File:Jl initialFH ex2.png

2018-05-17T16:46:34Z

Jl7816:

MRD:jl7816

2018-05-17T16:46:17Z

Jl7816: /* Exercise 2 */

==Molecular reaction dynamics==
===Exercise 1===
'''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.'''

At a minimum, the gradient only has one component, either ∂V/∂rAB= 0 or ∂V/∂rBC= 0.
The curvature at the minima is positive and the secondary derivative is positive.
At a transition structure, the gradients of both components rAB and rBC are equal to 0.
The secondary derivative of one component is positive and the other is negative, this is called a saddle point.

'''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.'''

rts= 0.9076
In the figure 1, it is an Internuclear Distances vs Time plot, the curves of A-B, B-C and A-C are flat, indicating that there is no vibration. The lines of A-B and B-C are overlapped.
[[File:jl_rts.png|500px|thumb|center|Figure 1]]

'''Comment on how the mep and the trajectory you just calculated differ.'''
[[File:jl_dynamics_ex1.png|300px|thumb|left|Figure 2-trajectory]]
[[File:jl_mep_ex1.png|300px|thumb|centre|Figure 3-MEP]]
Figure 2 and 3 show the Internuclear Distances vs Time plot.
{| class="wikitable" border="1"
|+ Difference in Internuclear Distances
! !! Distance (trajectory) !! Distance (MEP)
|-
| A-B || 0.79 || 0.74
|-
| B-C || 9.96 || 1.72
|-
| A-C || 10.73 || 2.46
|}
[[File:jl_dynamics_p_ex1.png|300px|thumb|left|Figure 4-trajectory]]
[[File:jl_mep_p_ex1.png|300px|thumb|centre|Figure 5-MEP]]

Figure 4 and 5 show the Internuclear Momenta vs Time plot.
{| class="wikitable" border="1"
|+ Difference in Internuclear momenta
! !! Average Momenta (trajectory) !! Average Momenta (MEP)
|-
| A-B || 1.24 || 0.00
|-
| B-C || 2.76 || 0.00
|-
| A-C || 1.86 || 0.00
|}

'''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 !!Total Energy!! Reactive or Unreactive? !!plot of the trajectory !!Description
|-
| -1.25 || -2.5 || -99.018 || Reactive ||[[File:jl_1_ex1.png|300px|thumb|centre]] || It is a reactive trajectory that starts in the region of the reactants and passes through the TS region and towards the products.
|-
| -1.5 || -2.0|| -100.456 || Unreactive ||[[File:jl_2_ex1.png|300px|thumb|centre]] || It is an unreactive trajectory that starts in the region of the reactants and returns near the TS region.
|-
| -1.5 || -2.5|| -98.956 || Reactive ||[[File:jl_3_ex1.png|300px|thumb|centre]] || It is a reactive trajectory that starts in the region of the reactants and passes through the TS region and towards the products.
|-
| -2.5 || -5.0|| -84.956 || Unreactive ||[[File:jl_4_ex1.png|300px|thumb|centre]] || It is an unreactive trajectory that starts in the region of the reactants and crosses the TS region, bonds are formed but then it reverts back to reactants.
|-
| -2.5 || -5.2|| -83.416 || Reactive ||[[File:jl_5_ex1.png|300px|thumb|centre]] || It is a reactive trajectory that starts in the region of the reactants and crosses around the TS region and reverts to the reactants, then moves towards the products.
|}

'''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?'''
<pre>
1. The reactants (or products) are in equilibrium with the activated complex.[1]
2. Quantum-tunneling effects are negligible and the Born-Oppenheimer approximation is invoked.[1]
3. Once the system attains the transition state, with a velocity towards the product configuration,
it will not reenter the initial state region again.[1]

Based on the Transition State Theory, there is no quantum-tunneling. However, according to quantum mechanics,
there is a possibility that particles can still tunnel across the barrier with a low activation energy,
which means that there is a chance that molecules will react by tunneling. Therefore, Transition State
Theory predictions for reaction rate values would be higher than the experimental values.[2]

Reference:
[1] Heterogeneous Catalysis, T. Bligaard, J.K. Nørskov, in Chemical Bonding at Surfaces and Interfaces, 2008
[2] Masel, R. (1996). Principles of Adsorption and Reactions on Solid Surfaces. New York: Wiley.
</pre>

===Exercise 2===
'''Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

F + H2 reaction is exothermic and H + HF reaction is endothermic. H-F bond energy is greater than H-H bond, so energy is released as H-F bond is formed, so it is an exothermic reaction, whereas the formation of H-H bond would be endothermic.
[[File:jl_f_hh_ex2.png|300px|thumb|left|Potential surface of F+HH reaction]]
[[File:jl_h_hf_ex2.png|300px|thumb|centre|Potential surface of H+HF reaction]]

In the F+HH reaction, the products have a lower energy than reactants, so it is exothermic, whereas in H+HF reaction, it is the other way round.

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

The position of the transition state is: rFH=1.81, rHH=0.745. The position is determined by the graph below, the lines are flat, indicating that the vibration of the bonds are minimum.
[[File:jl_rts_ex2.png|400px|thumb|centre|Transition state]]

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

AE of F + H2 reaction = -103.752 -(-100.049)= -3.703

AE of H + HF reaction = -103.752 -(-126.517)= 22.765

'''Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?'''

{| class="wikitable" border="1"
|+ Initial Conditions
|-
| FH distance || 2.30
|-
| HH distance || 0.74
|-
| FH momentum || -2.70
|-
| HH momentum || 0
|}
{| class="wikitable" border="1"
|+ Captures of an animation of a reactive trajectory for the F + H2
|-
| 1 || [[File:jl_animation_ex2.png|200px|thumb|left]]
|-
| 2 || [[File:jl_animation2_ex2.png|200px|thumb|left]]
|-
| 3 || [[File:jl_animation3_ex2.png|200px|thumb|left]]
|}
[[File:jl_momentum_ex2.png|400px|thumb|centre|Internuclear Momenta vs Time graph]]

From the captures of animation, after the formation of the H-F bond, energy is released as it is an exothermic reaction, and this reaction energy is converted to kinetic energy of the H atom, which is moving away from the HF molecule. There should be an increase in temperature if the reaction energy is converted.

'''Calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3:'''

{| class="wikitable" border="1"
|+ Contour Plots
! pHH !! Contour Plot !! pHH !! Contour Plot
|-
| -3 || [[File:jl_phh_n3_ex2.png|400px|thumb|left]] || 3 || [[File:jl_phhp3_ex2.png|400px|thumb|left]]
|-
| -2 || [[File:jl_phh_n2_ex2.png|400px|thumb|left]] || 2 || [[File:jl_phhp2_ex2.png|400px|thumb|left]]
|-
| -1 || [[File:jl_phh_n1_ex2.png|400px|thumb|left]] || 1 || [[File:jl_phhp1_ex2.png|400px|thumb|left]]
|-
| 0 || [[File:jl_phhp0_ex2.png|400px|thumb|left]] || ||The energy put in is to high, the vibration of H-H bond is too great that it cannot form a bond with F atom. It is unreactive.
|}

For the same initial position, increase slightly the momentum pFH = -0.8 and reduce the momentum pHH = 0.1, the HH approaches the F atom gradually.
[[File:jl_p_reduced2_ex2.png|300px|thumb|centre]]

The reaction is reactive by increasing the steps to 1200.

'''The reverse reaction, H + HF'''

{| class="wikitable" border="1"
|+ Initial Conditions
|-
| FH distance || 0.92
|-
| HH distance || 2.3
|-
| FH momentum || -4.2
|-
| HH momentum || -4.6
|}
[[File:jl_initialFH_ex2.png|300px|thumb|left]]

These conditions give a low vibrational motion on on the H-F bond, but with a high kinetic energy in H atom.

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

MRD:jl7816

2018-05-17T16:42:56Z

Jl7816: /* Exercise 2 */

==Molecular reaction dynamics==
===Exercise 1===
'''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.'''

At a minimum, the gradient only has one component, either ∂V/∂rAB= 0 or ∂V/∂rBC= 0.
The curvature at the minima is positive and the secondary derivative is positive.
At a transition structure, the gradients of both components rAB and rBC are equal to 0.
The secondary derivative of one component is positive and the other is negative, this is called a saddle point.

'''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.'''

rts= 0.9076
In the figure 1, it is an Internuclear Distances vs Time plot, the curves of A-B, B-C and A-C are flat, indicating that there is no vibration. The lines of A-B and B-C are overlapped.
[[File:jl_rts.png|500px|thumb|center|Figure 1]]

'''Comment on how the mep and the trajectory you just calculated differ.'''
[[File:jl_dynamics_ex1.png|300px|thumb|left|Figure 2-trajectory]]
[[File:jl_mep_ex1.png|300px|thumb|centre|Figure 3-MEP]]
Figure 2 and 3 show the Internuclear Distances vs Time plot.
{| class="wikitable" border="1"
|+ Difference in Internuclear Distances
! !! Distance (trajectory) !! Distance (MEP)
|-
| A-B || 0.79 || 0.74
|-
| B-C || 9.96 || 1.72
|-
| A-C || 10.73 || 2.46
|}
[[File:jl_dynamics_p_ex1.png|300px|thumb|left|Figure 4-trajectory]]
[[File:jl_mep_p_ex1.png|300px|thumb|centre|Figure 5-MEP]]

Figure 4 and 5 show the Internuclear Momenta vs Time plot.
{| class="wikitable" border="1"
|+ Difference in Internuclear momenta
! !! Average Momenta (trajectory) !! Average Momenta (MEP)
|-
| A-B || 1.24 || 0.00
|-
| B-C || 2.76 || 0.00
|-
| A-C || 1.86 || 0.00
|}

'''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 !!Total Energy!! Reactive or Unreactive? !!plot of the trajectory !!Description
|-
| -1.25 || -2.5 || -99.018 || Reactive ||[[File:jl_1_ex1.png|300px|thumb|centre]] || It is a reactive trajectory that starts in the region of the reactants and passes through the TS region and towards the products.
|-
| -1.5 || -2.0|| -100.456 || Unreactive ||[[File:jl_2_ex1.png|300px|thumb|centre]] || It is an unreactive trajectory that starts in the region of the reactants and returns near the TS region.
|-
| -1.5 || -2.5|| -98.956 || Reactive ||[[File:jl_3_ex1.png|300px|thumb|centre]] || It is a reactive trajectory that starts in the region of the reactants and passes through the TS region and towards the products.
|-
| -2.5 || -5.0|| -84.956 || Unreactive ||[[File:jl_4_ex1.png|300px|thumb|centre]] || It is an unreactive trajectory that starts in the region of the reactants and crosses the TS region, bonds are formed but then it reverts back to reactants.
|-
| -2.5 || -5.2|| -83.416 || Reactive ||[[File:jl_5_ex1.png|300px|thumb|centre]] || It is a reactive trajectory that starts in the region of the reactants and crosses around the TS region and reverts to the reactants, then moves towards the products.
|}

'''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?'''
<pre>
1. The reactants (or products) are in equilibrium with the activated complex.[1]
2. Quantum-tunneling effects are negligible and the Born-Oppenheimer approximation is invoked.[1]
3. Once the system attains the transition state, with a velocity towards the product configuration,
it will not reenter the initial state region again.[1]

Based on the Transition State Theory, there is no quantum-tunneling. However, according to quantum mechanics,
there is a possibility that particles can still tunnel across the barrier with a low activation energy,
which means that there is a chance that molecules will react by tunneling. Therefore, Transition State
Theory predictions for reaction rate values would be higher than the experimental values.[2]

Reference:
[1] Heterogeneous Catalysis, T. Bligaard, J.K. Nørskov, in Chemical Bonding at Surfaces and Interfaces, 2008
[2] Masel, R. (1996). Principles of Adsorption and Reactions on Solid Surfaces. New York: Wiley.
</pre>

===Exercise 2===
'''Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

F + H2 reaction is exothermic and H + HF reaction is endothermic. H-F bond energy is greater than H-H bond, so energy is released as H-F bond is formed, so it is an exothermic reaction, whereas the formation of H-H bond would be endothermic.
[[File:jl_f_hh_ex2.png|300px|thumb|left|Potential surface of F+HH reaction]]
[[File:jl_h_hf_ex2.png|300px|thumb|centre|Potential surface of H+HF reaction]]

In the F+HH reaction, the products have a lower energy than reactants, so it is exothermic, whereas in H+HF reaction, it is the other way round.

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

The position of the transition state is: rFH=1.81, rHH=0.745. The position is determined by the graph below, the lines are flat, indicating that the vibration of the bonds are minimum.
[[File:jl_rts_ex2.png|400px|thumb|centre|Transition state]]

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

AE of F + H2 reaction = -103.752 -(-100.049)= -3.703

AE of H + HF reaction = -103.752 -(-126.517)= 22.765

'''Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?'''

{| class="wikitable" border="1"
|+ Initial Conditions
|-
| FH distance || 2.30
|-
| HH distance || 0.74
|-
| FH momentum || -2.70
|-
| HH momentum || 0
|}
{| class="wikitable" border="1"
|+ Captures of an animation of a reactive trajectory for the F + H2
|-
| 1 || [[File:jl_animation_ex2.png|200px|thumb|left]]
|-
| 2 || [[File:jl_animation2_ex2.png|200px|thumb|left]]
|-
| 3 || [[File:jl_animation3_ex2.png|200px|thumb|left]]
|}
[[File:jl_momentum_ex2.png|400px|thumb|centre|Internuclear Momenta vs Time graph]]

From the captures of animation, after the formation of the H-F bond, energy is released as it is an exothermic reaction, and this reaction energy is converted to kinetic energy of the H atom, which is moving away from the HF molecule. There should be an increase in temperature if the reaction energy is converted.

'''Calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3:'''

{| class="wikitable" border="1"
|+ Contour Plots
! pHH !! Contour Plot !! pHH !! Contour Plot
|-
| -3 || [[File:jl_phh_n3_ex2.png|400px|thumb|left]] || 3 || [[File:jl_phhp3_ex2.png|400px|thumb|left]]
|-
| -2 || [[File:jl_phh_n2_ex2.png|400px|thumb|left]] || 2 || [[File:jl_phhp2_ex2.png|400px|thumb|left]]
|-
| -1 || [[File:jl_phh_n1_ex2.png|400px|thumb|left]] || 1 || [[File:jl_phhp1_ex2.png|400px|thumb|left]]
|-
| 0 || [[File:jl_phhp0_ex2.png|400px|thumb|left]] || ||The energy put in is to high, the vibration of H-H bond is too great that it cannot form a bond with F atom. It is unreactive.
|}

For the same initial position, increase slightly the momentum pFH = -0.8 and reduce the momentum pHH = 0.1, the HH approaches the F atom gradually.
[[File:jl_p_reduced2_ex2.png|300px|thumb|centre]]

The reaction is reactive by increasing the steps to 1200.

'''The reverse reaction, H + HF'''

{| class="wikitable" border="1"
|+ Initial Conditions
|-
| FH distance || 0.92
|-
| HH distance || 2.3
|-
| FH momentum || -4.2
|-
| HH momentum || -4.6
|}

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

MRD:jl7816

2018-05-17T16:32:56Z

Jl7816: /* Exercise 2 */

==Molecular reaction dynamics==
===Exercise 1===
'''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.'''

At a minimum, the gradient only has one component, either ∂V/∂rAB= 0 or ∂V/∂rBC= 0.
The curvature at the minima is positive and the secondary derivative is positive.
At a transition structure, the gradients of both components rAB and rBC are equal to 0.
The secondary derivative of one component is positive and the other is negative, this is called a saddle point.

'''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.'''

rts= 0.9076
In the figure 1, it is an Internuclear Distances vs Time plot, the curves of A-B, B-C and A-C are flat, indicating that there is no vibration. The lines of A-B and B-C are overlapped.
[[File:jl_rts.png|500px|thumb|center|Figure 1]]

'''Comment on how the mep and the trajectory you just calculated differ.'''
[[File:jl_dynamics_ex1.png|300px|thumb|left|Figure 2-trajectory]]
[[File:jl_mep_ex1.png|300px|thumb|centre|Figure 3-MEP]]
Figure 2 and 3 show the Internuclear Distances vs Time plot.
{| class="wikitable" border="1"
|+ Difference in Internuclear Distances
! !! Distance (trajectory) !! Distance (MEP)
|-
| A-B || 0.79 || 0.74
|-
| B-C || 9.96 || 1.72
|-
| A-C || 10.73 || 2.46
|}
[[File:jl_dynamics_p_ex1.png|300px|thumb|left|Figure 4-trajectory]]
[[File:jl_mep_p_ex1.png|300px|thumb|centre|Figure 5-MEP]]

Figure 4 and 5 show the Internuclear Momenta vs Time plot.
{| class="wikitable" border="1"
|+ Difference in Internuclear momenta
! !! Average Momenta (trajectory) !! Average Momenta (MEP)
|-
| A-B || 1.24 || 0.00
|-
| B-C || 2.76 || 0.00
|-
| A-C || 1.86 || 0.00
|}

'''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 !!Total Energy!! Reactive or Unreactive? !!plot of the trajectory !!Description
|-
| -1.25 || -2.5 || -99.018 || Reactive ||[[File:jl_1_ex1.png|300px|thumb|centre]] || It is a reactive trajectory that starts in the region of the reactants and passes through the TS region and towards the products.
|-
| -1.5 || -2.0|| -100.456 || Unreactive ||[[File:jl_2_ex1.png|300px|thumb|centre]] || It is an unreactive trajectory that starts in the region of the reactants and returns near the TS region.
|-
| -1.5 || -2.5|| -98.956 || Reactive ||[[File:jl_3_ex1.png|300px|thumb|centre]] || It is a reactive trajectory that starts in the region of the reactants and passes through the TS region and towards the products.
|-
| -2.5 || -5.0|| -84.956 || Unreactive ||[[File:jl_4_ex1.png|300px|thumb|centre]] || It is an unreactive trajectory that starts in the region of the reactants and crosses the TS region, bonds are formed but then it reverts back to reactants.
|-
| -2.5 || -5.2|| -83.416 || Reactive ||[[File:jl_5_ex1.png|300px|thumb|centre]] || It is a reactive trajectory that starts in the region of the reactants and crosses around the TS region and reverts to the reactants, then moves towards the products.
|}

'''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?'''
<pre>
1. The reactants (or products) are in equilibrium with the activated complex.[1]
2. Quantum-tunneling effects are negligible and the Born-Oppenheimer approximation is invoked.[1]
3. Once the system attains the transition state, with a velocity towards the product configuration,
it will not reenter the initial state region again.[1]

Based on the Transition State Theory, there is no quantum-tunneling. However, according to quantum mechanics,
there is a possibility that particles can still tunnel across the barrier with a low activation energy,
which means that there is a chance that molecules will react by tunneling. Therefore, Transition State
Theory predictions for reaction rate values would be higher than the experimental values.[2]

Reference:
[1] Heterogeneous Catalysis, T. Bligaard, J.K. Nørskov, in Chemical Bonding at Surfaces and Interfaces, 2008
[2] Masel, R. (1996). Principles of Adsorption and Reactions on Solid Surfaces. New York: Wiley.
</pre>

===Exercise 2===
'''Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

F + H2 reaction is exothermic and H + HF reaction is endothermic. H-F bond energy is greater than H-H bond, so energy is released as H-F bond is formed, so it is an exothermic reaction, whereas the formation of H-H bond would be endothermic.
[[File:jl_f_hh_ex2.png|300px|thumb|left|Potential surface of F+HH reaction]]
[[File:jl_h_hf_ex2.png|300px|thumb|centre|Potential surface of H+HF reaction]]

In the F+HH reaction, the products have a lower energy than reactants, so it is exothermic, whereas in H+HF reaction, it is the other way round.

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

The position of the transition state is: rFH=1.81, rHH=0.745. The position is determined by the graph below, the lines are flat, indicating that the vibration of the bonds are minimum.
[[File:jl_rts_ex2.png|400px|thumb|centre|Transition state]]

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

AE of F + H2 reaction = -103.752 -(-100.049)= -3.703

AE of H + HF reaction = -103.752 -(-126.517)= 22.765

'''Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?'''

{| class="wikitable" border="1"
|+ Initial Conditions
|-
| FH distance || 2.30
|-
| HH distance || 0.74
|-
| FH momentum || -2.70
|-
| HH momentum || 0
|}
{| class="wikitable" border="1"
|+ Captures of an animation of a reactive trajectory for the F + H2
|-
| 1 || [[File:jl_animation_ex2.png|200px|thumb|left]]
|-
| 2 || [[File:jl_animation2_ex2.png|200px|thumb|left]]
|-
| 3 || [[File:jl_animation3_ex2.png|200px|thumb|left]]
|}
[[File:jl_momentum_ex2.png|400px|thumb|centre|Internuclear Momenta vs Time graph]]

From the captures of animation, after the formation of the H-F bond, energy is released as it is an exothermic reaction, and this reaction energy is converted to kinetic energy of the H atom, which is moving away from the HF molecule. There should be an increase in temperature if the reaction energy is converted.

'''Calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3:'''

{| class="wikitable" border="1"
|+ Contour Plots
! pHH !! Contour Plot !! pHH !! Contour Plot
|-
| -3 || [[File:jl_phh_n3_ex2.png|400px|thumb|left]] || 3 || [[File:jl_phhp3_ex2.png|400px|thumb|left]]
|-
| -2 || [[File:jl_phh_n2_ex2.png|400px|thumb|left]] || 2 || [[File:jl_phhp2_ex2.png|400px|thumb|left]]
|-
| -1 || [[File:jl_phh_n1_ex2.png|400px|thumb|left]] || 1 || [[File:jl_phhp1_ex2.png|400px|thumb|left]]
|-
| 0 || [[File:jl_phhp0_ex2.png|400px|thumb|left]] || ||The energy put in is to high, the vibration of H-H bond is too great that it cannot form a bond with F atom. It is unreactive.
|}

For the same initial position, increase slightly the momentum pFH = -0.8 and reduce the momentum pHH = 0.1, the HH approaches the F atom gradually.
[[File:jl_p_reduced2_ex2.png|300px|thumb|centre]]

The reaction is reactive by increasing the steps to 1200.

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

MRD:jl7816

2018-05-17T16:28:09Z

Jl7816: /* Exercise 2 */

==Molecular reaction dynamics==
===Exercise 1===
'''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.'''

At a minimum, the gradient only has one component, either ∂V/∂rAB= 0 or ∂V/∂rBC= 0.
The curvature at the minima is positive and the secondary derivative is positive.
At a transition structure, the gradients of both components rAB and rBC are equal to 0.
The secondary derivative of one component is positive and the other is negative, this is called a saddle point.

'''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.'''

rts= 0.9076
In the figure 1, it is an Internuclear Distances vs Time plot, the curves of A-B, B-C and A-C are flat, indicating that there is no vibration. The lines of A-B and B-C are overlapped.
[[File:jl_rts.png|500px|thumb|center|Figure 1]]

'''Comment on how the mep and the trajectory you just calculated differ.'''
[[File:jl_dynamics_ex1.png|300px|thumb|left|Figure 2-trajectory]]
[[File:jl_mep_ex1.png|300px|thumb|centre|Figure 3-MEP]]
Figure 2 and 3 show the Internuclear Distances vs Time plot.
{| class="wikitable" border="1"
|+ Difference in Internuclear Distances
! !! Distance (trajectory) !! Distance (MEP)
|-
| A-B || 0.79 || 0.74
|-
| B-C || 9.96 || 1.72
|-
| A-C || 10.73 || 2.46
|}
[[File:jl_dynamics_p_ex1.png|300px|thumb|left|Figure 4-trajectory]]
[[File:jl_mep_p_ex1.png|300px|thumb|centre|Figure 5-MEP]]

Figure 4 and 5 show the Internuclear Momenta vs Time plot.
{| class="wikitable" border="1"
|+ Difference in Internuclear momenta
! !! Average Momenta (trajectory) !! Average Momenta (MEP)
|-
| A-B || 1.24 || 0.00
|-
| B-C || 2.76 || 0.00
|-
| A-C || 1.86 || 0.00
|}

'''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 !!Total Energy!! Reactive or Unreactive? !!plot of the trajectory !!Description
|-
| -1.25 || -2.5 || -99.018 || Reactive ||[[File:jl_1_ex1.png|300px|thumb|centre]] || It is a reactive trajectory that starts in the region of the reactants and passes through the TS region and towards the products.
|-
| -1.5 || -2.0|| -100.456 || Unreactive ||[[File:jl_2_ex1.png|300px|thumb|centre]] || It is an unreactive trajectory that starts in the region of the reactants and returns near the TS region.
|-
| -1.5 || -2.5|| -98.956 || Reactive ||[[File:jl_3_ex1.png|300px|thumb|centre]] || It is a reactive trajectory that starts in the region of the reactants and passes through the TS region and towards the products.
|-
| -2.5 || -5.0|| -84.956 || Unreactive ||[[File:jl_4_ex1.png|300px|thumb|centre]] || It is an unreactive trajectory that starts in the region of the reactants and crosses the TS region, bonds are formed but then it reverts back to reactants.
|-
| -2.5 || -5.2|| -83.416 || Reactive ||[[File:jl_5_ex1.png|300px|thumb|centre]] || It is a reactive trajectory that starts in the region of the reactants and crosses around the TS region and reverts to the reactants, then moves towards the products.
|}

'''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?'''
<pre>
1. The reactants (or products) are in equilibrium with the activated complex.[1]
2. Quantum-tunneling effects are negligible and the Born-Oppenheimer approximation is invoked.[1]
3. Once the system attains the transition state, with a velocity towards the product configuration,
it will not reenter the initial state region again.[1]

Based on the Transition State Theory, there is no quantum-tunneling. However, according to quantum mechanics,
there is a possibility that particles can still tunnel across the barrier with a low activation energy,
which means that there is a chance that molecules will react by tunneling. Therefore, Transition State
Theory predictions for reaction rate values would be higher than the experimental values.[2]

Reference:
[1] Heterogeneous Catalysis, T. Bligaard, J.K. Nørskov, in Chemical Bonding at Surfaces and Interfaces, 2008
[2] Masel, R. (1996). Principles of Adsorption and Reactions on Solid Surfaces. New York: Wiley.
</pre>

===Exercise 2===
'''Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

F + H2 reaction is exothermic and H + HF reaction is endothermic. H-F bond energy is greater than H-H bond, so energy is released as H-F bond is formed, so it is an exothermic reaction, whereas the formation of H-H bond would be endothermic.
[[File:jl_f_hh_ex2.png|300px|thumb|left|Potential surface of F+HH reaction]]
[[File:jl_h_hf_ex2.png|300px|thumb|centre|Potential surface of H+HF reaction]]

In the F+HH reaction, the products have a lower energy than reactants, so it is exothermic, whereas in H+HF reaction, it is the other way round.

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

The position of the transition state is: rFH=1.81, rHH=0.745. The position is determined by the graph below, the lines are flat, indicating that the vibration of the bonds are minimum.
[[File:jl_rts_ex2.png|400px|thumb|centre|Transition state]]

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

AE of F + H2 reaction = -103.752 -(-100.049)= -3.703

AE of H + HF reaction = -103.752 -(-126.517)= 22.765

'''Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?'''

{| class="wikitable" border="1"
|+ Initial Conditions
|-
| FH distance || 2.30
|-
| HH distance || 0.74
|-
| FH momentum || -2.70
|-
| HH momentum || 0
|}
{| class="wikitable" border="1"
|+ Captures of an animation of a reactive trajectory for the F + H2
|-
| 1 || [[File:jl_animation_ex2.png|200px|thumb|left]]
|-
| 2 || [[File:jl_animation2_ex2.png|200px|thumb|left]]
|-
| 3 || [[File:jl_animation3_ex2.png|200px|thumb|left]]
|}
[[File:jl_momentum_ex2.png|400px|thumb|centre|Internuclear Momenta vs Time graph]]

From the captures of animation, after the formation of the H-F bond, energy is released as it is an exothermic reaction, and this reaction energy is converted to kinetic energy of the H atom, which is moving away from the HF molecule. There should be an increase in temperature if the reaction energy is converted.

'''Calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3:'''

{| class="wikitable" border="1"
|+ Contour Plots
! pHH !! Contour Plot !! pHH !! Contour Plot
|-
| -3 || [[File:jl_phh_n3_ex2.png|400px|thumb|left]] || 3 || [[File:jl_phhp3_ex2.png|400px|thumb|left]]
|-
| -2 || [[File:jl_phh_n2_ex2.png|400px|thumb|left]] || 2 || [[File:jl_phhp2_ex2.png|400px|thumb|left]]
|-
| -1 || [[File:jl_phh_n1_ex2.png|400px|thumb|left]] || 1 || [[File:jl_phhp1_ex2.png|400px|thumb|left]]
|-
| 0 || [[File:jl_phhp0_ex2.png|400px|thumb|left]] || ||The energy put in is to high, the vibration of H-H bond is too great that it cannot form a bond with F atom. It is unreactive.
|}

For the same initial position, increase slightly the momentum pFH = -0.8 and reduce the momentum pHH = 0.1, the HH approaches the F atom gradually.
[[File:jl_p_reduced2_ex2.png|300px|thumb|centre]]

The reaction is reactive by increasing the steps to 1200.

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

File:Jl p reduced2 ex2.png

2018-05-17T16:26:09Z

Jl7816:

MRD:jl7816

2018-05-17T16:25:06Z

2018-05-17T16:20:55Z

Jl7816: /* Exercise 2 */

==Molecular reaction dynamics==
===Exercise 1===
'''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.'''

At a minimum, the gradient only has one component, either ∂V/∂rAB= 0 or ∂V/∂rBC= 0.
The curvature at the minima is positive and the secondary derivative is positive.
At a transition structure, the gradients of both components rAB and rBC are equal to 0.
The secondary derivative of one component is positive and the other is negative, this is called a saddle point.

'''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.'''

rts= 0.9076
In the figure 1, it is an Internuclear Distances vs Time plot, the curves of A-B, B-C and A-C are flat, indicating that there is no vibration. The lines of A-B and B-C are overlapped.
[[File:jl_rts.png|500px|thumb|center|Figure 1]]

'''Comment on how the mep and the trajectory you just calculated differ.'''
[[File:jl_dynamics_ex1.png|300px|thumb|left|Figure 2-trajectory]]
[[File:jl_mep_ex1.png|300px|thumb|centre|Figure 3-MEP]]
Figure 2 and 3 show the Internuclear Distances vs Time plot.
{| class="wikitable" border="1"
|+ Difference in Internuclear Distances
! !! Distance (trajectory) !! Distance (MEP)
|-
| A-B || 0.79 || 0.74
|-
| B-C || 9.96 || 1.72
|-
| A-C || 10.73 || 2.46
|}
[[File:jl_dynamics_p_ex1.png|300px|thumb|left|Figure 4-trajectory]]
[[File:jl_mep_p_ex1.png|300px|thumb|centre|Figure 5-MEP]]

Figure 4 and 5 show the Internuclear Momenta vs Time plot.
{| class="wikitable" border="1"
|+ Difference in Internuclear momenta
! !! Average Momenta (trajectory) !! Average Momenta (MEP)
|-
| A-B || 1.24 || 0.00
|-
| B-C || 2.76 || 0.00
|-
| A-C || 1.86 || 0.00
|}

'''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 !!Total Energy!! Reactive or Unreactive? !!plot of the trajectory !!Description
|-
| -1.25 || -2.5 || -99.018 || Reactive ||[[File:jl_1_ex1.png|300px|thumb|centre]] || It is a reactive trajectory that starts in the region of the reactants and passes through the TS region and towards the products.
|-
| -1.5 || -2.0|| -100.456 || Unreactive ||[[File:jl_2_ex1.png|300px|thumb|centre]] || It is an unreactive trajectory that starts in the region of the reactants and returns near the TS region.
|-
| -1.5 || -2.5|| -98.956 || Reactive ||[[File:jl_3_ex1.png|300px|thumb|centre]] || It is a reactive trajectory that starts in the region of the reactants and passes through the TS region and towards the products.
|-
| -2.5 || -5.0|| -84.956 || Unreactive ||[[File:jl_4_ex1.png|300px|thumb|centre]] || It is an unreactive trajectory that starts in the region of the reactants and crosses the TS region, bonds are formed but then it reverts back to reactants.
|-
| -2.5 || -5.2|| -83.416 || Reactive ||[[File:jl_5_ex1.png|300px|thumb|centre]] || It is a reactive trajectory that starts in the region of the reactants and crosses around the TS region and reverts to the reactants, then moves towards the products.
|}

'''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?'''
<pre>
1. The reactants (or products) are in equilibrium with the activated complex.[1]
2. Quantum-tunneling effects are negligible and the Born-Oppenheimer approximation is invoked.[1]
3. Once the system attains the transition state, with a velocity towards the product configuration,
it will not reenter the initial state region again.[1]

Based on the Transition State Theory, there is no quantum-tunneling. However, according to quantum mechanics,
there is a possibility that particles can still tunnel across the barrier with a low activation energy,
which means that there is a chance that molecules will react by tunneling. Therefore, Transition State
Theory predictions for reaction rate values would be higher than the experimental values.[2]

Reference:
[1] Heterogeneous Catalysis, T. Bligaard, J.K. Nørskov, in Chemical Bonding at Surfaces and Interfaces, 2008
[2] Masel, R. (1996). Principles of Adsorption and Reactions on Solid Surfaces. New York: Wiley.
</pre>

===Exercise 2===
'''Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

F + H2 reaction is exothermic and H + HF reaction is endothermic. H-F bond energy is greater than H-H bond, so energy is released as H-F bond is formed, so it is an exothermic reaction, whereas the formation of H-H bond would be endothermic.
[[File:jl_f_hh_ex2.png|300px|thumb|left|Potential surface of F+HH reaction]]
[[File:jl_h_hf_ex2.png|300px|thumb|centre|Potential surface of H+HF reaction]]

In the F+HH reaction, the products have a lower energy than reactants, so it is exothermic, whereas in H+HF reaction, it is the other way round.

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

The position of the transition state is: rFH=1.81, rHH=0.745. The position is determined by the graph below, the lines are flat, indicating that the vibration of the bonds are minimum.
[[File:jl_rts_ex2.png|400px|thumb|centre|Transition state]]

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

AE of F + H2 reaction = -103.752 -(-100.049)= -3.703

AE of H + HF reaction = -103.752 -(-126.517)= 22.765

'''Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?'''

{| class="wikitable" border="1"
|+ Initial Conditions
|-
| FH distance || 2.30
|-
| HH distance || 0.74
|-
| FH momentum || -2.70
|-
| HH momentum || 0
|}
{| class="wikitable" border="1"
|+ Captures of an animation of a reactive trajectory for the F + H2
|-
| 1 || [[File:jl_animation_ex2.png|200px|thumb|left]]
|-
| 2 || [[File:jl_animation2_ex2.png|200px|thumb|left]]
|-
| 3 || [[File:jl_animation3_ex2.png|200px|thumb|left]]
|}
[[File:jl_momentum_ex2.png|400px|thumb|centre|Internuclear Momenta vs Time graph]]

From the captures of animation, after the formation of the H-F bond, energy is released as it is an exothermic reaction, and this reaction energy is converted to kinetic energy of the H atom, which is moving away from the HF molecule. There should be an increase in temperature if the reaction energy is converted.

'''Calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3:'''

{| class="wikitable" border="1"
|+ Contour Plots
! pHH !! Contour Plot !! pHH !! Contour Plot
|-
| -3 || [[File:jl_phh_n3_ex2.png|400px|thumb|left]] || 3 || [[File:jl_phhp3_ex2.png|400px|thumb|left]]
|-
| -2 || [[File:jl_phh_n2_ex2.png|400px|thumb|left]] || 2 || [[File:jl_phhp2_ex2.png|400px|thumb|left]]
|-
| -1 || [[File:jl_phh_n1_ex2.png|400px|thumb|left]] || 1 || [[File:jl_phhp1_ex2.png|400px|thumb|left]]
|-
| 0 || [[File:jl_phhp0_ex2.png|400px|thumb|left]] || ||The energy put in is to high, the vibration of H-H bond is too great that it cannot form a bond with F atom.
|}

For the same initial position, increase slightly the momentum pFH = -0.8 and reduce the momentum pHH = 0.1, the HH approaches the F atom gradually.
[[File:jl_p_reduced_ex2.png|300px|thumb|centre]]

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

MRD:jl7816

2018-05-17T16:13:23Z

Jl7816: /* Exercise 2 */

==Molecular reaction dynamics==
===Exercise 1===
'''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.'''

At a minimum, the gradient only has one component, either ∂V/∂rAB= 0 or ∂V/∂rBC= 0.
The curvature at the minima is positive and the secondary derivative is positive.
At a transition structure, the gradients of both components rAB and rBC are equal to 0.
The secondary derivative of one component is positive and the other is negative, this is called a saddle point.

'''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.'''

rts= 0.9076
In the figure 1, it is an Internuclear Distances vs Time plot, the curves of A-B, B-C and A-C are flat, indicating that there is no vibration. The lines of A-B and B-C are overlapped.
[[File:jl_rts.png|500px|thumb|center|Figure 1]]

'''Comment on how the mep and the trajectory you just calculated differ.'''
[[File:jl_dynamics_ex1.png|300px|thumb|left|Figure 2-trajectory]]
[[File:jl_mep_ex1.png|300px|thumb|centre|Figure 3-MEP]]
Figure 2 and 3 show the Internuclear Distances vs Time plot.
{| class="wikitable" border="1"
|+ Difference in Internuclear Distances
! !! Distance (trajectory) !! Distance (MEP)
|-
| A-B || 0.79 || 0.74
|-
| B-C || 9.96 || 1.72
|-
| A-C || 10.73 || 2.46
|}
[[File:jl_dynamics_p_ex1.png|300px|thumb|left|Figure 4-trajectory]]
[[File:jl_mep_p_ex1.png|300px|thumb|centre|Figure 5-MEP]]

Figure 4 and 5 show the Internuclear Momenta vs Time plot.
{| class="wikitable" border="1"
|+ Difference in Internuclear momenta
! !! Average Momenta (trajectory) !! Average Momenta (MEP)
|-
| A-B || 1.24 || 0.00
|-
| B-C || 2.76 || 0.00
|-
| A-C || 1.86 || 0.00
|}

'''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 !!Total Energy!! Reactive or Unreactive? !!plot of the trajectory !!Description
|-
| -1.25 || -2.5 || -99.018 || Reactive ||[[File:jl_1_ex1.png|300px|thumb|centre]] || It is a reactive trajectory that starts in the region of the reactants and passes through the TS region and towards the products.
|-
| -1.5 || -2.0|| -100.456 || Unreactive ||[[File:jl_2_ex1.png|300px|thumb|centre]] || It is an unreactive trajectory that starts in the region of the reactants and returns near the TS region.
|-
| -1.5 || -2.5|| -98.956 || Reactive ||[[File:jl_3_ex1.png|300px|thumb|centre]] || It is a reactive trajectory that starts in the region of the reactants and passes through the TS region and towards the products.
|-
| -2.5 || -5.0|| -84.956 || Unreactive ||[[File:jl_4_ex1.png|300px|thumb|centre]] || It is an unreactive trajectory that starts in the region of the reactants and crosses the TS region, bonds are formed but then it reverts back to reactants.
|-
| -2.5 || -5.2|| -83.416 || Reactive ||[[File:jl_5_ex1.png|300px|thumb|centre]] || It is a reactive trajectory that starts in the region of the reactants and crosses around the TS region and reverts to the reactants, then moves towards the products.
|}

'''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?'''
<pre>
1. The reactants (or products) are in equilibrium with the activated complex.[1]
2. Quantum-tunneling effects are negligible and the Born-Oppenheimer approximation is invoked.[1]
3. Once the system attains the transition state, with a velocity towards the product configuration,
it will not reenter the initial state region again.[1]

Based on the Transition State Theory, there is no quantum-tunneling. However, according to quantum mechanics,
there is a possibility that particles can still tunnel across the barrier with a low activation energy,
which means that there is a chance that molecules will react by tunneling. Therefore, Transition State
Theory predictions for reaction rate values would be higher than the experimental values.[2]

Reference:
[1] Heterogeneous Catalysis, T. Bligaard, J.K. Nørskov, in Chemical Bonding at Surfaces and Interfaces, 2008
[2] Masel, R. (1996). Principles of Adsorption and Reactions on Solid Surfaces. New York: Wiley.
</pre>

===Exercise 2===
'''Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

F + H2 reaction is exothermic and H + HF reaction is endothermic. H-F bond energy is greater than H-H bond, so energy is released as H-F bond is formed, so it is an exothermic reaction, whereas the formation of H-H bond would be endothermic.
[[File:jl_f_hh_ex2.png|300px|thumb|left|Potential surface of F+HH reaction]]
[[File:jl_h_hf_ex2.png|300px|thumb|centre|Potential surface of H+HF reaction]]

In the F+HH reaction, the products have a lower energy than reactants, so it is exothermic, whereas in H+HF reaction, it is the other way round.

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

The position of the transition state is: rFH=1.81, rHH=0.745. The position is determined by the graph below, the lines are flat, indicating that the vibration of the bonds are minimum.
[[File:jl_rts_ex2.png|400px|thumb|centre|Transition state]]

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

AE of F + H2 reaction = -103.752 -(-100.049)= -3.703

AE of H + HF reaction = -103.752 -(-126.517)= 22.765

'''Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?'''

{| class="wikitable" border="1"
|+ Initial Conditions
|-
| FH distance || 2.30
|-
| HH distance || 0.74
|-
| FH momentum || -2.70
|-
| HH momentum || 0
|}
{| class="wikitable" border="1"
|+ Captures of an animation of a reactive trajectory for the F + H2
|-
| 1 || [[File:jl_animation_ex2.png|200px|thumb|left]]
|-
| 2 || [[File:jl_animation2_ex2.png|200px|thumb|left]]
|-
| 3 || [[File:jl_animation3_ex2.png|200px|thumb|left]]
|}
[[File:jl_momentum_ex2.png|400px|thumb|centre|Internuclear Momenta vs Time graph]]

From the captures of animation, after the formation of the H-F bond, energy is released as it is an exothermic reaction, and this reaction energy is converted to kinetic energy of the H atom, which is moving away from the HF molecule. There should be an increase in temperature if the reaction energy is converted.

'''Calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3:'''

{| class="wikitable" border="1"
|+ Contour Plots
! pHH !! Contour Plot !! pHH !! Contour Plot
|-
| -3 || [[File:jl_phh_3_ex2.png|400px|thumb|left]] || 3 || [[File:jl_phh3_ex2.png|400px|thumb|left]]
|-
| -2 || [[File:jl_phh_2_ex2.png|400px|thumb|left]] || 2 || [[File:jl_phh2_ex2.png|400px|thumb|left]]
|-
| -1 || [[File:jl_phh_1_ex2.png|400px|thumb|left]] || 1 || [[File:jl_phh1_ex2.png|400px|thumb|left]]
|-
| 0 || [[File:jl_phh0_ex2.png|400px|thumb|left]] || ||The energy put in is to high, the vibration of H-H bond is too great that it cannot form a bond with F atom.
|}

For the same initial position, increase slightly the momentum pFH = -0.8 and reduce the momentum pHH = 0.1, the HH approaches the F atom gradually.
[[File:jl_p_reduced_ex2.png|300px|thumb|centre]]

'''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.'''

2018-05-17T14:53:45Z

Jl7816: /* Exercise 2 */

File:Jl animation3 ex2.png

2018-05-17T14:53:00Z

Jl7816:

File:Jl animation2 ex2.png

2018-05-17T14:52:43Z

Jl7816: