==Exercise 1==
===Finding the Transition State===
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 a potential energy surface diagram, which is defined mathematically as a two-variable function generically Φ(x,y), the TS can be identified as a saddle point, that is, a point in which ∂(Φ)/∂(x) = ∂(Φ)/∂(y) = 0 and ∂(Φ)/∂2(x) > 0 , ∂(Φ)/∂2(y) < 0 or viceversa.

[[File:jcostoedit.png|400px]]
[[File:jcostoedit2.PNG|400px]]
[[File:jcostoedit3.PNG|400px]]

Observing this surface energy plot it can be seen that the TS is actually the saddle point that defined by ∂(V)/∂(r1+r2) = ∂(V)/∂(r1-r2) = 0 and ∂(V)/∂2(r2+r1) > 0 , ∂(V)/∂2(r1-r2) < 0

Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.
[[File:jcosindvst.png|thumb|300px|Internuclear Distance vs Time plot at the TS]]
By iterating for several different values of r, TS r was found to be r1=r2=0.90775, with the force along AB and along BC ≈ 0.
The Internuclear Distances vs. Time plot, shows all internuclear distances to be constant, confirming that we have found the coordinates of the Transition State.

===Reaction Trajectories===
Due to the symmetrical nature of the Potential Surface, the TS is found at r1=r2=req. A small displacement from this metastable equilibrium leads to the reaction taking place. Several trajectories will now be calculated, starting with a minimum energy potential (mep) and a regular dynamic trajectory, both resulting from slightly displacing r1. Then, a trajectory resulting from a displacing r2. Lastly, inverting the momenta of a system that has reached the products.
Comment on how the mep and the trajectory you just calculated differ.
{| class="wikitable" border=1
! Title !! Image !! Description!!
|-
| MEP || [[File:jcosmep_mep.png|250px]] || The mep for this triatomic system, beginning from r1 slightly displaced from the TS.
|-
| Dynamics Trajectory || [[File:jcosmep_dynam.png|250px]] || Similarly as for the mep, by using the dynamic calculation type.
|-
| Dynamics Trajectory (displacing r2 from equilibrium) || [[File:jcosgoback.png|250px]] || As above, only r2 is slightly displaced in this case, which makes the reaction proceed in the other direction
|-
| Trajectory inverting momentum from products || [[File:jcosrev.png|250px]] || Taking the end coordinates of the H-H + H -> H + H-H reaction and inverting the momentum, the system is made to undergo the reverse reaction.
|-
|}
The ''mep'' follows the ridge at the bottom of the potential energy well, with no oscillation in neither A-B or B-C distances, as there is no momentum in the ''mep''. The Dynamic trajectory shows oscillation, and contrary to the ''mep'' total energy stays constant as the potential energy is transformed into kinetic energy.

===Reactive and Unreactive Trajectories===
Trajectories that lead to reactions taking place are termed reactive, on the contrary, those that do not are termed unreactive. Initially, it would be assumed this is determined simply by if the trajectory has enough energy to proceed to products. The table below shows several trajectories that have been computed.
{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018|| Y || The BC distance steadily decreases as C approaches B, reaching the TS, past the TS, the BC distance oscillates slightly while AB distance increases as A and the new molecule move away from eachother. || [[File:Jcostable1.png|200px]]
|-
| -1.5 || -2.0 || -100.643|| N || The BC distance decreases (approach of C), however, the AB distance is strongly oscillating (vibrational motion). The system does not reach the TS due to this and returns to the reactants.|| [[File:Jcostable2.png|200px]]
|-
| -1.5 || -2.5 || -98.956|| Y || The BC distance decreases (approach of C) and AB oscillates. The system passes through the transition state and to the products, with A and B-C getting further apart and the B-C distance oscillating. || [[File:Jcostable3.png|200px]]
|-
| -2.5 || -5.0 || -84.956|| N || The BC distance decreases (approach of C), the reaction makes it past the transition state, however it crosses back into the reactants side, with A-B oscillating (due to A-B having absorbed the energy into vibrational energy.)|| [[File:Jcostable4.png|200px]]
|-
| -2.5 || -5.2 || -83.416|| Y || As above, the system crosses back into the reactants, only it crosses again into the products, with the product B-C vibrating strongly. || [[File:Jcostable5.png|200px]]
|}

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? 
What can be concluded from this set of trajectories is a system being higher in KE (with more that enough energy to reach the transition state) does not imply that the trajectory will eventually lead to the products side. 
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 (and the calculations that are carried out here) assumes reactions work according to Classical Mechanics, which at the scale of chemical reactions, is inaccurate due to orbital interactions (which arise from quantum mechanics) being the basis of chemical reactions. TS theory assumes simple nuclear collisions and electrons are not considered, electron mass or interactions are not accounted for due to electrons being much smaller the nucleus.

==Exercise 2==
===PES analysis===
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:jcoshf23d.png|thumb|HF PES]]
 
''F + H2 -> F-H + H'' Exothermic 
''F-H + H -> F + H2'' Endothermic 
 
This implies that the F-H bond is much stronger than the H-H and much difficult to break (needing an input of energy (+ΔG, thus endothermic) to do so). 
Locate the approximate position of the transition state 
By the same method used for the H--H--H potential energy surface, with the intuition (from Hammond's postulate) that the TS is going to be closer to the H-H + F side, the TS coordinates are: 
r1 = 0.74485 
r2 = 1.81050 

Report the activation energy for both reactions

''F + H2 -> F-H + H'' Activation E: 0.096 kcal/mol 
''F-H + H -> F + H2'' Activation E: 29.82 kcal/mol 
Calculated by slight displacement from the TS in an ''mep'' trajectory. 
 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. 
[[File:jcosyay.png|left]]
Initially, the H2 bond vibrates,then the bond breaks and the H-F bond forms and vibrates very strongly, some of the energy trapped in the H2 bond has gone into the H-F bond vibration. As mentioned before the reaction is exothermic and it would liberate heat.

Additionally, several calculations starting from coordinates r1=0.74, on the ''F + H2'' side, with low approaching F momentum (p2=-0.5) and a range of H-H momenta from -3 to +3 lead to no reactive trajectories, while increasing p2 to -0.8 while overall lowering the total energy of the system lead to trajectories being reactive.

For a H-F + H reaction, starting at that the H-F end of the PES(r1=2.4, r2=0.9);
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. 
Conditions in which momentum in the H-F bond is small and momentum of the approaching H is large lead to unreactive trajectories. Conversely, the opposite is true for high H-F momentum and low approaching H momentum, leading to reactive trajectories. As it was seen before, the TS is similar to the ''F + H2'' side, meaning that it is an early TS for ''F + H2 -> F-H + H'' and a late TS for ''F-H + H -> F + H2'', which would lead to their opposite behaviours.

2019-05-23T11:07:06Z

2019-05-22T18:20:01Z

Jc8717:

=====Finding the Transition State=====
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 a potential energy surface diagram, which is defined mathematically as a two-variable function generically Φ(x,y), the TS can be identified as a saddle point, that is, a point in which ∂(Φ)/∂(x) = ∂(Φ)/∂(y) = 0 and ∂(Φ)/∂2(x) > 0 , ∂(Φ)/∂2(y) < 0 or viceversa.

[[File:jcostoedit.png|300px]]
[[File:jcostoedit2.PNG|300px]]
[[File:jcostoedit3.PNG|300px]]

Observing this surface energy plot it can be seen that the TS is actually the saddle point that defined by ∂(V)/∂(r1+r2) = ∂(V)/∂(r1-r2) = 0 and ∂(V)/∂2(r2+r1) > 0 , ∂(V)/∂2(r1-r2) < 0

[[File:jcosindvst.png|thumb|300px|Internuclear Distance vs Time plot at the TS]]
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.

By iterating for several different values of r, TS r was found to be r1=r2=0.90775, with the force along AB and along BC ≈ 0.
The Internuclear Distances vs. Time plot, shows all internuclear distances to be constant, confirming that we have found the coordinates of the Transition State.
=====Reaction Trajectories=====
Due to the symmetrical nature of the Potential Surface, the TS is found at r1=r2=req. A small displacement from this metastable equilibrium leads to the reaction taking place. Several trajectories will now be calculated, starting with a minimum energy potential (mep) and a regular dynamic trajectory, both resulting from slightly displacing r1. Then, a trajectory resulting from a displacing r2. Lastly, inverting the momenta of a system that has reached the products.
Comment on how the mep and the trajectory you just calculated differ.
{| class="wikitable" border=1
! Title !! Image !! Description!!
|-
| MEP || [[File:jcosmep_mep.png|250px]] || The mep for this triatomic system, beginning from r1 slightly displaced from the TS.
|-
| Dynamics Trajectory || [[File:jcosmep_dynam.png|250px]] || Similarly as for the mep, by using the dynamic calculation type.
|-
| Dynamics Trajectory (displacing r2 from equilibrium) || [[File:jcosgoback.png|250px]] || As above, only r2 is slightly displaced in this case, which makes the reaction proceed in the other direction
|-
| Trajectory inverting momentum from products || [[File:jcosrev.png|250px]] || Taking the end coordinates of the H-H + H -> H + H-H reaction and inverting the momentum, the system is made to undergo the reverse reaction.
|-
|}

=====Reactive and Unreactive Trajectories=====

{| class="wikitable" border=1
! p1 !! p2 !! Etot !! Reactive? !! Description of the dynamics !! Illustration of the trajectory
|-
| -1.25 || -2.5 || -99.018|| Y || The BC distance steadily decreases as C approaches B, reaching the TS, past the TS, the BC distance oscillates slightly while AB distance increases as A and the new molecule move away from eachother. || [[File:Jcostable1.png|200px]]
|-
| -1.5 || -2.0 || -100.643|| N || The BC distance decreases (approach of C), however, the AB distance is strongly oscillating (vibrational motion). The system does not reach the TS due to this and returns to the reactants.|| [[File:Jcostable2.png|200px]]
|-
| -1.5 || -2.5 || -98.956|| Y || The BC distance decreases (approach of C) and AB oscillates. The system passes through the transition state and to the products, with A and B-C getting further apart and the B-C distance oscillating. || [[File:Jcostable3.png|200px]]
|-
| -2.5 || -5.0 || -84.956|| N || The BC distance decreases (approach of C), the reaction makes it past the transition state, however it crosses back into the reactants side, with A-B oscillating (due to A-B having absorbed the energy into vibrational energy.)|| [[File:Jcostable4.png|200px]]
|-
| -2.5 || -5.2 || -83.416|| Y || As above, the system crosses back into the reactants, only it crosses again into the products, with the product B-C vibrating strongly. || [[File:Jcostable5.png|200px]]
|}

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?

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?


=====fhh=====

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 -> F-H + H Exothermic
F-H + H -> F + H2 Endothermic

Locate the approximate position of the transition state
r1 = 0.74485
r2 = 1.81050
Hammond's postulate
Report the activation energy for both reactions

 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.

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.