ChemWiki - User contributions [en]

MRD:01541238

2020-05-22T16:39:39Z

Yt6718: /* Question 8. Report the activation energy for both reactions. */

= The Molecular reaction dynamics computational lab report =

=== Question 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? ===
[[File:1.png]][[File:2.png]]

These two diagrams are the counter plot and skew plot from a three H atoms model. A H + H2 system.

The transition state is the saddle point on the potential energy surface diagram (gallery 2). The black reaction trajectory line shows the minimum energy path of the reactants to the products.

When the trajectory line pass through the transition structure it shows a wavy line. And the transition state is defined as the maximum on the black trajectory minimum energy path.

Mathematically the transition state point has the special property that ∂V('''ri''')/∂'''ri'''=0 and the energy is higher than any other local minimum points on the minimum energy path.

And the transition state point is a critical point which can be identified by secondary derivative =0. The difference between local minimum is that they have a non zero second derivative number.

=== Question 2. 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. ===
My best estimate of the transition state position '''r1''' = '''r2 '''= 90.8 pm and '''p1''' = '''p2''' = 0.0 g.mol-1.pm.fs-1

According to the theory, when the structure is at transition structure, the trajectory point will only oscillate at on point.

[[File:Dist vs time2.png]]

The diagram above is the 'internuclear distance vs time' plot for my best estimation of the positions. As shown on the diagram, there is almost no oscillation of the bond distances between B-C and A-B which means

the structure is under a 'stable' state and this structure is called transition state structure.

=== Question 3. Comment on how the ''mep'' and the trajectory you just calculated differ. ===
In my case, positions '''r1''' = '''91.8''' pm, '''r2''' = '''90.8 '''pm and the momenta '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1 for the new trajectory calculated.

[[File:6_01541238.png]][[File:4_2_01541238.png]]

Compare with two diagrams, the MEP doesn't have the vibrations of the atoms during the process while the trajectory shows oscillation to the products. And other remains the same, both direct to the same product.

=== Question 4. 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? ===
For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|reactive
|<gallery>
7_01541238.png
</gallery>
|Enough kinetic energy to overcome activation barrier.
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|unreactive
|<gallery>
8_01541238.png
</gallery>
|Don't have enough energy to overcome the activation barrier, bounce
back to the more energy preferred state.
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|reactive
|<gallery>
9_01541238.png
</gallery>
|Enough kinetic energy to overcome activation barrier.
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|reactive
|<gallery>
10_2_01541238.png
</gallery>
|Enough kinetic energy to overcome activation barrier.
The energy is much higher than the previous ones.

Cause the reaction reacts back and overcomes the transition states

twice and back to the initial conditions.
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|reactive
|<gallery>
11_01541238.png
</gallery>
|Enough kinetic energy to overcome activation barrier.
The energy is higher than the previous one.

Cause the reaction reacts back and overcomes the transition states

three times and complete the reaction.
|}
The table above list some occasions for the different initial energies for the system. The table shows the energy is critically controlled in order to make the reaction complete.

The total energy can't be too low so that can't pass though the energy barrier or too high to bounce back even through the transition state point twice.

=== Question 5. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
With compare with experimental values, the Transition State Theory predictions will be over-estimated. Transition State Theory assumes the transition state structure will not go back to the initial reactants since the transition

structure is formed. An equilibrium will form between the reactants and the transition states. However, in real life there is barrier recrossing cause the transition state structure goes back.

Moreover, the Transition State Theory is treated classically Quantum mechanical effect like tunnelling effect is ignored. In real situations, particles with lower kinetic energy might able to cross the high energy barrier caused

by tunnelling effect. Transition State Theory may overestimate the rate because it requires more energy to overcome the barrier.

=== Question 6. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===
[[File:12_01541238.png]]

As shown in the diagram above, it is a potential energy surfaces for F + H2 system. The LHS is for HF energy level which is lower than H2 energy level on the RHS. Because the F-H bond is stronger than H-H bond for

F + H2 reaction is endothermic as stronger F-H bond formation required energy. Vice Versa, H + HF reaction is exothermic for strong H-F bond is broke.

=== Question 7. Locate the approximate position of the transition state. ===

[[File:13_01541238.png]]

Diagram above is the 'internuclear distance vs time' for my estimated transition state structure.

My best estimated transition state positions are F-H = 180.8 pm and H-H = 74.5 pm when '''p1''' = '''p2''' = 0.0 g.mol-1.pm.fs-1

=== Question 8. Report the activation energy for both reactions. ===
For F + H2 reaction, the activation energy is the energy differences between reactant and transition state structure.

Ea=1.075 kJ/mol

For H + HF reaction,

Ea=126.682 kJ/mol

=== Question 9. 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:14_01541238.png]][[File:15_01541238.png]]

[[File:16_01541238.png]][[File:17_01541238.png]]

For the F + H2 reaction. It is an exothermic reaction as the strong F-H bond formed. The potential energy will transfer to kinetic energy in order to complete the reaction.

