MRD:Jsr3217

2019-06-03T14:27:47Z

== H + H2 System ==

If we define H and H2 as A and BC in our system, as the hydrogen atom collides with the hydrogen molecule, provided H has enough momentum and thus enough energy, the system changes and is described as AB and C where AB is our new diatomic product (A,B and C are the same type of atom).

=== Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===

The transition state for this triatomic collision is a saddle point located at the intersection of the two potential energy surfaces, one PES for the Hydrogen atom and 1 PES for the Hydrogen molecule. The transition state is also a maximum on the trajectory pathway.

Firstly the transition state being a turning point has a derivative of 0 with respect to q. This is shown by the following equation: (∂V(q1)/∂q1) and (∂V(q2)/∂q2)=0. However being a saddle point, the transition state is also defined as the point in the surface plot where there is a minimum in the q1 direction but also a maximum in the q2 direction, where q1 and q2 are orthogonal to each other . This is where we have to take into account the second derivative of the potential energy V with respect to q. Mathematically these results are expressed as ∂2V(q1)/∂q12 >0 and ∂2V(q1)/∂q12 <0. The former second order differential equation refers to the minimum point in the q1 direction and the latter second order differential equation refers to the maximum point in the q2 direction.

The transition state can be distinguished from a local minimum of the potential energy surface as by definition a local minimum will only have one 2nd derivative of potential energy (with respect to q in this case) which is greater than 0. However the transition state, being a saddle point, is a point with 2 second derivatives of potential energy in two different directions, one which is greater than 0 and one which is less than 0.

{{fontcolor|pink|I can tell from this explanation that you understand the shape of the TS on the PES however your discussion here is not quite complete as you haven't defined the directions q1 and q2 so you haven't fully defined the shape of the TS. A diagram with the vectors included would be really helpful here. In general diagrams are useful! [[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 15:23, 3 June 2019 (BST)}}

Original Surface Plot showing both Potential Energy Surfaces and the Trajectory Pathway:

[[File:1st surface plot.PNG]]

The transition state is located at the point where the interatomic distances are equal i.e. when the bond length of A-B is equal to the bond length of B-C. This is when the A-B bond is forming to make the diatomic product AB and the B-C bond is breaking leaving C on its own. This is shown below in the contour plot:
{{fontcolor|pink|This statement about the equal distances is true of this system only because it is a special case where A=B=C. Be careful using this statement as it is not a generic description of a TS.[[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 15:27, 3 June 2019 (BST)}}

[[File:1st contour plot.PNG]]

=== Question 2: What is 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 position of the transition state can be found by varying the AB and BC distances, with both momentum values of AB and BC being 0, until no variation in the Internuclear Distances vs Time Plot is noticed or until a straight line in the plot is seen. No variation or oscillation in the plot means there are no vibrations which is what happens in the transition state. The distance at which no oscillation was seen was found to be at 0.90774 Å.

Internuclear Distance vs Time Plot at the transition state:

[[File:2nd question graph 1.PNG]]

=== Question 3: Comment on how the MEP and the trajectory you just calculated differ. ===

The plot with the MEP calculation shows a relatively short line with little variation in the y-axis (AB distance) however, the plot with the Dynamics calculation shows an oscillating line. The trajectory for the dynamics calculated plot is also longer in length than the MEP calculated plot. This is due to the MEP calculation not considering any vibrational modes therefore no oscillations are shown on the plot. In actuality, atoms have mass and momentum and so the vibrations of the hydrogen molecule are taken into account shown by the oscillations in the dynamics calculation plot. Thus the model with the dynamics calculation is the more accurate one.

Plot with MEP Calculation:

[[File:ACTUALMEP1.PNG]]

Plot with Dynamics Calculation:

[[File:ACTUALDYNAMIC.PNG]]

=== Question 4: 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
|+ Reactive or Unreactive Trajectories?
! p1 !! p2 !! Etot !! Reactive? !! Illustration of the Trajectory !! Description
|-
| -1.25 || -2.5 ||-99.018 || Yes || [[File:1T.PNG]] ||Contour Plot shows a successful reaction with reactants, having passed through the transition state shown by the trajectory, fully converted into products.||
|-
| -1.5 || -2.0 || -100.456 || No || [[File:2T.PNG]] || Contour Plot shows that reaction does not occur as reactants don't have enough initial momentum and thus required (kinetic) energy to get over the activation energy barrier and reach the energy of the transition state. This is shown by the trajectory not reaching the transition state point and tailing back to the reactants. As the transition state isn't reached, the products aren't formed. ||
|-
| -1.5 || -2.5 || -98.956|| Yes ||[[File:3T.PNG]] || Contour Plot shows a successful reaction with reactants, having passed through the transition state shown by the trajectory, fully converted into products. The reaction is successful as the reactants have enough initial momentum and thus energy to reach the energy of the transition state. ||
|-
| -2.5 || -5.0 || -84.956 || No || [[File:4T.PNG]] || Contour Plot shows although he reactants have enough initial momentum and thus energy to go through the transition state, the transition state collapses back into the reactants. This happens because of barrier recrossing. ||
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:5T.PNG]] || Trajectory from contour plot shows that the reaction goes through/reaches the transition state point more than once. This occurs because of barrier recrossing again. However with these specific reaction conditions, the reaction does take place and the products are formed from the transition state. ||
|}

=== Question 5: 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?===

Transition State Theory assumes and tells us that states that 1 the transition state or activated complex in a reaction is in equilibrium with the reactants and once the reactants turn into the activated complex, the complex doesn't return to the reactants. It also assumes the energy of the particles follows a Boltzmann distribution1 but in this system since there are only 3 atoms in total for this triatomic collision there is no boltzmann distribution. As just mentioned, transition state theory assumes once the reactants turn into the activated complex, the complex doesn't return to the reactants. However in contour plot 4 above where the trajectory shows that although the reaction goes through the transition state, the reactants are still reformed. This is because of barrier recrossing as mentioned above; with this present, the rate of this reaction, experimentally determined, would be lower than the reaction rate that the transition state theory would predict.

== F - H - H system ==

=== Question 1: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===

[[File:D1.PNG]]

The above surface plot shows the potential energy surface for the exothermic conversion of F and H2 to HF and H. The potential energy well slopes downwards from the reactants to the products showing that the energy of the products is lower than the energy of reactants. This reaction is an exothermic one as the bond strength of HF, the product, is stronger than the bond strength of H2, the reactant. This means that the energy released from the formation of the HF bond is greater than the energy required to break the H-H bond in the hydrogen reactant molecule making the overall reaction exothermic.

[[File:D2endo.PNG]]

The above surface plot shows the potential energy surface for the endothermic conversion of HF and H to H2 and F. This reaction is the reverse reaction of the exothermic conversion we've just considered. This potenital energy well slopes upwards from the reactants to the products showing that the energy of the products is higher than the energy of reactants making the reaction endothermic. The energy released from the hydrogen molecule, the product, is less than the energy taken in to break in the H-F bond in the reactant making the overall reaction endothermic.

=== Question 2: Locate the approximate position of the transition state. ===

Hammond's postulate essentially states that if 2 species are of similar energy they are of a similar structure. Thus in this reaction we can deduce that the transition state will be similar in structure to the reactants of the forward exothermic reaction, F and H2. By testing and varying the bond distances of F-H and H-H until the forces exterted on each other were 0 (characteristic in the transition state) the internuclear distances found were 1.811 Å for F-H and 0.745 for H-H Å.

=== Question 3: Report the activation energy for both reactions. ===

The activation energy can be easily calculated by taking away the reactant energy from the transition state energy. Using the internuclear distances of the atoms in the transition state calculated above in the settings function in the programme, the energy of the transition state was found to be -103.752 kcal mol1. The energy of the reactants, for the exothermic reaction for example, was found by inputting the normal bond length for H2 and making the distance between F and H relatively large so they're not interacting yielding the lowest energy for the reactants possible. From this the activation energies for the exothermic reaction was found to be 30.267 kcal mol1 and the activation energy for the endothermic reaction was found to be 0.242 kcal mol1

=== Question 4: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy for the reaction F + H2 -> HF + H. Explain how this could be confirmed experimentally. ===

Due to the conservation of energy, all initial kinetic energy from the reactants' momentum is converted and released as heat in the reaction once the products HF and H are formed. Due to the release of heat the surroundings temperature would increase as this is an exothermic reaction. This temperature change or increase in this case can be recorded by bomb calorimetry.

=== Question 5: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction for the reaction HF + H -> F + H2, and how this is influenced by the position of the transition state. ===

For this endothermic reaction (HF + H -> F + H2) there are two main modes of reactant energy to consider as the reactants approach the energy barrier/transition state point to form the products; these modes of energy are translational and vibrational energy. 2Polanyi stated that for simple reactions described as an A + BC system, such as this H + HF system, if they have a late transition state, reactants with high vibrational energy is more effective for the reaction and its success, than reactants with high translational energy. This is because for a late transition state reaction, reactant molecules with high vibrational energy, if in the correct orientation, can vibrate their way to the top of the barrier, passing through the transition state forming the products successfully; however, if the reactants had high translational energy, in trying to reach the top of the energy barrier, they would just collide into the inner wall of the potential energy well and return to where they came from. This reaction being considered here (HF + H -> F + H2) is an endothermic reaction with a late transition state as shown by the surface plot discussed earlier. Therefore to maximise the efficiency and the success of this reaction, reactants should have high vibrational energy. Polanyi also implied, that in this type of reaction (endothermic and late transition state), the products are likely to have high translational energy if the reaction effectively occurs with high vibrational energy reactants.

The reaction showing highest efficiency and most effectiveness for its success was then modelled; high vibrational energy was inputted in the settings function for the molecule HF and high initial momentum of the incoming hydrogen atom for the collision was also inputted. However, unfortunately no such combination of momenta and bond length values were found that yielded a successful reaction, but below shows a contour plot with high vibrational energy in the reactants as shown by the oscillating trajectory which is what is needed to ensure high efficiency of this late transition state endothermic reaction.

[[File:Loooool.png]]

== References ==

1 P. Atkins, J. De Paula, (n.d.). Physical chemistry. 7th edition.

2 J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics Prentice-Hall.

MRD:Jsr3217

2019-06-03T14:26:56Z

== H + H2 System ==

If we define H and H2 as A and BC in our system, as the hydrogen atom collides with the hydrogen molecule, provided H has enough momentum and thus enough energy, the system changes and is described as AB and C where AB is our new diatomic product (A,B and C are the same type of atom).

=== Question 1: On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface? ===

The transition state for this triatomic collision is a saddle point located at the intersection of the two potential energy surfaces, one PES for the Hydrogen atom and 1 PES for the Hydrogen molecule. The transition state is also a maximum on the trajectory pathway.

Firstly the transition state being a turning point has a derivative of 0 with respect to q. This is shown by the following equation: (∂V(q1)/∂q1) and (∂V(q2)/∂q2)=0. However being a saddle point, the transition state is also defined as the point in the surface plot where there is a minimum in the q1 direction but also a maximum in the q2 direction, where q1 and q2 are orthogonal to each other . This is where we have to take into account the second derivative of the potential energy V with respect to q. Mathematically these results are expressed as ∂2V(q1)/∂q12 >0 and ∂2V(q1)/∂q12 <0. The former second order differential equation refers to the minimum point in the q1 direction and the latter second order differential equation refers to the maximum point in the q2 direction.

The transition state can be distinguished from a local minimum of the potential energy surface as by definition a local minimum will only have one 2nd derivative of potential energy (with respect to q in this case) which is greater than 0. However the transition state, being a saddle point, is a point with 2 second derivatives of potential energy in two different directions, one which is greater than 0 and one which is less than 0.

{{fontcolor|pink|I can tell from this explanation that you understand the shape of the TS on the PES however your discussion here is not quite complete as you haven't defined the directions q1 and q2 so you haven't fully defined the shape of the TS. A diagram with the vectors included would be really helpful here. In general diagrams are useful! [[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 15:23, 3 June 2019 (BST)}}

Original Surface Plot showing both Potential Energy Surfaces and the Trajectory Pathway:

[[File:1st surface plot.PNG]]

The transition state is located at the point where the interatomic distances are equal i.e. when the bond length of A-B is equal to the bond length of B-C. This is when the A-B bond is forming to make the diatomic product AB and the B-C bond is breaking leaving C on its own. This is shown below in the contour plot:

[[File:1st contour plot.PNG]]

{{fontcolor|pink|This statement about the equal distances is true of this system only because it is a special case where A=B=C. Be careful using this statement as it is not a generic description of a TS. [[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 15:26, 3 June 2019 (BST)}}

=== Question 2: What is 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 position of the transition state can be found by varying the AB and BC distances, with both momentum values of AB and BC being 0, until no variation in the Internuclear Distances vs Time Plot is noticed or until a straight line in the plot is seen. No variation or oscillation in the plot means there are no vibrations which is what happens in the transition state. The distance at which no oscillation was seen was found to be at 0.90774 Å.

Internuclear Distance vs Time Plot at the transition state:

[[File:2nd question graph 1.PNG]]

=== Question 3: Comment on how the MEP and the trajectory you just calculated differ. ===

The plot with the MEP calculation shows a relatively short line with little variation in the y-axis (AB distance) however, the plot with the Dynamics calculation shows an oscillating line. The trajectory for the dynamics calculated plot is also longer in length than the MEP calculated plot. This is due to the MEP calculation not considering any vibrational modes therefore no oscillations are shown on the plot. In actuality, atoms have mass and momentum and so the vibrations of the hydrogen molecule are taken into account shown by the oscillations in the dynamics calculation plot. Thus the model with the dynamics calculation is the more accurate one.

Plot with MEP Calculation:

[[File:ACTUALMEP1.PNG]]

Plot with Dynamics Calculation:

[[File:ACTUALDYNAMIC.PNG]]

=== Question 4: 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
|+ Reactive or Unreactive Trajectories?
! p1 !! p2 !! Etot !! Reactive? !! Illustration of the Trajectory !! Description
|-
| -1.25 || -2.5 ||-99.018 || Yes || [[File:1T.PNG]] ||Contour Plot shows a successful reaction with reactants, having passed through the transition state shown by the trajectory, fully converted into products.||
|-
| -1.5 || -2.0 || -100.456 || No || [[File:2T.PNG]] || Contour Plot shows that reaction does not occur as reactants don't have enough initial momentum and thus required (kinetic) energy to get over the activation energy barrier and reach the energy of the transition state. This is shown by the trajectory not reaching the transition state point and tailing back to the reactants. As the transition state isn't reached, the products aren't formed. ||
|-
| -1.5 || -2.5 || -98.956|| Yes ||[[File:3T.PNG]] || Contour Plot shows a successful reaction with reactants, having passed through the transition state shown by the trajectory, fully converted into products. The reaction is successful as the reactants have enough initial momentum and thus energy to reach the energy of the transition state. ||
|-
| -2.5 || -5.0 || -84.956 || No || [[File:4T.PNG]] || Contour Plot shows although he reactants have enough initial momentum and thus energy to go through the transition state, the transition state collapses back into the reactants. This happens because of barrier recrossing. ||
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:5T.PNG]] || Trajectory from contour plot shows that the reaction goes through/reaches the transition state point more than once. This occurs because of barrier recrossing again. However with these specific reaction conditions, the reaction does take place and the products are formed from the transition state. ||
|}

=== Question 5: 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?===

Transition State Theory assumes and tells us that states that 1 the transition state or activated complex in a reaction is in equilibrium with the reactants and once the reactants turn into the activated complex, the complex doesn't return to the reactants. It also assumes the energy of the particles follows a Boltzmann distribution1 but in this system since there are only 3 atoms in total for this triatomic collision there is no boltzmann distribution. As just mentioned, transition state theory assumes once the reactants turn into the activated complex, the complex doesn't return to the reactants. However in contour plot 4 above where the trajectory shows that although the reaction goes through the transition state, the reactants are still reformed. This is because of barrier recrossing as mentioned above; with this present, the rate of this reaction, experimentally determined, would be lower than the reaction rate that the transition state theory would predict.

== F - H - H system ==

=== Question 1: By inspecting the potential energy surfaces, classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? ===

[[File:D1.PNG]]

The above surface plot shows the potential energy surface for the exothermic conversion of F and H2 to HF and H. The potential energy well slopes downwards from the reactants to the products showing that the energy of the products is lower than the energy of reactants. This reaction is an exothermic one as the bond strength of HF, the product, is stronger than the bond strength of H2, the reactant. This means that the energy released from the formation of the HF bond is greater than the energy required to break the H-H bond in the hydrogen reactant molecule making the overall reaction exothermic.

[[File:D2endo.PNG]]

The above surface plot shows the potential energy surface for the endothermic conversion of HF and H to H2 and F. This reaction is the reverse reaction of the exothermic conversion we've just considered. This potenital energy well slopes upwards from the reactants to the products showing that the energy of the products is higher than the energy of reactants making the reaction endothermic. The energy released from the hydrogen molecule, the product, is less than the energy taken in to break in the H-F bond in the reactant making the overall reaction endothermic.

=== Question 2: Locate the approximate position of the transition state. ===

Hammond's postulate essentially states that if 2 species are of similar energy they are of a similar structure. Thus in this reaction we can deduce that the transition state will be similar in structure to the reactants of the forward exothermic reaction, F and H2. By testing and varying the bond distances of F-H and H-H until the forces exterted on each other were 0 (characteristic in the transition state) the internuclear distances found were 1.811 Å for F-H and 0.745 for H-H Å.

=== Question 3: Report the activation energy for both reactions. ===

The activation energy can be easily calculated by taking away the reactant energy from the transition state energy. Using the internuclear distances of the atoms in the transition state calculated above in the settings function in the programme, the energy of the transition state was found to be -103.752 kcal mol1. The energy of the reactants, for the exothermic reaction for example, was found by inputting the normal bond length for H2 and making the distance between F and H relatively large so they're not interacting yielding the lowest energy for the reactants possible. From this the activation energies for the exothermic reaction was found to be 30.267 kcal mol1 and the activation energy for the endothermic reaction was found to be 0.242 kcal mol1

=== Question 4: In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy for the reaction F + H2 -> HF + H. Explain how this could be confirmed experimentally. ===

Due to the conservation of energy, all initial kinetic energy from the reactants' momentum is converted and released as heat in the reaction once the products HF and H are formed. Due to the release of heat the surroundings temperature would increase as this is an exothermic reaction. This temperature change or increase in this case can be recorded by bomb calorimetry.

=== Question 5: Discuss how the distribution of energy between different modes (translation and vibration) affect the efficiency of the reaction for the reaction HF + H -> F + H2, and how this is influenced by the position of the transition state. ===

For this endothermic reaction (HF + H -> F + H2) there are two main modes of reactant energy to consider as the reactants approach the energy barrier/transition state point to form the products; these modes of energy are translational and vibrational energy. 2Polanyi stated that for simple reactions described as an A + BC system, such as this H + HF system, if they have a late transition state, reactants with high vibrational energy is more effective for the reaction and its success, than reactants with high translational energy. This is because for a late transition state reaction, reactant molecules with high vibrational energy, if in the correct orientation, can vibrate their way to the top of the barrier, passing through the transition state forming the products successfully; however, if the reactants had high translational energy, in trying to reach the top of the energy barrier, they would just collide into the inner wall of the potential energy well and return to where they came from. This reaction being considered here (HF + H -> F + H2) is an endothermic reaction with a late transition state as shown by the surface plot discussed earlier. Therefore to maximise the efficiency and the success of this reaction, reactants should have high vibrational energy. Polanyi also implied, that in this type of reaction (endothermic and late transition state), the products are likely to have high translational energy if the reaction effectively occurs with high vibrational energy reactants.

The reaction showing highest efficiency and most effectiveness for its success was then modelled; high vibrational energy was inputted in the settings function for the molecule HF and high initial momentum of the incoming hydrogen atom for the collision was also inputted. However, unfortunately no such combination of momenta and bond length values were found that yielded a successful reaction, but below shows a contour plot with high vibrational energy in the reactants as shown by the oscillating trajectory which is what is needed to ensure high efficiency of this late transition state endothermic reaction.

[[File:Loooool.png]]

== References ==

1 P. Atkins, J. De Paula, (n.d.). Physical chemistry. 7th edition.

2 J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics Prentice-Hall.

2019-06-03T12:24:22Z

Rk2918: /* 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. */

== 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]]

{{fontcolor|pink|Be careful to makesure you say "potential energy" not just "potential". And yes this is all correct and clearly explained but you are missing part of the definition. The TS is a saddle point because it is the maximum along the reaction trajectory and a minimum in the orthogonal direction, so which vector r1 or r2 will be positive or negative? [[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 12:47, 3 June 2019 (BST)}}

===='''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.

{{fontcolor|pink|Great explanation - very clear and well understood.[[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 12:50, 3 June 2019 (BST)}}

===='''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.

{{fontcolor|pink|True, can you think of why? In terms of the calculation process itself? [[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 12:54, 3 June 2019 (BST)}}

==='''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.

{{fontcolor|pink|Yes, or you could also say that there is not a simple threshold energy to determine whether or not the system will be reactive. It is not as simple as molecules that do have the activation energy will react, and molecules that don't won't react. That would be the simplest form of conclusion to be drawn from this. Your observations are clear and insightful. [[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 12:59, 3 June 2019 (BST)}}

==='''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.

{{fontcolor|pink| Great![[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 13:00, 3 June 2019 (BST)}}

== 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

{{fontcolor|pink|All correct and clearly presented with appropriate diagrams. [[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 13:01, 3 June 2019 (BST)}}

==='''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.

{{fontcolor|pink|Yes, that's right, can you think of a particular experimental setup/technique which can detect such a change in conditions accurately?[[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 13:02, 3 June 2019 (BST)}}

==='''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]]
|}

{{fontcolor|pink|I don't know whether I am just not managing to clearly decipher your figure, but the left diagram doesn't look reactive to me and the right one does. You haven't done any calculations wrong but you've mixed up your order of Polanyi's rules. In an early barrier reaction, kinetic energy is effective in overcoming the barrier but vibrational energy is ineffective in overcoming the barrier. The distinction between the two systems would be clearer if you had one with high vibrational energy and low kinetic and vice versa. These both look like they have quite high vibrational energy. Also, TST and Polanyi's rules should both be quoted with an accompanying literature reference. [[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 13:10, 3 June 2019 (BST)}}

013366322

2019-06-03T12:10:16Z

== 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]]

{{fontcolor|pink|Be careful to makesure you say "potential energy" not just "potential". And yes this is all correct and clearly explained but you are missing part of the definition. The TS is a saddle point because it is the maximum along the reaction trajectory and a minimum in the orthogonal direction, so which vector r1 or r2 will be positive or negative? [[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 12:47, 3 June 2019 (BST)}}

===='''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.

{{fontcolor|pink|Great explanation - very clear and well understood.[[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 12:50, 3 June 2019 (BST)}}

===='''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.

{{fontcolor|pink|True, can you think of why? In terms of the calculation process itself? [[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 12:54, 3 June 2019 (BST)}}

==='''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.

{{fontcolor|pink|Yes, or you could also say that there is not a simple threshold energy to determine whether or not the system will be reactive. It is not as simple as molecules that do have the activation energy will react, and molecules that don't won't react. That would be the simplest form of conclusion to be drawn from this. Your observations are clear and insightful. [[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 12:59, 3 June 2019 (BST)}}

==='''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.

{{fontcolor|pink| Great![[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 13:00, 3 June 2019 (BST)}}

== 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

{{fontcolor|pink|All correct and clearly presented with appropriate diagrams. [[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 13:01, 3 June 2019 (BST)}}

==='''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.

{{fontcolor|pink|Yes, that's right, can you think of a particular experimental setup/technique which can detect such a change in conditions accurately?[[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 13:02, 3 June 2019 (BST)}}

==='''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]]
|}

{{fontcolor|pink|I don't know whether I am just not managing to clearly decipher your figure, but the left diagram doesn't look reactive to me and the right one does. You haven't done any calculations wrong but you've mixed up your order of Polanyi's rules. In an early barrier reaction, kinetic energy is effective in overcoming the barrier but vibrational energy is ineffective in overcoming the barrier. The distinction between the two systems would be clearer if you had one with high vibrational energy and low kinetic and vice versa. These both look like they have quite high vibrational energy. [[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 13:10, 3 June 2019 (BST)}}

2019-05-28T10:53:16Z

Rk2918: /* 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. */

== '''Question 1''' ==

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

The transition state is defined as the maximum on the minimum energy path linking reactants and the products.It is located at the saddle point in the potential energy surface diagram, which defined as : ∂V(ri)/∂ri=0 and H < 0, where H = frr(r0,V0)fVV(r0,V0)−f2rV(r0,V0) .

{{fontcolor|pink|The first sentence is correct however your mathematical definition isn't very clear. Because the TS is a maximum in one direction, the second partial derivative of that function with respect to the TS will be negative, as the second partial derivative of a maximum is always negative. However, it is the maximum on the trajectory of the minimum energy pathway and therefore in the direction that is orthogonal to the reaction trajectory, it is a minimum (in the direction of a cross-section of the well of the pathway) and therefore the second partial derivative will be positive as a second partial derivative of a minimum is always positive. [[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 11:42, 28 May 2019 (BST)}}

The transition state is the specific combination of internuclear distances AB and BC at which the trajectory is a point on the contour plot.

{{fontcolor|pink|Really not sure what this means, sorry! [[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 11:42, 28 May 2019 (BST)}}

=== 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. ===
When r(AB)=r(BC)=0.907 Å, p1 = p2 = 0.0, the transition state occurs. The internuclear distance vs time graph gives a relatively straight line which indicates a constant distance.

{{fontcolor|pink| And why would a "constant distance" (of what exactly - be explicit and specific) mean that you are at the TS? Make sure you always explain your answers thoroughly in order to demonstrate your understanding of the depths of the theory. [[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 11:44, 28 May 2019 (BST)}}

=== Comment on how the mep and the trajectory you just calculated differ. ===
The mep trajectory keeps the potential energy at the minimum in a straight line, whereas the dynamic trajectory is wavy.

{{fontcolor|pink|Exactly, but why? What does it mean, in terms of energy, momenta and the calculations themselves? [[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 11:48, 28 May 2019 (BST)}}

r1 = 0.907, r2 = 0.917

Molecules and atome moving away from each other.

Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed.

r1 = rts+δ, r2 = rts

final values of the positions r1(t) = 3.45, r2(t) = 0.72 and the average momenta p1(t) = -1.45, p2(t) = -2.48 at large t.

The molecule and the atom moving close together instead of moving apart.

=== 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="5"
! P1!!P2 !! Etot!! Reactive?!!Description of the dyanmics!!Illustration of the trajectory
|-
| -1.25 || -2.5|| -99.08|| yes||The trajectory goes through the transition state and also go into the products. ||[[File:Surface01_ql2817.png]]
|-
| -1.5|| -2.0|| -100.22|| no||The trajectory falls back down into the reactants because there was not enough energy for reactants to pass over the transition state. || [[File:Surface02_ql2817.png]]
|-
| -1.5|| -2.5|| -98.91|| yes||The trajectory goes through the transition state and also go into the products with enough energy. || [[File:Surface03_ql2817.png]]
|-
| -2.5|| -5.0|| -84.62|| no||The trajectory was carrying enough energy towards products but then it shifts back into the reactants. || [[File:Surface04_ql2817.png]]
|-
| -2.5|| -5.2|| -83.40|| yes||The trajectory goes through the transition state and also go into the products with enough energy. || [[File:Surface05_ql2817.png]]
|}

{{fontcolor|pink|So what can you conclude from these observations? Is the trend in total energy enough to dictate whether or not a reaction will be successful? (It's not, but why?) [[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 11:50, 28 May 2019 (BST)}}

=== 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? ===
Assumptions <ref name="main">https://en.wikibooks.org/wiki/Statistical_Thermodynamics_and_Rate_Theories/Eyring_Transition_State_Theory#Assumptions_of_Transition_State_Theory
</ref>

*Reactants are in constant equilibrium with the transition state structure.
*The energy of the particles follow a Boltzmann distribution.
*Once reactants become the transition state, the transition state structure does not collapse back to the reactants.

Transition state theory will predict a higher reaction rate than experimental one.

{{fontcolor|pink|This is all correct, but why? Which of these 3 points is untrue and how does it not reflect real life reactions, and why does this mean that the rate will be overpredicted? [[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 11:50, 28 May 2019 (BST)}}

== '''Exercise 2: F-H-H system''' ==

===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? ===
'''F+H2:'''

r(AB)=1.5,r(BC)=0.74, AB,BC m=0

[[File:ql2817-01.png]]

exothermic, because the trajectory goes from a higher PE to a lower PE.

'''H+F2:'''

r(AB)=1.5,r(BC)=0.74, AB,BC m=0

[[File:ql2817-02.png]]

endothermic, because the trajectory goes from a lower PE to a higher PE.

{{fontcolor|pink|You haven't commented on the relative bond strengths of the species involved. [[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 11:50, 28 May 2019 (BST)}}

===Locate the approximate position of the transition state. ===
'''F+H2:'''

r(AB): 1.8108
r(BC): 0.744877
AB BC m: 0

[[File:ql2817-03.png]]

'''H+HF:'''

r(AB): 0.744877
r(BC): 1.8108
AB BC m: 0

[[File:ql2817-04.png]]

===Report the activation energy for both reactions. ===
'''F+H2:'''

activation energy: 30 kcal/mol

[[File:ql2817-05.png]]

'''H+FH::'''

activation energy: 0.44kcal/mol

[[File:ql2817-06.png]]

=== 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.===

reactive trajectory:

r(AB):1.6, r(BC):0.74, P AB:-0.5, P BC:0

[[File:ql2817-07.png]]

[[File:ql2817-08.png]]

From the graph above, it is clearly shown that total energy remains the same. Kinetic energy increases to the same extent as potential energy decreases.
This proves that energy is conserved. This can be confirmed experimentally by measuring the kinetic and potential energy in the real experiment, ie measure temperature change and potential energy can be measured using this software.

{{fontcolor|pink|Can you highlight a particular experiment that could do this? [[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 11:51, 28 May 2019 (BST)}}

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

[[File:ql2817-09.png]]

high translational energy, r(AB)=2, r(BC)=0.74, P AB=-7, P BC=0.5

[[File:ql2817-10.png]]

high vibrational energy, r(AB)=2, r(BC)=0.74, P AB=-1, P BC=6

The reaction is a late TS reaction, also Polanyi's empirical rules tells us that vibrational energy is more important in compare to vibrational energy in a late transition state reaction, which can also be proved in the graphs above.

{{fontcolor|pink|This answer is not sufficient to demonstrate what the question asks. You need to be a bit more thorough i.e. explain Polanyi's rules more clearly and in more detail, and use several examples to demostrate the rules in reactions. You also need to annotate diagrams with explanations that discuss exactly how that proves Polanyi's rules. [[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 11:53, 28 May 2019 (BST)}}

=='''References'''==
<references/>:

2019-05-28T10:25:15Z

Rk2918: /* Reaction Dynamics */

== Introduction to Molecular Reaction Dynamics ==

'''Potential Energy Curve'''

The relative motions of the atoms in a molecule can be described by using an anharmonic oscillator which is derived from a simple harmonic oscillaton. The potential energy curve shows how the potential of the molecule (ex. diatomic molecule) changes as the nuclei are displaced from the equilibrium position which is located at the minimum point of the potential curve. In the diatomic case, the potential energy of the molecule increases exponentially as the separation between the nuclei decreases. As the distance between the nuclei increases, the potential energy of the molecule increases until it reaches the plateau. The difference between the energy of the plateau and the energy of the equilibrium position is the dissociation energy. When the separation is really small, the oscillator follows the simple harmonic oscillator but as the nuclei are further apart, they deviate from the ideality more and follows the anharmonic oscillator.

[[File:JEL3117_Morse_potential.png|frame|none|Figure 1. A potential energy curve of a diatomic molecule.<ref name="MorsePotential"/>]]

== Exercise 1 : H + H2 System ==

In order for H2 to react with H, the H atom should approach to the H2 with a sufficient amount of energy to overcome the activation energy. At the initial condition where time = 0. the bond distance between B and C is represented as r2 and the distance between A and B is represented as r1 as shown in the diagrams below

[[File:JEL_3117_animation_reactant.png|frame|none|Figure 2. positions of the H and H2 in the reactant state.]]

[[File:JEL_3117_animation_product.png|frame|none|Figure 3. positions of the H and H2 in the product state.]]

For the transition state of a symmetric molecule, r1 and r2 should be equal. If there is a sufficient energy to overcome the activation energy, A and B will form a new bond and C will move away as shown above in figure 3.

[[File:JEL_3117_animation_transition_statte.png|frame|none|Figure 4. positions of the H and H2 in the transition state.]]

=== Defining transition state in a potential energy surface diagram ===

On a potential energy surface diagram, the transition state is mathematically defined by:

[[File:JEL3117_transition_state.png]] - (1)

[[File:JEL3117_transition_state1.png]] - (2)

[[File:JEL3117_transition_state2.png]] - (3)

All three equations above should be satisfied to identify the transition state. The first condition states that the product of the first derivative of the potential energy with respect to r1 and the first derivative of the potential energy with respect to r2 should be zero. The second condition states that the second derivative of the potential energy with respect to r1 should be less than zero (meaning maximum point). The third condition states that the second derivative of the potential energy with respect to r2 should me more than zero (meaning minimum point). Thus, the transition state is defined as the maximum point on the minimum energy path linking reactants and the products.

{{fontcolor|pink|This is a really good description. You did not need to include the extra introduction but it is clear that you understand the theory behind the exercise, which is good. [[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 11:06, 28 May 2019 (BST)}}

The transition states are graphically represented below.

[[File:JEL3117_transition_graph1.png|framed|none|300px|Figure 5. the potential energy curve where r1 is constant and r2 is a variable. It shows the minimum energy path along the black line.]]

[[File:JEL3117_transition_graph2.png|frame|none|Figure 6. the potential energy curve where r2 is constant and r1 is a variable. It shows the maximum energy path.]]

The transition states can be distinguished from a local minimum of the potential energy surface because at the local minimum the second derivative of the potential energy with respect to one of the distances would not give positive value (not a maximum point).

=== Trajectories from r1 = r2: locating the transition state ===

The estimated distance transition state position rts is 0.908 Å and this can be shown by "internuclear Distance vs Time" plot as shown below

[[File:JEL3117_transition_0.98.png|frame|none|Figure 7. internuclear Distance vs Time plot at the transition state position rts.]]

At the transition state, I expect the the atoms to vibrate minimum near to zero because the transition state is the stationary maximum point on the minimum energy path. As the they vibrate more and more, they will slide down the potential surface curve and separate to form either product or reactant.

=== Trajectories from r1 = rts + δ, r2 = rts===

[[File:JEL3117_transition1_MEP.png|frame|none|Figure 8. contour plot of the reaction H + H2 when r1 = 0.909, r2 = 0.908 in MEP.]]

[[File:JEL3117_transition_MEP_momentavstime_graph.png|frame|none|Figure 9. momentum vs time graph of the reaction H + H2 when r1 = 0.909, r2 = 0.908 in MEP.]]

[[File:JEL3117_transition1_DYNAMIC.png|frame|none|Figure 10. contour plot of the reaction H + H2 when r1 = 0.909, r2 = 0.908 in Dynamic.]]

[[File:JEL3117_transition_DYNAMIC_momentavstime_graph.png|frame|none|Figure 11. surface plot of the reaction H + H2 when r1 = 0.909, r2 = 0.908 in Dynamic.]]

The above graphs were plotted by setting r1 = 0.909 Å and r2 = 0.908 Å and p1 = p2 = 0 kg m/s.

the first two graphs show the reaction path of the reaction with the MEP mode. MEP sets the momenta and velocities of the molecule to zero (shown in figure 9) in each time every step and process the reaction in a infinitely slow motion. The last two graphs show the reaction path of the reaction with the dynamics mode. Dynamics shows the vibrational motion of the molecule which is derived from the internal energy of the molecule. Thus, oscillation of the energy of the reaction path is shown in the potential energy surface. By setting the momentum to zero at every point, the reaction path will stop when the exchange of the atoms is completed. However, when the momentum is not zero, the products will move away from each other continuously due to conservation of momentum. This is shown in figure 8 and 10 where the reaction path ends earlier for MEP than Dynamics.

If r1 = 0.908 Å and r2 = 0.909 Å, A and B would form the H2 molecule and C would move away from the molecule. Thus, the reaction path would head towards the other end of the contour plot.

[[File:JEL3117_transition_DYNAMIC_intermolecular.png|frame|none|Figure 12. Intermolecular distance vs time graph of the reaction H + H2 when r1 = 0.908, r2 = 0.909.]]

[[File:JEL3117_transition_DYNAMIC_Momenta.png|frame|none|Figure 13. momenta vs time graph of the reaction H + H2 when r1 = 0.908, r2 = 0.909.]]

Figure 12 shows that the distance between B and C increases but the distance between A and B decreases with a little oscillation. Moreover, Figure 13 shows that the momentum of BC increases then reaches the plateau which means that the B and C are separated and the momentum of AB increases then start to oscillate as it plateaus. It means that AB bonds are formed and they are oscillating. Thus, the above data supports that AB bond is formed instead of BC bond when r1 and r2 are reversed

=== Reactive and unreactive trajectories ===

{| class="wikitable" border="1"
|+ Reactive and unreactive trajectories
! p1 !! p2 !! Etot / KJ mol-1 !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive || Initially, C approaches AB molecule. As C approaches, the distance between A and B remains almost constant with no oscillations and the potential energy between B and C increases. As it reaches the transition state, the system has an enough kinetic energy to overcome the activation energy. Therefore, a new BC bond is formed and the AB bond is broken. Then, as the reaction path passes the transition state, the BC molecule oscillates and A moves away from the new product BC. || [[File:JEL3117_reactive_nonreactive_contour_first.png]]
|-
| -1.5 || -2.0 || -100.456 || Unreactive || Initially, the AB molecule oscillates and C approaches AB molecule. As the reaction path reaches the transition state, the system does not have an enough amount of energy to overcome the activation energy and the new BC bond is not formed. Thus, C moves away from the AB molecule and AB molecule remains oscillating and keep their bonds. ||[[File:JEL3117_reactive_nonreactive_contour_second.png]]
|-
| -1.5 || -2.5 || -98.956 || Reactive || Initially, the AB molecule oscillates and C approaches the AB molecule. As the reaction path reaches the transition state, the system has the enough energy to overcome the activation energy and a new BC bond is formed and A moves away from the product as the reaction proceeds. || [[File:JEL3117_reactive_nonreactive_contour_third.png]]
|-
| -2.5 || -5.0 || -84.956 || Unreactive || Initially, the AB molecule does not oscillate and C approaches the molecule AB. As the reaction approaches the transition state, the system has an enough energy to overcome the transition state and the new BC bond is formed and A moves away from the BC molecule. However, the oscillation of the BC bond is too big such that the BC bond is broken and AB bond is formed again. As a result, the AB molecule remains as it is and C moves away from the AB molecule. || [[File:JEL3117_reactive_nonreactive_contour_fourth.png]]
|-
| -2.5 || -5.2 || -83.416 || Reactive || Initially, the AB molecules does not oscillate and C approaches the molecule. As the reaction approaches the transition state, the AB molecule starts to oscillate dramatically such that AB bond is broken. As a result,the BC bond is formed and A moves away from the the molecule BC || [[File:JEL3117_reactive_nonreactive_contour_fifth.png]]
|}

From the table, it can be concluded that activation energy is not the only factor that affects the success of the reaction but other factors such as collision in a correct geometry also contributes to the success of the reaction.

{{fontcolor|pink|OK, this is the right idea - total energy isn't enough to dictate whether or not a system will be reactive. However, where have you got this idea of geometry from? It's true that geometry can affect reaction success but your table doesn't show this. You either need to explain what you mean if this is from your own scientific instinct or you need to cite a literature reference. [[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 11:13, 28 May 2019 (BST)}}

=== Transition state Theory ===

The main assumptions of transition state theory are as follows<ref name="transition state"/> :

1, For each elementary step in a multi-step reaction, the intermediates are long lived enough to reach a Boltzmann distribution of energies before proceeding to the next step.

2, The atomic nuclei behave according to the classic mechanics.

3, The reaction is successful if the reaction path passes over the lowest energy saddle point on the potential energy surface (transition state).

However, in reality, the assumptions start to break down. The transition states are very short lived and the Boltzmann distribution of energies at each step near activated complexes is hard to be obtained. Moreover, in reality, atoms behave in a quantum mechanical manner and the reaction might be successful to form the product even though the energy of the reactants is not enough to overcome the activation energy via quantum tunnelling. Moreover, in reality, the transition state is not necessarily at the lowest saddle point at high temperatures. Thus, there are alternative pathway to the reaction.

As a result, the predicted reaction rate values by using Transition State Theory would be lower than the experimental values due to limitations of the assumptions.

{{fontcolor|pink| You are very close to the right idea in point 3 but you haven't quite followed the thought through to completion. This assumption (point 3) means that TST does not account for the idea of barrier recrossing. I.e. in TST once a molecule has hit the TS, that's it, the reaction is successful. In reality, the molecule can go back to the reactants side, and then over the TS to the product side, and dance around over the TS for a bit. This means that TST predicts that the rate from TST would be higher than in real life (reaction occurs quicker) as it doesn't account for this extra time spent at the TS. So TST overpredicts the rate constant.

Also note that tunelling is not accounted for in this experiment, and also tunneling would accelerate the reaction relative to theoretical prediction (no tunneling) and therefore would have the opposite effect to the barrier recrossing scenario. [[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 11:22, 28 May 2019 (BST)}}

== Exercise 2 : F - H - H system ==

=== PES inspection ===

[[File:JEL3117_F_HH_potential_curve1.png|frame|none|Figure 14. the potential energy surface of F + H2 reaction]]

[[File:JEL3117_H_HF_potential_curve1.png|frame|none|Figure 15. the potential energy surface of H + HF reaction]]

Based on the potential energy surfaces, the F + H2 reaction is exothermic because the energy of the reactant is higher than the energy of the product and the H + HF reaction is endothermic because the energy of the product is higher than the energy of the reactant.

This means that the bond strength of H2 is weaker than the bond strength of HF because the energy released from breaking H-H bond is lower than the energy required to form the H-F bond for the endothermic F + HF reaction and the energy released from breaking H-F bond is higher than the energy required to form the H-H bond for the exothermic F + H2 reaction .

=== Approximate Position of the transition state ===

[[File:JEL3117_F_H2_approximate_TS_intermoleculardistance2.png|frame|none|Figure 16. The intermolecular distance vs time graph of F + H2 reaction]]

[[File:JEL3117_F_H2_approximate_TS_potential_surface2.png|frame|none|Figure 17. The potential energy surface of F + H2 reaction indicating the approximate position of the transition state.]]

Based on the Hammond's postulate, the transition state of the exothermic reaction F + H2 lies near the reactant. Looking at F + H2 graph, the transition state therefore lies near the reactant and it is supported by the intermolecular vs time graph as shown above (No change in the intermolecular distances for a subsequent period of time). Likewise, the transition state of the endothermic H + HF reaction lies near the product.

With trial and errors, the position of the transition state for the exothermic F + H2 reaction was found at r1(AB Distance) = 1.8106929 Å and r2(BC Distance) = 0.745 Å.

The position of the transition state for the endothermic H + HF reaction was found at r1(AB Distance) = 0.745 Å and r2(BC Distance) = 1.8106929 Å.

The activation energies of both of the reactions were found by displacing the activated complex slightly from the transition state.

[[File:JEL_3117_energy_time_graph_exothermic.png|frame|none|Figure 18. Energy vs time graph of the exothermic reaction which illustrates the activation energy of the reaction.]]

[[File:JEL_3117_energy_time_graph_endothermic.png|frame|none|Figure 19. Energy vs time graph of the endothermic reaction which illustrates the activation energy of the reaction.]]

The activation energy of the exothermic reaction F + H2 was calculated by displacing r1 to 1.8107 Å. However, the activation energy of the exothermic reaction was negligible such that was hard to be observed from the energy vs time graph.

The activation energy of the endothermic reaction was calculated by displacing r2 to 1.8106 Å. The activation energy was found as 29.5 KJ/mol.

=== Reaction Dynamics ===

By the conservation of energy, the total energy is always conserved. At the start, H2 and and F atom are stationary thus they only have potential energies. As they are brought together, the potential energies between them decrease and the kinetic energies of H2 and F atom subsequently increase. As they collide, they start to oscillate and H-H bond is broken and new H-F bond is formed. Since the reaction is exothermic, the excess energy is released from the reaction as heat and the remaining energy is converted to the vibrational energy of HF molecule and kinetic energy of H atom.

{{fontcolor|pink|How can the potential energy of the system be decreasing as it goes up the hill of the TS? [[User:Rk2918|Rk2918]] ([[User talk:Rk2918|talk]]) 11:25, 28 May 2019 (BST)}}

Experimentally, the above mechanism can be confirmed by measuring the heat released during the reaction with a calorimeter and the vibrational energy of HF molecule can be measured by using IR.

[[File:JEL_3117_momentum_time_graph_exothermic.png|frame|none|Figure 19. Energy vs time graph of the endothermic reaction which illustrates the activation energy of the reaction.]]

=== Polanyi's empirical rule ===

The position of the transition state is important in determining the distribution of energy between different modes. For the early transition (exothermic reaction), having translational energy in majority is sufficient to form the product because having vibrational energy would cause the molecule to slide the potential energy surface side to side such that it does not have enough energy to overcome the activation energy. However, for the late transition (endothermic reaction, having vibrational energy helps to climb the potential energy surface up to the transition state. Having too much vibrational energy bounces back the product to the reactant.<ref name="polany's rule"/>

== References ==

<references>
<ref name="MorsePotential"> Chemistry LibreTexts. ''The Morse Potential graph''. Available from : https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_(McQuarrie_and_Simon)/05._The_Harmonic_Oscillator_and_the_Rigid_Rotator%3A_Two_Spectroscopic_Models/5.3%3A_The_Harmonic_Oscillator_Approximates_Vibrations </ref>
<ref name="transition state">
Atkins, P.W., and Paula J. De. ''Atkin's Physical Chemistry''. Oxford: Oxford University Press. 2006</ref>
<ref name="polany's rule"> Jeffrey I.Steinfeld, Joseph S. Francisco, William L.Hase. ''Chemical Kinetics and Dynamics''. United States: A Paramount Communications Company, 1989 </ref>
</references>