ChemWiki - User contributions [en]

013366322

2019-05-24T15:44:13Z

Do2617: /* Q7 - Report the activation energy for both reactions. */

== EXERCISE 1: H + H2 system ==

===='''Q1 - On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''====

On the potential energy surface, the transition state is the position along the minimum energy pathway, linking reactants and products, where potential is highest. The transition structure exists at a saddle point of the potential energy surface - that is when ∂V('''ri''')/∂('''ri''')=0. Furthermore, the 2nd partial derivatives with respect to the orthogonal vectors r1 and r2 have opposite signs; one is positive and the other negative. This tells us that this is a saddle point.

[[File:Screenshot_2019-05-21_at_14.27.22.png|400px]]

===='''Q2 - 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.'''====

{|style="margin: 0 auto;"
| [[File:Screenshot 2019-05-21 at 14.41.20.png|thumb|upright|400x400px|A graph showing Intermolecular distance Vs Time ]]
| [[File:Screenshot_2019-05-21_at_16.25.22.png|thumb|upright|400x400px|Contour Plot to find rts]]
|}

Firstly, we set AB=BC and momentum equal to 0. Then, plotting a graph of Internuclear Distance vs Time, we can find the transition state. The optimised distance is the distance at which there is no trajectory towards either reactants or products and at which oscillations are reduced producing a straight line as seen in the plot above. As the graph shows straight lines, we know there is no vibration, and the atoms stop moving. That is, potential energy is maximum and kinetic energy zero. As such, the optimised distance is found to be 0.9079 Å. Looking at the contour plot above, the transition state can also be seen in approximately the correct location, backing up the result.

===='''Q3 - Comment on how the MEP and the trajectory you just calculated differ.'''====

{|style="margin: 0 auto;"
| [[File:Screenshot_2019-05-21_at_15.06.08.png|thumb|upright|400x400px|MEP Calculation]]
| [[File:Screenshot 2019-05-21 at 15.07.56.png|thumb|upright|400x400px|Dynamics Calculation]]
|}

The MEP graph shows no molecular vibration (as velocity=0) with momentum being equal to 0 in every time step. In comparison, the dynamics plot shows the vibrational energy of the molecule as seen by the non-straight line.

==='''Q4 - Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''===

{| class="wikitable" border="1"

! p1 !! p2 !! Etot !! Reactive? !! Illustration of the trajectory !! Description of the dynamics
|-
| -1.25 || -2.5 || -99.018 || YES || [[File:Screenshot_dex_1.png|400x400px]] || Both reactants collide and have sufficient energy to surpass the activation barrier. Products are formed. Vibrational motion is also seen evident by the oscillations.
|-
| -1.5 || -2.0 || -100.456 || NO || [[File:Screenshot_dex_2.png|400x400px]] || Atom C collides molecule AB but has insufficient kinetic energy to overcome the activation energy, thus not forming products. Molecules do not pass through the transition state.
|-
| -1.5 || -2.5 || -98.956 || YES || [[File:Screenshot_dex_3.png|400x400px]] || Both molecules have sufficient energy to surpass the activation energy and so pass though the transition state, forming the products.
|-
| -2.5 || -5.0 || -84.956 || NO || [[File:Screenshot_dex_4.png|400x400px]] || Both molecules have large energies and are able to overcome the activation energy. However, their energy is so great, that they recross the barrier and reform reactants.
|-
| -2.5 || -5.2 || -83.416 || YES || [[File:Screenshot_dex_5.png|400x400px]] || Both molecules have sufficient energy to overcome the activation energy barrier, but have the potential to cross back over the barrier, and reform products.
|}

It can be concluded that if reactants do not have sufficient kinetic energy to surpass the activation barrier, then the reaction will not occur. Furthermore, if the reactants have an excess of kinetic energy, the reaction also may not proceed as the products formed turn back into reactants.

==='''Q5 - 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?'''===

Electron and nuclear motion are independent
Energy of the particles follows the Boltzmann distribution
Once reactants begin to combine and form the transition state, the transition state structure does not collapse to form reactants again

Theory states that transition state structure does not collapse to reform the reactants. However, as experimental procedure shows, this can occur. As theory assumes that all reactions with sufficient energy will go onto form products, the real rate of reaction is much slower than in theory as reactants can be reformed.

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

==='''Q6 - By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''===

{|style="margin: 0 auto;"
| [[File:Screenshot_dex_f_h2.png|thumb|upright|400x400px|Surface Plot for F + H2]]
| [[File:Screenshot_dex_h_hf.png|thumb|upright|400x400px|Surface Plot for H + HF]]
|}

As seen above, the surface plot for F + H2 starts at a high energy position dropping down lower energy after the transition state. Therefore, this is an exothermic reaction. In contrast, for the surface plot of HF + H, the reaction starts at a low energy position and needs a large amount of energy to over come the activation energy barrier. After passing the transition state, the energy falls a small amount. This reaction is therefore endothermic. From this, we can conclude that the HF bond is stronger than the HH bond since HF is lower in energy than HH.

Furthermore, Hammond's Postulate states that the transition state will resemble either the product or reactants, whichever it is closer to in energy. To find the transition state, optimisation was completed as before with both momenta set to zero. Optimisation was completed so that there was minimal oscillation (gradient of line =0) seen for the Internuclear Distance vs Time graph, as shown below. The optimised distances were found to be rts(H-F)=1.813 Å and rts(H-H)=0.7445 Å.

[[File:Screenshot_2019-05-21_at_18.21.10.png|thumb|upright|400x400px|Graph showing Internuclear Distance vs Time for F + H2]]

==='''Q7 - Report the activation energy for both reactions.'''===
Below shows two graphs depicting the total energy plot of both reactions. The activation energy for both reactions can be calculated as the difference between the highest and lowest energy levels.

{| class="wikitable"
![[File:Screenshot_2019-05-24_at_16.37.01.png|500px|left|thumb|Graph of Energy vs Steps for F + H2.]]
![[File:Screenshot_2019-05-24_at_16.41.09.png|500px|right|thumb|Graph of Energy vs Steps for HF + H.]]
|}

'''Activation Energy(H + HF)'''= +31.22 kcalmol-1
'''Activation Energy(F + H2)'''= +0.130 kcal mol-1

==='''Q8 - In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''===

For the reaction of F + H2 a reactive trajectory was found using the following set of initial conditions: AB = 0.74 Å, BC = 1.80 Å, '''pAB''' = -2.5, and '''pBC''' = -1.70. Using these intial conditions the graph of Momentum vs Time was produced, as seen below.

{|style="margin: 0 auto;"
| [[File:Screenshot_2019-05-21_at_18.43.01.png|thumb|upright|400x400px|Graph of Momentum vs Time for F + H2]]
| [[File:Screenshot_2019-05-21_at_18.43.11.png|thumb|upright|400x400px|Graph of Energy vs Time ]]
|}

As you can see, whenever potential energy is lost, kinetic energy is gained by the same amount. Due to the exothermic nature of this reactrion, this kinetic energy is then converted into heat and can experimentally be seen by the increase in temperature.

==='''Q9 - 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 Rule states that vibrational energy is much less successful in promoting an early transition state than translational energy is. Additionally, the opposite is true - translational energy is more efficient in promoting a late transition state than vibrational energy.

As seen above when calculating the activation energy of reaction, the reaction of F + H2 is exothermic. Hammond's Postulate states that this reaction therefore has an early transition state. Using Polanyi's Rules, this reaction would be most effectively activated by translational energy. Similarly, for the reaction of H + HF (endothermic), this would be most efficiently activated by vibrational energy.

For the first reaction of F + H2, I used the values of rHH=0.74 Å, and rHF=1.81 Å. Then i looked at the contour plots for when '''p'''AB=-2.5 and '''p'''BC=-0.5 (Left diagram), and for when '''p'''AB=-0.5 and '''p'''BC=-2.5 (right diagram). For the initial set of conditions, there is greater vibrational energy, and the 2nd set of conditions, greater translational energy.

According to Polanyi's Rules, we would expect this reaction to be best activated by vibrational energy. The two graphs below agree with these findings.

{|style="margin: 0 auto;"
| [[File:Screenshot_2019-05-24_at_13.32.46.png|thumb|upright|330x330px|Contour plot for greater vibrational energy - F + H2]]
| [[File:Screenshot_2019-05-24_at_13.33.04.png|thumb|upright|330x330px|Contour plot for greater translational energy - F + H2]]
|}

013366322

2019-05-24T15:41:51Z

Do2617: /* Q7 - Report the activation energy for both reactions. */

== EXERCISE 1: H + H2 system ==

===='''Q1 - On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''====

On the potential energy surface, the transition state is the position along the minimum energy pathway, linking reactants and products, where potential is highest. The transition structure exists at a saddle point of the potential energy surface - that is when ∂V('''ri''')/∂('''ri''')=0. Furthermore, the 2nd partial derivatives with respect to the orthogonal vectors r1 and r2 have opposite signs; one is positive and the other negative. This tells us that this is a saddle point.

[[File:Screenshot_2019-05-21_at_14.27.22.png|400px]]

===='''Q2 - 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.'''====

{|style="margin: 0 auto;"
| [[File:Screenshot 2019-05-21 at 14.41.20.png|thumb|upright|400x400px|A graph showing Intermolecular distance Vs Time ]]
| [[File:Screenshot_2019-05-21_at_16.25.22.png|thumb|upright|400x400px|Contour Plot to find rts]]
|}

Firstly, we set AB=BC and momentum equal to 0. Then, plotting a graph of Internuclear Distance vs Time, we can find the transition state. The optimised distance is the distance at which there is no trajectory towards either reactants or products and at which oscillations are reduced producing a straight line as seen in the plot above. As the graph shows straight lines, we know there is no vibration, and the atoms stop moving. That is, potential energy is maximum and kinetic energy zero. As such, the optimised distance is found to be 0.9079 Å. Looking at the contour plot above, the transition state can also be seen in approximately the correct location, backing up the result.

===='''Q3 - Comment on how the MEP and the trajectory you just calculated differ.'''====

{|style="margin: 0 auto;"
| [[File:Screenshot_2019-05-21_at_15.06.08.png|thumb|upright|400x400px|MEP Calculation]]
| [[File:Screenshot 2019-05-21 at 15.07.56.png|thumb|upright|400x400px|Dynamics Calculation]]
|}

The MEP graph shows no molecular vibration (as velocity=0) with momentum being equal to 0 in every time step. In comparison, the dynamics plot shows the vibrational energy of the molecule as seen by the non-straight line.

==='''Q4 - Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''===

{| class="wikitable" border="1"

! p1 !! p2 !! Etot !! Reactive? !! Illustration of the trajectory !! Description of the dynamics
|-
| -1.25 || -2.5 || -99.018 || YES || [[File:Screenshot_dex_1.png|400x400px]] || Both reactants collide and have sufficient energy to surpass the activation barrier. Products are formed. Vibrational motion is also seen evident by the oscillations.
|-
| -1.5 || -2.0 || -100.456 || NO || [[File:Screenshot_dex_2.png|400x400px]] || Atom C collides molecule AB but has insufficient kinetic energy to overcome the activation energy, thus not forming products. Molecules do not pass through the transition state.
|-
| -1.5 || -2.5 || -98.956 || YES || [[File:Screenshot_dex_3.png|400x400px]] || Both molecules have sufficient energy to surpass the activation energy and so pass though the transition state, forming the products.
|-
| -2.5 || -5.0 || -84.956 || NO || [[File:Screenshot_dex_4.png|400x400px]] || Both molecules have large energies and are able to overcome the activation energy. However, their energy is so great, that they recross the barrier and reform reactants.
|-
| -2.5 || -5.2 || -83.416 || YES || [[File:Screenshot_dex_5.png|400x400px]] || Both molecules have sufficient energy to overcome the activation energy barrier, but have the potential to cross back over the barrier, and reform products.
|}

It can be concluded that if reactants do not have sufficient kinetic energy to surpass the activation barrier, then the reaction will not occur. Furthermore, if the reactants have an excess of kinetic energy, the reaction also may not proceed as the products formed turn back into reactants.

==='''Q5 - 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?'''===

Electron and nuclear motion are independent
Energy of the particles follows the Boltzmann distribution
Once reactants begin to combine and form the transition state, the transition state structure does not collapse to form reactants again

Theory states that transition state structure does not collapse to reform the reactants. However, as experimental procedure shows, this can occur. As theory assumes that all reactions with sufficient energy will go onto form products, the real rate of reaction is much slower than in theory as reactants can be reformed.

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

==='''Q6 - By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''===

{|style="margin: 0 auto;"
| [[File:Screenshot_dex_f_h2.png|thumb|upright|400x400px|Surface Plot for F + H2]]
| [[File:Screenshot_dex_h_hf.png|thumb|upright|400x400px|Surface Plot for H + HF]]
|}

As seen above, the surface plot for F + H2 starts at a high energy position dropping down lower energy after the transition state. Therefore, this is an exothermic reaction. In contrast, for the surface plot of HF + H, the reaction starts at a low energy position and needs a large amount of energy to over come the activation energy barrier. After passing the transition state, the energy falls a small amount. This reaction is therefore endothermic. From this, we can conclude that the HF bond is stronger than the HH bond since HF is lower in energy than HH.

Furthermore, Hammond's Postulate states that the transition state will resemble either the product or reactants, whichever it is closer to in energy. To find the transition state, optimisation was completed as before with both momenta set to zero. Optimisation was completed so that there was minimal oscillation (gradient of line =0) seen for the Internuclear Distance vs Time graph, as shown below. The optimised distances were found to be rts(H-F)=1.813 Å and rts(H-H)=0.7445 Å.

[[File:Screenshot_2019-05-21_at_18.21.10.png|thumb|upright|400x400px|Graph showing Internuclear Distance vs Time for F + H2]]

==='''Q7 - Report the activation energy for both reactions.'''===
Below shows two graphs depicting the total energy plot of both reactions. The activation energy for both reactions can be calculated as the difference between the highest and lowest energy levels.

{| class="wikitable"
![[File:Screenshot_2019-05-24_at_16.37.01.png|500px|left|thumb|Graph of Energy vs Steps for F + H2.]]
![[File:Screenshot_2019-05-24_at_16.41.09.png|500px|right|thumb|Graph of Energy vs Steps for HF + H.]]
|}

'''Activation Energy(H + HF)'''= +31.22 kcalmol-1
'''Activation Energy(F + H2)'''= -+0.130 kcal mol-1

==='''Q8 - In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''===

For the reaction of F + H2 a reactive trajectory was found using the following set of initial conditions: AB = 0.74 Å, BC = 1.80 Å, '''pAB''' = -2.5, and '''pBC''' = -1.70. Using these intial conditions the graph of Momentum vs Time was produced, as seen below.

{|style="margin: 0 auto;"
| [[File:Screenshot_2019-05-21_at_18.43.01.png|thumb|upright|400x400px|Graph of Momentum vs Time for F + H2]]
| [[File:Screenshot_2019-05-21_at_18.43.11.png|thumb|upright|400x400px|Graph of Energy vs Time ]]
|}

As you can see, whenever potential energy is lost, kinetic energy is gained by the same amount. Due to the exothermic nature of this reactrion, this kinetic energy is then converted into heat and can experimentally be seen by the increase in temperature.

==='''Q9 - 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 Rule states that vibrational energy is much less successful in promoting an early transition state than translational energy is. Additionally, the opposite is true - translational energy is more efficient in promoting a late transition state than vibrational energy.

As seen above when calculating the activation energy of reaction, the reaction of F + H2 is exothermic. Hammond's Postulate states that this reaction therefore has an early transition state. Using Polanyi's Rules, this reaction would be most effectively activated by translational energy. Similarly, for the reaction of H + HF (endothermic), this would be most efficiently activated by vibrational energy.

For the first reaction of F + H2, I used the values of rHH=0.74 Å, and rHF=1.81 Å. Then i looked at the contour plots for when '''p'''AB=-2.5 and '''p'''BC=-0.5 (Left diagram), and for when '''p'''AB=-0.5 and '''p'''BC=-2.5 (right diagram). For the initial set of conditions, there is greater vibrational energy, and the 2nd set of conditions, greater translational energy.

According to Polanyi's Rules, we would expect this reaction to be best activated by vibrational energy. The two graphs below agree with these findings.

{|style="margin: 0 auto;"
| [[File:Screenshot_2019-05-24_at_13.32.46.png|thumb|upright|330x330px|Contour plot for greater vibrational energy - F + H2]]
| [[File:Screenshot_2019-05-24_at_13.33.04.png|thumb|upright|330x330px|Contour plot for greater translational energy - F + H2]]
|}

File:Screenshot 2019-05-24 at 16.41.09.png

2019-05-24T15:41:28Z

Do2617:

013366322

2019-05-24T15:40:11Z

Do2617: /* Q7 - Report the activation energy for both reactions. */

== EXERCISE 1: H + H2 system ==

===='''Q1 - On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''====

On the potential energy surface, the transition state is the position along the minimum energy pathway, linking reactants and products, where potential is highest. The transition structure exists at a saddle point of the potential energy surface - that is when ∂V('''ri''')/∂('''ri''')=0. Furthermore, the 2nd partial derivatives with respect to the orthogonal vectors r1 and r2 have opposite signs; one is positive and the other negative. This tells us that this is a saddle point.

[[File:Screenshot_2019-05-21_at_14.27.22.png|400px]]

===='''Q2 - 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.'''====

{|style="margin: 0 auto;"
| [[File:Screenshot 2019-05-21 at 14.41.20.png|thumb|upright|400x400px|A graph showing Intermolecular distance Vs Time ]]
| [[File:Screenshot_2019-05-21_at_16.25.22.png|thumb|upright|400x400px|Contour Plot to find rts]]
|}

Firstly, we set AB=BC and momentum equal to 0. Then, plotting a graph of Internuclear Distance vs Time, we can find the transition state. The optimised distance is the distance at which there is no trajectory towards either reactants or products and at which oscillations are reduced producing a straight line as seen in the plot above. As the graph shows straight lines, we know there is no vibration, and the atoms stop moving. That is, potential energy is maximum and kinetic energy zero. As such, the optimised distance is found to be 0.9079 Å. Looking at the contour plot above, the transition state can also be seen in approximately the correct location, backing up the result.

===='''Q3 - Comment on how the MEP and the trajectory you just calculated differ.'''====

{|style="margin: 0 auto;"
| [[File:Screenshot_2019-05-21_at_15.06.08.png|thumb|upright|400x400px|MEP Calculation]]
| [[File:Screenshot 2019-05-21 at 15.07.56.png|thumb|upright|400x400px|Dynamics Calculation]]
|}

The MEP graph shows no molecular vibration (as velocity=0) with momentum being equal to 0 in every time step. In comparison, the dynamics plot shows the vibrational energy of the molecule as seen by the non-straight line.

==='''Q4 - Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''===

{| class="wikitable" border="1"

! p1 !! p2 !! Etot !! Reactive? !! Illustration of the trajectory !! Description of the dynamics
|-
| -1.25 || -2.5 || -99.018 || YES || [[File:Screenshot_dex_1.png|400x400px]] || Both reactants collide and have sufficient energy to surpass the activation barrier. Products are formed. Vibrational motion is also seen evident by the oscillations.
|-
| -1.5 || -2.0 || -100.456 || NO || [[File:Screenshot_dex_2.png|400x400px]] || Atom C collides molecule AB but has insufficient kinetic energy to overcome the activation energy, thus not forming products. Molecules do not pass through the transition state.
|-
| -1.5 || -2.5 || -98.956 || YES || [[File:Screenshot_dex_3.png|400x400px]] || Both molecules have sufficient energy to surpass the activation energy and so pass though the transition state, forming the products.
|-
| -2.5 || -5.0 || -84.956 || NO || [[File:Screenshot_dex_4.png|400x400px]] || Both molecules have large energies and are able to overcome the activation energy. However, their energy is so great, that they recross the barrier and reform reactants.
|-
| -2.5 || -5.2 || -83.416 || YES || [[File:Screenshot_dex_5.png|400x400px]] || Both molecules have sufficient energy to overcome the activation energy barrier, but have the potential to cross back over the barrier, and reform products.
|}

It can be concluded that if reactants do not have sufficient kinetic energy to surpass the activation barrier, then the reaction will not occur. Furthermore, if the reactants have an excess of kinetic energy, the reaction also may not proceed as the products formed turn back into reactants.

==='''Q5 - 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?'''===

Electron and nuclear motion are independent
Energy of the particles follows the Boltzmann distribution
Once reactants begin to combine and form the transition state, the transition state structure does not collapse to form reactants again

Theory states that transition state structure does not collapse to reform the reactants. However, as experimental procedure shows, this can occur. As theory assumes that all reactions with sufficient energy will go onto form products, the real rate of reaction is much slower than in theory as reactants can be reformed.

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

==='''Q6 - By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''===

{|style="margin: 0 auto;"
| [[File:Screenshot_dex_f_h2.png|thumb|upright|400x400px|Surface Plot for F + H2]]
| [[File:Screenshot_dex_h_hf.png|thumb|upright|400x400px|Surface Plot for H + HF]]
|}

As seen above, the surface plot for F + H2 starts at a high energy position dropping down lower energy after the transition state. Therefore, this is an exothermic reaction. In contrast, for the surface plot of HF + H, the reaction starts at a low energy position and needs a large amount of energy to over come the activation energy barrier. After passing the transition state, the energy falls a small amount. This reaction is therefore endothermic. From this, we can conclude that the HF bond is stronger than the HH bond since HF is lower in energy than HH.

Furthermore, Hammond's Postulate states that the transition state will resemble either the product or reactants, whichever it is closer to in energy. To find the transition state, optimisation was completed as before with both momenta set to zero. Optimisation was completed so that there was minimal oscillation (gradient of line =0) seen for the Internuclear Distance vs Time graph, as shown below. The optimised distances were found to be rts(H-F)=1.813 Å and rts(H-H)=0.7445 Å.

[[File:Screenshot_2019-05-21_at_18.21.10.png|thumb|upright|400x400px|Graph showing Internuclear Distance vs Time for F + H2]]

==='''Q7 - Report the activation energy for both reactions.'''===
Below shows two graphs depicting the total energy plot of both reactions. The activation energy for both reactions can be calculated as the difference between the highest and lowest energy levels.

{| class="wikitable"
![[File:Screenshot_2019-05-24_at_16.37.01.png|500px|left|thumb|Graph of Energy vs Steps for F + H2.]]
![[File:Screenshot_2019-05-24_at_16.37.14.png|500px|right|thumb|Graph of Energy vs Steps for HF + H.]]
|}

'''Activation Energy(H + HF)'''= +31.22 kcalmol-1
'''Activation Energy(F + H2)'''= -+0.130 kcal mol-1

==='''Q8 - In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''===

For the reaction of F + H2 a reactive trajectory was found using the following set of initial conditions: AB = 0.74 Å, BC = 1.80 Å, '''pAB''' = -2.5, and '''pBC''' = -1.70. Using these intial conditions the graph of Momentum vs Time was produced, as seen below.

{|style="margin: 0 auto;"
| [[File:Screenshot_2019-05-21_at_18.43.01.png|thumb|upright|400x400px|Graph of Momentum vs Time for F + H2]]
| [[File:Screenshot_2019-05-21_at_18.43.11.png|thumb|upright|400x400px|Graph of Energy vs Time ]]
|}

As you can see, whenever potential energy is lost, kinetic energy is gained by the same amount. Due to the exothermic nature of this reactrion, this kinetic energy is then converted into heat and can experimentally be seen by the increase in temperature.

==='''Q9 - 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 Rule states that vibrational energy is much less successful in promoting an early transition state than translational energy is. Additionally, the opposite is true - translational energy is more efficient in promoting a late transition state than vibrational energy.

As seen above when calculating the activation energy of reaction, the reaction of F + H2 is exothermic. Hammond's Postulate states that this reaction therefore has an early transition state. Using Polanyi's Rules, this reaction would be most effectively activated by translational energy. Similarly, for the reaction of H + HF (endothermic), this would be most efficiently activated by vibrational energy.

For the first reaction of F + H2, I used the values of rHH=0.74 Å, and rHF=1.81 Å. Then i looked at the contour plots for when '''p'''AB=-2.5 and '''p'''BC=-0.5 (Left diagram), and for when '''p'''AB=-0.5 and '''p'''BC=-2.5 (right diagram). For the initial set of conditions, there is greater vibrational energy, and the 2nd set of conditions, greater translational energy.

According to Polanyi's Rules, we would expect this reaction to be best activated by vibrational energy. The two graphs below agree with these findings.

{|style="margin: 0 auto;"
| [[File:Screenshot_2019-05-24_at_13.32.46.png|thumb|upright|330x330px|Contour plot for greater vibrational energy - F + H2]]
| [[File:Screenshot_2019-05-24_at_13.33.04.png|thumb|upright|330x330px|Contour plot for greater translational energy - F + H2]]
|}

013366322

2019-05-24T15:39:54Z

Do2617: /* Q7 - Report the activation energy for both reactions. */

== EXERCISE 1: H + H2 system ==

===='''Q1 - On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''====

On the potential energy surface, the transition state is the position along the minimum energy pathway, linking reactants and products, where potential is highest. The transition structure exists at a saddle point of the potential energy surface - that is when ∂V('''ri''')/∂('''ri''')=0. Furthermore, the 2nd partial derivatives with respect to the orthogonal vectors r1 and r2 have opposite signs; one is positive and the other negative. This tells us that this is a saddle point.

[[File:Screenshot_2019-05-21_at_14.27.22.png|400px]]

===='''Q2 - 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.'''====

{|style="margin: 0 auto;"
| [[File:Screenshot 2019-05-21 at 14.41.20.png|thumb|upright|400x400px|A graph showing Intermolecular distance Vs Time ]]
| [[File:Screenshot_2019-05-21_at_16.25.22.png|thumb|upright|400x400px|Contour Plot to find rts]]
|}

Firstly, we set AB=BC and momentum equal to 0. Then, plotting a graph of Internuclear Distance vs Time, we can find the transition state. The optimised distance is the distance at which there is no trajectory towards either reactants or products and at which oscillations are reduced producing a straight line as seen in the plot above. As the graph shows straight lines, we know there is no vibration, and the atoms stop moving. That is, potential energy is maximum and kinetic energy zero. As such, the optimised distance is found to be 0.9079 Å. Looking at the contour plot above, the transition state can also be seen in approximately the correct location, backing up the result.

===='''Q3 - Comment on how the MEP and the trajectory you just calculated differ.'''====

{|style="margin: 0 auto;"
| [[File:Screenshot_2019-05-21_at_15.06.08.png|thumb|upright|400x400px|MEP Calculation]]
| [[File:Screenshot 2019-05-21 at 15.07.56.png|thumb|upright|400x400px|Dynamics Calculation]]
|}

The MEP graph shows no molecular vibration (as velocity=0) with momentum being equal to 0 in every time step. In comparison, the dynamics plot shows the vibrational energy of the molecule as seen by the non-straight line.

==='''Q4 - Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''===

{| class="wikitable" border="1"

! p1 !! p2 !! Etot !! Reactive? !! Illustration of the trajectory !! Description of the dynamics
|-
| -1.25 || -2.5 || -99.018 || YES || [[File:Screenshot_dex_1.png|400x400px]] || Both reactants collide and have sufficient energy to surpass the activation barrier. Products are formed. Vibrational motion is also seen evident by the oscillations.
|-
| -1.5 || -2.0 || -100.456 || NO || [[File:Screenshot_dex_2.png|400x400px]] || Atom C collides molecule AB but has insufficient kinetic energy to overcome the activation energy, thus not forming products. Molecules do not pass through the transition state.
|-
| -1.5 || -2.5 || -98.956 || YES || [[File:Screenshot_dex_3.png|400x400px]] || Both molecules have sufficient energy to surpass the activation energy and so pass though the transition state, forming the products.
|-
| -2.5 || -5.0 || -84.956 || NO || [[File:Screenshot_dex_4.png|400x400px]] || Both molecules have large energies and are able to overcome the activation energy. However, their energy is so great, that they recross the barrier and reform reactants.
|-
| -2.5 || -5.2 || -83.416 || YES || [[File:Screenshot_dex_5.png|400x400px]] || Both molecules have sufficient energy to overcome the activation energy barrier, but have the potential to cross back over the barrier, and reform products.
|}

It can be concluded that if reactants do not have sufficient kinetic energy to surpass the activation barrier, then the reaction will not occur. Furthermore, if the reactants have an excess of kinetic energy, the reaction also may not proceed as the products formed turn back into reactants.

==='''Q5 - 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?'''===

Electron and nuclear motion are independent
Energy of the particles follows the Boltzmann distribution
Once reactants begin to combine and form the transition state, the transition state structure does not collapse to form reactants again

Theory states that transition state structure does not collapse to reform the reactants. However, as experimental procedure shows, this can occur. As theory assumes that all reactions with sufficient energy will go onto form products, the real rate of reaction is much slower than in theory as reactants can be reformed.

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

==='''Q6 - By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''===

{|style="margin: 0 auto;"
| [[File:Screenshot_dex_f_h2.png|thumb|upright|400x400px|Surface Plot for F + H2]]
| [[File:Screenshot_dex_h_hf.png|thumb|upright|400x400px|Surface Plot for H + HF]]
|}

As seen above, the surface plot for F + H2 starts at a high energy position dropping down lower energy after the transition state. Therefore, this is an exothermic reaction. In contrast, for the surface plot of HF + H, the reaction starts at a low energy position and needs a large amount of energy to over come the activation energy barrier. After passing the transition state, the energy falls a small amount. This reaction is therefore endothermic. From this, we can conclude that the HF bond is stronger than the HH bond since HF is lower in energy than HH.

Furthermore, Hammond's Postulate states that the transition state will resemble either the product or reactants, whichever it is closer to in energy. To find the transition state, optimisation was completed as before with both momenta set to zero. Optimisation was completed so that there was minimal oscillation (gradient of line =0) seen for the Internuclear Distance vs Time graph, as shown below. The optimised distances were found to be rts(H-F)=1.813 Å and rts(H-H)=0.7445 Å.

[[File:Screenshot_2019-05-21_at_18.21.10.png|thumb|upright|400x400px|Graph showing Internuclear Distance vs Time for F + H2]]

==='''Q7 - Report the activation energy for both reactions.'''===
Below shows two graphs depicting the total energy plot of both reactions. The activation energy for both reactions can be calculated as the difference between the highest and lowest energy levels.

{| class="wikitable"
![[File:Screenshot_2019-05-24_at_16.37.01.png|500px|left|thumb|Graph of Energy vs Steps for F + H2.]]
![[File:Screenshot_2019-05-24_at_16.37.14.png|500px|right|thumb|Graph of Energy vs Steps for HF + H.]]
|}

'''Activation Energy(H + HF)'''= +31.22 kcalmol-1
'''Activation Energy(F + H2)'''= -+0.130 kcal mol-1

==='''Q8 - In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''===

For the reaction of F + H2 a reactive trajectory was found using the following set of initial conditions: AB = 0.74 Å, BC = 1.80 Å, '''pAB''' = -2.5, and '''pBC''' = -1.70. Using these intial conditions the graph of Momentum vs Time was produced, as seen below.

{|style="margin: 0 auto;"
| [[File:Screenshot_2019-05-21_at_18.43.01.png|thumb|upright|400x400px|Graph of Momentum vs Time for F + H2]]
| [[File:Screenshot_2019-05-21_at_18.43.11.png|thumb|upright|400x400px|Graph of Energy vs Time ]]
|}

As you can see, whenever potential energy is lost, kinetic energy is gained by the same amount. Due to the exothermic nature of this reactrion, this kinetic energy is then converted into heat and can experimentally be seen by the increase in temperature.

==='''Q9 - 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 Rule states that vibrational energy is much less successful in promoting an early transition state than translational energy is. Additionally, the opposite is true - translational energy is more efficient in promoting a late transition state than vibrational energy.

As seen above when calculating the activation energy of reaction, the reaction of F + H2 is exothermic. Hammond's Postulate states that this reaction therefore has an early transition state. Using Polanyi's Rules, this reaction would be most effectively activated by translational energy. Similarly, for the reaction of H + HF (endothermic), this would be most efficiently activated by vibrational energy.

For the first reaction of F + H2, I used the values of rHH=0.74 Å, and rHF=1.81 Å. Then i looked at the contour plots for when '''p'''AB=-2.5 and '''p'''BC=-0.5 (Left diagram), and for when '''p'''AB=-0.5 and '''p'''BC=-2.5 (right diagram). For the initial set of conditions, there is greater vibrational energy, and the 2nd set of conditions, greater translational energy.

According to Polanyi's Rules, we would expect this reaction to be best activated by vibrational energy. The two graphs below agree with these findings.

{|style="margin: 0 auto;"
| [[File:Screenshot_2019-05-24_at_13.32.46.png|thumb|upright|330x330px|Contour plot for greater vibrational energy - F + H2]]
| [[File:Screenshot_2019-05-24_at_13.33.04.png|thumb|upright|330x330px|Contour plot for greater translational energy - F + H2]]
|}

013366322

2019-05-24T15:39:29Z

Do2617:

== EXERCISE 1: H + H2 system ==

===='''Q1 - On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?'''====

On the potential energy surface, the transition state is the position along the minimum energy pathway, linking reactants and products, where potential is highest. The transition structure exists at a saddle point of the potential energy surface - that is when ∂V('''ri''')/∂('''ri''')=0. Furthermore, the 2nd partial derivatives with respect to the orthogonal vectors r1 and r2 have opposite signs; one is positive and the other negative. This tells us that this is a saddle point.

[[File:Screenshot_2019-05-21_at_14.27.22.png|400px]]

===='''Q2 - 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.'''====

{|style="margin: 0 auto;"
| [[File:Screenshot 2019-05-21 at 14.41.20.png|thumb|upright|400x400px|A graph showing Intermolecular distance Vs Time ]]
| [[File:Screenshot_2019-05-21_at_16.25.22.png|thumb|upright|400x400px|Contour Plot to find rts]]
|}

Firstly, we set AB=BC and momentum equal to 0. Then, plotting a graph of Internuclear Distance vs Time, we can find the transition state. The optimised distance is the distance at which there is no trajectory towards either reactants or products and at which oscillations are reduced producing a straight line as seen in the plot above. As the graph shows straight lines, we know there is no vibration, and the atoms stop moving. That is, potential energy is maximum and kinetic energy zero. As such, the optimised distance is found to be 0.9079 Å. Looking at the contour plot above, the transition state can also be seen in approximately the correct location, backing up the result.

===='''Q3 - Comment on how the MEP and the trajectory you just calculated differ.'''====

{|style="margin: 0 auto;"
| [[File:Screenshot_2019-05-21_at_15.06.08.png|thumb|upright|400x400px|MEP Calculation]]
| [[File:Screenshot 2019-05-21 at 15.07.56.png|thumb|upright|400x400px|Dynamics Calculation]]
|}

The MEP graph shows no molecular vibration (as velocity=0) with momentum being equal to 0 in every time step. In comparison, the dynamics plot shows the vibrational energy of the molecule as seen by the non-straight line.

==='''Q4 - Complete the table above by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?'''===

{| class="wikitable" border="1"

! p1 !! p2 !! Etot !! Reactive? !! Illustration of the trajectory !! Description of the dynamics
|-
| -1.25 || -2.5 || -99.018 || YES || [[File:Screenshot_dex_1.png|400x400px]] || Both reactants collide and have sufficient energy to surpass the activation barrier. Products are formed. Vibrational motion is also seen evident by the oscillations.
|-
| -1.5 || -2.0 || -100.456 || NO || [[File:Screenshot_dex_2.png|400x400px]] || Atom C collides molecule AB but has insufficient kinetic energy to overcome the activation energy, thus not forming products. Molecules do not pass through the transition state.
|-
| -1.5 || -2.5 || -98.956 || YES || [[File:Screenshot_dex_3.png|400x400px]] || Both molecules have sufficient energy to surpass the activation energy and so pass though the transition state, forming the products.
|-
| -2.5 || -5.0 || -84.956 || NO || [[File:Screenshot_dex_4.png|400x400px]] || Both molecules have large energies and are able to overcome the activation energy. However, their energy is so great, that they recross the barrier and reform reactants.
|-
| -2.5 || -5.2 || -83.416 || YES || [[File:Screenshot_dex_5.png|400x400px]] || Both molecules have sufficient energy to overcome the activation energy barrier, but have the potential to cross back over the barrier, and reform products.
|}

It can be concluded that if reactants do not have sufficient kinetic energy to surpass the activation barrier, then the reaction will not occur. Furthermore, if the reactants have an excess of kinetic energy, the reaction also may not proceed as the products formed turn back into reactants.

==='''Q5 - 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?'''===

Electron and nuclear motion are independent
Energy of the particles follows the Boltzmann distribution
Once reactants begin to combine and form the transition state, the transition state structure does not collapse to form reactants again

Theory states that transition state structure does not collapse to reform the reactants. However, as experimental procedure shows, this can occur. As theory assumes that all reactions with sufficient energy will go onto form products, the real rate of reaction is much slower than in theory as reactants can be reformed.

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

==='''Q6 - By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''===

{|style="margin: 0 auto;"
| [[File:Screenshot_dex_f_h2.png|thumb|upright|400x400px|Surface Plot for F + H2]]
| [[File:Screenshot_dex_h_hf.png|thumb|upright|400x400px|Surface Plot for H + HF]]
|}

As seen above, the surface plot for F + H2 starts at a high energy position dropping down lower energy after the transition state. Therefore, this is an exothermic reaction. In contrast, for the surface plot of HF + H, the reaction starts at a low energy position and needs a large amount of energy to over come the activation energy barrier. After passing the transition state, the energy falls a small amount. This reaction is therefore endothermic. From this, we can conclude that the HF bond is stronger than the HH bond since HF is lower in energy than HH.

Furthermore, Hammond's Postulate states that the transition state will resemble either the product or reactants, whichever it is closer to in energy. To find the transition state, optimisation was completed as before with both momenta set to zero. Optimisation was completed so that there was minimal oscillation (gradient of line =0) seen for the Internuclear Distance vs Time graph, as shown below. The optimised distances were found to be rts(H-F)=1.813 Å and rts(H-H)=0.7445 Å.

[[File:Screenshot_2019-05-21_at_18.21.10.png|thumb|upright|400x400px|Graph showing Internuclear Distance vs Time for F + H2]]

==='''Q7 - Report the activation energy for both reactions.'''===
Below shows two graphs depicting the total energy plot of both reactions. The activation energy for both reactions can be calculated as the difference between the highest and lowest energy levels.

{| class="wikitable"
![[File:Screenshot_2019-05-24_at_16.37.01.png|500px|left|thumb|Graph of Energy vs Steps for F + H2.]]
![[File:Screenshot_2019-05-24_at_16.37.14.png|500px|right|thumb|Graph of Energy vs Steps for HF + H.]]
|}

'''Activation Energy(H + HF)'''= +31.22 kcalmol-1
'''Activation Energy(F + H2)'''= -+0.130 kcal mol-1

==='''Q8 - In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. Explain how this could be confirmed experimentally.'''===

For the reaction of F + H2 a reactive trajectory was found using the following set of initial conditions: AB = 0.74 Å, BC = 1.80 Å, '''pAB''' = -2.5, and '''pBC''' = -1.70. Using these intial conditions the graph of Momentum vs Time was produced, as seen below.

{|style="margin: 0 auto;"
| [[File:Screenshot_2019-05-21_at_18.43.01.png|thumb|upright|400x400px|Graph of Momentum vs Time for F + H2]]
| [[File:Screenshot_2019-05-21_at_18.43.11.png|thumb|upright|400x400px|Graph of Energy vs Time ]]
|}

As you can see, whenever potential energy is lost, kinetic energy is gained by the same amount. Due to the exothermic nature of this reactrion, this kinetic energy is then converted into heat and can experimentally be seen by the increase in temperature.

==='''Q9 - 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 Rule states that vibrational energy is much less successful in promoting an early transition state than translational energy is. Additionally, the opposite is true - translational energy is more efficient in promoting a late transition state than vibrational energy.

As seen above when calculating the activation energy of reaction, the reaction of F + H2 is exothermic. Hammond's Postulate states that this reaction therefore has an early transition state. Using Polanyi's Rules, this reaction would be most effectively activated by translational energy. Similarly, for the reaction of H + HF (endothermic), this would be most efficiently activated by vibrational energy.

For the first reaction of F + H2, I used the values of rHH=0.74 Å, and rHF=1.81 Å. Then i looked at the contour plots for when '''p'''AB=-2.5 and '''p'''BC=-0.5 (Left diagram), and for when '''p'''AB=-0.5 and '''p'''BC=-2.5 (right diagram). For the initial set of conditions, there is greater vibrational energy, and the 2nd set of conditions, greater translational energy.

According to Polanyi's Rules, we would expect this reaction to be best activated by vibrational energy. The two graphs below agree with these findings.

{|style="margin: 0 auto;"
| [[File:Screenshot_2019-05-24_at_13.32.46.png|thumb|upright|330x330px|Contour plot for greater vibrational energy - F + H2]]
| [[File:Screenshot_2019-05-24_at_13.33.04.png|thumb|upright|330x330px|Contour plot for greater translational energy - F + H2]]
|}

File:Screenshot 2019-05-24 at 16.37.14.png

2019-05-21T17:26:56Z

Do2617:

013366322

2019-05-21T17:22:54Z

013366322

2019-05-21T15:00:01Z

Do2617: