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JN1316Lab2

2018-05-11T16:59:18Z

Jn1316:

All distances are in Angstroms and energies in kCal/mol

== Exercise 1: H + H 2 ==

 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. 

The potential energy along a minima line has dV/dr = 0 and because it is a maxima d^2V/dr^2 is <0. A transition state also has dV/dr = 0 however d^2V/dr^2 is in contrast >0.

 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. 

running an MEP calculation with both bond lengths set arbitrarily to 1 and exporting the data finds the saddle point to be where both bond lengths are 0.907741. The literature [1] is in agreement with this value.

[[File:IND-T.PNG]]

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

MEP plots correspond to infinitely slow motion with the velocity resetting to 0 in each step. This is not the case in dynamic plots. This explains why in MEP the molecular bond length does not vibrate and the free hydrogen atom does not escape to infinity but rather it's distance from the other two atoms plateaus.

[[File:JN idk 6.PNG]]
[[File:JN MEPcalc.PNG]]

The momentum - time graph below shows that one hydrogen leaves as a free atom and the other 2 as a vibrating molecule with translational and vibrational energy modes.
[[File:JN moment 1.PNG]]

If it were switched so that r1 = rts and r2 = rts + 0.01 the central hydrogen would bond with the the hydrogen to its right as opposed to its left and the free hydrogen will be switched. This is because asides from the slight displacement in distance the system is symmetrical.

 For the initial positions r1 = 0.74 Angstroms and r2 = 2.0 Angstroms, run trajectories with the following momenta combination: 

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy kCal/mol !! Reactive? !! Picture !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:JN TS 1.PNG]] || A successful reaction with a vibrating reactant transferring vibrational energy to a vibrating product.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:JN TS 2.PNG]] || There is not sufficient total energy for the reactants to overcome the activation energy.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:JN TS 3.PNG]] || This simple reactive trajectory passes once simply over the transition state.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:JN TS 4.PNG]] || This system has sufficient energy to overcome the activation energy. it does so but then 'bounces' of a steep wall of the potential well back to the reactants.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:JN TS 5.PNG]] || The transition state theory assumes the potential barrier cannot be recrossed. This example violates that assumption as the potential barrier is crossed 3 times
|}

 State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? 
The main assumption to the TST is that once you have left the maxima that is the transition state you cannot go back to it or past it. This assumption fails as we have seen in the above trajectories some of the minimum energy paths reach the transition state and then return backwards. The TST also neglects the effects of quantum mechanics. Particularly for light particles such as hydrogen the non-zero probability of tunneling through a classically forbidden potential barrier with quantum potential is significant. Transition state theory will accurately predict reactions that pass only once over the transition state however it could incorrectly predict a path will be reactive or vice versa.

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

 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 as H-F bond is stronger than H-H bond. As seen in the picture below the energy of the products is less than that of reactants.

[[File:F and H2 JN.PNG]]

H + HF: Reaction is endothermic as H-H bond is weaker than H-F bond. As seen in the picture below the energy of the products is greater than that of reactants. This can be found by inputting the output values from the previous computation

[[File:H and HF JN.PNG]]

 Locate the approximate position of the transition state. 

Hammond's postulate states that the energy of the TS can be approximated as being that of the state nearest to it, because H2 + F is exothermic that means it is nearer the reactants therefore by inputting the distance values of the reactants and making the momenta = 0 the transition state can be found. AB = 0.74 and BC = 1.815. The energy of this state is -103.743.
Therefore the activation energy of:

- H2 + F energy is -103.904 therefore activation energy is: 0.161

- HF + H energy is -137.063 therefore activation energy is: 33.32

[[File:JN TS NEW.PNG]]

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

The momenta of the system increases after the reaction compared to before it. This can be seen in the animation of the interaction. Therefore the energy released in this reaction is converted to kinetic energy as the relationship between momentum and energy is Ek = p^2 / 2m. By measuring the temperature before and after and using the relationship Ek = 3/2*kT this can be confirmed empirically. While this investigates the translational modes, Raman spectroscopy investigates the vibrational energy modes. Other infrared methods including chemiluminescene can confirm this finding.

[[File:JN pbeforeafter.PNG]]

[[File:JN animation.PNG]]

 Setup a 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 (explore values also close to these limits). What do you observe? 

Despite the fact that there is sufficient energy in H-H to overcome the activation energy the H-H molecule collides and bounces off of the fluorine atom when P(H-H) is -3. This can be rationalized as an H atom having enough energy to reach the bottom of the H-F potential well and then exit it again, the same is true for the momentum being +3.

 For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now? 
Under these conditions the reaction proceeds successfully

 for H + HF Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy). 

Under these conditions the incoming high energy Hydrogen collides with the HF molecule and bounces off in the direction it came from taking the other Hydrogen with it, in this way the reaction proceeds.

 Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.) 

Using the inversion of momentum procedure it can be found that the the highest p(HF) and lowest p(HH) that gives a reactive trajectory is 2 and -6.8 respectively. The result of this is pictured below.

[[File:JN traject.PNG]]

 The cases studied are an illustration of Polanyi's empirical rules. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. 

Polanyi's emperical rules state that for an early barrier reactions the translational energy mode is more efficient than vibrational energy in achieving a reactive trajectory, while the opposite is true for late barrier reactions [2] . Exothermic reactions have early transition states while endothermic reactions have late transition state. With that in mind the reaction of F with H-H which is exothermic readily becomes reactive with a small increase in translational energy whereas an increase in vibrational energy does little to progress the reaction. In contrast H with H-F responds to a small increase vibrational energy very strongly and will overcome the activation energy, whereas an increase in translational will have a less efficient effect on the system achieving a reactive trajectory. The 4 examples below demonstrate this with H + H-F and F + H-H systems:

[[File:JN idk 1.PNG]]

[[File:JN idk 2.PNG]]

[[File:JN idk 3.PNG]]

[[File:JN idk 4.PNG]]

References:
1: he Journal of Chemical Physics 22, 1142 (1954)

2: G. Czakó, J. M. Bowman, Science 2011, 334(6054), 343–346

JN1316Lab2

2018-05-11T16:55:22Z

Jn1316:

All distances are in Angstroms and energies in kCal/mol

== Exercise 1: H + H 2 ==

 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. 

The potential energy along a minima line has dV/dr = 0 and because it is a maxima d^2V/dr^2 is <0. A transition state also has dV/dr = 0 however d^2V/dr^2 is in contrast >0.

 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. 

running an MEP calculation with both bond lengths set arbitrarily to 1 and exporting the data finds the saddle point to be where both bond lengths are 0.907741. The literature [1] is in agreement with this value.

[[File:IND-T.PNG]]

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

MEP plots correspond to infinitely slow motion with the velocity resetting to 0 in each step. This is not the case in dynamic plots. This explains why in MEP the molecular bond length does not vibrate and the free hydrogen atom does not escape to infinity but rather it's distance from the other two atoms plateaus.

[[File:JN idk 6.PNG]]
[[File:JN MEPcalc.PNG]]

The momentum - time graph below shows that one hydrogen leaves as a free atom and the other 2 as a vibrating molecule with translational and vibrational energy modes.
[[File:JN moment 1.PNG]]

If it were switched so that r1 = rts and r2 = rts + 0.01 the central hydrogen would bond with the the hydrogen to its right as opposed to its left and the free hydrogen will be switched. This is because asides from the slight displacement in distance the system is symmetrical.

 For the initial positions r1 = 0.74 Angstroms and r2 = 2.0 Angstroms, run trajectories with the following momenta combination: 

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy kCal/mol !! Reactive? !! Picture !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:JN TS 1.PNG]] || A successful reaction with a vibrating reactant transferring vibrational energy to a vibrating product.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:JN TS 2.PNG]] || There is not sufficient total energy for the reactants to overcome the activation energy.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:JN TS 3.PNG]] || This simple reactive trajectory passes once simply over the transition state.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:JN TS 4.PNG]] || This system has sufficient energy to overcome the activation energy. it does so but then 'bounces' of a steep wall of the potential well back to the reactants.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:JN TS 5.PNG]] || The transition state theory assumes the potential barrier cannot be recrossed. This example violates that assumption as the potential barrier is crossed 3 times
|}

 State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? 
The main assumption to the TST is that once you have left the maxima that is the transition state you cannot go back to it or past it. This assumption fails as we have seen in the above trajectories some of the minimum energy paths reach the transition state and then return backwards. The TST also neglects the effects of quantum mechanics. Particularly for light particles such as hydrogen the non-zero probability of tunneling through a classically forbidden potential barrier with quantum potential is significant. Transition state theory will accurately predict reactions that pass only once over the transition state however it could incorrectly predict a path will be reactive or vice versa.

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

 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 as H-F bond is stronger than H-H bond. As seen in the picture below the energy of the products is less than that of reactants.

[[File:F and H2 JN.PNG]]

H + HF: Reaction is endothermic as H-H bond is weaker than H-F bond. As seen in the picture below the energy of the products is greater than that of reactants. This can be found by inputting the output values from the previous computation

[[File:H and HF JN.PNG]]

 Locate the approximate position of the transition state. 

Hammond's postulate states that the energy of the TS can be approximated as being that of the state nearest to it, because H2 + F is exothermic that means it is nearer the reactants therefore by inputting the distance values of the reactants and making the momenta = 0 the transition state can be found. AB = 0.74 and BC = 1.815. The energy of this state is -103.743.
Therefore the activation energy of:
- H2 + F energy is -103.904 therefore activation energy is: 0.161

- HF + H energy is -137.063 therefore activation energy is: 33.32

[[File:JN TS NEW.PNG]]

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

The momenta of the system increases after the reaction compared to before it. This can be seen in the animation of the interaction. Therefore the energy released in this reaction is converted to kinetic energy as the relationship between momentum and energy is Ek = p^2 / 2m. By measuring the temperature before and after and using the relationship Ek = 3/2*kT this can be confirmed empirically. While this investigates the translational modes, Raman spectroscopy investigates the vibrational energy modes. Other infrared methods including chemiluminescene can confirm this finding.

[[File:JN pbeforeafter.PNG]]

[[File:JN animation.PNG]]

 Setup a 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 (explore values also close to these limits). What do you observe? 

Despite the fact that there is sufficient energy in H-H to overcome the activation energy the H-H molecule collides and bounces off of the fluorine atom when P(H-H) is -3. This can be rationalized as an H atom having enough energy to reach the bottom of the H-F potential well and then exit it again, the same is true for the momentum being +3.

 For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now? 
Under these conditions the reaction proceeds successfully

 for H + HF Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy). 

Under these conditions the incoming high energy Hydrogen collides with the HF molecule and bounces off in the direction it came from taking the other Hydrogen with it, in this way the reaction proceeds.

 Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.) 

Using the inversion of momentum procedure it can be found that the the highest p(HF) and lowest p(HH) that gives a reactive trajectory is 2 and -6.8 respectively. The result of this is pictured below.

[[File:JN traject.PNG]]

 The cases studied are an illustration of Polanyi's empirical rules. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. 

Polanyi's emperical rules state that for an early barrier reactions the translational energy mode is more efficient than vibrational energy in achieving a reactive trajectory, while the opposite is true for late barrier reactions [2] . Exothermic reactions have early transition states while endothermic reactions have late transition state. With that in mind the reaction of F with H-H which is exothermic readily becomes reactive with a small increase in translational energy whereas an increase in vibrational energy does little to progress the reaction. In contrast H with H-F responds to a small increase vibrational energy very strongly and will overcome the activation energy, whereas an increase in translational will have a less efficient effect on the system achieving a reactive trajectory. The 4 examples below demonstrate this with H + H-F and F + H-H systems:

[[File:JN idk 1.PNG]]

[[File:JN idk 2.PNG]]

[[File:JN idk 3.PNG]]

[[File:JN idk 4.PNG]]

References:
1: he Journal of Chemical Physics 22, 1142 (1954)

2: G. Czakó, J. M. Bowman, Science 2011, 334(6054), 343–346

JN1316Lab2

2018-05-11T16:37:24Z

Jn1316:

All distances are in Angstroms and energies in kCal/mol

== Exercise 1: H + H 2 ==

 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. 

The potential energy along a minima line has dV/dr = 0 and because it is a maxima d^2V/dr^2 is <0. A transition state also has dV/dr = 0 however d^2V/dr^2 is in contrast >0.

 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. 

running an MEP calculation with both bond lengths set arbitrarily to 1 and exporting the data finds the saddle point to be where both bond lengths are 0.907741. The literature [1] is in agreement with this value.

[[File:IND-T.PNG]]

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

MEP plots correspond to infinitely slow motion with the velocity resetting to 0 in each step. This is not the case in dynamic plots. This explains why in MEP the molecular bond length does not vibrate and the free hydrogen atom does not escape to infinity but rather it's distance from the other two atoms plateaus.

[[File:JN idk 6.PNG]]
[[File:JN MEPcalc.PNG]]

The momentum - time graph below shows that one hydrogen leaves as a free atom and the other 2 as a vibrating molecule with translational and vibrational energy modes.
[[File:JN moment 1.PNG]]

If it were switched so that r1 = rts and r2 = rts + 0.01 the central hydrogen would bond with the the hydrogen to its right as opposed to its left and the free hydrogen will be switched. This is because asides from the slight displacement in distance the system is symmetrical.

 For the initial positions r1 = 0.74 Angstroms and r2 = 2.0 Angstroms, run trajectories with the following momenta combination: 

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy kCal/mol !! Reactive? !! Picture !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:JN TS 1.PNG]] || A successful reaction with a vibrating reactant transferring vibrational energy to a vibrating product.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:JN TS 2.PNG]] || There is not sufficient total energy for the reactants to overcome the activation energy.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:JN TS 3.PNG]] || This simple reactive trajectory passes once simply over the transition state.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:JN TS 4.PNG]] || This system has sufficient energy to overcome the activation energy. it does so but then 'bounces' of a steep wall of the potential well back to the reactants.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:JN TS 5.PNG]] || The transition state theory assumes the potential barrier cannot be recrossed. This example violates that assumption as the potential barrier is crossed 3 times
|}

 State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? 
The main assumption to the TST is that once you have left the maxima that is the transition state you cannot go back to it or past it. This assumption fails as we have seen in the above trajectories some of the minimum energy paths reach the transition state and then return backwards. The TST also neglects the effects of quantum mechanics. Particularly for light particles such as hydrogen the non-zero probability of tunneling through a classically forbidden potential barrier with quantum potential is significant. Transition state theory will accurately predict reactions that pass only once over the transition state however it could incorrectly predict a path will be reactive or vice versa.

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

 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 as H-F bond is stronger than H-H bond. As seen in the picture below the energy of the products is less than that of reactants.

[[File:F and H2 JN.PNG]]

H + HF: Reaction is endothermic as H-H bond is weaker than H-F bond. As seen in the picture below the energy of the products is greater than that of reactants. This can be found by inputting the output values from the previous computation

[[File:H and HF JN.PNG]]

 Locate the approximate position of the transition state. 

Hammond's postulate states that the energy of the TS can be approximated as being that of the state nearest to it, because H2 + F is exothermic that means it is nearer the reactants therefore by inputting the distance values of the reactants and making the momenta = 0 the transition state can be found. AB = 0.74 and BC = 1.815. The energy of this state is -103.743.
Therefore the activation energy of:
H2 + F energy is 103.904 therefore activation energy is: 0.161
HF + H: 33.32

[[File:JN TS NEW.PNG]]

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

The momenta of the system increases after the reaction compared to before it. This can be seen in the animation of the interaction. Therefore the energy released in this reaction is converted to kinetic energy as the relationship between momentum and energy is Ek = p^2 / 2m. By measuring the temperature before and after and using the relationship Ek = 3/2*kT this can be confirmed empirically.

[[File:JN pbeforeafter.PNG]]

[[File:JN animation.PNG]]

 Setup a 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 (explore values also close to these limits). What do you observe? 

Despite the fact that there is sufficient energy in H-H to overcome the activation energy the H-H molecule collides and bounces off of the fluorine atom when P(H-H) is -3. This can be rationalized as an H atom having enough energy to reach the bottom of the H-F potential well and then exit it again, the same is true for the momentum being +3.

 For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now? 
Under these conditions the reaction proceeds successfully

 for H + HF Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy). 

Under these conditions the incoming high energy Hydrogen collides with the HF molecule and bounces off in the direction it came from taking the other Hydrogen with it, in this way the reaction proceeds.

 Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.) 
Using the inversion of momentum procedure it can be found that the the highest p(HF) and lowest p(HH) that gives a reactive trajectory is 2 and -6.8 respectively. The result of this is pictured below.

[[File:JN traject.PNG]]

 The cases studied are an illustration of Polanyi's empirical rules. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. 

Polanyi's emperical rules state that for an early barrier reactions the translational energy mode is more efficient than vibrational energy in achieving a reactive trajectory, while the opposite is true for late barrier reactions [2] . Exothermic reactions have early transition states while endothermic reactions have late transition state. With that in mind the reaction of F with H-H which is exothermic readily becomes reactive with a small increase in translational energy whereas an increase in vibrational energy does little to progress the reaction. In contrast H with H-F responds to a small increase vibrational energy very strongly and will overcome the activation energy, whereas an increase in translational will have a less efficient effect on the system achieving a reactive trajectory. The 4 examples below demonstrate this with H + H-F and F + H-H systems:

[[File:JN idk 1.PNG]]

[[File:JN idk 2.PNG]]

[[File:JN idk 3.PNG]]

[[File:JN idk 4.PNG]]

References:
1: he Journal of Chemical Physics 22, 1142 (1954)
2: G. Czakó, J. M. Bowman, Science 2011, 334(6054), 343–346

JN1316Lab2

2018-05-11T16:33:50Z

Jn1316:

All distances are in Angstroms and energies in kCal/mol

== Exercise 1: H + H 2 ==

 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. 

The potential energy along a minima line has dV/dr = 0 and because it is a maxima d^2V/dr^2 is <0. A transition state also has dV/dr = 0 however d^2V/dr^2 is in contrast >0.

 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. 

running an MEP calculation with both bond lengths set arbitrarily to 1 and exporting the data finds the saddle point to be where both bond lengths are 0.907741. The literature [1] is in agreement with this value.

[[File:IND-T.PNG]]

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

MEP plots correspond to infinitely slow motion with the velocity resetting to 0 in each step. This is not the case in dynamic plots. This explains why in MEP the molecular bond length does not vibrate and the free hydrogen atom does not escape to infinity but rather it's distance from the other two atoms plateaus.

[[File:JN idk 6.PNG]]
[[File:JN MEPcalc.PNG]]

The momentum - time graph below shows that one hydrogen leaves as a free atom and the other 2 as a vibrating molecule with translational and vibrational energy modes.
[[File:JN moment 1.PNG]]

If it were switched so that r1 = rts and r2 = rts + 0.01 the central hydrogen would bond with the the hydrogen to its right as opposed to its left and the free hydrogen will be switched. This is because asides from the slight displacement in distance the system is symmetrical.

 For the initial positions r1 = 0.74 Angstroms and r2 = 2.0 Angstroms, run trajectories with the following momenta combination: 

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy kCal/mol !! Reactive? !! Picture !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:JN TS 1.PNG]] || A successful reaction with a vibrating reactant transferring vibrational energy to a vibrating product.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:JN TS 2.PNG]] || There is not sufficient total energy for the reactants to overcome the activation energy.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:JN TS 3.PNG]] || This simple reactive trajectory passes once simply over the transition state.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:JN TS 4.PNG]] || This system has sufficient energy to overcome the activation energy. it does so but then 'bounces' of a steep wall of the potential well back to the reactants.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:JN TS 5.PNG]] || The transition state theory assumes the potential barrier cannot be recrossed. This example violates that assumption as the potential barrier is crossed 3 times
|}

 State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? 
The main assumption to the TST is that once you have left the maxima that is the transition state you cannot go back to it or past it. This assumption fails as we have seen in the above trajectories some of the minimum energy paths reach the transition state and then return backwards. The TST also neglects the effects of quantum mechanics. Particularly for light particles such as hydrogen the non-zero probability of tunneling through a classically forbidden potential barrier with quantum potential is significant. Transition state theory will accurately predict reactions that pass only once over the transition state however it could incorrectly predict a path will be reactive or vice versa.

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

 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 as H-F bond is stronger than H-H bond. As seen in the picture below the energy of the products is less than that of reactants.

[[File:F and H2 JN.PNG]]

H + HF: Reaction is endothermic as H-H bond is weaker than H-F bond. As seen in the picture below the energy of the products is greater than that of reactants. This can be found by inputting the output values from the previous computation

[[File:H and HF JN.PNG]]

 Locate the approximate position of the transition state. 

Hammond's postulate states that the energy of the TS can be approximated as being that of the state nearest to it, because H2 + F is exothermic that means it is nearer the reactants therefore by inputting the distance values of the reactants and making the momenta = 0 the transition state can be found. AB = 0.74 and BC = 1.815. The energy of this state is -103.743.
Therefore the activation energy of:
H2 + F energy is 103.904 therefore activation energy is: 0.161
HF + H: 33.32

[[File:JN TS NEW.PNG]]

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

The momenta of the system increases after the reaction compared to before it. This can be seen in the animation of the interaction. Therefore the energy released in this reaction is converted to kinetic energy as the relationship between momentum and energy is Ek = p^2 / 2m. By measuring the temperature before and after and using the relationship Ek = 3/2*kT this can be confirmed empirically.

[[File:JN pbeforeafter.PNG]]
[[File:JN animation.PNG]]

 Setup a 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 (explore values also close to these limits). What do you observe? 

Despite the fact that there is sufficient energy in H-H to overcome the activation energy the H-H molecule collides and bounces off of the fluorine atom when P(H-H) is -3. This can be rationalized as an H atom having enough energy to reach the bottom of the H-F potential well and then exit it again, the same is true for the momentum being +3.

 For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now? 
Under these conditions the reaction proceeds successfully

 for H + HF Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy). 

Under these conditions the incoming high energy Hydrogen collides with the HF molecule and bounces off in the direction it came from taking the other Hydrogen with it, in this way the reaction proceeds.

 Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.) 
Using the inversion of momentum procedure it can be found that the the highest p(HF) and lowest p(HH) that gives a reactive trajectory is 2 and -6.8 respectively. The result of this is pictured below.

[[File:JN traject.PNG]]

 The cases studied are an illustration of Polanyi's empirical rules. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. 

Polanyi's emperical rules state that for an early barrier reactions the translational energy mode is more efficient than vibrational energy in achieving a reactive trajectory, while the opposite is true for late barrier reactions [2] . Exothermic reactions have early transition states while endothermic reactions have late transition state. With that in mind the reaction of F with H-H which is exothermic readily becomes reactive with a small increase in translational energy whereas an increase in vibrational energy does little to progress the reaction. In contrast H with H-F responds to a small increase vibrational energy very strongly and will overcome the activation energy, whereas an increase in translational will have a less efficient effect on the system achieving a reactive trajectory. The 4 examples below demonstrate this with H + H-F and F + H-H systems:

[[File:JN idk 1.PNG]]
[[File:JN idk 2.PNG]]
[[File:JN idk 3.PNG]]
[[File:JN idk 4.PNG]]

References:
1: he Journal of Chemical Physics 22, 1142 (1954)
2: G. Czakó, J. M. Bowman, Science 2011, 334(6054), 343–346

JN1316Lab2

2018-05-11T16:30:41Z

Jn1316:

All distances are in Angstroms and energies in kCal/mol

== Exercise 1: H + H 2 ==

 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. 

The potential energy along a minima line has dV/dr = 0 and because it is a maxima d^2V/dr^2 is <0. A transition state also has dV/dr = 0 however d^2V/dr^2 is in contrast >0.

 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. 

running an MEP calculation with both bond lengths set arbitrarily to 1 and exporting the data finds the saddle point to be where both bond lengths are 0.907741.

[[File:IND-T.PNG]]

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

MEP plots correspond to infinitely slow motion with the velocity resetting to 0 in each step. This is not the case in dynamic plots. This explains why in MEP the molecular bond length does not vibrate and the free hydrogen atom does not escape to infinity but rather it's distance from the other two atoms plateaus.

[[File:JN idk 6.PNG]]
[[File:JN MEPcalc.PNG]]

The momentum - time graph below shows that one hydrogen leaves as a free atom and the other 2 as a vibrating molecule with translational and vibrational energy modes.
[[File:JN moment 1.PNG]]

If it were switched so that r1 = rts and r2 = rts + 0.01 the central hydrogen would bond with the the hydrogen to its right as opposed to its left and the free hydrogen will be switched. This is because asides from the slight displacement in distance the system is symmetrical.

 For the initial positions r1 = 0.74 Angstroms and r2 = 2.0 Angstroms, run trajectories with the following momenta combination: 

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy kCal/mol !! Reactive? !! Picture !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:JN TS 1.PNG]] || A successful reaction with a vibrating reactant transferring vibrational energy to a vibrating product.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:JN TS 2.PNG]] || There is not sufficient total energy for the reactants to overcome the activation energy.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:JN TS 3.PNG]] || This simple reactive trajectory passes once simply over the transition state.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:JN TS 4.PNG]] || This system has sufficient energy to overcome the activation energy. it does so but then 'bounces' of a steep wall of the potential well back to the reactants.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:JN TS 5.PNG]] || The transition state theory assumes the potential barrier cannot be recrossed. This example violates that assumption as the potential barrier is crossed 3 times
|}

 State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? 
The main assumption to the TST is that once you have left the maxima that is the transition state you cannot go back to it or past it. This assumption fails as we have seen in the above trajectories some of the minimum energy paths reach the transition state and then return backwards. The TST also neglects the effects of quantum mechanics. Particularly for light particles such as hydrogen the non-zero probability of tunneling through a classically forbidden potential barrier with quantum potential is significant. Transition state theory will accurately predict reactions that pass only once over the transition state however it could incorrectly predict a path will be reactive or vice versa.

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

 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 as H-F bond is stronger than H-H bond. As seen in the picture below the energy of the products is less than that of reactants.

[[File:F and H2 JN.PNG]]

H + HF: Reaction is endothermic as H-H bond is weaker than H-F bond. As seen in the picture below the energy of the products is greater than that of reactants. This can be found by inputting the output values from the previous computation

[[File:H and HF JN.PNG]]

 Locate the approximate position of the transition state. 

Hammond's postulate states that the energy of the TS can be approximated as being that of the state nearest to it, because H2 + F is exothermic that means it is nearer the reactants therefore by inputting the distance values of the reactants and making the momenta = 0 the transition state can be found. AB = 0.74 and BC = 1.815. The energy of this state is -103.743.
Therefore the activation energy of:
H2 + F energy is 103.904 therefore activation energy is: 0.161
HF + H: 33.32

[[File:JN TS NEW.PNG]]

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

The momenta of the system increases after the reaction compared to before it. This can be seen in the animation of the interaction. Therefore the energy released in this reaction is converted to kinetic energy as the relationship between momentum and energy is Ek = p^2 / 2m. By measuring the temperature before and after and using the relationship Ek = 3/2*kT this can be confirmed empirically.

[[File:JN pbeforeafter.PNG]]
[[File:JN animation.PNG]]

 Setup a 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 (explore values also close to these limits). What do you observe? 

Despite the fact that there is sufficient energy in H-H to overcome the activation energy the H-H molecule collides and bounces off of the fluorine atom when P(H-H) is -3. This can be rationalized as an H atom having enough energy to reach the bottom of the H-F potential well and then exit it again, the same is true for the momentum being +3.

 For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now? 
Under these conditions the reaction proceeds successfully

 for H + HF Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy). 

Under these conditions the incoming high energy Hydrogen collides with the HF molecule and bounces off in the direction it came from taking the other Hydrogen with it, in this way the reaction proceeds.

 Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.) 
Using the inversion of momentum procedure it can be found that the the highest p(HF) and lowest p(HH) that gives a reactive trajectory is 2 and -6.8 respectively. The result of this is pictured below.

[[File:JN traject.PNG]]

 The cases studied are an illustration of Polanyi's empirical rules. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. 

Polanyi's emperical rules state that for an early barrier reactions the translational energy mode is more efficient than vibrational energy in achieving a reactive trajectory, while the opposite is true for late barrier reactions [1] . Exothermic reactions have early transition states while endothermic reactions have late transition state. With that in mind the reaction of F with H-H which is exothermic readily becomes reactive with a small increase in translational energy whereas an increase in vibrational energy does little to progress the reaction. In contrast H with H-F responds to a small increase vibrational energy very strongly and will overcome the activation energy, whereas an increase in translational will have a less efficient effect on the system achieving a reactive trajectory. The 4 examples below demonstrate this with H + H-F and F + H-H systems:

[[File:JN idk 1.PNG]]
[[File:JN idk 2.PNG]]
[[File:JN idk 3.PNG]]
[[File:JN idk 4.PNG]]

References:
1: G. Czakó, J. M. Bowman, Science 2011, 334(6054), 343–346

JN1316Lab2

2018-05-11T16:21:39Z

Jn1316:

All distances are in Angstroms and energies in kCal/mol

== Exercise 1: H + H 2 ==

 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. 

The potential energy along a minima line has dV/dr = 0 and because it is a maxima d^2V/dr^2 is <0. A transition state also has dV/dr = 0 however d^2V/dr^2 is in contrast >0.

 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. 

running an MEP calculation with both bond lengths set arbitrarily to 1 and exporting the data finds the saddle point to be where both bond lengths are 0.907741.

[[File:IND-T.PNG]]

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

MEP plots correspond to infinitely slow motion with the velocity resetting to 0 in each step. This is not the case in dynamic plots. This explains why in MEP the molecular bond length does not vibrate and the free hydrogen atom does not escape to infinity but rather it's distance from the other two atoms plateaus.

[[File:JN idk 6.PNG]]
[[File:JN MEPcalc.PNG]]

The momentum - time graph below shows that one hydrogen leaves as a free atom and the other 2 as a vibrating molecule with translational and vibrational energy modes.
[[File:JN moment 1.PNG]]

If it were switched so that r1 = rts and r2 = rts + 0.01 the central hydrogen would bond with the the hydrogen to its right as opposed to its left and the free hydrogen will be switched. This is because asides from the slight displacement in distance the system is symmetrical.

 For the initial positions r1 = 0.74 Angstroms and r2 = 2.0 Angstroms, run trajectories with the following momenta combination: 

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy kCal/mol !! Reactive? !! Picture !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:JN TS 1.PNG]] || A successful reaction with a vibrating reactant transferring vibrational energy to a vibrating product.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:JN TS 2.PNG]] || There is not sufficient total energy for the reactants to overcome the activation energy.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:JN TS 3.PNG]] || This simple reactive trajectory passes once simply over the transition state.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:JN TS 4.PNG]] || This system has sufficient energy to overcome the activation energy. it does so but then 'bounces' of a steep wall of the potential well back to the reactants.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:JN TS 5.PNG]] || The transition state theory assumes the potential barrier cannot be recrossed. This example violates that assumption as the potential barrier is crossed 3 times
|}

 State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? 
The main assumption to the TST is that once you have left the maxima that is the transition state you cannot go back to it or past it. This assumption fails as we have seen in the above trajectories some of the minimum energy paths reach the transition state and then return backwards. The TST also neglects the effects of quantum mechanics. Particularly for light particles such as hydrogen the non-zero probability of tunneling through a classically forbidden potential barrier with quantum potential is significant. Transition state theory will accurately predict reactions that pass only once over the transition state however it could incorrectly predict a path will be reactive or vice versa.

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

 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 as H-F bond is stronger than H-H bond. As seen in the picture below the energy of the products is less than that of reactants.

[[File:F and H2 JN.PNG]]

H + HF: Reaction is endothermic as H-H bond is weaker than H-F bond. As seen in the picture below the energy of the products is greater than that of reactants. This can be found by inputting the output values from the previous computation

[[File:H and HF JN.PNG]]

 Locate the approximate position of the transition state. 

Hammond's postulate states that the energy of the TS can be approximated as being that of the state nearest to it, because H2 + F is exothermic that means it is nearer the reactants therefore by inputting the distance values of the reactants and making the momenta = 0 the transition state can be found. AB = 0.74 and BC = 1.815. The energy of this state is -103.743.
Therefore the activation energy of:
H2 + F: 1.2
HF + H: 30

[[File:JN TS NEW.PNG]]

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

The momenta of the system increases after the reaction compared to before it. This can be seen in the animation of the interaction. Therefore the energy released in this reaction is converted to kinetic energy as the relationship between momentum and energy is Ek = p^2 / 2m. By measuring the temperature before and after and using the relationship Ek = 3/2*kT this can be confirmed empirically.

[[File:JN pbeforeafter.PNG]]
[[File:JN animation.PNG]]

 Setup a 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 (explore values also close to these limits). What do you observe? 

Despite the fact that there is sufficient energy in H-H to overcome the activation energy the H-H molecule collides and bounces off of the fluorine atom when P(H-H) is -3. This can be rationalized as an H atom having enough energy to reach the bottom of the H-F potential well and then exit it again, the same is true for the momentum being +3.

 For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now? 
Under these conditions the reaction proceeds successfully

 for H + HF Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy). 

Under these conditions the incoming high energy Hydrogen collides with the HF molecule and bounces off in the direction it came from taking the other Hydrogen with it, in this way the reaction proceeds.

 Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.) 
Using the inversion of momentum procedure it can be found that the the highest p(HF) and lowest p(HH) that gives a reactive trajectory is 2 and -6.8 respectively. The result of this is pictured below.

[[File:JN traject.PNG]]

 The cases studied are an illustration of Polanyi's empirical rules. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. 

Polanyi's emperical rules state that for an early barrier reactions the translational energy mode is more efficient than vibrational energy in achieving a reactive trajectory, while the opposite is true for late barrier reactions [1] . Exothermic reactions have early transition states while endothermic reactions have late transition state. With that in mind the reaction of F with H-H which is exothermic readily becomes reactive with a small increase in translational energy whereas an increase in vibrational energy does little to progress the reaction. In contrast H with H-F responds to a small increase vibrational energy very strongly and will overcome the activation energy, whereas an increase in translational will have a less efficient effect on the system achieving a reactive trajectory. The 4 examples below demonstrate this with H + H-F and F + H-H systems:

[[File:JN idk 1.PNG]]
[[File:JN idk 2.PNG]]
[[File:JN idk 3.PNG]]
[[File:JN idk 4.PNG]]

References:
1: G. Czakó, J. M. Bowman, Science 2011, 334(6054), 343–346

JN1316Lab2

2018-05-11T16:14:09Z

Jn1316:

All distances are in Angstroms and energies in kCal/mol

== Exercise 1: H + H 2 ==

 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. 

The potential energy along a minima line has dV/dr = 0 and because it is a maxima d^2V/dr^2 is <0. A transition state also has dV/dr = 0 however d^2V/dr^2 is in contrast >0.

 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. 

running an MEP calculation with both bond lengths set arbitrarily to 1 and exporting the data finds the saddle point to be where both bond lengths are 0.907741.

[[File:IND-T.PNG]]

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

MEP plots correspond to infinitely slow motion with the velocity resetting to 0 in each step. This is not the case in dynamic plots. This explains why in MEP the molecular bond length does not vibrate and the free hydrogen atom does not escape to infinity but rather it's distance from the other two atoms plateaus.

[[File:JN idk 6.PNG]]
[[File:JN MEPcalc.PNG]]

The momentum - time graph below shows that one hydrogen leaves as a free atom and the other 2 as a vibrating molecule with translational and vibrational energy modes.
[[File:JN moment 1.PNG]]

If it were switched so that r1 = rts and r2 = rts + 0.01 the central hydrogen would bond with the the hydrogen to its right as opposed to its left and the free hydrogen will be switched. This is because asides from the slight displacement in distance the system is symmetrical.

 For the initial positions r1 = 0.74 Angstroms and r2 = 2.0 Angstroms, run trajectories with the following momenta combination: 

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy kCal/mol !! Reactive? !! Picture !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:JN TS 1.PNG]] || A successful reaction with a vibrating reactant transferring vibrational energy to a vibrating product.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:JN TS 2.PNG]] || There is not sufficient total energy for the reactants to overcome the activation energy.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:JN TS 3.PNG]] || This simple reactive trajectory passes once simply over the transition state.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:JN TS 4.PNG]] || This system has sufficient energy to overcome the activation energy. it does so but then 'bounces' of a steep wall of the potential well back to the reactants.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:JN TS 5.PNG]] || The transition state theory assumes the potential barrier cannot be recrossed. This example violates that assumption as the potential barrier is crossed 3 times
|}

 State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? 
The main assumption to the TST is that once you have left the maxima that is the transition state you cannot go back to it or past it. This assumption fails as we have seen in the above trajectories some of the minimum energy paths reach the transition state and then return backwards. The TST also neglects the effects of quantum mechanics. Particularly for light particles such as hydrogen the non-zero probability of tunneling through a classically forbidden potential barrier with quantum potential is significant. Transition state theory will accurately predict reactions that pass only once over the transition state however it could incorrectly predict a path will be reactive or vice versa.

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

 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 as H-F bond is stronger than H-H bond. As seen in the picture below the energy of the products is less than that of reactants.
[[File:F and H2 JN.PNG]]

H + HF: Reaction is endothermic as H-H bond is weaker than H-F bond. As seen in the picture below the energy of the products is greater than that of reactants. This can be found by inputting the output values from the previous computation

[[File:H and HF JN.PNG]]

 Locate the approximate position of the transition state. 

Hammond's postulate states that the energy of the TS can be approximated as being that of the state nearest to it, because H2 + F is exothermic that means it is nearer the reactants therefore by inputting the distance values of the reactants and making the momenta = 0 the transition state can be found. AB = 0.74 and BC = 1.815. The energy of this state is -103.743.
Therefore the activation energy of:
H2 + F: 1.2
HF + H: 30

[[File:JN TS NEW.PNG]]

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

The momenta of the system increases after the reaction compared to before it. This can be seen in the animation of the interaction. Therefore the energy released in this reaction is converted to kinetic energy as the relationship between momentum and energy is Ek = p^2 / 2m. By measuring the temperature before and after and using the relationship Ek = 3/2*kT this can be confirmed empirically.

[[File:JN pbeforeafter.PNG]]
[[File:JN animation.PNG]]

 Setup a 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 (explore values also close to these limits). What do you observe? 

Despite the fact that there is sufficient energy in H-H to overcome the activation energy the H-H molecule collides and bounces off of the fluorine atom when P(H-H) is -3. This can be rationalized as an H atom having enough energy to reach the bottom of the H-F potential well and then exit it again, the same is true for the momentum being +3.

 For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now? 
Under these conditions the reaction proceeds successfully

 for H + HF Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy). 

Under these conditions the incoming high energy Hydrogen collides with the HF molecule and bounces off in the direction it came from taking the other Hydrogen with it, in this way the reaction proceeds.

 Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.) 
Using the inversion of momentum procedure it can be found that the the highest p(HF) and lowest p(HH) that gives a reactive trajectory is 2 and -6.8 respectively. The result of this is pictured below.

[[File:JN traject.PNG]]

 The cases studied are an illustration of Polanyi's empirical rules. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. 

Polanyi's emperical rules state that for an early barrier reactions the translational energy mode is more efficient than vibrational energy in achieving a reactive trajectory, while the opposite is true for late barrier reactions [1] . Exothermic reactions have early transition states while endothermic reactions have late transition state. With that in mind the reaction of F with H-H which is exothermic readily becomes reactive with a small increase in translational energy whereas an increase in vibrational energy does little to progress the reaction. In contrast H with H-F responds to a small increase vibrational energy very strongly and will overcome the activation energy, whereas an increase in translational will have a less efficient effect on the system achieving a reactive trajectory. The 4 examples below demonstrate this with H + H-F and F + H-H systems:

[[File:JN idk 1.PNG]]
[[File:JN idk 2.PNG]]
[[File:JN idk 3.PNG]]
[[File:JN idk 4.PNG]]

References:
1: G. Czakó, J. M. Bowman, Science 2011, 334(6054), 343–346

JN1316Lab2

2018-05-11T16:11:40Z

Jn1316:

All distances are in Angstroms and energies in kCal/mol

== Exercise 1: H + H 2 ==

 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. 

The potential energy along a minima line has dV/dr = 0 and because it is a maxima d^2V/dr^2 is <0. A transition state also has dV/dr = 0 however d^2V/dr^2 is in contrast >0.

 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. 

running an MEP calculation with both bond lengths set arbitrarily to 1 and exporting the data finds the saddle point to be where both bond lengths are 0.907741.

[[File:IND-T.PNG]]

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

MEP plots correspond to infinitely slow motion with the velocity resetting to 0 in each step. This is not the case in dynamic plots. This explains why in MEP the molecular bond length does not vibrate and the free hydrogen atom does not escape to infinity but rather it's distance from the other two atoms plateaus.

[[File:JN idk 6.PNG]]
[[File:JN MEPcalc.PNG]]

The momentum - time graph below shows that one hydrogen leaves as a free atom and the other 2 as a vibrating molecule with translational and vibrational energy modes.
[[File:JN moment 1.PNG]]

If it were switched so that r1 = rts and r2 = rts + 0.01 the central hydrogen would bond with the the hydrogen to its right as opposed to its left and the free hydrogen will be switched. This is because asides from the slight displacement in distance the system is symmetrical.

 For the initial positions r1 = 0.74 Angstroms and r2 = 2.0 Angstroms, run trajectories with the following momenta combination: 

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy kCal/mol !! Reactive? !! Picture !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:JN TS 1.PNG]] || A successful reaction with a vibrating reactant transferring vibrational energy to a vibrating product.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:JN TS 2.PNG]] || There is not sufficient total energy for the reactants to overcome the activation energy.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:JN TS 3.PNG]] || This simple reactive trajectory passes once simply over the transition state.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:JN TS 4.PNG]] || This system has sufficient energy to overcome the activation energy. it does so but then 'bounces' of a steep wall of the potential well back to the reactants.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:JN TS 5.PNG]] || The transition state theory assumes the potential barrier cannot be recrossed. This example violates that assumption as the potential barrier is crossed 3 times
|}

 State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? 
The main assumption to the TST is that once you have left the maxima that is the transition state you cannot go back to it or past it. This assumption fails as we have seen in the above trajectories some of the minimum energy paths reach the transition state and then return backwards. The TST also neglects the effects of quantum mechanics. Particularly for light particles such as hydrogen the non-zero probability of tunneling through a classically forbidden potential barrier with quantum potential is significant. Transition state theory will accurately predict reactions that pass only once over the transition state however it could incorrectly predict a path will be reactive or vice versa.

 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 as H-F bond is stronger than H-H bond. As seen in the picture below the energy of the products is less than that of reactants.
[[File:F and H2 JN.PNG]]

H + HF: Reaction is endothermic as H-H bond is weaker than H-F bond. As seen in the picture below the energy of the products is greater than that of reactants. This can be found by inputting the output values from the previous computation

[[File:H and HF JN.PNG]]

 Locate the approximate position of the transition state. 

Hammond's postulate states that the energy of the TS can be approximated as being that of the state nearest to it, because H2 + F is exothermic that means it is nearer the reactants therefore by inputting the distance values of the reactants and making the momenta = 0 the transition state can be found. AB = 0.74 and BC = 1.815. The energy of this state is -103.743.
Therefore the activation energy of:
H2 + F: 1.2
HF + H: 30

[[File:JN TS NEW.PNG]]

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

The momenta of the system increases after the reaction compared to before it. This can be seen in the animation of the interaction. Therefore the energy released in this reaction is converted to kinetic energy as the relationship between momentum and energy is Ek = p^2 / 2m. By measuring the temperature before and after and using the relationship Ek = 3/2*kT this can be confirmed empirically.

[[File:JN pbeforeafter.PNG]]
[[File:JN animation.PNG]]

 Setup a 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 (explore values also close to these limits). What do you observe? 

Despite the fact that there is sufficient energy in H-H to overcome the activation energy the H-H molecule collides and bounces off of the fluorine atom when P(H-H) is -3. This can be rationalized as an H atom having enough energy to reach the bottom of the H-F potential well and then exit it again, the same is true for the momentum being +3.

 For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now? 
Under these conditions the reaction proceeds successfully

 for H + HF Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy). 

Under these conditions the incoming high energy Hydrogen collides with the HF molecule and bounces off in the direction it came from taking the other Hydrogen with it, in this way the reaction proceeds.

 Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.) 
Using the inversion of momentum procedure it can be found that the the highest p(HF) and lowest p(HH) that gives a reactive trajectory is 2 and -6.8 respectively. The result of this is pictured below.

[[File:JN traject.PNG]]

 The cases studied are an illustration of Polanyi's empirical rules. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. 

Polanyi's emperical rules state that for an early barrier reactions the translational energy mode is more efficient than vibrational energy in achieving a reactive trajectory, while the opposite is true for late barrier reactions [1] . Exothermic reactions have early transition states while endothermic reactions have late transition state. With that in mind the reaction of F with H-H which is exothermic readily becomes reactive with a small increase in translational energy whereas an increase in vibrational energy does little to progress the reaction. In contrast H with H-F responds to a small increase vibrational energy very strongly and will overcome the activation energy, whereas an increase in translational will have a less efficient effect on the system achieving a reactive trajectory. The 4 examples below demonstrate this with H + H-F and F + H-H systems:

[[File:JN idk 1.PNG]]
[[File:JN idk 2.PNG]]
[[File:JN idk 3.PNG]]
[[File:JN idk 4.PNG]]

References:
1: G. Czakó, J. M. Bowman, Science 2011, 334(6054), 343–346

JN1316Lab2

2018-05-11T16:08:48Z

Jn1316:

All distances are in Angstroms and energies in kCal/mol

== Exercise 1: H + H 2 ==

 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. 

The potential energy along a minima line has dV/dr = 0 and because it is a maxima d^2V/dr^2 is <0. A transition state also has dV/dr = 0 however d^2V/dr^2 is in contrast >0.

 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. 

running an MEP calculation with both bond lengths set arbitrarily to 1 and exporting the data finds the saddle point to be where both bond lengths are 0.907741.

[[File:IND-T.PNG]]

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

MEP plots correspond to infinitely slow motion with the velocity resetting to 0 in each step. This is not the case in dynamic plots. This explains why in MEP the molecular bond length does not vibrate and the free hydrogen atom does not escape to infinity but rather it's distance from the other two atoms plateaus.

[[File:JN idk 6.PNG]]
[[File:JN MEPcalc.PNG]]

The momentum - time graph below shows that one hydrogen leaves as a free atom and the other 2 as a vibrating molecule with translational and vibrational energy modes.
[[File:JN moment 1.PNG]]

If it were switched so that r1 = rts and r2 = rts + 0.01 the central hydrogen would bond with the the hydrogen to its right as opposed to its left and the free hydrogen will be switched. This is because asides from the slight displacement in distance the system is symmetrical.

 For the initial positions r1 = 0.74 Angstroms and r2 = 2.0 Angstroms, run trajectories with the following momenta combination: 

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy kCal/mol !! Reactive? !! Picture !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:JN TS 1.PNG]] || A successful reaction with a vibrating reactant transferring vibrational energy to a vibrating product.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:JN TS 2.PNG]] || There is not sufficient total energy for the reactants to overcome the activation energy.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:JN TS 3.PNG]] || This simple reactive trajectory passes once simply over the transition state.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:JN TS 4.PNG]] || This system has sufficient energy to overcome the activation energy. it does so but then 'bounces' of a steep wall of the potential well back to the reactants.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:JN TS 5.PNG]] || The transition state theory assumes the potential barrier cannot be recrossed. This example violates that assumption as the potential barrier is crossed 3 times
|}

 State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? 
The main assumption to the TST is that once you have left the maxima that is the transition state you cannot go back to it or past it. This assumption fails as we have seen in the above trajectories some of the minimum energy paths reach the transition state and then return backwards. The TST also neglects the effects of quantum mechanics. Particularly for light particles such as hydrogen the non-zero probability of tunneling through a classically forbidden potential barrier with quantum potential is significant.

 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 as H-F bond is stronger than H-H bond. As seen in the picture below the energy of the products is less than that of reactants.
[[File:F and H2 JN.PNG]]

H + HF: Reaction is endothermic as H-H bond is weaker than H-F bond. As seen in the picture below the energy of the products is greater than that of reactants. This can be found by inputting the output values from the previous computation

[[File:H and HF JN.PNG]]

 Locate the approximate position of the transition state. 

Hammond's postulate states that the energy of the TS can be approximated as being that of the state nearest to it, because H2 + F is exothermic that means it is nearer the reactants therefore by inputting the distance values of the reactants and making the momenta = 0 the transition state can be found. AB = 0.74 and BC = 1.815. The energy of this state is -103.743.
Therefore the activation energy of:
H2 + F: 1.2
HF + H: 30

[[File:JN TS NEW.PNG]]

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

The momenta of the system increases after the reaction compared to before it. This can be seen in the animation of the interaction. Therefore the energy released in this reaction is converted to kinetic energy as the relationship between momentum and energy is Ek = p^2 / 2m. By measuring the temperature before and after and using the relationship Ek = 3/2*kT this can be confirmed empirically.

[[File:JN pbeforeafter.PNG]]
[[File:JN animation.PNG]]

 Setup a 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 (explore values also close to these limits). What do you observe? 

Despite the fact that there is sufficient energy in H-H to overcome the activation energy the H-H molecule collides and bounces off of the fluorine atom when P(H-H) is -3. This can be rationalized as an H atom having enough energy to reach the bottom of the H-F potential well and then exit it again, the same is true for the momentum being +3.

 For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now? 
Under these conditions the reaction proceeds successfully

 for H + HF Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy). 

Under these conditions the incoming high energy Hydrogen collides with the HF molecule and bounces off in the direction it came from taking the other Hydrogen with it, in this way the reaction proceeds.

 Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.) 
Using the inversion of momentum procedure it can be found that the the highest p(HF) and lowest p(HH) that gives a reactive trajectory is 2 and -6.8 respectively. The result of this is pictured below.

[[File:JN traject.PNG]]

 The cases studied are an illustration of Polanyi's empirical rules. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. 

Polanyi's emperical rules state that for an early barrier reactions the translational energy mode is more efficient than vibrational energy in achieving a reactive trajectory, while the opposite is true for late barrier reactions [1] . Exothermic reactions have early transition states while endothermic reactions have late transition state. With that in mind the reaction of F with H-H which is exothermic readily becomes reactive with a small increase in translational energy whereas an increase in vibrational energy does little to progress the reaction. In contrast H with H-F responds to a small increase vibrational energy very strongly and will overcome the activation energy, whereas an increase in translational will have a less efficient effect on the system achieving a reactive trajectory. The 4 examples below demonstrate this with H + H-F and F + H-H systems:

[[File:JN idk 1.PNG]]
[[File:JN idk 2.PNG]]
[[File:JN idk 3.PNG]]
[[File:JN idk 4.PNG]]

References:
1: G. Czakó, J. M. Bowman, Science 2011, 334(6054), 343–346

File:JN moment 1.PNG

2018-05-11T16:06:45Z

Jn1316:

JN1316Lab2

2018-05-11T16:03:52Z

Jn1316:

All distances are in Angstroms and energies in kCal/mol

== Exercise 1: H + H 2 ==

 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. 

The potential energy along a minima line has dV/dr = 0 and because it is a maxima d^2V/dr^2 is <0. A transition state also has dV/dr = 0 however d^2V/dr^2 is in contrast >0.

 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. 

running an MEP calculation with both bond lengths set arbitrarily to 1 and exporting the data finds the saddle point to be where both bond lengths are 0.907741.

[[File:IND-T.PNG]]

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

MEP plots correspond to infinitely slow motion with the velocity resetting to 0 in each step. This is not the case in dynamic plots. This explains why in MEP the molecular bond length does not vibrate and the free hydrogen atom does not escape to infinity but rather it's distance from the other two atoms plateaus.

[[File:JN idk 6.PNG]]
[[File:JN MEPcalc.PNG]]

 For the initial positions r1 = 0.74 Angstroms and r2 = 2.0 Angstroms, run trajectories with the following momenta combination: 

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy kCal/mol !! Reactive? !! Picture !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:JN TS 1.PNG]] || A successful reaction with a vibrating reactant transferring vibrational energy to a vibrating product.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:JN TS 2.PNG]] || There is not sufficient total energy for the reactants to overcome the activation energy.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:JN TS 3.PNG]] || This simple reactive trajectory passes once simply over the transition state.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:JN TS 4.PNG]] || This system has sufficient energy to overcome the activation energy. it does so but then 'bounces' of a steep wall of the potential well back to the reactants.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:JN TS 5.PNG]] || The transition state theory assumes the potential barrier cannot be recrossed. This example violates that assumption as the potential barrier is crossed 3 times
|}

 State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? 
The main assumption to the TST is that once you have left the maxima that is the transition state you cannot go back to it or past it. This assumption fails as we have seen in the above trajectories some of the minimum energy paths reach the transition state and then return backwards. The TST also neglects the effects of quantum mechanics. Particularly for light particles such as hydrogen the non-zero probability of tunneling through a classically forbidden potential barrier with quantum potential is significant.

 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 as H-F bond is stronger than H-H bond. As seen in the picture below the energy of the products is less than that of reactants.
[[File:F and H2 JN.PNG]]

H + HF: Reaction is endothermic as H-H bond is weaker than H-F bond. As seen in the picture below the energy of the products is greater than that of reactants. This can be found by inputting the output values from the previous computation

[[File:H and HF JN.PNG]]

 Locate the approximate position of the transition state. 

Hammond's postulate states that the energy of the TS can be approximated as being that of the state nearest to it, because H2 + F is exothermic that means it is nearer the reactants therefore by inputting the distance values of the reactants and making the momenta = 0 the transition state can be found. AB = 0.74 and BC = 1.815. The energy of this state is -103.743.
Therefore the activation energy of:
H2 + F: 1.2
HF + H: 30

[[File:JN TS NEW.PNG]]

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

The momenta of the system increases after the reaction compared to before it. This can be seen in the animation of the interaction. Therefore the energy released in this reaction is converted to kinetic energy as the relationship between momentum and energy is Ek = p^2 / 2m. By measuring the temperature before and after and using the relationship Ek = 3/2*kT this can be confirmed empirically.

[[File:JN pbeforeafter.PNG]]
[[File:JN animation.PNG]]

 Setup a 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 (explore values also close to these limits). What do you observe? 

Despite the fact that there is sufficient energy in H-H to overcome the activation energy the H-H molecule collides and bounces off of the fluorine atom when P(H-H) is -3. This can be rationalized as an H atom having enough energy to reach the bottom of the H-F potential well and then exit it again, the same is true for the momentum being +3.

 For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now? 
Under these conditions the reaction proceeds successfully

 for H + HF Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy). 

Under these conditions the incoming high energy Hydrogen collides with the HF molecule and bounces off in the direction it came from taking the other Hydrogen with it, in this way the reaction proceeds.

 Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.) 
Using the inversion of momentum procedure it can be found that the the highest p(HF) and lowest p(HH) that gives a reactive trajectory is 2 and -6.8 respectively. The result of this is pictured below.

[[File:JN traject.PNG]]

 The cases studied are an illustration of Polanyi's empirical rules. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. 

Polanyi's emperical rules state that for an early barrier reactions the translational energy mode is more efficient than vibrational energy in achieving a reactive trajectory, while the opposite is true for late barrier reactions [1] . Exothermic reactions have early transition states while endothermic reactions have late transition state. With that in mind the reaction of F with H-H which is exothermic readily becomes reactive with a small increase in translational energy whereas an increase in vibrational energy does little to progress the reaction. In contrast H with H-F responds to a small increase vibrational energy very strongly and will overcome the activation energy, whereas an increase in translational will have a less efficient effect on the system achieving a reactive trajectory. The 4 examples below demonstrate this with H + H-F and F + H-H systems:

[[File:JN idk 1.PNG]]
[[File:JN idk 2.PNG]]
[[File:JN idk 3.PNG]]
[[File:JN idk 4.PNG]]

References:
1: G. Czakó, J. M. Bowman, Science 2011, 334(6054), 343–346

File:JN idk 6.PNG

2018-05-11T16:03:12Z

Jn1316:

JN1316Lab2

2018-05-11T15:56:45Z

Jn1316:

All distances are in Angstroms and energies in kCal/mol

== Exercise 1: H + H 2 ==

 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. 

The potential energy along a minima line has dV/dr = 0 and because it is a maxima d^2V/dr^2 is <0. A transition state also has dV/dr = 0 however d^2V/dr^2 is in contrast >0.

 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. 

running an MEP calculation with both bond lengths set arbitrarily to 1 and exporting the data finds the saddle point to be where both bond lengths are 0.907741.

[[File:IND-T.PNG]]

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

MEP plots correspond to infinitely slow motion with the velocity resetting to 0 in each step. This is not the case in dynamic plots. This explains why in MEP the molecular bond length does not vibrate and the free hydrogen atom does not escape to infinity but rather it's distance from the other two atoms plateaus.

[[File:JN Dyncalc.PNG]]
[[File:JN MEPcalc.PNG]]

 For the initial positions r1 = 0.74 Angstroms and r2 = 2.0 Angstroms, run trajectories with the following momenta combination: 

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy kCal/mol !! Reactive? !! Picture !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:JN TS 1.PNG]] || A successful reaction with a vibrating reactant transferring vibrational energy to a vibrating product.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:JN TS 2.PNG]] || There is not sufficient total energy for the reactants to overcome the activation energy.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:JN TS 3.PNG]] || This simple reactive trajectory passes once simply over the transition state.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:JN TS 4.PNG]] || This system has sufficient energy to overcome the activation energy. it does so but then 'bounces' of a steep wall of the potential well back to the reactants.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:JN TS 5.PNG]] || The transition state theory assumes the potential barrier cannot be recrossed. This example violates that assumption as the potential barrier is crossed 3 times
|}

 State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? 
The main assumption to the TST is that once you have left the maxima that is the transition state you cannot go back to it or past it. This assumption fails as we have seen in the above trajectories some of the minimum energy paths reach the transition state and then return backwards. The TST also neglects the effects of quantum mechanics. Particularly for light particles such as hydrogen the non-zero probability of tunneling through a classically forbidden potential barrier with quantum potential is significant.

 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 as H-F bond is stronger than H-H bond. As seen in the picture below the energy of the products is less than that of reactants.
[[File:F and H2 JN.PNG]]

H + HF: Reaction is endothermic as H-H bond is weaker than H-F bond. As seen in the picture below the energy of the products is greater than that of reactants. This can be found by inputting the output values from the previous computation

[[File:H and HF JN.PNG]]

 Locate the approximate position of the transition state. 

Hammond's postulate states that the energy of the TS can be approximated as being that of the state nearest to it, because H2 + F is exothermic that means it is nearer the reactants therefore by inputting the distance values of the reactants and making the momenta = 0 the transition state can be found. AB = 0.74 and BC = 1.815. The energy of this state is -103.743.
Therefore the activation energy of:
H2 + F: 1.2
HF + H: 30

[[File:JN TS NEW.PNG]]

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

The momenta of the system increases after the reaction compared to before it. This can be seen in the animation of the interaction. Therefore the energy released in this reaction is converted to kinetic energy as the relationship between momentum and energy is Ek = p^2 / 2m. By measuring the temperature before and after and using the relationship Ek = 3/2*kT this can be confirmed empirically.

[[File:JN pbeforeafter.PNG]]
[[File:JN animation.PNG]]

 Setup a 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 (explore values also close to these limits). What do you observe? 

Despite the fact that there is sufficient energy in H-H to overcome the activation energy the H-H molecule collides and bounces off of the fluorine atom when P(H-H) is -3. This can be rationalized as an H atom having enough energy to reach the bottom of the H-F potential well and then exit it again, the same is true for the momentum being +3.

 For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now? 
Under these conditions the reaction proceeds successfully

 for H + HF Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy). 

Under these conditions the incoming high energy Hydrogen collides with the HF molecule and bounces off in the direction it came from taking the other Hydrogen with it, in this way the reaction proceeds.

 Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.) 
Using the inversion of momentum procedure it can be found that the the highest p(HF) and lowest p(HH) that gives a reactive trajectory is 2 and -6.8 respectively. The result of this is pictured below.

[[File:JN traject.PNG]]

 The cases studied are an illustration of Polanyi's empirical rules. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. 

Polanyi's emperical rules state that for an early barrier reactions the translational energy mode is more efficient than vibrational energy in achieving a reactive trajectory, while the opposite is true for late barrier reactions [1] . Exothermic reactions have early transition states while endothermic reactions have late transition state. With that in mind the reaction of F with H-H which is exothermic readily becomes reactive with a small increase in translational energy whereas an increase in vibrational energy does little to progress the reaction. In contrast H with H-F responds to a small increase vibrational energy very strongly and will overcome the activation energy, whereas an increase in translational will have a less efficient effect on the system achieving a reactive trajectory. The 4 examples below demonstrate this with H + H-F and F + H-H systems:

[[File:JN idk 1.PNG]]
[[File:JN idk 2.PNG]]
[[File:JN idk 3.PNG]]
[[File:JN idk 4.PNG]]

References:
1: G. Czakó, J. M. Bowman, Science 2011, 334(6054), 343–346

JN1316Lab2

2018-05-11T15:43:41Z

Jn1316:

== Exercise 1: H + H 2 ==

 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. 

The potential energy along a minima line has dV/dr = 0 and because it is a maxima d^2V/dr^2 is <0. A transition state also has dV/dr = 0 however d^2V/dr^2 is in contrast >0.

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

The internuclear distance - time graph shows the A-B bond length = B-C bond length = a sinusoidal wave from 1.03 Angstroms to 0.773 Angstroms therefore the average is 0.88 Angstroms

[[File:IND-T.PNG]]

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

MEP plots correspond to infinitely slow motion with the velocity resetting to 0 in each step. This is not the case in dynamic plots. This explains why in MEP the molecular bond length does not vibrate and the free hydrogen atom does not escape to infinity but rather it's distance from the other two atoms plateaus.

[[File:JN Dyncalc.PNG]]
[[File:JN MEPcalc.PNG]]

 For the initial positions r1 = 0.74 Angstroms and r2 = 2.0 Angstroms, run trajectories with the following momenta combination: 

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy kCal/mol !! Reactive? !! Picture !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:JN TS 1.PNG]] || A successful reaction with a vibrating reactant transferring vibrational energy to a vibrating product.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:JN TS 2.PNG]] || There is not sufficient total energy for the reactants to overcome the activation energy.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:JN TS 3.PNG]] || This simple reactive trajectory passes once simply over the transition state.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:JN TS 4.PNG]] || This system has sufficient energy to overcome the activation energy. it does so but then 'bounces' of a steep wall of the potential well back to the reactants.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:JN TS 5.PNG]] || The transition state theory assumes the potential barrier cannot be recrossed. This example violates that assumption as the potential barrier is crossed 3 times
|}

 State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? 
The main assumption to the TST is that once you have left the maxima that is the transition state you cannot go back to it or past it. This assumption fails as we have seen in the above trajectories some of the minimum energy paths reach the transition state and then return backwards. The TST also neglects the effects of quantum mechanics. Particularly for light particles such as hydrogen the non-zero probability of tunneling through a classically forbidden potential barrier with quantum potential is significant.

 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 as H-F bond is stronger than H-H bond. As seen in the picture below the energy of the products is less than that of reactants.
[[File:F and H2 JN.PNG]]

H + HF: Reaction is endothermic as H-H bond is weaker than H-F bond. As seen in the picture below the energy of the products is greater than that of reactants. This can be found by inputting the output values from the previous computation

[[File:H and HF JN.PNG]]

 Locate the approximate position of the transition state. 

Hammond's postulate states that the energy of the TS can be approximated as being that of the state nearest to it, because H2 + F is exothermic that means it is nearer the reactants therefore by inputting the distance values of the reactants and making the momenta = 0 the transition state can be found. AB = 0.74 and BC = 1.815. The energy of this state is -103.743.
Therefore the activation energy of:
H2 + F: 1.2
HF + H: 30

[[File:JN TS NEW.PNG]]

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

The momenta of the system increases after the reaction compared to before it. This can be seen in the animation of the interaction. Therefore the energy released in this reaction is converted to kinetic energy as the relationship between momentum and energy is Ek = p^2 / 2m. By measuring the temperature before and after and using the relationship Ek = 3/2*kT this can be confirmed empirically.

[[File:JN pbeforeafter.PNG]]
[[File:JN animation.PNG]]

 Setup a 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 (explore values also close to these limits). What do you observe? 

Despite the fact that there is sufficient energy in H-H to overcome the activation energy the H-H molecule collides and bounces off of the fluorine atom when P(H-H) is -3. This can be rationalized as an H atom having enough energy to reach the bottom of the H-F potential well and then exit it again, the same is true for the momentum being +3.

 For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now? 
Under these conditions the reaction proceeds successfully

 for H + HF Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy). 

Under these conditions the incoming high energy Hydrogen collides with the HF molecule and bounces off in the direction it came from taking the other Hydrogen with it, in this way the reaction proceeds.

 Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.) 
Using the inversion of momentum procedure it can be found that the the highest p(HF) and lowest p(HH) that gives a reactive trajectory is 2 and -6.8 respectively. The result of this is pictured below.

[[File:JN traject.PNG]]

 The cases studied are an illustration of Polanyi's empirical rules. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. 

Polanyi's emperical rules state that for an early barrier reactions the translational energy mode is more efficient than vibrational energy in achieving a reactive trajectory, while the opposite is true for late barrier reactions [1] . Exothermic reactions have early transition states while endothermic reactions have late transition state. With that in mind the reaction of F with H-H which is exothermic readily becomes reactive with a small increase in translational energy whereas an increase in vibrational energy does little to progress the reaction. In contrast H with H-F responds to a small increase vibrational energy very strongly and will overcome the activation energy, whereas an increase in translational will have a less efficient effect on the system achieving a reactive trajectory. The 4 examples below demonstrate this with H + H-F and F + H-H systems:

[[File:JN idk 1.PNG]]
[[File:JN idk 2.PNG]]
[[File:JN idk 3.PNG]]
[[File:JN idk 4.PNG]]

References:
1: G. Czakó, J. M. Bowman, Science 2011, 334(6054), 343–346

JN1316Lab2

2018-05-11T15:43:08Z

Jn1316:

<title> Exercise 1: H + H 2 </title>

 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. 

The potential energy along a minima line has dV/dr = 0 and because it is a maxima d^2V/dr^2 is <0. A transition state also has dV/dr = 0 however d^2V/dr^2 is in contrast >0.

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

The internuclear distance - time graph shows the A-B bond length = B-C bond length = a sinusoidal wave from 1.03 Angstroms to 0.773 Angstroms therefore the average is 0.88 Angstroms

[[File:IND-T.PNG]]

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

MEP plots correspond to infinitely slow motion with the velocity resetting to 0 in each step. This is not the case in dynamic plots. This explains why in MEP the molecular bond length does not vibrate and the free hydrogen atom does not escape to infinity but rather it's distance from the other two atoms plateaus.

[[File:JN Dyncalc.PNG]]
[[File:JN MEPcalc.PNG]]

 For the initial positions r1 = 0.74 Angstroms and r2 = 2.0 Angstroms, run trajectories with the following momenta combination: 

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy kCal/mol !! Reactive? !! Picture !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:JN TS 1.PNG]] || A successful reaction with a vibrating reactant transferring vibrational energy to a vibrating product.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:JN TS 2.PNG]] || There is not sufficient total energy for the reactants to overcome the activation energy.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:JN TS 3.PNG]] || This simple reactive trajectory passes once simply over the transition state.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:JN TS 4.PNG]] || This system has sufficient energy to overcome the activation energy. it does so but then 'bounces' of a steep wall of the potential well back to the reactants.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:JN TS 5.PNG]] || The transition state theory assumes the potential barrier cannot be recrossed. This example violates that assumption as the potential barrier is crossed 3 times
|}

 State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? 
The main assumption to the TST is that once you have left the maxima that is the transition state you cannot go back to it or past it. This assumption fails as we have seen in the above trajectories some of the minimum energy paths reach the transition state and then return backwards. The TST also neglects the effects of quantum mechanics. Particularly for light particles such as hydrogen the non-zero probability of tunneling through a classically forbidden potential barrier with quantum potential is significant.

 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 as H-F bond is stronger than H-H bond. As seen in the picture below the energy of the products is less than that of reactants.
[[File:F and H2 JN.PNG]]

H + HF: Reaction is endothermic as H-H bond is weaker than H-F bond. As seen in the picture below the energy of the products is greater than that of reactants. This can be found by inputting the output values from the previous computation

[[File:H and HF JN.PNG]]

 Locate the approximate position of the transition state. 

Hammond's postulate states that the energy of the TS can be approximated as being that of the state nearest to it, because H2 + F is exothermic that means it is nearer the reactants therefore by inputting the distance values of the reactants and making the momenta = 0 the transition state can be found. AB = 0.74 and BC = 1.815. The energy of this state is -103.743.
Therefore the activation energy of:
H2 + F: 1.2
HF + H: 30

[[File:JN TS NEW.PNG]]

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

The momenta of the system increases after the reaction compared to before it. This can be seen in the animation of the interaction. Therefore the energy released in this reaction is converted to kinetic energy as the relationship between momentum and energy is Ek = p^2 / 2m. By measuring the temperature before and after and using the relationship Ek = 3/2*kT this can be confirmed empirically.

[[File:JN pbeforeafter.PNG]]
[[File:JN animation.PNG]]

 Setup a 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 (explore values also close to these limits). What do you observe? 

Despite the fact that there is sufficient energy in H-H to overcome the activation energy the H-H molecule collides and bounces off of the fluorine atom when P(H-H) is -3. This can be rationalized as an H atom having enough energy to reach the bottom of the H-F potential well and then exit it again, the same is true for the momentum being +3.

 For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now? 
Under these conditions the reaction proceeds successfully

 for H + HF Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy). 

Under these conditions the incoming high energy Hydrogen collides with the HF molecule and bounces off in the direction it came from taking the other Hydrogen with it, in this way the reaction proceeds.

 Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.) 
Using the inversion of momentum procedure it can be found that the the highest p(HF) and lowest p(HH) that gives a reactive trajectory is 2 and -6.8 respectively. The result of this is pictured below.

[[File:JN traject.PNG]]

 The cases studied are an illustration of Polanyi's empirical rules. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. 

Polanyi's emperical rules state that for an early barrier reactions the translational energy mode is more efficient than vibrational energy in achieving a reactive trajectory, while the opposite is true for late barrier reactions [1] . Exothermic reactions have early transition states while endothermic reactions have late transition state. With that in mind the reaction of F with H-H which is exothermic readily becomes reactive with a small increase in translational energy whereas an increase in vibrational energy does little to progress the reaction. In contrast H with H-F responds to a small increase vibrational energy very strongly and will overcome the activation energy, whereas an increase in translational will have a less efficient effect on the system achieving a reactive trajectory. The 4 examples below demonstrate this with H + H-F and F + H-H systems:

[[File:JN idk 1.PNG]]
[[File:JN idk 2.PNG]]
[[File:JN idk 3.PNG]]
[[File:JN idk 4.PNG]]

References:
1: G. Czakó, J. M. Bowman, Science 2011, 334(6054), 343–346

JN1316Lab2

2018-05-11T15:42:58Z

Jn1316:

<title> Exercise 1: H + H 2

 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. 

The potential energy along a minima line has dV/dr = 0 and because it is a maxima d^2V/dr^2 is <0. A transition state also has dV/dr = 0 however d^2V/dr^2 is in contrast >0.

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

The internuclear distance - time graph shows the A-B bond length = B-C bond length = a sinusoidal wave from 1.03 Angstroms to 0.773 Angstroms therefore the average is 0.88 Angstroms

[[File:IND-T.PNG]]

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

MEP plots correspond to infinitely slow motion with the velocity resetting to 0 in each step. This is not the case in dynamic plots. This explains why in MEP the molecular bond length does not vibrate and the free hydrogen atom does not escape to infinity but rather it's distance from the other two atoms plateaus.

[[File:JN Dyncalc.PNG]]
[[File:JN MEPcalc.PNG]]

 For the initial positions r1 = 0.74 Angstroms and r2 = 2.0 Angstroms, run trajectories with the following momenta combination: 

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy kCal/mol !! Reactive? !! Picture !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:JN TS 1.PNG]] || A successful reaction with a vibrating reactant transferring vibrational energy to a vibrating product.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:JN TS 2.PNG]] || There is not sufficient total energy for the reactants to overcome the activation energy.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:JN TS 3.PNG]] || This simple reactive trajectory passes once simply over the transition state.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:JN TS 4.PNG]] || This system has sufficient energy to overcome the activation energy. it does so but then 'bounces' of a steep wall of the potential well back to the reactants.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:JN TS 5.PNG]] || The transition state theory assumes the potential barrier cannot be recrossed. This example violates that assumption as the potential barrier is crossed 3 times
|}

 State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? 
The main assumption to the TST is that once you have left the maxima that is the transition state you cannot go back to it or past it. This assumption fails as we have seen in the above trajectories some of the minimum energy paths reach the transition state and then return backwards. The TST also neglects the effects of quantum mechanics. Particularly for light particles such as hydrogen the non-zero probability of tunneling through a classically forbidden potential barrier with quantum potential is significant.

 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 as H-F bond is stronger than H-H bond. As seen in the picture below the energy of the products is less than that of reactants.
[[File:F and H2 JN.PNG]]

H + HF: Reaction is endothermic as H-H bond is weaker than H-F bond. As seen in the picture below the energy of the products is greater than that of reactants. This can be found by inputting the output values from the previous computation

[[File:H and HF JN.PNG]]

 Locate the approximate position of the transition state. 

Hammond's postulate states that the energy of the TS can be approximated as being that of the state nearest to it, because H2 + F is exothermic that means it is nearer the reactants therefore by inputting the distance values of the reactants and making the momenta = 0 the transition state can be found. AB = 0.74 and BC = 1.815. The energy of this state is -103.743.
Therefore the activation energy of:
H2 + F: 1.2
HF + H: 30

[[File:JN TS NEW.PNG]]

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

The momenta of the system increases after the reaction compared to before it. This can be seen in the animation of the interaction. Therefore the energy released in this reaction is converted to kinetic energy as the relationship between momentum and energy is Ek = p^2 / 2m. By measuring the temperature before and after and using the relationship Ek = 3/2*kT this can be confirmed empirically.

[[File:JN pbeforeafter.PNG]]
[[File:JN animation.PNG]]

 Setup a 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 (explore values also close to these limits). What do you observe? 

Despite the fact that there is sufficient energy in H-H to overcome the activation energy the H-H molecule collides and bounces off of the fluorine atom when P(H-H) is -3. This can be rationalized as an H atom having enough energy to reach the bottom of the H-F potential well and then exit it again, the same is true for the momentum being +3.

 For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now? 
Under these conditions the reaction proceeds successfully

 for H + HF Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy). 

Under these conditions the incoming high energy Hydrogen collides with the HF molecule and bounces off in the direction it came from taking the other Hydrogen with it, in this way the reaction proceeds.

 Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.) 
Using the inversion of momentum procedure it can be found that the the highest p(HF) and lowest p(HH) that gives a reactive trajectory is 2 and -6.8 respectively. The result of this is pictured below.

[[File:JN traject.PNG]]

 The cases studied are an illustration of Polanyi's empirical rules. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. 

Polanyi's emperical rules state that for an early barrier reactions the translational energy mode is more efficient than vibrational energy in achieving a reactive trajectory, while the opposite is true for late barrier reactions [1] . Exothermic reactions have early transition states while endothermic reactions have late transition state. With that in mind the reaction of F with H-H which is exothermic readily becomes reactive with a small increase in translational energy whereas an increase in vibrational energy does little to progress the reaction. In contrast H with H-F responds to a small increase vibrational energy very strongly and will overcome the activation energy, whereas an increase in translational will have a less efficient effect on the system achieving a reactive trajectory. The 4 examples below demonstrate this with H + H-F and F + H-H systems:

[[File:JN idk 1.PNG]]
[[File:JN idk 2.PNG]]
[[File:JN idk 3.PNG]]
[[File:JN idk 4.PNG]]

References:
1: G. Czakó, J. M. Bowman, Science 2011, 334(6054), 343–346

JN1316Lab2

2018-05-11T15:40:58Z

Jn1316:

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. 

The potential energy along a minima line has dV/dr = 0 and because it is a maxima d^2V/dr^2 is <0. A transition state also has dV/dr = 0 however d^2V/dr^2 is in contrast >0.

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

The internuclear distance - time graph shows the A-B bond length = B-C bond length = a sinusoidal wave from 1.03 Angstroms to 0.773 Angstroms therefore the average is 0.88 Angstroms

[[File:IND-T.PNG]]

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

MEP plots correspond to infinitely slow motion with the velocity resetting to 0 in each step. This is not the case in dynamic plots. This explains why in MEP the molecular bond length does not vibrate and the free hydrogen atom does not escape to infinity but rather it's distance from the other two atoms plateaus.

[[File:JN Dyncalc.PNG]]
[[File:JN MEPcalc.PNG]]

 For the initial positions r1 = 0.74 Angstroms and r2 = 2.0 Angstroms, run trajectories with the following momenta combination: 

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy kCal/mol !! Reactive? !! Picture !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:JN TS 1.PNG]] || A successful reaction with a vibrating reactant transferring vibrational energy to a vibrating product.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:JN TS 2.PNG]] || There is not sufficient total energy for the reactants to overcome the activation energy.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:JN TS 3.PNG]] || This simple reactive trajectory passes once simply over the transition state.
|-
| -2.5 || -5.0 || -84.956 || No || [[File:JN TS 4.PNG]] || This system has sufficient energy to overcome the activation energy. it does so but then 'bounces' of a steep wall of the potential well back to the reactants.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:JN TS 5.PNG]] || The transition state theory assumes the potential barrier cannot be recrossed. This example violates that assumption as the potential barrier is crossed 3 times
|}

 State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? 
The main assumption to the TST is that once you have left the maxima that is the transition state you cannot go back to it or past it. This assumption fails as we have seen in the above trajectories some of the minimum energy paths reach the transition state and then return backwards. The TST also neglects the effects of quantum mechanics. Particularly for light particles such as hydrogen the non-zero probability of tunneling through a classically forbidden potential barrier with quantum potential is significant.

 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 as H-F bond is stronger than H-H bond. As seen in the picture below the energy of the products is less than that of reactants.
[[File:F and H2 JN.PNG]]

H + HF: Reaction is endothermic as H-H bond is weaker than H-F bond. As seen in the picture below the energy of the products is greater than that of reactants. This can be found by inputting the output values from the previous computation

[[File:H and HF JN.PNG]]

 Locate the approximate position of the transition state. 

Hammond's postulate states that the energy of the TS can be approximated as being that of the state nearest to it, because H2 + F is exothermic that means it is nearer the reactants therefore by inputting the distance values of the reactants and making the momenta = 0 the transition state can be found. AB = 0.74 and BC = 1.815. The energy of this state is -103.743.
Therefore the activation energy of:
H2 + F: 1.2
HF + H: 30

[[File:JN TS NEW.PNG]]

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

The momenta of the system increases after the reaction compared to before it. This can be seen in the animation of the interaction. Therefore the energy released in this reaction is converted to kinetic energy as the relationship between momentum and energy is Ek = p^2 / 2m. By measuring the temperature before and after and using the relationship Ek = 3/2*kT this can be confirmed empirically.

[[File:JN pbeforeafter.PNG]]
[[File:JN animation.PNG]]

 Setup a 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 (explore values also close to these limits). What do you observe? 

Despite the fact that there is sufficient energy in H-H to overcome the activation energy the H-H molecule collides and bounces off of the fluorine atom when P(H-H) is -3. This can be rationalized as an H atom having enough energy to reach the bottom of the H-F potential well and then exit it again, the same is true for the momentum being +3.

 For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now? 
Under these conditions the reaction proceeds successfully

 for H + HF Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy). 

Under these conditions the incoming high energy Hydrogen collides with the HF molecule and bounces off in the direction it came from taking the other Hydrogen with it, in this way the reaction proceeds.

 Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.) 
Using the inversion of momentum procedure it can be found that the the highest p(HF) and lowest p(HH) that gives a reactive trajectory is 2 and -6.8 respectively. The result of this is pictured below.

[[File:JN traject.PNG]]

 The cases studied are an illustration of Polanyi's empirical rules. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. 

Polanyi's emperical rules state that for an early barrier reactions the translational energy mode is more efficient than vibrational energy in achieving a reactive trajectory, while the opposite is true for late barrier reactions [1] . Exothermic reactions have early transition states while endothermic reactions have late transition state. With that in mind the reaction of F with H-H which is exothermic readily becomes reactive with a small increase in translational energy whereas an increase in vibrational energy does little to progress the reaction. In contrast H with H-F responds to a small increase vibrational energy very strongly and will overcome the activation energy, whereas an increase in translational will have a less efficient effect on the system achieving a reactive trajectory. The 4 examples below demonstrate this with H + H-F and F + H-H systems:

[[File:JN idk 1.PNG]]
[[File:JN idk 2.PNG]]
[[File:JN idk 3.PNG]]
[[File:JN idk 4.PNG]]

References:
1: G. Czakó, J. M. Bowman, Science 2011, 334(6054), 343–346

JN1316Lab2

2018-05-11T15:31:42Z

Jn1316:

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. 

The potential energy along a minima line has dV/dr = 0 and because it is a maxima d^2V/dr^2 is <0. A transition state also has dV/dr = 0 however d^2V/dr^2 is in contrast >0.

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

The internuclear distance - time graph shows the A-B bond length = B-C bond length = a sinusoidal wave from 1.03 Angstroms to 0.773 Angstroms therefore the average is 0.88 Angstroms

[[File:IND-T.PNG]]

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

MEP plots correspond to infinitely slow motion with the velocity resetting to 0 in each step. This is not the case in dynamic plots. This explains why in MEP the molecular bond length does not vibrate and the free hydrogen atom does not escape to infinity but rather it's distance from the other two atoms plateaus.

[[File:JN Dyncalc.PNG]]
[[File:JN MEPcalc.PNG]]

 For the initial positions r1 = 0.74 Angstroms and r2 = 2.0 Angstroms, run trajectories with the following momenta combination: 

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy kCal/mol !! Reactive? !! Picture !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:JN TS 1.PNG]] ||
|-
| -1.5 || -2.0 || -100.456 || No || [[File:JN TS 2.PNG]] ||
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:JN TS 3.PNG]] ||
|-
| -2.5 || -5.0 || -84.956 || No || [[File:JN TS 4.PNG]] ||
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:JN TS 5.PNG]] ||
|}

 State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? 
The main assumption to the TST is that once you have left the maxima that is the transition state you cannot go back to it or past it. This assumption fails as we have seen in the above trajectories some of the minimum energy paths reach the transition state and then return backwards. The TST also neglects the effects of quantum mechanics. Particularly for light particles such as hydrogen the non-zero probability of tunneling through a classically forbidden potential barrier with quantum potential is significant.

 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 as H-F bond is stronger than H-H bond. As seen in the picture below the energy of the products is less than that of reactants.
[[File:F and H2 JN.PNG]]

H + HF: Reaction is endothermic as H-H bond is weaker than H-F bond. As seen in the picture below the energy of the products is greater than that of reactants. This can be found by inputting the output values from the previous computation

[[File:H and HF JN.PNG]]

 Locate the approximate position of the transition state. 

Hammond's postulate states that the energy of the TS can be approximated as being that of the state nearest to it, because H2 + F is exothermic that means it is nearer the reactants therefore by inputting the distance values of the reactants and making the momenta = 0 the transition state can be found. AB = 0.74 and BC = 1.815. The energy of this state is -103.743.
Therefore the activation energy of:
H2 + F: 1.2
HF + H: 30

[[File:JN TS NEW.PNG]]

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

The momenta of the system increases after the reaction compared to before it. This can be seen in the animation of the interaction. Therefore the energy released in this reaction is converted to kinetic energy as the relationship between momentum and energy is Ek = p^2 / 2m. By measuring the temperature before and after and using the relationship Ek = 3/2*kT this can be confirmed empirically.

[[File:JN pbeforeafter.PNG]]
[[File:JN animation.PNG]]

 Setup a 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 (explore values also close to these limits). What do you observe? 

Despite the fact that there is sufficient energy in H-H to overcome the activation energy the H-H molecule collides and bounces off of the fluorine atom when P(H-H) is -3. This can be rationalized as an H atom having enough energy to reach the bottom of the H-F potential well and then exit it again, the same is true for the momentum being +3.

 For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now? 
Under these conditions the reaction proceeds successfully

 for H + HF Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy). 

Under these conditions the incoming high energy Hydrogen collides with the HF molecule and bounces off in the direction it came from taking the other Hydrogen with it, in this way the reaction proceeds.

 Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.) 
Using the inversion of momentum procedure it can be found that the the highest p(HF) and lowest p(HH) that gives a reactive trajectory is 2 and -6.8 respectively. The result of this is pictured below.

[[File:JN traject.PNG]]

 The cases studied are an illustration of Polanyi's empirical rules. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. 

Polanyi's emperical rules state that for an early barrier reactions the translational energy mode is more efficient than vibrational energy in achieving a reactive trajectory, while the opposite is true for late barrier reactions [1] . Exothermic reactions have early transition states while endothermic reactions have late transition state. With that in mind the reaction of F with H-H which is exothermic readily becomes reactive with a small increase in translational energy whereas an increase in vibrational energy does little to progress the reaction. In contrast H with H-F responds to a small increase vibrational energy very strongly and will overcome the activation energy, whereas an increase in translational will have a less efficient effect on the system achieving a reactive trajectory. The 4 examples below demonstrate this with H + H-F and F + H-H systems:

[[File:JN idk 1.PNG]]
[[File:JN idk 2.PNG]]
[[File:JN idk 3.PNG]]
[[File:JN idk 4.PNG]]

References:
1: G. Czakó, J. M. Bowman, Science 2011, 334(6054), 343–346

JN1316Lab2

2018-05-11T15:29:45Z

Jn1316:

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. 

The potential energy along a minima line has dV/dr = 0 and because it is a maxima d^2V/dr^2 is <0. A transition state also has dV/dr = 0 however d^2V/dr^2 is in contrast >0.

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

The internuclear distance - time graph shows the A-B bond length = B-C bond length = a sinusoidal wave from 1.03 Angstroms to 0.773 Angstroms therefore the average is 0.88 Angstroms

[[File:IND-T.PNG]]

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

MEP plots correspond to infinitely slow motion with the velocity resetting to 0 in each step. This is not the case in dynamic plots. This explains why in MEP the molecular bond length does not vibrate and the free hydrogen atom does not escape to infinity but rather it's distance from the other two atoms plateaus.

[[File:JN Dyncalc.PNG]]
[[File:JN MEPcalc.PNG]]

 For the initial positions r1 = 0.74 Angstroms and r2 = 2.0 Angstroms, run trajectories with the following momenta combination: 

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy kCal/mol !! Reactive? !! Picture !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:JN TS 1.PNG]] ||
|-
| -1.5 || -2.0 || -100.456 || No || [[File:JN TS 2.PNG]] ||
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:JN TS 3.PNG]] ||
|-
| -2.5 || -5.0 || -84.956 || No || [[File:JN TS 4.PNG]] ||
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:JN TS 5.PNG]] ||
|}

 State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? 
The main assumption to the TST is that once you have left the maxima that is the transition state you cannot go back to it. This assumption fails as we have seen in the above trajectories some of the minimum energy paths reach the transition state and then return backwards.

 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 as H-F bond is stronger than H-H bond. As seen in the picture below the energy of the products is less than that of reactants.
[[File:F and H2 JN.PNG]]

H + HF: Reaction is endothermic as H-H bond is weaker than H-F bond. As seen in the picture below the energy of the products is greater than that of reactants. This can be found by inputting the output values from the previous computation

[[File:H and HF JN.PNG]]

 Locate the approximate position of the transition state. 

Hammond's postulate states that the energy of the TS can be approximated as being that of the state nearest to it, because H2 + F is exothermic that means it is nearer the reactants therefore by inputting the distance values of the reactants and making the momenta = 0 the transition state can be found. AB = 0.74 and BC = 1.815. The energy of this state is -103.743.
Therefore the activation energy of:
H2 + F: 1.2
HF + H: 30

[[File:JN TS NEW.PNG]]

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

The momenta of the system increases after the reaction compared to before it. This can be seen in the animation of the interaction. Therefore the energy released in this reaction is converted to kinetic energy as the relationship between momentum and energy is Ek = p^2 / 2m. By measuring the temperature before and after and using the relationship Ek = 3/2*kT this can be confirmed empirically.

[[File:JN pbeforeafter.PNG]]
[[File:JN animation.PNG]]

 Setup a 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 (explore values also close to these limits). What do you observe? 

Despite the fact that there is sufficient energy in H-H to overcome the activation energy the H-H molecule collides and bounces off of the fluorine atom when P(H-H) is -3. This can be rationalized as an H atom having enough energy to reach the bottom of the H-F potential well and then exit it again, the same is true for the momentum being +3.

 For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now? 
Under these conditions the reaction proceeds successfully

 for H + HF Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy). 

Under these conditions the incoming high energy Hydrogen collides with the HF molecule and bounces off in the direction it came from taking the other Hydrogen with it, in this way the reaction proceeds.

 Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.) 
Using the inversion of momentum procedure it can be found that the the highest p(HF) and lowest p(HH) that gives a reactive trajectory is 2 and -6.8 respectively. The result of this is pictured below.

[[File:JN traject.PNG]]

 The cases studied are an illustration of Polanyi's empirical rules. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. 

Polanyi's emperical rules state that for an early barrier reactions the translational energy mode is more efficient than vibrational energy in achieving a reactive trajectory, while the opposite is true for late barrier reactions [1] . Exothermic reactions have early transition states while endothermic reactions have late transition state. With that in mind the reaction of F with H-H which is exothermic readily becomes reactive with a small increase in translational energy whereas an increase in vibrational energy does little to progress the reaction. In contrast H with H-F responds to a small increase vibrational energy very strongly and will overcome the activation energy, whereas an increase in translational will have a less efficient effect on the system achieving a reactive trajectory. The 4 examples below demonstrate this with H + H-F and F + H-H systems:

[[File:JN idk 1.PNG]]
[[File:JN idk 2.PNG]]
[[File:JN idk 3.PNG]]
[[File:JN idk 4.PNG]]

References:
1: G. Czakó, J. M. Bowman, Science 2011, 334(6054), 343–346

File:JN idk 4.PNG

2018-05-11T15:28:05Z

Jn1316:

File:JN idk 3.PNG

2018-05-11T15:28:04Z

Jn1316:

File:JN idk 2.PNG

2018-05-11T15:28:04Z

Jn1316:

File:JN idk 1.PNG

2018-05-11T15:28:03Z

Jn1316:

JN1316Lab2

2018-05-11T15:13:34Z

Jn1316:

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. 

The potential energy along a minima line has dV/dr = 0 and because it is a maxima d^2V/dr^2 is <0. A transition state also has dV/dr = 0 however d^2V/dr^2 is in contrast >0.

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

The internuclear distance - time graph shows the A-B bond length = B-C bond length = a sinusoidal wave from 1.03 Angstroms to 0.773 Angstroms therefore the average is 0.88 Angstroms

[[File:IND-T.PNG]]

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

MEP plots correspond to infinitely slow motion with the velocity resetting to 0 in each step. This is not the case in dynamic plots. This explains why in MEP the molecular bond length does not vibrate and the free hydrogen atom does not escape to infinity but rather it's distance from the other two atoms plateaus.

[[File:JN Dyncalc.PNG]]
[[File:JN MEPcalc.PNG]]

 For the initial positions r1 = 0.74 Angstroms and r2 = 2.0 Angstroms, run trajectories with the following momenta combination: 

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy kCal/mol !! Reactive? !! Picture !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:JN TS 1.PNG]] ||
|-
| -1.5 || -2.0 || -100.456 || No || [[File:JN TS 2.PNG]] ||
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:JN TS 3.PNG]] ||
|-
| -2.5 || -5.0 || -84.956 || No || [[File:JN TS 4.PNG]] ||
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:JN TS 5.PNG]] ||
|}

 State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? 
The main assumption to the TST is that once you have left the maxima that is the transition state you cannot go back to it. This assumption fails as we have seen in the above trajectories some of the minimum energy paths reach the transition state and then return backwards.

 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 as H-F bond is stronger than H-H bond. As seen in the picture below the energy of the products is less than that of reactants.
[[File:F and H2 JN.PNG]]

H + HF: Reaction is endothermic as H-H bond is weaker than H-F bond. As seen in the picture below the energy of the products is greater than that of reactants. This can be found by inputting the output values from the previous computation

[[File:H and HF JN.PNG]]

 Locate the approximate position of the transition state. 

Hammond's postulate states that the energy of the TS can be approximated as being that of the state nearest to it, because H2 + F is exothermic that means it is nearer the reactants therefore by inputting the distance values of the reactants and making the momenta = 0 the transition state can be found. AB = 0.74 and BC = 1.815. The energy of this state is -103.743.
Therefore the activation energy of:
H2 + F: 1.2
HF + H: 30

[[File:JN TS NEW.PNG]]

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

The momenta of the system increases after the reaction compared to before it. This can be seen in the animation of the interaction. Therefore the energy released in this reaction is converted to kinetic energy as the relationship between momentum and energy is Ek = p^2 / 2m. By measuring the temperature before and after and using the relationship Ek = 3/2*kT this can be confirmed empirically.

[[File:JN pbeforeafter.PNG]]
[[File:JN animation.PNG]]

 Setup a 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 (explore values also close to these limits). What do you observe? 

Despite the fact that there is sufficient energy in H-H to overcome the activation energy the H-H molecule collides and bounces off of the fluorine atom when P(H-H) is -3. This can be rationalized as an H atom having enough energy to reach the bottom of the H-F potential well and then exit it again, the same is true for the momentum being +3.

 For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now? 
Under these conditions the reaction proceeds successfully

 for H + HF Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy). 

Under these conditions the incoming high energy Hydrogen collides with the HF molecule and bounces off in the direction it came from taking the other Hydrogen with it, in this way the reaction proceeds.

 Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.) 
Using the inversion of momentum procedure it can be found that the the highest p(HF) and lowest p(HH) that gives a reactive trajectory is 2 and -6.8 respectively. The result of this is pictured below.

[[File:JN traject.PNG]]

 The cases studied are an illustration of Polanyi's empirical rules. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. 

Polanyi's emperical rules state that for an early barrier reactions the translational energy mode is more efficient than vibrational energy in achieving a reactive trajectory, while the opposite is true for late barrier reactions [1] . Exothermic reactions have early transition states while endothermic reactions have late transition state. With that in mind the reaction of F with H-H which is exothermic readily becomes reactive with a small increase in translational energy whereas an increase in vibrational energy does little to progress the reaction. In contrast H with H-F responds to a small increase vibrational energy very strongly and will overcome the activation energy, whereas an increase in translational

References:
1: G. Czakó, J. M. Bowman, Science 2011, 334(6054), 343–346

JN1316Lab2

2018-05-11T14:31:30Z

Jn1316:

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. 

The potential energy along a minima line has dV/dr = 0 and because it is a maxima d^2V/dr^2 is <0. A transition state also has dV/dr = 0 however d^2V/dr^2 is in contrast >0.

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

The internuclear distance - time graph shows the A-B bond length = B-C bond length = a sinusoidal wave from 1.03 Angstroms to 0.773 Angstroms therefore the average is 0.88 Angstroms

[[File:IND-T.PNG]]

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

MEP plots correspond to infinitely slow motion with the velocity resetting to 0 in each step. This is not the case in dynamic plots. This explains why in MEP the molecular bond length does not vibrate and the free hydrogen atom does not escape to infinity but rather it's distance from the other two atoms plateaus.

[[File:JN Dyncalc.PNG]]
[[File:JN MEPcalc.PNG]]

 For the initial positions r1 = 0.74 Angstroms and r2 = 2.0 Angstroms, run trajectories with the following momenta combination: 

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy kCal/mol !! Reactive? !! Picture !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:JN TS 1.PNG]] ||
|-
| -1.5 || -2.0 || -100.456 || No || [[File:JN TS 2.PNG]] ||
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:JN TS 3.PNG]] ||
|-
| -2.5 || -5.0 || -84.956 || No || [[File:JN TS 4.PNG]] ||
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:JN TS 5.PNG]] ||
|}

 State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? 
The main assumption to the TST is that once you have left the maxima that is the transition state you cannot go back to it. This assumption fails as we have seen in the above trajectories some of the minimum energy paths reach the transition state and then return backwards.

 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 as H-F bond is stronger than H-H bond. As seen in the picture below the energy of the products is less than that of reactants.
[[File:F and H2 JN.PNG]]

H + HF: Reaction is endothermic as H-H bond is weaker than H-F bond. As seen in the picture below the energy of the products is greater than that of reactants. This can be found by inputting the output values from the previous computation

[[File:H and HF JN.PNG]]

 Locate the approximate position of the transition state. 

Hammond's postulate states that the energy of the TS can be approximated as being that of the state nearest to it, because H2 + F is exothermic that means it is nearer the reactants therefore by inputting the distance values of the reactants and making the momenta = 0 the transition state can be found. AB = 0.74 and BC = 1.815. The energy of this state is -103.743.
Therefore the activation energy of:
H2 + F: 1.2
HF + H: 30

[[File:JN TS NEW.PNG]]

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

The momenta of the system increases after the reaction compared to before it. This can be seen in the animation of the interaction. Therefore the energy released in this reaction is converted to kinetic energy as the relationship between momentum and energy is Ek = p^2 / 2m. By measuring the temperature before and after and using the relationship Ek = 3/2*kT this can be confirmed empirically.

[[File:JN pbeforeafter.PNG]]
[[File:JN animation.PNG]]

 Setup a 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 (explore values also close to these limits). What do you observe? 

Despite the fact that there is sufficient energy in H-H to overcome the activation energy the H-H molecule collides and bounces off of the fluorine atom when P(H-H) is -3. This can be rationalized as an H atom having enough energy to reach the bottom of the H-F potential well and then exit it again, the same is true for the momentum being +3.

 For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now? 
Under these conditions the reaction proceeds successfully

 for H + HF Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy). 

Under these conditions the incoming high energy Hydrogen collides with the HF molecule and bounces off in the direction it came from taking the other Hydrogen with it, in this way the reaction proceeds.

 Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.) 
Using the inversion of momentum procedure it can be found that the the highest p(HF) and lowest p(HH) that gives a reactive trajectory is 2 and -6.8 respectively. The result of this is pictured below.

[[File:JN traject.PNG]]

 The cases studied are an illustration of Polanyi's empirical rules. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. 

Polanyi's emperical rules state that for an early barrier reactions the translational energy mode is more efficient than vibrational energy in achieving a reactive trajectory, while the opposite is true for late barrier reactions. [1] 

References:
1: G. Czakó, J. M. Bowman, Science 2011, 334(6054), 343–346

JN1316Lab2

2018-05-11T14:27:40Z

Jn1316:

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. 

The potential energy along a minima line has dV/dr = 0 and because it is a maxima d^2V/dr^2 is <0. A transition state also has dV/dr = 0 however d^2V/dr^2 is in contrast >0.

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

The internuclear distance - time graph shows the A-B bond length = B-C bond length = a sinusoidal wave from 1.03 Angstroms to 0.773 Angstroms therefore the average is 0.88 Angstroms

[[File:IND-T.PNG]]

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

MEP plots correspond to infinitely slow motion with the velocity resetting to 0 in each step. This is not the case in dynamic plots. This explains why in MEP the molecular bond length does not vibrate and the free hydrogen atom does not escape to infinity but rather it's distance from the other two atoms plateaus.

[[File:JN Dyncalc.PNG]]
[[File:JN MEPcalc.PNG]]

 For the initial positions r1 = 0.74 Angstroms and r2 = 2.0 Angstroms, run trajectories with the following momenta combination: 

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy kCal/mol !! Reactive? !! Picture !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:JN TS 1.PNG]] ||
|-
| -1.5 || -2.0 || -100.456 || No || [[File:JN TS 2.PNG]] ||
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:JN TS 3.PNG]] ||
|-
| -2.5 || -5.0 || -84.956 || No || [[File:JN TS 4.PNG]] ||
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:JN TS 5.PNG]] ||
|}

 State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? 
The main assumption to the TST is that once you have left the maxima that is the transition state you cannot go back to it. This assumption fails as we have seen in the above trajectories some of the minimum energy paths reach the transition state and then return backwards.

 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 as H-F bond is stronger than H-H bond. As seen in the picture below the energy of the products is less than that of reactants.
[[File:F and H2 JN.PNG]]

H + HF: Reaction is endothermic as H-H bond is weaker than H-F bond. As seen in the picture below the energy of the products is greater than that of reactants. This can be found by inputting the output values from the previous computation

[[File:H and HF JN.PNG]]

 Locate the approximate position of the transition state. 

Hammond's postulate states that the energy of the TS can be approximated as being that of the state nearest to it, because H2 + F is exothermic that means it is nearer the reactants therefore by inputting the distance values of the reactants and making the momenta = 0 the transition state can be found. AB = 0.74 and BC = 1.815. The energy of this state is -103.743.
Therefore the activation energy of:
H2 + F: 1.2
HF + H: 30

[[File:JN TS NEW.PNG]]

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

The momenta of the system increases after the reaction compared to before it. This can be seen in the animation of the interaction. Therefore the energy released in this reaction is converted to kinetic energy as the relationship between momentum and energy is Ek = p^2 / 2m. By measuring the temperature before and after and using the relationship Ek = 3/2*kT this can be confirmed empirically.

[[File:JN pbeforeafter.PNG]]
[[File:JN animation.PNG]]

 Setup a 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 (explore values also close to these limits). What do you observe? 

Despite the fact that there is sufficient energy in H-H to overcome the activation energy the H-H molecule collides and bounces off of the fluorine atom when P(H-H) is -3. This can be rationalized as an H atom having enough energy to reach the bottom of the H-F potential well and then exit it again, the same is true for the momentum being +3.

 For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now? 
Under these conditions the reaction proceeds successfully

 for H + HF Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy). 

Under these conditions the incoming high energy Hydrogen collides with the HF molecule and bounces off in the direction it came from taking the other Hydrogen with it, in this way the reaction proceeds.

 Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.) 
Using the inversion of momentum procedure it can be found that the the highest p(HF) and lowest p(HH) that gives a reactive trajectory is 2 and -6.8 respectively. The result of this is pictured below.

[[File:JN traject.PNG]]

 The cases studied are an illustration of Polanyi's empirical rules. Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction, and how this is influenced by the position of the transition state. 

Polanyi's emperical rules state that for an early barrier reactions the translational energy mode is more efficient than vibrational energy in achieving a reactive trajectory, while the opposite is true for late barrier reactions.

JN1316Lab2

2018-05-11T14:23:31Z

Jn1316:

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. 

The potential energy along a minima line has dV/dr = 0 and because it is a maxima d^2V/dr^2 is <0. A transition state also has dV/dr = 0 however d^2V/dr^2 is in contrast >0.

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

The internuclear distance - time graph shows the A-B bond length = B-C bond length = a sinusoidal wave from 1.03 Angstroms to 0.773 Angstroms therefore the average is 0.88 Angstroms

[[File:IND-T.PNG]]

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

MEP plots correspond to infinitely slow motion with the velocity resetting to 0 in each step. This is not the case in dynamic plots. This explains why in MEP the molecular bond length does not vibrate and the free hydrogen atom does not escape to infinity but rather it's distance from the other two atoms plateaus.

[[File:JN Dyncalc.PNG]]
[[File:JN MEPcalc.PNG]]

 For the initial positions r1 = 0.74 Angstroms and r2 = 2.0 Angstroms, run trajectories with the following momenta combination: 

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy kCal/mol !! Reactive? !! Picture !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:JN TS 1.PNG]] ||
|-
| -1.5 || -2.0 || -100.456 || No || [[File:JN TS 2.PNG]] ||
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:JN TS 3.PNG]] ||
|-
| -2.5 || -5.0 || -84.956 || No || [[File:JN TS 4.PNG]] ||
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:JN TS 5.PNG]] ||
|}

 State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? 
The main assumption to the TST is that once you have left the maxima that is the transition state you cannot go back to it. This assumption fails as we have seen in the above trajectories some of the minimum energy paths reach the transition state and then return backwards.

 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 as H-F bond is stronger than H-H bond. As seen in the picture below the energy of the products is less than that of reactants.
[[File:F and H2 JN.PNG]]

H + HF: Reaction is endothermic as H-H bond is weaker than H-F bond. As seen in the picture below the energy of the products is greater than that of reactants. This can be found by inputting the output values from the previous computation

[[File:H and HF JN.PNG]]

 Locate the approximate position of the transition state. 

Hammond's postulate states that the energy of the TS can be approximated as being that of the state nearest to it, because H2 + F is exothermic that means it is nearer the reactants therefore by inputting the distance values of the reactants and making the momenta = 0 the transition state can be found. AB = 0.74 and BC = 1.815. The energy of this state is -103.743.
Therefore the activation energy of:
H2 + F: 1.2
HF + H: 30

[[File:JN TS NEW.PNG]]

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

The momenta of the system increases after the reaction compared to before it. This can be seen in the animation of the interaction. Therefore the energy released in this reaction is converted to kinetic energy as the relationship between momentum and energy is Ek = p^2 / 2m. By measuring the temperature before and after and using the relationship Ek = 3/2*kT this can be confirmed empirically.

[[File:JN pbeforeafter.PNG]]
[[File:JN animation.PNG]]

 Setup a 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 (explore values also close to these limits). What do you observe? 

Despite the fact that there is sufficient energy in H-H to overcome the activation energy the H-H molecule collides and bounces off of the fluorine atom when P(H-H) is -3. This can be rationalized as an H atom having enough energy to reach the bottom of the H-F potential well and then exit it again, the same is true for the momentum being +3.

 For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now? 
Under these conditions the reaction proceeds successfully

 for H + HF Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy). 

Under these conditions the incoming high energy Hydrogen collides with the HF molecule and bounces off in the direction it came from taking the other Hydrogen with it, in this way the reaction proceeds.

 Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.) 
Using the inversion of momentum procedure it can be found that the the highest p(HF) and lowest p(HH) that gives a reactive trajectory is 2 and -6.8 respectively.
[[File:JN traject.PNG]]

File:JN traject.PNG

2018-05-11T14:23:07Z

Jn1316:

JN1316Lab2

2018-05-11T14:22:57Z

Jn1316:

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. 

The potential energy along a minima line has dV/dr = 0 and because it is a maxima d^2V/dr^2 is <0. A transition state also has dV/dr = 0 however d^2V/dr^2 is in contrast >0.

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

The internuclear distance - time graph shows the A-B bond length = B-C bond length = a sinusoidal wave from 1.03 Angstroms to 0.773 Angstroms therefore the average is 0.88 Angstroms

[[File:IND-T.PNG]]

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

MEP plots correspond to infinitely slow motion with the velocity resetting to 0 in each step. This is not the case in dynamic plots. This explains why in MEP the molecular bond length does not vibrate and the free hydrogen atom does not escape to infinity but rather it's distance from the other two atoms plateaus.

[[File:JN Dyncalc.PNG]]
[[File:JN MEPcalc.PNG]]

 For the initial positions r1 = 0.74 Angstroms and r2 = 2.0 Angstroms, run trajectories with the following momenta combination: 

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy kCal/mol !! Reactive? !! Picture !! Description
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:JN TS 1.PNG]] ||
|-
| -1.5 || -2.0 || -100.456 || No || [[File:JN TS 2.PNG]] ||
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:JN TS 3.PNG]] ||
|-
| -2.5 || -5.0 || -84.956 || No || [[File:JN TS 4.PNG]] ||
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:JN TS 5.PNG]] ||
|}

 State what are the main assumptions of Transition State Theory. Given the results you have obtained, how will Transition State Theory predictions for reaction rate values compare with experimental values? 
The main assumption to the TST is that once you have left the maxima that is the transition state you cannot go back to it. This assumption fails as we have seen in the above trajectories some of the minimum energy paths reach the transition state and then return backwards.

 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 as H-F bond is stronger than H-H bond. As seen in the picture below the energy of the products is less than that of reactants.
[[File:F and H2 JN.PNG]]

H + HF: Reaction is endothermic as H-H bond is weaker than H-F bond. As seen in the picture below the energy of the products is greater than that of reactants. This can be found by inputting the output values from the previous computation

[[File:H and HF JN.PNG]]

 Locate the approximate position of the transition state. 

Hammond's postulate states that the energy of the TS can be approximated as being that of the state nearest to it, because H2 + F is exothermic that means it is nearer the reactants therefore by inputting the distance values of the reactants and making the momenta = 0 the transition state can be found. AB = 0.74 and BC = 1.815. The energy of this state is -103.743.
Therefore the activation energy of:
H2 + F: 1.2
HF + H: 30

[[File:JN TS NEW.PNG]]

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

The momenta of the system increases after the reaction compared to before it. This can be seen in the animation of the interaction. Therefore the energy released in this reaction is converted to kinetic energy as the relationship between momentum and energy is Ek = p^2 / 2m. By measuring the temperature before and after and using the relationship Ek = 3/2*kT this can be confirmed empirically.

[[File:JN pbeforeafter.PNG]]
[[File:JN animation.PNG]]

 Setup a 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 (explore values also close to these limits). What do you observe? 

Despite the fact that there is sufficient energy in H-H to overcome the activation energy the H-H molecule collides and bounces off of the fluorine atom when P(H-H) is -3. This can be rationalized as an H atom having enough energy to reach the bottom of the H-F potential well and then exit it again, the same is true for the momentum being +3.

 For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now? 
Under these conditions the reaction proceeds successfully

 for H + HF Setup initial conditions starting at the bottom of the entry channel, with very low vibrational motion on on the H - F bond, and an arbitrarily high value of pHH above the activation energy (an H atom colliding with a high kinetic energy). 

Under these conditions the incoming high energy Hydrogen collides with the HF molecule and bounces off in the direction it came from taking the other Hydrogen with it, in this way the reaction proceeds.

 Try to obtain a reactive trajectory by decreasing the momentum of the incoming H atom and increasing the energy of the H - F vibration. (It may be difficult to find the initial conditions that generate a reactive trajectory for this reaction. Using the inversion of momentum procedure for a trajectory starting on the transition state can be useful in this case. Working on the Skew Plot framework could also be helpful.) 
Using the inversion of momentum procedure it can be found that the the highest p(HF) and lowest p(HH) that gives a reactive trajectory is 2 and -6.8 respectively.

JN1316Lab2

2018-05-11T13:53:43Z

Jn1316:

JN1316Lab2

2018-05-11T13:49:53Z

Jn1316:

JN1316Lab2

2018-05-11T13:28:59Z

Jn1316:

File:JN animation.PNG

2018-05-11T13:28:25Z

Jn1316:

JN1316Lab2

2018-05-11T13:28:05Z

Jn1316:

File:JN pbeforeafter.PNG

2018-05-11T13:27:24Z

Jn1316:

JN1316Lab2

2018-05-11T13:26:44Z

Jn1316:

JN1316Lab2

2018-05-10T20:45:24Z

Jn1316:

JN1316Lab2

2018-05-10T20:32:52Z

Jn1316:

File:JN TS NEW.PNG

2018-05-10T20:29:36Z

Jn1316:

JN1316Lab2

2018-05-10T20:10:37Z

Jn1316:

File:H and HF JN.PNG

2018-05-10T20:09:54Z

Jn1316:

JN1316Lab2

2018-05-10T19:51:06Z

Jn1316:

File:F and H2 JN.PNG

2018-05-10T19:50:34Z

Jn1316:

JN1316Lab2

2018-05-10T19:33:49Z

Jn1316:

JN1316Lab2

2018-05-10T19:28:23Z

Jn1316:

JN1316Lab2

2018-05-10T19:12:28Z

Jn1316:

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.

The potential energy along a minima line has dV/dr = 0 and because it is a maxima d^2V/dr^2 is <0. A transition state also has dV/dr = 0 however d^2V/dr^2 is in contrast >0.

Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
The internuclear distance - time graph shows the A-B bond length = B-C bond length = a sinusoidal wave from 1.03 Angstroms to 0.773 Angstroms therefore the average is 0.88 Angstroms

[[File:IND-T.PNG]]

Comment on how the mep and the trajectory you just calculated differ.
MEP plots correspond to infinitely slow motion with the velocity resetting to 0 in each step. This is not the case in dynamic plots. This explains why in MEP the molecular bond length does not vibrate and the free hydrogen atom does not escape to infinity but rather it's distance from the other two atoms plateaus.

[[File:JN Dyncalc.PNG]]
[[File:JN MEPcalc.PNG]]

For the initial positions r1 = 0.74 Angstroms and r2 = 2.0 Angstroms, run trajectories with the following momenta combination:

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy kCal/mol !! Reactive? !! Picture !! Description
|-
| -1.25 || -2.5 || -99.119 || Yes || [[File:JN TS 1.PNG]] ||
|-
| -1.5 || -2.0 || -99.119 || No || [[File:JN TS 2.PNG]] ||
|-
| -1.5 || -2.5 || -99.119 || Yes || [[File:JN TS 3.PNG]] ||
|-
| -2.5 || -5.0 || -99.119 || No || [[File:JN TS 4.PNG]] ||
|-
| -2.5 || -5.2 || -99.119 || Yes || [[File:JN TS 5.PNG]] ||
|}

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

The main assumption to the TST is that once you have left the maxima that is the transition state you cannot go back to it. This assumption fails as we have seen in the above trajectories some of the minimum energy paths reach the transition state and then return backwards

JN1316Lab2

2018-05-08T14:24:58Z

Jn1316:

JN1316Lab2

2018-05-08T14:22:16Z

Jn1316:

File:JN TS 5.PNG

2018-05-08T14:21:22Z

Jn1316:

File:JN TS 4.PNG

2018-05-08T14:21:21Z

Jn1316: