HA is approaching HBHC with an initial momentum. r1 is the distance between HA and HB and r2 is the distance between HB and HC. The total potential energy of the system (<math>V(r_1,r_2)</math>) is changing with respect to both r1 and r2. Hence the gradient of potential energy consists of two components:<math> { \partial V(r_1,r_2)\over \partial r_1}</math> and <math> { \partial V(r_1,r_2)\over \partial r_1}</math>. The minima correspond to the minimum potential energy of the oscillating system HBHC (before collision)or HAHB (after collision). Mathematically, the minimum can be expressed as <math> { \partial V(r_1,r_2)\over \partial r_2}</math>=0 and <math> { \partial V(r_1,r_2)\over \partial r_1}</math>=0 respectively. <math> { \partial V(r_1,r_2)\over \partial r_2}</math> is partial differentiation, it differentiates V with respect to r2 while treating r1 as constant. The curvatures of the minima are described by the second partial derivative: Vr1r1 and Vr1r1 and smaller 0. They are the minimum point viewing from cross sections along r1 and r2 respectively. It is the same when Hc is leaving after the collision.

However, as said at the beginning while HBHC are oscillating, HA is approaching and this increases the potential energy of the system. The different minima at different r1 values increases to a saddle point. This is the maximum point in the trough created by joining the minima when viewing from potential surface. At this point both <math> { \partial V(r_1,r_2)\over \partial r_2}</math> and <math> { \partial V(r_1,r_2)\over \partial r_1}</math> are 0. The curvature at this point can be described by the determinant: Vr1r1•Vr1r1-V2r1r2. This point corresponds to the transition state of the system.

===Question 2: estimation of transition state===

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

The transition state is estimated to be at r1=r2=0.90773 as shown in figure 1.

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

figure 1. “Inter-nuclear Distances vs Time” plot at stationary point

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

figure2. “Inter-nuclear Distances vs Time” plot under one initial condition

Figure 1 shows under one initial condition when HA collides with HB HC the intercept of the A-B and B-C line gives a rough estimation of where the transition would be. A more accurate estimation can be done by setting the momentum to zero while changing values for distances (keep r1=r2 as the potential surface of H + H2 is symmetric. The trajectory would be stationary i.e. distance does not vary with time for a transition state as suggested by the definition of a stationary point(saddle as an example). while at other point it would produce a oscillation pattern over time as shown in figure 3.

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

figure 3. “Inter-nuclear Distances vs Time” plot at stationary point at other r1=r2 points other then the stationary points

===Question 3: comparing dynamics and MEP===

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

Figure 4 and 5 shows the reaction path calculation under dynamic method and MEP method respectively. Both of them have initial condition r1=0.91773, r2=0.90773, p1=p2=0. The pathway calculated with dynamic method is wavy, while the pathway calculated with MEP method is a smooth curve. MEP stands for minimum energy pathway, in each step of MEP the momentum was reset to zero, hence the inter-conversion between kinetic and the oscillation are ignored. HBH1 of the molecule was ignored and was set at minimum potential energy along the pathway. The step number chosen for dynamics was 150 steps while that of the MEP calculation was chosen to be 10000. The reason for MEP has a much larger step number is also because that in each step of MEP the momentum was reset to zero and the inertia of the particles are ignored, while the momentum in the dynamic method was cumulative. Hence, if the same step number was chosen for both method, MEP result would give a much shorter of path length due to this reset of momentum.

[[File:Yu_dynamic_150steps.PNG|400px]]

Figure 4. Pathway calculation with dynamic method

[[File:Yu_MEP_10000_steps.PNG|400px]]

Figure 5. Pathway calculation with MEP method

If the initial conditions was set to r1 = rts and r2 = rts+0.01 instead, the reaction pathways would be a mirror image of r1=rts+0.01 and r2=rts condition with orthogonal plane r1=r2 as the mirror plane as shown in figure 6. This is due to the fact that the initial position of the trajectory was reversed.
If the initial positions was set to be the same as the final positions of the previous trajectory, the new resultant trajectory would represent the reverse reaction of the previous trajectory.

[[File:yu_mirror1.PNG|400px]]
[[File:yu_mirror2.PNG|400px]]

Figure 6. mirror image to each other: r1=rts+0.01 and r2=rts(left) r1 = rts and r2 = rts+0.01 (right) with steps extended to 500

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

Figure 7. pathway initial positions correspond to the final positions with opposite momenta

=== Question 4: reactive and unreactive trajectory ===

''Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.''

