ChemWiki - User contributions [en]

Domimi

2019-05-23T15:34:14Z

Yc7817:

== EXERCISE 1: H + H2 system ==
'''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?'''

On the potential energy surface diagram, the transition point is at the saddle point, where the derivatives are zero. Therefore, it is mathematically defined as: ∂V(ri)/∂ri=0 and H < 0, where H = Vr1r1(r10r20)Vr2r2(r10r20) - V2r1r2(r10r20), which is the determinant of the second partial derivative at the saddle point.

The transition state is defined as the ''maximum'' on the minimum energy path linking reactants and the products. So it can be identified by finding the maximum point on the minimum energy path.

It can be distinguished from a local minimum of the potential energy surface by comparing the second partial derivative determinant as the local minimum will have H > 0.

'''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.'''[[File:0.9078 internuclear distance.png|201x201px|thumb|Figure.1 Plot of internuclear vs time]]

Starting with r(AB) = r(BC) = 0.74 Å, p1 = p2 = 0.0, after varying the internuclear distance, the transition state was observed where the internuclear distance equal to 0.9078 Å. At this distance, no oscillation is observed on the plot shown in Figure.1.

'''3. Comment on how the ''mep'' and the trajectory you just calculated differ.'''
[[File:Mep.png|201x201px|left|thumb|Figure.2 Surface plot by Mep calculation]]
[[File:Dynamics.png|201x201px|right|thumb|Figure.3 Surface plot by Dynamics calculation]]
As shown in Figure.2 and Figure.3, the mep trajectory is smooth and dynamic one is wavy. The potential energy of mep trajectory keeps at the minimum and does not show any oscillation, whereas the dynamic trajectory shows oscillation. This is because mep is the minimum energy path where the momenta are always reset to zero in each time step which corresponds to infinitely slow motion of species.

<nowiki> </nowiki>

'''4. Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.What do you observe?'''
[[File:Distance vs t.png|201x201px|left|thumb|Figure.4 Plot of distance vs time]]
[[File:Momenta vs t.png|201x201px|right|thumb|Figure.5 Plot of momenta vs time]]

'''5. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''
{| class="wikitable"
!p1
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|This trajectory illustrates the reactants go through the transition state and turn into products, which means the reactants have enough energy to overcome the energy barrier to become products. [[File:12525.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.0</nowiki>
|No
|This trajectory shows the reactants do not have enough energy to pass through the transition state into the products, they fall back to reactants after reaching transition state.[[File:15200.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.4</nowiki>
|Yes
|This trajectory illustrates the reactants go through the transition state and turn into products, which means the reactants have enough energy to overcome the energy barrier to become products.[[File:15250.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.8</nowiki>
|No
|This trajectory shows the reactants do not have enough energy to pass through the transition state into the products, they fall back to reactants after reaching transition state.[[File:25500.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.5</nowiki>
|Yes
|This trajectory illustrates the reactants reach the transition state but fall back, however they eventually become products which means the reactants have enough energy to overcome the energy barrier to turn into products.[[File:25520.png]]
|}
A higher momenta of the species does not always produce reactive trajectory.

When the initial momenta is too high, reactants pass through products then fall back towards reactants. Because it has enough energy to reach transition state, it would reform products in the end.

'''6. State what are the main assumptions of Transition State Theory1. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values?'''
# Reactants are in constant equilibrium with the transition state structure.
# The energy of the particles follow a Boltzmann distribution.
# Once reactants become the transition state, the transition state structure does not collapse back to the reactants.
Trial 1,2 and 3 follow the transition state theory as either the reactants have enough energy to turn into products or do not have enough energy to overcome energy barrier where they reach transition state and fall back towards reactants.

Trail 4 and 5 do not match the prediction as reactants pass through products and fall back towards reactants.

== EXERCISE 2: F - H - H system ==
'''1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''
{| class="wikitable"
|F + H2
r(AB)=1.37

r(BC)=0.74

p(AB)=0

p(BC)=-1.5
|[[File:Fh2.png|thumb|Figure.6 Surface plot of F+H2]]
|This reaction is exothermic as the products is lower in energy than the reactants, therefore energy is released during reaction, hence exothermic.
|-
|H + HF
r(AB)=1.54

r(BC)=0.91

p(AB)=0

p(BC)=-10
|[[File:Hhf.png|thumb|Figure.7 Surface plot of H+HF]]
|This reaction is endothermic as the products is higher in energy than the reactants, therefore energy is absorbed during reaction, hence endothermic.
|}
'''2.Locate the approximate position of the transition state.'''
{| class="wikitable"
|F + H2
r(AB)=1.82

r(BC)=0.74

p(AB)=0

p(BC)=0
|[[File:Fh2Exo ts.png|thumb|Figure.8 Contour plot for Transition State of Exothermic Reaction]]
|[[File:Fh2exots d vs t.png|thumb|Figure.9 Internuclear distance vs time plot for Transition State of Exothermic Reaction]]
|-
|H + HF
r(AB)=0.74

r(BC)=1.82

p(AB)=0

p(BC)=0
|[[File:HhfEndo ts.png|thumb|Figure.10 Contour plot for Transition State of Endothermic Reaction]]
|[[File:Hhfendots d vs t.png|thumb|Figure.11 Internuclear distance vs time plot for Transition State of Endothermic Reaction]]
|}
The transition state is where no reaction trajectory is observed and lines on internuclear distance vs time plot are all linear.

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

The activation energy of the two systems are found by varying the distance slightly from the distance for transition state on the plot energy vs time. The activation energy is the energy difference between the reactant and product.
{| class="wikitable"
!System
!Plot
!Activation energy/ kcal mol-1
!Comment
|-
|F + H2
r(AB)=1.7

r(BC)=0.74

p(AB)=0

p(BC)=0
|[[File:Fh2 actE.png|thumb|Figure.12 Energy vs time plot for Exothermic Reaction]]
|27.8
|r(AB) for this system is 1.7 which is less than it of transition state 1.82, this is because this system is an exothermic reaction, early transition state which is closer to reactant is involved.
|-
|H + HF
r(AB)=0.74

r(BC)=1.9

p(AB)=0

p(BC)=0
||[[File:Hhf actE.png|thumb|Figure.13 Energy vs time plot for Endothermic Reaction]]
|0.23
|r(AB) for this system is 1.9 which is greater than it of transition state 1.82, this is because this system is an endothermic reaction, late transition state which is closer to product is involved.
|}
'''4.In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''
{| class="wikitable"
|F + H2
r(AB)=1.82

r(BC)=0.74

p(AB)=-0.5

p(BC)=-3.0
|[[File:Contour conseved E.png|200x200px|thumb|Figure.14 Contour plot]]
|[[File:Energy vs time conserved E.png|200x200px|thumb|Figure.15 Energy vs time plot ]]
|As shown in Figure.15, the kinetic energy of the system goes up with the same amount of energy as the potential energy decreases over time when the reactants pass through products, this confirm that the energy is conserved. The increase in kinetic energy could be confirmed by recording the difference in temperature of the reaction.
|}
'''5.Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''
{| class="wikitable"
!System
!Plot
!System
!Plot
|-
|F + H2
r(AB)=1.82

r(BC)=0.74

p(AB)=-0.5

p(BC)=-3.0
|[[File:Fh2 productive.png|200x200px|thumb|Figure.16 Contour plot of translational E > vibrational E]]
|H + HF
r(AB)=0.74

r(BC)=1.82

p(AB)=-3

p(BC)=-0.5
|[[File:Hhf okproductive.png|200x200px|thumb|Figure.18 Contour plot of vibrational E > translational E]]
|-
|F + H2
r(AB)=1.82

r(BC)=0.74

p(AB)=-3.0

p(BC)=-0.5
|[[File:Fh2 unproductive.png|200x200px|thumb|Figure.17 Contour plot of vibrational E > translational E]]
|H + HF
r(AB)=0.74

r(BC)=1.82

p(AB)=-0.5

p(BC)=-3
|[[File:Hhf unproductive.png|200x200px|thumb|Figure.17 Contour plot of translational E > vibrational E]]
|}
p(AB) and p(BC) for system F+H2 are translational energy and vibrational energy respectively. As shown in table above, for this exothermic reaction, translational energy outweighing vibrational energy give a reactive trajectory, this is result from the fact that exothermic reaction has an early transition state.

However, p(AB) and p(BC) for system H+HF are opposite. In this system, translational energy outweighing vibrational energy does not give a reactive trajectory, this is result from the fact that a late transition state is involved in endothermic reaction.

The results agree with 2Polanyi's rules.

== References ==
1.Atkins, P., de Paula, J., ''Elements of Physical Chemistry,'' 5th ed.; Oxford University Press, 2009.

2. Guo, H., & Liu, K. (2016). Control of chemical reactivity by transition-state and beyond. Chemical Science. https://doi.org/10.1039/c6sc01066k

Domimi

2019-05-23T15:30:04Z

Yc7817:

== EXERCISE 1: H + H2 system ==
'''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?'''

On the potential energy surface diagram, the transition point is at the saddle point, where the derivatives are zero. Therefore, it is mathematically defined as: ∂V(ri)/∂ri=0 and H < 0, where H = Vr1r1(r10r20)Vr2r2(r10r20) - V2r1r2(r10r20), which is the determinant of the second partial derivative at the saddle point.

The transition state is defined as the ''maximum'' on the minimum energy path linking reactants and the products. So it can be identified by finding the maximum point on the minimum energy path.

It can be distinguished from a local minimum of the potential energy surface by comparing the second partial derivative determinant as the local minimum will have H > 0.

'''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.'''[[File:0.9078 internuclear distance.png|201x201px|thumb|Figure.1 Plot of internuclear vs time]]

Starting with r(AB) = r(BC) = 0.74 Å, p1 = p2 = 0.0, after varying the internuclear distance, the transition state was observed where the internuclear distance equal to 0.9078 Å. At this distance, no oscillation is observed on the plot shown in Figure.1.

'''3. Comment on how the ''mep'' and the trajectory you just calculated differ.'''
[[File:Mep.png|201x201px|left|thumb|Figure.2 Surface plot by Mep calculation]]
[[File:Dynamics.png|201x201px|right|thumb|Figure.3 Surface plot by Dynamics calculation]]
As shown in Figure.2 and Figure.3, the mep trajectory is smooth and dynamic one is wavy. The potential energy of mep trajectory keeps at the minimum and does not show any oscillation, whereas the dynamic trajectory shows oscillation. This is because mep is the minimum energy path where the momenta are always reset to zero in each time step which corresponds to infinitely slow motion of species.

<nowiki> </nowiki>

'''4. Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.What do you observe?'''
[[File:Distance vs t.png|201x201px|left|thumb|Figure.4 Plot of distance vs time]]
[[File:Momenta vs t.png|201x201px|right|thumb|Figure.5 Plot of momenta vs time]]

'''5. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''
{| class="wikitable"
!p1
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|This trajectory illustrates the reactants go through the transition state and turn into products, which means the reactants have enough energy to overcome the energy barrier to become products. [[File:12525.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.0</nowiki>
|No
|This trajectory shows the reactants do not have enough energy to pass through the transition state into the products, they fall back to reactants after reaching transition state.[[File:15200.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.4</nowiki>
|Yes
|This trajectory illustrates the reactants go through the transition state and turn into products, which means the reactants have enough energy to overcome the energy barrier to become products.[[File:15250.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.8</nowiki>
|No
|This trajectory shows the reactants do not have enough energy to pass through the transition state into the products, they fall back to reactants after reaching transition state.[[File:25500.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.5</nowiki>
|Yes
|This trajectory illustrates the reactants reach the transition state but fall back, however they eventually become products which means the reactants have enough energy to overcome the energy barrier to turn into products.[[File:25520.png]]
|}
A higher momenta of the species does not always produce reactive trajectory.

When the initial momenta is too high, reactants pass through products then fall back towards reactants. Because it has enough energy to reach transition state, it would reform products in the end.

'''6. 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?'''
# Reactants are in constant equilibrium with the transition state structure.
# The energy of the particles follow a Boltzmann distribution.
# Once reactants become the transition state, the transition state structure does not collapse back to the reactants.
Trial 1,2 and 3 follow the transition state theory as either the reactants have enough energy to turn into products or do not have enough energy to overcome energy barrier where they reach transition state and fall back towards reactants.

Trail 4 and 5 do not match the prediction as reactants pass through products and fall back towards reactants.

== EXERCISE 2: F - H - H system ==
'''1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''
{| class="wikitable"
|F + H2
r(AB)=1.37

r(BC)=0.74

p(AB)=0

p(BC)=-1.5
|[[File:Fh2.png|thumb|Figure.6 Surface plot of F+H2]]
|This reaction is exothermic as the products is lower in energy than the reactants, therefore energy is released during reaction, hence exothermic.
|-
|H + HF
r(AB)=1.54

r(BC)=0.91

p(AB)=0

p(BC)=-10
|[[File:Hhf.png|thumb|Figure.7 Surface plot of H+HF]]
|This reaction is endothermic as the products is higher in energy than the reactants, therefore energy is absorbed during reaction, hence endothermic.
|}
'''2.Locate the approximate position of the transition state.'''
{| class="wikitable"
|F + H2
r(AB)=1.82

r(BC)=0.74

p(AB)=0

p(BC)=0
|[[File:Fh2Exo ts.png|thumb|Figure.8 Contour plot for Transition State of Exothermic Reaction]]
|[[File:Fh2exots d vs t.png|thumb|Figure.9 Internuclear distance vs time plot for Transition State of Exothermic Reaction]]
|-
|H + HF
r(AB)=0.74

r(BC)=1.82

p(AB)=0

p(BC)=0
|[[File:HhfEndo ts.png|thumb|Figure.10 Contour plot for Transition State of Endothermic Reaction]]
|[[File:Hhfendots d vs t.png|thumb|Figure.11 Internuclear distance vs time plot for Transition State of Endothermic Reaction]]
|}
The transition state is where no reaction trajectory is observed and lines on internuclear distance vs time plot are all linear.

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

The activation energy of the two systems are found by varying the distance slightly from the distance for transition state on the plot energy vs time. The activation energy is the energy difference between the reactant and product.
{| class="wikitable"
!System
!Plot
!Activation energy/ kcal mol-1
!Comment
|-
|F + H2
r(AB)=1.7

r(BC)=0.74

p(AB)=0

p(BC)=0
|[[File:Fh2 actE.png|thumb|Figure.12 Energy vs time plot for Exothermic Reaction]]
|27.8
|r(AB) for this system is 1.7 which is less than it of transition state 1.82, this is because this system is an exothermic reaction, early transition state which is closer to reactant is involved.
|-
|H + HF
r(AB)=0.74

r(BC)=1.9

p(AB)=0

p(BC)=0
||[[File:Hhf actE.png|thumb|Figure.13 Energy vs time plot for Endothermic Reaction]]
|0.23
|r(AB) for this system is 1.9 which is greater than it of transition state 1.82, this is because this system is an endothermic reaction, late transition state which is closer to product is involved.
|}
'''4.In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''
{| class="wikitable"
|F + H2
r(AB)=1.82

r(BC)=0.74

p(AB)=-0.5

p(BC)=-3.0
|[[File:Contour conseved E.png|200x200px|thumb|Figure.14 Contour plot]]
|[[File:Energy vs time conserved E.png|200x200px|thumb|Figure.15 Energy vs time plot ]]
|As shown in Figure.15, the kinetic energy of the system goes up with the same amount of energy as the potential energy decreases over time when the reactants pass through products, this confirm that the energy is conserved. The increase in kinetic energy could be confirmed by recording the difference in temperature of the reaction.
|}
'''5.Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''
{| class="wikitable"
!System
!Plot
!System
!Plot
|-
|F + H2
r(AB)=1.82

r(BC)=0.74

p(AB)=-0.5

p(BC)=-3.0
|[[File:Fh2 productive.png|200x200px|thumb|Figure.16 Contour plot of translational E > vibrational E]]
|H + HF
r(AB)=0.74

r(BC)=1.82

p(AB)=-3

p(BC)=-0.5
|[[File:Hhf okproductive.png|200x200px|thumb|Figure.18 Contour plot of vibrational E > translational E]]
|-
|F + H2
r(AB)=1.82

r(BC)=0.74

p(AB)=-3.0

p(BC)=-0.5
|[[File:Fh2 unproductive.png|200x200px|thumb|Figure.17 Contour plot of vibrational E > translational E]]
|H + HF
r(AB)=0.74

r(BC)=1.82

p(AB)=-0.5

p(BC)=-3
|[[File:Hhf unproductive.png|200x200px|thumb|Figure.17 Contour plot of translational E > vibrational E]]
|}
p(AB) and p(BC) for system F+H2 are translational energy and vibrational energy respectively. As shown in table above, for this exothermic reaction, translational energy outweighing vibrational energy give a reactive trajectory, this is result from the fact that exothermic reaction has an early transition state.

However, p(AB) and p(BC) for system H+HF are opposite. In this system, translational energy outweighing vibrational energy does not give a reactive trajectory, this is result from the fact that a late transition state is involved in endothermic reaction.

The results agree with Polanyi's rules.

== References ==
1.Atkins, P., de Paula, J., ''Elements of Physical Chemistry,'' 5th ed.; Oxford University Press, 2009.

Domimi

2019-05-23T15:21:41Z

Yc7817:

== EXERCISE 1: H + H2 system ==
'''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?'''

On the potential energy surface diagram, the transition point is at the saddle point, where the derivatives are zero. Therefore, it is mathematically defined as: ∂V(ri)/∂ri=0 and H < 0, where H = Vr1r1(r10r20)Vr2r2(r10r20) - V2r1r2(r10r20), which is the determinant of the second partial derivative at the saddle point.

The transition state is defined as the ''maximum'' on the minimum energy path linking reactants and the products. So it can be identified by finding the maximum point on the minimum energy path.

It can be distinguished from a local minimum of the potential energy surface by comparing the second partial derivative determinant as the local minimum will have H > 0.

'''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.'''[[File:0.9078 internuclear distance.png|201x201px|thumb|Figure.1 Plot of internuclear vs time]]

Starting with r(AB) = r(BC) = 0.74 Å, p1 = p2 = 0.0, after varying the internuclear distance, the transition state was observed where the internuclear distance equal to 0.9078 Å. At this distance, no oscillation is observed on the plot shown in Figure.1.

'''3. Comment on how the ''mep'' and the trajectory you just calculated differ.'''
[[File:Mep.png|201x201px|left|thumb|Figure.2 Surface plot by Mep calculation]]
[[File:Dynamics.png|201x201px|right|thumb|Figure.3 Surface plot by Dynamics calculation]]
As shown in Figure.2 and Figure.3, the mep trajectory is smooth and dynamic one is wavy. The potential energy of mep trajectory keeps at the minimum and does not show any oscillation, whereas the dynamic trajectory shows oscillation. This is because mep is the minimum energy path where the momenta are always reset to zero in each time step which corresponds to infinitely slow motion of species.

<nowiki> </nowiki>

'''4. Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.What do you observe?'''
[[File:Distance vs t.png|201x201px|left|thumb|Figure.4 Plot of distance vs time]]
[[File:Momenta vs t.png|201x201px|right|thumb|Figure.5 Plot of momenta vs time]]

'''5. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''
{| class="wikitable"
!p1
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|This trajectory illustrates the reactants go through the transition state and turn into products, which means the reactants have enough energy to overcome the energy barrier to become products. [[File:12525.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.0</nowiki>
|No
|This trajectory shows the reactants do not have enough energy to pass through the transition state into the products, they fall back to reactants after reaching transition state.[[File:15200.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.4</nowiki>
|Yes
|This trajectory illustrates the reactants go through the transition state and turn into products, which means the reactants have enough energy to overcome the energy barrier to become products.[[File:15250.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.8</nowiki>
|No
|This trajectory shows the reactants do not have enough energy to pass through the transition state into the products, they fall back to reactants after reaching transition state.[[File:25500.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.5</nowiki>
|Yes
|This trajectory illustrates the reactants reach the transition state but fall back, however they eventually become products which means the reactants have enough energy to overcome the energy barrier to turn into products.[[File:25520.png]]
|}
A higher momenta of the species does not always produce reactive trajectory.

When the initial momenta is too high, reactants pass through products then fall back towards reactants. Because it has enough energy to reach transition state, it would reform products in the end.

'''6. 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?'''
# Reactants are in constant equilibrium with the transition state structure.
# The energy of the particles follow a Boltzmann distribution.
# Once reactants become the transition state, the transition state structure does not collapse back to the reactants.
Trial 1,2 and 3 follow the transition state theory as either the reactants have enough energy to turn into products or do not have enough energy to overcome energy barrier where they reach transition state and fall back towards reactants.

Trail 4 and 5 do not match the prediction as reactants pass through products and fall back towards reactants.

== EXERCISE 2: F - H - H system ==
'''1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''
{| class="wikitable"
|F + H2
r(AB)=1.37

r(BC)=0.74

p(AB)=0

p(BC)=-1.5
|[[File:Fh2.png|thumb|Figure.6 Surface plot of F+H2]]
|This reaction is exothermic as the products is lower in energy than the reactants, therefore energy is released during reaction, hence exothermic.
|-
|H + HF
r(AB)=1.54

r(BC)=0.91

p(AB)=0

p(BC)=-10
|[[File:Hhf.png|thumb|Figure.7 Surface plot of H+HF]]
|This reaction is endothermic as the products is higher in energy than the reactants, therefore energy is absorbed during reaction, hence endothermic.
|}
'''2.Locate the approximate position of the transition state.'''
{| class="wikitable"
|F + H2
r(AB)=1.82

r(BC)=0.74

p(AB)=0

p(BC)=0
|[[File:Fh2Exo ts.png|thumb|Figure.8 Contour plot for Transition State of Exothermic Reaction]]
|[[File:Fh2exots d vs t.png|thumb|Figure.9 Internuclear distance vs time plot for Transition State of Exothermic Reaction]]
|-
|H + HF
r(AB)=0.74

r(BC)=1.82

p(AB)=0

p(BC)=0
|[[File:HhfEndo ts.png|thumb|Figure.10 Contour plot for Transition State of Endothermic Reaction]]
|[[File:Hhfendots d vs t.png|thumb|Figure.11 Internuclear distance vs time plot for Transition State of Endothermic Reaction]]
|}
The transition state is where no reaction trajectory is observed and lines on internuclear distance vs time plot are all linear.

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

The activation energy of the two systems are found by varying the distance slightly from the distance for transition state on the plot energy vs time. The activation energy is the energy difference between the reactant and product.
{| class="wikitable"
!System
!Plot
!Activation energy/ kcal mol-1
!Comment
|-
|F + H2
r(AB)=1.7

r(BC)=0.74

p(AB)=0

p(BC)=0
|[[File:Fh2 actE.png|thumb|Figure.12 Energy vs time plot for Exothermic Reaction]]
|27.8
|r(AB) for this system is 1.7 which is less than it of transition state 1.82, this is because this system is an exothermic reaction, early transition state which is closer to reactant is involved.
|-
|H + HF
r(AB)=0.74

r(BC)=1.9

p(AB)=0

p(BC)=0
||[[File:Hhf actE.png|thumb|Figure.13 Energy vs time plot for Endothermic Reaction]]
|0.23
|r(AB) for this system is 1.9 which is greater than it of transition state 1.82, this is because this system is an endothermic reaction, late transition state which is closer to product is involved.
|}
'''4.In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''
{| class="wikitable"
|F + H2
r(AB)=1.82

r(BC)=0.74

p(AB)=-0.5

p(BC)=-3.0
|[[File:Contour conseved E.png|200x200px|thumb|Figure.14 Contour plot]]
|[[File:Energy vs time conserved E.png|200x200px|thumb|Figure.15 Energy vs time plot ]]
|As shown in Figure.15, the kinetic energy of the system goes up with the same amount of energy as the potential energy decreases over time when the reactants pass through products, this confirm that the energy is conserved. The increase in kinetic energy could be confirmed by recording the difference in temperature of the reaction.
|}
'''5.Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''
{| class="wikitable"
!System
!Plot
!System
!Plot
|-
|F + H2
r(AB)=1.82

r(BC)=0.74

p(AB)=-0.5

p(BC)=-3.0
|[[File:Fh2 productive.png|200x200px|thumb|Figure.16 Contour plot of translational E > vibrational E]]
|H + HF
r(AB)=0.74

r(BC)=1.82

p(AB)=-3

p(BC)=-0.5
|[[File:Hhf okproductive.png|200x200px|thumb|Figure.18 Contour plot of vibrational E > translational E]]
|-
|F + H2
r(AB)=1.82

r(BC)=0.74

p(AB)=-3.0

p(BC)=-0.5
|[[File:Fh2 unproductive.png|200x200px|thumb|Figure.17 Contour plot of vibrational E > translational E]]
|H + HF
r(AB)=0.74

r(BC)=1.82

p(AB)=-0.5

p(BC)=-3
|[[File:Hhf unproductive.png|200x200px|thumb|Figure.17 Contour plot of translational E > vibrational E]]
|}

== References ==
1.Atkins, P., de Paula, J., ''Elements of Physical Chemistry,'' 5th ed.; Oxford University Press, 2009.

File:Hhf okproductive.png

2019-05-23T15:20:43Z

Yc7817:

File:Hhf unproductive.png

2019-05-23T15:13:45Z

Yc7817:

File:Hhf productive.png

2019-05-23T15:13:10Z

Yc7817:

File:Fh2 unproductive.png

2019-05-23T15:05:11Z

Yc7817:

File:Fh2 productive.png

2019-05-23T15:01:49Z

Yc7817:

Domimi

2019-05-23T14:55:23Z

Yc7817:

== EXERCISE 1: H + H2 system ==
'''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?'''

On the potential energy surface diagram, the transition point is at the saddle point, where the derivatives are zero. Therefore, it is mathematically defined as: ∂V(ri)/∂ri=0 and H < 0, where H = Vr1r1(r10r20)Vr2r2(r10r20) - V2r1r2(r10r20), which is the determinant of the second partial derivative at the saddle point.

The transition state is defined as the ''maximum'' on the minimum energy path linking reactants and the products. So it can be identified by finding the maximum point on the minimum energy path.

It can be distinguished from a local minimum of the potential energy surface by comparing the second partial derivative determinant as the local minimum will have H > 0.

'''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.'''[[File:0.9078 internuclear distance.png|201x201px|thumb|Figure.1 Plot of internuclear vs time]]

Starting with r(AB) = r(BC) = 0.74 Å, p1 = p2 = 0.0, after varying the internuclear distance, the transition state was observed where the internuclear distance equal to 0.9078 Å. At this distance, no oscillation is observed on the plot shown in Figure.1.

'''3. Comment on how the ''mep'' and the trajectory you just calculated differ.'''
[[File:Mep.png|201x201px|left|thumb|Figure.2 Surface plot by Mep calculation]]
[[File:Dynamics.png|201x201px|right|thumb|Figure.3 Surface plot by Dynamics calculation]]
As shown in Figure.2 and Figure.3, the mep trajectory is smooth and dynamic one is wavy. The potential energy of mep trajectory keeps at the minimum and does not show any oscillation, whereas the dynamic trajectory shows oscillation. This is because mep is the minimum energy path where the momenta are always reset to zero in each time step which corresponds to infinitely slow motion of species.

<nowiki> </nowiki>

'''4. Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.What do you observe?'''
[[File:Distance vs t.png|201x201px|left|thumb|Figure.4 Plot of distance vs time]]
[[File:Momenta vs t.png|201x201px|right|thumb|Figure.5 Plot of momenta vs time]]

'''5. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''
{| class="wikitable"
!p1
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|This trajectory illustrates the reactants go through the transition state and turn into products, which means the reactants have enough energy to overcome the energy barrier to become products. [[File:12525.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.0</nowiki>
|No
|This trajectory shows the reactants do not have enough energy to pass through the transition state into the products, they fall back to reactants after reaching transition state.[[File:15200.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.4</nowiki>
|Yes
|This trajectory illustrates the reactants go through the transition state and turn into products, which means the reactants have enough energy to overcome the energy barrier to become products.[[File:15250.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.8</nowiki>
|No
|This trajectory shows the reactants do not have enough energy to pass through the transition state into the products, they fall back to reactants after reaching transition state.[[File:25500.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.5</nowiki>
|Yes
|This trajectory illustrates the reactants reach the transition state but fall back, however they eventually become products which means the reactants have enough energy to overcome the energy barrier to turn into products.[[File:25520.png]]
|}
A higher momenta of the species does not always produce reactive trajectory.

When the initial momenta is too high, reactants pass through products then fall back towards reactants. Because it has enough energy to reach transition state, it would reform products in the end.

'''6. 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?'''
# Reactants are in constant equilibrium with the transition state structure.
# The energy of the particles follow a Boltzmann distribution.
# Once reactants become the transition state, the transition state structure does not collapse back to the reactants.
Trial 1,2 and 3 follow the transition state theory as either the reactants have enough energy to turn into products or do not have enough energy to overcome energy barrier where they reach transition state and fall back towards reactants.

Trail 4 and 5 do not match the prediction as reactants pass through products and fall back towards reactants.

== EXERCISE 2: F - H - H system ==
'''1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''
{| class="wikitable"
|F + H2
r(AB)=1.37

r(BC)=0.74

p(AB)=0

p(BC)=-1.5
|[[File:Fh2.png|thumb|Figure.6 Surface plot of F+H2]]
|This reaction is exothermic as the products is lower in energy than the reactants, therefore energy is released during reaction, hence exothermic.
|-
|H + HF
r(AB)=1.54

r(BC)=0.91

p(AB)=0

p(BC)=-10
|[[File:Hhf.png|thumb|Figure.7 Surface plot of H+HF]]
|This reaction is endothermic as the products is higher in energy than the reactants, therefore energy is absorbed during reaction, hence endothermic.
|}
'''2.Locate the approximate position of the transition state.'''
{| class="wikitable"
|F + H2
r(AB)=1.82

r(BC)=0.74

p(AB)=0

p(BC)=0
|[[File:Fh2Exo ts.png|thumb|Figure.8 Contour plot for Transition State of Exothermic Reaction]]
|[[File:Fh2exots d vs t.png|thumb|Figure.9 Internuclear distance vs time plot for Transition State of Exothermic Reaction]]
|-
|H + HF
r(AB)=0.74

r(BC)=1.82

p(AB)=0

p(BC)=0
|[[File:HhfEndo ts.png|thumb|Figure.10 Contour plot for Transition State of Endothermic Reaction]]
|[[File:Hhfendots d vs t.png|thumb|Figure.11 Internuclear distance vs time plot for Transition State of Endothermic Reaction]]
|}
The transition state is where no reaction trajectory is observed and lines on internuclear distance vs time plot are all linear.

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

The activation energy of the two systems are found by varying the distance slightly from the distance for transition state on the plot energy vs time. The activation energy is the energy difference between the reactant and product.
{| class="wikitable"
!System
!Plot
!Activation energy/ kcal mol-1
!Comment
|-
|F + H2
r(AB)=1.7

r(BC)=0.74

p(AB)=0

p(BC)=0
|[[File:Fh2 actE.png|thumb|Figure.12 Energy vs time plot for Exothermic Reaction]]
|27.8
|r(AB) for this system is 1.7 which is less than it of transition state 1.82, this is because this system is an exothermic reaction, early transition state which is closer to reactant is involved.
|-
|H + HF
r(AB)=0.74

r(BC)=1.9

p(AB)=0

p(BC)=0
||[[File:Hhf actE.png|thumb|Figure.13 Energy vs time plot for Endothermic Reaction]]
|0.23
|r(AB) for this system is 1.9 which is greater than it of transition state 1.82, this is because this system is an endothermic reaction, late transition state which is closer to product is involved.
|}
'''4.In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''
{| class="wikitable"
|F + H2
r(AB)=1.82

r(BC)=0.74

p(AB)=-0.5

p(BC)=-3.0
|[[File:Contour conseved E.png|200x200px|thumb|Figure.14 Contour plot]]
|[[File:Energy vs time conserved E.png|200x200px|thumb|Figure.15 Energy vs time plot ]]
|As shown in Figure.15, the kinetic energy of the system goes up with the same amount of energy as the potential energy decreases over time when the reactants pass through products, this confirm that the energy is conserved. The increase in kinetic energy could be confirmed by recording the difference in temperature of the reaction.
|}
'''5.Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state.'''

== References ==
1.Atkins, P., de Paula, J., ''Elements of Physical Chemistry,'' 5th ed.; Oxford University Press, 2009.

Domimi

2019-05-23T14:19:29Z

Yc7817:

== EXERCISE 1: H + H2 system ==
'''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?'''

On the potential energy surface diagram, the transition point is at the saddle point, where the derivatives are zero. Therefore, it is mathematically defined as: ∂V(ri)/∂ri=0 and H < 0, where H = Vr1r1(r10r20)Vr2r2(r10r20) - V2r1r2(r10r20), which is the determinant of the second partial derivative at the saddle point.

The transition state is defined as the ''maximum'' on the minimum energy path linking reactants and the products. So it can be identified by finding the maximum point on the minimum energy path.

It can be distinguished from a local minimum of the potential energy surface by comparing the second partial derivative determinant as the local minimum will have H > 0.

'''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.'''[[File:0.9078 internuclear distance.png|201x201px|thumb|Figure.1 Plot of internuclear vs time]]

Starting with r(AB) = r(BC) = 0.74 Å, p1 = p2 = 0.0, after varying the internuclear distance, the transition state was observed where the internuclear distance equal to 0.9078 Å. At this distance, no oscillation is observed on the plot shown in Figure.1.

'''3. Comment on how the ''mep'' and the trajectory you just calculated differ.'''
[[File:Mep.png|201x201px|left|thumb|Figure.2 Surface plot by Mep calculation]]
[[File:Dynamics.png|201x201px|right|thumb|Figure.3 Surface plot by Dynamics calculation]]
As shown in Figure.2 and Figure.3, the mep trajectory is smooth and dynamic one is wavy. The potential energy of mep trajectory keeps at the minimum and does not show any oscillation, whereas the dynamic trajectory shows oscillation. This is because mep is the minimum energy path where the momenta are always reset to zero in each time step which corresponds to infinitely slow motion of species.

<nowiki> </nowiki>

'''4. Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.What do you observe?'''
[[File:Distance vs t.png|201x201px|left|thumb|Figure.4 Plot of distance vs time]]
[[File:Momenta vs t.png|201x201px|right|thumb|Figure.5 Plot of momenta vs time]]

'''5. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''
{| class="wikitable"
!p1
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|This trajectory illustrates the reactants go through the transition state and turn into products, which means the reactants have enough energy to overcome the energy barrier to become products. [[File:12525.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.0</nowiki>
|No
|This trajectory shows the reactants do not have enough energy to pass through the transition state into the products, they fall back to reactants after reaching transition state.[[File:15200.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.4</nowiki>
|Yes
|This trajectory illustrates the reactants go through the transition state and turn into products, which means the reactants have enough energy to overcome the energy barrier to become products.[[File:15250.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.8</nowiki>
|No
|This trajectory shows the reactants do not have enough energy to pass through the transition state into the products, they fall back to reactants after reaching transition state.[[File:25500.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.5</nowiki>
|Yes
|This trajectory illustrates the reactants reach the transition state but fall back, however they eventually become products which means the reactants have enough energy to overcome the energy barrier to turn into products.[[File:25520.png]]
|}
A higher momenta of the species does not always produce reactive trajectory.

When the initial momenta is too high, reactants pass through products then fall back towards reactants. Because it has enough energy to reach transition state, it would reform products in the end.

'''6. 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?'''
# Reactants are in constant equilibrium with the transition state structure.
# The energy of the particles follow a Boltzmann distribution.
# Once reactants become the transition state, the transition state structure does not collapse back to the reactants.
Trial 1,2 and 3 follow the transition state theory as either the reactants have enough energy to turn into products or do not have enough energy to overcome energy barrier where they reach transition state and fall back towards reactants.

Trail 4 and 5 do not match the prediction as reactants pass through products and fall back towards reactants.

== EXERCISE 2: F - H - H system ==
'''1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''
{| class="wikitable"
|F + H2
r(AB)=1.37

r(BC)=0.74

p(AB)=0

p(BC)=-1.5
|[[File:Fh2.png|thumb|Figure.6 Surface plot of F+H2]]
|This reaction is exothermic as the products is lower in energy than the reactants, therefore energy is released during reaction, hence exothermic.
|-
|H + HF
r(AB)=1.54

r(BC)=0.91

p(AB)=0

p(BC)=-10
|[[File:Hhf.png|thumb|Figure.7 Surface plot of H+HF]]
|This reaction is endothermic as the products is higher in energy than the reactants, therefore energy is absorbed during reaction, hence endothermic.
|}
'''2.Locate the approximate position of the transition state.'''
{| class="wikitable"
|F + H2
r(AB)=1.82

r(BC)=0.74

p(AB)=0

p(BC)=0
|[[File:Fh2Exo ts.png|thumb|Figure.8 Contour plot for Transition State of Exothermic Reaction]]
|[[File:Fh2exots d vs t.png|thumb|Figure.9 Internuclear distance vs time plot for Transition State of Exothermic Reaction]]
|-
|H + HF
r(AB)=0.74

r(BC)=1.82

p(AB)=0

p(BC)=0
|[[File:HhfEndo ts.png|thumb|Figure.10 Contour plot for Transition State of Endothermic Reaction]]
|[[File:Hhfendots d vs t.png|thumb|Figure.11 Internuclear distance vs time plot for Transition State of Endothermic Reaction]]
|}
The transition state is where no reaction trajectory is observed and lines on internuclear distance vs time plot are all linear.

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

The activation energy of the two systems are found by varying the distance slightly from the distance for transition state on the plot energy vs time. The activation energy is the energy difference between the reactant and product.
{| class="wikitable"
!System
!Plot
!Activation energy/ kcal mol-1
!Comment
|-
|F + H2
r(AB)=1.7

r(BC)=0.74

p(AB)=0

p(BC)=0
|[[File:Fh2 actE.png|thumb|Figure.12 Energy vs time plot for Exothermic Reaction]]
|27.8
|r(AB) for this system is 1.7 which is less than it of transition state 1.82, this is because this system is an exothermic reaction, early transition state which is closer to reactant is involved.
|-
|H + HF
r(AB)=0.74

r(BC)=1.9

p(AB)=0

p(BC)=0
||[[File:Hhf actE.png|thumb|Figure.13 Energy vs time plot for Endothermic Reaction]]
|0.23
|r(AB) for this system is 1.9 which is greater than it of transition state 1.82, this is because this system is an endothermic reaction, late transition state which is closer to product is involved.
|}
'''4.In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''
{| class="wikitable"
|F + H2r(AB)=1.82

r(BC)=0.74

p(AB)=-0.5

p(BC)=-3.0
|[[File:Contour conseved E.png|200x200px|thumb|Figure.14 Contour plot]]
|[[File:Energy vs time conserved E.png|200x200px|thumb|Figure.15 Energy vs time plot ]]
|
|}

== References ==
1.Atkins, P., de Paula, J., ''Elements of Physical Chemistry,'' 5th ed.; Oxford University Press, 2009.

Domimi

2019-05-23T14:17:57Z

Yc7817:

== EXERCISE 1: H + H2 system ==
'''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?'''

On the potential energy surface diagram, the transition point is at the saddle point, where the derivatives are zero. Therefore, it is mathematically defined as: ∂V(ri)/∂ri=0 and H < 0, where H = Vr1r1(r10r20)Vr2r2(r10r20) - V2r1r2(r10r20), which is the determinant of the second partial derivative at the saddle point.

The transition state is defined as the ''maximum'' on the minimum energy path linking reactants and the products. So it can be identified by finding the maximum point on the minimum energy path.

It can be distinguished from a local minimum of the potential energy surface by comparing the second partial derivative determinant as the local minimum will have H > 0.

'''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.'''[[File:0.9078 internuclear distance.png|201x201px|thumb|Figure.1 Plot of internuclear vs time]]

Starting with r(AB) = r(BC) = 0.74 Å, p1 = p2 = 0.0, after varying the internuclear distance, the transition state was observed where the internuclear distance equal to 0.9078 Å. At this distance, no oscillation is observed on the plot shown in Figure.1.

'''3. Comment on how the ''mep'' and the trajectory you just calculated differ.'''
[[File:Mep.png|201x201px|left|thumb|Figure.2 Surface plot by Mep calculation]]
[[File:Dynamics.png|201x201px|right|thumb|Figure.3 Surface plot by Dynamics calculation]]
As shown in Figure.2 and Figure.3, the mep trajectory is smooth and dynamic one is wavy. The potential energy of mep trajectory keeps at the minimum and does not show any oscillation, whereas the dynamic trajectory shows oscillation. This is because mep is the minimum energy path where the momenta are always reset to zero in each time step which corresponds to infinitely slow motion of species.

<nowiki> </nowiki>

'''4. Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.What do you observe?'''
[[File:Distance vs t.png|201x201px|left|thumb|Figure.4 Plot of distance vs time]]
[[File:Momenta vs t.png|201x201px|right|thumb|Figure.5 Plot of momenta vs time]]

'''5. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''
{| class="wikitable"
!p1
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|This trajectory illustrates the reactants go through the transition state and turn into products, which means the reactants have enough energy to overcome the energy barrier to become products. [[File:12525.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.0</nowiki>
|No
|This trajectory shows the reactants do not have enough energy to pass through the transition state into the products, they fall back to reactants after reaching transition state.[[File:15200.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.4</nowiki>
|Yes
|This trajectory illustrates the reactants go through the transition state and turn into products, which means the reactants have enough energy to overcome the energy barrier to become products.[[File:15250.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.8</nowiki>
|No
|This trajectory shows the reactants do not have enough energy to pass through the transition state into the products, they fall back to reactants after reaching transition state.[[File:25500.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.5</nowiki>
|Yes
|This trajectory illustrates the reactants reach the transition state but fall back, however they eventually become products which means the reactants have enough energy to overcome the energy barrier to turn into products.[[File:25520.png]]
|}
A higher momenta of the species does not always produce reactive trajectory.

When the initial momenta is too high, reactants pass through products then fall back towards reactants. Because it has enough energy to reach transition state, it would reform products in the end.

'''6. 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?'''
# Reactants are in constant equilibrium with the transition state structure.
# The energy of the particles follow a Boltzmann distribution.
# Once reactants become the transition state, the transition state structure does not collapse back to the reactants.
Trial 1,2 and 3 follow the transition state theory as either the reactants have enough energy to turn into products or do not have enough energy to overcome energy barrier where they reach transition state and fall back towards reactants.

Trail 4 and 5 do not match the prediction as reactants pass through products and fall back towards reactants.

== EXERCISE 2: F - H - H system ==
'''1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''
{| class="wikitable"
|F + H2
r(AB)=1.37

r(BC)=0.74

p(AB)=0

p(BC)=-1.5
|[[File:Fh2.png|thumb|Figure.6 Surface plot of F+H2]]
|This reaction is exothermic as the products is lower in energy than the reactants, therefore energy is released during reaction, hence exothermic.
|-
|H + HF
r(AB)=1.54

r(BC)=0.91

p(AB)=0

p(BC)=-10
|[[File:Hhf.png|thumb|Figure.7 Surface plot of H+HF]]
|This reaction is endothermic as the products is higher in energy than the reactants, therefore energy is absorbed during reaction, hence endothermic.
|}
'''2.Locate the approximate position of the transition state.'''
{| class="wikitable"
|F + H2
r(AB)=1.82

r(BC)=0.74

p(AB)=0

p(BC)=0
|[[File:Fh2Exo ts.png|thumb|Figure.8 Contour plot for Transition State of Exothermic Reaction]]
|[[File:Fh2exots d vs t.png|thumb|Figure.9 Internuclear distance vs time plot for Transition State of Exothermic Reaction]]
|-
|H + HF
r(AB)=0.74

r(BC)=1.82

p(AB)=0

p(BC)=0
|[[File:HhfEndo ts.png|thumb|Figure.10 Contour plot for Transition State of Endothermic Reaction]]
|[[File:Hhfendots d vs t.png|thumb|Figure.11 Internuclear distance vs time plot for Transition State of Endothermic Reaction]]
|}
The transition state is where no reaction trajectory is observed and lines on internuclear distance vs time plot are all linear.

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

The activation energy of the two systems are found by varying the distance slightly from the distance for transition state on the plot energy vs time. The activation energy is the energy difference between the reactant and product.
{| class="wikitable"
!System
!Plot
!Activation energy/ kcal mol-1
!Comment
|-
|F + H2
r(AB)=1.7

r(BC)=0.74

p(AB)=0

p(BC)=0
|[[File:Fh2 actE.png|thumb|Figure.12 Energy vs time plot for Exothermic Reaction]]
|27.8
|r(AB) for this system is 1.7 which is less than it of transition state 1.82, this is because this system is an exothermic reaction, early transition state which is closer to reactant is involved.
|-
|H + HF
r(AB)=0.74

r(BC)=1.9

p(AB)=0

p(BC)=0
||[[File:Hhf actE.png|thumb|Figure.13 Energy vs time plot for Endothermic Reaction]]
|0.23
|r(AB) for this system is 1.9 which is greater than it of transition state 1.82, this is because this system is an endothermic reaction, late transition state which is closer to product is involved.
|}
'''4.In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''
{| class="wikitable"
|F + H2r(AB)=1.82

r(BC)=0.74

p(AB)=-0.5

p(BC)=-3.0
|[[File:Contour conseved E.png|thumb|Figure.14 Contour plot]]
|[[File:Energy vs time conserved E.png|thumb|Figure.15 Energy vs time plot ]]
|
|}

== References ==
1.Atkins, P., de Paula, J., ''Elements of Physical Chemistry,'' 5th ed.; Oxford University Press, 2009.

File:Energy vs time conserved E.png

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== EXERCISE 1: H + H2 system ==
'''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?'''

On the potential energy surface diagram, the transition point is at the saddle point, where the derivatives are zero. Therefore, it is mathematically defined as: ∂V(ri)/∂ri=0 and H < 0, where H = Vr1r1(r10r20)Vr2r2(r10r20) - V2r1r2(r10r20), which is the determinant of the second partial derivative at the saddle point.

The transition state is defined as the ''maximum'' on the minimum energy path linking reactants and the products. So it can be identified by finding the maximum point on the minimum energy path.

It can be distinguished from a local minimum of the potential energy surface by comparing the second partial derivative determinant as the local minimum will have H > 0.

'''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.'''[[File:0.9078 internuclear distance.png|201x201px|thumb|Figure.1 Plot of internuclear vs time]]

Starting with r(AB) = r(BC) = 0.74 Å, p1 = p2 = 0.0, after varying the internuclear distance, the transition state was observed where the internuclear distance equal to 0.9078 Å. At this distance, no oscillation is observed on the plot shown in Figure.1.

'''3. Comment on how the ''mep'' and the trajectory you just calculated differ.'''
[[File:Mep.png|201x201px|left|thumb|Figure.2 Surface plot by Mep calculation]]
[[File:Dynamics.png|201x201px|right|thumb|Figure.3 Surface plot by Dynamics calculation]]
As shown in Figure.2 and Figure.3, the mep trajectory is smooth and dynamic one is wavy. The potential energy of mep trajectory keeps at the minimum and does not show any oscillation, whereas the dynamic trajectory shows oscillation. This is because mep is the minimum energy path where the momenta are always reset to zero in each time step which corresponds to infinitely slow motion of species.

<nowiki> </nowiki>

'''4. Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.What do you observe?'''
[[File:Distance vs t.png|201x201px|left|thumb|Figure.4 Plot of distance vs time]]
[[File:Momenta vs t.png|201x201px|right|thumb|Figure.5 Plot of momenta vs time]]

'''5. Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''
{| class="wikitable"
!p1
!p2
!Etot
!Reactive?
!Description of the dynamics and Illustration of the trajectory
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.6</nowiki>
|Yes
|This trajectory illustrates the reactants go through the transition state and turn into products, which means the reactants have enough energy to overcome the energy barrier to become products. [[File:12525.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.0</nowiki>
|No
|This trajectory shows the reactants do not have enough energy to pass through the transition state into the products, they fall back to reactants after reaching transition state.[[File:15200.png]]
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.4</nowiki>
|Yes
|This trajectory illustrates the reactants go through the transition state and turn into products, which means the reactants have enough energy to overcome the energy barrier to become products.[[File:15250.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.8</nowiki>
|No
|This trajectory shows the reactants do not have enough energy to pass through the transition state into the products, they fall back to reactants after reaching transition state.[[File:25500.png]]
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.5</nowiki>
|Yes
|This trajectory illustrates the reactants reach the transition state but fall back, however they eventually become products which means the reactants have enough energy to overcome the energy barrier to turn into products.[[File:25520.png]]
|}
A higher momenta of the species does not always produce reactive trajectory.

When the initial momenta is too high, reactants pass through products then fall back towards reactants. Because it has enough energy to reach transition state, it would reform products in the end.

'''6. 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?'''
# Reactants are in constant equilibrium with the transition state structure.
# The energy of the particles follow a Boltzmann distribution.
# Once reactants become the transition state, the transition state structure does not collapse back to the reactants.
Trial 1,2 and 3 follow the transition state theory as either the reactants have enough energy to turn into products or do not have enough energy to overcome energy barrier where they reach transition state and fall back towards reactants.

Trail 4 and 5 do not match the prediction as reactants pass through products and fall back towards reactants.

== EXERCISE 2: F - H - H system ==
'''1. By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''
{| class="wikitable"
|F + H2
r(AB)=1.37

r(BC)=0.74

p(AB)=0

p(BC)=-1.5
|[[File:Fh2.png|thumb|Figure.6 Surface plot of F+H2]]
|This reaction is exothermic as the products is lower in energy than the reactants, therefore energy is released during reaction, hence exothermic.
|-
|H + HF
r(AB)=1.54

r(BC)=0.91

p(AB)=0

p(BC)=-10
|[[File:Hhf.png|thumb|Figure.7 Surface plot of H+HF]]
|This reaction is endothermic as the products is higher in energy than the reactants, therefore energy is absorbed during reaction, hence endothermic.
|}
'''2.Locate the approximate position of the transition state.'''
{| class="wikitable"
|F + H2
r(AB)=1.82

r(BC)=0.74

p(AB)=0

p(BC)=0
|[[File:Fh2Exo ts.png|thumb|Figure.8 Contour plot for Transition State of Exothermic Reaction]]
|[[File:Fh2exots d vs t.png|thumb|Figure.9 Internuclear distance vs time plot for Transition State of Exothermic Reaction]]
|-
|H + HF
r(AB)=0.74

r(BC)=1.82

p(AB)=0

p(BC)=0
|[[File:HhfEndo ts.png|thumb|Figure.10 Contour plot for Transition State of Endothermic Reaction]]
|[[File:Hhfendots d vs t.png|thumb|Figure.11 Internuclear distance vs time plot for Transition State of Endothermic Reaction]]
|}
The transition state is where no reaction trajectory is observed and lines on internuclear distance vs time plot are all linear.

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

The activation energy of the two systems are found by varying the distance slightly from the distance for transition state on the plot energy vs time. The activation energy is the energy difference between the reactant and product.
{| class="wikitable"
!System
!Plot
!Activation energy/ kcal mol-1
|-
|F + H2
r(AB)=1.7

r(BC)=0.74

p(AB)=0

p(BC)=0
|[[File:Fh2 actE.png|thumb|Figure.12 Energy vs time plot for Exothermic Reaction]]
|27.8
|-
|H + HF
r(AB)=0.74

r(BC)=1.9

p(AB)=0

p(BC)=0
||[[File:Hhf actE.png|thumb|Figure.13 Energy vs time plot for Endothermic Reaction]]
|0.23
|}

== References ==
1.Atkins, P., de Paula, J., ''Elements of Physical Chemistry,'' 5th ed.; Oxford University Press, 2009.

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Yc7817: /* EXERCISE 2: F - H - H system */

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Yc7817: /* EXERCISE 2: F - H - H system */

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