On diagram 1, the set is F-H = 230 pm and H-H = 74 pm when '''p1''' = -1.0 '''p2''' = 6 g.mol-1.pm.fs-1. It shows that F-H bond is formed then break apart. Potential energy is transferred to kinetic energy that cause the recrossing

barrier and go back to initial structure as the kinetic energy is too large.

On diagram 2, the set is F-H = 230 pm and H-H = 74 pm when '''p1''' = -1.0 '''p2''' = 0 g.mol-1.pm.fs-1. The diagram clearly shows 0.526 kJ kinetic energy is transferred which is lower than Activation energy and the reaction can't

happen. F-H bond not formed.

On diagram 3, the set is F-H = 230 pm and H-H = 74 pm when '''p1''' = -1.0 '''p2''' = -6 g.mol-1.pm.fs-1. F-H bond is formed after several formation and breakage. A significant amount of potential energy transfer into kinetic energy

cause the oscillation for the middle H atom during the reaction.

On diagram 4, the set is F-H = 230 pm and H-H = 74 pm when '''p1''' = -1.6 '''p2''' = 0.2 g.mol-1.pm.fs-1. Kinetic energy is 1.707 kJ/mol overcomes the activation barrier and from the F-H bond.

[[File:18_01541238.png]]

On the reverse reaction H + HF, the initial set is H-H = 230 pm and H-F = 100 pm when '''p1''' = -20 '''p2''' = -1 g.mol-1.pm.fs-1. This diagram is one possible reactive trajectory for the reaction to complete. H atom carries 380.526 kJ/mol

to collide with H-F molecule and form new H-H bond.

In conclusion, the kinetic energy can be expressed as temperature. As the potential energy is transferred into kinetic energy, temperature will raise as the result of kinetic energy increases. Monitoring the temperature can help us

to confirm this mechanism processes.

=== Question 10. 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. ===
Translational energy is due to the change in positions and states while vibrational energy is due to the change in structure of molecules. In fact, the translation energy is more effective in promoting reactions for system with early

barrier, whereas vibrational energy is more effective for reactions with late barrier. The position of the transition state is at the top of the activation energy barrier curve on a energy-process diagram. As a result, if the transition state

structure is similar to reactant (early barrier) translational energy is more effective. Vice Versa, if the transition state structure is more similar to the product (late barrier) vibrational energy is more effective. Moreover, symmetric

structure always have higher reaction rate and reactivity than unsymmetric structure.

MRD:01541238

2020-05-22T14:57:10Z

Yt6718: /* Question 10. 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 Molecular reaction dynamics computational lab report =

=== Question 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? ===
[[File:1.png]][[File:2.png]]

These two diagrams are the counter plot and skew plot from a three H atoms model. A H + H2 system.

The transition state is the saddle point on the potential energy surface diagram (gallery 2). The black reaction trajectory line shows the minimum energy path of the reactants to the products.

When the trajectory line pass through the transition structure it shows a wavy line. And the transition state is defined as the maximum on the black trajectory minimum energy path.

Mathematically the transition state point has the special property that ∂V('''ri''')/∂'''ri'''=0 and the energy is higher than any other local minimum points on the minimum energy path.

And the transition state point is a critical point which can be identified by secondary derivative =0. The difference between local minimum is that they have a non zero second derivative number.

=== Question 2. 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. ===
My best estimate of the transition state position '''r1''' = '''r2 '''= 90.8 pm and '''p1''' = '''p2''' = 0.0 g.mol-1.pm.fs-1

According to the theory, when the structure is at transition structure, the trajectory point will only oscillate at on point.

[[File:Dist vs time2.png]]

The diagram above is the 'internuclear distance vs time' plot for my best estimation of the positions. As shown on the diagram, there is almost no oscillation of the bond distances between B-C and A-B which means

the structure is under a 'stable' state and this structure is called transition state structure.

=== Question 3. Comment on how the ''mep'' and the trajectory you just calculated differ. ===
In my case, positions '''r1''' = '''91.8''' pm, '''r2''' = '''90.8 '''pm and the momenta '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1 for the new trajectory calculated.

[[File:6_01541238.png]][[File:4_2_01541238.png]]

Compare with two diagrams, the MEP doesn't have the vibrations of the atoms during the process while the trajectory shows oscillation to the products. And other remains the same, both direct to the same product.

=== Question 4. 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? ===
For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|reactive
|<gallery>
7_01541238.png
</gallery>
|Enough kinetic energy to overcome activation barrier.
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|unreactive
|<gallery>
8_01541238.png
</gallery>
|Don't have enough energy to overcome the activation barrier, bounce
back to the more energy preferred state.
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|reactive
|<gallery>
9_01541238.png
</gallery>
|Enough kinetic energy to overcome activation barrier.
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|reactive
|<gallery>
10_2_01541238.png
</gallery>
|Enough kinetic energy to overcome activation barrier.
The energy is much higher than the previous ones.

Cause the reaction reacts back and overcomes the transition states

twice and back to the initial conditions.
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|reactive
|<gallery>
11_01541238.png
</gallery>
|Enough kinetic energy to overcome activation barrier.
The energy is higher than the previous one.

Cause the reaction reacts back and overcomes the transition states

three times and complete the reaction.
|}
The table above list some occasions for the different initial energies for the system. The table shows the energy is critically controlled in order to make the reaction complete.

The total energy can't be too low so that can't pass though the energy barrier or too high to bounce back even through the transition state point twice.

=== Question 5. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
With compare with experimental values, the Transition State Theory predictions will be over-estimated. Transition State Theory assumes the transition state structure will not go back to the initial reactants since the transition

structure is formed. An equilibrium will form between the reactants and the transition states. However, in real life there is barrier recrossing cause the transition state structure goes back.

Moreover, the Transition State Theory is treated classically Quantum mechanical effect like tunnelling effect is ignored. In real situations, particles with lower kinetic energy might able to cross the high energy barrier caused

by tunnelling effect. Transition State Theory may overestimate the rate because it requires more energy to overcome the barrier.

=== Question 6. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===
[[File:12_01541238.png]]

As shown in the diagram above, it is a potential energy surfaces for F + H2 system. The LHS is for HF energy level which is lower than H2 energy level on the RHS. Because the F-H bond is stronger than H-H bond for

F + H2 reaction is endothermic as stronger F-H bond formation required energy. Vice Versa, H + HF reaction is exothermic for strong H-F bond is broke.

=== Question 7. Locate the approximate position of the transition state. ===

[[File:13_01541238.png]]

Diagram above is the 'internuclear distance vs time' for my estimated transition state structure.

My best estimated transition state positions are F-H = 180.8 pm and H-H = 74.5 pm when '''p1''' = '''p2''' = 0.0 g.mol-1.pm.fs-1

=== Question 8. Report the activation energy for both reactions. ===
For F + H2 reaction, the activation energy is the energy differences between reactant and transition state structure.

Ea=1.075 kJ/mol

For H + HF reaction,

Ea=173.682 kJ/mol

=== Question 9. 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:14_01541238.png]][[File:15_01541238.png]]

[[File:16_01541238.png]][[File:17_01541238.png]]

For the F + H2 reaction. It is an exothermic reaction as the strong F-H bond formed. The potential energy will transfer to kinetic energy in order to complete the reaction.

On diagram 1, the set is F-H = 230 pm and H-H = 74 pm when '''p1''' = -1.0 '''p2''' = 6 g.mol-1.pm.fs-1. It shows that F-H bond is formed then break apart. Potential energy is transferred to kinetic energy that cause the recrossing

barrier and go back to initial structure as the kinetic energy is too large.

On diagram 2, the set is F-H = 230 pm and H-H = 74 pm when '''p1''' = -1.0 '''p2''' = 0 g.mol-1.pm.fs-1. The diagram clearly shows 0.526 kJ kinetic energy is transferred which is lower than Activation energy and the reaction can't

happen. F-H bond not formed.

On diagram 3, the set is F-H = 230 pm and H-H = 74 pm when '''p1''' = -1.0 '''p2''' = -6 g.mol-1.pm.fs-1. F-H bond is formed after several formation and breakage. A significant amount of potential energy transfer into kinetic energy

cause the oscillation for the middle H atom during the reaction.

On diagram 4, the set is F-H = 230 pm and H-H = 74 pm when '''p1''' = -1.6 '''p2''' = 0.2 g.mol-1.pm.fs-1. Kinetic energy is 1.707 kJ/mol overcomes the activation barrier and from the F-H bond.

[[File:18_01541238.png]]

On the reverse reaction H + HF, the initial set is H-H = 230 pm and H-F = 100 pm when '''p1''' = -20 '''p2''' = -1 g.mol-1.pm.fs-1. This diagram is one possible reactive trajectory for the reaction to complete. H atom carries 380.526 kJ/mol

to collide with H-F molecule and form new H-H bond.

In conclusion, the kinetic energy can be expressed as temperature. As the potential energy is transferred into kinetic energy, temperature will raise as the result of kinetic energy increases. Monitoring the temperature can help us

to confirm this mechanism processes.

=== Question 10. 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. ===
Translational energy is due to the change in positions and states while vibrational energy is due to the change in structure of molecules. In fact, the translation energy is more effective in promoting reactions for system with early

barrier, whereas vibrational energy is more effective for reactions with late barrier. The position of the transition state is at the top of the activation energy barrier curve on a energy-process diagram. As a result, if the transition state

structure is similar to reactant (early barrier) translational energy is more effective. Vice Versa, if the transition state structure is more similar to the product (late barrier) vibrational energy is more effective. Moreover, symmetric

structure always have higher reaction rate and reactivity than unsymmetric structure.

MRD:01541238

2020-05-22T14:36:19Z

Yt6718:

= The Molecular reaction dynamics computational lab report =

=== Question 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? ===
[[File:1.png]][[File:2.png]]

These two diagrams are the counter plot and skew plot from a three H atoms model. A H + H2 system.

The transition state is the saddle point on the potential energy surface diagram (gallery 2). The black reaction trajectory line shows the minimum energy path of the reactants to the products.

When the trajectory line pass through the transition structure it shows a wavy line. And the transition state is defined as the maximum on the black trajectory minimum energy path.

Mathematically the transition state point has the special property that ∂V('''ri''')/∂'''ri'''=0 and the energy is higher than any other local minimum points on the minimum energy path.

And the transition state point is a critical point which can be identified by secondary derivative =0. The difference between local minimum is that they have a non zero second derivative number.

=== Question 2. 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. ===
My best estimate of the transition state position '''r1''' = '''r2 '''= 90.8 pm and '''p1''' = '''p2''' = 0.0 g.mol-1.pm.fs-1

According to the theory, when the structure is at transition structure, the trajectory point will only oscillate at on point.

[[File:Dist vs time2.png]]

The diagram above is the 'internuclear distance vs time' plot for my best estimation of the positions. As shown on the diagram, there is almost no oscillation of the bond distances between B-C and A-B which means

the structure is under a 'stable' state and this structure is called transition state structure.

=== Question 3. Comment on how the ''mep'' and the trajectory you just calculated differ. ===
In my case, positions '''r1''' = '''91.8''' pm, '''r2''' = '''90.8 '''pm and the momenta '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1 for the new trajectory calculated.

[[File:6_01541238.png]][[File:4_2_01541238.png]]

Compare with two diagrams, the MEP doesn't have the vibrations of the atoms during the process while the trajectory shows oscillation to the products. And other remains the same, both direct to the same product.

=== Question 4. 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? ===
For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|reactive
|<gallery>
7_01541238.png
</gallery>
|Enough kinetic energy to overcome activation barrier.
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|unreactive
|<gallery>
8_01541238.png
</gallery>
|Don't have enough energy to overcome the activation barrier, bounce
back to the more energy preferred state.
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|reactive
|<gallery>
9_01541238.png
</gallery>
|Enough kinetic energy to overcome activation barrier.
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|reactive
|<gallery>
10_2_01541238.png
</gallery>
|Enough kinetic energy to overcome activation barrier.
The energy is much higher than the previous ones.

Cause the reaction reacts back and overcomes the transition states

twice and back to the initial conditions.
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|reactive
|<gallery>
11_01541238.png
</gallery>
|Enough kinetic energy to overcome activation barrier.
The energy is higher than the previous one.

Cause the reaction reacts back and overcomes the transition states

three times and complete the reaction.
|}
The table above list some occasions for the different initial energies for the system. The table shows the energy is critically controlled in order to make the reaction complete.

The total energy can't be too low so that can't pass though the energy barrier or too high to bounce back even through the transition state point twice.

=== Question 5. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
With compare with experimental values, the Transition State Theory predictions will be over-estimated. Transition State Theory assumes the transition state structure will not go back to the initial reactants since the transition

structure is formed. An equilibrium will form between the reactants and the transition states. However, in real life there is barrier recrossing cause the transition state structure goes back.

Moreover, the Transition State Theory is treated classically Quantum mechanical effect like tunnelling effect is ignored. In real situations, particles with lower kinetic energy might able to cross the high energy barrier caused

by tunnelling effect. Transition State Theory may overestimate the rate because it requires more energy to overcome the barrier.

=== Question 6. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===
[[File:12_01541238.png]]

As shown in the diagram above, it is a potential energy surfaces for F + H2 system. The LHS is for HF energy level which is lower than H2 energy level on the RHS. Because the F-H bond is stronger than H-H bond for

F + H2 reaction is endothermic as stronger F-H bond formation required energy. Vice Versa, H + HF reaction is exothermic for strong H-F bond is broke.

=== Question 7. Locate the approximate position of the transition state. ===

[[File:13_01541238.png]]

Diagram above is the 'internuclear distance vs time' for my estimated transition state structure.

My best estimated transition state positions are F-H = 180.8 pm and H-H = 74.5 pm when '''p1''' = '''p2''' = 0.0 g.mol-1.pm.fs-1

=== Question 8. Report the activation energy for both reactions. ===
For F + H2 reaction, the activation energy is the energy differences between reactant and transition state structure.

Ea=1.075 kJ/mol

For H + HF reaction,

Ea=173.682 kJ/mol

=== Question 9. 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:14_01541238.png]][[File:15_01541238.png]]

[[File:16_01541238.png]][[File:17_01541238.png]]

For the F + H2 reaction. It is an exothermic reaction as the strong F-H bond formed. The potential energy will transfer to kinetic energy in order to complete the reaction.

On diagram 1, the set is F-H = 230 pm and H-H = 74 pm when '''p1''' = -1.0 '''p2''' = 6 g.mol-1.pm.fs-1. It shows that F-H bond is formed then break apart. Potential energy is transferred to kinetic energy that cause the recrossing

barrier and go back to initial structure as the kinetic energy is too large.

On diagram 2, the set is F-H = 230 pm and H-H = 74 pm when '''p1''' = -1.0 '''p2''' = 0 g.mol-1.pm.fs-1. The diagram clearly shows 0.526 kJ kinetic energy is transferred which is lower than Activation energy and the reaction can't

happen. F-H bond not formed.

On diagram 3, the set is F-H = 230 pm and H-H = 74 pm when '''p1''' = -1.0 '''p2''' = -6 g.mol-1.pm.fs-1. F-H bond is formed after several formation and breakage. A significant amount of potential energy transfer into kinetic energy

cause the oscillation for the middle H atom during the reaction.

On diagram 4, the set is F-H = 230 pm and H-H = 74 pm when '''p1''' = -1.6 '''p2''' = 0.2 g.mol-1.pm.fs-1. Kinetic energy is 1.707 kJ/mol overcomes the activation barrier and from the F-H bond.

[[File:18_01541238.png]]

On the reverse reaction H + HF, the initial set is H-H = 230 pm and H-F = 100 pm when '''p1''' = -20 '''p2''' = -1 g.mol-1.pm.fs-1. This diagram is one possible reactive trajectory for the reaction to complete. H atom carries 380.526 kJ/mol

to collide with H-F molecule and form new H-H bond.

In conclusion, the kinetic energy can be expressed as temperature. As the potential energy is transferred into kinetic energy, temperature will raise as the result of kinetic energy increases. Monitoring the temperature can help us

to confirm this mechanism processes.

=== Question 10. 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. ===
Translational energy is due to the change in positions and states while vibrational energy is due to the change in structure of molecules.

MRD:01541238

2020-05-22T14:34:34Z

Yt6718: /* Question 9. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally. */

= The Molecular reaction dynamics computational lab report =

=== Question 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? ===
[[File:1.png]][[File:2.png]]

These two diagrams are the counter plot and skew plot from a three H atoms model. A H + H2 system.

The transition state is the saddle point on the potential energy surface diagram (gallery 2). The black reaction trajectory line shows the minimum energy path of the reactants to the products.

When the trajectory line pass through the transition structure it shows a wavy line. And the transition state is defined as the maximum on the black trajectory minimum energy path.

Mathematically the transition state point has the special property that ∂V('''ri''')/∂'''ri'''=0 and the energy is higher than any other local minimum points on the minimum energy path.

And the transition state point is a critical point which can be identified by secondary derivative =0. The difference between local minimum is that they have a non zero second derivative number.

=== Question 2. 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. ===
My best estimate of the transition state position '''r1''' = '''r2 '''= 90.8 pm and '''p1''' = '''p2''' = 0.0 g.mol-1.pm.fs-1

According to the theory, when the structure is at transition structure, the trajectory point will only oscillate at on point.

[[File:Dist vs time2.png]]

The diagram above is the 'internuclear distance vs time' plot for my best estimation of the positions. As shown on the diagram, there is almost no oscillation of the bond distances between B-C and A-B which means

the structure is under a 'stable' state and this structure is called transition state structure.

=== Question 3. Comment on how the ''mep'' and the trajectory you just calculated differ. ===
In my case, positions '''r1''' = '''91.8''' pm, '''r2''' = '''90.8 '''pm and the momenta '''p1''' = '''p2''' = 0 g.mol-1.pm.fs-1 for the new trajectory calculated.

[[File:6_01541238.png]][[File:4_2_01541238.png]]

Compare with two diagrams, the MEP doesn't have the vibrations of the atoms during the process while the trajectory shows oscillation to the products. And other remains the same, both direct to the same product.

=== Question 4. 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? ===
For the initial positions '''r1''' = 74 pm and '''r2''' = 200 pm
{| class="wikitable"
!p1/ g.mol-1.pm.fs-1
!p2/ g.mol-1.pm.fs-1
!Etot/kJ mol-1
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|reactive
|<gallery>
7_01541238.png
</gallery>
|Enough kinetic energy to overcome activation barrier.
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|unreactive
|<gallery>
8_01541238.png
</gallery>
|Don't have enough energy to overcome the activation barrier, bounce
back to the more energy preferred state.
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|reactive
|<gallery>
9_01541238.png
</gallery>
|Enough kinetic energy to overcome activation barrier.
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|reactive
|<gallery>
10_2_01541238.png
</gallery>
|Enough kinetic energy to overcome activation barrier.
The energy is much higher than the previous ones.

Cause the reaction reacts back and overcomes the transition states

twice and back to the initial conditions.
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|reactive
|<gallery>
11_01541238.png
</gallery>
|Enough kinetic energy to overcome activation barrier.
The energy is higher than the previous one.

Cause the reaction reacts back and overcomes the transition states

three times and complete the reaction.
|}
The table above list some occasions for the different initial energies for the system. The table shows the energy is critically controlled in order to make the reaction complete.

The total energy can't be too low so that can't pass though the energy barrier or too high to bounce back even through the transition state point twice.

=== Question 5. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? ===
With compare with experimental values, the Transition State Theory predictions will be over-estimated. Transition State Theory assumes the transition state structure will not go back to the initial reactants since the transition

structure is formed. An equilibrium will form between the reactants and the transition states. However, in real life there is barrier recrossing cause the transition state structure goes back.

Moreover, the Transition State Theory is treated classically Quantum mechanical effect like tunnelling effect is ignored. In real situations, particles with lower kinetic energy might able to cross the high energy barrier caused

by tunnelling effect. Transition State Theory may overestimate the rate because it requires more energy to overcome the barrier.

=== Question 6. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===
[[File:12_01541238.png]]

As shown in the diagram above, it is a potential energy surfaces for F + H2 system. The LHS is for HF energy level which is lower than H2 energy level on the RHS. Because the F-H bond is stronger than H-H bond for

F + H2 reaction is endothermic as stronger F-H bond formation required energy. Vice Versa, H + HF reaction is exothermic for strong H-F bond is broke.

=== Question 7. Locate the approximate position of the transition state. ===

[[File:13_01541238.png]]

Diagram above is the 'internuclear distance vs time' for my estimated transition state structure.

My best estimated transition state positions are F-H = 180.8 pm and H-H = 74.5 pm when '''p1''' = '''p2''' = 0.0 g.mol-1.pm.fs-1

=== Question 8. Report the activation energy for both reactions. ===
For F + H2 reaction, the activation energy is the energy differences between reactant and transition state structure.

Ea=1.075 kJ/mol

For H + HF reaction,

Ea=173.682 kJ/mol

=== Question 9. 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:14_01541238.png

File:15_01541238.png

File:16_01541238.png

File:17_01541238.png

For the F + H2 reaction. It is an exothermic reaction as the strong F-H bond formed. The potential energy will transfer to kinetic energy in order to complete the reaction.

On diagram 1, the set is F-H = 230 pm and H-H = 74 pm when '''p1''' = -1.0 '''p2''' = 6 g.mol-1.pm.fs-1. It shows that F-H bond is formed then break apart. Potential energy is transferred to kinetic energy that cause the recrossing

barrier and go back to initial structure as the kinetic energy is too large.

On diagram 2, the set is F-H = 230 pm and H-H = 74 pm when '''p1''' = -1.0 '''p2''' = 0 g.mol-1.pm.fs-1. The diagram clearly shows 0.526 kJ kinetic energy is transferred which is lower than Activation energy and the reaction can't

happen. F-H bond not formed.

On diagram 3, the set is F-H = 230 pm and H-H = 74 pm when '''p1''' = -1.0 '''p2''' = -6 g.mol-1.pm.fs-1. F-H bond is formed after several formation and breakage. A significant amount of potential energy transfer into kinetic energy

cause the oscillation for the middle H atom during the reaction.

On diagram 4, the set is F-H = 230 pm and H-H = 74 pm when '''p1''' = -1.6 '''p2''' = 0.2 g.mol-1.pm.fs-1. Kinetic energy is 1.707 kJ/mol overcomes the activation barrier and from the F-H bond.

File:18_01541238.png

On the reverse reaction H + HF, the initial set is H-H = 230 pm and H-F = 100 pm when '''p1''' = -20 '''p2''' = -1 g.mol-1.pm.fs-1. This diagram is one possible reactive trajectory for the reaction to complete. H atom carries 380.526 kJ/mol

to collide with H-F molecule and form new H-H bond.

In conclusion, the kinetic energy can be expressed as temperature. As the potential energy is transferred into kinetic energy, temperature will raise as the result of kinetic energy increases. Monitoring the temperature can help us

to confirm this mechanism processes.

=== Question 10. 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. ===
Translational energy is due to the change in positions and states while vibrational energy is due to the change in structure of molecules.

File:18 01541238.png

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Yt6718: /* Question 7. Locate the approximate position of the transition state. */

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Yt6718: /* Question 8. Report the activation energy for both reactions. */

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Yt6718: /* Question 3. Comment on how the mep and the trajectory you just calculated differ. */

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Yt6718: /* Question 2. 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. */

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Yt6718: /* The Molecular reaction dynamics computational lab report */

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Yt6718: /* Question 5. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? */

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Yt6718: /* Question 3.Comment on how the mep and the trajectory you just calculated differ. */

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Yt6718: /* Question 3.Comment on how the mep and the trajectory you just calculated differ. */

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Yt6718: /* The Molecular reaction dynamics computational lab report */

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Yt6718: /* The Molecular reaction dynamics computational lab report */

= The Molecular reaction dynamics computational lab report =
<gallery>
Surface_Plot.png

</gallery>

MRD:01541238

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Yt6718: Created page with "= The Molecular reaction dynamics computational lab report ="

= The Molecular reaction dynamics computational lab report =

Yt6718

2019-03-14T19:14:29Z

Yt6718: /* CH4 optimisation */

=NH3 optimisation=
Calculation Method = RB3LYP

Basis Set = 6-31G(d,p)

E(RB3LYP) = -56.55776871 a.u.

RMS Gradient Norm = 0.00004473 a.u.

Point Group = C3V

optimised N-H bond distance=1.01806Å

optimised H-N-H bond angle=105.747°
{| class="wikitable"
!Item
!Value
!Threshold
!Converged?
|-
|Maximum Force
|0.000080
|0.000450
|YES
|-
|RMS Force
|0.000053
|0.000300
|YES
|-
|Maximum Displacement
|0.000190
|0.001800
|YES
|-
|RMS Displacement
|0.000124
|0.001200
|YES
|}
Predicted change in Energy=-2.197407D-08

<jmol><jmolApplet>
<title>test molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>Yt6718 NH3.LOG</uploadedFileContents>
</jmolApplet></jmol>

[[File:01541238-2.png]]
{| class="wikitable"
|wavenumber cm-1
|1089.6540
|1694.0484
|1694.0484
|3460.3171
|3588.8744
|3588.8744
|-
|symmetry
|A1
|E
|E
|A1
|E
|E
|-
|intensity
(arbitrary units)
|145.4055
|13.5543
|13.5544
|1.0615
|0.2712
|0.2712
|-
|image
|[[File:01541238-3.png]]
|[[File:01541238-4.png]]
|[[File:01541238-5.png]]
|[[File:01541238-6.png]]
|[[File:01541238-7.png]]
|[[File:01541238-8.png]]
|}
how many modes do you expect from the 3N-6 rule?
6

which modes are degenerate (ie have the same energy)?
two modes of frequency at 1694

which modes are "bending" vibrations and which are "bond stretch" vibrations?
first three modes are bending vibrations and last three modes are bond stretch vibrations

which mode is highly symmetric?
No.4 mode

one mode is known as the "umbrella" mode, which one is this?
mode 1

how many bands would you expect to see in an experimental spectrum of gaseous ammonia?
4

[[File:01541238-9.png]]
charge positive for H and charge negative for N atom. Because N atom is more electronegativity than H atom and it will attract electron pairs closer to it.

== N2 optimisation ==
Calculation Method = RB3LYP

Basis Set = 6-31G(d,p)

E(RB3LYP) = -109.52359111 a.u.

RMS Gradient Norm = 0.02473091 a.u.

Point Group = D∞h

optimised N-N bond distance=1.10550Å

{| class="wikitable"
!Item
!Value
!Threshold
!Converged?
|-
|Maximum Force
|0.042835
|0.000450
|NO
|-
|RMS Force
|0.042835
|0.000300
|NO
|-
|Maximum Displacement
|0.011569
|0.001800
|NO
|-
|RMS Displacement
|0.016360
|0.001200
|NO
|}
Predicted change in Energy=-4.958069D-04

<jmol><jmolApplet>
<title>test molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>01541238-N2.LOG</uploadedFileContents>
</jmolApplet></jmol>
[[File:01541238-8.png]]
{| class="wikitable"
|wavenumber cm-1
|2457.3283
|-
|symmetry
|SGG
|-
|intensity
(arbitrary units)
|0.0000
|-
|image
|[[File:01541238-11.png]]
|}
[[File:01541238-15.png]]

== H2 optimisation ==
Calculation Method = RB3LYP

Basis Set = 6-31G(d,p)

E(RB3LYP) = -1.17853936 a.u.

RMS Gradient Norm = 0.00000017 a.u.

Point Group = D∞h

optimised H-H bond distance=0.74279Å

{| class="wikitable"
!Item
!Value
!Threshold
!Converged?
|-
|Maximum Force
|0.168347
|0.000450
|NO
|-
|RMS Force
|0.168347
|0.000300
|NO
|-
|Maximum Displacement
|0.119698
|0.001800
|NO
|-
|RMS Displacement
|0.169278
|0.001200
|NO
|}
Predicted change in Energy=-2.130559D-02

<jmol><jmolApplet>
<title>test molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>01541238-H2.LOG</uploadedFileContents>
</jmolApplet></jmol>
[[File:01541238-12.png]]
{| class="wikitable"
|wavenumber cm-1
|4465.6824
|-
|symmetry
|SGG
|-
|intensity
(arbitrary units)
|0.0000
|-
|image
|[[File:01541238-13.png]]
|}
[[File:01541238-14.png]]

Sturcture ABASUR[[https://www.ccdc.cam.ac.uk/structures/Search?Ccdcid=ABASUR&DatabaseToSearch=Published here]] has H-H bond length 0.99435Å and computer simulation gets 0.74279Å.
This difference may because of the angle strain fron the compound and different chemical environments. Also, in the compound H forms two bounds while pure H2 gas only forms one bond.

E(NH3)= -56.55776871 a.u.
2*E(NH3)= -113.1155374 a.u.
E(N2)= -109.52359111 a.u.
E(H2)= -1.17853936 a.u.
3*E(H2)= -3.53561808 a.u.
ΔE=2*E(NH3)-[E(N2)+3*E(H2)]= -148.2 kJ/mol

The energy for converting hydrogen and nitrogen gas into ammonia gas is -148.2 kJ/mol. Ammonia productis are more stable because it is an exothermic reaction the product is lower in energy states and thermodinamicly more stable.

== CO optimisation ==
Calculation Method = RB3LYP

Basis Set = 6-31G(d,p)

E(RB3LYP) = -113.30945314 a.u.

RMS Gradient Norm = 0.00001828 a.u.

Point Group = C∞V

optimised C=O bond distance=1.13793Å

{| class="wikitable"
!Item
!Value
!Threshold
!Converged?
|-
|Maximum Force
|0.191096
|0.000450
|NO
|-
|RMS Force
|0.191096
|0.000300
|NO
|-
|Maximum Displacement
|0.113036
|0.001800
|NO
|-
|RMS Displacement
|0.159857
|0.001200
|NO
|}
Predicted change in Energy=-2.270461D-02

<jmol><jmolApplet>
<title>test molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>01541238-CO.LOG</uploadedFileContents>
</jmolApplet></jmol>
[[File:01541238-16.png]]
{| class="wikitable"
|wavenumber cm-1
|2209.1389
|-
|symmetry
|SG
|-
|intensity
(arbitrary units)
|67.9587
|-
|image
|[[File:01541238-17.png]]
|}
[[File:01541238-18.png]]

[[File:01541238-19.png]]

This HOMO gerada sigma bonding forms by one electron from C's 2P orbital and one electron from O's 2P.

[[File:01541238-20.png]]

This gerada pi bonding forms by one electron from C's 2P orbital and three electrons from O's 2P.

[[File:01541238-21.png]]

This ungerada sigma anti-bonding forms by one electron from C's 2S orbital and one electron from O's 2S.

[[File:01541238-22.png]]

This gerada sigma bonding forms by one electron from C's 2S orbital and one electron from O's 2S.

[[File:01541238-23.png]]

This ungerada sigma anti-bonding forms by one electron from C's 1S orbital and one electron from O's 1S.

== CH4 optimisation ==
Calculation Method = RB3LYP

Basis Set = 6-31G(d,p)

E(RB3LYP) = -40.52401404 a.u.

RMS Gradient Norm = 0.00003263 a.u.

Point Group = TD

optimised C-H bond distance=1.09197Å

optimised H-C-H bond angle=109.471
{| class="wikitable"
!Item
!Value
!Threshold
!Converged?
|-
|Maximum Force
|0.015565
|0.000450
|NO
|-
|RMS Force
|0.008320
|0.000300
|NO
|-
|Maximum Displacement
|0.041520
|0.001800
|NO
|-
|RMS Displacement
|0.022194
|0.001200
|NO
|}
Predicted change in Energy=-1.301482D-03

<jmol><jmolApplet>
<title>test molecule</title>
<color>black</color>
<size>200</size>
<uploadedFileContents>01541238-CH4.LOG</uploadedFileContents>
</jmolApplet></jmol>
[[File:01541238-24.png]]
{| class="wikitable"
|wavenumber cm-1
|1356.2043
|1356.2043
|1356.2043
|1578.5779
|1578.5779
|3046.4628
|3162.3267
|3162.3267
|3162.3267
|-
|symmetry
|T2
|T2
|T2
|E
|E
|A1
|T2
|T2
|T2
|-
|intensity
(arbitrary units)
|14.1008
|14.1008
|14.1008
|0.0000
|0.0000
|0.0000
|25.3343
|25.3343
|25.3343
|-
|image
|[[File:01541238-25.png]]
|[[File:01541238-26.png]]
|[[File:01541238-27.png]]
|[[File:01541238-28.png]]
|[[File:01541238-29.png]]
|[[File:01541238-30.png]]
|[[File:01541238-31.png]]
|[[File:01541238-32.png]]
|[[File:01541238-33.png]]
|}
[[File:01541238-34.png]]

[[File:01541238-35.png]]

This LOMO unoccupied gerada pi bonding bond.

[[File:01541238-36.png]]

This HOMO gerada sigma bonding forms by two electron from H and four electrons from C's 2P orbital.

[[File:01541238-37.png]]

This ungerada sigma anti-bonding forms by one electron from C's 1S orbital and one electron from H.

[[File:01541238-38.png]]

This gerada sigma bonding forms by one electron from C's 1S orbital and one electron from one of the H's 1S.

[[File:01541238-39.png]]

This un occupied ungerada pi anti-bonding..

File:01541238-39.png

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