{| class="wikitable" border=1
! p1 !! p2!!Energy
|-
| -1.25 || -2.5 || -99.018
|-
| -1.5 || -2.0 || -100.456
|-
| -1.5 || -2.5 || -98.956
|-
| -2.5 || -5.0 || -84.956
|-
| -2.5 || -5.2 || -83.416
|}

table 1. Energy tabel with various initial monenta when r1=0.74 and r2= 2.0

[[File:yu_q3_1.PNG|400px]]

Figure 7. trajectory with p1=-1.25 and p2=-2.5

'''Comment:''' Reactive

[[File:yu_q3_2.PNG|400px]]

Figure 8. trajectory with p1=-1.5 and p2=-2.0

'''Comment:''' Unreactive

[[File:yu_q3_3.PNG|400px]]

Figure 9. trajectory with p1=-1.5 and p2=-2.5

'''Comment:''' Reactive

[[File:yu_q3_4.PNG|400px]]

Figure 10. trajectory with p1=-2.5 and p2=-5.0

'''Comment:''' Unreactive

[[File:yu_q3_5.PNG|400px]]

Figure 11. trajectory with p1=-2.5 and p2=-5.2

'''Comment:''' Reactive

=== Question 5: 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?''

My answer:

The main assumption of transition state theory are give as:

1. The reactants (or products) are in equilibrium with the activated complex.

2. Quantum-tunneling effects are negligible and the Born-Oppenheimer approximation is invoked.

3. Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.

According to the assumptions, the transition state prediction for rate of reaction should be slower than the experimental value. It can be rationalized by the following two reasons:

1. As stated in the second assumption above. Transition State Theory predictions ignores quantum tunneling effects and treat them as noreactive, where in reality particles with energies lower than the activation energy could react by tunneling through the activation energy barrier.

2. The third assumption can be interpreted that the reaction trajectory are only allowed to cross the transition state once. However, the calculated reaction trajectory for last initial condition in question 4 shows that the reaction trajectory could cross the transition states multiple times and eventually reach the product side. This reactive condition is also ignored by the transition state prediction.

==The study F-H-H system==

===Question 6: energetic of the reactions===

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?

My answer:

Figure 12 and figure 13 shows the potential surface of a F + H2 reaction and a H + HF reaction respectively.

[[file: yu_FHH_condition.PNG|400px]]

Figure 12. Potential surface of Yu_HHF_condition.PNG reaction

In this condition the F atom is approaching the H2 molecule with the momentum showing in the figure 12. The F + H2 has a higher potential energy then HF+H. This shows that the reaction: F + H2 ---> HF + H is exothermic. This result indicates that H-F which is formed is bond is stronger than the H-H which is broken.

[[file: Yu_HHF_condition.PNG|400px]]

Figure 13. Potential surface of H + HF reaction

In this condition the H atom is approaching the HF molecule with the momentum showing in the figure 13. The H + HF has a lower potential energy then H2 + F. This shows that the reaction: H + HF ---> H2 + F is endothermic. This result agrees with the previous result that H-H which is formed is bond is weaker than the H-F which is broken.

===Question 7: Transition states===

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

My answer:
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 first reaction: F + H2 = HF + H is exothermic. Hence, the transition state is closer in energy to the reactants: F + H2 and it adopts the following asymmetric geometry:

F------H--H

•H-H bond is only slightly elongated

•F atom is much further away

[[file:yu_ts_FHH.PNG|400px]]
[[file:yu_stationary_1.PNG|400px]]

This is known as an early transition state. The method used to find this transition state is similar to the one used for H-H-H transition state. The H-H (BC) distance can be interpreted to be around 0.74 Å which is a typical H-H bond length. As shown in figure the transition state is at F-H=1.8085 Å and H-H=0.7509. Figure shows that the point is a stationary point and it lies closely to the reactants.

The second reaction: H + HF= H2 + F is endothermic. Hence, the transition state is closer in energy to the products: F + H2 and it adopts the following asymmetric geometry:

H--H-------F

•H-F bond is largely elongated

•H-H bond is almost formed

[[file:yu_ts_HHF.PNG|400px]]
[[file:yu_stationary_2.PNG|400px]]

This is known as a late transition state. The H-H (AB) distance can be interpreted to be around 0.74 Å. As shown in figure the transition state is at H-H=0.7495 Å and H-H=1.8080. Figure shows that the point is a stationary point and it lies closely to the products.

===Question 8: Activation energy===

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

My answer:

Activation energy is the energy between the reactant and the transition state. If the initial condition is set to be slightly displaced from the transition state and closer to the reactant position, the trajectory will end in the reactant condition. The energy vs time graph would start as a plateau which resembles the energy of the transition state and drops and finally reaches a plateau with resembles the energy of the reactant. The energy difference between the two plateau would give the activation energy of the reaction.

For F + H2 ---> HF + H:

{| class="wikitable"
|+
|-
| Energy of the transition state(kJ/mol) || Energy of the transition state(kJ/mol) || Activation energy(kJ/mol)
|-
| -103.752
| -103.987
| 0.235
|}

Table 3 Vibrational analysis

For H + HF= H2 + F:

===Question 9: the reaction trajectory===

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

My answer:

The following initial condition was set:

[[file: yu_initial_3.PNG]]

[[file: yu_fhh_reactive.PNG|400px]]

[[file: yu_traj_et.PNG|400px]]
[[file: yu_traj_pt.PNG|400px]]

The reaction is exothermic. The reaction energy are converted to kinetic energy and specifically vibrational energy of the newly formed H-F bond in this case. As shown on the contour plot, the oscillation along AB (HF in this case) distance axis is significantly larger than the oscillation along BC (HH in this case) distance. This can be confirmed by the energy versus time plot as it shows a constant conversion between kinetic and potential energy. The momentum versus time plot also show a vibration pattern for AB after the reaction and the amplitude is much larger than that of BC before the reaction.

The course of reaction can be studied by the flow method. The reaction mixture that traveled further down in reaction tube were left reacted for a longer time. Hence, the reaction mixture furthest away from the point of mixing resembles the product and the mixture at the point of mixing resemble the reactants, between these two points are the intermediates.

Carrying out IR spectroscopy measurement the points allows the study of the vibration of chemical bonds. As said above, the reaction is exothermic and allows higher vibrational energy levels to become populated. Therefore, besides the original Vn=0 to Vn=1 transition, hot band transitions such as transition from Vn=1 to Vn=2 become possible. This gives rise to overtone signals in the spectrum. Overtones peaks are lower in energy and intensity when compared to the absorption peak representing Vn=0 to Vn=1 transition. The more vibrational energy levels populated the more overtone bands can be observed. Hence the spectrum of the product would have more overtone signals compared to the reactant.

===Question 10: efficiency of the reaction===

''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.''

My answer:
As the higher translational and vibrational energy level occupied by the molecule the higher the efficiency of the reaction in general. However, that is not always the case. More detailed discussion can be achieved by quoting the Polanyi's rule:

The Polanyi's Empirical Rules states that translational energy is more efficient at promoting early transition state reactions, while vibrational energy is more efficient at promoting late transition state energy reactions.

We can look at specific examples of the reactions:

1.For a early transition state reaction: F + H2 ----> HF + H

[[file: yu_t_e.PNG|400px]]

As shown in the figure above the reaction trajectory is reactive. The H2 is vibrating with a small amplitude. The translation energy which can be roughly interpreted by PHF is large. The effect varying vibrational energy by varing PH2 value between -3 and 3 has a small impact. This agrees with the first part rules stated above.

2.For a late transition state reaction: H + HF ----> H2+ F

[[file: yu_v_l.PNG|400px]]

A reactive trajectory is show as the figure above. The H-F bond oscillates with a large amplitude this corresponds to the fact that the vibrational energy is more efficient in promoting this reaction. A much higher AB momentum is needed if BC vibration is set small, and will often cause the trajectory to bounce back resulting in a nonreactive condition.

Rep:YR316DYN

2018-05-25T16:02:13Z

Yr316: /* Question 9: the reaction trajectory */

=Molecular Reaction Dynamics=

== H + H2 system==

===Question 1: minimum and saddle===

''What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.''

In this triatomic system where the HA is approaching HBHC with an initial momentum. r1 is the distance between HA and HB and r2 is the distance between HB and HC. The total potential energy of the system (<math>V(r_1,r_2)</math>) is changing with respect to both r1 and r2. Hence the gradient of potential energy consists of two components:<math> { \partial V(r_1,r_2)\over \partial r_1}</math> and <math> { \partial V(r_1,r_2)\over \partial r_1}</math>. The minima correspond to the minimum potential energy of the oscillating system HBHC (before collision)or HAHB (after collision). Mathematically, the minimum can be expressed as <math> { \partial V(r_1,r_2)\over \partial r_2}</math>=0 and <math> { \partial V(r_1,r_2)\over \partial r_1}</math>=0 respectively. <math> { \partial V(r_1,r_2)\over \partial r_2}</math> is partial differentiation, it differentiates V with respect to r2 while treating r1 as constant. The curvatures of the minima are described by the second partial derivative: Vr1r1 and Vr1r1 and smaller 0. They are the minimum point viewing from cross sections along r1 and r2 respectively. It is the same when Hc is leaving after the collision.

However, as said at the beginning while HBHC are oscillating, HA is approaching and this increases the potential energy of the system. The different minima at different r1 values increases to a saddle point. This is the maximum point in the trough created by joining the minima when viewing from potential surface. At this point both <math> { \partial V(r_1,r_2)\over \partial r_2}</math> and <math> { \partial V(r_1,r_2)\over \partial r_1}</math> are 0. The curvature at this point can be described by the determinant: Vr1r1•Vr1r1-V2r1r2. This point corresponds to the transition state of the system.

===Question 2: estimation of transition state===

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

The transition state is estimated to be at r1=r2=0.90773 as shown in figure 1.

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

figure 1. “Inter-nuclear Distances vs Time” plot at stationary point

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

figure2. “Inter-nuclear Distances vs Time” plot under one initial condition

Figure 1 shows under one initial condition when HA collides with HB HC the intercept of the A-B and B-C line gives a rough estimation of where the transition would be. A more accurate estimation can be done by setting the momentum to zero while changing values for distances (keep r1=r2 as the potential surface of H + H2 is symmetric. The trajectory would be stationary i.e. distance does not vary with time for a transition state as suggested by the definition of a stationary point(saddle as an example). while at other point it would produce a oscillation pattern over time as shown in figure 3.

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

figure 3. “Inter-nuclear Distances vs Time” plot at stationary point at other r1=r2 points other then the stationary points

===Question 3: comparing dynamics and MEP===

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

Figure 4 and 5 shows the reaction path calculation under dynamic method and MEP method respectively. Both of them have initial condition r1=0.91773, r2=0.90773, p1=p2=0. The pathway calculated with dynamic method is wavy, while the pathway calculated with MEP method is a smooth curve. MEP stands for minimum energy pathway, in each step of MEP the momentum was reset to zero, hence the inter-conversion between kinetic and the oscillation are ignored. HBH1 of the molecule was ignored and was set at minimum potential energy along the pathway. The step number chosen for dynamics was 150 steps while that of the MEP calculation was chosen to be 10000. The reason for MEP has a much larger step number is also because that in each step of MEP the momentum was reset to zero and the inertia of the particles are ignored, while the momentum in the dynamic method was cumulative. Hence, if the same step number was chosen for both method, MEP result would give a much shorter of path length due to this reset of momentum.

[[File:Yu_dynamic_150steps.PNG|400px]]

Figure 4. Pathway calculation with dynamic method

[[File:Yu_MEP_10000_steps.PNG|400px]]

Figure 5. Pathway calculation with MEP method

If the initial conditions was set to r1 = rts and r2 = rts+0.01 instead, the reaction pathways would be a mirror image of r1=rts+0.01 and r2=rts condition with orthogonal plane r1=r2 as the mirror plane as shown in figure 6. This is due to the fact that the initial position of the trajectory was reversed.
If the initial positions was set to be the same as the final positions of the previous trajectory, the new resultant trajectory would represent the reverse reaction of the previous trajectory.

[[File:yu_mirror1.PNG|400px]]
[[File:yu_mirror2.PNG|400px]]

Figure 6. mirror image to each other: r1=rts+0.01 and r2=rts(left) r1 = rts and r2 = rts+0.01 (right) with steps extended to 500

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

Figure 7. pathway initial positions correspond to the final positions with opposite momenta

=== Question 4: reactive and unreactive trajectory ===

''Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.''

{| class="wikitable" border=1
! p1 !! p2!!Energy
|-
| -1.25 || -2.5 || -99.018
|-
| -1.5 || -2.0 || -100.456
|-
| -1.5 || -2.5 || -98.956
|-
| -2.5 || -5.0 || -84.956
|-
| -2.5 || -5.2 || -83.416
|}

table 1. Energy tabel with various initial monenta when r1=0.74 and r2= 2.0

[[File:yu_q3_1.PNG|400px]]

Figure 7. trajectory with p1=-1.25 and p2=-2.5

'''Comment:''' Reactive

[[File:yu_q3_2.PNG|400px]]

Figure 8. trajectory with p1=-1.5 and p2=-2.0

'''Comment:''' Unreactive

[[File:yu_q3_3.PNG|400px]]

Figure 9. trajectory with p1=-1.5 and p2=-2.5

'''Comment:''' Reactive

[[File:yu_q3_4.PNG|400px]]

Figure 10. trajectory with p1=-2.5 and p2=-5.0

'''Comment:''' Unreactive

[[File:yu_q3_5.PNG|400px]]

Figure 11. trajectory with p1=-2.5 and p2=-5.2

'''Comment:''' Reactive

=== Question 5: 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?''

My answer:

The main assumption of transition state theory are give as:

1. The reactants (or products) are in equilibrium with the activated complex.

2. Quantum-tunneling effects are negligible and the Born-Oppenheimer approximation is invoked.

3. Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.

According to the assumptions, the transition state prediction for rate of reaction should be slower than the experimental value. It can be rationalized by the following two reasons:

1. As stated in the second assumption above. Transition State Theory predictions ignores quantum tunneling effects and treat them as noreactive, where in reality particles with energies lower than the activation energy could react by tunneling through the activation energy barrier.

2. The third assumption can be interpreted that the reaction trajectory are only allowed to cross the transition state once. However, the calculated reaction trajectory for last initial condition in question 4 shows that the reaction trajectory could cross the transition states multiple times and eventually reach the product side. This reactive condition is also ignored by the transition state prediction.

==The study F-H-H system==

===Question 6: energetic of the reactions===

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?

My answer:

Figure 12 and figure 13 shows the potential surface of a F + H2 reaction and a H + HF reaction respectively.

[[file: yu_FHH_condition.PNG|400px]]

Figure 12. Potential surface of Yu_HHF_condition.PNG reaction

In this condition the F atom is approaching the H2 molecule with the momentum showing in the figure 12. The F + H2 has a higher potential energy then HF+H. This shows that the reaction: F + H2 ---> HF + H is exothermic. This result indicates that H-F which is formed is bond is stronger than the H-H which is broken.

[[file: Yu_HHF_condition.PNG|400px]]

Figure 13. Potential surface of H + HF reaction

In this condition the H atom is approaching the HF molecule with the momentum showing in the figure 13. The H + HF has a lower potential energy then H2 + F. This shows that the reaction: H + HF ---> H2 + F is endothermic. This result agrees with the previous result that H-H which is formed is bond is weaker than the H-F which is broken.

===Question 7: Transition states===

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

My answer:
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 first reaction: F + H2 = HF + H is exothermic. Hence, the transition state is closer in energy to the reactants: F + H2 and it adopts the following asymmetric geometry:

F------H--H

•H-H bond is only slightly elongated

•F atom is much further away

[[file:yu_ts_FHH.PNG|400px]]
[[file:yu_stationary_1.PNG|400px]]

This is known as an early transition state. The method used to find this transition state is similar to the one used for H-H-H transition state. The H-H (BC) distance can be interpreted to be around 0.74 Å which is a typical H-H bond length. As shown in figure the transition state is at F-H=1.8085 Å and H-H=0.7509. Figure shows that the point is a stationary point and it lies closely to the reactants.

The second reaction: H + HF= H2 + F is endothermic. Hence, the transition state is closer in energy to the products: F + H2 and it adopts the following asymmetric geometry:

H--H-------F

•H-F bond is largely elongated

•H-H bond is almost formed

[[file:yu_ts_HHF.PNG|400px]]
[[file:yu_stationary_2.PNG|400px]]

This is known as a late transition state. The H-H (AB) distance can be interpreted to be around 0.74 Å. As shown in figure the transition state is at H-H=0.7495 Å and H-H=1.8080. Figure shows that the point is a stationary point and it lies closely to the products.

===Question 8: Activation energy===

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

My answer:

Activation energy is the energy between the reactant and the transition state. If the initial condition is set to be slightly displaced from the transition state and closer to the reactant position, the trajectory will end in the reactant condition. The energy vs time graph would start as a plateau which resembles the energy of the transition state and drops and finally reaches a plateau with resembles the energy of the reactant. The energy difference between the two plateau would give the activation energy of the reaction.

For F + H2 = HF + H:

For H + HF= H2 + F:

===Question 9: the reaction trajectory===

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

My answer:

The following initial condition was set:

[[file: yu_initial_3.PNG]]

[[file: yu_fhh_reactive.PNG|400px]]

[[file: yu_traj_et.PNG|400px]]
[[file: yu_traj_pt.PNG|400px]]

The reaction is exothermic. The reaction energy are converted to kinetic energy and specifically vibrational energy of the newly formed H-F bond in this case. As shown on the contour plot, the oscillation along AB (HF in this case) distance axis is significantly larger than the oscillation along BC (HH in this case) distance. This can be confirmed by the energy versus time plot as it shows a constant conversion between kinetic and potential energy. The momentum versus time plot also show a vibration pattern for AB after the reaction and the amplitude is much larger than that of BC before the reaction.

The course of reaction can be studied by the flow method. The reaction mixture that traveled further down in reaction tube were left reacted for a longer time. Hence, the reaction mixture furthest away from the point of mixing resembles the product and the mixture at the point of mixing resemble the reactants, between these two points are the intermediates.

Carrying out IR spectroscopy measurement the points allows the study of the vibration of chemical bonds. As said above, the reaction is exothermic and allows higher vibrational energy levels to become populated. Therefore, besides the original Vn=0 to Vn=1 transition, hot band transitions such as transition from Vn=1 to Vn=2 become possible. This gives rise to overtone signals in the spectrum. Overtones peaks are lower in energy and intensity when compared to the absorption peak representing Vn=0 to Vn=1 transition. The more vibrational energy levels populated the more overtone bands can be observed. Hence the spectrum of the product would have more overtone signals compared to the reactant.

===Question 10: efficiency of the reaction===

''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.''

My answer:
As the higher translational and vibrational energy level occupied by the molecule the higher the efficiency of the reaction in general. However, that is not always the case. More detailed discussion can be achieved by quoting the Polanyi's rule:

The Polanyi's Empirical Rules states that translational energy is more efficient at promoting early transition state reactions, while vibrational energy is more efficient at promoting late transition state energy reactions.

We can look at specific examples of the reactions:

1.For a early transition state reaction: F + H2 ----> HF + H

[[file: yu_t_e.PNG|400px]]

As shown in the figure above the reaction trajectory is reactive. The H2 is vibrating with a small amplitude. The translation energy which can be roughly interpreted by PHF is large. The effect varying vibrational energy by varing PH2 value between -3 and 3 has a small impact. This agrees with the first part rules stated above.

2.For a late transition state reaction: H + HF ----> H2+ F

[[file: yu_v_l.PNG|400px]]

A reactive trajectory is show as the figure above. The H-F bond oscillates with a large amplitude this corresponds to the fact that the vibrational energy is more efficient in promoting this reaction. A much higher AB momentum is needed if BC vibration is set small, and will often cause the trajectory to bounce back resulting in a nonreactive condition.

Rep:YR316DYN

2018-05-25T16:01:49Z

Yr316: /* Question 9: the reaction trajectory */

=Molecular Reaction Dynamics=

== H + H2 system==

===Question 1: minimum and saddle===

''What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.''

In this triatomic system where the HA is approaching HBHC with an initial momentum. r1 is the distance between HA and HB and r2 is the distance between HB and HC. The total potential energy of the system (<math>V(r_1,r_2)</math>) is changing with respect to both r1 and r2. Hence the gradient of potential energy consists of two components:<math> { \partial V(r_1,r_2)\over \partial r_1}</math> and <math> { \partial V(r_1,r_2)\over \partial r_1}</math>. The minima correspond to the minimum potential energy of the oscillating system HBHC (before collision)or HAHB (after collision). Mathematically, the minimum can be expressed as <math> { \partial V(r_1,r_2)\over \partial r_2}</math>=0 and <math> { \partial V(r_1,r_2)\over \partial r_1}</math>=0 respectively. <math> { \partial V(r_1,r_2)\over \partial r_2}</math> is partial differentiation, it differentiates V with respect to r2 while treating r1 as constant. The curvatures of the minima are described by the second partial derivative: Vr1r1 and Vr1r1 and smaller 0. They are the minimum point viewing from cross sections along r1 and r2 respectively. It is the same when Hc is leaving after the collision.

However, as said at the beginning while HBHC are oscillating, HA is approaching and this increases the potential energy of the system. The different minima at different r1 values increases to a saddle point. This is the maximum point in the trough created by joining the minima when viewing from potential surface. At this point both <math> { \partial V(r_1,r_2)\over \partial r_2}</math> and <math> { \partial V(r_1,r_2)\over \partial r_1}</math> are 0. The curvature at this point can be described by the determinant: Vr1r1•Vr1r1-V2r1r2. This point corresponds to the transition state of the system.

===Question 2: estimation of transition state===

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

The transition state is estimated to be at r1=r2=0.90773 as shown in figure 1.

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

figure 1. “Inter-nuclear Distances vs Time” plot at stationary point

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

figure2. “Inter-nuclear Distances vs Time” plot under one initial condition

Figure 1 shows under one initial condition when HA collides with HB HC the intercept of the A-B and B-C line gives a rough estimation of where the transition would be. A more accurate estimation can be done by setting the momentum to zero while changing values for distances (keep r1=r2 as the potential surface of H + H2 is symmetric. The trajectory would be stationary i.e. distance does not vary with time for a transition state as suggested by the definition of a stationary point(saddle as an example). while at other point it would produce a oscillation pattern over time as shown in figure 3.

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

figure 3. “Inter-nuclear Distances vs Time” plot at stationary point at other r1=r2 points other then the stationary points

===Question 3: comparing dynamics and MEP===

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

Figure 4 and 5 shows the reaction path calculation under dynamic method and MEP method respectively. Both of them have initial condition r1=0.91773, r2=0.90773, p1=p2=0. The pathway calculated with dynamic method is wavy, while the pathway calculated with MEP method is a smooth curve. MEP stands for minimum energy pathway, in each step of MEP the momentum was reset to zero, hence the inter-conversion between kinetic and the oscillation are ignored. HBH1 of the molecule was ignored and was set at minimum potential energy along the pathway. The step number chosen for dynamics was 150 steps while that of the MEP calculation was chosen to be 10000. The reason for MEP has a much larger step number is also because that in each step of MEP the momentum was reset to zero and the inertia of the particles are ignored, while the momentum in the dynamic method was cumulative. Hence, if the same step number was chosen for both method, MEP result would give a much shorter of path length due to this reset of momentum.

[[File:Yu_dynamic_150steps.PNG|400px]]

Figure 4. Pathway calculation with dynamic method

[[File:Yu_MEP_10000_steps.PNG|400px]]

Figure 5. Pathway calculation with MEP method

If the initial conditions was set to r1 = rts and r2 = rts+0.01 instead, the reaction pathways would be a mirror image of r1=rts+0.01 and r2=rts condition with orthogonal plane r1=r2 as the mirror plane as shown in figure 6. This is due to the fact that the initial position of the trajectory was reversed.
If the initial positions was set to be the same as the final positions of the previous trajectory, the new resultant trajectory would represent the reverse reaction of the previous trajectory.

[[File:yu_mirror1.PNG|400px]]
[[File:yu_mirror2.PNG|400px]]

Figure 6. mirror image to each other: r1=rts+0.01 and r2=rts(left) r1 = rts and r2 = rts+0.01 (right) with steps extended to 500

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

Figure 7. pathway initial positions correspond to the final positions with opposite momenta

=== Question 4: reactive and unreactive trajectory ===

''Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.''

{| class="wikitable" border=1
! p1 !! p2!!Energy
|-
| -1.25 || -2.5 || -99.018
|-
| -1.5 || -2.0 || -100.456
|-
| -1.5 || -2.5 || -98.956
|-
| -2.5 || -5.0 || -84.956
|-
| -2.5 || -5.2 || -83.416
|}

table 1. Energy tabel with various initial monenta when r1=0.74 and r2= 2.0

[[File:yu_q3_1.PNG|400px]]

Figure 7. trajectory with p1=-1.25 and p2=-2.5

'''Comment:''' Reactive

[[File:yu_q3_2.PNG|400px]]

Figure 8. trajectory with p1=-1.5 and p2=-2.0

'''Comment:''' Unreactive

[[File:yu_q3_3.PNG|400px]]

Figure 9. trajectory with p1=-1.5 and p2=-2.5

'''Comment:''' Reactive

[[File:yu_q3_4.PNG|400px]]

Figure 10. trajectory with p1=-2.5 and p2=-5.0

'''Comment:''' Unreactive

[[File:yu_q3_5.PNG|400px]]

Figure 11. trajectory with p1=-2.5 and p2=-5.2

'''Comment:''' Reactive

=== Question 5: 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?''

My answer:

The main assumption of transition state theory are give as:

1. The reactants (or products) are in equilibrium with the activated complex.

2. Quantum-tunneling effects are negligible and the Born-Oppenheimer approximation is invoked.

3. Once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again.

According to the assumptions, the transition state prediction for rate of reaction should be slower than the experimental value. It can be rationalized by the following two reasons:

1. As stated in the second assumption above. Transition State Theory predictions ignores quantum tunneling effects and treat them as noreactive, where in reality particles with energies lower than the activation energy could react by tunneling through the activation energy barrier.

2. The third assumption can be interpreted that the reaction trajectory are only allowed to cross the transition state once. However, the calculated reaction trajectory for last initial condition in question 4 shows that the reaction trajectory could cross the transition states multiple times and eventually reach the product side. This reactive condition is also ignored by the transition state prediction.

==The study F-H-H system==

===Question 6: energetic of the reactions===

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?

My answer:

Figure 12 and figure 13 shows the potential surface of a F + H2 reaction and a H + HF reaction respectively.

[[file: yu_FHH_condition.PNG|400px]]

Figure 12. Potential surface of Yu_HHF_condition.PNG reaction

In this condition the F atom is approaching the H2 molecule with the momentum showing in the figure 12. The F + H2 has a higher potential energy then HF+H. This shows that the reaction: F + H2 ---> HF + H is exothermic. This result indicates that H-F which is formed is bond is stronger than the H-H which is broken.

[[file: Yu_HHF_condition.PNG|400px]]

Figure 13. Potential surface of H + HF reaction

In this condition the H atom is approaching the HF molecule with the momentum showing in the figure 13. The H + HF has a lower potential energy then H2 + F. This shows that the reaction: H + HF ---> H2 + F is endothermic. This result agrees with the previous result that H-H which is formed is bond is weaker than the H-F which is broken.

===Question 7: Transition states===

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

My answer:
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 first reaction: F + H2 = HF + H is exothermic. Hence, the transition state is closer in energy to the reactants: F + H2 and it adopts the following asymmetric geometry:

F------H--H

•H-H bond is only slightly elongated

•F atom is much further away

[[file:yu_ts_FHH.PNG|400px]]
[[file:yu_stationary_1.PNG|400px]]

This is known as an early transition state. The method used to find this transition state is similar to the one used for H-H-H transition state. The H-H (BC) distance can be interpreted to be around 0.74 Å which is a typical H-H bond length. As shown in figure the transition state is at F-H=1.8085 Å and H-H=0.7509. Figure shows that the point is a stationary point and it lies closely to the reactants.

The second reaction: H + HF= H2 + F is endothermic. Hence, the transition state is closer in energy to the products: F + H2 and it adopts the following asymmetric geometry:

H--H-------F

•H-F bond is largely elongated

•H-H bond is almost formed

[[file:yu_ts_HHF.PNG|400px]]
[[file:yu_stationary_2.PNG|400px]]

This is known as a late transition state. The H-H (AB) distance can be interpreted to be around 0.74 Å. As shown in figure the transition state is at H-H=0.7495 Å and H-H=1.8080. Figure shows that the point is a stationary point and it lies closely to the products.

===Question 8: Activation energy===

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

My answer:

Activation energy is the energy between the reactant and the transition state. If the initial condition is set to be slightly displaced from the transition state and closer to the reactant position, the trajectory will end in the reactant condition. The energy vs time graph would start as a plateau which resembles the energy of the transition state and drops and finally reaches a plateau with resembles the energy of the reactant. The energy difference between the two plateau would give the activation energy of the reaction.

For F + H2 = HF + H:

For H + HF= H2 + F:

===Question 9: the reaction trajectory===

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

My answer:

The following initial condition was set:

[[file: yu_initial_3.PNG]]

[[file: yu_fhh_reactive.PNG]]

[[file: yu_traj_et.PNG]]
[[file: yu_traj_pt.PNG]]

The reaction is exothermic. The reaction energy are converted to kinetic energy and specifically vibrational energy of the newly formed H-F bond in this case. As shown on the contour plot, the oscillation along AB (HF in this case) distance axis is significantly larger than the oscillation along BC (HH in this case) distance. This can be confirmed by the energy versus time plot as it shows a constant conversion between kinetic and potential energy. The momentum versus time plot also show a vibration pattern for AB after the reaction and the amplitude is much larger than that of BC before the reaction.

The course of reaction can be studied by the flow method. The reaction mixture that traveled further down in reaction tube were left reacted for a longer time. Hence, the reaction mixture furthest away from the point of mixing resembles the product and the mixture at the point of mixing resemble the reactants, between these two points are the intermediates.

Carrying out IR spectroscopy measurement the points allows the study of the vibration of chemical bonds. As said above, the reaction is exothermic and allows higher vibrational energy levels to become populated. Therefore, besides the original Vn=0 to Vn=1 transition, hot band transitions such as transition from Vn=1 to Vn=2 become possible. This gives rise to overtone signals in the spectrum. Overtones peaks are lower in energy and intensity when compared to the absorption peak representing Vn=0 to Vn=1 transition. The more vibrational energy levels populated the more overtone bands can be observed. Hence the spectrum of the product would have more overtone signals compared to the reactant.

===Question 10: efficiency of the reaction===

''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.''

My answer:
As the higher translational and vibrational energy level occupied by the molecule the higher the efficiency of the reaction in general. However, that is not always the case. More detailed discussion can be achieved by quoting the Polanyi's rule:

The Polanyi's Empirical Rules states that translational energy is more efficient at promoting early transition state reactions, while vibrational energy is more efficient at promoting late transition state energy reactions.

We can look at specific examples of the reactions:

1.For a early transition state reaction: F + H2 ----> HF + H

[[file: yu_t_e.PNG|400px]]

As shown in the figure above the reaction trajectory is reactive. The H2 is vibrating with a small amplitude. The translation energy which can be roughly interpreted by PHF is large. The effect varying vibrational energy by varing PH2 value between -3 and 3 has a small impact. This agrees with the first part rules stated above.

2.For a late transition state reaction: H + HF ----> H2+ F

[[file: yu_v_l.PNG|400px]]

A reactive trajectory is show as the figure above. The H-F bond oscillates with a large amplitude this corresponds to the fact that the vibrational energy is more efficient in promoting this reaction. A much higher AB momentum is needed if BC vibration is set small, and will often cause the trajectory to bounce back resulting in a nonreactive condition.

File:Yu v l.PNG

2018-05-25T16:00:28Z

Yr316:

Rep:YR316DYN

2018-05-22T15:34:51Z

Yr316: /* Question 6: energetic of the reactions */

Rep:YR316DYN

2018-05-22T15:19:33Z

Yr316: /* Question 6: energetic of the reactions */