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MRD:bc2116

2018-06-09T21:32:02Z

Fjs113: Marked by fjs113

= Molecular Reaction Dynamics: Boru Chen =

== <math>H_2</math> + H system ==

=== Dynamics from Transition State Region ===

The PES has two components:

<math>\frac{\partial V}{\partial r_1}</math> and <math>\frac{\partial V}{\partial r_2}</math>

Where V is the potential energy, and <math>r_1</math> is the intermolecular distance between <math>H_b</math> <math>H_c</math>, <math>r_2</math> is the intermolecular distance between <math>H_a</math> and <math>H_b</math>. At transition structure, both components have value of zero. At energy minima and transition structure, their first derivatives equal to zero. Transition state can be distinguished by taking the second derivative of potential energy. Transition state is at energy maximum along reaction coordinate, and at energy minimum in any other direction, whereas energy minimum is at a minimum in all directions. Mathematically speaking, energy minimum’s second derivative is always a positive value at any position, whereas transition state’s second derivative is negative along reaction coordinate, and positive everywhere else.

{{fontcolor1|gray|(Almost perfect! You need to make sure to say second '''partial''' derivative though. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 22:32, 9 June 2018 (BST))}}

==== Transition State Position Estimation ====
Best estimate = 0.9075 Armstrong. Inter nuclear distance vs time plot

[[File:Exercise 1 Internuclear distance to time1.png]]

Note: A-B line can be vaguely seen behind the yellow line.

{{fontcolor1|gray|(Ok, but how did you arrive at this value? [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 22:32, 9 June 2018 (BST))}}

==== MEP Model ====
[[File:Exercise 1 MEP plot.png]] [[File:12.png]] [[File:11.png]]

In MEP model, internuclear momentum remains constant, BC distance increases slowly and AB distance decreases slowly. This is because in MEP model, velocity is equated to 0 after every calculation. This means despite the fact that atoms should be accelerating as it picks up kinetic energy, its velocity is kept constant, hence its momentum is kept constant.

{{fontcolor1|gray|(Velocity is actually always 0 as you can see from your momentum plot. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 22:32, 9 June 2018 (BST))}}

==== Dynamic Model ====

[[File:Exercise 1 Dynamic plot.png]] [[File:0.png]] [[File:01.png]]

In this model, as distance between BC increase, and BC momentum increases with an increasing rate. This is because as reaction goes from transition state to AB, potential energy in transition state is released as kinetic energy. Since the system is isolated, this kinetic energy is captured by C, and it begins to accelerate. Since momentum is proportional to velocity, BC momentum increases.

AB momentum experiences periodic fluctuation after collision, because it is expressing the extra kinetic energy as vibrational energy between A and B.

==== MEP calculations with different initial conditions ====

{{fontcolor1|gray|(Why does this say MEP? These calculations are clearly not MEP. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 22:32, 9 June 2018 (BST))}}

1. If we changed the initial conditions (reverse)?

[[File:Surface Plot 1.png]] [[File:Figure 1.png]] [[File:Figure 1-1.png]]

In this scenario, transition state structure rolls back to BC + A structure. As distance between AB increase, and AB momentum increases with an increasing rate. This is because as reaction goes from transition state to BC, potential energy in transition state is released as kinetic energy. Since the system is isolated, this kinetic energy is captured by A, and it begins to accelerate. Since momentum is proportional to velocity, AB momentum increases.

BC momentum experiences periodic fluctuation after collision, because it is expressing the extra kinetic energy as vibrational energy between B and C.

2. If the new initial positions correspond to the final positions above and final momenta values but with their signs reversed?

[[File:Surface Plot 2.png]] [[File:Figure 2-1.png]] [[File:Figure 2-2.png]]

AB moves towards to and stops at transition state. AB distance is kept constant as BC distance decreases. AB momentum fluctuates periodically as it is vibrating. BC momentum is kept constant then experience sudden increase as it accelerates up the potential energy hill to reach transition state.

=== Reactive and Unreactive Trajectories Table ===

{| class="wikitable" border=1
! p1 !! p2 !! Etot <math>kcalmol^{-1}</math> !! Trajectory
|-
| -1.25 || -2.5 ||| -99.018 ||| [[File:1.png]] ||| Plot shows a reactive trajectory, as AB bond shortens and BC bond lengthens.
|-
| -1.5 || -2.0 ||| -100.456 ||| [[File:2.png]] ||| Plot shows an unreactive trajectory. BC bond never lengthens and AB bond never becomes close enough.
|-
| -1.5 || -2.5 ||| -98.956 ||| [[File:3.png]] ||| Plot shows a reactive trajectory, as AB bond shortens and BC bond lengthens.
|-
| -2.5 || -5.0 ||| -84.956 ||| [[File:444.png]] ||| Plot shows barrier crossing. Although bond does form for some time, however reaction reverts back to reactants.
|-
| -2.5 || -5.2 ||| -83.416 ||| [[File:5.png]] ||| Plot shows barrier crossing. Product is formed
|}

The main assumption is if the incoming atom has enough kinetic energy to overcome kinetic barrier, the reaction will occur. This model also assumes atoms follow newtonian physics.

{{fontcolor1|gray|(You need to elaborate more here. This is a very lacklustre answer. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 22:32, 9 June 2018 (BST))}}

== F + H + H System ==
=== PES Report ===
[[File:HHF PES.png]]

Plot above illustrate reaction H + H + F system potential energy surface, where A = F, B = H and C = H. Graph shows that HH appears to be deeper in energy valley, hence HH formation is exothermic, and HF formation is endothermic.

==== Position of transition state ====
HH distance 1.812 A

HF distance 0.744 A
{{fontcolor1|gray|(Again, how did you get these values? [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 22:32, 9 June 2018 (BST))}}

[[File:HHF transition state.png]]
==== Activation energy calculation: ====
Positions of these structures were estimated using MEP plot

Transition state potential energy: -103.749 <math>kcalmol^{-1}</math>

H2 potential energy: -133.871 <math>kcalmol^{-1}</math>

[[File:H2 potential energy.PNG]] [[File:H2 internuclear distance vs time.PNG]]

HF potential energy: -103.837 <math>kcalmol^{-1}</math>

[[File:HF_potential_energy.PNG]] [[File:HF internuclear distance vs time.PNG]]

HF + H activation energy = -103.749 - -103.837 = 0.088 <math>kcalmol^{-1}</math>

HH + F activation energy = -103.749 - -133.871 = 30.122 <math>kcalmol^{-1}</math>

=== Reaction Dynamics ===
==== Energy conservation ====

[[File:Skate park.png]]

Model sets reaction as an isolated system, and all atoms obey newtonian laws of physics. Once atom moves downhill, the potential energy is converted into kinetic energy, and is released as vibration of bonds. Temperature is directly proportional to the kinetic energy molecules have, hence as reaction progress, temperature of system would rise. This can be measured by calorimetry experimentally.

====Trajectories table ====
[[File:Trajectory table HF.PNG]]

[[File:Internuclear momentum HF=-0.5 HH=2.9.png]]

During the collision, HH momentum has relatively a small change in comparison with HF momentum After the collision, if the trajectory is reactive, HH momentum will have none to small fluctuation, vice versa for HF momentum. If the trajectory is unreactive, HH momentum will have very large fluctuation, vice versa for HF momentum.

[[File:Internuclear momentum HF=-0.8 HH=0.1.png]]

If pHH = 0.1 and pHF = -0.8, HH momentum has small periodic fluctuations and HF momentum has large periodic fluctuations.

{{fontcolor1|gray|(You're missing a few questions at the end! Overall, this is not a good report. You did not explain most of what you did. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 22:32, 9 June 2018 (BST))}}

MRD:BR516

2018-06-03T17:27:00Z

Fjs113: Marked by fjs113

==BR516 Molecular reaction dynamics==
===Excercise 1: H + H2 system===
====What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.====

The transition state is the maximum energy point on the minimum energy trough as a saddle point. This transition state has a potential energy gradient of 0 with respect to rAB and rBC. This transition point is a maximum in one direction yet a minimum in another.

{{fontcolor1|gray|(All correct, but what is "one direction" and "another"? What you mean is commonly called the reaction coordinate and its orthogonal coordinate. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 18:26, 3 June 2018 (BST))}}

The minima, however, are minimums in both directions as ∂2V/∂rAB > 0 and ∂2V/∂rBC > 0. This means any variance in the length of rAB or rBC will lead to an increase of energy, indicating that these regions represent the stable reactants and products.

====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.====
Due to the fact that this reaction involves 3 identical atoms, at the transtion state, rAB=rBC. Buy setting the reactants to have no initial momentum, they wont have nay energy to escape the minimum path. By varying the values of rAB and rBC, and observing the “Internuclear Distances vs Time” plot, the transition state distance can be found as this will have no variation in distance with time. rts was found to be 0.908

[[File:BR516_Q2_internuclear_distances_vs_time.PNG|center|thumb|“Internuclear Distances vs Time” plot for 3 H atoms at initial momentum = 0 and initial internuclear separation 0.908]]

====Comment on how the MEP and the trajectory you just calculated differ.====
The MEP (minimum energy path) shows the path of the reaction if the molecules have 0 inertia. This allows the calculation to trace the lowest energy path the reaction could possibly take, the "floor" of the trough in the potential energy surface. The dynamic trajectory takes into account the momentum of the particles. This causes the slight oscillation along the same rough path, as shown by the dynamic plot.

[[File:BR516_Q3_MEP.PNG|center|thumb|MEP contour plot]]
[[File:BR516_Q3_dynamic.PNG|center|thumb|Dynamic contour plot]]

====Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.====
{| class="wikitable" border=1
! p1 !! p2 !! Total Energy !! Reactive/Unreactive !! Contour plot !! Description
|-
| -1.25 || -2.5 || -99.018 || Reactive || [[File:BR516_Q4_-1.25_-2.5.PNG]] || The stationary H-H molecule reacts with another colliding H with the correct momentum, and a new H-H bond forms.
|-
| -1.5 || -2.0 || -100.456 || Unreactive || [[File:BR516_Q4_-1.5_-2.0.PNG]] || An oscillating H-H molecule collides with an H, but not with enough momentum to form a new bond so they are repelled.
|-
| -1.5 || -2.5 || -98.956 || Reactive || [[File:BR516_Q4_-1.5_-2.5.PNG]] || An oscilating H-H molecule collides with another H to form a new bond with a higher vibrational frequency
|-
| -2.5 || -5.0 || -84.956 || Unreactive || [[File:BR516_Q4_-2.5_-5.0.PNG]] || A stationary H-H molecule collides with another H to form a new bond, however the H has too much momentum, and is repelled by the B molecuels nucleus. This causes The AB molecule to remain intact with a much larger vibrational mode, and causes the BC distance to increase as it is repelled.
|-
| -2.5 || -5.2 || -83.416 || Reactive || [[File:BR516_Q4_-2.5_-5.2.PNG]] || The AB molecule of H-H collides with the C molecule of H, but unlike in the above condition, The C molecule has enough momentum to recross the transition state region. The BC bonds forms, Breaks into an AB bond then reforms into the BC bond.
|}

====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 splits the reaction into 2 distinct areas: the reactant and product space. It states that the reactants must cross the energy threshold of the saddle in the transition area to reach the product space. It also assumes atoms obey the laws of classical mechanics using the Born-Oppenheimer approximation, and that each reaction only has 1 saddle point and 1 transition state.

We can see from the above scenarios that for molecules with high momentum, the transition state can be recrossed, so the transition state prediction for reaction rate values will not provide an close approximation to experimental vales run at high temperatures. Therefore, the transition state theory should only be used for reactions at low temperature conditions for predicting reaction rates.

===EXERCISE 2: F - H - H system===
====Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?====
The F + H2 -> FH + H reaction is exothermic, while it's opposite reaction FH + H -> F + H2. This is due to the greater relative strength of the H-F bond (565 kj/mol) to the H-H (432 kj/mol)bond, meaning more energy has to be used to break the H-F bond than is gained forming an H-H bond[1].

{{fontcolor1|gray|(Would've been better if you used the potential energy surface generated by the program to show this. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 18:26, 3 June 2018 (BST))}}

====Locate the approximate position of the transition state.====
The transition state of the F - H - H system is found at a distance of 0.745 between H-H and 1.811 between H-F. The Momenta of all particles were set to 0 and the initial distance between them varied until an initial distance was found that would remain constant over time.

{{fontcolor1|gray|(Correct, but you did not explain why there is no movement at the TS. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 18:26, 3 June 2018 (BST))}}

[[File:BR516_Q7_Distance_time.PNG|center|thumb|Interatomic distance vs time with atom A H, atom B H and atom C F at rAB = 0.745 and rBC = 1.811]]

====Report the activation energy for both reactions====
To find the activation energy, the MEP was found for both the formation of H2 and HF from the H-H-F transition state. The minimum energy point from both products was found, and using this value the activation energy can be found (difference between the maximum energy of the saddle and the minimum energy of the products).
H-H-F to HF MEP: Initial energy -103.869, product energy -133.858. Activation energy of HF + H -> H2 + F = +29.989
H-H-F to H2 MEP: Initial energy -103.869, product energy -103.941. Activation energy of H2 + F -> HF + H = + 0.072

[[File:BR516_Q8_MEP_H2_to_HF.PNG|center|thumb|MEP of progression to HF product]]
[[File:BR516_Q8_energy_time_HF.PNG|center|thumb|Energy vs Time of H-H-F to HF product]]
[[File:BR516_Q8_energy_time_HH.PNG|center|thumb|Energy vs Time of H-H-F to H2 product]]
{{fontcolor1|gray|(You can zoom in on this graph to show the slight drop in potential energy! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 18:26, 3 June 2018 (BST))}}
[[File:BR516_Q8_MEP_HH.PNG|center|thumb|MEP of progression to H2 product]]

====In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?====
Using the reaction condition of A=H, B=H, C=F, rAB = 0.8, rBC = 2.0, pAB = 1, pBC = -1.5, a successful reaction can be seen of the formation of HF. The reaction energy is converted from potential to kinetic, as shown by the vast increase in magnitude of vibration of H-F as compared to the initial H2 molecule. Experimentally, this would be shown by a large increase in temperature of the reaction mixture caused by the increase in average kinetic energy.

[[File:BR516_Q9_contour_plot_reaction.PNG|center|thumb|Contour plot of successful reaction of formation of HF]]
[[File:BR516_Q9_momentum_graph.PNG|center|thumb|Momentum Vs. time of H2 + F to form HF + H]]

====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.====
Polyani's rules link the effect of vibrational and translational kinetic energy on a reaction. They state that if the reaction has an early transition state close to the energy of the products, translational energy is more important for promoting a successful reaction. Conversely, if a reaction has a late transition state, a higher vibrational energy will promote a more successful reaction.

For the reaction of HF + H -> H2 + F, a higher vibrational energy of the HF molecule is required, as stated by polyani's rules. This is due to the relatively late transition state of the endothermic reaction. Similarly for the reverse reaction, very little vibrational energy is required and only a slight amount of translational energy is required due to the low activation energy of the exothermic reaction.

{{fontcolor1|gray|(You should use some trajectories to show your point here. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 18:26, 3 June 2018 (BST))}}

===References===
[1] http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html, accessed 19/05/2018

MRD:IM915

2018-06-03T16:58:28Z

Fjs113: Marked by fjs113

==Theory of Molecular Dynamics==

Chemical reactions can be simulated by assuming motion of particles follows Newton's laws of motion (as atom mass is large enough to ignore relativistic effects without causing too great an error, although this neglects quantised vibrational effects), and describing their relative motion. Interatomic interactions govern the relative motion of particles and can be expressed as a potential energy surface V(r1,r2,....rn-1) where n is number of particles. In diatomic systems this is simply the potential energy curve described by the Lennard-Jones potential V(r), where r is the interatomic distance. Force will depend on the derivative with respect to time of the potential - therefore the force on the system will equal 0 at turning points (minima or maxima in the 2D case).

Diatom collisions can be studied using a 3D potential energy surface, where for the reaction A + BC → AB + C, the surface is defined by V(r1,r2) where r1 is the distance between A and B, and r2 is the distance between B and C. Initially, the distance r1 will be large and will decrease as molecule A approaches BC, while r2 will remain relatively constant, oscillating around the equilibrium bond length due to vibrational energy. After the reaction, r1 will now be small as the bond AB has been formed, while R2 will be large and increasing as atom C moves away from AB. During the reaction, the system will pass through a transition state A--B--C, which will be a saddle point on the potential energy surface in the minimum energy path (i.e. from one minimum energy channel at equilibrium bond length to the other).<ref>P. Atkins and J. de Paula, in Atkins’ Physical Chemistry, Oxford University Press, 10th edition, 2014.</ref>

As the triatomic system is expressed as a 3D surface, calculating the minima (equilibrium bond lengths) and the saddle point (transition state) is more complicated than for the simple 2D case. For a 3D surface, a stationary point is where the gradient in all directions is zero.

<math>\frac{dV}{dr_1} = \frac{dV}{dr_2} = 0</math>

Stationary points on a surface can be many types, however the most common are minima, maxima, and saddle. These can be differentiated from each other using a second derivative test.

<math>D(r_1,r_2) = \frac{dV^2}{dr_1^2}.\frac{dV^2}{dr_2^2} - [\frac{dV^2}{dr_1r_2}]^2</math>

For a maximum point; <math> D > 0 </math> and <math>\frac{dV^2}{dr_1^2} > 0 </math>

For a minimum point; <math> D > 0 </math> and <math>\frac{dV^2}{dr_1^2} < 0 </math>

For a saddle point; <math> D < 0 </math>

{{fontcolor1|gray|(Very nice writeup! Well done. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:58, 3 June 2018 (BST))}}

==H + H2 system==

===Potential Energy Surface===

The potential surface was generated for the H + H2 system. As the start and end products of the reaction are the same, the surface plot is perfectly symmetric. Plotted on the surface is the reactive trajectory for the initial conditions chosen (r1 = 2.3, r2 = 0.74, ρ1 = -2.7, ρ2 = 0). The surface is shown as both a contour plot and a 3D surface plot.

{| class = "wikitable"
|-
|[[File:IM Hydrogen system surface 1.png|300 px]]||[[File:IM Hydrogen system contour 1.png|300 px]]
|-
|Surface plot for H + H2
|Contour plot for H + H2
|}

===Determining the Transition State===

Finding the transition state requires finding the saddle point between the two minima channels of the start and end products. As the system is totally symmetric, in the transition state r1 = r2. The transition state can be determined by testing the trajectory for different values of r1, keeping both initial momenta equal to 0. As on the saddle point there is no gradient at right angles to the ridge, on the transition state ridge the trajectory will oscillate on the ridge but not 'fall off' into either product or reactant channels. At the exact transition state, there will be no oscillation and the trajectory will remain constant.

Using a simulation with 1000 steps, the transition state was determined to be at r1 = r2 = rTS = 0.90774 Å. At this rTS, the trajectory does not fall down to either products or reactants, and the graph of Internuclear Distances vs Time shows no oscillation and is a straight line.

{| class = "wikitable"
|-
|[[File:IM Hydrogen system TS surface.PNG|300 px]]||[[File:IM Hydrogen system TS contour.PNG|300 px]]||[[File:IM Hydrogen system TS dist v time.PNG|300 px]]
|-
|Surface plot showing transition state
|Contour plot showing transition state
|Plot of Internuclear Distances vs Time
|}

===Comparison of dynamics and mep===

The mep (minimum energy path) is a trajectory corresponding to infinitely slow motion. By setting the initial conditions to slightly displaced from the transition state (r1 = rTS + 0.01, r2 = rTS, the minimum energy path can be plotted.

The mep does not provide a realistic account of motion of atoms in a reaction- using the dynamics calculation type, mass of atoms (and therefore inertial motion) can be taken into account. The same initial conditions were used for a dynamics simulation as used for the mep.

{| class = "wikitable"
|-
|[[File:IM mep contour plot.png|300 px]]||[[File:IM dynamics contour plot.png|300 px]]
|-
|Contour plot of the minimum energy path
|Contour plot of the dynamics calculated trajectory
|}

The differences are best seen in the contour plot- the mep simply follows a straight path along the bottom of the channel, wile the dynamics trajectory results in an oscillation in the final trajectory so it oscillates up and down the sides of the product channel.

{| class = "wikitable"
|-
|[[File:IM dynamics distances vs time.png|300 px]]||[[File:IM dynamics momenta vs time.png|300 px]]
|-
|Intermolecular Distances vs Time
|Intermolecular Momenta vs Time
|}

===Trajectories===

The final value of r2 oscillates around r2 = 0.74 Å, which is the equilibrium bond length for the H-H molecule formed. The final value of r1 = 9.01 Å, although this value is steadily increasing and will continue to increase as the newly formed H-H molecule and H atom move away from each other.

The average momentum ρ2 oscillates around roughly ρ2 = 1.24, while ρ1 is constant at ρ1 = 2.48.

If the inital conditions were swapped (i.e. r1 = rTS and r2 = rTS + 0.01), the final values for r1 and r2 would be swapped, as would the momentum values. The trajectory would be down the other channel, i.e. the bond formed would be AB rather than BC- the starting point of the trajectory would be offset from the transition state towards the other channel from that of the first set of conditions.

{|class = "wikitable"
|-
|[[File:IM dynamics swapped conditions.png|300 px]]
|-
|Dynamics for swapped initial conditions
|}

Using the final positions of the previous trajectories as the new initial positions, and using the final momenta as initial momenta with reversed sign, a new simulation was run. This simulation shows a trajectory that reaches almost the top of the transition state but does not have enough energy to make it over the barrier- it returns to the point at which the previous trajectory was started (slightly post barrier) before falling back down to the starting channel.

{|class = "wikitable"
|-
|[[File:IM using end traj for start.png|300 px]]||[[File:IM using end traj for start dist v time.png|300 px]]
|-
|Trajectory
|Distance vs Time
|}

===Reactive and Unreactive Trajectories===

From previous calculations it can be concluded that trajectories with initial conditions in the range r1 = 0.74 Å, r2 = 2.0 Å with -1.5 < ρ1 < -0.8 and p2 = -2.5 are reactive. Hypothetically all trajectories with the same initial positions and higher momenta would be reactive as they possess enough kinetic energy to overcome the activation barrier.

{|class = "wikitable"
|-
!ρ1
!ρ2
!Total energy kcal/mol
!Trajectory plot
!Reactive?
!Description
|-
| -1.25
| -2.5
| -99.018
|[[File:IM trajectory combination 1.png|300 px]]
|Yes
|The trajectory approaches the barrier and passes over the transition state barrier, falling down to the products with a small amount of vibrational energy. In reality this would be an H atom approaching the non-oscillatory H2 molecule and reacting to form an oscillating H2 molecule and an H atom.
|-
| -1.5
| -2.0
| -100.456
|[[File:IM trajectory combination 2.png|300 px]]
|No
|The trajectory approaches the transition state barrier but has too little kinetic energy to pass over the transition state, and so returns to the initial reactants. In reality this would be the H atom approaching the H2 and then moving away again without reacting.
|-
| -1.5
| -2.5
| -98.956
|[[File:IM trajectory combination 3.png|300 px]]
|Yes
|The trajectory approaches the barrier with a small amount of vibrational energy and passes over the transition state barrier, falling down to the products with a larger amount of vibrational energy. In reality this would be an H atom approaching the slightly
oscillating H2 molecule and reacting to form a more oscillatory H2 molecule and an H atom.
|-
| -2.5
| -5.0
| -84.956
|[[File:IM trajectory combination 4.png|300 px]]
|No
|The trajectory approaches the barrier with enough kinetic energy to pass over the transition state and forms a very highly oscillatory state that passes back over the transition barrier and reforms the initial molecules. In reality this would be an H atom approaching the H2 molecule with high velocity, forming the product H2 molecule which vibrates hard enough that it dissociates and the initial H2 molecule reforms.
|-
| -2.5
| -5.2
| -83.416
|[[File:IM trajectory combination 5.png|300 px]]
|Yes
|The trajectory approaches the barrier with enough kinetic energy to pass over the transition state and forms a very highly oscillatory state that passes back over the transition barrier but then crosses back adn forms the products. In reality this would be an H atom approaching the H2 molecule with high velocity, forming the product H2 molecule which vibrates hard enough that it dissociates and the initial H2 molecule reforms, which then vibrates and dissociates again and the product H2 is formed.
|}

===Transition state theory and it's limitations===

Transition state theory is a highly useful tool for analysis of chemical reaction dynamics. Also referred to as activated complex theory, it provides a way to relate rate constants of reactions to models of the cluster of atoms formed when reactants come together (the transition state / activated complex) and the rate at which this transition state is formed.<ref>P. Atkins and J. de Paula, in Atkins’ Physical Chemistry, Oxford University Press, 10th edition, 2014.</ref>

The main assumptions of transition state theory are;

• Intermediates are long-lived and reach a Boltzmann distribution of energies before continuing to the next step of the reaction. Short-lived intermediates result in deviations from transition state theory.

• Atomic nuclei follow newton's laws of motion and can be described by classical mechanics. This ignores quantum tunneling effects (especially for reactions with low energy barriers (low activation energy) as tunneling probability is higher for lower barrier heights). Classical mechanics also ignores the quantisation of vibrations.

• The reaction is carried out at low temperature. At high temperatures, molecules have a large amount of vibrational energy and can proceed via transition states far from the lowest energy saddle point, resulting in very different trajectories to those predicted.

Transition state theory will compare best with experimental values when activation energy of the studied reactions is high (to reduce tunnelling effects), when the transition state is long-lived (slow reactions), and when the reaction is carried out at a low temperature.

{{fontcolor1|gray|(Very good! One more thing is important: TST ignore barrier recrossing effects, which you can see happening in the previous question in cases 4&5. This will lead to overestimating the reaction rate. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:58, 3 June 2018 (BST))}}

==F-H-H system==

===Potential Energy Surface===

The potential surface was generated for the F-H-H system. The surface is shown as both a contour plot and a 3D surface plot. (AB = H-F, BC = H-H)

{| class = "wikitable"
|-
|[[File:IM Fluorine surface plot.png|300 px]]||[[File:IM Fluorine contour plot.png|300 px]]
|-
|Surface plot for F-H-H system
|Contour plot for F-H-H system
|}

The energy of the AB bond channel is lower than the energy of the BC bond channel- i.e. the energy of the system with an H-F bond and lone H is lower than that of the system with an H-H bond and lone F. Therefore it can be concluded that the reaction F + H2 → F-H + H is '''exothermic''', while the reverse reaction F-H + H → F + H2 is '''endothermic'''. Therefore the H-F bond is stronger than the H-H bond as the forward reaction is exothermic. This is due to the high electronegativity of fluorine resulting in a polar bond with a great amount of ionic character.

====Locating the Transition State====

This reaction is asymmetric therefore the transition state is harder to find than for the symmetric H + H2 system. As the activation energy of the forward reaction is so low, the transition state will be an early transition state and thus by Hammond's postulate the transition state will be very similar to the reactants i.e. the H-H distance (r2) will be very close to equilibrium bond length while the H-F distance (r1) will be far larger than the equilibrium bond length.

By observation of the potential energy surface, an initial guess could be made for the location of the transition state. By then adjusting the distances r1 and r2 the transition state was located as the initial conditions which did not result in any change in trajectory. A dynamics simulation with 1000 steps was used to determine the position at which the trajectory remained constant and did not fall down into either the product or reactant channels, and internuclear distances did not oscillate. The transition state was determined to be at r1 = 1.8107 Å and r2 = 0.745 Å.

{| class = "wikitable"
|-
|[[File:IM Fluorine TS surface.png|300 px]]||[[File:IM Fluorine TS contour.png|300 px]]||[[File:IM Fluorine TS nucl vs time.png|300 px]]
|-
|Surface plot showing transition state
|Contour plot showing transition state
|Plot of Internuclear Distances vs Time
|}

====Determining activation energies====

The activation energy for each reaction can be determined by calculating the difference in energy between the transition state and either the product or reactants. The energy of the transition state is -103.752 kcal/mol.

The energy of the products and reactants can be determined by choosing a position on the potential energy surface far enough from the transition state in the relevant channel. The energy of H2 + F is found by using r1 = 10 Å and r2 = 0.74 Å and is determined to be -104.020 kcal/mol; while the energy of H + H-F is found using r1 = 0.92 Å and r2 = 10 Å and is determined to be -134.025 kcal/mol.

{|class = "wikitable"
|-
!System
!Energy (kcal/mol)
|-
|F + H2
| -104.020
|-
|F-H + H
| -134.025
|-
|F--H--H
| -103.752
|}

Therefore the activation energy for H2 + F → H + H-F is 0.268 kcal/mol and the activation energy for H + H-F → H2 + F is 30.273 kcal/mol.

===Reaction Dynamics===

====F + H2 → F-H + H====

A reactive trajectory was found for the F + H2 → F-H + H reaction using the initial conditions r1 = 2 Å, r2 = 0.74 Å, ρ1 = -0.8 and ρ2 = 0.5 (where r1 = F-H, r2 = H-H).

{|class = "wikitable"
|-
|[[File:IM Fluorine 1strxn reactive contour.png|300 px]]
|[[File:IM Fluorine 1strxn reactive momvtime.png|300 px]]
|-
|Reactive trajectory for F + H2 → F-H + H
|Internuclear momentum vs time
|}

This reaction is exothermic- energy is released in the reaction. This energy can be calculated as the difference between the energy of the reactants and products, which is equal to -30.005 kcal/mol. Therefore around 30 kcal/mol worth of energy is released in the reaction, mostly as vibrational and kinetic energy in the products- this is predicted in the trajectory which has a high vibration in the products. This increased kinetic and vibrational energy results in increased temperature of the system, so energy is dissipated as heat.

Using the initial conditions of rHH = 0.74 Å, rFH = 2 Å and ρFH = -0.5, the effect of changing ρHH within the range -3 < ρFH < 3 was investigated.

{|class = "wikitable"
|-
|[[File:IM changing pHH neg 3.png|300 px]]||[[File:IM changing pHH neg 2.9.png|300 px]]||[[File:IM changing pHH neg 2.5.png|300 px]]||[[File:IM changing pHH neg 2.png|300 px]]||[[File:IM changing pHH neg 1.5.png|300 px]]
|-
|ρHH = -3||ρHH = -2.9||ρHH = -2.5||ρHH = -2||ρHH = -1.5
|-
|[[File:IM changing pHH neg one.png|300 px]]||[[File:IM changing pHH neg 0.5.png|300 px]]||[[File:IM changing pHH 0.png|300 px]]||[[File:IM changing pHH 0.5.png|300 px]]||[[File:IM changing pHH 1.png|300 px]]
|-
|ρHH = -1||ρHH = -0.5||ρHH = 0||ρHH = 0.5||ρHH = 1
|-
|[[File:IM changing pHH 1.5.png|300 px]]||[[File:IM changing pHH 2.png|300 px]]||[[File:IM changing pHH 2.5.png|300 px]]||[[File:IM changing pHH 2.9.png|300 px]]||[[File:IM changing pHH 3.png|300 px]]
|-
|ρHH = 1.5||ρHH = 2||ρHH = 2.5||ρHH = 2.9||ρHH = 3
|}

Varying ρHH gives various reactive and unreactive trajectories. From -3 to -2.5, the trajectories are unreactive as they pass cross over the barrier but then cross back as the molecule is shaken apart by a large amount of vibrational motion. ρHH = -2 and -1.5 are reactive, while between ρHH= -1 and ρHH = 1.5 the trajectories are non-reactive as the trajectory crosses the barrier multiple times but finishes in the reactants channel. ρHH = 2 is a reactive trajectory but ρHH = 2.5 crosses back over the barrier and so is unreactive. ρHH = 3 is unreactive as it crosses back over the barrier after reacting and finishes in the reactants channel. A large amount of vibrational motion is needed to give a reactive trajectory : |ρHH| > 1.5

For the same initial positions, the momenta were changed to ρHH = 0.1 and ρFH = -0.8.

{|class = "wikitable"
|-
|[[File:IM Fluoring changed mometa.png|300 px]]
| rowspan = "2"| There is very littile vibrational motion in this case, but a relatively small amount of translational motion (ρFH = -0.8) will still give a reactive trajectory. This can be explained by Polanyi's rules (see section 3.2.3).
|-
| style="text-align: center;"|ρHH = 0.1, ρFH = -0.8
|}

====F-H + H → F + H2====

Initial conditions were chosen at the bottom of the entry channel with a low HF vibration and high HH momentum to give a reactive trajectory: rFH = 0.92 Å, rHH = 2 Å, ρFH = 0.1 and ρHH = -9.

{|class = "wikitable"
|-
|style="text-align: center;"|[[File:IM initial reactive traj high mom.png|350 px]]
|-
| style="text-align: center;"|Initial conditions: high momentum, low vibrational energy
|}

Decreasing the momentum of the incoming H atom to ρHH = -7 results in a non-reactive trajectory, however then increasing vibration of F-H to ρFH = 1.5 results in a reactive trajectory again.

{|class = "wikitable"
|-
|[[File:IM non reactive traj 7 mom 0.1vib.png|300 px]]
|[[File:IM reactive traj 7 mom 1.5vib.png|300 px]]
|-
|ρHH = -7, ρFH = 0.1
|ρHH = -7, ρFH = 1.5
|}

====Polanyi's Empirical Rules====

An asymmetric reaction such as the H2 + F → H + H-F system will have an asymmetric potential energy surface and depending on the energies of each side of the system the transition state will either resemble the reactants or the products of a system (predicted using Hammond's Postulate). If the transition state resembles the reactants, the reaction is said to be an early barrier reaction; if the transition state resembles the products the reaction is said to be a late barrier reaction. For the forward reaction H2 + F → H + H-F, the transition state resembles the reactants and so this is an early barrier reaction. The reverse reaction is therefore a late barrier reaction.

In a typical chemical reaction there is both translational and vibrational energy, both of which can contribute to crossing the energy barrier. The Polanyi rules, based off theoretical studies of triatomic systems (like those studied in this report) relate the type of barrier to which type of energy is most important for crossing this barrier.

Polanyi's rules
<ref>Z. Zhang, Y. Zhou, D. H. Zhang, G. Czakó and J. M. Bowman, Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD 3 Reaction, J. Phys. Chem. Lett., 2012, 3, 3416–3419. {{DOI|10.1021/jz301649w}}
</ref>
* Vibrational energy is more efficient in promoting a late-barrier reaction
* Translational energy is more efficient in promoting an early-barrier reaction

Therefore, the forward reaction will be promoted more by increasing translational energy (rFH) than by increasing vibrational energy (rHH), while the reverse reaction will be promoted more by increasing vibrational energy (rFH than by increasing translational energy (rHH).

This is seen above- for the forwards reaction a large magnitude of vibrational energy was need to give a reactive trajectory when translational energy was small, however a relatively small increase in translational energy with a low vibrational energy was sufficient to give a reactive trajectory.

{{fontcolor1|gray|(Overall, a very impressive report! Very well done indeed. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:58, 3 June 2018 (BST))}}

==References==
<references/>

MRD:TKkrn

2018-06-03T16:25:06Z

Fjs113: Marked by fjs113

===EXERCISE 1: H + H2 system===
====Dynamics from the transition state region====

On a potential energy surface, both the first derivatives (with respect to r1 and r2) of the minimum and the transition state equal 0. This means that they're either minimum, maximum or a saddle point. The minimum then has a positive value for both the second derivatives(with respect to r1 and r2) as this indicates that this is a minimum. The transition state is the maximum of the minimum energy path. This means that one of its second derivatives has a positive value, that corresponds to the minimum and the other one has a negative value, which corresponds to a maximum. and combing those 2 together results in a maximum of the minimum energy path.

{{fontcolor1|gray|(Be careful with terminology - what you mean here are '''partial''' derivatives! Furthermore, you should always avoid colloquial English when writing reports like this. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:25, 3 June 2018 (BST))}}

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

My best guess for the rts is 0.908 at 500 steps. This value was achieved by taking the r for the bond length(0.74) and making it larger step by step. At a certain point the “Internuclear Distances vs Time” plot showed two constant values for both distances without any oscillations and both neighboring distances being the same and this is the true transition state. When the value is higher than that, it will show a reactive trajectory and if the value is lower than that, it shows an unreactive trajectory.

[[File:TK_ex2.png]]

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

The first plot showed below is the calculation of the trajectory that correspond to the minimum energy path. It only show the movement of the atoms from the position that was inputed until the transition state. The input is a position a bit off the transition state and the mep shows the trajectory to the transition state at infinitesimally small momentum. At transition state for H + H-H the both distances reach the same value and that's where the mep plot stops. On the second plot, the lines continue to effectively infinity(if enough steps), where the B-C and the A-C distances keep increasing and the A-B distance remains constant with some fluctuations due to vibrations.

[[File:TK_ex3_mep.png]]

[[File:TK_ex3_dyn.png]]

Looking at the momenta of the internuclear distances, using the mep calculation shows a constant of 0 for all three values, which is correct as this is the trajectory at infinitely slow motion. For the dynamics calculation, all three momenta rise as they are leaving the transition state. They reach the platue and some of the oscillate due to vibrations.

[[File:TK_ex3_mep_mom.png]]

[[File:TK_ex3_dyn_mom.png]]

Final positions and average momenta at large t in the calculations:

r1(t)= 9.03
r2(t)= 0.75
p1(t)= 2.49
p2(t)= 1.26

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead?
Only the signs of the atoms change, all the plots remain the same.

Another calculation was performed, using tha values above, with both momenta's signs reversed and it result in the plot below.

[[File:TK_reverse.png]]

The trajectory in this case is unreactive.
====Reactive and unreactive trajectories====
* For the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, run trajectories with the following momenta combination:
{| class="wikitable" border=1
! p1 !! p2 !! Total energy !! !!
|-
| -1.25 || -2.5 || -99.0 || reactive
|-
| -1.5 || -2.0 || -100.5 || unreactive
|-
| -1.5 || -2.5 || -99.0 || reactive
|-
| -2.5 || -5.0 || -85.0 || unreactive
|-
| -2.5 || -5.2 || -83.0 || reactive
|}

[[File:TK_ex4_1.png]]

The reactants pass through the transition state normally and then the new molecule starts to vibrate.

[[File:TK_ex4_2.png]]

The reactants approach each other at the transition state, but the energy is not large enough to pass the energy barrier, so they just collide and start moving into opposite directions.

[[File:TK_ex4_3.png]]

The reactants pass through the transition state normally and then the new molecule starts to vibrate.

[[File:TK_ex4_4.png]]

The molecule and the atom approach each other and pass the transition state to form a new molecule, which then vibrates extensively and repasses the transition state to form back the reactants.

[[File:TK_ex4_5.png]]

The reactants approach the transition state, pass through it to form a new molecule, then repass it and pass through it again, so in the end you get the new molecule formed, with extensive vibrations.

The general assumptions of the transition state theory are the Born-Oppenheimer approximation and that the states of the reactants are distributed according to the Maxwell-Boltzmann distribution. Additional assumptions are as follows:
-Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants.
-In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as translation.
-The transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann distribution.
It also assumes a pseudo equilibrium between reactants and products.
One of the main limitations is that the atoms are treated classically. That means that only the reactants that have a high enough collision energy can react and form products. For reactions that have a low activation energy, this often fails, because quantum tunneling can occur and reaction proceeds without reaching that required energy.
The assumptions that once transition states proceeds to products, they cannot reform reactants also, evidently, fails sometimes as we can see in some of the reactions above. Where the products are formed and then rebound to form back the reactants due to a high vibrational energy.

{{fontcolor1|gray|(All correct, but how does this affect predictions made about the reaction rate? [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:25, 3 June 2018 (BST))}}

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

The reaction H-H + F = H-F + H is exothermic, as the products on the potential energy surface are located lower than the reactants in terms of the energy. The reverse reaction is exothermic as the products are located higher in terms of energy. This means that the H-F in stronger and energetically more favored than the H-H bond.

{{fontcolor1|gray|(Would be nice to see a picture for this. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:25, 3 June 2018 (BST))}}

The transition state position was determined by applying Hammond's Postulate and tweaking the values of r to obtain a stable structure without any fluctuations, as seen in the Internuclear distances vs. time plots provided below. The positions were then moved slightly out of the transition state and mep calculations were performed with 800000 steps to obtain the activation energies, this is also plotted on a Energy vs. time plot below.

'''Transition state for H-H + F:'''
r1 = 0.744891
r2 = 1.810746699
Activation energy = 0.248 kcal/mol

[[File:TK_HF-H_mep.png]]
[[File:TK_ex5_HH-F.png]]

'''Transition state for H-F + H:'''
r1 = 0.744891
r2 = 1.8107467
Activation energy = 32.612 kcal/mol

[[File:TK_HH-F_mep.png]]
[[File:TK_ex5_FH-H.png]]

The system passes the transition state, repasses it and passes it again. The momentum of the newly formed bond H-F initially quickly drops and the quickly rises each time it passes the transition state as the atoms are pulled together and then pulled apart again. Each time after passing the transition state this momentum rises as the atoms are closes together. After passing the transition state the final time, the momentum oscillates due to vibrations. Each time they pass the transition state the kinetic energy converts to the potential and then right after vice versa. This could be measured with some kind of time resolved IR spectroscopy, because when the kinetic energy is higher the molecule is in a vibrationally excited state, has more active IR stretches and would show more bands on the spectrum (overtones).

[[File:TK_ex6_1.png]]

[[File:TK_ex6_2.png]]

[[File:TK_ex6_3.png]]

====Reaction dynamics====

{| class="wikitable" border=1
! !! r1 !! r2 !! p1 !! p2 !! !!
|-
|[[File:TK_ex7_1.png]]|| 0.74 || 2.5 || -3 || -0.5 || reactive
|-
|[[File:TK_ex7_2.png]]|| 0.74 || 2.5 || -2.75 || -0.5 || reactive
|-
|[[File:TK_ex7_3.png]]|| 0.74 || 2.5 || -2.7503 || -0.5 || unreactive
|-
|[[File:TK_ex7_4.png]]|| 0.74 || 2.5 || -1 || -0.5 || unreactive
|-
|[[File:TK_ex7_5.png]]|| 0.74 || 2.5 || 0 || -0.5 || unreactive
|-
|[[File:TK_ex7_6.png]]|| 0.74 || 2.5 || 2 || -0.5 || unreactive
|-
|[[File:TK_ex7_7.png]]|| 0.74 || 2.5 || 2.4975 || -0.5 || unreactive
|-
|[[File:TK_ex7_8.png]]|| 0.74 || 2.5 || 2.4976 || -0.5 || reactive
|-
|[[File:TK_ex7_9.png]]|| 0.74 || 2.5 || 3 || -0.5 || reactive
|}
'''Different trajectories of the H-H + F reaction by adjusting the pHH value.'''

When the momentum of the F-H distance is changed to -0.8, the trajectory now becomes reactive.

{| class="wikitable" border=1
! !! r1 !! r2 !! p1 !! p2 !! !!
|-
|[[File:TK_ex8_1.png]]|| 0.9 || 2.1 || 0.5 || -3 || unreactive
|-
|[[File:TK_ex8_2.png]]|| 0.9 || 2.1 || 3 || -2 || unreactive
|-
|[[File:TK_ex8_3.png]]|| 0.9 || 2.1 || 5 || -1 || unreactive
|-
|[[File:TK_ex8_4.png]]|| 0.9 || 2.1 || 8 || -0.5 || unreactive
|-
|[[File:TK_ex8_5.png]]|| 0.9 || 2.1 || 12 || -0.2 || reactive
|-
|[[File:TK_ex8_6.png]]|| 0.9 || 2.1 || 15 || -1.5 || reactive
|}
'''Different trajectories of the H-F + H reaction by adjusting the pHF and the pHF value.'''

Looking at all the data provided in this section, some conclusions can be drawn. The first table shows data for when the H-H vibrational energy is changed in the F + H-H reaction. The trajectory is unreactive in a gap between ~-2.7503 and ~2.4976 with respect to the momentum of the H-H bond distance. This is done for the Value of the H-F momentum of -0.5. But when this momentum is changed to -0.8, most of that gap becomes reactive as well. Because that gap is not that wide and a small change in the momentum of H-F makes trajectories reactive, we can assume that tha translational energy of the collision is more important for this reaction to proceed than the vibrational energy of the starting molecule. Polanyi's empirical rules predict such behavior for early transition states. Indeed if we look at the transition state for this reaction, it resembles the starting material, not the product.

[[File:TK_ex9_1.png]]

Looking at the data for the reverse reaction H-F + F, both the momenta were adjusted to obtain a reactive trajectory. This turned out to be quite difficult to do. The vibrational energy of the H-F had to be increased significantly for the system to pass the activation barrier. At the same time the incoming H---H-F momentum had to be lowered. This shows that the vibrational energy in this case has a much higher contribution to the reaction than the translational energy. Polanyi's rules predict such behavior for early transition states. And again, this reaction indeed has an early transition state, which look similar to the one for the first reaction. The two hydrogens are close together and in this case, that structure resembles the products.

{{fontcolor1|gray|(Overall, well done. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:25, 3 June 2018 (BST))}}

MRD:sc3916

2018-06-03T16:02:24Z

Fjs113: Marked by fjs113

=Molecular Reaction Dynamics Report=

==Potential Energy Surfaces==

'''What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.'''

The transition structure is at a saddle point - the intersection of the minimum of a curve in one plane, and the maximum of a curve in the other. In this case, the minimum curve is on the plane corresponding to RAB=RBC, and the maximum is on the perpendicular plane. The gradient of both these curves is equal to zero.

This is different from when the potential energy surface is at a minimum - in one plane the gradient = 0, but not in the other. This allows us to distinguish between the transition structure and minima by looking at the curvature in '''both''' planes.

{{fontcolor1|gray|(Not quite - both the minimum and the saddle point have 0 gradient in all directions. You need to use second order partial derivatives to distinguish between minima and saddle points. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:02, 3 June 2018 (BST))}}

[[File:Saddle_Point_Visualisation_sc3916.png|thumb|none|500px|Saddle Point Visualisation]]

'''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.'''

'''rts= 0.9075 angstroms =rAB=rBC'''

To estimate this, a rough estimate was found by looking at the saddle point. At the saddle point, gradient = 0 in both planes - so if the system is at exactly this point, the atoms will remain stationary if it has no initial momentum. If it isn't at this point, it will start to oscillate - and the further away from the saddle point, the greater the oscillation. Therefore by plotting this internuclear distance vs time, and adjusting the distance by small amounts until there was very little oscillation seen, a more precise value was found.

[[File: Sc3916_1.PNG|thumb|300px|none|Internuclear Distances vs Time]]

==Reaction trajectories==

[[File:Sc3916_mep.PNG|thumb|400px|left|Minimum Energy Path from from r1 = rts+δ, r2 = rts]]

[[File:Sc3916_dynamics.PNG|thumb|400px|center|Dynamic Energy Path from r1 = rts+δ, r2 = rts]]

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

The minimum energy path represents an infinitely slow reaction trajectory. We start off a very small distance from the transition structure, and then the trajectory "rolls" down the potential energy surface, following where the gradient is most negative. Since velocity resets to 0 every time interval, so does momentum (p=mv). This means at every infinitesimally small step, this pathway will minimise its energy. This is not how it works in reality.

The dynamic trajectory is much closer to reality. As the trajectory starts to move down the potential energy surface it gains momentum. This means it will carry on rolling up the curve on the other side of the minimum (which increases its potential energy), leading to oscillations. This reflects the vibrational motion of atoms, which MEP does not include.

Dynamics Internuclear distance vs time plot: 
Final AB distance = 0.8 
Final BC distance = 9.0 

Dynamics Internuclear Momenta vs time plot: 
Final average momentum for AB = 1.25 
Final average momentum for BC = 2.5 

This shows the reaction path from the above calculation leads to A-B + C

If we switch around initial values, so now rAB=rts+δ and rBC=rts, the reaction path will output the same values, except the products are B-C + A.

'''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.'''

The reaction proceeds from left to right across the below plot, A-B + C -> A + B-C. This reaction is essentially the reverse of the previous trajectory (allowing the pathway to "roll" down from the transition state), however it crosses over, to form the products.

[[File:Sc3916_dynamics2.PNG|thumb|400px|none| initial conditions: 
rAB = 0.8 
rBC = 9.0 
pAB = -1.25 
pBC = -2.5]]

==Reactive and unreactive trajectories==
The following table is for the initial positions '''r1''' = 0.74 and '''r2''' = 2.0, with varying momentum.

{| class="wikitable" border=1
! p1 !! p2 !! E(kcal/mol) !! Reactive?
|-
| -1.25 || -2.5 || -99.119 || Yes
|-
| -1.5 || -2.0 || -100.456 || No
|-
| -1.5 || -2.5 || -98.956 || Yes
|-
| -2.5 || -5.0 || -84.956 || No
|-
| -2.5 || -5.2 || -83.416 || Yes
|}

[[File:Sc3916_rxn1.PNG|thumb|300px|none|Initial p1=-1.25, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn2.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.0 
the trajectory goes up towards the transition state smoothly (but not very far), and then rolls back down the way it came, with oscillating]]

[[File:Sc3916_rxn3.PNG|thumb|300px|none|Initial p1=-1.5, r2=-2.5 
the trajectory goes smoothly up to the transition state, then starts oscillating as it rolls down to the products]]

[[File:Sc3916_rxn4b.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.0 
the trajectory goes smoothly up to the transition state, crosses it, but recrosses it and reverts back to the reactants, with very large oscillations]]

[[File:Sc3916_rxn5.PNG|thumb|300px|none|Initial p1=-2.5, r2=-5.2 
the trajectory crosses the transition state, forming the products, then recrosses it to the reactant side, and then recrosses it again, forming the products finally]]

'''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?'''

4 major assumptions are used. Firstly, the Born-Oppenheimer approximation is used, and that the effects of quantum tunneling are negligible (there is a non-zero probability that reactants can convert to products without reaching the activation energy). The reactants are assumed to have a Boltzmann distribution - which is perfectly valid at low temperatures. The final assumption is that "once the system attains the transition state, with a velocity towards the product configuration, it will not reenter the initial state region again."[https://doi.org/10.1016/B978-044452837-7.50005-8 1]

This last assumption does not fit with our simulations. Both the 4th, and 5th plots contain the trajectory crossing over the transition state and back again. Since there will be more trajectories (like the 4th) which involve crossing over and are still unreactive - which are not predicted by Transition State Theory - experimental rate values will be slower than those predicted by the theory.

{{fontcolor1|gray|(Very good! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:02, 3 June 2018 (BST))}}

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

'''Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''

F + H2 is exothermic - the products have a more negative energy than the reactants. Because the enthalpy change only involves breaking a H-H bond and making a H-F bond, ΔH = +H(H-H) - H(H-F).

Since ΔH < 0:

H(H-F)>H(H-H)

This means HF has a greater bond enthalpy (and therefore strength) than H2. It can also be seen that H + HF is endothermic, which leads to the same conclusion.

{{fontcolor1|gray|(Would've been nice to see you using the PES given by the programme to explain this as well. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:02, 3 June 2018 (BST))}}

'''Locate the approximate position of the transition state.'''

Since F + H2 is exothermic, we can assume the transition state is close in structure to the reactants (according to Hammond's postulate). Starting from near the reactants, MEP calculations with a large number of steps were done, varying the starting position to try minimise how far the trajectory moves from that position. This is because at the transition point, gradient will equal zero in both planes, so the MEP will not increase from there - so the shorter the MEP (from a constant number of steps), the closer we are to the transition state. With this method, a good approximation was found.

rts: 
RHH = 0.7449 angstroms 
RHF = 1.8108 angstroms

(0.7448778, 1.810754) is the super accurate value, but since this is just a model, this won't exactly match the real transition state position, so there is not much point doing it to this many decimal places.

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

By performing an MEP from next to the transition structure, the path from the TS to the reactants, or products, was found. By subtracting the final energy from the TS, the Ea was obtained. Since the activation energy for H2 + F is very small, the MEP has a very small gradient, and as such the calculation required an order of magnitude more steps in the calculation (than HF + H) to reach the reactants. Conversely, in HF + H a step change is seen.

[[File:Mep_energyvstime_sc3916.PNG|thumb|left|initial conditions: 
RHH=0.745 
RHF=1.83]]

[[File:3916_energyvstime_hf2.PNG|thumb|center|initial conditions: 
RHH=0.745 
RHF=1.8]]

Ea(H2 + F Ea) = 0.256 kcal/mol 
Ea(HF + H Ea) = 30.202 kcal/mol 

{{fontcolor1|gray|(Well done for labelling the graphs! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:02, 3 June 2018 (BST))}}

=== Reaction dynamics ===
* Identify a set of initial conditions that results in a reactive trajectory for the F + H2, and look at the “Animation” and “Internuclear Momenta vs Time”. {{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

[[File:H2+f_surface__3916.PNG|thumb|none|F + H2, initial conditions: 
RHH (AB) = 0.74 angstroms 
RHF (BC) = 2 angstroms 
AB momentum = +3 
BC momentum = -5]]

[[File:H2+F_MOMENTAVTIME3916.PNG|thumb|none|F + H2 Momentum vs Time Plot]]

The animation of this reaction clearly shows the velocity of the lone H atom formed has increased, and vibrations occurring in HF. This can be seen in the momentum vs time plot - with the increased AB and AC momentum reflecting the increase in velocity, and BC momentum oscillation reflecting the vibrations of the H-F bond. Since translational and vibrational energy contribute to kinetic energy, we know that the energy released by this exothermic reaction (a decrease in enthalpy) is conserved in the form of increased kinetic energy. This can be seen in the below graph, and is easily shown experimentally by the resultant increase in temperature. To determine the extent vibrational energy contributes to this, we can use IR spectroscopy. If there is a lot of vibrational energy, there will be a relatively high population in the first vibrational level, and we will see overtone bands in the IR spectrum.

[[File:H2+F_ENERGYVSTIME3916.PNG|thumb|none|F + H2 Energy vs Time Plot]]

* Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration.

[[File:SC3916-3CONTOUR.PNG|thumb|none|F + H2, initial conditions A: 
rHH=0.74 rHF=1.83 
pHH=-3.0 pHF=-0.5 ]]

[[File:SC3916-2CONTOUR.PNG|thumb|none|F + H2, initial conditions B: 
rHH=0.74 rHF=1.83 
pHH=-2.0 pHF=-0.5 ]]

[[File:SC3916-1.5CONTOUR.PNG|thumb|none|F + H2, initial conditions C: 
rHH=0.74 rHF=1.83 
pHH=-1 pHF=-0.5 ]]

[[File:SC39160CONTOUR.PNG|thumb|none|F + H2, initial conditions D: 
rHH=0.74 rHF=1.83 
pHH=0.0 pHF=-0.5 ]]

[[File:Sc3916_redone1.PNG|thumb|none|F + H2, initial conditions E: 
rHH=0.74 rHF=1.83 
pHH=+1.0 pHF=-0.5 ]]

[[File:Sc3916_redone2.PNG|thumb|none|F + H2, initial conditions F: 
rHH=0.74 rHF=1.83 
pHH=+3.0 pHF=-0.5 ]]

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

[[File:Sc3916_redonelast.PNG|thumb|none|F + H2, initial conditions G: 
rHH=0.74 rHF=1.83 
pHH=+0.1 pHF=-0.8 ]]

*Let us now focus on the reverse reaction, H + HF (an H atom with high kinetic energy colliding with HF).

[[File:Sc3916_hh-10hf-.5.PNG|thumb|none|HF + H, initial conditions H: 
rHH=0.9 rHF=1.83 
pHH=-10 pHF=-5 ]]

[[File:Sc3916_hh-1.1h4-13.PNG|thumb|none|HF + H, initial conditions I: 
rHH=0.9 rHF=1.83 
pHH=-1.1 pHF=-13 ]]

The cases studied are an illustration of Polanyi's empirical rules. {{fontcolor|blue|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 rules state, for a reaction with an early transition state, vibrational energy is more effective in overcoming the TS - so initial conditions with higher vibrational energy are more likely to result in reaction, so are more efficient. For a late transition state, translational energy is more effective in overcoming it - so atoms/molecules with a higher translational energy create a more efficient reaction.[http://doi.org/10.1021/jz301649w 2]

This can be seen in the above reaction trajectories. H2 + F has an early transition state (Hammond's postulate tells us that in an endothermic reaction, the transition state most resembles the reactants). Reaction trajectories with high initial vibrational energy (shown by pHH, conditions A, B, F) are unreactive trajectories. Conversely, those with low vibrational energy (C, D, E) are reactive. F also has increased kinetic energy on top of low vibrational energy, and this too is reactive.

The reverse reaction, HF + H, has a late transition state (from Hammond's postulate) - so initial conditions with greater vibrational energy will be more efficient. This can be seen in the reactive trajectory with conditions I. Conversely, conditions H has a much greater translational energy, and is unreactive.

{{fontcolor1|gray|(Overall, well done. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:02, 3 June 2018 (BST))}}

==References==

1. T. Bligaard, J.K. Nørskov, Chemical Bonding at Surfaces and Interfaces, 2008, p.255–321, https://doi.org/10.1016/B978-044452837-7.50005-8 [accessed 24/5/18]

2. Z. Zhang, Y. Zhou, D.Zhang, G.Czakó, J. Bowman, J. Phys. Chem. Lett., 2012, 3 (23), pp 3416–3419
https://pubs.acs.org/doi/abs/10.1021/jz301649w [accessed 25/5/18]

MRD:01208568

2018-06-03T15:37:40Z

Fjs113: Marked by fjs113

== H + H2 reaction dynamics ==
 What value do the different components of the gradient of the potential energy surface have at a minimum transition structure? Briefly explain how minima and transition structures can distinguished using the curvature of the potential energy surface. 

[[File:H+H2_potential_energy_trajectory_plot_01208568.png|thumb|center|The potential energy curve for the reaction between H and H2]]

The transition point can be viewed by rotating the image of the plot to show the maximum and minimum at the same time as follows:
[[File:Potential_energy_surface_showing_maximum_01208568.png|thumb|left|The perspective of the maximum of the transition state orthogonal to the reaction pathway.]]
[[File:surface_plot_showing_the_minimum_of_saddle_point_01208568.png|thumb|center|The perspective of the minimum of the transition state parallel to the reaction pathway.]]

These two different perspectives of the potential energy curve at the transition state show that indeed there is a maximum in the direction orthogonal to the reaction pathway and there is a minimum in the direction parallel to the reaction pathway confirming there is a saddle point.

The start and end point of the reactive pathway can be best viewed by following the path of the black dots and since the first derivative is negative and the second derivative is positive, it shows there is a minimum. (This can be viewed from the direction is either axis.)

You can distinguish the minima and transition states by evaluating the curvature of the potential energy curve plot by taking the second derivative and the sign determining whether there is maximum or minimum. The first derivative will be equal to zero. If the second derivative can have both positive and negative values, the reaction is in its transition state, whereas if the potential energy curve can be viewed as only a minimum from the view of either axis, the reaction is in its reactant or state or it has completed and is in its product state.

{{fontcolor1|gray|(Saying that the second derivative can have multiple values here is wrong. What you mean are the '''partial''' derivates with respect to the reaction coordinate (along the reaction pathway) as well as its orthogonal counterpart. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:37, 3 June 2018 (BST))}}

The reaction coordinate can be used to explain this, if the potential energy curve plot is in its transition state, viewing in the diagonal direction representing the minimum, moving in any direction equates to the distance between AB and BC changing. When the distance becomes shorter, it means that the hydrogen atoms are moving closer together and therefore the potential energy would increase exponentially as the system would be unstable. If the distance becomes larger, it equates to overcoming the attraction between the AB and BC species in the transition state, thus increasing the potential energy again.

In the view showing the maximum, going diagonally left would mean the distance of AB would be getting short and the distance of BC would be getting larger, up until the point at which the two bonds are equidistant.

{{fontcolor1|gray|(Using mathematical expressions here would have made your answer a lot clearer. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:37, 3 June 2018 (BST))}}

=== Reporting the best estimate of the transition state position (rts) with the explanation ===

In this transition state the bond distance of AB and BC would be equal and this would be at the saddle point of the potential energy curve plot. If the position of the transition state, is not found to a good enough approximation, the inter-nuclear vs time plot would show large oscillations between the distance of AB and BC, as the reaction pathway would be moving up and down in the potential well. However when this is approximated well, less oscillation is observed.

Using the inter-nuclear distance vs time plot, the approximate value of the the distance between AB and BC in its transition state was recorded.

[[File:Internucleardistancevstimeplot01208568.png|thumb|center|The inter-nuclear distance vs time plot showing a distance of around 0.9 angstroms for the transition state.]]

The distance at which the transition state occurs would be when the distance between AB and BC would be equal. This was found by setting the momentum of each species to zero in the initial conditions and just changing the distances. Using this method, it was determined the approximate distance was around 0.9078 angstroms.

[[File:Internucleardistancevstimeplot0.9078angstroms01208568.png|thumb|left|The inter-nuclear distance vs time plot showing a distance of 0.9078 angstroms for the transition state showing there is no oscillation moving up and down the sides of the potential well.]]

[[File:Internucleardistancevstimeplot0.93angstroms01208568.png|thumb|center|The inter-nuclear distance vs time plot showing a distance of 0.93 angstroms for the transition state showing there is some oscillation moving up and down the sides of the potential well, therefore this is not a good approximation.]]

In the initial conditions the distance between AB and BC are set to have the same value, therefore the orange line is completely covering the blue line. In the transition state the distance between AC should be double the distance between AB and this value is 1.8156 angstroms.
[[File:Surfaceplotshowingtransitionstatedistance01208568.png|thumb|center|The surface plot of the potential energy curve showing there is very little oscillation up and down the sides of the potential well at 0.9078 angstroms.]]

This shows a single black dot at the bottom of the potential well showing that 0.9078 is a good approximation of the inter-nuclear distance of the transition state.

=== The difference of the MEP and the trajectory calculated ===

[[File:MEPplotalongvalleyfloor01208568.png|thumb|left|The MEP calculation showing the reaction going smoothly along the valley floor]]

[[File:Dynamicplotbouncingalongvalleyfloor01208568.png|thumb|center|The dynamic calculation showing oscillating along the valley floor.]]

The dynamic calculation shows the reaction pathway oscillating along the valley floor, while the MEP calculation shows a smooth line along the valley floor. To give a full reaction pathway the MEP calculation took 8000 steps whereas the dynamic calculation took 553 steps.
[[File:Internuclearmomentavstimedynamic01208568.png|thumb|left|The dynamic calculation showing the momentum oscillating as the reaction proceeds.]]
[[File:Internuclearmomentavstimemep01208568.png|thumb|center|The MEP calculation showing the momentum is zero throughout the reaction.]]

As the momentum is zero for the MEP calculation it shows that this form of calculation does not provide a realistic view of the reaction as the masses of each atoms are assumed to be zero therefore the momenta of each atom is zero.

[[File:Inetrnucleardistancevstimedyanmicsplot201208568.png|thumb|left|The dynamic calculation showing the vibrational energy is taken into account via the oscillating lines.]]
[[File:Internucleardistancevstimemepcalcuation01208568.png ‎|thumb|center|The MEP calculation showing the vibrational energy is ignoreda as the lines representing the inter-nuclear distances is a straight line.]]

=== Reactive and unreactive trajectories ===
{| class="wikitable" border="1"
|+ caption
! p(AB) !! p(BC) !! Total energy (Kcal/mol) !! Plot of trajectory !! Reactivity
|-
| -1.25|| -2.5 || -99.018 || [[File:Surfaceplot1-01208568.png|300px]] || Reactive: There is a smooth trajectory until the transition state is reached, then there is some oscillation as the reaction proceeds towards the product state.
|-
| -1.5 || -2.0 || -100.456|| [[File:Surfaceplot2-01208568.png|300px]] || Unreactive: The reaction does not proceed with this trajectory as it doesn't reach the transition state.
|-
| -1.5 || -2.5 || -98.956 || [[File:Surfaceplot3-1208568.png|300px]] || Reactive: There is a smooth trajectory up until the transition state is reached, then there is some oscillation as the reaction proceeds towards the end product.
|-
| -2.5 || -5.0 || -84.956 ||[[File:Plotnonreactive1-01208568.png|300px]] [[File:Surfaceplot4-01208568.png|300px]] || Unreactive: There are very large oscillation up until the transition state is reached, then the reaction does not proceed further and returns to the reactant state.
|-
| -2.5 || -5.2 || -83.416 || [[File:Plotreactive-01208568.png|300px]] [[File:Surfaceplot5-01208568.png|300px]] || Reactive: There are very large oscillations up until the transition state is reached, then the reaction proceeds succesfully to reach the product state.
|}

 The main assumptions of Transition State Theory 

The assumptions made in transition state theory are as follows:

-Once the transition state is reached it will proceed to the product state. This means that it cannot reverse back into the initial state

-The energy of the atoms in the reactant state have that of a boltzmann distribution.

In the table it is clear to see that the first assumption is not correct in all cases as in the fourth set of conditions, the reaction did not proceed and reverted back to the initial state.At a high momentum, the plot shows that the reaction does not go along the potential energy surface plot, therefore the reaction has a possibility of returning to the initial state.

The rate of reaction would be faster using transition state theory than experimental data suggests due to the fact that the reactant pathway may reverse on itself acting as a barrier. Some experiments would not even produce results as no products are formed such as shown above in the table.

= H-H-F reaction system =
===F + H2 reaction system ===
The potential energy surface curve was plotted using the initial conditions rAB= 2.30 angrstoms, rBC= 0.74 angstroms, p1(AB) = -2.7 p2(BC)= 0

[[File:FHHsystemPEScurve.png|300px]]

The reaction of F and H2 is exothermic as show by the reaction trajectory that the product is lower in energy than the reactant therefore energy is released into the surroundings. This is also confirmed by that fact that the bond strength of H-F is much stronger than that of a diatomic hydrogen molecule. According to Hammond's postulate, the structure of the transition is closer to that in which it is closer in energy. For an exothermic reaction such as this one, the transition state is closer in energy to the reactant , therefore the distance at which the transition state occurs can be approximated along the reaction trajectory as the reaction just begins to proceed. In addition, the activation energy for this reaction is subsequently very low.

The values for the inter-nuclear distances in the transition state approximation were calculated using the same method as previously described in the H + H2 system.

[[File:FHHinternucleardistance-01208568.png|300px]]

This plot shows that at the distances F-H= 1.810 angstroms and H-H= 0.745 angstroms, the lines are straight indicating that the approximation is a good one. This is consistent with theory as the value for the bond distance between the two hydrogens should be slightly displaced from 0.74 angstroms, due to the repulsion of the electronegative fluorine atom and is also consistent with Hammond's postulate.

In order to approximate the activation energy, the initial conditions were changed to the point at which the reaction pathway did not proceed to the product state. This was done by evaluating the contour plots at different distances.

[[File:Activationenergyplot-1208568.png|300px]]

From the plot above, it shows that reactant particle starts at around the transition state and reverses back towards the reactant state. The conditions set were (rAB = 1.843 angstroms and rBC= 0.743 angstroms). To figure out the activation energy the energy difference between the transition and reactant states can be calculated to be approximately 0.207 Kcal/mol. However This could be better approximated by calculating a larger number of steps and furthermore trying to find a distances closer to the transition state but in which the reaction pathway is still sloping towards the reactant state. With a 50,000 step calculation, the activation energy was found to be 0.224 Kcal/mol which is 0.017 Kcal/mol higher than the first calculation carried out.

Another method to calculate the activation energy, is to set the inital conditions to where the distance between the fluorine and hydrogen molecules is very large and where the bond length of the hydrogen molecule is the normal length. The potential energy at these distances taking a calculation with only one step would be theoretically the energy of the reactant state. Then taking the difference between this energy and the transition state energy would be approximately equal to the activation energy. Using this method the approximate activation energy was calculated to be 0.261 Kcal/mol.
===H +HF reaction system ===
This reaction is the reverse reaction the reaction described above. Therefore the potential energy surface is identical with the axis swapped round and the reaction is endothermic. By applying Hammond's postulate, the transition state would be expected to be found closer to the product state rather than the reactant state.
The potential energy of the products and reactants have a lower bond energies than the respective opposites.
The approximate bond enthalpies are 565 kJ.mol-1 for the HF bond and 436 kJ.mol-1. <ref> Atkins, Pauli (2014) Atkin's Physical Chemistry 10th edition, 986 </ref>. The enthalpy changes for the H + HF --> HH + F are ΔH = +133 kJ.mol-1 and for the F + HH --> HF + H, ΔH = -133 kJ.mol-1.

The transition state was located via trial and error and was found where the distance rAB = 1.8114 angstroms and rBC = 0.7442 angstroms.

[[File:FHHinternucleardistance-01208568.png|300px]]

The activation for each reaction was found to be 0.26 Kcal/mol-1 for the first reaction and 30.23 Kcal.mol-1 for the reverse reaction.

Using the law of conservation of energy which states that the energy cannot be created or destroyed but only converted into different forms, it can be concluded that the total energy in the system is constant but is being converted from potential to kinetic energy and vice versa.

When the fluorine atom approaches the hydrogen molecule the speed of the fluorine atom decreases as the kinetic energy is converted into potential energy because of the interaction between the atoms. This potential energy increases up until a maximum of where the transition state is, (kinetic energy is zero at this point). After passing this point of transition, the formation of HF because the reaction is exothermic, causes the potential energy between the atoms to drop immensely which is transferred to the hydrogen atom that leaves increasing its kinetic energy and temperature.

Polanyi's rules can be used to determine to what extent a source of energy contributes to a reaction depending on where the transition state is in relation to the reactant or products. If the transition state is closer to the products, the vibrational energy contributes more the transition structure and so a large amount of vibrational energy will encourage this reaction. Conversely if the transition state is closer to the reactant state, the translational energy of the atom that is approaching will control the efficieny of the reaction.

Therefore for the F + HH --> reaction the needs a small amount of vibrational energy to achieve the transition state and largely requires translational energy for the reaction to fully proceed. Whereas the H + HF --> HH + F reaures a large vibrational energy to achieve the transition state as the reactants lie lower in potential energy.

{{fontcolor1|gray|(The last bit looks very rushed and you missed out on a few bits of the last few questions. Otherwise, well done overall. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:37, 3 June 2018 (BST))}}

MRD:mjb216

2018-06-03T15:06:04Z

Fjs113: Marked by fjs113

==EXERCISE 1: H + H2 system==

'''''What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.
'''''

The surface energy plot can be used to identify the minimum pathway. By taking the first derivative, minimum points on the surface can be calculated, for example the reactant and product channels. ∂V(rAB)/∂rAB=0 and ∂V(rBC)/∂rBC=0 express this mathematically.

The transition state is a saddle point in which it is a maximum in relation to the minimum reaction pathway and a minimum orthogonal to the reaction pathway. This can be obtained by taking the second derivative; upon doing this you will get a maximum and minimum. Maximum points can be identified when the second derivative is less than 0; minimum points have a second derivative greater than 0. A saddle point satisfies both conditions.

[[File:MJB H3BondDistance.PNG]]

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

The transition state position can be located at a bond distance of rAB = rBC. {{fontcolor1|gray|(Since in this case, the PES is symmetric along the diagonal, the TS '''must''' be located at equal distances. The TS is unique! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:06, 3 June 2018 (BST))}}
The Internuclear Distances vs. Time graph (displayed below) show that a value of 0.9077 Å has minimised the oscillation, as the line is flat. A minimum value corresponds to the transition state, as there would be no movement between A and C around B. If this was not a minimum, movement would take place and oscillation would be observed.

[[File:MJB H3Oscillation1.PNG]]

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

The MEP displays the minimum energy path, in which there is no oscillation due to the trajectory having infinitely slow motion. This means that after each step, the momenta is reset to 0, hence the zero oscillation as there is no accumulation of kinetic energy allowing it to move up the well. This differs from the dynamic plot, in which the motion is not infinitely slow and therefore oscillation can be observed. The plots were obtained by using the parameters rAB = 0.9087 Å | rBC = 0.9077 Å and 0 momenta.

The Energy vs. Time graphs further show this; in the dynamic plot, there is a clear displacement in the potential and kinetic energies which corresponds to oscillations. In the MEP graph, the drop in potential energy is not matched with an increase in kinetic energy, due to the infinitely slow motion.

{| class="wikitable" border"!"
! || Dynamic || MEP
|-
| Surface Plot || [[File:MJB H3Dynamic.PNG|600px]] || [[File:MJB H3MEP.png|600px]]
|-
| Energy vs. Time Plot || [[File:MJB H3Dynamic ET.PNG|600px]] || [[File:MJB H3MEP ET.PNG|600px]]
|}

If the parameters were reversed, i.e. rAB = 0.9077 Å, rBC = 0.9087 Å and 0 momenta, the reaction would proceed back to the reactant states, down the channel perpendicular to the one observed above.

'''''Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.'''''

Initial positions r1 = 0.74 and r2 = 2.0

{| class="wikitable" border"!"
! Condition || p1 || p2 || Total Energy / kcalmol-1 || Surface Plot || Contour Plot || Description
|-
| 1 || -1.25 || -2.5 || -99.018 || [[File:MJB H3surfaceplot p1.PNG|600px]] || [[File:MJB H3contour p1.PNG|600px]] || For the initial positions of the two particles, the momentum is sufficient so that the reaction has enough energy to proceed over the transition state. Once over the transition state, oscillations can be seen (very clear in the contour plot). This is due to the reaction proceeding down a negative gradient, causing the accumulation of kinetic energy which leads to oscillations.

AB + C --> A + BC.
|-
| 2 || -1.5 || -2.0 || -100.456 || [[File:MJB H3surfaceplot p2.PNG|600px]] || [[File:MJB H3contour p2.PNG|600px]] || For the initial positions, the momentum is insufficient to overcome the energy barrier to proceed over the transition state. As a result, no reaction occurs. Oscillation can be seen at all points of the reaction trajectory as the momentum p1 is greater than half of p2. As a result, AB vibrates.

AB + C -/-> A + BC.
|-
| 3 || -1.5 || -2.5 || -98.956 || [[File:MJB H3surfaceplot p3.PNG|600px]] || [[File:MJB H3contour p3.PNG|600px]] || For the initial position, the particles have sufficient momentum to overcome the energy barrier of the transition state and proceed to the products.

AB + C --> A + BC.
|-
| 4 || -2.5 || -5.0 || -84.956 || [[File:MJB H3surfaceplot p4v3.PNG|600px]] || [[File:MJB H3contour p4v2.PNG|600px]] || For the initial positions, the particles have sufficient energy to overcome the transition state. However, in this case, the new H2 molecule (BC) has very high kinetic energy, and proceeds to collide with A once again, reforming the initial molecule of AB.
The recrossing of the transition state can be seen clearly in the distance vs. time graph below:
[[File:MJB H3distancesplot p4.PNG|400px]]
The first crossing of the transition state can be seen at 0.163 , representing the formation of BC, this quickly reacts to reform AB at 0.541 . The two particles still have very high kinetic energy after the second collision, which is displayed in the contour plot.
AB + C --> A + BC --> AB + C.

|-
| 5 || -2.5 || -5.2 || -83.416 || [[File:MJB H3surfaceplot p5.PNG|600px]] || [[File:MJB H3contour p5.PNG|600px]] || For the initial positions, the particles have sufficient energy to overcome the transition state. There is then 2 subsequent recrossings of the transition state, firstly to reform AB and the finally to form BC. BC still has high kinetic energy, however, due to 3 collisions taking place the oscillations have lower amplitude and frequency than in the condition 4 above. This can be seen in the contour plot.
The recrossings of the transition state can be seen in the Distance vs. time graph below:

[[File:MJB H3distancesplot p5.PNG|400px]]

At 0.156 the first crossing of the transition state to form BC takes place, recrossing the occurs at 0.312 reforming AB. The transition state is crossed for the last time at 0.534 and is the formation of BC for the second time.
|}

{{fontcolor1|gray|(This is very well done. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:06, 3 June 2018 (BST))}}

===Transition state theory===

Transition state theory makes the assumption that nuclei behave under classical mechanical conditions. It states that reactants must proceed over an energy barrier, in the form of a saddle point (transition state) to reach the products. In addition, once the reactants have passed over the transition state, they must form products.<ref>M. J. Pilling, P. W. Seakins Reaction Kinetics, 2nd edition, OUP, 1995</ref>

Looking at the 5 different calculations from above; it is clear that in the case of 4 crossing of the transition state occurs, however, instead of resulting in product formation the reactants are reformed. This breaks the assumption that once the reactants have passed over the transition state they will convert to products.

Transition state theory is also contradicted by the quantum mechanical process of tunneling. Instead of gaining enough energy to overcome the transition state, the reactants tunnel through the energy barrier quantum mechanically; this clearly breaks the idea that reactants must pass through the transition state to reach products.

Finally, the theory breaks down at high temperatures. The theory assumes that the reactants will pass over the lowest energy point, i.e. the saddle point (transition state). This is clearly not the case, as at higher temperatures higher vibrational energy levels are populated (giving the molecule more vibrational energy). This can be seen by the reaction trajectories of calculation 4 & 5 above.

{{fontcolor1|gray|(This is all very good, but you have not said how these effects will affect predictions of the reaction rate. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:06, 3 June 2018 (BST))}}

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

'''''Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?'''''

H - H Bond enthalpy: 435 KJmol-1
H - F Bond enthalpy: 569 kJmol-1<ref>http://www4.ncsu.edu/~shultz/Common_Bond_enthalpies.pdf</ref>

{| class="wikitable" border"!"
! || H-H + F --> H + H-F || H-F + H --> H-H + F
|-
| Initial Conditions || [[File:MJB H2F atoms.PNG]] A-B Bond distance = 0.74 Å ; B-C Bond distance = 2.0 Å ; p1 = 0 ; p2 = -1.5 || [[File:MJB HFH atoms.PNG]] A-B Bond distance = 0.917 Å ; B-C Bond distance = 2.0 Å ; p1 = 0 ; p2 = -5.0
|-
| Surface plot || [[File:MJB H2F surfaceplot.PNG|600px]] || [[File:MJB HFH surfaceplot.PNG|600px]]
|-
| Description || The surface plot clearly shows that this is an exothermic reaction as A-B is higher in energy than B-C. This can be rationalised by considering the H-H & H-F bond enthalpies; the H-F bond is stronger than that of H-H, therefore the reaction has a ΔH = -134 kJmol-1. || The surface plot clearly shows that this is an endothermic process, as A-B is lower in energy than B-C. As a result, the reaction must proceed uphill in order for B-C formation. This can be seen, as p2 has a much higher momentum than in the H-H + F plot, however, it is still unable to overcome the transition state. This can be rationalised by considering the bond enthalpies to calculate the enthalpy of the reaction. It is the reverse of H-F formation; for this reaction ΔH = -134 kJmol-1.
|}

'''''Locate the approximate position of the transition state.'''''

The transition state was located using Hammond's postulate to determine the approximate region of the transition state. In this case, due to the exothermic nature of the H-H + F --> H-F + H reaction the transition state will resemble the reactants. The distances calculated were r1 (AB) = 0.744 Å ; r2 (BC) = 1.811 Å. The graph of distances vs. time shows that the distances obtained correspond to that of the transition state using the same rationale as for the H-H-H system above (no oscillations).

[[File:MJB HFH TSdistances.PNG]]

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

Once the transition state is located, the A-B and B-B distances can be distorted slightly and then an MEP calculation can be run to determine values for the activation energy of the two reactions.

The reaction of H-H + F --> H + H-F has an activation energy of ~0.2 kcalmol-1. The parameters used for the calculation were as followed: r1 (H-H) = 0.744 Å ; r2 (H-F) = 1.8111 Å ; pAB=pBC=0 ; step = 200000. The information to calculate Ea was extracted from the Energy vs. time graph below; (-103.95) - (-103.75) = 0.2 kcalmol-1. The activation is very low as H-F formation is very favorable from H-H.

[[File:MJB HFH Ea1 v2.PNG]]

The reaction of H + H-F --> H-H + F has an activation energy of ~30 kcalmol-1. The parameters used for the calculation were as followed: r1 (H-H) = 0.744 Å ; r2 (H-F) = 1.801 Å ; pAB=pBC=0 ; step = 200000. The information to calculate Ea was extracted from the Energy vs. time graph below; (-103.75 - (-134) = 30.25 kcalmol-1. The activation is very much higher for H-H formation from H-F as this is an unfavourable reaction thermodynamically, as shown above using bond enthalpies.

[[File:MJB HFH Ea2.PNG]]

===Reaction Dynamics===

'''''Identify a set of initial conditions that results in a reactive trajectory for the F + H2'''''

The initial conditions selected for the reaction of H2 + F are as follows: AB distance = 0.74 Å ; BC distance = 2.3 Å ; pAB = -1.0 ; pBC = -1.0
The contour and surface plots displayed below shown that a reactive trajectory has been achieved.

{| class="wikitable" border"!"
! [[File:MJB Initial surfaceplot.PNG|600px]] || [[File:MJB Initial contour.PNG|600px]]
|-
| Surface Plot || Contour Plot
|}

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

The reaction of F + H2 --> H-F + H is highly exothermic, with a ΔH = -134 kJmol-1. The reason for this large negative enthalpy is because H-F is a more stable molecule than H2, with the H-F bond formation being the driving force of the reaction. The loss of potential energy after the collision of H2 with F is converted into kinetic energy in the form of vibrational energy (satisfying the first law of thermodynaics: energy can neither be created nor destroyed merely converted from one for to another). This large increase in vibrational energy can be seen in the plot of Internuclear momenta vs. time below. At approximately time = 2, after the collapse of the transition state, the amplitude of the oscillations of the H-F bond increase significantly corresponding to the large increase in kinetic energy. The high kinetic energy of the molecule will cause an increase in temperature due to the transfer of energy either to other molecules (via collisions with H-F), or relaxation down to lower vibrational states. This increase in temperature indicates the release of potential energy.

[[File:MJB Initial Momenta.PNG]]

'''''F + H2''
'''

Conditions: rHH = 0.74 Å ; rHF = 2.3 Å ; pHF = -0.5
{| class="wikitable" border"!"
! pHH || Surface Plot || Contour Plot || Reactivity
|-
| -3.0 || [[File:MJB FHH surface-3.PNG|600px]] || [[File:MJB FHH contour-3.PNG|600px]] || Unreactive
|-
| -2.0 || [[File:MJB FHH surface-2.PNG|600px]] || [[File:MJB FHH contour-2.PNG|600px]] || Unreactive
|-
| -1.0 || [[File:MJB FHH surface-1.PNG|600px]] || [[File:MJB FHH contour-1.PNG|600px]] || Unreactive
|-
| 0.0 || [[File:MJB FHH surface0.PNG|600px]] || [[File:MJB FHH contour0.PNG|600px]] || Unreactive
|-
| 1.0 || [[File:MJB FHH surface1.PNG|600px]] || [[File:MJB FHH contour1.PNG|600px]] || Unreactive
|-
| 2.0 || [[File:MJB FHH surface2.PNG|600px]] || [[File:MJB FHH contour2.PNG|600px]] || Unreactive
|-
| 2.0 || [[File:MJB FHH surface3.PNG|600px]] || [[File:MJB FHH contour3.PNG|600px]] || Unreactive
|}

The plots above vary the H-H momentum for the reaction of F + H-H which shows that this does not allow the reaction to proceed to completion.

'''''Comparison of F + H-H and H-F + H reactions'''''

{| class="wikitable" border"!"
! H-H + F --> H + H-F || H + H-F --> H-H + F
|-
| [[File:MJB FHH contour01.PNG]] || [[File:MJB HF Hcontourv2.PNG]]
|-
| Conditions: rHH = 0.74 Å ; rHF = 2.3 Å ; pHF = 0.1 ; pHF = -0.8 || Conditions: rHH = 2.3 Å ; rHF = 0.91 Å ; pHF = -1.0 ; pHF = -9.0
|}

Polyani's empirical rules state that for an early transition state, translational energy is more efficient in facilitating the reaction (overcoming the transition state). On the other hand, a late transition state requires more vibrational energy to overcome the transition state.<ref>J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics Prentice-Hall</ref><ref>Atkins and de Paula: Physical Chemistry, 7th Edition</ref>

The reaction of H2 + F --> H + H-F only proceeds to HF formation when pHF is increased as the reaction has an early transition state, which can be concluded using Hammond's postulate as the transition state resembles the reactants; the reaction is exothermic. Using the diagram below, it can be seen that starting from the orange cross, which is the reaction start for H2 + F, translational energy is required to overcome the transition state. This explains why changing the values of pHH does not facilitate the reaction, as vibrational energy is not efficient for facilitating an early transition state as explained by Polyani's rules.

[[File:MJB polyaniv2.PNG]]

{{fontcolor1|gray|(This is great! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:06, 3 June 2018 (BST))}}

The reaction of H + H-F --> H-H + F only proceeds to H-H formation when pHH is increased, as this increases the vibrational motion of the system. This is important as the transition state for this reaction is late; the reaction is endothermic. The green cross on the diagram above shows the starting point of this reaction; an increase in vibrational energy will allow the reaction trajectory to reach the transition state and proceed to the products, as stated by Polyani's rules.

==References==

MRD:nk2018

2018-06-03T13:38:57Z

Fjs113:

=Molecular Reaction Dynamics Lab=
==Exercise 1: H + H2 System==
===What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.===
[[File:Contour_Map_nk2018.png|frameless|centre|400px]]

The energy surface shown above represents the relative energy of the molecules as the reaction progresses. It can be seen that for the transition state and minima the potential energy gradient with respect to r1 and r2 is 0. The transition state is a saddle point, hence it is a local minimum in one plane but a local maximum in another plane. {{fontcolor1|gray|(Using planes here is not correct. A TS is a maximum in one direction (the reaction coordinate) and a minimum in the orthogonal coordinate. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 14:38, 3 June 2018 (BST))}}
However for the minima any perturbation will cause an increase in energy, hence the minima correspond to stable species, i.e. the reactants and products.

The minima and transition state can be distinguished mathematically. The first derivative of both the minima and the transition state will be equal to zero. However if you take the second partial derivatives ∂2V('''r1''')/∂'''r1''' and ∂2V('''r2''')/∂'''r2'''. For the minima the second derivatives in both directions will be greater than zero (minimum). However for the transition state one derivative will be greater than zero (minimum) and one will be less than zero (maximum).

===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.===

After setting the momentum to zero and setting the distances r1 equal to r2 the distance can then be varied until the lines in the internuclear distance vs time plot flatten. This is significant because at the transition state the atoms will not be oscillating. My value for the transition state position, rts, is 0.9075 Å.

[[File:rts_nk2018.png|frameless|centre|400px]]

===Comment on how the mep and the trajectory you just calculated differ.===
{| class="wikitable" border=1
! Calculation Type !! Dynamics !! MEP
|-
| Contour Plot || [[File:contour_Dynamics_nk2018.png|frameless|centre|400px]]
|| [[File:contour_MEP_nk2018.png|frameless|centre|400px]]

|-
| Internuclear Distances vs Time || [[File:IDT_Dynamics_nk2018.png|frameless|centre|400px]]
|| [[File:IDT_MEP_nk2018.png|frameless|centre|400px]]

|-
| Internuclear Momenta vs Time || [[File:IMT_Dynamics_nk2018.png|frameless|centre|400px]]
|| [[File:IMT_MEP_nk2018.png|frameless|centre|400px]]
|}

The reaction path (minimum energy path or mep) is a type of trajectory that corresponds to infinitely slow motion, this means the velocity always reset to zero in each time step. So for the MEP you do not observe the vibrational energies and hence the oscillations and not observed. The MEP shows the lowest energy path whereas the dynamic shows a truer path taken including the vibratonal energies. This can be seen in the contour plots as for the dynamics plot it oscillates whereas in the MEP it is a smooth line.
If you change the & final positions around, the result would the a mirror the graphs shown above. The graphs would just run in the opposite direction.

{{fontcolor1|gray|(This is all ok, but you really need to proof read your report. There is a staggering number of typos here! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 14:38, 3 June 2018 (BST))}}

===Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.===

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy / kcal !! Reactive? !! Contour Plot !! Comments
|-
| -1.25 || -2.5 || -99.018 || Reactive || [[File:table1_contour_nk2018.png|frameless|centre|300px]] || The atoms are moving with the correct momentum to pass the transition state and form the products
|-
| -1.5 || -2.0 || -100.456 || Non-reactive || [[File:table2_contour_nk2018.png|frameless|centre|300px]] || The atoms are not moving with enough kinetic energy in order to overcome the activation energy and hence simply collide but do not react.
|-
| -1.5 || -2.5 || -98.956 || Reactive || [[File:table3_contour_nk2018.png|frameless|centre|300px]] || The atoms are moving with enough energy in order to overcome the activation energy and hence can pass the transition state to form the products.
|-
| -2.5 || -5.0 || -84.956 || Non-reactive || [[File:table4_contour_nk2018.png|frameless|centre|300px]] || The molecules momentum in this case is too large. In the collision the atoms form a bond, but repels due to coulombic repulsion. Hence the molecule crosses the transition state again in a process known as 'recrossing'. It results in large oscillations.
|-
| -2.5 || -5.2 || -83.416 || Reactive || [[File:table5_contour_nk2018.png|frameless|centre|300px]] || In this case again the molecules collide with high momentum and crosses the transition state twice but in this it results in a reaction and the products are formed. It also results in large oscillations.
|}

===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 that the nuclei behave according to classical mechanics. It also assumes that the reactions have one saddle point, and hence one transition state, that must be passed over. The theory works well at lower temperatures but at the higher temperatures due to recrossing some reactions that pass the transition state it still may not react. Hence the observed rate of reaction is slightly lower than that predicted by the theory. The theory also does not consider quantum mechanical effects such as tunneling.

<ref> M. J. Pilling, P. W. Seakins Reaction Kinetics, 2nd edition, OUP, 1995</ref>

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

===Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?===

The reaction between F and H2 is exothermic, this can be seen in the downhill gradient of the surface plot. The total energy decreases from the reactants to the products. The bond dissociation energy of H2 is 432 kJ mol-1 whereas the bond dissociation energy of HF 562 kJ mol-1 <ref>http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html</ref>
. Hence the HF molecule is more stable than the H2 and hence energy is released and the reaction is exothermic. For H + HF to form H2 and F is a very endothermic reaction. However a proton transfer reaction given enough momentum is feasible.

{| class="wikitable" border=1
! F + H2 → HF + H !! H'F + H → H' + HF
|-
| [[File:FHH_Surface_nk2018.png|frameless|centre|400px]] || [[File:proton_transfer_nk2018.png|frameless|centre|400px]]
|}

{{fontcolor1|gray|(What you're meant to explore here is H + HF -> H2 + F and not the proton transfer. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 14:38, 3 June 2018 (BST))}}

===Locate the approximate position of the transition state===

Again the values for r1 and r2 were varied until no oscillation was observed in the internuclear distances vs time plot, ie. the lines were flat. The values for r1 was 0.744 Å and the value for r2 was 1.811 Å. It can be seen that the r1 bond length similar to that of the H-H, this is because the transition state energy is more similar to the reactants than the products. Hence the structure is more similar to that of the reactants than the product.

[[File:TS_FHH_nk2018.png|frameless|centre|400px]]

===Report the activation energy for both reactions.===

Using an MEP calculation type by slightly displacing the transition state the activation energy could be calculated. The relative activation energies show the relative feasibility of the reaction.

{| class="wikitable" border=1
! Reaction !! Approximate Activation Energy (kcal mol-1)
|-
| F + H2 → HF + H || 0.2
|-
| HF + H → F + H2 || 30
|}

{{fontcolor1|gray|(How did you calculate these numbers? You need to show your working. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 14:38, 3 June 2018 (BST))}}

{| class="wikitable" border=1
! F + H2 → HF + H !! HF + H → F + H2
|-
| [[File:FHH_GRAPH_1_nk2018.png|frameless|centre|400px]] || [[File:FHH_GRAPH_2_nk2018.png|frameless|centre|400px]]
|}

===Identify a set of initial conditions that results in a reactive trajectory for the F + H2===
The set of conditions used were A-B= 0.74, B-C= 2.3, pAB= 0 pBC=-2, it can be seen from the contour plot that under these conditions it is reactive.

[[File:FHH_GRAPH_3_nk2018.png|frameless|centre|400px]]

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

[[File:FHH_Reaction_dynamics_nk2018.png|frameless|centre|400px]]
The conservation of energy can be seen from the internuclear momenta vs time graph. As the reaction is very exothermic energy must released in the form of kinetic energy. This can be seen in the larger oscillations after the reaction. This can be confirmed experimentally via both temperature change and IR.

{{fontcolor1|gray|(You need to give some more detail here. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 14:38, 3 June 2018 (BST))}}

{| class="wikitable" border=1
! PFH !! PHH !! Contour !! Reactive
|-
| -0.5 || -3 || [[File:CON_1_nk2018.png|frameless|centre|400px]] || Unreactive
|-
| -0.5 || -2 || [[File:CON_2_nk2018.png|frameless|centre|400px]] ||Unreactive
|-
| -0.5 || -1 || [[File:CON_3_nk2018.png|frameless|centre|400px]] || Unreactive
|-
| -0.5 || 0 || [[File:CON_4_nk2018.png|frameless|centre|400px]] || Unreactive
|-
| -0.5 || 1 || [[File:CON_5_nk2018.png|frameless|centre|400px]] || Unreactive
|-
| -0.5 || 2 ||[[File:CON_6_nk2018.png|frameless|centre|400px]] || Unreactive
|-
| -0.5 || 3 || [[File:CON_7_nk2018.png|frameless|centre|400px]] || Unreactive
|}

{| class="wikitable" border=1
! PFH !! PHH !! Contour !! Reactive
|-
| -0.8 || 0.1 || [[File:CON_15_nk2018.png|frameless|centre|400px]] || Reactive
|}

As this is an exothermic reaction it has an early transition state, and hence it can be seen that according to Polanyi's rules it is more dependent upon the translational energy than the vibrational energy <ref>Atkins and de Paula: Physical Chemistry, 7th Edition Chp</ref>. This can be seen as only slightly altering the translational momentum resulted in a reaction whereas drastically changing the vibrational energy had no effect, as seen in the contour plots above.

=== H + HF ===

{| class="wikitable" border=1
! PFH !! PHH !! Contour !! Reactive
|-
| -8.5 || -0.5 || [[File:CON_16_nk2018.png|frameless|centre|400px]] || Reactive
|}

The F + H2 → H-F + H is endothermic and hence has a late transitional state. Again according to Polanyi's rules as this is a late transitional state and hence a lot of vibrational energy is required.

{{fontcolor1|gray|(Again, you need to elaborate a bit more. You also missed out on the last question! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 14:38, 3 June 2018 (BST))}}

==References==

MRD:NW716

2018-05-28T17:11:42Z

Fjs113: Marked by fjs113

=H + H2 System=

==Potential Energy Surface==

===Transition State and Minima===

'''Qn: What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.'''

{| class="wikitable" border="1"
|-
| [[File:NW716-MRD-PES.png|600px|left]]
|-
| Figure 1 - Transition State and Minimum <ref name="TS"/>
|}

The gradients of the potential energy surface at a minimum and at a transition structure are both 0. At a local minimum, the distance between two bonded atoms is a constant, hence, one of the component, eg. <math>{ \partial V\over \partial r_1}</math>, is zero and <math>{ \partial V^2\over \partial^2 r_1}</math> > 0. The other component, eg. <math>{ \partial V^2\over \partial^2 r_2}</math>, is increasing as r2 decreases, i.e. the single atom approaches the diatomic molecule. Transition state linking the two minima represents a maximum along the direction of the reaction coordinate, but along all other directions, it is a minimum. At the transition state, which is the saddle point of the graph, both <math>{ \partial V\over \partial q_1}</math> and <math>{ \partial V\over \partial q_2}</math> are zero. However, for the two reaction coordinates, one of the second derivatives is negative and the other is positive. Hence, if the point is a minimum in one direction and does not decrease in the orthogonal direction, it is a minimum. However, if the point is a minimum in one direction but a maximum in the orthogonal direction, it is a saddle point, which corresponds to the transition state.

{{fontcolor1|gray|(Perfect! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 18:11, 28 May 2018 (BST))}}

'''Qn: 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.'''

{| class="wikitable" border="1"
|-
| [[File:NW716-MRD-001.png|500px|left]] || [[File:NW716-MRD-002.png|500px|left]]
|-
| Figure 2 - Plot of Internuclear Distance VS Time || Figure 3 - Plot of Energy VS Time
|}

rts is estimated to be 0.9077 Å. Since r1 = r2 and there is no momentum, the distances of A-B and B-C are the same and should not vary. Hence, only two lines are observed in the Internuclear Distances vs Time plot as two lines overlap and the lines are perfectly horizontal. This can also be confirmed using the Energy vs Time plot, Fig 3. At the transition state position, the kinetic energy is zero and potential energy should be a constant.

==Reaction Trajectories==

===Minimum Energy Pathway and Dynamics Calculations===

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

{| class="wikitable" border="1"
|-
| [[File:NW716-MRD-003.png|500px|left]] || [[File:NW716-MRD-004.png|500px|left]]
|-
| Figure 4 - Minimum Energy Path (MEP) Calculation || Figure 5 - Dynamics Calculation
|}

The MEP trajectory is a smooth line without oscillation. The trajectory calculated using Dynamics shows the vibration of HB-Hc bond. This difference is owing to the different calculation methods. MEP corresponds to an infinitely slow motion. Each step is extremely small and the velocity is set to zero after each step. Hence, the motion of the molecule at each step is independent of the previous step and is a trajectory connecting all of the lowest energy points for each step. Therefore, MEP is a smooth and non-oscillatory line. On the contrary, Dynamics calculation corresponds to a continuous motion and every step is dependent upon the previous step (motion of atoms is inertial). Hence, the molecule possesses a velocity to climb up the potential energy surface and results in the oscillation. Moreover, to obtain the length of MEP shown in Fig 4 above, the Steps set for calculation is 50000. However, the Steps set for Dynamics calculation is only 500. Since each step for MEP is extremely small, more steps are required to obtain the same length of the trajectory with the same amount of time.

{{fontcolor1|gray|(This is perfect! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 18:11, 28 May 2018 (BST))}}

{| class="wikitable" border="1"
|-
| [[File:NW716-MRD-005.png|500px|left]] || [[File:NW716-MRD-006.png|500px|left]]
|-
| Figure 6 - Dynamics Calculation: Plot of Internuclear Momenta VS Time || Figure 7 - Dynamics Calculation: Plot of Internuclear Distance VS Time
|}

Using the Dynamics calculation, at large time, r1 increases linearly as HB and HC separates and r2 decreases slightly to 0.74 as HA-HB bond forms, seen in Fig 7. At large time, p1 increases to 2.5 and p2 increases to 1.25 (on average). These values indicate that once the reactants surpass the transition state, even slightly, the reaction proceeds readily.

When the parameters of the final position are used as the initial conditions and the signs of the momentum are reversed, the reaction will proceed and end at the transition state, as seen in Fig 8 and 9 below. The final position of this reaction is the red cross at the transition state. This is confirmed by the plot of Internuclear Distance VS Time as r1 is equal to r2 at the end of the reaction.

{| class="wikitable" border="1"
|-
| [[File:NW716-MRD-007.png|500px|left]] || [[File:NW716-MRD-008.png|500px|left]]
|-
| Figure 8 - Contour Plot || Figure 9 - Plot of Internuclear Distance VS Time
|}

===Reaction with Different Momenta Combinations===

'''Qn: Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.'''

{| class="wikitable" border="1"
|+ Table 1 - Trajectories with Various Momenta Combination
|-
| '''Reaction No.''' || '''p1''' || '''p2''' || '''Total Energy / kcal mol-1''' ||''' Kinetic Energy / kcal mol-1''' || '''Reactivity'''
|-
| 1 || - 1.25 || - 2.5 || - 99.018 || + 4.687 || Reactive
|-
| 2 || - 1.5 || - 2.0 || - 100.456 || + 3.250 || Unreactive
|-
| 3 || - 1.5 || - 2.5 || - 98.956 || + 4.750 || Reactive
|-
| 4 || - 2.5 || - 5.0 || - 84.956 || + 18.750 || Unreactive
|-
| 5 || - 2.5 || - 5.2 || - 83.416 || + 20.290 || Reactive
|}

====Reaction 1: p1 = -1.25, p2 = -2.5====

{| class="wikitable" border="1"
|-
| [[File:NW716-MRD-009.png|500px|left]] || [[File:NW716-MRD-014.png|500px|left]] || HA moves towards HB with HB-HC distance increases slightly. The reactants then reach the transition state structure and the reaction proceeds with HA-HB bond formation and HC moves away. The initial reaction path is smooth without oscillation as p1 is much smaller than p2. The translational kinetic energy in the system is dominantly in the AB distance coordinate. Once the reaction completes, the oscillation in AB coordinate indicates the vibration of HA-HB bond. This is because part of the kinetic energy and convert to vibrational energy of the bond.
|-
| Figure 10 - Surface Plot || Figure 11 - Contour Plot || Description
|}

====Reaction 2: p1 = -1.5, p2 = -2.0 ====

{| class="wikitable" border="1"
|-
| [[File:NW716-MRD-010.png|500px|left]] || [[File:NW716-MRD-015.png|500px|left]] || An increase in p1 with a decrease in p2 from the conditions in Reaction 1 result in the initial system with relatively more kinetic energy in BC distance coordinate which can be observed from the oscillation of HB-HC bond. However, the atoms do not possess sufficient kinetic energy to climb up the energy surface and reach the transition state, this is owing to the decrease in p2, which leads to a decrease in translational energy in AB coordinate. Hence, HA moves away from HB with HB-HC bond retains. No reaction takes place.
|-
| Figure 12 - Surface Plot || Figure 13 - Contour Plot || Description
|}

====Reaction 3: p1 = -1.5, p2 = -2.5====

{| class="wikitable" border="1"
|-
| [[File:NW716-MRD-011.png|500px|left]] || [[File:NW716-MRD-016.png|500px|left]] || An increase of p2 from Reaction 2 enables the atoms to climb up the energy surface to allow the reaction to proceed. The vibration of HB-HC bond is smaller compared to Reaction 2 as p2 increases which cancels out some of the kinetic energy in the BC coordinate. However, the vibration is greater compared to Reaction 1 with an increase of p1 only. This illustrates that the relative values of p1 and p2 affect the initial shape of the trajectory by altering the kinetic energy in the two coordinates of the system. The amplitude of the oscillation is greater after the reaction. This indicates that the release of vibrational energy from the reaction.
|-
| Figure 14 - Surface Plot || Figure 15 - Contour Plot || Description
|}

====Reaction 4: p1 = -2.5, p2 = -5.0====

{| class="wikitable" border="1"
|-
| [[File:NW716-MRD-012.png|500px|left]] || [[File:NW716-MRD-017.png|500px|left]] || There is a huge increase in both p1 and p2. The system does reach the transition state region but barrier recrossing takes place. The initial trajectory is smooth without oscillation as p1 is significantly lower than p2 (half of p2, similar to Reaction 1). Hence, the initial kinetic energy is mainly along the AB distance coordinate. With a high energy content, after collision, HB-HC bond vibrates more rigorously as shown with a large amplitude of oscillation owing to energy transfer from kinetic to vibrational.

|-
| Figure 16 - Surface Plot || Figure 17 - Contour Plot || Description
|}

====Reaction 5: p1 = -2.5, p2 = -5.2====

{| class="wikitable" border="1"
|-
| [[File:NW716-MRD-013.png|500px|left]] || [[File:NW716-MRD-018.png|500px|left]] || p2 increases slightly with p1 unchanged from conditions for Reaction 4. Barrier recrossing takes place but the reaction proceeds in this case. This means a small change in p2 will change the reactivity of the reaction. Initial trajectory is smooth without oscillation as p1 is relatively small comparing to p2. HA-HB bond formed vibrates more rigorously owing to energy transfer from kinetic to vibrational.
|-
| Figure 18 - Surface Plot || Figure 19 - Contour Plot || Description
|}

===Transition State Theory===

'''Qn: 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 (TST) assumes that the motion of the atoms obeys classic mechanics and with higher initial momenta, trajectories starting with the same position would be more likely to
form the transition state and react as the system contains more kinetic energy to overcome the activation energy barrier. Hence, the reaction should be more likely to proceed with a higher momenta combination as the system would possess a higher kinetic energy. However, from the experimental values above, it can be concluded that the system with a lower kinetic energy ('''lower than the activation energy of -99.305 - (-103.869) = 4.564 kcal mol-1, calculated using the programme''') would not be reactive, for instance, a kinetic energy of 3.25 kcal mol-1 in Reaction 2. However, a system would be unreactive despite a very high kinetic energy which is higher than the activation energy, for instance, Reaction 4 above. This is because as the momenta increases, the atoms populate higher energy vibrational modes. Their motion becomes more complex and collisions might cause the transition state to deviate from the lowest energy saddle point. Hence, even with sufficient energy, a reaction does not occur. This means TST will fail at high temperatures when more reactant molecules occupy higher energy vibrational modes.

TST also assumes that transitions from a reactant state to a product state occur without barrier recrossings.<ref name='BR'/> However, Reaction 4 and 5 above show barrier recrossing which does not agree with TST.

Moreover, since the TST is based on the assumption that nuclei behave according to classic mechanics, it does not describe the quantum effects, specifically, quantum tunnelling. There is always a possibility that the reactants will react even if they do not collide to form the transition state and cross the activation barrier. Quantum tunnelling is significant if the activation barrier is low as the tunnelling probability increases with decreasing barrier height.

TST also assumes that the transition state is long-lived so that the reaction continues. It fails if the transition state is short-lived and could affect product selectivity.<ref name="TST"/>
Therefore, whether a reaction will take place does not solely depend on the initial momenta of the reactants, which illustrates the kinetic energy the system possessed. The Transition State Theory is not accurate to predict the reaction rate with a low activation barrier, a short-lived transition state and at high temperatures.

=F - H - H System=
==Potential Energy Surface==

{| class="wikitable" border="1"
|-
| [[File:NW716-MRD-019.png|500px]] || [[File:NW716-MRD-020.png|500px]]
|-
| Figure 20 - Surface Plot of F + H2, reactants on the left || Figure 21 - Surface Plot of H + HF, reactants on the left
|}

Setting the reactants to be 2.3 Å away from each other, B-C distance to be H-H and H-F bond length respectively and both momenta to be 0, the above two surface plots are obtained. From Fig 20, by observing the two minima, the reactants, H2 and F are at a higher minimum comparing to the product. Hence, the products are lower in potential energy which means that the reaction is exothermic.

Similarly, from Fig 21, the products are higher in potential energy and the reaction between H and HF is endothermic.

Formation of H-F bond and breaking of H-H bond releases energy to the surrounding. Formation of H-H bond and breaking of H-F bond need energy input. These results reflect that H-F bond is stronger than H-H bond, which agrees with a higher H-F bond energies. Bond energy of H-F is 565 kJ mol-1 and that of H-H is 432 kJ mol-1.

==Transition State Approximation==

===F + H2===
Transition state of F + H2 reaction should compose of longer H-H and H-F bond distances. Based on observation of Fig 20, distance AB is around 1.8 Å and distance BC is around 0.75 Å. This agrees with the Hammond postulate as the transition state of an exothermic reaction should resemble more closely to the reactants, H2 and a separate F atom. Hence, the H-H bond is only stretched a bit from the bond length of 0.74 Å. Using trial and error, F-H distance is estimated to be 1.8107 Å and H-H distance is 0.7450 Å for the transition state structure. From the contour plot, Fig 22, the reactants do not move along the PES and inter-atomic distances stay constant, shown in Fig 23.

{| class="wikitable" border="1"
|-
| [[File:NW716-MRD-021.png|500px]] || [[File:NW716-MRD-022.png|500px]]
|-
| Figure 22 - Surface Plot of Transition State || Figure 23 - Internuclear Distance VS Time of Transition State
|}

===H + HF===

The H-H and H-F distances for this reaction should be the same as the above reaction as this is the reverse reaction. The transition state in this endothermic reaction should resemble more closely to the products, which are H2 and a separate F atom, based on Hammond postulate. Hence, the F-H distance is 1.8107 Å and H-H distance is 0.7450 Å. The transition state is illustrated with Fig 24 an 25 below.

{| class="wikitable" border="1"
|-
| [[File:NW716-MRD-023.png|500px]] || [[File:NW716-MRD-024.png|500px]]
|-
| Figure 24 - Surface Plot of Transition State || Figure 25 - Internuclear Distance VS Time of Transition State
|}

===Activation Energy===

A MEP calculation from a structure neighbouring the transition state, H-F length used in F + H2 reaction 1.8207 Å and is 1.8007 Å in H + HF reaction. These distances are chosen so that the trajectory is towards the reactants and the change in potential energy is therefore the activation energy. The potential energy of the transition state was determined to be -103.752 kcal mol-1, which is the same for both reactions.

The potential energy of reactants in F + H2 reaction is -133.624 kcal mol-1 and that in H + HF reaction is -103.886 kcal mol-1. Hence the activation energies for the two reactions are:

F + H2 : Ea = -103.752 - (-103.886) = 0.134 kcal mol-1

H + HF : Ea = -103.752 - (-133.624) = 29.872 kcal mol-1

{| class="wikitable" border="1"
|-
| [[File:NW716-MRD-026.png|500px]] || [[File:NW716-MRD-027.png|500px]]
|-
| Figure 26 - Energy VS Time (F + H2) || Figure 27 - Energy VS Time (H + HF)
|}

==Reaction Dynamics==

===F + H2===

====Reactive Conditions====

{| class="wikitable" border="1"
|-
| '''Reaction No.''' || '''p1''' || '''p2''' || '''Contour Plot'''
|-
| 1 || -1.5 || 0 || [[File:NW716-MRD-028.png|500px|thumb|Figure 28]]
|-
| 2 || -1.5 || -0.25 || [[File:NW716-MRD-029.png|500px|thumb|Figure 29]]
|-
| 3 || -2.0 || -1.25 || [[File:NW716-MRD-030.png|500px|thumb|Figure 30]]
|-
| 4 || -2.2 || -1.25 || [[File:NW716-MRD-031.png|500px|thumb|Figure 31]]
|-
| 5 || -2.2 || -1.5 || [[File:NW716-MRD-032.png|500px|thumb|Figure 32]]
|}

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

From all of the five conditions above, it can be observed that the initial oscillation of the trajectory is extremely small and the final HF molecule contains great vibrational energy, as seen from the large amplitude of the oscillation once the reaction completes. Reaction energy released as F approaches H2 becomes the motion in HF, the product vibration, whereas energy released as HF separates from H becomes the motion along the BC distance coordinate, the product translation. The HF vibrational energy can be determined using IR and analyse the frequency of the vibrational band. Translational energy of H atom can be confirmed by measuring the scatter of the products. The energy distribution can be measured by recording the infrared chemiluminescence of the reaction under "arrested relaxation".<ref name="PER"/>

The above five conditions illustrate that a higher p1 (pFH) is always required for the reaction to be reactive, especially in Reaction 1 when p2 (pHH) is zero. Hence, one can assume that pFH, which corresponds to a high translational energy of the reactants, more effectively affects the reactivity of the reaction.

====Reactivity with Variation of pHH====

When rHH = 0.74 and momentum pFH = -0.5, values of pHH in the range from -3 to 3 were used for calculation. As pHH gets close to -3 or 3, barrier recrossing takes place and the reaction conditions render the reaction unreactive. Starting from pHH = -3, when pHH is greater than approximately -2.5, the reaction is reactive. When pHH reaches 1, or extremely close to 1, the reaction is unreactive and remains unreactive until pHH is above 1.5 and below approximately 2.4, although the reaction may be unreactive with pHH in between 1.5 and 2.4 (eg. 1.7 and 2.1). This shows that change in pHH can affect the reactivity of the reaction even with just a small change at constant pFH but there is no trend observed for whether the reaction is reactive or not.

However, increasing pFH slightly to -0.8, and reduce the overall energy of the system by reducing pHH to 0.1, the reaction is now reactive as shown in Fig 34. This means that a higher overall energy of the system does not mean that the reaction will be reactive, but a high pFH, which corresponds to a high translational energy, is required.

{| class="wikitable" border="1"
|-
| [[File:NW716-MRD-033.png|500px]]
|-
| Figure 33 - Contour Plot with pFH = -0.8 and pHH = 0.1
|}

The reaction between F and H2 is highly exothermic with an extremely low activation barrier. The transition state is in the entrance valley, hence, an early barrier. Using the assumption from the five reactive conditions above ( pFH more effectively affects the reactivity of the reaction.) and observations from various pHH values, pFH does have a significant effect on the reactivity of the reaction.

===H + HF===

{| class="wikitable" border="1"
|-
| [[File:NW716-MRD-034.png|500px]] || [[File:NW716-MRD-035.png|500px]]
|-
| Figure 34 - Contour Plot with pHH = -50.0 and pHF = 0.01 || Figure 35 - Contour Plot with pHH = -8.0 and pHF = 0.5
|}

pHH = -8.0 is still very large. This complies with the high activation energy of this reaction as the products formed are thermodynamically less stable than the reactants. It can be concluded that an increase in pHF, which corresponds to a high vibrational energy in H-F coordinate, is necessary for the reaction to take place. pHH, which corresponds to the translational energy in the H-H coordinate that is always very large, does not contribute too much to affect the reactivity of the reaction.

The transition state is present in the exit valley, hence, a late barrier for this endothermic reaction. The reactivity of endothermic reactions is thus dominated by the vibrational energy of the system from the results above.

==Conclusion from Reaction Dynamics==
'''Qn: 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.'''

For substantial exothermic reactions, such as F + H2, the transition state is located in the entrance valley, corresponds to an early barrier of the PES. For substantially endothermic reactions, such as H + HF, the transition state is in the exit valley, corresponds to a late barrier. The favoured degree of freedom for barrier crossing in exothermic reactions would be translation. This means that the momentum of the approaching atom and the atom which it is going to collide with has a greater impact on the rate of the reaction (eg. F atom in F + H2 reaction). However, reagent vibration, which is related to the momentum of the two bonded atoms (eg. HF in H + HF reaction) in the colliding particles, would be most effective in enabling endothermic reactions to take place.<ref name="PER"/> The above reaction results do agree with Polanyi's empirical rules. However, sometimes a small change in one of the momenta might change the reaction from reactive to unreactive. Hence, Polanyi's empirical rules are a guideline to describe general required excitation for exothermic and endothermic reactions, but unable to predict whether the reaction is reactive for a certain initial conditions.

{{fontcolor1|gray|(Overall, this is a pretty much perfect report. Very well done. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 18:11, 28 May 2018 (BST))}}

=References=

<references>
<ref name="TS">E. G. Lewars, Computational Chemistry (Springer Netherlands, Dordrecht, 2011; http://link.springer.com/10.1007/978-90-481-3862-3).</ref>
<ref name="TST">D. Dyson, Advanced Chemical Kinetics, World Technologies, 2012.</ref>
<ref name="PER">J. C. Polanyi, Some Concepts in Reaction Dynamics. Accounts of Chemical Research. 5, 161–168 (1972).</ref>
<ref name="BR">T. Komatsuzaki, M. Nagaoka, Study on “regularity” of barrier recrossing motion. Journal of Chemical Physics. 105, 10838–10848 (1996).</ref>
</references>

MRD:BPW16Y2

2018-05-28T16:41:08Z

Fjs113: Marked by fjs113

== Exercise 1 - HHH System ==

=== Distinguishing Between a Minimum and Transition State on a Potential Energy Surface ===

[[File:BPW16_Exercise1_Surface.PNG|thumb|centre|Figure 1 - Potential energy surface showing trajectory across lowest energy pathway.]]

[[File:BPW16_Exercise1_Surface2.PNG|centre|thumb|Figure 2 - Potential energy surface showing maximum character.]]

[[File:BPW16_Exercise1_Surface(showingrolltoreac).PNG|centre|thumb|Figure 3 - Potential energy surface showing minimum character of reactants.]]

These plots were obtained with the parameters of H1H2 distance = 0.74 Å, momentum = 0, H2H3 distance = 2.30 Å, momentum = -2.7.

At a transition state or a mimimum, the first derivative of the potential energy with respect to each distance (r1 and r2) will be equal to zero. To determine whether the point is a minimum or a transition state, the second derivative of the potential energy with respect to r1 and r2 must be taken (these are orthogonal axes). If the result of this gives two positive values, the point is a minimum (figure 3), while if the result of this gives one negative value and one positive value, the point is a local maximum, hence the transition state (figure 2).

{{fontcolor1|gray|(You need to clarify, that here we are taking the second '''partial''' derivative. Also, the transition state is not a local maximum, but rather a saddle point! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:41, 28 May 2018 (BST))}}

=== Determining the Transition State Trajectory Distance ===

The internuclear distance in the transition state can be determined through a plot of internuclear distance vs time, where we set both momenta to zero. The transition state can then be found my testing different values of the internuclear distances until there is no oscillation observed on the plot. This corresponds to a sphere sitting on the top of the maximum, and having no momentum, so it just 'sits there'. There would be no gradient in the direction orthogonal to the maximum point, hence the trajectory will just oscillate on the maximum point and never 'roll off'. The value determined was 0.9076 Å, to 4 d.p, as can be seen from figure 4.

[[File:BPW16_H2_H_TSDET.PNG|thumb|centre|Figure 4 - Internuclear distance vs time plot for the H + H2 trajectory with equal bond lengths.]]

=== Comparing MEP and Dynamics Calculation Methods ===

Figures 5 and 6 show the PES for when r1 = 0.9176 Å and r2 = 0.9076 Å (i.e. slightly displaced from each other).

[[File:BPW16Exercise1_MEP.PNG|thumb|centre|Figure 5 - Potential energy surface calculated with MEP.]]

[[File:BPW16Exercise1_Dynamics.PNG|thumb|centre|Figure 6 - Potential energy surface calculated with Dynamics.]]

MEP is the minimum energy pathway for the trajectory. It corresponds to infinitely slow motion of the species. However, the MEP does not give a realistic account of the motion of the atoms during the reaction since the atoms have a mass, and hence their motion will be inertial.

The MEP plot shows the pathway for the species coming together if we treat them as two solid entities. The dynamic plot treats the system as a diatomic molecule and a single atom. Hence, as can be seen from the figures 5 and 6, the MEP plot does not show any oscillation, as there is no vibrational energy, but the dynamic plot shows oscillations and hence vibration energy of the diatomic.

=== Testing Different Reaction Trajectories to Determine Reactivity ===

{| class="wikitable" border="1"
|+ Table 1 - Testing different reaction trajectories
! p1 !! p2 !! Reactivity !! Total Energy (kcal/mol) !! PES !! Comments
|-
| -1.25 || -2.5 || Reactive || -99.119 || [[File:BPW16Surface_Plot1.png]] || This trajectory is a productive reaction, with reactants passing through the transition state and becoming products, ie AB + C --> A + BC. For this to occur, the reactants must have had enough energy to overcome and traverse the transition state energy maximum. It can be seen that the reactant, AB, has no vibrational energy, as the trajectory does not oscillate in that region, however it does for the product, showing that BC had vibrational energy. This energy would have been converted from translational (kinetic) energy when the two reactant species collided.
|-
| -1.5 || -2.0 || Unreactive || -100.456 || [[File:BPW16Surface_Plot2.png]] || This trajectory is non-productive, with the reactants reaching the transition state but rather than forming products they go back on themselves. This corresponds to the reactants, AB and C, not having enough energy to overcome the transition state energy barrier and hence do not form products. The diatomic simply 'bounces off' the third atom, moving in the reverse direction to what it started of as. It can be seen that the diatomic AB has vibrational energy before the collision and after. AB will have gained some more vibrational energy from the collision as translational energy will have been converted into vibrational energy, in order to obey the conservation of energy. It has been calculated that this energy has a lower total energy than system one, and therefore does not have enough energy to overcome the transition state energy barrier.
|-
| -1.5 || -2.5 || Reactive || -98.956 || [[File:BPW16Surface_Plot3.png]] || This shows a productive reaction where AB + C turns into products A and BC after passing successfully through the transition state region. It has been calculated that this system has a higher energy that the first two systems studied, and hence can overcome the transition state energy barrier and cross it to form products. It can also be seen that both the reactant and product diatomics, AB and BC respectively, have vibrational energy and so the trajectory oscillates.
|-
| -2.5 || -5.0 || Unreactive || -84.956 || [[File:BPW16Surface_Plot4.png]] || The trajectory here is unproductive, with reactants not forming products at the end of the reaction. The reactants, AB and C are able, momentarily, to form products (A and BC). Therefore, the reactants must have contained enough energy to cross the transition state maximum. However, most likely due to the large vibrational energy gained from the transfer of translational energy into vibrational energy when the species collide, there is so much vibrational energy in the diatomic BC that the bond dissociates. This now produces three single atoms, A, B, and C. Atom A is moving in direction of atom B and hence they react to form the reactant species, AB, again, but now possessing a great deal of vibrational energy (but not so much that the AB bond breaks). The energy of the system is higher than any of the ones studied previously, showing that there is enough energy to break the BC bond when it forms.
|-
| -2.5 || -5.2 || Reactive || -83.416 || [[File:BPW16Surface_Plot5.png]] || This plot is a productive reaction. It can be seen that the reactants, possessing no vibrational energy in the diatomic, are able to reach and cross the transition state energy barrier to form products. However, the product diatomic, BC, then dissociates to reform the reactants, as it must have a high amount of vibrational energy, which has been converted into it from the high amount of translational energy in the reactants. When the BC molecule dissociates it sends atom B in the same direction as atom A, so these two can react together to reform the reactant diatomic, AB. It can be seen from the plot that when this occurs, AB is formed in a vibrationally excited state, having enough energy to break apart again and react successfully with atom C this time, to form products of BC and A. The BC product diatomic has a large amount of vibrational energy. This is an example of barrier recrossing.
|}

=== Transition State Theory ===

In Transition State Theory, a reaction is described by considering the motion of atoms. The Born-Oppenheimer approximation is applied and the motion of the nuclei and electrons are separated, given that nuclei are much heavier than electrons. Following this, the motion of the atoms can be treated classically using Newton's laws of motion. This neglects the fact that vibrations are quantised or that tunneling can happen through quantum mechanics. Another assumption is that the transition states obey a Maxwell-Boltzmann distribution shape and that the species have to collide with enough energy for a reaction to be successful, i.e. to pass over the transition state barrier, however this is not always the case in reality due to quantum tunneling effects.

Transition state theory says that activated complexes exist in a type of equilibrium, quasi-equilibrium, with the reactants, and that these complexes can convert into products. By using rate equations such as the Arrhenius or Eyring equations, the rate of reaction can be found. The transition state will be a saddle point on the PES. The quasi-equilibrium point means that reactants are in equilibrium with the activated complex, but the activated complex is not in equilibrium with the products, i.e. once the transition state has been traversed, the products can not reform the reactants. Another downfall of transition state theory is that, especially at higher temperatures, the trajectory does not always pass through the lowest energy saddle point in the PES, as higher energy vibrational modes are populated.

For systems 1 to 3 studied above, the predicted reaction rates through transition state theory are likely to match well with experimental values, as either the reactants simply cross the transition state barrier to form products, or do not have enough energy to react at all. For systems 4 and 5 however, the rates are likely to differ from predicted values as these systems show barrier-recrossing, where once products have formed, they actually cross back over the transition state barrier and reform the reactants, something that is not 'allowed' in transition state theory. The higher temperature idea could be seen through systems 4 and 5 having higher momenta than in 1 to 3. <ref name="TST" />

{{fontcolor1|gray|(Very good. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:41, 28 May 2018 (BST))}}

== Exercise 2 - FHH system ==

=== Classifying the Reaction Thermodynamics ===

From figures 7 and 8 it can be seen that the reaction of H2 + F is exothermic as the products, HF + F (with the diatomic distance of HF labelled as BC on the plot) is lower in energy than the reactant species, H2 and F. It is therefore the case that the reverse reaction, HF + H is an endothermic. This can be related to the relative bond strengths: clearly the HF bond is much stronger than an H-H bond. This is due to the difference in electronegativity in HF, creating an ionic contribution to the bond, making it stronger, hence energy is released when HF forms from H2 and F.

The parameters used were H1H2 (diatomic) distance = 0.74 Å, momentum = 0, H2F distance = 2.30 Å, momentum = -2.7.

[[File:BPW16_H2_F_Surfaceplot.png|thumb|centre|Figure 7 - PES showing overall exothermic reaction of H2 + F. ]]

[[File:BPW16_H2_F_Surfaceplot2.png|thumb|centre|Figure 8 - PES showing how the products HF + F are lower in energy than HH + F.]]

=== Determining the Transition State ===

To determine the transition state, I first looked at the internuclear distance vs time plot (figure 9) from the trajectory plotted in figures 7 and 8. The transition state will be in the region where the AB and BC lines cross over, i.e. between 0.7 and 0.8 seconds approximately. To determine the transition state accurately, The momenta for each species, H2 and F, were set to be zero, and the distances AB and BC varied until there was no oscillation observed in either line, corresponding to the H-H-F activated complex at the very top of the peak of the transition state, and not 'rolling' to either reactants or products. This can be seen in figure 10, where the HH (AB) distance is 0.745 Å and the HF (BC) distance is 1.811 Å.

[[File:BPW16_H2_F_TS_Det_first.png|thumb|centre|Figure 9 - Internuclear distance vs time for reactive trajectory]]

[[File:BPW16_H2_F_TS_Det_opt.png|thumb|centre|Figure 10 - Internuclear distance vs time for transition state]]

=== Determining the Activation Energy ===

[[File:BPW16_HF_H_AE.png|thumb|centre|Figure 11 - Energy vs time plot showing the energy change for HF + H.]]

[[File:BPW16_HH_F_AE.png|thumb|centre|Figure 12 - Energy vs time plot showing the energy change for HH + F.]]

To find the activation energy, I used an MEP calculation and displaced one of the distances slightly from the transition state distances determined previously. This meant that, depending on which distance was changed, the trajectory would fall from the transition state maximum and simply roll to the valley floor of either the reactants or products. The activation energy was then calculated from the difference in potential energy on an energy vs time graph.

HF + H --> H2 + F
Activation energy = 29.969 kcal/mol

H2 + F --> HF + H
Activation energy = 0.200 kcal/mol

=== Reaction Dynamics ===

A productive, reactive trajectory, where H2 + F turn into HF + F products was found with the initial conditions of HH(0.79 Å, momentum = 0.78), HF(1.8 Å, momentum = -0.5).

[[File:BPW16_H2_F_Surfaceplot_reactive.png|thumb|centre|Figure 13 - Contour plot showing productive trajecotry. ]]

[[File:BPW16_H2_F_Surfaceplot_reactive2.png|thumb|centre|Figure 14 - Surface plot showing productive trajectory. ]]

[[File:BPW16_H2_F_mom_time_reactive.png|thumb|centre|Figure 15 - Internuclear momenta vs time graph for reactive conditions. ]]

[[File:BPW16_H2_F_energy_time_reactive.png|thumb|centre|Figure 16 - Energy vs time graph for reactive conditions. ]]

It can be seen from the contour plot, that the product diatomic, HF, is formed in a vibrationally excited state, and continues to oscillate, possessing vibrational energy. From the energy vs time, graph it can be seen that the conservation of energy principle is obeyed, as when the potential energy is at a maximum in an oscillation, the kinetic energy is at a minimum, and vice versa. The vibrationally excited HF molecule will release its energy to the surrounds as heat most likely, colliding with other gas molecules. A way to experimentally test this prediction would be to obtain a gas-phase FTIR spectrum of the HF produced, and this should show relaxation from the vibrational excited state to the ground state. A photon would be emitted corresponding to the difference in energy between the first excited and ground electronic states.

=== Polanyi's Empirical Rules ===

Polanyi's empirical rules state that vibrational energy is much more efficient at activating a late transition state reaction than translational energy. The opposite is true for an early transition state, where translational energy is more effective than vibrational energy. In order to test these rules, different initial momenta were set up. The H2 + F --> HF + F reaction is exothermic, as seen in figure 7, and hence through the Hammond postulate will have an early transition state, with the transition state structure resembling the reactants. Therefore, the reaction should be initiated more effectively with translational energy, rather than vibrational energy. The opposite is true for the HF + H --> H2 + F reaction, which is endothermic and hence will have a late transition state, and being initiated more effectively with vibrational energy.The quick brown fox jumps over the lazy dog. {{fontcolor1|gray|(Not sure if meant to be here or not? ;) [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:41, 28 May 2018 (BST))}} <ref name="PR" />

{| class="wikitable" border="1"
|+ Table 2 - Testing different Reaction Momenta For H2 + F
! p(HH) (Vibrational Energy) !! p(HF) (Translational Energy) !! Plot
|-
| -2.8 || -0.5|| [[File:BPW16_H2_F_Surfaceplot_1.png]]
|-
| -0.5 || -2.0 || [[File:BPW16_H2_F_Surfaceplot_2.png]]
|}

{| class="wikitable" border="1"
|+ Table 3 - Testing different Reaction Momenta For HF + H
! p(HF) (Vibrational Energy) !! p(HH) (Translational Energy) !! Plot
|-
| -0.4 || -6.7|| [[File:BPW16_HF_H_Surfaceplot_1.png]]
|-
| -6.7 || -0.4|| [[File:BPW16_HF_H_Surfaceplot_2.png]]
|}

It can be seen from table 2 that a greater weighting of translational energy rather than vibrational energy drives the exothermic, H2 + F --> HF + F, reaction, with an early transition state, forming the products after successfully passing over the transition state barrier. However, a greater weighting of vibrational energy does not lead to a productive reaction. It can be seen from table 3 that the opposite is true, with an excess of vibrational energy driving the endothermic reaction, with a late transition state, forming products of H2 and F, whereas an excess of translational energy does not give a reactive, productive trajectory. These results are in agreement with Polanyi's empirical rules.

== References ==

<references>
<ref name="TST">https://en.wikipedia.org/wiki/Transition_state_theory, accessed May 2018</ref>
<ref name="PR">Science 21 Oct 2011: Vol. 334, Issue 6054, pp. 343-346, DOI: 10.1126/science.1208514</ref>
</references>

MRD:OPB16

2018-05-28T16:00:16Z

Fjs113: Marked by fjs113

=Second Year Computational Lab: Molecular Reaction Dynamics=
==Introduction==

'''introduce using r_1 and r_2 - and use vectors'''
'''force relation with potential'''
'''all simulations are using a collision angle on 180 (deg)

All systems in this report are triatomic and involve the collision between an atom and a diatomic molecule at a collision angle of π radians. Successful collisions take the general form:

A + B-C → A-B + C

For a reaction to be successful, it has to overcome the activation energy. If this energy is not met or exceeded, the reaction is unsuccessful leaving BC as the diatomic molecule. The activation energy is the energy required to form the transition state (TS):

[A--B--C]‡

{{fontcolor1|gray|(Nice little introduction! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:00, 28 May 2018 (BST))}}

The TS occurs at the maximum of the minimum energy path of the reaction and is a stationary point. This path also has two other stationary points which are minima. These represent the reactants and the products. The reactant minima is when the B-C bond length is at equilibrium and the distance between A and B is large. The product minima occurs for the opposite this.

'''The TS is at the saddle point on the potential surface'''

To determine between minima and the saddle point (TS), a Hessian Matrix for the potential surface, ''V('''r1''','''r2''')'' is developed:

:<math>H(r_1,r_2) = \begin{pmatrix}V_{r_1 r_1}(r_1,r_2) &V_{r_1 r_2}(r_1,r_2)\\V_{r_2 r_1}(r_1,r_2) &V_{r_2 r_2}(r_1,r_2)\end{pmatrix}</math>.

Let ''D(x,y)'' be determinant of the Hessian Matrix:

:<math>D(r_1,r_2)=\det(H(r_1,r_2)) = V_{r_1 r_1}(r_1,r_2)V_{r_2 r_2}(r_1,r_2) - \left( V_{r_1 r_2}(r_1,r_2) \right)^2 </math>

A stationary point on the surface occurs at point (''a,b'') when ''fr1(a,b)'' and ''Vr2(a,b)'' = 0. The type of stationary point is found using the rules below:

#If <math>D(a,b)>0</math> and <math>V_{r_1 r_1}(a,b)>0</math> then <math>(a,b)</math> is a local minimum of ''V''.
#If <math>D(a,b)>0</math> and <math>V_{r_1 r_1}(a,b)<0</math> then <math>(a,b)</math> is a local maximum of ''V''.
#If <math>D(a,b)<0</math> then <math>(a,b)</math> is a saddle point of ''V''.

If the determinant at ''(a,b)'' is equal to zero, then the nature of the stationary point is unknown.

==H + H2 system==

This system follows the reaction:

H + H2 → H2 + H

===Transition State Determination===

For the TS determination, the momentum of atom pair is set to zero and the outer hydrogens are set to be equidistant from the central one ('''r1''' = '''r2''').

For large distance values, a Lennard-Jones like potential can be seen on the contour plot (Shown in figure 1). The minima of this curve is along the reaction pathway and is the TS. As the three atom system oscillates along the curve the distance between the atoms is shortened till it no longer oscillates, and the TS if formed. This occurs at an atom separation of 0.90774 Å. Due to the symmetry of the TS, it is located at (0.90774,0.90774) along the '''r1''' and '''r2''' axis (Shown in Figure 2).

Once the separations were close to this value, the 'Internuclear Distances vs. Time' graphs generated from the simulation shown by Figure 3 were zoomed in and the oscillation more closely observed. If from time = 0, the distance increases, the separation value is too small, and the opposite is true for separation values too large. As can be seen by Figure 4, the value I have stated is too small, but looking at the vertical-axis of the graph the oscillation occurs over a 5*106 Å range which is incredibly small. To get a more accurate value for the position is redundant, if the value was change to being in the centre of the oscillation region, an insensible number of decimal places would have to be used to report the figure. As well, as this, further enhancement of the plot would show that there is still an oscillation.

The upper line on these two figures is the position of the third H atom so is equal to two times the TS H--H 'bond' length.

{|style="margin: 0 auto;"
| [[File:Opb_h3_ljlike.png|thumb| Figure 1: A Lennard-Jones potential for the oscillating H--H--H system about the minimum potential point.]]
| [[File:Opb_h3_ts_pos.png|thumb| Figure 2: The position of the transition state shown on a contour map.]]
| [[File:Opb_h3_TS_evt.png|thumb| Figure 3: Energy vs. Time graph for the H--H--H transition state.]]
| [[File:Opb_h3_TS_dvt_zoom.png|thumb| Figure 4: An enhanced part showing the inconsistencies of the potential energy of the TS in the Energy vs. Time graph.]]
|}

{{fontcolor1|gray|(Very good work. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:00, 28 May 2018 (BST))}}

===Minimum Energy Path===

The reaction pathway was calculated by offsetting one of the H-H separations by 0.1 Å from the transition state, allowing the reactants to 'roll' down to the valley floor towards the products. Since the minimum energy path (MEP) corresponds to infinitely slow motion, no oscillation occurs in the H-H bond of the product, thus the reaction path seen in Figure 5 is a straight line along the valley floor. If instead the distance between the opposite pair of neighboring hydrogens is increased by 0.1 Å, then the reaction is driven towards opposite side of the potential energy surface. Figures 6 and 7 illustrate the difference in Intermolecular momenta vs Time for the two simulations as there is no oscillation in the H-H product bond for the minimum energy path.

{|style="margin: 0 auto;"
| [[File:Opb_h3_mep.png|thumb| Figure 5: The minimum energy path of the H + H2 reaction]]
| [[File:Opb_mom_mep.png|thumb| Figure 6: Intermolecular momenta vs. Time (MEP)]]
| [[File:Opb_mom_dyn.png|thumb| Figure 7: Intermolecular momenta vs. Time (Dynamics)]]
|}

===Trajectories===

The following trajectories were test using '''r1''' and '''r2''' values of 0.74 and 2.0 respectively:

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy !! Trajectory
|-
| -1.25 || -2.5 || -99.018 || Reactive
|-
| -1.5 || -2.0 || -100.456 || Unreactive
|-
| -1.5 || -2.5 || -98.956 || Reactive
|-
| -2.5 || -5.0 || -84.956 || Unreactive
|-
| -2.5 || -5.2 || -83.416 || Reactive
|}

{|style="margin: 0 auto;"
| [[File:Opb_h3_traj1.png|thumb|left|Figure 8: Trajectory contour for Conditions 1]]
| [[File:Opb_h3_traj2.png|thumb|center|Figure 9: Trajectory contour for Conditions 2]]
| [[File:Opb_h3_traj3.png|thumb|left|Figure 10: Trajectory contour for Conditions 3]]
| [[File:Opb_h3_traj4.png|thumb|center|Figure 11: Trajectory contour for Conditions 4]]
| [[File:Opb_h3_traj5.png|thumb|left|Figure 12: Trajectory contour for Conditions 5]]
|}

For this reaction, A-B is the initial H2 molecule and C is the free hydrogen. The first reaction is successful as indicated by the increased distance between A and B and the slight oscillation of bond length for B-C once the trajectory has surpassed the transition state. The increased oscillation of the bond is from the vibrational energy gained by the molecule. In the second reaction, the trajectory fails to overcome the activation energy of the TS and is instead reflected back along the potential well. The third trajectory is similar to the first as the collision is successful as the trajectory passes over the transition state. One difference between the first and third trajectories is that the third has more initial vibrational energy in the B-C bond as evidenced by the larger B-C separation fluctuation before reaching the saddle point. The fourth reaction shows the trajectory crossing the TS, however due to the high vibrational energy, the trajectory strongly oscillates along the "walls" of the potential ridge till passes back through the TS, into the initial state but with larger vibrational energy in the B-C bond. In the final reaction, the trajectory surpasses the transition state, crosses back over the TS as in the previous trajectory, but then crosses the TS again leading to a successful reaction.

The table also shows that there is no general trend between energy and the success of the reaction. This is because there are different types of energies in the system that each effect the trajectory differently. This can be seen in the case of reactions 4 and 5 where the large amount of vibrational energy leads to the trajectory passing over the TS multiple times.

{{fontcolor1|gray|(Well done. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:00, 28 May 2018 (BST))}}

===Limitations of Transition State Theory===
Transition state theory (TST) is the idea that the border between the initial and final states of a reaction is given a transition state which occurs at a potential energy maximum along the reaction path and must be energetically overcome for the reaction to proceed. The following assumptions are made: <ref name="TST">T. Bilgaard and T. J. Nørskov, '' Chemical Bonding at Surfaces and Interfaces'', '''2008''', 143-254.</ref>

#The atomic nuclei behave classically
#If a product crosses the TS with velocity towards the final state, it will not cross the border again
#Only one reaction path - via lowest energy saddle point

The first assumption ignores possible quantum effects. These effects are important when simulating systems with low activation energies and low particle masses as there is increased chance of tunnelling through the energy barrier, meaning a lower energy than assumed is required for the trajectory to cross the TS. The affect of these properties on the transmission coefficient for a particle tunnelling through a barrier of energy ''U(x)'' is given by:

:<math>T(E) = e^{-2\int_{x_1}^{x_2} \mathrm{d}x \sqrt{\frac{2m}{\hbar^2} \left[ U(x) - E \right]}}</math>

This would result in a higher rate being given based on TST than for experimental values.
The second point was violated in the last two examples where the trajectory crossed the TS multiple times. Since this is not expected for TST, the experimental rate will be lower than predicted. The third point means that no other alternate reaction pathways are allowed, only the one that follows the minimum energy path. Since quantum effects are only likely to occur for very small mass systems, overall, the experimental rate constant for a given reaction will be larger than the value calculated according to TST.

==F--H--H System==

The are two possible reactions in this system:

F + H2 → HF + H '''(Reaction 1)'''

H + HF → H2 + F '''(Reaction 2)'''

The H-F bond strength is much stronger than that of H-H, and are reported in literature as 569.7 and 435.7 kJ mol-1 respectively.<ref name="Bonds">Y. R. Luo, ''Comprehensive Handbook of Chemical Bond Energies'', Cambridge , '''2007'''.</ref> Because of this, more energy will be required for a successful Reaction 2. These values show that 1 is exothermic and Reaction 2 is endothermic. Even though Reaction 1 will release energy, an activation energy will still be required to reach the [F--H--H]‡ TS.

{{fontcolor1|gray|(Would've been good if you explained this using the contour plot from the program rather than just using bond energies. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:00, 28 May 2018 (BST))}}

===Determination of the Transition State===

As similar method was employed as before for locating the TS where the momentum of each component was set to zero. In this case however, '''r1''' and '''r2''' are not equal, and due to the larger interaction between H-F as opposed to H-H, the distance between H-F will be larger in the TS than between the two hydrogen atoms. As before, the atom separations were altered till the potential of the system oscillated about a certain point. The bond separations were then further altered till this oscillation was infinitesimally small and the TS was left. This gave the coordinates of the TS as (1.81074, 0.74488) along the F-H and H-H axis respectively and can be seen in Figure 13.

{|style="margin: 0 auto;"
| [[File:Opb_fhh_ts.png|thumb|Figure 13: The position of the [F--H--H]‡ TS.]]
|}

Hammond's postulate states that the structure of the TS will most closely resemble the species in the reaction which is closest in free energy to it. This means that for an exothermic reaction (like reaction 1) the TS most closely resembles the reactants in structure, and that for endothermic reactions (reaction 2) the TS is closest to the products. Since this TS is non-symmetrical, the TS no longer occurs equidistant from the reactant and product minima. This results in early and late transition states, where an early TS is when the TS is closer to the initial state than the final state. From Hammond's postulate we can say that exothermic reactions will have an early TS, and endothermic reactions will have a late TS. Figure 13 shows that the TS lies closer to the higher in energy F + H2 side.

===Activation Energies===

The activation energy is found using the same method as used for calculating the MEP in the three hydrogen atom system. However, larger deviations of atom separations from the TS had to be used to make sure that the reaction occurred (± 0.05 Å). The activation energy of the reaction is not equal to the H-H bond enthalpy since this a concerted reaction and not an 'SN1 - like' reaction. <ref name="Levine">R. D. Levine, ''Molecular Reaction Dynamics'', Cambridge, '''2009''', 150-160.</ref> The activation is calculated from the Energy vs. Time plot where the change in potential for the reaction to take place is easily observed (shown in Figure 14). These were worked out to be 0.36 and 29.75 kcal mol-1 for reaction 1 and 2 respectively.

{|style="margin: 0 auto;"
| [[File:Opb_hhf_act.png|thumb|Figure 14: Energy vs. Time for the reaction pathway of reaction 2.]]
|}

===Reaction Dynamics===

As the total energy of the system is conserved, when the reactants pass over the TS in reaction 1, the loss of potential energy is converted in both vibrational and translational kinetic energy. The gain in translational kinetic energy effects the motion of the products whilst the vibrational kinetic energy increases the oscillation of the diatomic molecule bond in the product. As the kinetic energy of the system increases, the heat and thus the temperature of the system will increase:

:<math>E_k = \frac{3}{2} k_B T</math> (For Three Dimensions)

Because of this, the reaction can be measured experimentally using calorimetry.

For reaction 1, a set of high kinetic energy trajectories with rHH = 0.74, rFH = 1.5, pFH = -0.5 and a range of pHH of [-3,3]. Figure 11 shows the trajectory for the reactant H2 with a momentum of 0. Here the reaction is successful and once the trajectory has passed the TS, due to the large potential energy drop discussed earlier, very strong oscillation of the newly formed F-H bond can be seen. The reaction remains successful until the magnitude of the initial HH momentum reaches 1.4. At this point the vibrational energy of the initial bond is very large, this translates to the vibrational energy of the product bond, causing for the trajectory to pass back over the TS yielding the reaction unsuccessful. An example of this is shown in Figure 12'.

The system has enough total energy to overcome the potential barrier of the TS but is unsuccessful. These results confirm that the increasing of kinetic vibrational energy hinders the prospects of a succesful collision for this reaction and that kinetic translational energy is of more importance.
These results confirm that increasing the vibrational energy present in the system does not correlate an increased probability of reaction.

{|style="margin: 0 auto;"
| [[File:Opb_fhh_traj0.png|thumb|Figure 15: Trajectory for reaction 1 using pHH = 0.]]
| [[File:Opb_fhh_traj_2.png|thumb|Figure 16: Trajectory for reaction 1 using pHH = 2.]]
|}

For reaction 2, a series of simulations were run with rHH = 2 and rHF = 1 values with both of the momenta being altered. Unlike reaction 1, an increase in vibrational energy which can come from a more negative momentum of the initial HF molecule increases the chance of a reaction. Using a fixed initial hydrogen atom momentum of -1.25, a pHF value of -10 or lower was needed for the formation of H2. This is a large number but from the calculation earlier, the activation energy for this reaction is very high. Switching the values of the momentum around (i.e. pHF = -1.25, pHH = -10) does not overcome the TS. This proves that the kinetic energy factors affecting reaction 2 are the antithesis of those affecting reaction 1. Figures 17 and 18 shows the significance of a large vibrational energy indicated by the large bond oscillations, on the the reaction outcome.

{|style="margin: 0 auto;"
| [[File:Opb_hhf_two.png|thumb|Figure 17: Trajectory for reaction 2 using pHH = -10.]]
| [[File:Opb_hhf_two-1.png|thumb|Figure 18: Trajectory for reaction 2 using pHH = -5]]
|}

To clarify the momentum between the free atom and the first atom of the diatomic does not completely correspond to translational energy as there is there is a mixture of vibrational energy in there as well. Though it is true that increasing the momentum of the diatomic will substantially increase the vibrational energy of the bond, increasing the oscillation as seen by the contour plot.

These results are coherent with the empirical rules stated by Polanyi. These rules state that the translational energy of the reactants are more effective at overcoming an early potential barrier whereas vibrational kinetic energy is best for a late potential barrier.<ref name="Polanyi">S. Yan, Y. T. Wu and K. Liu, ''PNAS'', '''2008''', 105, (35), 12667-12672.</ref> Earlier, we recognised that due to the exothermic nature of reaction 1 it has an early state, and the endothermic nature of reaction 2 leads it to having a late transition state. This justifies that reaction 1 has a higher chance for a successful collision when the majority of the kinetic energy is translational since it has an early TS. And that the late TS of reaction 2 is best overcome with vibrational energy.

{{fontcolor1|gray|(Overall, a very good report. Very well done indeed. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:00, 28 May 2018 (BST))}}

==References==
<references>
</references>

MRD:AM9315 CP3MD

2018-05-28T15:30:54Z

Fjs113: Marked by fjs113

==Exercise 1: H + H2==

===Initial coordinates===
{| class="wikitable" border=3
! r1 !! r2 !! p1 !! p2 !
|-
| 0.91 || 1.824 || 0.1 || -7.5
|}

===Initial contour plot===
[[File:AM9315_initial_Cont_plt.JPG|thumbnail|left|Initial Contour Plot (dynamics)|150px]]

===Initial skew plot===
[[File:AM9315_initial_skew_plt.PNG|thumbnail|left|Initial Skew Plot (dynamics)|150px]]

===Initial surface plot===
[[File:AM9315_initial_surface_plt.PNG|thumbnail|left|Initial surface Plot (dynamics)|150px]]

===Initial Internuclear Distance vs time plot===
[[File:AM9315_int_dist_v._time.PNG|thumbnail|left|Initial surface Plot (dynamics)|150px]]

===Initial Internuclear Momenta vs Time plot===
[[File:AM9315_int_nuc_v._time.PNG|thumbnail|left|Initial surface Plot (dynamics)|150px]]

===Initial Energy vs Time Plot===
[[File:AM9315_EN_v._time.PNG|thumbnail|left|Initial surface Plot (dynamics)|150px]]

{{fontcolor1|gray|(Is there any point to including these initial condition plots? [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:30, 28 May 2018 (BST))}}

===Dynamics from the transition state region===

====What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.====

The value for the gradient at the potential energy surface is 0 both at the minimum and at the transition state as shown by the fact that the minima (see on graph below) is represented by ∂V/∂r1 = 0, and the transition state (see on graph below) is represented by ∂V/∂r2 = 0. However, the transition state can be distinguished from the minima via carrying out a 2nd derivative. A second derivative for minima would show that ∂2V/∂r12 > 0, thus demonstrating that a minima is a minimum energy point. A second derivative for the transition state would show that ∂2V/∂r12 < 0, thus demonstrating that the transition state is a maximum energy point, which would be expected. However, this is dependent on the direction, as the transition state can occassionaly be a minima. So long as the derivative in both directions shows one maxima, then the curve is a transition state.

===Trajectories from R1 = R2, locating the transition state===

====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.====

rts = 0.905653. This is demonstrated by the 'Internuclear Distance vs Time Plot' shown below where one can observe that the A-B and B-C curves are super imposed with one another, thus the averaging all the distances, and the transition state is the final value one gets.

[[File:AM9315_Mep_cont_plot_traj.PNG|thumbnail|left| Contour plot with transition state marked (dynamics)|150px]]

[[File:AM9315_Mep_dist_v._time_plot_traj.PNG ‎|thumbnail|left| Internuclear vs distance plot for r1 = r2 and p1 = p2 = 0 (dynamics)|150px]]

{{fontcolor1|gray|(Unfortunately, this does not show the transition state. At the TS, neither r1 or r2 should be varying, and thus the internuclear distances plot should show straight horizontal lines. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:30, 28 May 2018 (BST))}}

====Comment on how the mep and the trajectory you just calculated differ.====
The mep calculation depends solely on the trajectory corresponding to infinitely slow motion or in other words, the velocity is always reset to zero. The effect of this can be shown on the MEP contour plot shown below whereby the there are no fluctuations in the reaction pathway as the velocity is always considered to be reset to 0 prior to each step. Contrast this to the the step based dynamics contour plot, and it is clear to see how not resetting the velocity results in fluctuations for each step. What is also interesting note is that MEP requires nearly 32x the number steps that a regular dynamics process would require tor a similar trajectory. The dynamics oscillations are thus shown to be significantly faster. This is because it is only taking into account the vibrational motion in an MEP as compared to the vibrational and translational state motion. {{fontcolor1|gray|(The last sentence is wrong - the MEP does not take vibrational motion into account. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:30, 28 May 2018 (BST))}}

[[File:AM9315_dyn_contour.PNG ‎|thumbnail|left| Contour plot (dynamics)|150px]]

[[File:AM9315_mep_contour.PNG ‎‎|thumbnail|left| contour plot (mep)|150px]]

====Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.====

{| class="wikitable" border=3
! p1 !! p2 !! Total energy !! R or U !
|-
| -1.25 || -2.5 || -99.018 || R
|-
| -1.5 || -2.0 || -100.456 || U
|-
| -1.5 || -2.5 || -98.546 || R
|-
| -2.5 || -5.0 || -84.390 || U
|-
| -2.5 || -5.2 || -83.786 || U
|}

{| class="wikitable" border=3
! p1 !! p2 !! Surface Plots !! Detail !
|-
| -1.25 || -2.5 || [[File:AM9315_1st_surface_plt.PNG ‎‎|thumbnail|left| contour plot (mep)|150px]] || Reactive trajectory as the reaction proceeds across the transition state.
|-
| -1.5 || -2.0 || [[File:AM9315_2nd_surface_plt.PNG ‎‎|thumbnail|left| contour plot (mep)|150px]] || Unreactive trajectory as the reaction does not proceed across the transition state
|-
| -1.5 || -2.5 || [[File:AM9315_3rd_surface_plt.PNG|thumbnail|left| contour plot (mep)|150px]] || Reactive trajectory as the reaction proceeds across the transition state.
|-
| -2.5 || -5.0 || [[File:AM9315_4th_surface_plt.PNG ‎‎|thumbnail|left| contour plot (mep)|150px]] || Unreactive trajectory as the reaction does not proceed across the transition state
|-
| -2.5 || -5.2 || [[File:AM9315_5th_surface_plt.PNG ‎‎|thumbnail|left| contour plot (mep)|150px]] || Unreactive trajectory as the reaction does not proceed across the transition state {{fontcolor1|gray|(This is reactive! You should have a closer look. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:30, 28 May 2018 (BST))}}
|}

{{fontcolor1|gray|(This is too sporadic. You need to not only describe what is happening, but also why. You need to show understanding! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:30, 28 May 2018 (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?====

Transition state theory states that a hypothetical transition state occurs between reactants and the products. This special type of transition state known as an activated transition state will have a special type of chemical equilibrium with the reactant state known as quasi equilibrium. The species found in the activated transition state is known as the activated complex. There are some basic theory points that one can make with regard to transition state theory:

1) Rates of reaction can be studied by examining activated complexes near the transition state of a potential energy surface.

2)The activated complexes are in a special equilibrium (quasi-equilibrium) with the reactant molecules.

3)The activated complexes can convert into products, and kinetic theory can be used to calculate the rate of this conversion.

When applying transition state theory to all the examples applied above, given that only two of the reaction are reactive trajectories, it is likely that the activated complexes are all in quasi equilibrium with the reactants in at least these two trajectories. This is because Hammonds postulate states that the transition state resembles reactants for exothermic processes and resembles products for endothermic processes. Thus, it is unlikely that the unreactive trajectories be exothermic more likely that they are endothermic thus the transition states cannot be in quasi equilibrium with the reactants.

{{fontcolor1|gray|(In this exercise, we are looking at an isolated triatomic system, not a statistical ensemble, so considering equilibria makes less sense in this case! One assumption of TST is that once the trajectory crosses the transition state, the reaction has been successful. However, as you can see in examples 4&5, this is not always the case. This effect is sometimes called barrier recrossing. Furthermore, you have not given any references. Did you come up with these facts entirely by yourself?? [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:30, 28 May 2018 (BST))}}

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

====Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? Locate the approximate position of the transition state. Report the activation energy for both reactions.====

{| class="wikitable" border=3
! Reaction !! Exo or Endo !! TS position (r1) (r2) !! Activation Energy !
|-
| F + H2 || Exothermic|| 1.810692, 0.745 || 0.175
|-
| H + HF || Endothermic || 1.810692, 0.745 || 30.103
|}

[[File:AM9315_TS_HF_F_moleclue.PNG ‎‎|thumbnail|left| Transition state for HH + HF (mep)|150px]]

{{fontcolor1|gray|(Where did these values come from?? You just submitted values without showing either working or understanding. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:30, 28 May 2018 (BST))}}

====Initial conditions for reactive trajectory HH + F =====
{| class="wikitable" border=3
! rFH!! rHH !! pFH !! pHH !
|-
| 1.824 || 0.745 || -2.5 || 0
|}

{{fontcolor1|gray|(Again, how did you arrive at these values? [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:30, 28 May 2018 (BST))}}

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

As the F and H2 molecules approach one another, there is an initial increase in kinetic energy as there is no potential energy barrier. As they collide dead center, the initial kinetic energy of both F and H2 is converted into potential energy as shown by the sudden spike in the potential energy as shown on the energy vs time graph.1 This is followed by a transition state between time 0.5 and 1.0, followed by continual vibrations as a result of the the HF bond vibrating thus influencing the potential curves. This can be confirmed by the Internuclear Momenta vs Time AB curve as the momentum is initially flat as the bond has not formed. The momenta drops initially as the direction is reversed and then increases significantly as the F and H2 molecules continue to colide. The momentum once again drops during the transition state before fluctuating indicating clear molecular vibration thus the formation of the HF molecule. Contrasts this to the BC curve where the momentum fluctuates throughout the transition state but after time 1.4, it proceeds to level out indicating that the energy that was released to produce the now fluctuating HF bond was used to break the HH bond thus the energy was conserved.

{{fontcolor1|gray|(This is just a bunch of observations without any explanation. You haven't shown that you understood as to why any of these phenomena occur. Neither have you answered the question of how to confirm this experimentally. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:30, 28 May 2018 (BST))}}

[[File:AM9315_int_mom_v._time_HF.PNG ‎‎|thumbnail|left| Internuclear distance vs time for release of reactoin energy|150px]]

[[File:AM9315_EN_v._time_HF.PNG ‎ ‎‎|thumbnail|left| Energy vs time graph for energy conservation|150px]]

====Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (explore values also close to these limits). What do you observe?====

The internuclear momenta vs time graph below demonstrates that when the HH momenta is close to -3, the HH momenta initially fluctuates in a symmetric manner and then proceeds into a transition state whereby the fluctuations of the HH momenta decreases whilst the HF bond increases indicating the potential formation of a HF bond. After the transition state, HF fluctuations increase whereas the HH momenta eventually becomes constant thus indicating that the HF molecule forms. Thus, it can be concluded that when the HH momenta is close to -3 there is more than enough energy to overcome the transition state and form the relevant product.

A completely different observation is made for the internuclear vs time graph with a HH momenta close to 3, with there being an exceptionally small transition state is even smaller, indicating that there is not enough energy in the system, thus the reaction does not proceed to form a HF molecule.

However, in comparison, when the HH momenta equals 0, there is clearly enough enough energy for the reaction to form a HF molecule as demonstrated by the continual fluctuating HF momenta line following the transition state between periods.

The limit for momenta which seems to result in a continual transiotn state is when the HH momenta is equal to 0.48874 as shown in the graph below.

{{fontcolor1|gray|(There are no figures here to support your observations. Also, there is no such thing as a continual transition state - it is always one single point on the potential energy sufarce. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:30, 28 May 2018 (BST))}}

====For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?====
The molecule follows a similar fluctuation to what is shown for when pFH = -3 and pHH = 0, whereby there is a small transition state caused by the collision of the HH and F particles but there is not enough energy to overcome the transition state thus the HH continues to fluctuate whereas the HF becomes a constant momenta.

{{fontcolor1|gray|(Figure! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:30, 28 May 2018 (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.====

====Initial conditions for reactive trajectory HF + H =====
{| class="wikitable" border=3
! rFH!! rHH !! pFH !! pHH !
|-
| 0.91 || 1.824 || 0.1 || -7.5
|}

The Polanyi Rules state that the vibrational energy used for a reaction is more efficient translational energy in promoting a late barrier/endothermic reaction i.e. where the transition state more closely resembles the products whereas the reverse is true for an earlier barrier/exothermic reaction i.e. where the transition state more closely resembles the reactants. This rule can be compared to our reaction because on the one hand, HH + F is exothermic and the reverse is endothermic. When viewing the exothermic process, it s clear to see that the forward reaction clearly depends more on translational energy than vibrational energy as whilst it is vibrationally active in that it crosses many vibrational energy barriers, it still appears to be mostly translational energy focused as it follows the reaction path one would expect to generate the HF molecule.

{{fontcolor1|gray|(How exactly can this be seen for the skew plot? [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:30, 28 May 2018 (BST))}}

[[File:Skew_plot_for_fwd.PNG‎‎|thumbnail|left| Skew plot for forward reaction|150px]]

However, when viewing a contour plot for the reverse reaction, the reaction path seems to match the reaction path for the reverse reaction in that whilst it is vibrationally efficient, it is more efficient in a translational sense in that it follows the translational path. This actually disagrees with Polanyis rules. A potential reason could be down to the energies. The total energy for the reverse reaction was still found to be -76.008 kcal/mol {{fontcolor1|gray|(How did you arrive at this value?? [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:30, 28 May 2018 (BST))}} which is significantly and theoretically, Polanyis rules hold unless the collisions occur at very low energies thus, this could be a potential reason as to why the revese reaction appears to be more translationally efficient than vibrationally efficient.

[[File:Skew_plot_for_reverse.PNG ‎‎|thumbnail|left| Skew plot for reversereaction|150px]]

MRD:MS6216

2018-05-28T14:06:06Z

Fjs113: Marked by fjs113

== H + H2 System ==
=== Differentiating between Saddle Points ===
In these calculations r1 = 0.74 Å, p1 = 0.0 kg.m/s, r2 = 2.30 Å and p2 = -2.7 kg.m/s
[[File:MS6216_TS_minimum.PNG|thumb|left|Figure 1 - Potential energy plot viewed from the point of view of the transition state.]]
[[File:MS6216_potential_energy_surface_showing_minimum.PNG|thumb|center|Figure 2 - Potential energy plot showing minimum.]]

From the two plots above it can be seen that both for the transition state and for the reactants before they react there is no gradient therefore the first derivative of the graph will be equal to 0 with respect to r1 and r2. Then to differentiate between the two you must consider the second derivative of local minima and local maxima of saddle points. For the transition state (a local maxima) the second derivative will yield two values one of which will be negative and one of which will be positive. Whereas for the reactants (a local minima) both of the values will be positive.

{{fontcolor1|gray|(This is a very "wishy washy" answer. The second derivative will not just yield two values. It will give one value when differentiating with respect to one coordinate and another value with respect to another coordinate. This is done by partial differentiation. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 15:06, 28 May 2018 (BST))}}

=== Transition State determination ===
[[File:MS6216_reaction_TS_radius.PNG|thumb|left|Figure 3 - A graph showing inter-nuclear transition state distances.]]

rts=0.90775 Å

This was found by equating AB and BC distances between hydrogen atoms whilst setting the momenta of both AB and BC to 0. When the inter-nuclear distances no longer oscillated this was taken to be the point of transition state as the distance between A-B and B-C did not change with time indicating equal interactions of A-B and B-C.

{{fontcolor1|gray|(Whilst the value is correct, you did not give any reason as to why the transition is found when the inter-nuclear distances no longer oscillate. You need to show understanding as well as give the correct answer! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 15:06, 28 May 2018 (BST))}}

=== Reaction Path: MEP vs Dynamics Calculation Type ===
[[File:MS6216_H2+H_MEP_reaction_path.PNG|thumb|left|Figure 4 - Reaction path calculated by MEP calculation type.]]
[[File:MS6216_H2+H_Dynamics_reaction_path.PNG|thumb|center|Figure 5 - Reaction path calculated by Dynamics calculation type.]]

An MEP calculation is a minimum energy path calculation therefore in this method the velocity is always reset to zero in each time step. Whereas in the dynamics calculation type the velocity is not always set to zero hence oscillations are observed for the dynamics calculation type whereas they are not for the MEP calculation as seen in the two diagrams above. Here r1 =0.90775 Å and r2=0.90875 Å.

=== Reactive and Unreactive Trajectories ===
{| class="wikitable" border="1"
|+ Table 1 - Reactive and Unreactive Trajectories - r1 = 0.74 Å and r2 = 2.0 Å
! p2 !! p1 !! Total Energy (kCal/mol) !! Potential Energy Surface !! Trajectory Reactivity !!
|-
| -1.25 || -2.5 || -99.018 || [[File:MS6216_-1.25_-2.5.png]] || From the potential energy surface it can be seen that the trajectory is reactive as the trajectory passes over the transition state to form the products. By closer inspection of the diagram it can also be seen that initially the H-H bond between atoms A and B do not have vibrational energy as there is not oscillation whereas following collision translational energy is converted to vibrational energy resulting in oscillation in the H-H bond between atoms B and C.
|-
| -1.5 || -2.0 || -100.456 || [[File:MS6216_-1.5_-2.0.png]] || Unlike previously, this time atoms A and B in the H-H bond do have vibrational energy however the total energy of the system this time is insufficient to overcome the transition state and so reaction trajectory is unreactive. This is seen in the picture whereby no product H-H where the atoms are B and C are formed.
|-
| -1.5 || -2.5 || -98.956 || [[File:MS6216_-1.5_-2.5.png]] || Initially it can be seen there is slight oscillation in the H-H bond between atoms A and B but it is very limited. However, this time collision is successful and the reaction trajectory is reactive resulting in the formation of a H-H bond between atoms B and C. Again as can be seen there is a significant increase in translational energy as oscillation of H-H for atoms B and C substantially increases.
|-
| -2.5 || -5.0 || -84.956 || [[File:MS6216_-2.5_-5.0.png]] || As can be seen in the graph initially the H-H bond consisting of atoms A and B has no vibrational energy however the system as a whole does have sufficient energy to overcome the transition state resulting in a reactive reaction trajectory. However, this time the formed H-H bond of atoms B and C has so much vibrational energy that the bond then breaks again and atom B reforms a bond to atom A creating a molecule with vibrational energy.
|-
| -2.5 || -5.2 || -83.416 || [[File:MS6216_-2.5_-5.2.png]] || In the potential energy surface we see that initially the H-H bond of atoms A and B has little vibrational energy if any however it does, as a system, have sufficient energy to overcome the transition state leading to a reactive reaction trajectory. However, following the transition state the H-H bond of atoms B and C has so much vibrational energy that the initial oscillation leads to the reformation of the A-B bond. This too has a lot of vibrational energy such that this bond then breaks again leading to the final formation of the H-H bond of atoms B and c which have a lot of vibrational energy.
|}

{{fontcolor1|gray|(Using the contour plots would've have made this a lot clearer. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 15:06, 28 May 2018 (BST))}}

===Transition State Theory ===

The main assumptions of the transition state theory are that the reaction rates can be determined by looking at the saddle points of a potential energy surface of a reaction (known as transition state), the activated complexes are in an equilibrium called quasi-equilibrium (different from classical) and that then these activated complexes can form products. Quasi-equilibrium is different from a classical equilibrium as it assumes that the reactants are constantly in equilibrium with the transition state and so at any one point in a mixture there will be a proportion of the transition state configuration present, however, the reactants and products themselves are not equilibrium (i.e. an irreversible reaction).

Above the first 3 systems appear to agree with the transition state theory as either the reaction has sufficient energy to go to completion or it does not and the reactants do not form products remaining as reactants. However, the final 2 systems appear to show deviations from this theory as the reaction coordinate appear to cross over the activation energy saddle point contradicting the theory that once you pass over the saddle point to form products you cannot return to reactants. System 4 disagrees as although it does pass over the saddle to form products it then returns back over the saddle to reform reactants. System 5 also initially passes over the saddle to form products before returning over the saddle to form reactants before again passing back over the saddle point to form products.

{{fontcolor1|gray|(Yes, but how does this influence predictions of the reaction rates made by TST as compared to experimental values? [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 15:06, 28 May 2018 (BST))}}

== Exercise 2: F-H-H system==
===Potential energy Inspection===
[[File:MS6216_H2+F_PES.png|thumb|left|Figure 6 - Potential energy plot for the reaction H2 + F.]] [[File:MS6216_HF+H_PES.png|thumb|centre|Figure 7 - Potential energy plot for the reaction HF + H.]]

For the reaction H2 + F it was seen that the overall energetics were exothermics with a value of -100.049 kCal/mol. {{fontcolor1|gray|(Where did this value come from?? [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 15:06, 28 May 2018 (BST))}}
This suggests that the H-F bond formed is stronger than the H-H bond broken as the overall energetics of a reaction is equal to the difference in energy of the energy released on bond formation and the energy required on bond breakage. It can be seen in the potential energy surface that energetically the reaction is going downhill in energy showing it is exotherimc.
For the reaction HF + H it can be seen that this is the inverse of the reaction above as it can be seen that in the formation of products energy is required by the reactants indicating an endothermic process.

===Transition State Location ===
[[File:MS6216_TS_location.png|thumb|left|Figure 8 - Transition State Location.]] The transition state was found, like before, by setting the HF and HH momentum to zero and altering the bond lengths of H-F and H-H slightly. This was done until the distance no longer oscillated or changed in time and yielded values of 1.811 Å for F-H distance and 0.745 Å. for H-H distance.

===Activation Energy ===

Activation energies were found by equating the 2 momenta at 0 and altering the radii of F-H and H-H atoms as done for H2+H slightly from the transition state lengths in order to push the reaction either towards HF + H or H2 + F.

{{fontcolor1|gray|(You're not altering the radii here! It is the interatomic distances that you're changing. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 15:06, 28 May 2018 (BST))}}

[[File:MS6216_H2+F_AE2.png|thumb|left|Figure 9 - Activation energy graph for H2+F.]] [[File:MS6216_H+HF_AE.png|thumb|centre|Figure 10 - Activation energy graph for H2+F.]]

The activation energy of the reaction H2 + F is very small (0.189 kCal/mol). In this calculation r(H-H)= 0.745 Å and r(F-H) = 1.812 Å.

The activation energy of the reaction HF + H is very small (28.719 kCal/mol). In this calculation r(H-H)= 0.745 Å and r(F-H) = 1.805 Å. {{fontcolor1|gray|(Are you sure you meant "very small" here? [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 15:06, 28 May 2018 (BST))}}

=== Reaction Dynamics ===
For the following graphs the F-H distance was set to 1.7 Å and the H-H distance was set to 0.8 Å.
[[File:MS6216_F+H2_momentavstime.png|thumb|left|Figure 11 - H2+F Momenta vs Time.]] [[File:MS6216_F+H2_energyvstime.png|thumb|centre|Figure 12 - H2+F Energy vs Time.]]

In the mechanism as H2 approaches the F atom initially the H with strong interactions between both the F and H is pulled strongly towards the F atom (due to electrostatics) however this results in a strong oscillation which then results in a slight interaction of the H again with the H atom it was initially bound to before being pulled back towards the F atom again.
The release of energy from the reaction then comes from the vibrationally hot state which then oscillates to release the energy. This could be determined experimentally by performing infra-red analysis of the molecule to determine the vibrational relaxation wavenumbers and therefore energies. Rotational spectra (rovibrational) would probably be observed showing rotational excitations and therefore relaxations.

=== Polanyi's empirical rules ===

The Polanyi rules are a set of rules that suggest that in a reaction that has a significant energy barrier (transition state) there will be a saddle point which must be overcome to for products. The rules stated that the type of energy which is most efficient in overcoming this barrier would ultimately depend on the position of the barrier. If the transition state was early on (resembling reactants more than products) then the most effective form of energy would be translational energy. Whereas for a late transition state (resembling products more than reactants) the most effective form of energy would be vibrational energy.<ref name="Polanyi Rules" />
The following tables demonstrate this as H2+F is exothermic therefore has an early transition state and so translational energy has the dominant effect. Whereas HF+H is endothermic and therefore has a late transition state so vibrational energy has the dominant effect. From table 2 it can be seen that as the percentage of translational:vibrational energy decreases the reaction no longer occurs emphasizing the effect of translational energy. Similarly in table 3 as the percentage of vibrational:translational energy decreases this stops the reaction emphasizing the vibrational energy dependence.

{| class="wikitable" border="1"
|+ Table 2- H2+F
! Initial Atom Conditions !! Contour Plot !! Reaction Efficieny
|-
| F-H momentum= -6.4 kg.m/s, F-H distance = 1.7 Å, H-H momentum = -0.5 kg.m/s, H-H distance = 0.8 Å || [[File:MS6216_H2+F_reactivecontour1.png]] || Reactive
|-
|-
| F-H momentum= -5.4 kg.m/s, F-H distance = 1.7 Å, H-H momentum = -0.5 kg.m/s, H-H distance = 0.8 Å || [[File:MS6216_H2+F_unreactivecontour1.png]] || Unreactive
|}

{| class="wikitable" border="1"
|+ Table 3 - HF+H
! Initial Atom Conditions !! Contour Plot !! Reaction Efficieny
|-
| F-H momentum= -6.4 kg.m/s, F-H distance = 1.7 Å, H-H momentum = -0.5 kg.m/s, H-H distance = 0.8 Å || [[File:MS6216_HF+H_reactivecontour.png]] || Reactive
|-
|-
| F-H momentum= -5.4 kg.m/s, F-H distance = 1.7 Å, H-H momentum = -0.5 kg.m/s, H-H distance = 0.8 Å || [[File:MS6216_HF+H_unreactivecontour.png]] || Unreactive
|}

==References==
<references>
<ref name="Polanyi Rules">dx.doi.org/10.1021/jz301649w | J. Phys. Chem. Lett. 2012, 3, 3416−3419</ref>
</references>

MRD:RCR16

2018-05-28T13:40:00Z

Fjs113: Marked by fjs113

== EXERCISE 1: H + H2 system ==

=== Dynamics from the transition state region ===
{{fontcolor|blue|What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.}}
The partial derivative of V with respect to r1 and r2 must be 0 at local extremum point. To determine whether this point is a minimum, maximum or a saddle point(transition state), the second partial derivatives test can be used. The second derivative test discriminant D is calculated from the second order derivatives of the potential energy surface with respect to r1, r2 and both r1 and r2. If D is higher than 0 and the second order derivative with respect to r1 is higher than 0, the point is a minimum. If D is smaller than 0, then it is a saddle point.

{{fontcolor1|gray|(Good. But what is D? Use of some equations/inequalities would've been better here. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 14:40, 28 May 2018 (BST))}}

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

{{fontcolor|blue|Report your best estimate of the transition state position ('''rts''') and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.}}
The transition state distance has been estimated to be around 0.908 Å for both A-B and B-C distances by testing of range of equal distances between the H atoms with 0 momentum and checking if the reaction trajectory still goes to the bottom of the well or stays still. This is easily seen in the internuclear distance vs time plot for the transition state, where the A-B and B-C distance lines overlap completely and show very little oscillation, as can be seen in Figure 1.

[[File:RCR16_2x.png|thumb|350x350px|center|Figure 1: Internuclear distance vs time plot for transition state]]

==== Trajectories from r1 = rts+δ, r2 = rts ====
{{fontcolor1|blue|Comment on how the ''mep'' and the trajectory you just calculated differ.}}

The mep line towards the products is quite straight, whereas the dynamic trajectory has a sinusoidal aspect. The mep line, in other words, shows the potential energy with the oscillation term completely removed {{fontcolor1|gray|(This is not the whole story. In the MEP, kinetic energy is reset to 0 at every time step. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 14:40, 28 May 2018 (BST))}} while the dynamic trajectory accounts for bond oscillation. This can be seen in Figure 2 and 3.

[[File:RCR16_3.png|thumb|300x300px|left|Figure 2: MEP contour plot]][[File:RCR16_4.png|thumb|300x300px|center|Figure 3: Dynamics contour plot]]

{{fontcolor1|blue| What would change if we used the initial conditions were reversed?}} The reaction would take place in the opposite direction.

{{fontcolor1|blue| Setup a calculation where the initial positions correspond to the final positions of the trajectory you calculated above, the same final momenta values but with their signs reversed. What do you observe?}} The reaction system reaches very close to the transition state but doesn't have enough energy to cross barrier, so it returns back to products, as seen in Figure 4.
[[File:RCR16_11.png|350x350px|thumb|centre|Figure 4: Reversal of final conditions]]

==== Reactive and unreactive trajectories ====
{{fontcolor1|blue|Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.}}

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy(kcal/mol) !! Plot !! Reactive/Unreactive
|-
| -1.25 || -2.5 || -99.018 || [[File:RCR16_5.png|thumb]] || Reactive - initial trajectory is linear so molecule doesn't oscillate before collision.
|-
| -1.5 || -2.0 || -100.456 || [[File:RCR16_6.png|thumb]] || Unreactive - system does not have sufficient energy to reach the activation barrier and oscillates back.
|-
| -1.5 || -2.5 || -98.956 || [[File:RCR16_7.png|thumb]] || Reactive - oscillating initial trajectory proves the H-H bond starts deforming before the attack of the other atom.
|-
| -2.5 || -5.0 || -84.956 || [[File:RCR16_8.png|thumb]] || Unreactive - the 3 atom complex oscillates around the transition state and crosses the barrier but ultimately turns back to reactants.
|-
| -2.5 || -5.2 || -83.416 || [[File:RCR16_9.png|thumb]] || Reactive - the small additional momentum gives the complex enough energy now to cross to products.
|}

{{fontcolor|blue|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 has 3 main assumptions stated slightly differently in different sources<ref>Atkins, P., de Paula, J., Elements of Physical Chemistry, 5th ed.; Oxford University Press, 2009.</ref>:
1. The transition state is in quasi-equilibrium with the reactants but not the products.
2. The particles involved follow Boltzmann's distribution and obey classical mechanics.
3. Once the reactants reach the minimum energy saddle point, the transition state does not collapse back to reactants.
The simulations that have been made prove that barrier recrossing is possible: the activated complex would have enough energy to turn into products, but not all collisions with good energy will result in a successful reaction since the transition state is not in equilibrium with the reactants. Tunneling in low activation barrier systems is not accounted for either in the transition state theory, but it is difficult to predict its influence on the simulations made. Considering these factors, experimental rates should be usually slower than the ones predicted by theory.
{{fontcolor1|gray|(Very good. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 14:40, 28 May 2018 (BST))}}

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

===PES inspection===
{{fontcolor1|blue|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?}}

The F + H2 reaction is exothermic while the H + HF reaction is endothermic, since the H-F bond dissociation energy(565 kJ/mol) is higher compared to the H-H one(432 kJ/mol), meaning that HF has a stronger bond and is more stable.

{{fontcolor1|gray|(You should've used the program here, not just bond dissociation energies. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 14:40, 28 May 2018 (BST))}}

{{fontcolor1|blue|Locate the approximate position of the transition state.}}

Following Hammond's postulate, the transition state will be closer in structure to the F+ H-H system rather than F-H + H system since it is higher in energy. Thus, by testing different F-H distances while the H-H distance is kept approximately at equilibrium, a stable transition state has been found at r(H-H) of 0.745 Å and r(F-H) of 1.8115 Å, and its total energy is -103.752 kcal/mol. Figure 5 indicates the stability of these coordinates.
[[File:RCR16_10.png|thumb|350x350px|center|Figure 5: Internuclear distance vs time plot for F H H transition state]]

{{fontcolor1|blue|Report the activation energy for both reactions.}}

For the F + H-H reaction, a MEP calculation has been made from r(F-H)=1.9 Å and r(H-H)=0.75 Å, and the final energy after 2500 has been determined to be -104.008 kcal/mol, which would lead to an activation energy in this direction of 0.256 kcal/mol.
In an analogue manner, a calculation was run from r(F-H)=1.8 Å and r(H-H)=0.8 Å, and the final energy after 25 seconds was -133.727 kcal/mol. The activation energy in this direction is then 29.975 kcal/mol.

{{fontcolor1|blue|In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?}}

The reaction is exothermic, so an excess of energy is produced. From the Animation result, it is quite obvious that the system contains more energy {{fontcolor1|gray|(No, the system contains the same amount of energy, since energy is conserved. Whilst I know what you mean, this can be misunderstood badly. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 14:40, 28 May 2018 (BST))}} after the H-F molecule is formed, since it oscillates much quicker than the H-H molecule. Since this is quite a simple system and energy must be conserved, it is then obvious that the excess reaction energy is present in the system as vibrational energy. The momentum vs time plot in Figure 6 shows the H-H (B-C) and H-F (A-B) being quickly deformed at the transition state, after which the oscillating H-F momentum is large and constant while the H-H momentum is lost.
This gain of vibrational energy is also evident in the energy vs time plot in Figure 7, where this excess energy is interplayed between the kinetic and the potential terms periodically. Experimentally, this could be determined using calorimetry to determined the amount of heat energy lost to the environment due to vibrations or through IR spectroscopy, following the intensity of the H-F band over time.

{{fontcolor1|gray|(Good suggestion. Would've loved to see some more details. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 14:40, 28 May 2018 (BST))}}

[[File:RCR16_12.png|thumb|300x300px|left|Figure 6: F+H-H momentum vs time plot]][[File:RCR16_13.png|thumb|300x300px|center|Figure 7:F+H-H energy vs time plot]]

=== Reaction dynamics ===

{{fontcolor|blue|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 empirical rules state that reaching a late transition state can be reached more efficiently by increasing the vibrational energy of the system rather than its translational energy, while the opposite applies for early transition state<ref>Z. Zhang, Y. Zhou, D. H. Zhang, G. Czakó and J. M. Bowman, ''J. Phys. Chem. Lett.'', 2012, '''3''', 3416–3419.</ref>.

Since the F+ H-H is exothermic, it has an early transition state. Setting r(F-H)=2 Å and r(H-H)=0.74 Å, p(F-H)=-0.5 and p(H-H)=0.1, the reaction goes smoothly to products. This is the case for high translational energy and low vibrational energy and can be seen in Figure 8. When we alter p(H-H) to 2.4 or more, the system gains so much vibrational energy that the reaction goes back to reagents even though the translational energy was enough to cross the energy barrier. Figure 9 shows this case, and it proves the Polanyi assumptions about an early transition state correct.

[[File:RCR16_15.png|thumb|300x300px|left|Figure 8: Surface plot of high translational E, low vibrational E for F + H-H]][[File:RCR16_16.png|thumb|300x300px|center|Figure 9: Surface plot for low translational E, low vibrational E for F + H-H]]

For the H-F + H reaction, the situation should be reversed, since it has a late transition state. Setting r(F-H)=0.92 Å and r(H-H)=2.0 Å, a low p(H-F)=-0.1, no matter how high the absolute value of p(H-H) is chosen, the reaction won't take place. Figure 10 shows the case for p(H-H)=-5, a very high translational energy leading to no reaction. Now setting a very low p(H-H)=-0.1 and increasing the vibrational energy by setting p(H-F) up to -10, successful reaction trajectories can be obtained, one of them being shown in Figure 11. Thus, the Polanyi predictions for a late transition state have been confirmed as well.

[[File:RCR16_18.png|thumb|300x300px|left|Figure 10: Surface plot of high translational E, low vibrational E for H + H-F]][[File:RCR16_17.png|thumb|350x350px|center|Figure 11: Surface plot for low translational E, low vibrational E for H + H-F]]

== References ==

MRD:dmd216 01198401

2018-05-28T13:12:40Z

Fjs113: Marked by fjs113

== Molecular Reaction Dynamics ==

===H + H2===

'''''What value do the different components of the gradient of the potential energy surface have at the minimum and at a transition structure? Briefly explain how minima and transition states can be distinguished using the curvature of the potential energy surface.'''''

At the minimum of a potential energy surface, the gradient will be ∂V(ri)/∂ri=0. {{fontcolor1|gray|(This is not true. ∂V(ri)/∂ri>0 [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 14:12, 28 May 2018 (BST))}} At a transition structure, the gradient of the potential energy surface will be ∂V(r1)/∂r1=0 and ∂V(r2)/∂r2=0. These two second partial derivatives depend on the reaction direction.

Minima are found when ∂2V(ri)/∂2ri=0.
The transition state is defined as the maximum on the minimum energy path linking reactants and the products. It can be found when the gradients are 0 and ∂2V(r1)/∂2r1>0 and ∂2V(r2)/∂2r2>0 i.e. a saddle point.

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

The best estimate of the transition state position was 0.908Å. This was determined by varying inter-nuclear distances using dynamics calculations. There was a period of no oscillation which is found at the transition state positon so 0.908Å was a good estimate which showed this as shown below.
[[FIle: Dmd_internuclear_distances_vs_time_estimate.png]]

When a trajectory with no initial momentum is started at the transition state where ∂V(ri)/∂ri=0 and dpi/dt=0 and hence the trajectory will remain there forever. As seen in the above graph, the inter-nuclear distance does not change for ~ 5 seconds and the trajectory is a point at the transition state but then suddenly the inter-nuclear distances begin to increase. This is due to the trajectory changing slightly either in the products or reactants direction, resulting in a molecule of hydrogen (B-C) being produced as shown by the oscillation in the above figure. The hydrogen atom A is produced which moves in the opposite direction to B-C, causing increases in A-C and A-B distances.

{{fontcolor1|gray|(Good explanation. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 14:12, 28 May 2018 (BST))}}

{| class="wikitable" border="1"
|+ Surface Plot
! Angle 1 !! Angle 2
|-
| [[File:Dmd_TS_surface_q2_angle1.png]] || [[File:Dmd216_TS_surface_q2_angle2.png]]
|}

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

The minimum energy path (mep) is a specific trajectory that corresponds to infinitely slow motion - at each time step, the velocity always resets to zero. The initial conditions for mep and dynamics were setting p1=p2=0, r1=ts+0.01 and p2=Pts. {{fontcolor1|gray|(What do you mean by this last equation? [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 14:12, 28 May 2018 (BST))}}
Mep is useful for characterising chemical reactions but does not provide a realistic account of the motion of atoms during the reaction. This is apparent in the below table. Oscillation is shown in dynamics calculations but not in mep calculations as in mep calculations, the velocity is reset to zero along the minima of the surface. In dynamics calculations, the pathway of minimum energy is followed. Oscillations are expected from the formation of a hydrogen molecule so dynamics is better for realistic results.

The below table includes surface plots but also contour plots to emphasise the oscillations shown in the dynamics calculations.

{| class="wikitable" border="1"
|+ Surface Plot
! Dynamics !! MEP
|-
| [[File:Dmd_Surface_Plot_dynamics.png]] || [[File:Dmd_Surface_Plot_MEP.png]]
|-
| Surface plot A || Surface plot B
|-
| [[File:Dmd_Contour_Plot_dynamic.png]] || [[File:Dmd_contour_Plot_MEP.png]]
|-
| Contour plot A || Contour plot B
|}

'''''Look at the “Internuclear Distances vs Time” and “Internuclear Momenta vs Time”. Take note of the final values of the positions r1(t) r2(t) and the average momenta p1(t) p2(t) at large t.'''''
<pre>
r1(t) = 0.7570
r2(t) = 10.4475
p1(t) = 1.3509
p2(t) = 1.9030
</pre>

'''''What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead?'''''

The momentum values p2(t) and p1(t) would swap values.

{| class="wikitable" border="1"
|+ Momentum graphs
! Original values !! r1 = rts and r2 = rts+0.01
|-
| [[File:Dmd_Intermomenta_dynamic_o.png]] || [[File:Dmd_Intermomenta_dynamic.png ]]
|-
| Momentum graph A || Momentum graph B
|}

'''''Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.'''''

{| class="wikitable" border="1"
|+ Reactive and Unreactive trajectories where inter-nuclear distances AB = 0.74Å and BC = 2.0Å
! p1 !! p2 !! Total energy !! Reactive/Unreactive !! Dynamics contour plot !! Comment on Trajectory
|-
| -1.25 || -2.5 || -98.4330 || Reactive || [[File: Dmd_Contour_RU_1.png]] || B-C bond formation occurs as there is enough energy to overcome the transition state. Hydrogen atom C approaches with enough momentum to produce the product B-C following collision with A-B.
|-
| -1.5 || -2.0 || -99.8835 || Unreactive || [[File:Dmd_Contour_RU_2.png ]] || There is not enough energy produced from the momentum of atom C to produce the new bond B-C from the reactant A-B. The AB bond vibrates more than the above contour plot but there is not enough momentum to overcome the transition state so the atom C moves away from A-B in the opposite direction.
|-
| -1.5 || -2.5 || -99.0225 || Reactive || [[File:Dmd_Contour_RU_3.png ]] || B-C bond formation occurs as there is enough momentum from atom C to produce a collision with the energy to overcome the transition state.
|-
| -2.5 || -5.0 || -84.6913 || Unreactive || [[File:Dmd_Contour_RU_4.png ]] || The bond formation of B-C does actually occur as there is enough momentum to overcome the transition state. However atom C approaches the A-B with high energy to produce B-C but as a result, the product B-C now has a high energy and re-collides with atom A, overcoming the transition state again to reproduce the reactants. This high energy collision produces a highly oscillating A-B bond and a high energy atom C. This is known as a re-colliding process. {{fontcolor1|gray|(Almost. It's called barrier re-crossing. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 14:12, 28 May 2018 (BST))}}
|-
| -2.5 || -5.2 || -83.8566 || Reactive || [[File:Contour_RU_5.png ]] || Overall, B-C formation is permanent. Hydrogen atom C collides with A-B with very high energy to produce B-C. This formed B-C bond has such a high energy that it breaks and the resultant atom B collides with atom A to reproduce the original reactants. These newly reformed original reactants now collide again, crossing the transition state for the third time to produce the bond B-C.
|}

''''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 (TST) is based on the assumption that atomic nuclei behave according to classic mechanics. It is assumed that reactions will not occur unless atoms or molecules collide with enough energy to form the transition structure. However, according to quantum mechanics, there is a possibility for any barrier with a finite amount of energy that quantum tunneling can occur. This is where particles can tunnel across an energy barrier without actually having the required energy to cross it; as barrier height decreases, tunneling probability increases.

TST assumes recrossing cannot occur but as shown in the calculations, re-crossings is possible. The fourth and fifth set of the above table display recrossing. Recrossing is a passage over the potential energy barrier which separates two species followed by a return to the original side.

TST fails at high temperatures. Theory assumes reaction system will pass over the lowest energy saddle point on the potential energy surface. However, at high temperatures, molecules populate higher energy vibrational modes, which leads to more collisions which can lead to transition states far away from the lowest energy saddle point. Experimental energy values at higher temperatures would be higher than the theoretical values.

{{fontcolor1|gray|(This is good. However, ignoring re-crossing will give higher predicted values, which you haven't mentioned here. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 14:12, 28 May 2018 (BST))}}

===F-H-H===

'''''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:Dmd_F_+_H2_surface_plot.png ‎]]

F + H2 --> H-F + H is an exothermic reaction as can be seen in the energy surface plot above. The products are at a lower energy than the reactants which is characteristic of an exothermic reaction. The fluorine atom collides with the hydrogen molecule to produce F-H and a hydrogen atom - the system decreases in energy to become more stable.

H-F + H --> F + H2 is an endothermic reaction as it is the reverse process of the above reaction and forming the reactants again. As seen in the above energy surface plot, the reactants are found at a higher energy than the products which is characteristic for an endothermic reaction.

{{fontcolor1|gray|(Whilst you refer back to the plot, a layperson would not be able to understand this. Which channel represents the reactants and which the products? Labelling the plot would've helped. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 14:12, 28 May 2018 (BST))}}

''''Locate the approximate position of the transition state.''''

[[File: Dmd_Transition_state_internuclear_time_HF.png]]

The F-H bond distance was estimated at 1.8099Å and the H-H bond distance was estimated to be 0.7475Å at the transition state. This was determined by varying inter-nuclear distances in dynamics calculations. As shown in the inter-nuclear distances vs time graph above, the transition state exists for ~ 5 seconds as all the inter-nuclear distances remain the same for this period of time. However, after 5 seconds the transition state begins to collapse as A-B and A-C increases as time increases.

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

Hammond's postulate is where a transition state and an unstable intermediate occur consecutively during a reaction process and have nearly the same energy; their inter-conversion only requires a small reorganization of the molecular structures. Hammond's postulate was used to predict structures either side of the transition state to estimate the activation energies of the following reactions:

F + H2 --> H-F + H
H-F + H --> F + H2

The initial and final states were found for both reactions and the energy difference was found to find the activation energy. MEP calculations were performed.

For the F + H2 --> H-F + H reaction, an activation energy of -103.75 - -103.982= 0.23 KCal/mol was obtained (A-B=1.81107, B-C=0.745, pAB=pBC=0 steps=200000).

[[File:Dmd_Energy_time_small.png]]

For the H-F + H --> H2 + F reaction, an activation energy of -103.75 - -133.96= 30.21 KCal/mol was obtained (A-B=1.80, B-C=0.745,pAB=pBC=0, steps=70000).

[[File:Dmd_Energy_time_F_H2.png]]

The F + H2 --> H-F + H reaction is more likely to occur and dominate because it has a much lower activation energy - it is also an exothermic reaction and therefore the products of the reaction are much more stable than the reactants.

'''''Identify a set of initial conditions that results in a reactive trajectory for the F + H2'''''

A-B= 2.3, B-C= 0.74, pAB= -1 pBC=-0.5
The below plots show this trajectory is reactive.

{| class="wikitable" border="1"
|+ Reactive Trajectory
|-
| [[File:Dmd_Contour_plot_reactive.png]] || [[File:DmdSurface_Plot_reactive.png]]
|-
| Contour plot || Surface plot
|}

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

F + H2 --> H-F + H
When this reaction occurs there is a large release of reaction energy and so is an exothermic reaction. The products are much lower in energy compared to the reactants and hence are more stable. However, due to the 1st Law of Thermodynamics, energy cannot be created or destroyed but can be transformed from one form to another. The loss in potential energy is transformed into an increase in kinetic energy. This can be observed in the below inter-nuclear momenta vs time graph. After the transition state collapses at time 2, there is a large increase in the amplitude of the oscillations of the A-B bond (H-F). The H-F bond has a very high vibrational energy and this causes the release of heat.
[[File: Dmd_Momentum_graph_hf.png]]

If radiation is observed from excited product molecules then this is clear indication that the products from the reaction are made in the excited state. Experimentally, infra-red spectroscopy can be used to confirm the mechanism because if the excitation energy is mainly vibrational then IR peaks will be produced. <ref name="ckd" />

{{fontcolor1|gray|(This is a good suggestion. Some more details would have been useful, e.g. mentioning vibrational modes etc. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 14:12, 28 May 2018 (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.'''''

{| class="wikitable" border="1"
|+ Reactive trajectories
! A-B= 0.88, B-C= 2.2, pHF= -0.5 pH=-9 !! A-B= 2.3, B-C= 0.74, pHF= -4.9 pH=2
|-
| [[File: Dmd_Contour_plot_actuallyreactive_HF_to_h2.png]] || [[File:Dmd_Contour_plot_h2_to_F.png ]]
|-
| H-F + H --> F + H2 || F + H2 --> H-F + H
|}

Polanyi's rules state that the vibrational energy is more efficient at promoting a late-barrier reaction than translational.<ref name="polanyi" /> It also states the opposite; translational energy is more efficient at promoting an early-barrier reaction than vibrational energy. A late-barrier reaction is one where the transition state is late and resembles the products whilst an early-barrier reaction is one where the transition state is early and resembles the reactants.

The reaction of F + H2 --> H-F + H is exothermic and therefore resembles the early transition state as per Hammond's postulate. To overcome the activation energy barrier, not much vibrational energy is required. This is shown in the above inter-nuclear momenta vs time graph where only a little momentum is required to set off the reaction.

The reaction of H-F + H --> F + H2 is endothermic and therefore resembles the late transition state, again predicted by Hammond's postulate. To overcome the activation energy barrier, a lot of vibrational energy is required.

==References==

<references>
<ref name="polanyi">Z.Zhang, Y.Zhou, D.Zhang, G.Czako, J.Bowman, J. Phys. Chem, 1989, 3, 3416-3419, DOI: 10.1021/jz301649w </ref>

<ref name="ckd">J. Steinfeld, J.Francisco and W. Hase, Chemical Kinetics and Dynamics, 2012, ch9, 291

</references>

MRD:psl1618

2018-05-17T21:30:43Z

Fjs113: Marked by fjs113

==Introduction==
The reactivity of triatomic systems can be studied using molecular reaction dynamics. For a successful reaction, the reactants must collide with sufficient energy to overcome the energy barrier, and the energy has to be in the appropriate vibrational modes. In this report, the linear collision between an atom and a diatomic molecule in the gas phase was studied with an H + H2 system and an F-H-F system.

{{fontcolor1|gray|(Very nice. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 22:30, 17 May 2018 (BST))}}

==Exercise 1: H + H2 system==
===Locating the transition state===
The transition state is found at the maximum of the lowest energy pathway linking the reactants and products and is a saddle point on the potential energy surface. If the system was placed at the TS with no momentum, it would stay at that point.

'''''1. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.'''''

The gradient of the potential energy surface is 0 at all minima and at a transition structure. To distinguish between the two, we can evaluate the second derivatives of the potential energy with respect to r, a given interatomic distance, to find the curvature. A positive curvature indicates a minimum whereas at a transition structure, the curvature is positive with respect to one interatomic distance and negative respect to the orthogonal interatomic distance, representing a saddle point in the PE surface.

{{fontcolor1|gray|(Careful - you're referring to second order ''partial'' derivatives! Also, the TS is a saddle point with respect to the reaction coordinate and its orthogonal counterpart, not with respect to the interatomic distances. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 22:30, 17 May 2018 (BST))}}

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

At the transition state, rAB = rBC since the system is symmetric. With the conditions in part one, the internuclear distance vs time plot showed that the transition state was around 0.91 Å. Through further trial and error, the AB/BC distance at the transition was determined to be '''0.90777 Å'''. The plot of internuclear distance against time showed that the rAB and rBC stayed constant until around 13 s, suggesting that this is still not the optimised TS, but given the relatively long lifetime of the transition structure, it can be said that this is a very close estimate. The constant A-B/B-C distance shows that the trajectory oscillates around that point, and the fact that A-C is double that of A-B shows that the atoms are equidistant, which is as expected for a transition state for this system.

[[File:Psl_h3_ts_distanceplot_new.png|center]]

{{fontcolor1|gray|(Very good! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 22:30, 17 May 2018 (BST))}}

===Calculating the reaction path===
====MEP vs. Dynamics calculation====
To make the system slightly displaced from the transition state, r1 was set to be 0.91777 Å (rts + 0.01) while r2 remained at the rts value of 0.90777 Å. The following contour plots were obtained using MEP and dynamics calculations respectively:

Contour plot obtained using MEP:

[[File:Psl_h3_displaced_mep_contourplot.png]]

Contour plot obtained using dynamics:

[[File:Psl_h3_displaced_dynamics_contourplot.png]]

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

The MEP plot shows the trajectory as a straight line, demonstrating that the trajectory progresses along the lowest potential energy pathway. This is because after each step, the velocity is set to 0 and it does not accumulate as the reaction occurs so the system is not able to go via a higher energy reaction coordinate. However, in the dynamics plot, the trajectory oscillates, showing that the molecule vibrates. This can happen because in the dynamics calculation, the velocity accumulates as the reaction continues, and the molecule has enough energy to deviate slightly from the lowest energy pathway.

====Analysis of the internuclear distances and momenta====
For r1 = 0.91777 Å and r2 = 0.90777 Å:
* At large t, rAB = 9.98317239895753 Å and rBC = 0.7561648286099723 Å
* At large t, the average AB momentum = 2.4808930094594905 and the average BC momentum = 1.304330465173753

If we had r1 = rts = 0.90777 Å and r2 = rts + 0.01 = 0.91777 Å instead, the same internuclear distances and average momenta as above were obtained, except the values for A-B and B-C were swapped. This is as expected as if you start at a point that was displaced towards the reactants, the reaction trajectory would go towards the reactants, and vice versa if you started at a point displaced towards the products.

If we set the starting distances as 9.98317239895753 Å and 0.7561648286099723 Å (output distances from above) and the momenta as -2.4808930094594905 and -1.304330465173753 (negative of the output momenta from above) respectively, the following plots of internuclear distance and internuclear momentum against time is obtained. Since the momenta are reversed, the system travels back towards the transition state. At roughly 2.7s, rAB = rBC and the momenta = 0, as if the reaction was going to happen. However the trajectory travels back towards where it started, showing that a reaction did not occur, perhaps because the energy was not in the correct vibrational modes.

[[File:Psl_h3_distance_reverse.png]]
[[File:Psl_h3_momentum_reverse.png]]

===Reactive and Unreactive trajectories===
[[File:Psl_h3_trajectory1a.png|thumb|Trajectory 1]]
[[File:Psl_h3_trajectory2_2.png|thumb|Trajectory 2]]
[[File:Psl_h3_trajectory3_1.png|thumb|Trajectory 3]]
[[File:Psl_h3_trajectory4_1.png|thumb|Trajectory 4]]
[[File:Psl_h3_trajectory5_1.png|thumb|Trajectory 5]]
'''''4. Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.'''''

With r1 = 0.74 Å and r2 = 2.0 Å, calculations with the following momenta values were run and the results summarised:

{| class="wikitable" border=1
! Trajectory !! p1 !! p2 !! Total Energy !! Reactive?
|-
| 1 || -1.25 || -2.5 || -99.018 || Yes
|-
| 2 || -1.5 || -2.0 || -100.456 || No
|-
| 3 || -1.5 || -2.5 || -98.956 || Yes
|-
| 4 || -2.5 || -5.0 || -84.956 || No
|-
| 5 || -2.5 || -5.2 || -83.416 || Yes
|}

As seen in the contour plots for the trajectories:
* Trajectory 1: The trajectory passes through the transition state and forms products, showing a successful reaction. Atom C approaches the vibrating molecule A-B, they collide and form molecule B-C which also vibrates, while atom A moves away.
* Trajectory 2: The trajectory approaches the transition state but does not pass through it, instead doubling back towards the reactants, showing an unsuccessful reaction. Atom C does not collide with molecule A-B with sufficient energy to overcome the energy barrier so the reactants move back towards the respective directions from which they came.
* Trajectory 3: The trajectory passes through the transition state towards the products, showing a successful reaction. Atom C collides with the vibrating molecule A-B and forms products molecule B-C (also vibrating) and atom A and they move away from each other.
* Trajectory 4: The trajectory passes through the transition state region and briefly forms a new molecule, but then loops back towards the reactants, showing an unsuccessful reaction. Atom C collides with non-vibrating molecule A-B and forms a new molecule B-C via a high energy transition state. However, molecule B-C vibrates once and is quickly dissociated again, reforming A-B which vibrates strongly and atom C.
* Trajectory 5: The trajectory passes through the transition state region and doubles back towards the reactants, before going towards the products, showing a successful reaction. Atom C collides with non-vibrating molecule A-B, passes through a transition state to reform A-B, which vibrates once and dissociates, finally forming molecule B-C (strongly vibrating) and atom A.

===The Main Assumptions of Transition State Theory===
'''''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 is based on the principles of classical statistical mechanics, where the transition state is found at the saddle point on the potential energy surface, and does not consider the microscopic picture of chemical collisions. Its main assumptions are<ref name="Chemical Kinetics and Dynamics" />:
* The electron and nuclear motions are separated
* The reactants, as well as the transition states that are becoming products, are distributed according to the Maxwell-Boltzmann distribution
* If the system has crossed the transition state towards the product side, it cannot go backwards to reform the reactants
* The system's motion in the transition state is treated classically as a translation and can be separated from the other motions
However, as shown in trajectories 4 and 5, systems can recross the transition state and potentially reform reactants, leading to a rate constant that is much lower than transition state theory predicts. This theory also neglects quantum effects such as tunnelling, which is especially significant for reactions that involve light species like Hydrogen atoms. Finally, the theory ignores the possibility of the reaction progressing through a pathway that is not the lowest energy configuration such that the effective activation energy is higher than that predicted. Nonetheless, transition state theory is a useful estimate for rate constants and equilibrium partition functions for a wide range of chemical systems.

{{fontcolor1|gray|(Perfect! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 22:30, 17 May 2018 (BST))}}

==Exercise 2: F-H-F system==
===PES inspection===
'''''6. 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?'''''

For F + H2 -> H + HF (H-H = B-C distance), the potential energy surface indicates that the reactants are at a higher energy than the products, showing that the reaction is exothermic. For F + HF+ H2 (H-F = B-C distance), the opposite is observed, showing that the reaction is endothermic. This is as expected since the first reaction involves the breaking of a weaker H-H bond (436.0 kJ/mol<ref name="bond_energies" />) to form a stronger H-F bond (568.6 kJ/mol<ref name="bond_energies" />) i.e more energy is released from the formation of the product bond than was put into breaking the reactant bond, giving an overall exothermic processes.

''PES for F + H2 (right) and H + HF (left):''

[[File:Psl_hhf_PES.png|PES for F + H2]][[File:Psl_fh2_PES.png|PES for H + HF]]

'''''7. Locate the approximate position of the transition state.'''''

Through trial and error, the transition state was determined to be at '''AB = 1.8112 Å''', '''BC = 0.7440 Å''' (AB and BC momenta set to 0). As can be seen in the graph of internuclear distance vs. time, the system remains at this position for nearly 10s before deviating, showing that it is a very close approximate to the transition state.

[[File:Psl_hhf_tsdetermination.png|center]]

{{fontcolor1|gray|(Correct, but how did you arrive at these values? Please show some working, even if it's just an educated guess! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 22:30, 17 May 2018 (BST))}}

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

Ea = ETS - Ereactant

The total energy of the approximated transition state is -103.752 kcal/mol.
By doing an MEP calculation starting at a position slightly displaced towards the H-F (AB = 1.75 Å, BC = 0.7440 Å), the following graph of energy vs. time was obtained, giving the E(H-H) to be -103.856 kcal/mol and E(H-F) = -133.601 kcal/mol.

[[File:Psl hhf displaced energy.png|center]]

* For F + H2: Ea = -103.752 - (-103.856) = '''0.104 kcal/mol'''
* For H + HF: Ea = -103.752 - (-133.601) = '''30.081 kcal/mol'''
The activiation energy for the first reaction is much lower, showing that it is a more facile reaction. This is as expected since the reaction involves breaking a weaker H-H bond to form a stronger H-F bond, but the reverse reaction requires the breaking of a stronger bond to form a weaker bond.

===Reaction Dynamics===
====H2 + F====

Through trial and error, the following initial conditions were found to lead to a successful reaction: '''rAB = 2.4 Å, rBC = 0.75 Å, pAB = -3.0, pBC = 2.99.'''

Contour plot for the reaction:
[[File:Psl_hhf_dynamics_contour.png|center]]

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

As seen in the graph of energy vs. time below, the KE increases after the system passes through the transition state at around 0.6s and the product molecule oscillates with a bigger amplitude. This is as expected since the reaction is exothermic, meaning that some of the the 'excess' energy produced is converted into the vibrational energy of the products. The PE mirrors the KE, showing that the overall energy in the system is conserved. To confirm this experimentally, we could use bomb calorimetry to measure the amount of energy released in the reaction. We can also analyse overtone bands in the IR spectrum which would be more significant for the product molecule since it has more vibrational states.

[[File:Psl_hhf_dynamics_energy.png|center]]

====Polanyi's Empirical Rules====
'''''10. 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.'''''

For a successful reaction, the system must have enough energy to cross the transition state. Polanyi's empirical rules state that a reaction with an early transition state (i.e in the reactant channel) is favoured by a high translational energy, whereas a reaction with a late transition state (i.e in the product channel) is favoured by high vibrational energy.<ref name="Polanyi paper" /><ref name="Polanyi_book" /> By Hammond's postulate, an exothermic reaction has an early transition state and an endothermic reaction has a late transition state.<ref name="Polanyi_book" />

Since the F + H2 reaction is exothermic, a high translational energy (large pHF, small pHH) should lead to a successful reaction.

The following conditions were set up: '''rHH = 0.74 Å''', '''rHF = 2.1 Å''', '''pHF = -0.5''' and '''-3 < pHH < 3''' and the results summarised in the table:
{| class="wikitable" border=1
! pHH !! Contour Plot !! Momentum !! Reactive?
|-
| -3 || [[File:Psl_forward_bc-3_contour.png]] || [[File:Psl_forward_bc-3_momentum.png]] || Yes
|-
| -2 || [[File:Psl_forward_bc-2_contour.png]] || [[File:Psl_forward_bc-2_momentum.png]] || No
|-
| -1 || [[File:Psl_forward_bc-1_contour.png]] || [[File:Psl_forward_bc-1_momentum.png]] || Yes
|-
| 1 || [[File:Psl_forward_bc1_contour.png]] || [[File:Psl_forward_bc1_momentum.png]] || No
|-
| 2 || [[File:Psl_forward_bc2_contour.png]] || [[File:Psl_forward_bc2_momentum.png]] || No
|-
| 3 || [[File:Psl_forward_bc3_contour.png]] || [[File:Psl_forward_bc3_momentum.png]] || No
|-
|}

If we increased the translational energy slightly such that '''pHF = -0.8''' and decreased the vibrational energy such that '''pHH = 0.1''', the reaction was successful as seen below. This is in accordance with Polanyi's rules.

[[File:Psl_forward_reducedmomentum_contour.png]] [[File:Psl_forward_reducedmomentum_momentum.png]]

However, as seen in the first and third case in the table, the reaction can still occur if there was a high vibrational energy, showing that although it may be more favourable to have higher translational energy for this system, it is possible to surmount the energy barrier and cross over to the product channel even if the vibrational energy was larger than the translational energy.

For the reverse reaction H + HF -> H2 + F, it is expected that higher vibrational energy would favour the reaction since it is endothermic. Indeed, in scenario 1 where there is a high vibrational energy, the reaction was successful, whereas in scenario 1 where there is a low vibrational energy, the reaction does not cross over the transition state successfully and instead reforms the reactants.
{| class="wikitable" border=1
!Scenario !! rHH !! rHF !! pHH !! pHF !! Contour Plot !! Momentum !! Reactive?
|-
| 1 || 2.4 || 0.75 || -0.5 || 2 || [[File:Psl_backward_contour.png]] || [[File:Psl_backward_momentum.png]] || Yes
|-
| 2 || 2.4 || 0.75 || -2 || 0.1 || [[File:Psl_backward_unsuccessful_contour.png]] || [[File:Psl_backward_unsuccessful_momentum.png]] || No
|-
|}

{{fontcolor1|gray|(Very good! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 22:30, 17 May 2018 (BST))}}

==References==
<references>
<ref name="Polanyi_book">Steinfeld, J.; Francisco, J.S.; Hase, W.L. ''Chemical Kinetics and Dynamics''; New Jersey, 1989; pp 297-299.</ref>
<ref name="Chemical Kinetics and Dynamics">Steinfeld, J.; Francisco, J.S.; Hase, W.L. ''Chemical Kinetics and Dynamics''; New Jersey, 1989; pp 308-337.</ref>
<ref name="bond_energies">T. L. Cottrell. ''The Strengths of Chemical Bonds''; 2d ed.; Butterworth, London, 1958</ref>
<ref name="Polanyi paper">Zhang, Z.; Zhou, Y.; Zhang, D.; Czakó, G.; Bowman, J. Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction. ''J. Phys. Chem. Lett.'' '''2012''', 3 (23), 3416-3419
</references>

MRD:OliviaPini

2018-05-17T21:15:40Z

Fjs113: Marked by fjs113

===Exercise 1===
====Question 1 = What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.====

At a minimum the gradient of the potential energy surface is zero. At a transition state, as it is a maximum on the minimum energy path, it also has a differential of 0. However, the second differential of the transition state (maximum) is negative, whereas that of the minimum would be positive and so the curvature of the potential energy surface can be used to distinguish them. This potential energy surface below clearly shows the maximum of the minimum and so the position of the transition state.

[[File:PES.max.min.oliviapinijpg|400px]]

''Fig. 1 - Potential energy surface of H + H2 with the transition state illustrated.''

{{fontcolor1|gray|(The correct mathematical term is derivative, not differential. The TS is not a maximum! It's a saddle point, which means that it is a maximum in one direction and a minimum in another direction orthogonal to the first. Brownie points for the rather obscure drawing. In that perspective, the TS is actually covered up by the reaction path. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 22:15, 17 May 2018 (BST))}}

====Question 2 = Report your best estimate of the transition state position (rts) and explain your reasoning illustrating it with a “Internuclear Distances vs Time” plot for a relevant trajectory.====

The best estimate for the transition state position to 2 d.p. is 0.91 Å for both AB and BC distance. This is because at 0.91 Å on the contour plot featured below the trajectory can no longer be seen as just a cross is present.This is expected at the transition state as this is a stationary point with gradient 0 and so if the reaction is started here with momentum 0 it will stay here forever. However, if an infinitesimally small change is made to the geometry then the reaction will fall into the products or reactants.

[[File:contourploth2.oliviapinijpg|400px]]

''Fig.2 - Contour plot showing that at 0.91 Å the transition state is present as the trajectory is no longer seen.''

This is supported by the inter-nuclear distance vs. time plot, which shows that AB and BC distances are the same, which is known to be a condition that must be fulfilled for a transition state to be present.

[[File:internucdistanceoliviapini.jpg|400px]]

''Fig. 3 - Internuclear distance vs time plot showing AB and BC distances to be the same, showing the transition state must be present.''

{{fontcolor1|gray|(Ideally, at the TS this plot should not display any oscillations whatsoever. This can only be achieved by going to 3 decimal places. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 22:15, 17 May 2018 (BST))}}

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

Both show that if an infinitesimally small change is made to the geometry in the direction of the products, then the trajectory will fall into the product section of the potential energy surface. The main differences between these two graphs are due to velocity. In the MEP graph the velocity at each time step is zero and so the line is shown to stop as soon as it reaches an energy minimum after the transition state. Whereas in the dynamic graph where velocity varies, the trajectory is shown to continue on to infinity.

{{fontcolor1|gray|(This is unfortunately not correct. How far the trajectory (MEP or Dynamics) goes is solely depending on the number of steps calculated by the simulation. Zooming out on the dynamics trajectory would show you that it also ends somewhere. Likewise, increasing the steps would cause the MEP to continue further. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 22:15, 17 May 2018 (BST))}}

[[File:Surface_Plotmepoliviapini.jpg|600px]]

''Fig. 4 - MEP potential energy surface, illustrating the trajectory to stop at an energy minimum''

[[File:Surface_Plotdynamicoliviapini.jpg|600px]]

''Fig. 5 - Dynamic potential energy surface, illustrating the trajectory to carry on to infinity.''

====Question 4 = Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.====

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy !! Reactive/Unreactive !! Contour Plot !! Reasoning
|-
| -1.25 kgm/s || -2.5 kgm/s || -99.01 Kcal/mol || Reactive || [[File:contorplot1reactive.oliviapini.jpg|400px]] || From this contour plot it is known that the reaction is reactive as the blue trajectory line continues onto the product section of the plot after the transition state.
|-
| -1.5 kgm/s || -2.0 kgm/s || -100.46 Kcal/mol || Unreactive || [[File:contorplot2reactive.oliviapini.jpg|400px]] || From this contour plot it is known that the reaction is unreactive as the blue trajectory line bounces off from the transition state and goes back into the reactants.
|-
| -1.5 kgm/s || -2.5 kgm/s || -98.96 Kcal/mol|| Reactive || [[File:contorplot3reactive.oliviapini.jpg|400px]] || From this contour plot it is known that the reaction is reactive as the blue trajectory line continues onto the product section of the plot after the transition state.
|-
| -2.5 kgm/s || -5.0 kgm/s || -84.96 Kcal/mol || Unreactive || [[File:contorplot4reactive.oliviapini.jpg|400px]] || From this contour plot it is known that the reaction is unreactive as the blue trajectory line bounces off from the transition state and goes back into the reactants. {{fontcolor1|gray|(This is not the whole story though... For a short period, the products are formed but then barrier is recrossed. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 22:15, 17 May 2018 (BST))}}
|-
| -2.5 kgm/s || -5.2 kgm/s || -83.42 Kcal/mol || Reactive || [[File:contorplot5reactive.oliviapini.jpg|400px]] || From this contour plot it is known that the reaction is reactive as the blue trajectory line continues onto the product section of the plot after 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? ====

The assumptions of the Transition State Theory are:

1) Molecular systems that have gone over the transition state towards the products cannot go back to reform the reactants.

2) In the transition state, motion along the reaction coordinate can be separated from all other motion and treated classically as translation.

3) In the absence of equilibrium between reactants and products, the transition states becoming products are distributed among their states according to the Maxwell-Boltzmann laws.

4) The electronic and nuclear motions are separated (Born-Oppenheimer).

5) Reactant molecules are distributed among their states in accordance with Maxwell-Boltzmann distribution.

The first assumption does not match the results obtained here as in the unreactive reactions featured above the reactants reach the transition state but then are unable to turn into products and so go back to being reactants. Also the second approximation does not fit the data and graphs here, as vibrations also cause oscillations which affect the reaction trajectory and these cannot be treated quantum mechanically as they appear as quantized levels. Additionally, even if not shown here, Transition State Theory doesn't consider the fact that there is the possibility for particles to quantum tunnel through any barrier of finite energy, meaning some molecules will react even if they don't collide with enough energy to over come the transition state and activation energy. This all means that Transition State Theory will underestimate reaction rate values. (1)

{{fontcolor1|gray|(Not quite - tunnelling is noticable at very specific conditions. TST will usually overestimate the reaction rate as it does not account for barrier recrossing, see assumption 1). [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 22:15, 17 May 2018 (BST))}}

===Exercise 2===

====Question 6 = 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 is an exothermic reaction, as the products are lower in energy then the reactants. The reaction also takes place and this all means that more energy is given out as the H-F bond forms then is needed to break the H-H bond.

[[File:Surface_PlotH2+F.oliviapini.jpg|800px]]

''Fig. 6 - Potential energy surface plot for H2 + F.''

Whereas, H + HF is an endothermic reaction, as the product energy is higher then the reactant energy. Also, the reaction doesn't occur, meaning under these conditions the energy given out as the H2 bond forms isn't enough to compensate for the energy needed to break the HF bond.

{{fontcolor1|gray|(Would've been good if you could relate your answer above to the picture. Which is the product channel and which the reactant channel? Talk about the drop in potential energy. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 22:15, 17 May 2018 (BST))}}

[[File:Surface_PlotH+HF.oliviapini.jpg|800px]]

''Fig. 7 - Potential energy surface plot for HF + F''

Overall this shows that the HF bond is stronger then then the H2 as it takes more energy to break and releases more energy as it forms.

====Question 7 = Locate the approximate position of the transition state.====

The approximate position of the transition state is at HF distance = 1.810 Å and HH distance = 0.746 Å, as at these bond lengths the trajectory has again disappeared. As stated above, this is expected for the transition state because it is a stationary point with gradient 0 that will remain forever unless an infinitesimally small change is made to the geometry to favor the reactants or products.

[[File:contourplothftsoliviapini.jpg|450px]]

''Fig. 8 - Potential energy surface plot illustrating a transition state as reaction trajectory is not visible.''

{{fontcolor1|gray|(Yes, but simply stating bare values does not show understanding. You need to show your working, even if it was just an educated guess! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 22:15, 17 May 2018 (BST))}}

====Question 8 = Report the activation energy for both reactions.====

For the reaction of HF + H the energy of the transition state is -104.664 Kcal/mol and for the reactants is -132.841 kcal/mol. This was worked out from the graph below where the initial potential energy value is the energy of the transition state, whereas the final potential energy value on the plateu relates to the energy of the reactants. The difference between these values then gives an activation energy of 28.177 Kcal/mol.

[[File:SurfacePlotactivationenergyreverseoliviapini.jpg|400px]]

''Fig. 9 - Energy vs time plot for HF + H to show difference between TS and reactant energy.''

For the reaction of F + H2 the energy of the transition state is -103.001 Kcal/mol and that for the reactants is -103.744, obtained using the graph below in the same way as above. This gives a difference and so activation energy 0.773 Kcal/mol. This also matches the literature value of 0.74 Kcal/mol for the barrier height of this reaction.(2)

[[File:SurfacePlotactivationenergyforwardoliviapini.jpg|450px]]

''Fig. 10 - Energy vs time plot for F + H2 to show difference between TS and reactant energy.''

====Question 9 = In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?====

When the F atom approaches the H2 it has potential not kinetic energy as it is only a single atom and so cannot vibrate. However, when it bonds to the hydrogen atom there is now also kinetic energy in the form of vibrations and as energy is conserved it means some of the potential energy must become kinetic energy. This is seen in the inter-nuclear momentum vs time graph where the blue line representing the F atom is originally straight, illustrating purely potential energy but once it bonds to hydrogen it begins to show oscialltions, illustrating the conversion to kinetic energy as the bond vibrates. The opposite effect is seen for the BC bond (orange line) but on a much smaller scale, which starts with oscillations meaning it has some kinetic energy due to bond vibrations and ends having only potential energy, as the line platues off. This is because at first you have the H 2 bond which is capable of vibrations but it then breaks and becomes a single H atom which can't vibrate and so energy is converted into purely a potential energy form. {{fontcolor1|gray|(It's not just potential energy. Non-zero momentum still means it has kinetic energy. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 22:15, 17 May 2018 (BST))}} This can be confirmed experimentally carrying out reactions like calorimetry, as an increase in vibrations and so kinetic energy will lead to more heat being released.

[[File:question9oliviapini.jpg|500px]]

''Fig. 11 - Internuclear momentum vs time plot for F + H2, illustrating conversion of translational to vibrational energy.''

This idea is supported by the contour plot shown below, as it starts with very little oscillation, illustrating that the total energy of the system is mostly potential energy as the single F atom approaches the slightly vibrating H2 molecule. However, then the oscillations become much larger as the reaction progresses, showing most energy has been converted into kinetic energy due to the much larger vibrations of the HF bond.

[[File:question9distanceoliviapini.jpg|500px]]

''Fig. 12 - Contour plot for F + H2, illustrating conversion of translational to vibrational energy.''

====Question 10 = 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. ====

The Polanyi rules state that vibrational energy is the more efficient kinetic energy mode in promoting a late-transition state than translational energy. (3)

For the H2 + F reaction the transtion state is very early and so will have a structure similar to that of the reactants. This means it should not be promoted by an increase in vibrational energy of the H2 bond. Whereas the reverse reaction (HF + F), occurs via a late transition state which will have a structure resembling that of that products. In this case an increase in the vibrational energy should in fact promote the reaction.

This is supported by the fact that if both the forward and reverse reactions are given the same very small translational energy (AB momentum = -0.5 Kgm/s) it takes the forward reaction with an early transition state a vibrational (BC) momentum of -5.5 Kgm/s, whereas it takes the reverse reaction, with a late transition state only -2.9 kgm/s. This demonstrates that the late transition state is promoted more by vibrational energy then the early transition state.

[[File:conoutplot5.5momentumoliviapini.jpg|400px]]

''Fig. 13 - Contour plot for F + H2, illustrating reaction occurs at -5.5 Kgm/s.''

[[File:conoutplot2.9momentumoliviapini.jpg|400px]]

''Fig. 14 - Contour plot for HF + H, illustrating reaction occurs at -2.9 Kgm/s.''

It can also be proven that translational energy is more efficient in promoting an early transition state then a late transition state. When both the reactions are given a fixed vibrational (BC) momentum of -1.0 kgm/s, it takes the forward reaction a translational momentum (AB) of -7.7 kgm/s to occur. Whereas, the reverse reaction, with a late transition state still did not appear to go with a translational momentum up to -10kgm/s. This shows translational energy does promote an early transition state more then a late one.

{{fontcolor1|gray|(These are astronomical values to be inputting! At such high momenta and velocities, the reaction depends on different things. These effects can usually be seen best using values of up to 2 for the momenta. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 22:15, 17 May 2018 (BST))}}

[[File:conoutplot7.7momentumoliviapini.jpg|400px]]

''Fig. 15 - Contour plot for F + H2, illustrating reaction occurs at -7.7 Kgm/s.''

[[File:conoutplot10momentumoliviapini.jpg|400px]]

''Fig. 16 - Contour plot for HF + H, illustrating reaction does not occur up to -10 Kgm/s.''

===References===

1) J. I. Steinfeld, J.S. Francisco, William I. Hase, ''Chemical Kinetics and Dynamics'', Pearson, England, 1998.

2) D. G. Truhlar, B.C. Garret, N. C. Blais, ''The Journal of Chemical Physics'', 1984, 80, 232.

3) Z. Zhang, Y. Zhou, D. H. Zhang, ''J. Phys. Chem. Lett.'', 2012, 23, 3416-3419.

Hhc16Y2

2018-05-17T20:52:31Z

Fjs113: Marked by fjs113

=Molecular Reaction Dynamics=
==Introduction==
Triatomic reactions have the general form:

<math>\textrm{A + BC} \rightarrow \textrm{AB + C}</math>

Molecular dynamics in triatomic reactions can be approximated in terms of a potential surface <math> V(x,y) </math> generated (partly) by superimposing two orthogonal potential energy curves ('Leonard-Jones'-like potentials) complemented with quantum mechanical calculations, with one axis representing the incoming diatomic bond and the other representing the resulting diatomic bond. The potential energy surface contains three stationary points;

*the reactants at the beginning of the reaction;
*the products after the reaction and;
*the transition state.

The two former structures exist as local minima, where the the potential gradients along the initial bond and formed bond are both zero. The latter exist as a saddle point, where the gradients of the two components are equal and opposite thus cancelling out the overall gradient to zero. Once the potentials are known and stationary points found (by computing <math> \nabla V(x,y) = 0 </math>), the two types of structures can be distinguished using the second partial differential test:

*If structure is a minima, then the Hessian of the potential function:<math>V_{xx}(x,y)V_{yy}(x,y) - \left( V_{xy}(x,y) \right)^2 > 0</math> and <math>V_{xx}(x,y) > 0</math>
*If structure is a saddle, then the Hessian of the potential function: <math>V_{xx}(x,y)V_{yy}(x,y) - \left( V_{xy}(x,y) \right)^2 < 0</math>

Where the Hessian is the determinant of the Hessian matrix applied to the potential function[https://en.wikipedia.org/wiki/Second_partial_derivative_test]. The reaction will proceed as described if the species collide with an energy greater than the local potential barrier at the transition saddle point.

{{fontcolor1|gray|(Very nice to see a concise introduction! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:52, 17 May 2018 (BST))}}

==H-H-H System==
Consider a three hydrogen triatomic reaction:

<math>\mathrm{H_2 + H}\rightarrow \mathrm{H + H_2}</math>

As the three atoms are identical, the potential surface is symmetrical. Hence the transition state is expected to occur when the three atoms are equidistant to each other. Using the simulator ''lepsgui.py'', the transition state was found by setting the particle momenta to 0 and locating a bond distance on the saddle structure where the particle remains stationary on the potential surface. It is as though the molecule has potential to collapse into either reactant or product, yet does not have the inertia to do so. This distance was found to be 0.9075 A. The distance between the two end nuclei was found to be 1.815 A, consistent with double of the previous observation. Neither kinetic nor potential energy showed any change over time, consistent with theoretical behaviour at a stationary transition state.

<gallery class="center" heights="300px" widths="300px">
File:1.jpeg|<center>Figure 1: Transition state on the saddle point of the hydrogen triatomic reaction potential.</center>
File:2.jpeg|<center>Figure 2: Internuclear distances at the transition state.</center>
File:3.jpeg|<center>Figure 3: Energy of the system.</center>
</gallery>

===Reaction Paths===
From the transition state it was also possible to find the minimum energy path (MEP) for the transition state to collapse into the product. This was achieved by starting the simulation with the particle displaced at a small distance (0.01 A) from the transition state and having no momentum. The particle effectively 'rolled off' the saddle point and travelled along the bottom of the potential valley, mapping out the path of minimum energy. It is useful for extracting information about the reaction: if the one dimensional path was mapped relative to only the height, a reaction profile plot can be obtained. Equilibrium geometries can also be found this way.

<gallery class="center" heights="300px" widths="300px">
File:4MEP.jpeg|<center>Figure 4a: Minimum energy path.</center>
File:4distance.jpeg|<center>Figure 4b: Internuclear distance when reaction proceeds via minimum energy path.</center>
</gallery>

The MEP assumes an infinitely slow reaction where the velocity self adjusts. A more realistic representation would be one where the bond in the product oscillates. The path generated still follows that of the potential valley, but oscillates within the well. In this model the reaction collapses into the product much quicker as evident in the internuclear distance plot, where the dynamic system diverges further in shorter time.

<gallery class="center" heights="300px" widths="300px">
File:5Dynamic.jpeg|<center>Figure 5a: Dynamic reaction path. Notice the oscillation in the bond.</center>
File:5distance.jpeg|<center>Figure 5b: Internuclear distance when reaction proceeds via the dynamic energy path. Components move away much faster than previous example.</center>
File:5energy.jpeg|<center>Figure 5c: Internuclear momenta shows that the product is oscillating.</center>
</gallery>

As the potential surface is symmetrical, regardless of which bond has an added small displacement the result will be the same but the reaction will proceed backwards.

<gallery class="center" heights="300px" widths="300px">
File:6.jpeg|<center>Figure 6a: A reaction trajectory going the other way.</center>
File:6.1.jpeg|<center>Figure 6b: The internuclear distance is the same but with labels swapped.</center>
</gallery>

The final momenta and positions from the dynamic simulation were fed back into the simulator with reversed momentum. The resulting reaction path is the same but in the opposite direction, where the products return to the transition state and stays there.

<gallery class="center" heights="300px" widths="300px">
File:7.jpeg|<center>Figure 7: Reverse reaction trajectory with system returning to transition state.</center>
</gallery>

{{fontcolor1|gray|(Very good! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:52, 17 May 2018 (BST))}}

===Path Energy Considerations and Reactivity===
Not all paths will lead to the crossing of the energy barrier via the transition state and result in a reaction. By fixing reactant H2 bond at 0.72 A and distance between H and H2 at 2.0 A, a range of momenta were tested to investigate whether or not they are reactive.
{| class="wikitable" border=1
! p1 !! p2 !! ETotal !! Reactive !! Path !! Comments
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:8.1.jpeg|thumb|250px|centre|]]
|| Trajectory passes over transition state.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:8.2.jpeg|thumb|250px|centre|]] || Trajectory bounces off the energy barrier and returns to the well.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:8.3.jpeg|thumb|250px|centre|]] || Trajectory passes over transition state with larger initial oscillation than case one.
|-
| -2.5 || -5.0 || -83.956 || No || [[File:8.4.jpeg|thumb|250px|centre|]] || Although the trajectory passes over the energy barrier, the high momenta hence energy made the particle pass it again and return to the reactant state.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:8.5.jpeg|thumb|250px|centre|]] || Trajectory passes the barrier multiple times before collapsing into product.
|}

===Transition State Theory===
The Transition State Theory[https://en.wikipedia.org/wiki/Transition_state_theory] assumes that:
#Particles behave classically.
#Reactions proceed via the lowest energy saddle point
#Once a system reaches the transition state, it collapses to products and reaction goes to completion.
In essence, all molecules must have enough energy to overcome the saddle point in order to generate product, hence the rate of product generation must be determined in some way by how high the energy barrier is.

This however breaks down when reactions occur via a much higher energy barrier instead of the transition state, or proceeds in some obscure way like crossing back to the reactant state (as the last case above). The rate of reaction will therefore be lower than predicted by the theory. It also fails to account for quantum mechanical effects such as tunneling, which will allow the particle to bypass the energy barrier without having the energy required to overcome it. Given tunneling only occurs with small masses and low energy barriers, the transition state theory will in most cases overestimate the rate of reaction.

{{fontcolor1|gray|(Barrier recrossing is definitely not "some obscure way". ;) Otherwise very good! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:52, 17 May 2018 (BST))}}

==F-H-H System==
Consider a triatomic reaction between a fluorine atom and hydrogen:

<math>\mathrm{F + H_2} \rightarrow \mathrm{HF + H}</math>

The forward reaction (generation of HF) is exothermic, and the backward reaction (generation of hydrogen from hydrogen fluoride) is endothermic. The thermodynamics is evidenced by the potential surface for this reaction: the potential well for the H-F bond is deeper than that for H2, in agreement with the former having a stronger bond than the latter. Using the same method for the triatomic hydrogen, the transition state bond distance was found to be 1.8112 A for H-F and 0.7452 A for H-H.

{{fontcolor1|gray|(Simply stating the TS initial conditions is not quite enough. You need to show your working, even if it's just an educated guess! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:52, 17 May 2018 (BST))}}

<gallery class="center" heights="300px" widths="300px">
File:9.jpeg|<center>Figure 8a: Potential surface for this system. Notice the deeper valley for the hydrogen fluoride product.</center>
File:9.1.jpeg|<center>Figure 8b: Transition state for this system.</center>
</gallery>

===Activation Energy===
The activation energy was determined by subtracting the reactant energies with the transition state energy. For the forward hydrogen fluoride generation reaction the activation was 0.23 kcal/mol, and the backward hydrogen generation reaction was 30.25 kcal/mol. There is some discrepancy between that reported in literature[http://science.sciencemag.org/content/311/5766/1440.full], possibly due to quantum effects lowering the energy requirement.

{{fontcolor1|gray|(Where did these values come from? What did you use to determing the different energy levels? Again, you need to show your working. Bare values don't show understanding. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:52, 17 May 2018 (BST))}}

===Reaction Dynamics===
====Forward Reaction====
The forward reaction involves crossing a low transition barrier relative to the reactants (an early transition state). One of the various reaction trajectories leading to a successful forward reaction is H-H = 0.74 A, F - H2 = 2.3 A, fluorine hydrogen momentum -1.5, hydrogen bond momentum -0.1. As illustrated, the reaction almost returns to the reactants but crosses the potential barrier back to the products. Given the total energy of the system must stay the same by virtue of energy conservation, when the reactants collapse to the product, the drop in potential must be converted into both vibrational kinetic energy observed in the higher frequency oscillation of the product bond, and translational kinetic energy as the product moves away, as seen below. If the resulting oscillation is too rigorous or movement too rapid, it can cross the potential barrier again and return to the reactant. The nature of the kinetic energy evolution being a reflection of the potential energy evolution evidenced the interchange hence conservation of energy. Eventually the reaction proceeded into higher energy oscillations in the product. This can be verified experimentally using calorimetry - the higher kinetic energy will manifest itself as heat in the system, thus should be detectable as a temperature increase and in line with the exothermic nature of the forward reaction.

<gallery class="center" heights="300px" widths="300px">
File:New8.1.jpeg|<center>Figure 9a: Forward reaction trajectory.</center>
File:New8.2.jpeg|<center>Figure 9b: Energy evolution of the forward reaction. As the system collapses into the product notice the increase in kinetic energy and decrease in potential energy. Fluctuations are due to bond vibrations, with oscillations interchanging with potential stored in bond stretching/compression.</center>
File:New8.3.jpeg|<center>Figure 9c: Internuclear momenta evolution. The drop in potential after crossing the energy barrier makes the oscillation of the H-F bond large.</center>
</gallery>

For hydrogen bond momenta around 3 or -3, with hydrogen fluorine momentum -0.5, H-H = 0.74 A and F - H2 = 2.3 A, the system pertained high vibrational but low translational energy. Despite total energy exceeding that of the activation energy, the reaction crossed the potential barrier but returned to products. It would appear that high vibrational energy does not favour the forward reaction as much as high translational energy.

<gallery class="center" heights="300px" widths="300px">
File:new8.4.jpeg|<center>Figure 10: Reaction does not proceed with p(F - H2) = -0.5, p(H-H) = 3. </center>
</gallery>

However when the bond momentum is reduced, there are certain values that allowed the reaction to still proceed. For instance at bond momentum 2.29, the reaction was able to go to completion. This is in agreement with Polyani's Rules, which states that crossing of a late transition state is favoured by high vibrational energy modes than translational modes, whereas higher translational modes favour the crossing of early transition states. Below 1.48 reaction does not have enough energy to reach the transition state.

<gallery class="center" heights="300px" widths="300px">
File:9.9.jpeg|<center>Figure 11: Reaction proceeds for p(F - H2) = -0.5 and p(H-H) = 3. </center>
File:9.8.jpeg|<center>Figure 12: At or below 1.48 reaction does not pass transition state.</center>
</gallery>

When the positions were kept the same but fluorine hydrogen momentum at -0.8 and hydrogen bond momentum at 0.1 the system goes through again. This is consistent with recent theoretical revisiting of the Rules which said that the Rules apply except when collision energies are low[https://pubs.acs.org/doi/abs/10.1021/jz301649w].

<gallery class="center" heights="300px" widths="300px">
File:9.5.jpeg|<center>Figure 13: Low collision energy allows reaction to proceed as well.</center>
</gallery>

====Backward Reaction====
The backward reaction involves crossing a high transition barrier relative to the reactants (a late transition state). One of the various reaction trajectories leading to a successful backward reaction is FH - H = 2.0 A, F-H = 0.925 A, hydrogen fluoride bond momentum -12.0, hydrogen fluorine momentum -4.0, obtained by setting low H-F bond momentum and high incident hydrogen momentum then tuning the two until a reaction occurs.

<gallery class="center" heights="300px" widths="300px">
File:10.jpeg|<center>Figure 14a: Successful reaction.</center>
File:10.1.jpeg|<center>Figure 14b: Kinetic energy drops and potential energy increases.</center>
</gallery>

Note that the high vibrational energy of the H-F bond aids the crossing of the transition state more so than high translational energy. This again appears consistent with Polanyi's Rules.

<gallery class="center" heights="300px" widths="300px">
File:10.2.jpeg|<center>Figure 15: Reaction does not proceed at p(H-F) = -0.05, p(H - HF) = -4.</center>
</gallery>

{{fontcolor1|gray|(It seems that you never actually stated Polanyi's rules. But good that you provided results supporting them. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:52, 17 May 2018 (BST))}}

{{fontcolor1|gray|(Would've been better if you had provided a list of references at the end. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:52, 17 May 2018 (BST))}}

MRD:kkz16

2018-05-17T20:32:30Z

Fjs113:

=Molecular Reaction Dynamics=
=EXERCISE 1: H + H2 system=
==Dynamics of the Transition State Region==
[[File:Transition State Maximum.png|200x200px|thumb|Clearly visible maximum transition state and minima reactants and products]]
Q. What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

Along the minimum energy pathway there are 2 minima points of in energy (at the reactants and at the products). These correspond to the points at which 𝛿V/𝛿r1=0 and 𝛿V/𝛿r2=0, as these are minima, the second derivatives (𝛿2V/𝛿r12 and 𝛿2V/𝛿2r2) would be greater than 0.
Along the minimum energy pathway there is a maximum produced at the transition state, this would be a saddle point overall. At this point both 𝛿V/𝛿r1 and 𝛿V/𝛿r2 are equal to 0, and the second derivative test must be used in order to confirm the saddle point. This is done using the equation H=fr1r1(r1,r2)fr2r2(r1,r2)−fr1r2(r1,r2)2, where fr1r1(r1,r2) represents the second partial derivative with respect to r1r1 etc. When H<0 then the stationary point is a saddle point, and therefore, in this case, the transition state.

{{fontcolor1|gray|(This is almost perfect. You need to be careful about the coordinate system that you're using for the partial derivatives. At the TS, the partial derivatives should be with respect to the reaction coordinate and its orthogonal counterpart. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:31, 17 May 2018 (BST))}}

The transition state could therefore be found by providing 0 momentum and finding the point at which no change in energy/movement occurs, if the transition state is not immediately found, the trajectories of the atoms/molecules would 'roll' either towards the reactants or the products (down the energy gradients; towards the minima) depending on which side of the transition state they were on.

==Trajectories from r1 = r2: Locating the Transition State==
Q. 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.

r1=r2=1.0 was used as a starting point for the trajectory, and was found to be too high, as the internuclear distance against time graph produced a strong curve when using the calculation method of MEP. The position of the plateau of the A-B and B-C bonds were noted and trialed. The process was repeated with increasing accuracy until the forces were observed to be 0.000 to the nearest 3dp. This point was found to be r1=r2=0.9077425, and continued to provide forces of 0.000 to 3dp when the number of steps was increased to 1000 (from 500) showing good accuracy. This produced a plot of what appeared to be 2 parallel lines, however the A-B and B-C lines are overlapped (blue under orange), so it is in fact 3 lines. The distances did not change over time, showing no change in trajectory, and therefore that this is a fairly good estimate of the transition state position. When the graph at this position was changed back to the dynamic calculation type, no change was observed (remained as 2 parallel likes) showing that there are no oscillations at the transition state.

{|style="margin: 0 auto;"
| [[File:Locating the Transition State - Initial.png|thumb|upright|alt=Intial|Internuclear distance against time plot for r1=r2=1.0 (using MEP)]]
| [[File:Locating the Transition State.png|thumb|upright|alt=Estimate|Graph of the internuclear distances at the estimated transition state position: r1=r2=0.9077425 (using dynamic calculation method)]]
|}

{{fontcolor1|gray|(Impressive. Loved that you showed your working and how you determined that it was indeed the TS! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:31, 17 May 2018 (BST))}}

==Trajectories from r1 = rts+δ, r2 = rts==
Q. Comment on how the mep and the trajectory you just calculated differ.
The MEP sets the velocity to 0 for each time interval, resulting in the equivalent of multiple snapshots/pictures being pieced together. This means that oscillations are not observed on the MEP plot, however they are on the dynamic plot. The dynamic plot does not reset the velocity to zero in each time interval, and so oscillations are observed due to the acceleration and deceleration due to the spring like forces in the bond.
While the dynamic plot looks almost linear overall (ignoring/averaging the oscillations), the MEP plot produces curves. This seems to suggest that there is a deceleration over time, indicating a loss of energy. The distances on the MEP plot are also much smaller than those observed on the axis of the dynamic plot.

{|style="margin: 0 auto;"
| [[File:Calculating the Reaction Path - MEP.png|thumb|upright|alt=Intial|Internuclear distance against time MEP plot for r1=0.9077425, r2=0.9177425]]
| [[File:Calculating the Reaction Path - Dynamic.png|thumb|upright|alt=Estimate|Internuclear distance against time dynamic plot for r1=0.9077425, r2=0.9177425]]
|}

==Reactive and unreactive trajectories==
Q.Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

r1=0.74, and r2=2.0 were used to compile the following table:

{| class="wikitable" style="margin-left: auto; margin-right: auto; border: none;"
|+ '''Cells left-aligned, table centered'''
|-
! p1
! p2
!Total Energy
! Reactivity
!Trajectory
!Description
|-
| -1.25
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|Reactive
|[[File:Kkz16 Reactive or not -1.25-2.5.png]]
|Trajectory goes towards the products, through the transition state
. Vibrations/oscillations increase after transition.
|-
|<nowiki>-1.5</nowiki>
| -2.0
|<nowiki>-100.456</nowiki>
| Unreactive
|[[File:Reactive or not -1.5-2.0.png]]
|Trajectory returns towards the reactants before reaching the transition state.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|Reactive
| [[File:Reactive or not -1.5-2.5.png]]
|Trajectory goes towards the products, through the transition state.

|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|Unreactive
|[[File:Reactive or not -2.5-5.0.png]]
|Trajectory returns towards the reactants, after the transition state has been formed.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|Reactive
|[[File:Reactive or not -2.5-5.2.png]]
|Trajectory goes towards the products, through the transition state.
After passing through transition state, trajectory returns back towards reactants,

before turning again towards the products
|-
|}

Q.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?

As the transition state theory (TST) is a model based on classical mechanics, 2 main assumptions are made<ref> J. I. Steinfeld, J. S. Francisco, W. L. Hase, (1989), Chemical Kinetic and Dynamics, Chapter 10, p.311 </ref>:

1) The separation of electronic and nuclear motions

2) The reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

Additionally 3 other assumptions are made which as specific to this theory:

1) Molecular systems that have crossed the transition state in the direction of products cannot turn around and reform reactants

2) In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation

3) Even in the absence of an equilibrium between reactant and product molecules, the transition states that are becoming products are distributed among their states according to the Maxwell-Boltzmann laws

These assumptions would mean that the rate would be overestimated, and the reality of the 4th reaction shows that the additional assumption 1) is not true in reality as it shows the system reforming the reactants after the formation of the transition state. Graphs like reaction 4 are observed in reality due to tunneling which enables the the activation barrier to be re-crossed, even after the formation of a transition state. Tunneling is a quantum effect and so is not accounted for in the classical model of the TST.

{{fontcolor1|gray|(Perfect! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:31, 17 May 2018 (BST))}}

=EXERCISE 2: F - H - H system=
==PES inspection==
Q.Classify the F + H2 and H + HF reactions according to their energetic (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved?

The F + H2 reaction to H + HF is shown to be exothermic due to the products being at a lower energy than the reactants (energy is released in the form of heat). This would mean that the reverse reaction (H + HF --> F + H2) would be endothermic (requires energy). This means that the H-F bond stronger than the H-H bond.

{{fontcolor1|gray|(That is correct. How does this relate to the picture you've shown? Would've just been nice to relate your answer to the depth of the wells etc. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:31, 17 May 2018 (BST))}}

[[File:Exothermic F-H-H system.png|300x300px|thumb|A surface plot of the F-H-H system reacting|centre]]

Q.Locate the approximate position of the transition state.

Using Hammond's Postulate, since the reaction is known to be exothermic, it can be assumed that the transition state will be towards the reactants. A point (r1=0.75 and r2=1.75. p1=p2=0) was chosen as a starting point for this reason, and improved through trial and error until the state of 0 kinetic energy and minimum forced was found.
The transition state of the reaction was found to be when the H-H bond length was 0.746 and the distance between the H-H and the F as 1.810.

The transition state of the exothermic reaction was found to have a potential energy of -103.751, while the reactants were found to be -104.020, (H-F distance=6, H-H bond length=0.74, p1=p2=0)so the activation energy was calculated to be 0.269.
The transition state of the endothermic reaction is the same, while the reactants were found to be -134.025 (H-F bond length= 0.92, H-H distance = 6), therefore the activation energy was 30.274.

{{fontcolor1|gray|(Very good! Maybe a figure or two could've illustrated this? The MEP energy vs time graph shows the drop in potential energy very nicely. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:31, 17 May 2018 (BST))}}

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

The reaction coordinates used to model this successful reaction of F and H2 were:

r1=0.69, r2=2.5, p1=-1.25, p2=-2.5

The reaction is an exothermic reaction so there is a release in energy, from the graphs it can be observed that the energy is released in another form. This is true due to the first law of thermodynamics which states that energy must be conserved as it cannot be created or destroyed, and the conservation of momentum, which would suggest that the same momentum must exist before and after the collision. The graphs below show that the kinetic energy increases, not just from the kinetic energy graph (average energy increases) but also the increase observed in the momentum of the H-F bond, while the H-H bond vibrations and momentum decreased. The increase in amplitude of the graphs shows this increase in momentum. The kinetic energy is also seen to increase in the energy-time graph, where the potential energy is observed to decrease by the same amount, showing that energy is converted from potential energy to kinetic energy.

{|style="margin: 0 auto;"
| [[File:Kkz16 Surface Successful F H-H reaction.png|thumb|upright|alt=Surface Plot|Surface Plot of the Successful Reaction of F with H2]]
| [[File:Kkz16 Successful F H-H reaction Momenta.png|thumb|upright|alt=Momentum|Momentum-Time Plot of the Successful Reaction of F with H2]]
| [[File:Kkz16 Successful F H-H reaction Energy.png|thumb|upright|alt=Energy|Energy-Time Plot of the Successful Reaction of F with H2]]
|}

These can be experimentally confirmed using calorimetry experiments, as the increase in kinetic energy of the molecules should result in an increase in temperature of the reaction mixture. IR could also be used to measure the increase in the vibrations as overtones would be produced.

Q. 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.

The Polanyi empirical rules state that translational energy is more efficient than vibrational energy for early-barrier reactions, where as vibrational energy is more efficient at activating late-barrier reactions. <ref> Polanyi, J.C. Some Concepts in Reaction Dynamics, Nobel Lectures Including Presentation Speeches and Laureates. Biographies. Chemistry 1981-1990; Malmström, B.G., Ed.; World Scientific Publishing Co.: Singapore-River Edge, NJ-London-Hong Kong, 1992; p.359-407 </ref>

 Reaction of H2+F 

In this model, r1 = H-H distance, r2 = H-F distance, and similarly the magnitude of p1 is related to the magnitude of H-F energy (translational at start) and the magnitude of p2 is related to the H-H energy (vibrational at start). Using Hammond's Postulate on this reaction (which we know to be exothermic), it can be determined that the transition state should lie towards the reactants and therefore be an early barrier reaction and so a larger translational energy should make the reaction more efficient.

Using the reaction values r1=0.74 r2=2.3 p1=0.5 p2=-6:
Which represent a much greater translation energy than vibrational energy the reaction was found to occur due to the balance of energy being good enough to overcome the transition state, and proceed to the products.
As show in the graphs below, the A-B (H-F interaction) starts with predominantly translational energy (few oscillations and a slope observed) where as after the reaction, only vibrational energy (no sloping, however many oscillations) was observed. The B-C (H-H interaction) is shown to be predominantly vibrational to begin with, however gained a lot of translational energy after the interaction, which leas to the sloping as the distance between the atoms increased. In the momentum graph an increase in total momentum is observed in the molecule, this is due to the increase of energy due to gaining kinetic energy through the drop in potential energy from the transition state.

<gallery caption="Larger Translational Energy than Vibrational Energy" widths="250px" heights="250px" class="center" >
File:Polanyi H H-F Contour r1=0.74 r2=2.3 p1=0.5 p2=-6.png|Contour Plot
File:Polanyi F H-H 1b.png|Distance-Time Graph
File:Polanyi F H-H 1.png|Momentum-Time Graph
</gallery>

Using the reaction values r1=0.74 r2=2.3 p1=6 p2=-0.5:
Which represents a greater vibrational energy than translational (H-H bond greater momentum/energy than the H-F interaction), despite the transitiona state formation, as observed in the graphs below, the reactants are reformed due to the low translation energy component, which is more effective. The balance of energies is not enough to overcome the transition state to form the products. In this example, due to the lower translational energy, the H-H bond is consistently dominated by vibrational energy with little to no translational energy observed. There is also little to no overall change in momentum as the products as not formed and so there is no overall decrease in energy. There is no overall reaction and agrees with the Polanyi empirical rules.

<gallery caption="Larger Vibrational Energy than Translational Energy" widths="250px" heights="250px" class="center" >
File:Polanyi H H-F Contour r1=0.74 r2=2.3 p1=6 p2=-0.5.png|Contour Plot
File:Polanyi F H-H 2.png|Distance-Time Graph
File:Polanyi F H-H 2b.png|Momentum-Time Graph
</gallery>

 Reaction of HF+H 

In this model, r1 = H-F distance, r2 = H-H distance, and similarly the magnitude of p1 is related to the magnitude of H-H energy (translational at start) and the magnitude of p2 is related to the H-F energy (vibrational at start). As expected of the reverse reaction to the reaction above, it would be expected that the energy barrier would be late in the reaction and therefore according to Polanyi's rules vibrational energy should encourage the reaction.

Using the reaction values r1=0.92 r2=2.3 p1=1.5 p2=-7:
Which represents a greater translational energy, no reaction was observed, as the B-C (H-F bond) remained did not change in distance (not gained translational energy) and continued to vibrate in the same place (as seen from the distance-time graph).

<gallery caption="Larger Translational Energy than Vibrational Energy" widths="250px" heights="250px" class="center" >
File:Polanyi H H-F Contour r1=0.92 r2=2.3 p1=1.5 p2=-7.png|Contour Plot
File:Polanyi H H-F 1 r1=0.92 r2=2.3 p1=1.5 p2=-7.png|Distance-Time Graph
File:Polanyi H H-F Momentum r1=0.92 r2=2.3 p1=1.5 p2=-7.png|Momentum-Time Graph
</gallery>

Using the reaction values r1=0.92 r2=2.3 p1=7 p2=-1.5:
Which represents a greater vibrational energy than translational energy. The contour plot shows that the reaction goes to completion, through the transition state. The A-B (H-H interaction) loses translational energy, which the B-C (H-F interaction) gains which is shown in the distance time plot.

<gallery caption="Larger Vibrational Energy than Translational Energy" widths="250px" heights="250px" class="center" >
File:Polanyi H H-F Contour r1=0.92 r2=2.3 p1=7 p2=-1.5.png|Contour Plot
File:Polanyi H H-F 1 r1=0.92 r2=2.3 p1=7 p2=-1.5.png|Distance-Time Graph
File:Polanyi H H-F Momentum r1=0.92 r2=2.3 p1=7 p2=-1.5.png|Momentum-Time Graph
</gallery>

The results of all these graphs agree with the Polanyi Empirical Rules.

{{fontcolor1|gray|(Nothing to comment. Very well done indeed. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:32, 17 May 2018 (BST))}}

=References=
<references>
</references>

MRD:kkz16

2018-05-17T20:31:55Z

Fjs113: Marked by fjs113

MRD:rf116

2018-05-17T20:12:23Z

Fjs113: Marked by fjs113

=Molecular Reaction Dynamics =
==Introduction ==
In this experiment, two reaction scenarios are investigated. In the first case, the H + H2 reaction is investigated, and in the second case, both the F + H2 --> H + HF and the reverse reaction H + HF --> F + H2 is investigated. The potential surface of the reactions are plotted, and the conditions optimised to find reactive trajectories and the locations of the respective transition states. Differences in the momenta and position of the particles are shown to have an effect on the kinetic energy of the reactants and thus determine whether or not the trajectory is reactive.

{{fontcolor1|gray|(Great that you included an introduction! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:12, 17 May 2018 (BST))}}

==H + H2 System ==
[[File:1 dist time.jpg|200px|Image: 200 pixels|thumb|Figure 1ː Internuclear Distance vs Time at the Transition State]]
[[File:2 ts contourmap.jpg|200px|Image: 200 pixels|thumb|Figure 2ː Contour map showing the location of the transition state]]

There are two minima (potential wells) and one saddle point, and at each the gradient, the partial derivative of the potential with respect to the position, r1 or r2, becomes zero. If a molecule starts in the first potential well, the potential defined by the Lennard-Jones potential, it then reacts and gains enough energy to oscillate out of the potential well. The transition state is then a saddle point, and the products exist in another potential well minima. The curvature of the potential energy can be used to distinguish between the minima and transition state, using differential calculus.
If the potential is defined as V(r1, r2), then D is defined as D =∂V²/∂²r1.∂V²/∂²r2 - (∂V²∂r1∂r2)², which is the determinant of the Hessian. If D is less than zero, then the stationary point is a saddle point, if D is greater than zero, and ∂V²/∂²r1 is greater than zero, then the stationary point is an energy minima.

{{fontcolor1|gray|(Careful - the TS is not a saddle point with respect to r1 and r2, but rather q1 and q2 which denote the reaction coordinate as well as its orthogonal counterpart. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:12, 17 May 2018 (BST))}}

===Locating the Transition State ===
The transition state is located at the position where r1 = r2. This was determined to be rts = 0.9 Å, as this is the region where both potential and kinetic energy are constant, there is no oscillation and A-B and B-C overlap. The internuclear distance vs time, as shown in Figure 1, shows minimal oscillation of the bond distances and hence represents the saddle point, which is the transition state. The contour map shown in Figure 2 shows that the trajectory of the particle is minimal (as stationary as possible), corresponding to the transition state.

{{fontcolor1|gray|(As far as your plot goes, there is definitely still oscillation there. You could've gone to more decimal places to find a more accurate estimate of the TS. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:12, 17 May 2018 (BST))}}

===Calculating the Reaction Path ===
[[File:3 mep 2.jpg|200px|Image: 200 pixels|thumb|Figure 3ː Mep plot]]
[[File:3 mep.jpg|200px|Image: 200 pixels|thumb|Figure 4ː Dynamics plot]]
When r1 = 0.9 + 0.01 = 0.91 Å = rts + 0.01 and r2 = 0.9 Å = rts, and p1 = p2 = 0, the path from H1 to H2 - H3 as minimum energy path (mep) plot is observed as a smooth path, in comparison to the trajectory (calculated using dynamics) which is slightly oscillating as the particle oscillates along the well moving up and down the sides, as observed in Figures 3 and 4. The real trajectory is oscillating as a particle moves as a wave hence the AB and BC distance oscillates accordingly. Conversely, the minimum energy path does not oscillate in this way as the mep represents the most likely reaction mechanism, and therefore averages the motion, creating a smooth line. The minimum energy corresponds to the bottom of the potential well and therefore the particle does not move up and down the sides of the well, instead mapping a straight line along the base of the well, where the energy is at a minimum.

{{fontcolor1|gray|(What do you mean by "particle" here? Remember, this is a triatomic system. Speaking of particles is misleading. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:12, 17 May 2018 (BST))}}

[[File:5 distvstime.jpg|200px|Image: 200 pixels|thumb|Figure 5ː Internuclear Distance vs Time graph]]
[[File:6 momentatime.jpg|200px|Image: 200 pixels|thumb|Figure 6ː Internuclear Momenta vs Time]]
[[File:7 reverse.jpg|200px|Image: 200 pixels|thumb|Figure 7ː Contour Plot of particle trajectory when r1 and r2 are reversed]]
The Internuclear distance vs Time and Internuclear Momenta vs Time graphs, as shown in Figures 5 and 6 respectively, indicate that the final position of r1 is 0.73 Å and r2 is 9.00 Å. The internuclear distance vs time graph shows that the distance between H1 and H3 and the distances between H2 and H3 increase linearly with time, due to repulsion between particles. {{fontcolor1|gray|(Not quite - at large-ish distances, there is no repulsion. Instead, the particles travel at almost constant speeds apart because they have positive momentum with respect to each other. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:12, 17 May 2018 (BST))}} Similarly, the momenta is shown to also increase with time. Initially a smooth increase is observed, but after a short period of time, the momenta of the H1-H2 and H1-H3 particles begins to oscillate periodically, whereas the H2-H3 does not oscillate, but remains smooth and at the highest value. This is as the H2-H3 bond breaks, so there is no longer any oscillation between the atoms. Conversely, the momenta of the other molecules does oscillate as the distances between these particles is still changing - one represents the new bond formed, the bond vibration, and the other represents the changing distance between the particles. The final average momentum of p1(t) and p2(t) at large time, t is 1.74. In this order the reaction is shown as starting at the transition state and the trajectory of the particle moves towards the products. If the values of r1 and r2 are reversed, that is r1 = 0.9 Å and r2 = 0.91 Å, then the trajectory of the particle reverses, and approaches the transition state from the opposite direction. In this instance, the trajectory starts at the transition state, and proceeds to move towards the reactants, this is shown in Figure 7.

[[File:8 finalinitial.jpg|200px|Image: 200 pixels|thumb|Figure 8ː Internuclear Distance vs Time graph when the initial conditions are set to the final conditions]]
[[File:9 initialfinal.jpg|200px|Image: 200 pixels|thumb|Figure 9ː Internuclear Momenta vs Time when the initial conditions are set to the final conditions]]
When the initial conditions are set to the final conditions, and the sign of the momenta changed to negative, the spectra reverse and show that the original initial conditions now become the final conditions. This represents the reverse reaction, the final conditions start at the products and end at the reactants, when previously this was the opposite.

[[File:10 table.jpg|200px|Image: 200 pixels|thumb|Figure 10ː Contour plot when p1 = -1.25 and p2 = -2.5]]
[[File:11 table.jpg|200px|Image: 200 pixels|thumb|Figure 11ː Contour plot when p1 = -1.5 and p2 = -2.0]]
[[File:12 table.jpg|200px|Image: 200 pixels|thumb|Figure 12ː Contour plot when p1 = -1.5 and p2 = -2.5]]
[[File:13 table.jpg|200px|Image: 200 pixels|thumb|Figure 13ː Contour plot when p1 = -2.5 and p2 = -5.0]]
[[File:14 table.jpg|200px|Image: 200 pixels|thumb|Figure 14ː Contour plot when p1 = -2.5 and p2 = -5.2]]

If a trajectory has enough kinetic energy to overcome the activation energy barrier, then the trajectory will be reactive, and if it has less energy than this amount, then it will be unreactive. Kinetic energy is equal to the momentum squared divided by 2 x the mass of the particle. Hence the momenta of the particles can be used to determine whether or not the trajectory is reactive. The trajectories of the particle at different momenta are shown in Figures 10-14.

Figure 10 is representative of a reactive trajectory, which occurs when the values of the momenta are p1 = -1.25 and p2 = -2.5. The particle clearly begins in the first potential well representative of the reactants, before oscillating past the transition state and into the second potential well, which corresponds to the products.
Figure 11 shows the trajectory at p1 = -1.5 and p2 = -2.0, and shows an unreactive pathway. The particle begins in the reactants potential well, it oscillates closer to the transition state, but does not have sufficient energy to overcome the activation barrier, hence it oscillates back along the same path, and remains as unreacted reactants.
Figure 12, when p1 = -1.5 and p2 = -2.5, shows a similar trajectory to that of Figure 10, in which the value of p2 is the same and p1 differs by only -0.25. This trajectory is also reactive as the particle starts at the reactants, oscillates out of the potential well, past the transition state and into the products. The particle in this instance oscillates with a higher frequency than in Figure 10, as observed by a slight increase in the number of oscillations.
Figure 13 represents the trajectory when p1 = -2.5 and p2 = -5.0. This trajectory is slightly reactive, as the particle does oscillate slightly beyond the transition state and into the second potential well of the products, however the path does not extend very far into the products and it soon recrosses the barrier and circles back to the reactants.
Figure 14 shows a reactive trajectory, when p1 = -2.5 and p2 = -5.2. The trajectory again maps from the reactants but initially does not have sufficient energy to enter the potential well of the products, and briefly recrosses the barrier back to the reactants, before finally entering the product potential well, oscillating at a much higher frequency.

In conclusion, when p2 is decreased too much (made less negative), the trajectory becomes unreactive. When p2 is increased (made more negative) to a sufficiently large amount and p1 is increased to -2.5, the trajectory is reactive but does not map a straight path through the transition state to the products, and circles back to the potential well of the reactants before it reaches the products - the trajectory recrosses the barrier. When the total energy is large <-100 kcal/mol, the trajectory is unreactive. When the momenta is decreased, the kinetic energy is decreased also as they are directly proportional. It can be concluded that when the momenta of p2 is decreased to at least -2.0, the particle no longer has sufficient kinetic energy to overcome the activation energy barrier.

{| class="wikitable" border=1
| p1 || p2 || Total Energy || Is the trajectory reactive or unreactive?
|-
| -1.25 || -2.5 ||-99.0 || Reactive
|-
| -1.5 || -2.0 ||-100.5 || Unreactive
|-
| -1.5 || -2.5 ||-99.0 || Reactive
|-
| -2.5 || -5.0 ||-85.0 || Slightly Reactive
|-
| -2.5 || -5.2 ||-83.4 || Reactive
|}

===Transition State Theory ===
Transition state theory assumes an equilibrium between reactants and products, and that there is a quasi-equilibrium between the transition state and the products. An equilibrium constant for this equilibrium can then be determined using the equation K = (concentration of the transition state)/(concentration of reactant A)(concentration of reactant B) = [TS]++/[A][B]. The rate of reaction is therefore given by d[Products]/dt = k[Reactants]. Where k =k++K++. The value of k++ = κν, where κ is the proportionality constant and ν is the vibrational frequency. K++ = kBT/hν x (e-ΔG++/RT). The value of k is therefore dependent on the Gibbs free energy of activation, ΔG++. As ΔG++ increases and becomes more negative, the value of ΔG++/RT will increase, and of e-ΔG++/RT will decrease. Hence the value of the rate constant k will also decrease. The rate therefore decreases as ΔG++ increases. Transition state theory will therefore predict that when the energy difference between the reactants and the transition state is high - at high ΔG++, the rate of reaction decreases. When the particle has lots of excess energy the rate of reaction will be low, which would correspond to Figures 10 and 12.
However, the theory does not take into consideration any barrier recrossing or quantum tunnelling as it does not account for any quantum phenomena, which can be observed as shown in Figures 13 and 14. The particles here have sufficient energy to overcome the activation barrier, and using transition state theory they would be expected to have a quicker rate of reaction than the trajectories shown in Figures 10 and 12. The experimental values would therefore differ from theory in this instance as the experimental rate is likely to be lower than the predicted value, as at any time, some of the reactants that have already reacted to form the products may cross back over the barrier to form the reactants again, reducing the rate.

{{fontcolor1|gray|(The latter half is absolutely perfect! Talking about the quasi-equilibrium makes less sense in this context as we're looking at three isolated atoms rather than a statistical ensemble. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:12, 17 May 2018 (BST))}}

==F-H-H System ==
[[File:15 exoendo.jpg|200px|Image: 200 pixels|thumb|Figure 15ː Plot of F + H2 reaction]]
[[File:16 exoendo.jpg|200px|Image: 200 pixels|thumb|Figure 16ː H + HF reaction]]
A surface plot of both F + H2 and H + HF reactions are shown in Figures 15 and 16 respectively. The F + H2 reaction is exothermic. The surface plot of the reaction shows the particles oscillating along the Lennard-Jones potential, and the particle eventually goes off to infinity, which indicates that the bond has been broken. The trajectory is concentrated near the reactants, and hence the transition state is early, and is located nearer the reactants, which is formed when the reactants approach each other. This is indicative of an exothermic reaction.
Conversely, the H + HF reaction is endothermic. Figure 16 shows the trajectory, which begins nearer the products before oscillating out of the page past the reactants. This shows that the transition state is late and is formed nearer the products.
The H-F bond strength is greater than the H-H bond strength. When the H-F bond breaks, more energy is absorbed in comparison to when the H-H bond breaks and alternatively more energy is released to form the H-F bond than the H-H. Since energy is released when a new bond is formed, and energy is absorbed when a bond is broken, in the forwards reaction F + H2 --> H + HF, more energy is released from the forming of the H-F bond than is taken in when breaking the H-H bond. Therefore more energy is released than is absorbed, and the process is exothermic (releases energy) and the enthalpy change is negative.
For the reverse reaction H + HF --> F + H2, more energy is absorbed when breaking the H-F bond than is released when the H-H bond forms. Hence more energy is absorbed than released and the reaction is endothermic (absorbs energy) and the enthalpy change is positive.

{{fontcolor1|gray|(Very good! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:12, 17 May 2018 (BST))}}

[[File:17 transitionstate.jpg|200px|Image: 200 pixels|thumb|Figure 17ː Surface plot of the location of the transition state for the F-H-H system]]
[[File:18 transitionstate.jpg|200px|Image: 200 pixels|thumb|Figure 18ː Contour plot of the location of the transition state for the F-H-H system]]
===Locating the Transition State ===
The transition state for the F + H2 reaction is located midway through the Lennard-Jones potential at the coordinates rHH = 0.745 Å and rFH = 1.811 Å. The transition state for this reaction can be observed as a surface plot in Figure 17 and as a contour plot in Figure 18. The transition state of the reverse reaction is located in the potential well at the back of the potential surface (not the Lennard-Jones potential well), and it is located in the centre of this well for the H + HF reaction at alternate coordinates. If the transition state was any further left or right, it would oscillate towards the reactants or the products.

{{fontcolor1|gray|(Good, but would've loved to see some reasoning/derivation of how you got to the TS. Show your working, even it is just an educated guess. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:12, 17 May 2018 (BST))}}

===Activation Energy ===
The activation energy was determined by first finding the potential energy of the transition state, which was determined to be -103.752 kcal/mol. The Hammond postulate was then used, that states that the transition state and a reactive intermediate near to the transition state have similar energy. Therefore, finding the energy of the intermediate will also give the energy of the transition state.
For the forwards reaction, F + H2 --> H + HF, the trajectory was then modified to find the energy slightly nearer the reactants from the transition state; values of rFH = 0.745 Å and rHH = 11 Å were chosen and an mep plotted. The distance of rHH was extended so the trajectory of the particle is at a maximum. The potential energy at these coordinates was determined to be -134.025 kcal/mol. The activation energy is the difference between these two values, hence Ea = + 30.273 kcal/mol for the forwards reaction.

For the H + HF reaction, the coordinates were chosen such that rHF = 11 Å and rHH = 0.745 Å. The potential energy was determined to be -104.101 kcal/mol at this point, and therefore this reaction has an activation energy of Ea = + 0.349 kcal/mol, which is significantly smaller in comparison to the forwards reaction.

[[File:19 reactivetraj.jpg|200px|Image: 200 pixels|thumb|Figure 19ː Contour plot showing the reactive trajectory corresponding]]
[[File:20 reactivetraj.jpg|200px|Image: 200 pixels|thumb|Figure 20ː Internuclear momentum vs time for the initial reactive conditions]]

=== Finding Reactive Initial Conditions for F + H2===
A set of initial conditions was found that indicated a reactive trajectory when rHH = 0.799 Å, rHF = 1.74 Å, pHH = -1.7 and pHF = -2.6. Figure 19 shows the contour map of this trajectory, whilst Figure 20 shows the corresponding graph of internuclear momentum vs time.

In the exothermic reaction F + H2 --> H + HF, total energy is conserved, but the ratio of kinetic to potential energy is subject to change. The reactants initially have both kinetic energy, and start at a potential minima. As the particles move along the trajectory and collide, some of the kinetic energy is converted to potential energy due to the repulsion between the two particles and some remains as kinetic energy. This reaction is an exothermic reaction, hence there is an excess of energy at the end of the reaction, which is released in the form of heat energy. In order to confirm this experimentally, calorimetry at constant volume can be used to monitor the difference in temperature before and after the reaction to determine whether or not energy has been released. If the excess energy has been released as heat energy, the temperature of the calorimeter should increase after the reaction has completed.

[[File: 21 calculation3 3 plus 2.9.jpg|200px|Image: 200 pixels|thumb|Figure 21ː Contour map when pHH = 2.9]]
[[File:22 calculation3 3 plus 2.5.jpg|200px|Image: 200 pixels|thumb|Figure 22ː Contour map when pHH = 2.5]]
[[File: 23 calculation3 3 plus 2.jpg|200px|Image: 200 pixels|thumb|Figure 23ː Contour map when pHH = 2.0]]
[[File:24 calculation3 3 plus 1.5.jpg|200px|Image: 200 pixels|thumb|Figure 24ː Contour map when pHH = 1.5]]
[[File: 25 calculation3 3 plus 1.0.jpg|200px|Image: 200 pixels|thumb|Figure 25ː Contour map when pHH = 1.0]]
[[File:26 calculation3 3 plus 0.5.jpg|200px|Image: 200 pixels|thumb|Figure 26ː Contour map when pHH = 0.5]]
[[File: 27 calculation3 3 0.jpg|200px|Image: 200 pixels|thumb|Figure 27ː Contour map when pHH = 0]]
[[File:28 calculation3 3 minus0.5.jpg|200px|Image: 200 pixels|thumb|Figure 28ː Contour map when pHH = -0.5]]
[[File:29 calculation3 3 minus1.jpg|200px|Image: 200 pixels|thumb|Figure 29ː Contour map when pHH = -1.0]]
[[File:30 calculation3 3 minus1.5.jpg|200px|Image: 200 pixels|thumb|Figure 30ː Contour map when pHH = -1.5]]
[[File:31 calculation3 3 minus2.0.jpg|200px|Image: 200 pixels|thumb|Figure 31ː Contour map when pHH = -2.0]]
[[File:32 calculation3 3 minus2.5.jpg|200px|Image: 200 pixels|thumb|Figure 32ː Contour map when pHH = -2.5]]
[[File:33 calculation3 3 minus2.9.jpg|200px|Image: 200 pixels|thumb|Figure 33ː Contour map when pHH = -2.9]]

[[File:34 calculation pFH0.8 pHH0.1.jpg|200px|Image: 200 pixels|thumb|Figure 34ː Contour map when pHH = 0.1 and pHF = -0.8]]

A reaction was set up that began located on the reactants side, when rHH is equal to 0.74 Å, pFH = -0.5, and rHF = 2.0 Å. The following range of values of pHH were investigated; 2.9, 2.5, 2.0, 1.5, 1.0, 0.5, 0, -0.5, -1.0, -1.5, -2.0, -2.5, -2.9, as shown in Figures 21-33.

{{fontcolor1|gray|(That's a very comprehensive range of values. You did however forget to check what happens at 3 and -3. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:12, 17 May 2018 (BST))}}

When pHH = 2.9, the trajectory is reactive and the particle oscillates freely from the starting point past the transition state and through to the products. As pHH is decreased to 2.5, the trajectory becomes less reactive and recrosses the barrier, with a final position in the reactants region. When decreased again to 2.0, the particle does not recross the barrier, but instead is reactive, entering the products region. A similar pattern to 2.0 is observed when pHH = 1.5, the particle follows a similar trajectory, but oscillates with a higher frequency near the products in comparison. When pHH = 1.0, the initial oscillation has a much higher frequency, and the particle does not fully enter the potential well of the products. When pHH = 0.5, again the initial oscillation is of much higher frequency, but still crosses the transition state and into the products region. When pHH = 0 and -0.5, the particle moves similar to when pHH = 0.5, though they do not enter the products well as far. When pHH = -0.5 and -1.0, a similar pattern is observed - the particle slightly enters the potential well of the products. For pHH = -1.5, some barrier recrossing is observed. At pHH = -2.0, the trajectory becomes very reactive again, as the particle enters far into the products region. When pHH is decreased to -2.5 and -2.9, the trajectory becomes less reactive and barrier recrossing occurs, with the final position of the particle being located in the reactants potential well.
In summary, when pHH is close to the upper limit (3), the trajectory is very reactive and defined, when pHH is decreased but still remains positive, the trajectory becomes not as reactive and the frequency of the initial oscillation increases. When pHH becomes more negative, from -0.5 to -1.5, the trajectory reactivity decreases ultimately remains reactive. When pHH = -2.0, the trajectory becomes very reactive, but values more negative than this and close to the lower limit (-3), such as pHH = -2.5 and -2.9, show barrier recrossing and are less reactive trajectories.

As pHH is increased, so is the frequency of the vibration between the atoms. When the particles recross the barrier, the products only exist momentarily, and are then converted back to the products.

{{fontcolor1|gray|(Some very good observations, although you're not offering much in terms of explanation. Some text along the lines of why do you think that this happens at that value would've been even better. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:12, 17 May 2018 (BST))}}

The momentum pFH was then increased to -0.8, and the overall energy of the system was reduced by setting pHH = 0.1, as shown in Figure 34. This shows the particle oscillating, the particle oscillates through to the second potential well, the products, but with an increased wavelength (reduced frequency). The particle moves far into the products region and is therefore a very reactive trajectory.
The overall energy of the system is reduced, but since pFH is still relatively high at -0.8, the trajectory is still reactive.

[[File:35 final part 1 firstsetupinitialconditions.jpg|200px|Image: 200 pixels|thumb|Figure 35ː Contour map for the reverse reaction when pHH = 11 and pHF = -0.1]]

===Reverse Reaction H + HF ===
For the reverse reaction H + HF, initial conditions were chosen such that the start position was located at the bottom of the reactants potential well, as shown in Figure 35. The conditions were set to pHH = 11, and pHF = -0.1. pHH was chosen to be large, so that the energy is greater than the activation energy, and pHF was chosen to be low, to ensure a low vibration of the HF bond. The position was set to rHH = 2.0 Å and rHF = 0.92 Å. The particle is observed to oscillate through to the products, but re-enters the barrier and oscillates in the opposite direction towards the reactants.

The momentum of pHH was then decreased (decreasing the energy of the HH vibration) and the momentum of pFH was increased (increasing the energy of the HF vibration), and both distances kept constant to find a reactive trajectory. The distances used are rHH = 2.0 Å, rFH = 0.92 Å, and momenta pHH = -0.4 and pFH = 7.6, the trajectory is shown in Figures 36 and 37 as a contour map and skew plot, respectively. The plots show the particle oscillating at a high frequency initially, until in the product state, where the frequency decreases slightly.

[[File:36 reactive trajector final image.jpg|200px|Image: 200 pixels|thumb|Figure 36ː Contour map for the reactive trajectory of the reverse reaction when pHH = -0.4 and pHF = 7.6]]
[[File:37 final trajectory skew plot.jpg|200px|Image: 200 pixels|thumb|Figure 37ː Reactive trajectory viewed as a skew plot]]

===Energy Distribution ===
The total energy of the products and reactants is further constituted of both vibrational and translational energy. Polyani's empirical rules state that vibrational energy is more efficient than translational energy in promoting a reaction with a late transition state barrier, and that the translational energy promotes a more efficient reaction when there is an early transition state barrier than the vibrational energy. <ref>Z. Zhang, Y. Zhou, D. H. Zhang, G. Czakó and J. M. Bowman, Theoretical study of the validity of the polanyi rules for the late-barrier Cl + CHD3 reaction, J. Phys. Chem. Lett., 2012, 3, 3416–3419. </ref> Since F + H2 --> H + HF is an exothermic reaction, the energy barrier is located in the coordinates corresponding to the reactants, and the transition state barrier is early (nearer the reactants) as observed in this case. Hence the reaction proceeds more efficiently due to the translational energy of the reactants, as opposed to the vibrational energy of the reactants and so molecules with a higher degree of translational energy in comparison to vibrational are favoured, as they are more effective at crossing the barrier and promoting the reaction. <ref> L. Jerry, A Case Study in Reaction Dynamics </ref>

The set of conditions investigated above show the difference in translational and vibrational energy. For the F + H2 --> H + HF system, changing the value of the pHH momentum changes the amount of vibrational energy the molecule has - increasing the momentum increases the amount of vibrational energy between the two H atoms in the H2 molecule. When the value of the momentum pFH is increased, the amount of translational energy the molecule has is increased. It can be observed that when pFH is increased, but the overall energy of the system is greatly reduced (as shown in Figure 34), the trajectory is still reactive. Also, when pHH is varied between 3 and -3, the trajectories do not change as drastically. This is indicative that the translational energy is more efficient at promoting a reactive trajectory than the vibrational energy in this instance as the transition state is early, in accordance with Polyani's empirical rules.

For the reverse reaction H + HF --> F + H2, the pHH momentum is decreased, and the pHF momentum is increased to obtain a reactive trajectory (Figure 36). In this instance, the translational energy is decreased and the vibrational energy is increased. Since for the reverse reaction the transition state is late, Polyani's empirical rules states that vibrational energy will be more efficient at promoting the reaction than the translational energy, which is indeed the case.

Mrd:gm1616

2018-05-14T18:28:48Z

Fjs113: Marked by fjs113

==H + H2 system==

===Dynamics from the transition state region===
At the transition state, the PE surface has a gradient of 0 and it is a maximum point on the minimum energy pathway. A reaction must pass through this point to form products.

The minima of the potential energy surface changes value as the collision occurs, and the maximum point of this minimum PE path is the transition state. There is a the largest curvature along the minimum PE path at the TS, so if a small change in the reactions conditions occurs that favours either reactants or products, the point along the minimum energy pathway will quickly fall and the newly favoured molecule will form. At a minima, the potential energy will increase with any change in reaction conditions or change in the reactant/product molecule's geometry.

{{fontcolor1|gray|(This is ok. Would have been better to see mention of partial derivatives and how their values relate to different types of stationary points. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 19:28, 14 May 2018 (BST))}}

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

The best estimate for the transition state position is with the distance of AB=BC=0.9212 (4dp) because at this geometry the TS remains oscillating in the well only of the minimum energy pathway and has no gradient. This transition state can be seen below with the inter-nuclear distance-time graph which shows that the A-B and A-C are oscillating in phase, so A-B must also be oscillating in phase. This means that atoms A and C are vibrating periodically and in opposing directions around central atom B, so the system is not tending towards products or reactants.

[[File:Gm_h_h2_internuc_graph.PNG]]

{{fontcolor1|gray|(This is unfortunately not correct. At the TS, no vibrations should occur. Since the TS is an unstable stationary point, if the initial conditions are set to the TS with 0 momenta, the system should stay completely still. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 19:28, 14 May 2018 (BST))}}

===Calculating the reaction path===

The differences between the MEP and dynamics calculation types on the PE surface plot are that there is no oscillating bond length in the MEP plot. Also, the MEP plot does not reach as large a bond length for the same amount of steps as the dynamic calculation type.

{{fontcolor1|gray|(These are correct observations. But why is this the case? You need to provide more explanation and maybe a picture or two for evidence. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 19:28, 14 May 2018 (BST))}}

r1(t) = 0.5,
r2(t) = 8.75,
p1(t) = 1.1,
p2(t) = 2.4.

r1(t) = 8.75,
r2(t) = 0.5,
p1(t) = 2.4,
p2(t) = 1.1.

In the calculation where the final distances were set to the initial distances and the momenta were added with a negative sign, it could be seen that the final distances were much closer together, suggesting the system came back to the transition state conditions and neither reactants nor products predominated as the reaction proceeded. The final momenta was of the same magnitude but with a positive sign.

===Reactive and unreactive trajectories===
{| class="wikitable" border="1"
! p1 !! p2 !! Total Energy (kJ/mol) !! Reactive? !! Potential Energy Surface Plot !! Description of trajectory
|-
| -1.25 || -2.5 || -99.018 || Yes || [[File:Gm_h_h2_momenta_1.PNG]] ||Initially, B-C's bond length oscillates and A-B's bond length decreases up until the transition state at which the reaction occurs and A-B's bond length oscillates and C-B's bond length increases.
|-
| -1.5 || -2.0 || -100.456 || No || [[File:Gm_h_h2_momenta_2.PNG]] || The A-B bond oscillates as it vibrates and the B-C bond length decreases as C approaches AB but the transition state is never reached and the molecule remains unchanged and C moves away from AB.
|-
| -1.5 || -2.5 || -98.956 || Yes || [[File:Gm_h_h2_momenta_3.PNG]] || Initially, B-C's bond length oscillates and A-B's bond length decreases up until the transition state at which the reaction occurs and A-B's bond length oscillates and C-B's bond length increases.
|-
| -2.5 || -5.0 || -84.956 || Yes || [[File:Gm_h_h2_momenta_4.PNG]] || The barrier is recrossed so that the C-B bond is formed briefly after the transition state before the A-B bond reforms and the C-B bond distance increases as it move away from the A-B molecule that has an oscillating bond length. There is greater overall energy in the system and the vibrations are much larger in the final molecules.
|-
| -2.5 || -5.2 || -83.416 || Yes || [[File:Gm_h_h2_momenta_5.PNG]] || The barrier is recrossed so that the A-B bond is formed briefly after the transition state before the B-C bond reforms and the A-B bond distance increases once again as B-C's bond distance oscillates about a fixed length as it vibrates. There is greater overall energy in the system and the vibrations are much larger in the final molecules.
|-
|}

{{fontcolor1|gray|(Are you sure that case 4 is reactive? Otherwise good. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 19:28, 14 May 2018 (BST))}}

Transition state theory assumes that during a reaction, a quasi-equilibrium will be set up between the reactants and activated complexes in the transition state and the rate of reaction can be determined by the gradients and shape of the molecular potential profile that a moloecule follows as it crosses through the transition state <ref>Activated complex theory of bimolecular reactions. DOI: 10.1021/ed051p709</ref>. It assumes only classical behaviour of the molecules. The theory is good at predicting rates for generic reaction trajectories at high temperatures (since classical mechanics deviates less from quantum mechanics), but it assumes that the product state remains unchanged indefinitely and so it cannot be accurate for the barrier crossing trajectories since the product state exists temporarily for those conditions. Also, it does not take into account the concentration/surface area which can affect a reaction rate significantly.

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

===PES inspection===
The total energy for H2 + F is -98.5 kJ/mol (20.5 kJ/mol kinetic and -119.0 kJ/mol potential) when HH distance = 0.74, HF distance = 2, p(HH)=2 and p(FH)=-0.5.
The total energy for H-F + H is -78.0 kJ/mol (31.0 kJ/mol kinetic and -109.0 kJ/mol potential) when HH distance = 1, HF distance = 1.5, p(HH)=2 and p(FH)=-3.
This shows that the H2 + F is the exothermic process and the reverse reaction is endothermic.
The H-H bond strength is 432 kJ/mol and the H-F bond strength is 565 kJ/mol<ref>http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html</ref> so the H-F bond is stronger by 133 kJ/mol and hence the reaction which breaks the H-F bond will be unfavourable since the energy required to break the bond is not regained by the H-H bond formation.

{{fontcolor1|gray|(Good. Would have been even better with a picture, e.g. a surface plot or an MEP energy vs time plot showing the difference in potential energy. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 19:28, 14 May 2018 (BST))}}

The approximate transition state was when the atoms has a bond length of 1.58 for H-F and 0.76 for H-H at which point the H atom in between the H and F atoms oscillated periodically between the H-F bond and the H-H bond. For this asymmetric system, it is expected that the distance between all 3 atoms is not equal. The transition state is closer in energy to H2 + F and so, according to Hammond's Postulate, this suggests the transition state has a geometry which is similar to these reactants too.
Energy of the transition state is -104.5 kJ/mol (12.5 kJ/mol kinetic and -117 kJ/mol potential).

{{fontcolor1|gray|(As above, this is unfortunately '''not''' the TS. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 19:28, 14 May 2018 (BST))}}

The difference between the energy of the transition state and the energy of the reactant molecules gives the activation energy:

Activation Energy of H2 + F = -98.5 - (-104.5) = 6 kJ/mol

Activation Energy of H-F + H = -78.0 - (-104.5) = 26.5 kJ/mol

The difference in these activation energies proves that when H2 + F is the forward reaction, the reaction is exothermic since H2 + F has a higher energy than the H-F + H products, and so ΔHreaction = - 20.5 kJ/mol.

{{fontcolor1|gray|(Again, this could be illustrated better. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 19:28, 14 May 2018 (BST))}}

===Reaction dynamics===
In the reaction trajectory between F and H2, the reaction barrier is crossed three times as the F + H2 reforms and then H-F + H remains as the final product. This can be seen in the inter-nuclear momenta vs time graph below which shows the system being crossed when t=0.33, 0.42 and 0.57, of which the crossing at 0.33 and 0.57 correspond to the H-F bond forming and the H-F inter-nuclear momenta at these points increases to a displacement of 9 as the molecule vibrates periodically. Initially, the system has mainly kinetic energy as the H-H bond approaches the stationary F atom until the first H atom is closer to F than the equilibrium bond distance, at which point the nuclei from the atoms repel each other and most of the kinetic energy is converted to potential energy as the H atom is now stationary. However, as the H-F bond lengthens under the nuclei's repulsion, the second H atom is still approaching the first H atom until they reach the minimum separation distance and begin to repel. This repulsion moves the first H atom back towards the F atom and so the H-F bond reforms and the second H atom moves away from the H-F molecule as the potential energy it gained in the repulsion at the minimum separation distance is converted into kinetic energy. Throughout the reaction, the total energy never changes but its composition of kinetic energy and potential energy does change.

This could be proven experimentally by using heavier Deuterium to replace one of the hydrogens and then determining whether there is a change in activation energy or rate and hence confirming how the hydrogen is involved in the mechanism.

{{fontcolor1|gray|(Interesting and possibly plausible approach. You could have elaborated a little more, e.g. how to relate the results from this simulation to results from the experiment. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 19:28, 14 May 2018 (BST))}}

[[File:Gm_f_h2_momenta_graph.PNG]]

As the momentum of the H-H is varied between 3 and -3 for a reaction between F and H2, it can be seen that the momenta has an effect on whether the reaction is possible and how much inter-system crossing will occur. At the lowest momentum of -3, the system remains closer to the transition state and the system passes through the transition state many times. As the momenta is increased to a less negative value, the side of the reactants (H2) is favoured. However, by pHH=-1.3, the system once again remains close to the transition state and involves lots of crossing, before the product (HF + H) is favoured. AS momenta increases, there is no intersystem crossing and the products instantly form. This pattern repeats once more with pHH=1.5 favouring reactants, pHH=2.6 staying near the transition state and pHH=3 favouring products. All of the systems have a maximum potential energy of approximately -105 kJ/mol. The more momenta in the system, the more intersystem crossing occurs. This is because there is a much larger energy in all of the systems than the activation energy required for hydrogen, so the hydrogen has a large KE as it approaches the F atom and so it will have a larger energy when it is repelled from the Florine, meaning there is more energy for the hydrogens' bond to reform.

{| class="wikitable" border="1"
! p(H-H) !! Reactive? !! Potential Energy Surface Plot !! p(H-H) !! Reactive? !! Potential Energy Surface Plot
|-
| -3.0 || No || [[File:Gm_f_h2_momenta_-3.PNG]] || -2.8 || No || [[File:Gm_f_h2_momenta_-2.8.PNG]]
|-
| -2.4 || No || [[File:Gm_f_h2_momenta_-2.4.PNG]] || -1.3 || No || [[File:Gm_f_h2_momenta_-1.3.PNG]]
|-
| -1 || Yes || [[File:Gm_f_h2_momenta_-1.PNG]] || 1 || Yes || [[File:Gm_f_h2_momenta_1.PNG]] ||
|-
| 1.5 || No || [[File:Gm_f_h2_momenta_1.5.PNG]] || 2.6 || No || [[File:Gm_f_h2_momenta_2.6.PNG]] ||
|-
| 3.0 || Yes || [[File:Gm_f_h2_momenta_3.PNG]] ||
|-
|}

When the overall energy of the system was reduced with p(HF)=0.8 and p(HH)=-0.1 then the potential energy surface looked like the figure below and the lower PE H + HF products were formed with no intersystem crossing taking place. There is reduced total energy so it is expected that the molecules will have less potential energy and so there will be less repulsion when the H-F atoms reach the minimum distance of separation. The hydrogen's bond never reforms since the H atoms are never close enough to overcome the stronger H-F attraction and form a new bond.

[[File:Gm_f_h2_momenta_-0.1.PNG]]

For the inverse momenta method to find a reaction trajectory, the reaction proceeded to the products with the initial conditions of F-H=0.9, H-H=2.4, p(FH)=p-(HH)=5.45. Any conditions where p-(HH) decreases and p(FH) increases by the same amount from 5.45 also resulted in a reaction.

Polanyi's Empirical Rules <ref>Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction. DOI:10.1021/jz301649w</ref> state that vibrational energy is better at promoting a late energetic-barrier to reaction in comparison to translational energy which is more efficient at promoting an early energy barrier to reaction. If the barrier is late, than the transition state at that point will resemble the products more closely than the reactants and vice versa. This means that vibrational energy is better at promoting an endothermic reaction and translational energy is better at promoting an exothermic reaction.

{{fontcolor1|gray|(Would've been nice to see some evidence that this indeed holds, using plots of simulations etc. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 19:28, 14 May 2018 (BST))}}

<references/>

JN1316Lab2

2018-05-14T17:52:06Z

Fjs113: Marked by fjs113

All distances are in Angstroms and energies in kCal/mol

== Exercise 1: H + H 2 ==

 What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface. 

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

{{fontcolor1|gray|(This is mostly wrong. You need to refer to partial derivatives here. The TS is a saddle point where the first order partial derivative along any direction is 0, the second order partial derivative along the reaction coordinate is <0 and along the orthogonal coordinate is >0. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 18:52, 14 May 2018 (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. 

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

{{fontcolor1|gray|(How did you get from the aforementioned MEP calculation to the saddle point? Whilst the bare number is correct, there is not enough explanation and evidence in the form of plots provided here. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 18:52, 14 May 2018 (BST))}}

[[File:IND-T.PNG]]

{{fontcolor1|gray|(Does this plot have to do something with your answer? If so, this is most certainly not the distances vs time plot obtained from your bond length value at the saddle point. You need to explain your figures! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 18:52, 14 May 2018 (BST))}}

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

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

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

{{fontcolor1|gray|(Again, you need to label your figures! Also, neither plot shows any plateauing as time progresses. In fact, in an MEP, it is very much possible for the lone hydrogen to escape to "infinity" with enough time. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 18:52, 14 May 2018 (BST))}}

The momentum - time graph below shows that one hydrogen leaves as a free atom and the other 2 as a vibrating molecule with translational and vibrational energy modes.

{{fontcolor1|gray|(OK, but how does this tie in with your comparison of MEP vs Dynamics? [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 18:52, 14 May 2018 (BST))}}
[[File:JN moment 1.PNG]]

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

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

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

{{fontcolor1|gray|(The picture for case 1 is not correct. Are you sure you put in the correct initial conditions. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 18:52, 14 May 2018 (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? 
The main assumption to the TST is that once you have left the maxima that is the transition state you cannot go back to it or past it. This assumption fails as we have seen in the above trajectories some of the minimum energy paths reach the transition state and then return backwards. The TST also neglects the effects of quantum mechanics. Particularly for light particles such as hydrogen the non-zero probability of tunneling through a classically forbidden potential barrier with quantum potential is significant. Transition state theory will accurately predict reactions that pass only once over the transition state however it could incorrectly predict a path will be reactive or vice versa.

{{fontcolor1|gray|(Very good discussion, but how does this specifically affect predicted reactions '''rates'''? [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 18:52, 14 May 2018 (BST))}}

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

 Classify the F + H2 and H + HF reactions according to their energetics (endothermic or exothermic). How does this relate to the bond strength of the chemical species involved? 

F + H2: Reaction is exothermic as H-F bond is stronger than H-H bond. As seen in the picture below the energy of the products is less than that of reactants.

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

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

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

 Locate the approximate position of the transition state. 

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

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

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

[[File:JN TS NEW.PNG]]

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

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

{{fontcolor1|gray|(Please be more specific here. In this reaction, potential energy is being converted to vibrational energy. You also need to be more specific about the experiments. What results would you expect to see. How would you related that back to the results from the simulation? [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 18:52, 14 May 2018 (BST))}}

[[File:JN pbeforeafter.PNG]]

[[File:JN animation.PNG]]

 Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (explore values also close to these limits). What do you observe? 

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

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

{{fontcolor1|gray|(Please provide some pictures for evidence. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 18:52, 14 May 2018 (BST))}}

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

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

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

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

[[File:JN traject.PNG]]

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

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

[[File:JN idk 1.PNG]]

[[File:JN idk 2.PNG]]

[[File:JN idk 3.PNG]]

[[File:JN idk 4.PNG]]

{{fontcolor1|gray|(This is good! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 18:52, 14 May 2018 (BST))}}

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

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

Rep:Mod:einstein

2018-05-14T16:51:18Z

Fjs113: Marked by fjs113

==James Cochrane's Wiki==
===H + H2 <math>\rightarrow</math> H + H2===
In the following analysis of the collision between an H atom and an H2 molecule, the atoms are constrained to be collinear (collision angle is 180). It is important to realize that although the potential barrier is least for collinear attack, other lines of attack are feasible and contribute to the overall rate of reaction in reality. Additionally, effects of quantum mechanical tunnelling on reactivity are ignored; we just consider the classical trajectories of particles over surfaces.<ref name="e1" />

Two parameters are required to define the nuclear separation: the HA-HB separation <math>r_{AB}</math> and the HB-HC separation <math>r_{BC}</math>. A plot of the total energy of the system against <math>r_{AB}</math> and <math>r_{BC}</math> gives the potential energy surface of this reaction.

[[File:einsteinSurface1new.png|700px|thumb|none|Generic potential energy surface for the H + H2 <math>\rightarrow</math> H + H2 reaction.]]

Although the 3D surface provides a useful visualization tool, contour diagrams (with contours of equal potential energy) will be more suitable for analysis.

[[File:einsteinContour1new.png|700px|thumb|none|Generic contour diagram for the H + H2 <math>\rightarrow</math> H + H2 reaction. Equilibrium bond lengths are shown (strictly they represent the relevant bond lengths when the third atom is infinitely far away).]]

Here, I use the notation <math>x=r_{AB}</math>, <math>y=r_{BC}</math> and <math>V_{x}=\frac{\partial V}{\partial x}</math>, where <math>V</math> is the potential energy function.

Suppose that <math>V(x,y)</math> has a local extremum at a point <math>(x_0,y_0)</math>, then <math>V_{x}(x_0,y_0)=V_{y}(x_0,y_0)=0</math>. These properties hold true at both a transition state and at minimum. To distinguish between these two structures we must use the following formula:
<math>D \equiv V_{xx}V_{yy} - V_{xy}V_{yx}</math>. If <math>D>0</math> and <math>V_{xx}>0</math>, we are at a minimum. Whereas, if <math>D>0</math> and <math>V_{xx}<0</math>, we are at a transition state. Additionally, if <math>D<0</math>, we are at a saddle point.<ref name="e2" />

{{fontcolor1|gray|(Perfect! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:51, 14 May 2018 (BST))}}

[[File:Einstein1.png|700px|thumb|none|Successful reactive trajectory for the reaction of HA + HBHC <math>\rightarrow</math> HC + HAHB. Conditions: <math>r_{BC}=0.74</math>, <math>r_{AB}=2.3</math>, <math>p_{BC}=0</math> and <math>p_{AB}=-2.7</math>. The approximate location of the transition state (TS) is shown.]]

The transition state is located near <math>r_{AB}=r_{BC}=0.91</math>. To improve upon this value we test a range of values <math>r_{AB}=r_{BC}</math> and <math>p_{AB}=p_{BC}=0</math>: akin to seeing which side of the hill a stationary ball rolls down.

{| class="wikitable sortable" border="1"
! Trial !! <math>r_{AB}</math> !! <math>r_{BC}</math> !! <math>p_{AB}</math> !! <math>p_{BC}</math> !!class="unsortable" | Internuclear Distances vs Time!! class="unsortable" | Successful?
|-
| 1 || 0.9 || 0.9 || 0 || 0 ||[[File:Einsteintrail1.png|300px]] || ✗
|-
| 2 || 0.91 || 0.91 || 0 || 0 ||[[File:Einsteintrail2.png|300px]] || ✗
|-
| 3 || 0.92 || 0.92 || 0 || 0 ||[[File:Einsteintrail3.png|300px]] || ✓
|-
| 4 || 0.917 || 0.917 || 0 || 0 ||[[File:Einsteintrail4.png|300px]] || ✓
|-
|}

After 4 tested values (with 500 steps), the transition state occurs when the system has roughly the arrangement: <math>r_{AB}=r_{BC}=0.917</math>. Specifically, the transition state is a set of configurations (a line on the potential energy surface which passes through a saddle point) associated with a critical geometry such that every trajectory that goes through this geometry reacts.

{{fontcolor1|gray|(This is unfortunately wrong. The transition state '''is''' the saddle point. This is one single set of initial conditions at which the system ideally does not move at all. At 0.917, you can still see oscillations as your distances vs time plot shows. At the TS, this plot should give almost perfectly horizontal lines. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:51, 14 May 2018 (BST))}}

{| class="wikitable sortable" border="1"
! Calculation Type !! <math>r_{AB}</math> !! <math>r_{BC}</math> !! <math>p_{AB}</math> !! <math>p_{BC}</math> !!class="unsortable" | Contour Diagram
|-
| MEP || 0.917 || 0.927 || 0 || 0 ||[[File:Einsteinmep1.png|300px|thumb|none| This trajectory (towards the products) is of least energy along the valley floor because each iteration sets the velocity (and kinetic energy) to zero - preventing any movement up the valley walls.]]
|-
| Dynamics || 0.917 || 0.927 || 0 || 0 ||[[File:Einsteinmep2.png|300px|thumb|none| Whereas this trajectory (towards the products) does not take the route of least potential energy because the non-zero velocity of the reactants allows the valley walls to be traversed.]]
|-
| MEP || 0.927|| 0.917 || 0 || 0 ||[[File:Einsteinmep4.png|300px|thumb|none| When the initial conditions are moved onto the reactant "side" of the transition state the trajectory goes to reactants. It has the same properties as the other MEP calculation above. ]]
|-
| Dynamics || 0.927 || 0.917 || 0 || 0 ||[[File:Einsteinmep3.png|300px|thumb|none| Upon a change in conditions, the arrangement falls to the reactants. It has the same properties as the other Dynamics calculation above. ]]
|-
|}

{{fontcolor1|gray|(Despite your value for the TS being slightly off, this is very good. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:51, 14 May 2018 (BST))}}

In an attempt to run the minimal kinetic energy reverse reaction, we take the output average radii and momenta from the above calculations. It is expected that the reaction trajectory will almost get to the transition state and then return via the reactant channel. Here is the result of such a collision.

{{fontcolor1|gray|(Careful: Did you mean to say radii? In the classical interpretation used here, we treat atoms as point particles. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:51, 14 May 2018 (BST))}}

[[File:Einsteinreverse1.png|600px|thumb|none|The reverse reaction approaches from the "product" channel with no vibration energy, almost reaches the transition state (TS) and then returns via the "product" channel in a vibrationally excited state. Therefore, HAHB had insufficient kinetic energy. Conditions: <math>r_{BC}=9.06</math>, <math>r_{AB}=0.74</math>, <math>p_{BC}=-2.46</math> and <math>p_{AB}=-1.19</math>.]]

The following table shows the effect of higher collision momenta on the success of a reaction with fixed radii.

{| class="wikitable sortable" border="1"
! <math>p_{AB}</math> !! <math>p_{BC}</math> !!class="unsortable" | Contour Diagram (<math>r_{BC}=0.74</math>, <math>r_{AB}=2.0</math>) !! Total energy !! class="unsortable" |Reactive?
|-
| -2.5 || -1.25 ||[[File:Einsteinq1.png|400px|thumb|none| HA approaches a non-vibrating HBHC molecule, reacts successfully; a vibrationally excited HAHB molecule emerges. ]] || -99.02 || ✓
|-
| -2 || -1.5 ||[[File:Einsteinq2.png|400px|thumb|none| HA approaches a vibrating HBHC molecule, collides unsuccessfully; a vibrationally excited HBHC molecule re-emerges. The phase of vibration is insufficient for the reaction. ]] || -100.46 || ✗
|-
| -2.5 || -1.5 ||[[File:Einsteinq3.png|400px|thumb|none| HA approaches a slowly vibrating HBHC molecule, reacts successfully; a vibrationally excited HAHB molecule emerges. ]] || -98.96 || ✓
|-
| -5.0 || -2.5 ||[[File:Einsteinq4.png|400px|thumb|none| HA approaches a non-vibrating HBHC molecule, passes the transition state but recrosses this region; a vibrationally excited HBHC molecule re-emerges. ]] || -84.96 || ✗
|-
| -5.2 || -2.5 ||[[File:Einsteinq5.png|400px|thumb|none| HA approaches a non-vibrating HBHC molecule, passes the transition state and recrosses this region several times; a highly vibrationally excited HAHB molecule emerges. ]] || -83.42 || ✓
|-
|}

The main assumption of transition state theory (TST) is that there is a crucial configuration of no return called the transition state. If molecules pass through this spatial configuration then it is inevitable that they will form products from this encounter.<ref name="e1" /> In strict terms, this "inevitability" is only certain once the nascent products are significantly separated from each other.<ref name="e3" />

{{fontcolor1|gray|(This is otherwise known as barrier recrossing, which TST does not take into account. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:51, 14 May 2018 (BST))}}

The experimental result with a calculated energy of -84.96 (see above table) does not agree with TST predictions because this trajectory crosses through the transition state several times but does not afford products.

{{fontcolor1|gray|(Here, we asked you to comment on the reaction rate predicted by TST compared to experiments. You're on the right track, but haven't actually stated that TST will overestimate the reaction rates and why. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:51, 14 May 2018 (BST))}}

===F + H2 <math>\rightarrow</math> H + HF===
It is known that the F + H2 <math>\rightarrow</math> H + HF reaction is highly exoergic (exothermic) with a value of <math>\Delta\text{H}\simeq</math>-32 kcal/mol.<ref name="e4" /> The MEP calculated value (see later) of <math>\Delta\text{H}\simeq</math>-134.016-(-104.004)= -30.012 kcal/mol for the reaction is accurate and correctly implies that the H-F bond is much stronger than the H-H bond.

[[File:Einsteinfh2.png|700px|thumb|none|Successful reactive trajectory for the reaction of HCHB + FA <math>\rightarrow</math> HC + HBFA. Conditions: <math>r_{BC}=0.74</math>, <math>r_{AB}=2.3</math>, <math>p_{BC}=0</math> and <math>p_{AB}=-2.7</math>.]]

For the successful trajectory above, energy is channelled efficiently into high states of HBFA vibration (shown by the oscillating AB distance). This high vibrational excitation in product HBFA arises from the release of HC-HB bond repulsion while the HB-FA distance is still large; the large zero point vibration in HCHB introduces variability in the observed HBFA vibrational distribution.<ref name="e5" /> In simpler terms, trajectories that give rise to large vibrational energy of product(s) typically cut the corner of the energy surface and approach the exit valley from the side.

One way of confirming the formation of vibrationally excited HBFA is the method of ''infrared chemiluminescence'' in which emission of infrared radiation can be detected as HBFA returns to its ground state. The populations of the vibrational states of the product can then be determined. An alternative method relies on ''laser-induced flourescence'' in which HBFA is excited from a specific vibration-rotation level using lasers. ''Crossed molecular beams'' can also be employed.<ref name="e1" />

By running 20 Dynamics calculations (1000 steps, <math>p_{BC}=p_{AB}=0</math>, <math>r_{BC}=0.745</math> and <math>1.75<r_{AB}<1.85</math>), the approximate location of the early transition state for this reaction was determined to be <math>r_{BC}=0.745</math>, <math>r_{AB}=1.8107</math> (see above diagram). The early transition state shows that this reaction surface is ''attractive'' and it is well known that such surfaces proceed more efficiently if the energy supplied is largely translational.<ref name="e1" />

The energy of the products was determined using an MEP calculation (400,000 steps, <math>p_{BC}=p_{AB}=0</math>, <math>r_{BC}=0.745</math> and <math>r_{AB}=1.81</math>) to give -134.016 kcal/mol (see diagram below).

[[File:Einsteinfh2product.png|500px|thumb|none|(MEP) energy level diagram going from TS to products. ]]

The energy of the reactants was determined using an MEP calculation (400,000 steps, <math>p_{BC}=p_{AB}=0</math>, <math>r_{BC}=0.745</math> and <math>r_{AB}=1.82</math>) to give -104.004 kcal/mol (see diagram below).

[[File:Einsteinfh2product1.png|500px|thumb|none|(MEP) energy level diagram going from TS to reactants.]]

The energy of the transition state, <math>\Delta \text{G}^{\ddagger}</math>, in both energy level diagrams (above) is -103.752 kcal/mol. This means that the activation energy of the reaction, <math>E_\text{a}</math>= +0.252 kcal/mol. This very small activation energy is in accordance with the reaction's vigour.

The table below explores the effect of changing the amount of translational energy provided to the system.
{| class="wikitable sortable" border="1"
! <math>p_{BC}</math> !!class="unsortable" | Contour Diagram (<math>r_{BC}=0.74</math>, <math>r_{AB}=2.0</math>, <math>p_{AB}=-0.5</math>) !! Total energy !! class="unsortable" |Reactive?
|-
| -3 ||[[File:Einsteinfh2blue.png|400px ]] || -96.254 || ✗
|-
| -2 ||[[File:Einsteinfh2blue1.png|400px ]] || -100.754 || ✗
|-
| -1 ||[[File:Einsteinfh2blue2.png|400px ]] || -103.254 || ✗
|-
| 2 ||[[File:Einsteinfh2blue3.png|400px ]] || -98.754 || ✗
|-
| 2.3 ||[[File:Einsteinfh2blue4.png|400px ]] || -97.314 || ✓
|-
| 2.5 ||[[File:Einsteinfh2blue5.png|400px ]] || -96.254 || ✗
|-
| 3 ||[[File:Einsteinfh2blue6.png|400px ]] || -93.254 || ✗
|-
|}
It is clear from the above table that the amount of translational energy provided is not solely responsible for a successful trajectory. Rather, the dominant variable is suggested to be the vibrational phase in HCHB at the moment FA approaches closely.<ref name="e5" /> The following contour diagram proves this point.

[[File:Einsteinweird.png|700px|thumb|none|Low energy Dynamics trajectory with following conditions: <math>r_{BC}=0.74</math>, <math>r_{AB}=2.3</math>, <math>p_{AB}=-0.8</math> and <math>p_{BC}=0.1</math>. The success of this clearly due to factors other than initial translational energy.]]

Overall, this reaction of is an example of an attractive surface (with an early transition state) - the success of a trajectory on such a surface, according to ''Polanyi's rules'',<ref name="e6" /> is assured by high initial translational energy. However, the results show that other factors such as correct vibrational phase of HCHB are also necessary.

{{fontcolor1|gray|(Hugely impressive! Well done. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:51, 14 May 2018 (BST))}}

=== H + HF<math>\rightarrow</math> H2 + F===
The reverse reaction, H + HF<math>\rightarrow</math> H2 + F, is therefore endoergic with <math>\Delta\text{H}\simeq</math>-104.004-(-134.016)= +30.012 kcal/mol.

A successful reaction trajectory across this repulsive (late TS) surface is shown below.

[[File:Einsteinfinal.png|700px|thumb|none|Dynamics trajectory with following conditions: <math>r_{BC}=0.94</math>, <math>r_{AB}=3</math>, <math>p_{AB}=-5</math> and <math>p_{BC}=4.3</math>. The TS is encountered late into the reaction.]]

Using similar calculations as used earlier, the approximate location of the late transition state for this reaction was determined to be <math>r_{BC}=0.730</math>, <math>r_{AB}=1.021</math> (see above diagram).

{{fontcolor1|gray|(How come these values are different from your previous TS? The PES is the same, having just been mirrored along the diagonal, meaning that the TS coordinates are the same, just reversed. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:51, 14 May 2018 (BST))}}

The energy of the reactants was determined using an MEP calculation to give -133.781 kcal/mol (see diagram below).

{{fontcolor1|gray|(Diagram missing? [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:51, 14 May 2018 (BST))}}

The energy of the transition state, <math>\Delta \text{G}^{\ddagger}</math> is -74.1589 kcal/mol. This means that the activation energy of the reaction, <math>E_\text{a}</math>= +59.622 kcal/mol.

{{fontcolor1|gray|(The TS energy is the same regardless of forward or backward reaction. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 17:51, 14 May 2018 (BST))}}

==References==
<references>

<ref name= "e1">P. Atkins and J. de Paula, ''Atkins' Physical Chemistry'', Oxford University Press, Oxford, 10th edn., 2014, ch. 21, 881-928. </ref>

<ref name= "e2"> Wolfram Mathworld, http://mathworld.wolfram.com/SecondDerivativeTest.html, (accessed May 2018).</ref>

<ref name= "e3"> R. D Levine, ''Molecular Reaction Dynamics'', Cambridge University Press, Cambridge, 2009, 202-212. </ref>

<ref name= "e4"> J. Chen, Z. Sun and D. H. Zhang, ''J. Chem. Phys.'', 2015, '''142''', 1-11.</ref>

<ref name= "e5">J. C. Polanyi and J. L. Schreiber, ''Faraday Discuss. Chem. Soc.'', 1977, '''62''', 267-290.</ref>

<ref name= "e6">Z. Zhang, Y. Zhou, D. H. Zhang, G. Czakó and J. M. Bowman, ''J. Phys. Chem. Lett.'', 2012, '''3''', 3416-3419.</ref>
</references>

MRD:gianpiero

2018-05-14T15:40:31Z

Fjs113:

{{fontcolor1|gray|(Whilst we don't strictly ask for an intro and a conclusion, it would've been nice to see at least some sort of structural elements, e.g. headers and a table of contents. Remember: You're writing a wiki entry after all. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:40, 14 May 2018 (BST))}}

Q1:

At both a minimum and transition state the gradient will have a value of zero. The minima is an absolute minimum whilst the transition state is a saddle point. Using second derivatives will allow us to distinguish between the two; the second derivative at the transition state (saddle point) will give a positive and a negative value (representing the two orthogonal axes) whilst the second derivative at the minimum will give a single positive value.

{{fontcolor1|gray|(How can one derivative be both positive and negative? You need to mention partial derivatives as well as along which coordinates they take what values. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:40, 14 May 2018 (BST))}}

Q2:

The transition state position can be determined using an Internuclear Distance vs Time plot which will give a straight line at the transition state distance due to a lack of internuclear oscillations. The calculated value was '''rts''' = 0.9079 a.u.

[[File:plot-5.png|300px|thumb|center]]

{{fontcolor1|gray|(Why does the transition state lack oscillations? Also, how was the value calculated? A little more in-depth explanation would've shown understanding. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:40, 14 May 2018 (BST))}}

Q3:

MEP calculations gave a flat line with zero momentum whilst the dynamics calculation gave a reaction path that shows oscillations. So whilst the MEP shows the most direct path to the transition state, the dynamics calculation gives us an insight into molecular vibrations.

{{fontcolor1|gray|(No - the MEP is the path along the steepest descent on the PES, which is '''not''' the most direct path to the TS. In most cases, the MEP trajectory will move away from the TS, as the TS is an unstable stationary point. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:40, 14 May 2018 (BST))}}

[[File:surface_plot.png|300px|thumb|left|Contour Plot for the Dynamics Calculation]]

[[File:surface_plot-1.png|300px|thumb|center|Contour Plot for the MEP Calculation]]

Q4:
{| class="wikitable"
!
!p1
!p2
!E
!R/U
|-
|1
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|R
|-
|2
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|U
|-
|3
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|R
|-
|4
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|U
|-
|5
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|R
|}

[[File:plot.png|250px|thumb|center|1. This reactive trajectory shows the formation of diatomic BC and monatomic A from AB and C. The energy of the system is sufficient to overcome the transition state activation barrier; whilst the reactant AB has no vibrational energy, the product BC does due to the conversion of translational energy to vibrational energy upon collision. ]]

[[File:plot-1.png|250px|thumb|center|2. Here we have an unreactive trajectory where the energy of the system is not sufficient to overcome the transition state activation barrier. Rather than reacting upon collision, the diatomic AB simply returns with greater vibrational energy gained from the translational energy involved with the collision.]]

[[File:plot-2.png|250px|thumb|center|3. This is a reactive trajectory with both reactant and product diatomics acquiring vibrational energy.]]

[[File:plot-3.png|250px|thumb|center|4. This trajectory shows a brief conversion to products followed by a return to the original reactants, resulting in an overall unproductive reaction. This is due to the high vibrational energy meaning the product BC can immediately dissociate back into its constituents ]]

[[File:plot-4.png|250px|thumb|center|5. Overall this is a productive reaction; first the product is formed followed by reversal back into the reactants and then back to products. Again, when diatomic BC is initially formed, it has so much vibrational energy that it immediatley dissociates. B then reacts with A forming AB with enough vibrational energy to dissociate, finally reforming BC upon subsequent collision of B and C. ]]

Q5:

The main assumption of transition theory is that the transition state is in equilibrium with the reactants and products (Quasi-equilibrium assumption)<ref>D. C. Elton, Energy Barriers and Rates - Transition State Theory for Physicists, 2013, pp.5.</ref>. It is thus assumed that the rates of formation of the transition state from both reactant collisions and product degradation are equal. The theory also assumes that the reaction always proceeds via the lowest enrgy saddle point, but this is not always the case (especially at higher temperatures).

{{fontcolor1|gray|(Is the quasi-equilibrium assumption useful in this case? We're look at a triatomic system in isolation instead of a statistical ensemble. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:40, 14 May 2018 (BST))}}

For the results obtained here, we would expect transition state theory to give lower rates of conversion than experimental values, simply because the systems are not in true equilibrium.

{{fontcolor1|gray|(Unfortunately wrong: The main assumption of TST is to not take barrier re-crossing into account. You can see this effect happening in the last two cases in the previous question. The non-recrossing condition will lead to a higher predicted reaction rate. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:40, 14 May 2018 (BST))}}

Q6:

The reaction between hydrogen and F is exothermic and thus the reverse reaction is endothermic. This is in direct relation to bond strengths as the HF bond is stronger than the hydrogen bond; thus energy must be added to the system in order for the reverse reaction to occur.

{{fontcolor1|gray|(Please use the program to answer this question! Hint: look at the PES and notice the difference in potential energy in the reactant and product channels. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:40, 14 May 2018 (BST))}}

Q7:

The transition state can be identified by varying the internuclear distances until straight lines are obtained on the Internuclear Distance vs Time plot. The H-H transition state distance was found to be 0.744 a.u. whilst the H-F transition state distance was found to be 1.812 a.u.

{{fontcolor1|gray|(Simply stating the values is not enough! How did you obtain these numbers? Show some evidence that this is indeed the TS, e.g. a distances vs time plot. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:40, 14 May 2018 (BST))}}

Q8.

Perturbation of the transition state allows for calculation of the activation energy. Using an energy vs time graph, whilst running an MEP calculation, gave an activation energy =0.283 kcal/mol for diatomic hydrogen reacting with fluorine. Whilst the reverse reaction was found to have an activation energy = 29.980 kcal/mol using the same procedure. This is consistent with our previous findings that the reverse reaction is endothermic (due to the strong HF bond) and thus requires more energy to overcome the activation barrier.

{{fontcolor1|gray|(Again, how were these values obtained? Bare numbers are meaningless without evidence produced by the simulation software. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:40, 14 May 2018 (BST))}}

Q9:

A reactive set of initial conditions was determined: rF-H= 2.2, rH-H=0.84, pF-H=-0.75, pH-H=-1.5 a.u. It is evident, from the conservation of energy, that reaction energy is converted into vibrational energy upon formation of the products. IR analysis could be used to confirm this experimentally in which we would expect to see peaks corresponding to relaxation from the vibrationally excited state to the ground state.

{{fontcolor1|gray|(How were the initial conditions obtained? Please provide an explanation, even if it was just an educated gess. Regarding the experiment: This is a good start. Would've been nice if you elaborated a bit further. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:40, 14 May 2018 (BST))}}

Q10.

The Polanyi rules state that vibrational energy is more efficient in promoting a late-barrier reaction than translational energy.<ref>Z. Zhang, Y. Zhou , J. Phys. Chem. Lett., 2012, 3 (23), pp 3416–3419</ref> For the reaction of dihydrogen with F, we observe an early transition state which is similar in structure to the reactants. Conversely, the reverse reaction has a late transition state and hence is a more efficient reaction by Polanyi's rules.

{{fontcolor1|gray|(You should have supported this with some plots and elaboration how the different sets of initial conditions are consistent with Polanyi's rules. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:40, 14 May 2018 (BST))}}

References

MRD:01053372

2018-05-14T15:21:48Z

Fjs113: Marked by fjs113

= Molecular Reaction Dynamics: Applications to Triatomic systems =

==EXERCISE 1: H + H2 system==

===Potential Energy Surfaces===

'''Q1:''' What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

The Potential Energy surface of a system shows how the potential varies with distance in a 3D plot. Two points of interest are minima and transition states.

{{fontcolor1|gray|(Which distance do you mean exactly? You need to explain that this is a triatomic system and that we're looking at the interatomic distances. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:21, 14 May 2018 (BST))}}

At a minimum and at a transition structure the gradient has a 0 value. For a transition state there is a saddle point, this means in one direction there is a minimum and so 0 gradient, and in the orthogonal direction there is a maximum and so 0 gradient. They can be identified on a potential energy surface by finding the "flat" parts of the surface, for example by starting a trajectory near the transition state and seeing which way the reaction goes, to reactants or products (The transition state will be in the opposite direction).

{{fontcolor1|gray|(This is all correct. It would've been nice if you used mathematical terms such as first and second order partial derivatives to characterise the minimum and the saddle point. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:21, 14 May 2018 (BST))}}

===Estimating the Transition State===

'''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.

To find the TS first a surface plot of the system was used to identify where the likely positioning would be. Since the system is symmetric it was assumed the TS would lie along the XY diagonal, therefore r(AB) and r(BC) would be equal for the TS. Following this an initial trajectory was set up where p(AB) and p(BC) were 0, and the position along the XY diagonal was varied. Eventually when the 'Internuclear Distance' vs time graph showed no vibration, it was clear the TS had been reached (the system did not move, so no forces were acting on it and therefore the gradient on the PES was 0)

{{fontcolor1|gray|(Perfect explanation! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:21, 14 May 2018 (BST))}}

r(ts)= 0.908 to 3 s.f.

It can be seen from the Internuclear Distance vs Time graph that at 0.908 a.u for AB and AC there is no oscillation, which means we are on a point of 0 gradient on the potential energy surface, which is the TS saddle point.

[[File:Q2JNL01053372.png|thumb|centre| A Distance vs Time graph for a trajectory starting at the TS with 0 momentum]]

===MEP===

'''Q3:'''Comment on how the mep and the trajectory you just calculated differ.

An MEP was carried out with initial conditions of r1 = r(ts)+0.01, r2 = r(ts)

The MEP results is the minimum energy path, this is achieved by resetting the velocity to 0 at each time step. Therefore with the MEP no oscillations are observed in the BC bond distance. On the other hand the reaction path from the same starting position has an oscillating BC bond, as the initial inertia in BC direction sets up a vibration.

[[File:Mep01053372.png|thumb|centre| MEP Trajectory of HHH System]]

[[File:Dyn01053372.png|thumb|centre| Dynamic Trajectory of HHH System]]

An inversion of momentum procedure was then carried out, by taking the las Geometry from the MEP trajectory and reversing the momentum. This resulted in forming a system which ended in the TS.

Last Geometry:

Final r(AB): 9.004725826979717

Final r(BC): 0.7400803088831149

Final p(AB): -2.4806505081787176

Final p(BC): -0.9082751884031437

[[File:InverseMom01035772.png|thumb|centre|Inverse Momentum Trajectory]]

{{fontcolor1|gray|(Good! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:21, 14 May 2018 (BST))}}

===Reactive Trajectories===

'''Q4:'''Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

{| class="wikitable" border=1
! Set !! p1 !! p2 !! total energy !! reactivity
|-
| i || -1.25 ||-2.5 || -99.018 || Yes || [[File:IC_i_01053372.png|thumb]] || Atom A approaches the B-C molecule, they collide, and have enough energy to move over the transition state. An A-B bond is formed which oscillates due to the energy from the collsison. The B-C bond is broken so the C molecule atom away from A-B molecule.
|-
| ii || -1.5 || -2.0 || -100.456 || No || [[File:IC_ii_01053372.png|thumb]] || Atom A approaches an oscillating B-C molecule, they collide, but do not have enough energy to pass over the transition state. This results in the B-C bond remaning and the C atom and B-C molecule 'bouncing' away from each other
|-
| iii || -1.5 || -2.5 || -98.956 || Yes || [[File:Ic_iii_01053372.png|thumb]] || Atom A approaches an oscillating B-C molecule, they collide, and have enough energy to move over the transition state. An A-B bond is formed which oscillates due to the energy from the collsison. The B-C bond is broken so the C molecule atom away from A-B molecule.
|-
| iv || -2.5 || -5.0 || -84.956 || No || [[File:Ic_iv_01053372.png|thumb]] || Atom A approache the B-C molecule, they collide and form an A-B molecule and a C atom. However the A-B molecule has so much excess energy from the collision that it's oscillations lead to a reforming of the B-C bond, which moves away from the A atom. This is known as barrier crossing.
|-
| v || -2.5 || -5.2 || -83.416 || Yes || [[File:Ic_v_01053372.png|thumb]] || Atom A approaches the B-C molecule, they collide and form an A-B molecule and a C atom. The A-B molecule has so much excess energy it reforms the B-C molecule. The B-C molecule in turn vibrates with a great enough amplitude to reform the A-B molecule and C atom, which move away from eachother. This is double barrier crossing.
|}

===Transition State Theory===

'''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?

Transition state theory describes how a chemical reaction will occur based on the “transition state” that this reaction has and it is based upon collision theory.

There are three main assumptions in Transition State Theory:
1) In a multi step mechanism we assume each intermediate exist for long enough to achieve a Boltzmann distribution of energies before it continues. This may not be true, and the momentum of the reaction trajectory can lead to selectivity.
2) We assume that the atomic nuclei behave according to classical mechanics. However quantum mechanical tunneling can lead to a reaction even if the energy barrier is higher than the reactants energy. This affect can be important for low activation energy reactions.
3) We assume that the system passes over the lowest energy saddle point. For high temperatures with complex vibrations occurring, this may not be true. <ref name="TST" />

In General TST will over estimate the rate. This is because we are assuming in (1) that each intermediate exists in Boltzmann distribution quantities, when this is not really true. So there will be less intermediates present in reality and therefore a lower rate.
On top of the for more complex systems, we have not considered the angle of approach, If there is a required angle of approach some trajectories, although reactive from a kinetic perspective, will not actually lead to a reaction.
Finally by assuming in (2) that the systems behave classically, we have ignored the effects of Quantum tunneling.

{{fontcolor1|gray|(Well researched answer. However, you did not mention that classical TST does not consider barrier recrossing, an effect that you can see happening in the last two simulations in the previous question. This is mainly the reason why TST will over-estimate the reaction rate in classical conditions. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:21, 14 May 2018 (BST))}}

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

===Reaction Enthalpy===

'''Q1:''' 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?

The Bond energies for HF and HH are given in the following table:<ref name="Bond" />

{| class="wikitable" border=1
! Bond !! Energy
|-
| H-F || 565 kJ/mol
|-
| H-H || 432 kJ/mol
|}

F + HH is Exothermic, as can be seen from the decrease in energy of the Reactants to the products.
This makes sense because we form a stronger F-H bond and break a weaker H-H bond, releasing energy.
[[File:F+HH_Exo01053372.png|thumb|center|Exothermic Reaction]]

H + HF is endothermic, as can be seen from the increase in energy from the Reactants to the products.
This makes sense because form a weaker H-H bond and break a stronger H-F bond.
[[File:H+HF_endo01053372.png|thumb|center|Endothermic Reaction]]

===Estimating the Transition State===

'''Q2:''' Locate the approximate position of the transition state.

The transition state for the FHH system was not as easy to locate as the symmetric HHH system. Hammond's Postulate was used to make a guess at the position, and using the Contour Plot the initial starting position was moved opposite to the movement of the reactants, until no movement was seen.

'''r(ts) AB 1.806 a.u, BC 0.759 a.u'''

[[File:FHH_TS01035772.png|thumb|centre|Contour Plot of the TS]]

===Activation Energy===

'''Q3:''' Report the activation energy for both reactions

The Activation energy is the difference between the TS energy and the energy of the reactants, i.e the energy required to push the reactants over the TS. The energy of the TS was recorded as -103.676 from the r(ts). An MEP was then carried out with r(AB) = r(ts) ± 0.1 and the final geometry of those reactions was used to find the energy of the reactants of H+HF and F+HH.

[[File:TSMEP-1_01053372.png|thumb|centre| MEP with r(AB) = r(ts) - 0.1 ]]
[[File:TSMEP+1_01035572.png|thumb|centre| MEP with r(AB) = r(ts) + 0.1]]

 Activation Energy = T.S Energy - Reactant Energy 

H+HF: -103.676 + 133.941 = '''30.265'''

F+HH: -103.676 + 103.957 = '''0.281'''

===Reactive Trajectories===

'''Q3:''' In light of the fact that energy is conserved, discuss the mechanism of release of the reaction energy. How could this be confirmed experimentally?

The Initial Conditions which lead to a reactive trajectory where '''rF-H = 2.0, rH-H = 0.73, pF-H = -2.0, pH-H = -1.3'''

{{fontcolor1|gray|(Please explain/illustrate how you reached these initial conditions, even if it was just an educated guess! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:21, 14 May 2018 (BST))}}

[[File:FHHReacTraj01053372.png|thumb|center|F+HH Reactive Trajectory]]

The Mechanism for this process can be deduced from the Momentum vs Time graph. It can be seen that the momentum of the initial vibration of the H2 molecule and the momentum of the H2 and F species moving towards each other is very small compared to the final vibration of the F-H molecule formed. From the surface plots of this reaction we know it is exothermic, and due to the law of conservation of energy, it is logical to say that the released energy was converted to vibrational energy in the F-H molecule.

This could be confirmed through IR Spectroscopy. Where the F-H molecule which would be in a vibrationaly excited state would display overtone bands.

[[File:FHHReacTraj_Momentum01053372.png|thumb|center| F+HH Momentum vs Time Graph]]

===Polanyi's empirical rules===

'''Q4:'''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.

John Polanyi in 1972 proposed what is known as Polayni's rules, which determine whether vibrational or rotational energy is more likely to lead to a reactive trajectory. The rules apply to atom to di-atom collisions and states that vibrational energy is more efficient in promoting a late transition state reaction, whereas translational energy is more likely in promoting a early transition state reaction.<ref name = Pol />

Hammond's postulate states that a transition state will resemble the species closest in energy to it. Therefore an exothermic reaction has an early transition state resembling the high energy reactants, and an endothermic reaction has a late TS resembling the high energy products.

The F + HH reaction, is Exothermic (early TS) and therefore according to Polayni's rules a reactive trajectory is favoured by translational energy.
Initial Conditions r(FH)=2 r(HH)=0.74
It was observed that at lower values of p(HH) a reaction was more likely, and at higher values of p(HH) it was less likely.
[[File:F+HH_HighTrans01035772.png|thumb|center|F+HH with a High value of Translational Energy p(HH)=-7, P(FH)=-14]]
[[File:F+HH-3_01053372.png|thumb|centre|F+HH with a High value of Vibrational Energy p(HH)=-3, p(FH)=-0.5]]

The H + HF reaction is endothermic (Late TS) and therefore according to Polayni's rules a reactive trajectory is favoured by vibrational energy.
Inital Conditions r(HH) = 2, r(HF) = 0.9
It was observed that at higher values of p(HF), the reaction was likely to be reactive.
[[File:H+HF_HighVib01053372.png|thumb|center| H + HF with a high Value of Vibrational energy p(HH)=-2, P(HF)=8.4]]
[[file:H+HF_lowVib01053372.png|thumb|center| H + HF with a low value of Translational energy p(HH)=-8.4 p(HF)=-8.4]]

{{fontcolor1|gray|(Some great examples here! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 16:21, 14 May 2018 (BST))}}

<references>
<ref name="TST">https://en.wikipedia.org/wiki/Transition_state_theory.</ref>
<ref name="Bond">http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html</ref>
<ref name="Pol">Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction
Zhaojun Zhang, Yong Zhou, Dong H. Zhang, Gábor Czakó, and Joel M. Bowman
The Journal of Physical Chemistry Letters 2012 3 (23), 3416-3419
DOI: 10.1021/jz301649w</ref>

</references>

MRD:CH1516

2018-05-10T20:36:19Z

Fjs113: Marked by fjs113

== Exercise 1 ==

Q: What value do the different components of the gradient of the potential energy surface have at a minimum and at a transition structure? Briefly explain how minima and transition structures can be distinguished using the curvature of the potential energy surface.

A: r1 = distance of H1-H2; r2 = distance of H2-H3

There are two minima and one transition structure. Point 1 is the transition state, while point 2 and point 3 are minima. Because they are all stationary points, their first derivatives are always zero. The difference lies in the second derivatives.

For a minimum, both ∂2V/∂r12 and ∂2V/∂r22 are larger than 0, while for a transition structure, ∂2V/∂r12 > 0, and ∂2V/∂r22 < 0.

[[File:HCYex1q1.png|700px]]

{{fontcolor1|gray|(Almost perfect. At the TS, the second-order partial derivates have opposite signs along the reaction coordinate and its orthogonal coordinate. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:36, 10 May 2018 (BST))}}

Q: 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.

A: Steps = 830; Cutoff = -77 Kcal/mol.
My best estimate of transition structure in this situation is AB distance = BC distance = 0.91202188.

[[File:E1Q2ts.png|700px]]

The distance is chosen because as shown in the graph, the point does not roll down and is stay at a stationary point region. The reason why it does not roll down, after 830 steps, must because it does not experience a net force. Therefore, the force on it is zero. Also, force is the gradient of the potential energy diagram, so it can be concluded that the gradient at that point is also zero. Therefore, it is a stationary point, and also the transition state.

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

A:
[[File:E1Q3dynamics.png|700px]]

The trajectory under dynamics condition

[[File:E1Q3mep.png|700px]]

The trajectory under MEP condition

Firstly, the dynamics trajectory goes through a long journey while MEP only moves through a small distance. This is because is the MEP condition, the velocity at any moment is reset to zero, hence the velocity would not be accumulated. In other words, the inertia is removed. Secondly, the dynamics trajectory is wavey. This is because while move through the potential surface, the molecule is still vibrating and the wave is due to the change in bond length from vibration. However, there is no vibration is MEP because of the same reason: once the equilibrium is reached, there is no velocity in the system and the molecule simply could not vibrate anymore and just stays at the equilibrium position.

{{fontcolor1|gray|(Very good! [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:36, 10 May 2018 (BST))}}

Q: Look at the “Internuclear Distances vs Time” and “Internuclear Momenta vs Time”. Take note of the final values of the positions r1(t) r2(t) and the average momenta p1(t) p2(t) at large t.

What would change if we used the initial conditions r1 = rts and r2 = rts+0.01 instead?

A:
when steps = 200, and r1 = rts + 0.1

{| class="wikitable" border=1
! !!final distance !!final momentum
|-
| A-B||3.44 || 2.47
|-
|B-C|| 0.74 || 1.21
|-
| A-C||4.14 || 1.85
|}

when r2 = rts + 0.1

{| class="wikitable" border=1
! !!final distance !!final momentum
|-
| A-B||0.74 || 1.21
|-
|B-C|| 3.44 || 2.47
|-
| A-C||4.14 || 1.85
|}

When r2 is increased instead, the second situation just simply a reverse of the first situation - A-B forms the molecule instead of B-C. The final momentum is reversed too.

{{fontcolor1|gray|(A few plots would've illustrated this better. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:36, 10 May 2018 (BST))}}

Q: 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.

A:

[[File:zuihou.png]]

As can be seen, there momentum would cross off at around 0 at some time simultaneously!

{{fontcolor1|gray|(What are the initial conditions used to obtain this? And what does the plot mean? We're looking for an explanation here. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:36, 10 May 2018 (BST))}}

Q: Complete the table by adding a column with the total energy, and another column reporting if the trajectory is reactive or unreactive. For each set of initial conditions, provide a plot of the trajectory and a small description for what happens along the trajectory.

A:

{| class="wikitable" border=1
! set !!p1 !! p2!! energy !! reactive?
|-
| 1||-1.25 || -2.5 || -98.6|| Yes
|-
|2|| -1.5 || -2.0 || -100.1||No
|-
| 3||-1.5 || -2.5 || -98.75||Yes
|-
|4|| -2.5 || -5.0 || -84.8 ||No
|-
|5|| -2.5 || -5.2 || -83.0 ||Yes
|}

Set1

[[File:HCY_set1.png|400px]]

Description: This is a reactive system. C approaches A-B while A-B is not vibrating, then B and C formed a molecule, and moves away from A while vibrating.

Set2

[[File:HCY_set2.png|400px]]

Description: This is an unreactive system. C approaches A-B while A-B is vibrating, then C collides with A-B but does not have enough energy to overcome the energy barrier and react. Therefore, C is bounced off and moves in another direction.

Set3

[[File:HCY_set3.png|400px]]

Description: This is a reactive system. C approaches A-B while A-B is vibrating, then B and C formed a molecule, and moves away from A while vibrating. The difference from set 1 is the initial state of A-B.

Set4

[[File:HCY_set4.png|400px]]

Description: This is an unreactive system. C approaches A-B while A-B is not vibrating, then C collides with A-B and crosses the energy barrier. However, the system then crosses back the barrier and regenerates the reactants. Therefore, C is bounced off and moves in another direction in a slow speed. A-B is collided and starts to vibrate violently

Set5

[[File:HCY_set5.png|400px]]

Description: This is a reactive system. C approaches A-B at a fast speed, then C collides with A-B and crosses the energy barrier. However, the system then crosses back the barrier and regenerates the reactants. Therefore, C is bounced off and moves in another direction in a slow speed.

Q: 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?

A:
Main assumptions:
1. reactant molecules are distributed among their states in accordance with Boltzmann distribution.
2. Molecular systems that have crossed the trasition state in the direction of products cannot turn around and reform reactants.
3. In the transition state, motion along the reaction coordinate may be separated from the other motions and treated classically as a translation.

In the case of set 4 and set 5, transition state theory deviates from reality, and it over-estimate the rate of the reaction

{{fontcolor1|gray|(Please give some justification for this claim. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:36, 10 May 2018 (BST))}}

== Exercise 2 ==

In this exercise, A = F, B = H and C = H

Q: 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?

A:
The F + H2 reaction is exothermic and the H + HF is endothermic as can be seen in the diagram.

{{fontcolor1|gray|(How is this seen in the diagram? You need to explain the difference in the depth on the PES and relate that to exo/endothermic. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:36, 10 May 2018 (BST))}}

[[File:NMB1.png]]

The bond energy of H-H is 435 kJ/mol; the bond energy of H-F is 569 kJ/mol. Forming H-F would be exothermic.

{{fontcolor1|gray|(Where did these values come from? -> Reference. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:36, 10 May 2018 (BST))}}

Q: Locate the approximate position of the transition state.

A: Steps = 4000; Cutoff = -86 Kcal/mol; A = F, B = H, C = H
At transition state; AB distance = 1.8112 A; BC distance = 0.7440 A; AB momentum = BC momentum = 0

{{fontcolor1|gray|(Good. Would've been even better if you gave some justification/derivation. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:36, 10 May 2018 (BST))}}

[[File:EX2Q2.png]]

As can be seen form the graph, the system does not fall down the potential surface under dynamics condition after 1300 steps. This means it is at a transition state, as followed from the same argument from exercise 1.

[[File:E2Q21.8113.png]]

[[File:E2Q21.8111.png]]

To confirm that this is a transition state, the distances are slightly changed. When the A-B is increased from 1.8112 to 1.8113, the system rolls down to the left, while when the A-B is decreased to the 1.8111, the system rolls down to the right. This is an indication of transition state because if it is a minimum, the system will quickly revert back to the minimum, rather than rolling all the way down to two directions.

Q: Report the activation energy for both reactions.

A: The system is slightly displaced from equilibrium position to A-B = 1.8113. This is because I want it to roll down the hill from F-H-H transition state to F and H2 reactants, hence I can locate the energy of the transition state and the reactant. The potential energy vs. time is shown below.

[[File:E2Q3act.png|500px]]

As can be seen, because the system is only slightly displaced (from 1.8112 to 1.8113) from transition state, so where it starts is basically the energy of transition state, which is -103.75 kCal/mol to two decimal place. The reactants (F and H2) energy can also be read at -104.02 kCal/mol. The activation energy from F and H2 direction is hence 0.27 kCal/mol.

To determine the reactant energy of H and H-F, the initial distance of H-F is set at 1.1, which then rolls down the hill to the H and H-F side, and hence the potential energy could be read off. The graph is shown below.

[[File:E2Q3actf.png|500px]]

It can be easily seen that the reactant energy is -133.83 kCal/mol, hence the activation energy from H and H-F side is 30.07 kCal/mol, which is much larger than the other direction.

{{fontcolor1|gray|(Good explanation. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:36, 10 May 2018 (BST))}}

Q: Setup a calculation starting on the side of the reactants of F + H2, at the bottom of the well rHH = 0.74, with a momentum pFH = -0.5, and explore several values of pHH in the range -3 to 3 (explore values also close to these limits). What do you observe? Note that we are putting a significant amount of energy (much more than the activation energy) into the system on the H - H vibration.

A:

{| class="wikitable" border=1
! pHH value !!picture !!reactive?!! pHH value !!picture !!reactive?
|-
| 1.9||[[File:1.9.png|200px]] ||yes ||-1.9|| [[File:-1.9.png|200px]]||yes
|-
| 2.0||[[File:2.0a.png|200px]] ||yes ||-2.0|| [[File:-2.0.png|200px]]||yes
|-
|2.8|| [[File:2.8.png|200px]]||no ||-2.8|| [[File:-2.8.png|200px]]||no
|-
|2.9|| [[File:2.9.png|200px]]||yes ||-2.9|| [[File:-2.9.png|200px]]||no
|-
|3.0|| [[File:3.0.png|200px]]||no ||-3.0|| [[File:-3.0.png|200px]]||no
|-

|}

According to the results, it can be seen that the reaction is easier to happen when the pHH value is small. Another observation is that when pHH is positive, the reaction is easier to happen. This could be justified by a positive momentum means force in separating the H-H bond, which could help generating the product.

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

A:

When the condition is set as: A-B distance = 2, B-C distance = 0.74, AB momentum = -0.5, BC momentum = 2.0. The pathway is reactive as shown below.

[[File:hcygv.png|400px]]

The internuclear momentum vs. time graph is shown below. Two things can be seen. Firstly the bond vibration of the molecule get more violently after reaction. Secondly, the momentum of H-H becomes positive, which means whey are moving away from each other. This means traslational energy is being transferred to kinetic energy. In experiment, this could be measured by measuring the temperature of the reaction mixture. When translational motion is reduced, the temperature could be expected to decrease.
[[File:hcyav.png|400px]]

{{fontcolor1|gray|(Careful - not the translational energy is converted, but potential energy is converted to vibrational energy, the latter of which is a type of kinetic energy. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:36, 10 May 2018 (BST))}}

Q: For the same initial position, increase slightly the momentum pFH = -0.8, and considerably reduce the overall energy of the system by reducing the momentum pHH = 0.1. What do you observe now?

A: The profile is shown below. The total energy of the system has been reduced quite a lot, but it is still reactive. This means translational energy is probably more important in crossing the activation barrier.

[[File:nitama.png|400px]]

Q: 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.

A: The total energy was set to be fixed. Momentum of AB + BC = 8.5, different situations were tried out.

{| class="wikitable" border=1
! pHF value !!pHH value !! picture !!reactive?
|-
| 0.0|| -8.5||[[File:-8.5.png|300px]] ||yes
|-
| 0.5||-8.0||[[File:0.5-8.png|300px]] ||yes
|-
| 1.0||-7.5||[[File:1-7.5.png|300px]]||yes
|-
|1.5||-7.0|| [[File:1.5-7.png|300px]]||no
|-
|2.0|| -6.5||[[File:2-6.5.png|300px]]||no
|-

|}

It can be seen that when total energy is conserved, the larger the translational contribution, the more likely the reaction to react. Also, according to Polanyi's rule, vibrational energy is more efficient in promoting a late-barrier reaction than translational energy, and the reverse is true for a early transition state reaction. This reaction is a early transition state reaction, so the observation here fits in the rule.

MRD:sk716-2

2018-05-10T20:11:54Z

Fjs113: Marked by fjs113

== Exercise 1: H + H2 ==
[[File:SK - TS internuclear distance vs time.PNG|thumb|Plot of internuclear distance vs. time, showing a virtually linear lines for TS system]]

=== Potential energy vs. trajectory surface ===
A plot of potential energy vs. trajectory can be used to investigate the reaction with the above reactants. The gradients along both axes of a minimum in the aforementioned graph is 0; the gradient of a transition state in both directions is 0 too. A transition state would be a saddle point on this surface so they could be distinguished by calculating the second partial derivative: the second partial derivative of a transition state along one axis would be negative, and along the other axis it would be positive; whereas this value would be positive along both axes for a minimum.

{{fontcolor1|gray|(Overall, good discussion. However, the second partial derivatives at the TS are not calculated along either axes, but rather along the reaction coordinate and its orthogonal axis. E.g. in the case of HHH, this is turned by 45° with respect to the axes. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:11, 10 May 2018 (BST))}}

=== Transition state ===
Estimate transition state distance, rts, = 0.908 Å. The internuclear distance vs time plot shows that there is no oscillation in the lines representing the AB and BC distances: at the TS, there is no internuclear vibration as AB and BC are at the dissociation limit.

=== Changing the calculation type ===
* Dynamics: since the atoms have a momentum realistically, this method shows the molecular vibrations.
* MEP: since the velocity resets to 0 after each step, the H2 molecule formed shows no vibrations, indicated by the smooth A-B curve. Reversing the dynamics calculation doesn't reverse the reaction.
{| class="wikitable"
!Dynamics calculation
!MEP calculation
!Dynamics calculation, reversed trajectory
|-
|[[File:Sk - r1 = rts + delta, distance vs time, dynamics.PNG|frameless]]
|[[File:Sk - r1 = rts + delta, distance vs time, mep.PNG|frameless]]
|[[File:Sk - r1 = rts + delta, distance vs time, reversed.PNG|frameless]]
|}

=== Reactive and unreactive trajectories ===
{| class="wikitable"
!p1
!p2
!Reactive?
!Internuclear distance vs. time plot
!Description
|-
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|Yes
|[[File:Sk - Rxn test 1.PNG|frameless]]
|HC collides with initially stationary HA-HB molecule. After colliding, the new hydrogen shows oscillation.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|No
|[[File:Sk - Rxn test 2.PNG|frameless]]
|The H atom approaches the hydrogen molecule, but then is repelled away.
|-
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|Yes
|[[File:Rxn test 3.PNG|frameless]]
|The same thing occurs as in the first scenario.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|No
|[[File:Sk - Rxn test 4.PNG|frameless]]
|The reaction initially occurs, but then the barrier is crossed and the reactants are reformed.
|-
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|Yes
|[[File:Sk - Rxn test 5.PNG|frameless]]
|The reaction occurs as before. The hydrogen molecule has a noticeably higher vibrational energy, shown by the more frequent oscillations of the B-C line.
|}

{{fontcolor1|gray|(Missing total energy data. Also, a contour plot would've illustrated the results from the different initial conditions better. Lastly, a more in-depth discussion of each case, mentioning how the different momenta influence different types of energies, would've shown a deeper understanding. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:11, 10 May 2018 (BST))}}

=== Transition state theory ===
Transition State Theory assumes that there is a "quasi-equilibrium" between the reactants and activated complex and not between the latter and the products.<ref>''IUPAC Gold Book - transition state theory''. [online] Goldbook.iupac.org. Available at: <nowiki>http://goldbook.iupac.org/html/T/T06470.html</nowiki> [Accessed 4 May 2018].
</ref> It also assumes the if the reactants do not collide with an energy above the activation energy, the reaction will not occur. Activation energy predictions using this theory may slightly differ from experimental values for reactions with a low activation energy because it doesn't take quantum mechanics into account: quantum tunneling could lower the activation energy, so the reactants won't have to collide with as much force.

{{fontcolor1|gray|(Yes, but this is only half the story: Barrier recrossing (as seen in cases 4&5) is not taken into account by TST. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:11, 10 May 2018 (BST))}}

== Exercise 2: H2 + F ==

=== Classifying reactions ===
The trough corresponding to the H-F bond is deeper than the one for the H-H bond, meaning that the H-F bond is stronger than the H-H bond. In other words, the potential energy of the reactants is less negative than that of the products. Therefore, the F + H2 --> HF + H reaction is exothermic and the reverse reaction is endothermic.

{{fontcolor1|gray|(This should've been supported by a figure. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:11, 10 May 2018 (BST))}}

=== Transition state ===
Using Hammond's postulate, the transition state should be early as F is the most electronegative atom, i.e. the H-H bond distance at the transition state should be close to the bond distance of a free H2 molecule (0.74 A). The approximate calculated H-H bond length at the TS was 0.745 A and a H-F bond length of 1.8107 Å

{{fontcolor1|gray|(Some more details of the derivation are needed here. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:11, 10 May 2018 (BST))}}

=== Activation energy ===
(Parameters: MEP, F-H = 1.7507, H-H = 0.745, 10000 steps). TS = -104, H2 = -103.869 and HF = -135 kcal mol-1
* Activation energy of forward reaction = +0.131 kcal mol-1
* Activation energy of reverse reaction = +31.131 kcal mol-1.

{{fontcolor1|gray|(Again, a picture or two would've supported these rather bleak and bland numbers. Simply stating the values and some parameters is not enough: Why those particular parameters? [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 21:11, 10 May 2018 (BST))}}

=== Reaction dynamics of F + H2 ===
The pHF value reaches high values in the program, but since energy has to be conserved, this vibrational energy is released as heat through internal conversion, thus allowing HF to go to its ground vibrational state. You could confirm this by measuring the enthalpy change using calorimetry or using a thermometer.

With the parameters of rHH = 0.74, rFH = 2.3 and pFH = -0.5, the following was observed: a pHH value of -2.8 led to a reaction, but deviation of 0.5 or more away from this value led to barrier crossing and the reaction didn't occur, despite the high amount of energy going into the H-H vibration.
{| class="wikitable"
!pHH = -2.8
!pHH = -2.75
!pHH = -2.85
|-
|[[File:Sk - F + H2 contour, pHH -2.8.PNG|frameless]]
|[[File:Sk - F + H2 contour, pHH -2.75.PNG|frameless]]
|[[File:Sk - F + H2 contour, pHH -2.85.PNG|frameless]]
|}
Using the values of pHF = -0.8 and pHH = 0.1,

=== Reaction dynamics of H + HF ===
For rHF = 0.935 and rHH = 2.3, it was difficult to yield a reaction with a low vibrational energy and a high translational energy; a very high translational energy energy was needed for a reaction to occur, whereas a relatively lower vibrational energy was needed to the same effect. This concurs with Polanyi's rules of a high vibrational energy being more effective in allowing a reaction to progress than translational energy for a reaction with a high activation energy.<ref>Zhang, Z., Zhou, Y., Zhang, D., Czakó, G. and Bowman, J. (2012). Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction. ''The Journal of Physical Chemistry Letters'', 3(23), pp.3416-3419.
</ref>

An exothermic reaction leads to an early transition state and the converse is true for an endothermic reaction. According to Polanyi's rules, the translational energy doesn't have much of an effect in dictating whether a reaction occurs if it has an early transition state, unlike the vibrational energy (vice versa for a late transition state. Therefore, the program concurs with the predictions of these rules, and therefore for the forward reaction, the translational energy would be the more dominant factor.<ref>Zhang, W., Kawamata, H. and Liu, K. (2009). CH Stretching Excitation in the Early Barrier F + CHD3 Reaction Inhibits CH Bond Cleavage. ''Science'', 325(5938), pp.303-306.
</ref>
{| class="wikitable"
!pHF = -5
pHH = -10
!pHF = -15
pHH = -8
|-
|[[File:Sk - H + HF reaction 3 (-.5, -10).PNG|frameless]]
|[[File:Sk - H + HF reaction 3 (-15, -8).PNG|frameless]]
|}

MRD:jo416

2018-05-10T19:46:58Z

Fjs113: Marked by fjs113.

== H + H2 System ==

===Dynamics from the transition state region:===

In a potential energy surface, the transition state is located at the saddle point on the surface. This is the highest point on the lowest energy reaction path. Along two orthogonal directions, transition states can be distinguished from minima by their second derivatives. A saddle point is characterized by a negative value for the second derivative with respect to one component and a positive value with respect to the other component. A minima would have positive values for the second derivatives with respect to both components.

{{fontcolor1|gray|(Good discussion, but what do you mean by components? Use terms like the reaction coordinate along the reaction pathway and the orthogonal coordinate. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 20:46, 10 May 2018 (BST))}}

===Locating the transition state:===

The location of the transition state was determined by making an initial guess based on some of the properties inherent to the system and then minimizing fluctuations in internuclear distance vs. time plots. The H + H2 system is symmetrical, therefore (r1) and (r2) were kept equal throughout and the momenta were set to 0.

[[File:1.5r.png|400px|centre|thumb| Fig 1: Internuclear distance vs time where p=o and r=1.5 Å]]

By looking at the AB and BC vectors the second guess reduced the distance to r= 0.9 Å.

[[File:0.9r.png|400px|centre|thumb| Fig 2: Internuclear distance vs time where p=o and r=0.9 Å]]

This guess led to a substantial reduction in fluctuations in the plot. The best estimate for (rts) was r= 0.908 Å. This resulted in an almost flat plot:

[[File:0.908r.png|400px|centre|thumb| Fig 3: Internuclear distance vs time where p=o and r=0.908 Å]]

At this value the reaction pathway showed an almost completely static dot. Therefore, the transition state was found at r= 0.908 Å.

[[File:0.9075rc.png|400px|centre|thumb| Contour Plot where p=o and r=0.908 Å]]

{{fontcolor1|gray|(Perfect! Good derivation. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 20:46, 10 May 2018 (BST))}}

=== Calculating the Reaction Path: ===

After the transition state has been located, the trajectory the represents the reaction path or minimum energy pathway (MEP) can be determined. The MEP trajectory is determined by resetting the velocity of the system to zero at each time step. The MEP was investigated by setting the initial conditions of the system to: (r1)= (rts) Å , (r2)= (rts) + 0.1 Å, and both momenta still equal to 0. These adapted conditions result in the reaction no longer being static at TS but 'rolling' downhill.

This behavior is illustrated by the internuclear distance vs. time plot and the contour plot below:

[[File:mep1.png|400px|centre|thumb| Fig 4: MEP contour plot, showing reaction 'rolling']]

[[File:mep1i.png|400px|centre|thumb| Fig 5: MEP internuclear distance vs. time plot, showing reaction 'rolling' (B-C diverging) and the absence of internuclear vibrations]]

As the internuclear distance vs. time plot shows, the MEP trajectory halts any internuclear vibrations, this is an unrealistic model of the molecules.

[[File:mep1id.png|400px|centre|thumb| Fig 6: MEP contour Plot, showing reaction 'rolling' with internuclear vibrations present.]]

{{fontcolor1|gray|(This is neither a MEP nor a contour plot, but rather the Distances vs Time plot for the Dynamics equivalent. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 20:46, 10 May 2018 (BST))}}

Due to the symmetry of the reaction, changing the initial conditions to: (r2)= (rts) Å , (r1)= (rts) + 0.1 Å, has no effect on the reaction dynamics apart from different bonds vibrating and changing distance.

{{fontcolor1|gray|(Good discussion showing understanding. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 20:46, 10 May 2018 (BST))}}

=== Reactive and Unreactive Trajectories ===

With the knowledge that reactions with the initial conditions: r1 = 0.74 and r2 = 2.0, -1.5 < p1 <-0.8, and p2 = -2.5 will pass the transition state and go to completion, it would be intuitive to suggest that any reaction with the same initial values for internuclear distance but larger momenta, would also go to completion. To investigate this assumption, a series of starting conditions were tested and their results

{| class="wikitable" border=1
! p1 !! p2 !! Total Energy !! Outcome
|-
| -1.25 || -2.5 || -99.018 || Reactive
|-
| -1.5 || -2.0 || -100.456 || Unreactive
|-
| -1.5 || -2.5 || -98.956 || Reactive
|-
| -2.5 || -5.0 || -83.956 || Unreactive
|-
| -2.5 || -5.2 || -83.416 || Reactive
|}

[[File:RT1.png|400px|centre|thumb| Fig 7: Contour plot showing reaction crossing the energy barrier and going to completion.]]

[[File:RT2.png|400px|centre|thumb| Fig 8: Contour plot showing reaction with insufficient energy to overcome the barrier and returning to reactants.]]

[[File:RT3.png|400px|centre|thumb| Fig 9: Contour plot showing reaction crossing the energy barrier and going to completion.]]

[[File:RT4.png|400px|centre|thumb| Fig 10: Contour plot showing reaction crossing the energy barrier before returning to the reactants.]]

[[File:RT5.png|400px|centre|thumb| Fig 11: Contour plot showing reaction crossing the energy barrier before initially returning to the reactants and eventually going to completion, so called 'barrier recrossing'.]]

=== Transition State Theory ===

Transition state theory is the idea that a reaction can be described as a reactants forming an activated complex at a transition state, which then decays into products. It relies on a few major assumptions. The first major assumption is that the system can be treated with classical mechanics, this ignores and quantum effects. This results in the theory assuming that unless molecules collide with enough energy to reach the transition state, no reaction will occur. In reality tunneling can lead to reactions proceeding without meeting this energy requirement.

The second major assumption of the theory is that all reactions with the required energy will pass over the lowest energy saddle point and decay into products. As shown in the results above, this is not strictly true. For systems with low total energy (Fig. 7), this is the case. However, for systems with significantly higher total energies, behaviors such as barrier recrossing can be observed (Fig. 10) this is not accounted for in the theory. Furthermore, the theory does not account for the fact that there are reactions which are successful, despite not ever crossing the lowest energy saddle point (Fig. 11). This would lead to TS theory underestimating the reaction rate compared to experimental values.

{{fontcolor1|gray|(Good explanation. Well done for referring back to the previous figures. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 20:46, 10 May 2018 (BST))}}

== F-H-H system ==

=== Potential Energy Surface Inspection ===

By inspection of the potential energy surfaces for the F-H-H system, insight into the reaction energetics and bond strengths of the chemical species involved could be obtained. This was achieved by An arbitrary set of conditions were used for the first surface and then r1 and r2 were reversed.

[[File:H2hfs.png|400px|centre|thumb| Fig 12: Showing the potential surface for H2 + F -> HF + H. The deeper channel for the HF side shows the higher bond strength relative to H2. The overall downhill pathway of the trajectory shows the exothermic nature of the reaction.]]

[[File:Hfh2s.png|400px|centre|thumb| Fig 13: Showing the potential surface for HF + H -> H2 + F. The shallower channel for the H2 side shows the weaker bond strength relative to HF. The overall uphill pathway of the trajectory shows the exothermic nature of the reaction. Under the same conditions as before the the reaction is not successful in this reverse direction, as seen in the diagram.]]

The process of locating the TS is more difficult due to the asymmetry of the reaction (i.e. r1 =/= r1). {{fontcolor1|gray|(Surely you mean r2 here ;) [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 20:46, 10 May 2018 (BST))}}
This process was aided by consideration of Hammond's postulate. From the information in the surfaces (fig. 12, fig. 13) the first, exothermic reaction will have a transition state closer to the reactants and vice versa for the endothermic, reverse reaction. By trial and error the transition state for the reactions were as follows:

[H2 + F -> HF + H]; rab = 1.8098 Å, rbc = 0.7454 Å

[[File:H2hfts.png|400px|centre|thumb| Fig 14: Locating the H2 + F -> HF + H, TS]]

[HF + H -> F + H2]; rab = 0.7454 Å, rbc = 1.8098 Å

[[File:Hfh2ts.png|400px|centre|thumb| Fig 15: Locating the H2 + F -> HF + H, TS]]

===Activation Energy===

The activation energy for the reactions were determined by finding the difference between the energy at the transition state and the energy where rab was very large.

For the reaction to form HF: Ea = + 0.44 kcal/mol (lit. 0.7-3.4 kcal/mol, the inaccuracies reflect the limitations of transition state theory.)

For the reaction to form H2: Ea = + 17.1 kcal/mol

{{fontcolor1|gray|(Where did you get these values from? This could be illustrated using a MEP Energy vs Time plot showing the drop in potential energy. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 20:46, 10 May 2018 (BST))}}

===Reaction Dynamics===

Initial conditions of: rab = 2.0 Å, rbc = 0.74 Å; pAB = -0.40, pBC = -1.41, were determined to result in a successful reaction to form HF.

{{fontcolor1|gray|(Where did you get these initial conditions from? Was it an informed guess? Did you use the inversion of momentum procedure?[[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 20:46, 10 May 2018 (BST))}}

[[File:Succhf.png|400px|centre|thumb| Fig 16: Successful reaction trajectory surface plot]]

A plot of energy vs. time shows the conservation of energy through the reaction (symmetry) and that the reaction results in an increase in kinetic energy and a reduction in energy. An increase in kinetic energy means an increase in vibrational energy which is converted into heat energy. This could be measured by calorimetry.

{{fontcolor1|gray|(Did you really mean "reduction in energy"? It would've been better if you elaborated slightly more on the increase in kinetic energy -> translational/vibrational. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 20:46, 10 May 2018 (BST))}}

[[File:hfe.png|400px|centre|thumb| Fig 17: Energy vs. time plot]]

A new calculation was set up with the initial conditions rHH = 0.74, rHF = 2.40, pHF = -0.5, and -3< pHH <3. Several values for pHH in the range were investigated. Towards the extreme limits of the inequality it was found that the reactions had so much vibrational energy that they passed the transition state before returning to the reactants.

[[File:2.9.png|400px|centre|thumb| Fig 18: pHH = 2.9 ]]

{{fontcolor1|gray|(As far as I can see, this reaction does not return to the reactants. [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 20:46, 10 May 2018 (BST))}}

[[File:-2.9.png|400px|centre|thumb| Fig 19: pHH = -2.9 ]]

When the initial conditions were adjusted to pFH = -0.8 pHH = 0.1, the reaction went to completion.

[[File:-0.8.png|400px|centre|thumb| Fig 20: pHH = 0.1 ]]

Focusing now on the reverse reaction, to form H2. The initial conditions: rHH = 2.25, rHF = 0.74, pHH = -0.739, and pHF =0.098. This resulted in a reactive trajectory. This was obtained by reducing the value for pHH from an arbitrarily high value and increasing the other momentum from a low value.

This behavior illustrates how the vibrational energy of H-F follows Polanyi's empirical rules. The late transition state means that increasing the vibrational energy of H-F will make it easier to cross over the transition state barrier. High translational energy would result in the reaction initially crossing the TS before returning to reactants. For an early transition state the behavior is reversed.

[[File:finale.png|400px|centre|thumb| Fig 21: Successful trajectory]]

MRD:fjs113 testing

2018-05-09T17:00:06Z

Fjs113:

Hello World!

{{fontcolor1|gray|(write your comment in this space [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 18:28, 8 May 2018 (BST))}}

<nowiki>~~~~</nowiki>

MRD:fjs113 testing

2018-05-08T17:28:16Z

Fjs113:

Hello World!

{{fontcolor1|gray|(write your comment in this space [[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 18:28, 8 May 2018 (BST))}}

MRD:fjs113 testing

2018-05-08T13:23:45Z

Fjs113:

MRD:fjs113 testing

2018-05-08T13:23:26Z

Fjs113:

MRD:fjs113 testing

2018-05-08T13:22:58Z

Fjs113: Created page with "Hello World! ~~~~Comment try no1~~~~"

Hello World!

[[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 14:22, 8 May 2018 (BST)Comment try no1[[User:Fjs113|Fjs113]] ([[User talk:Fjs113|talk]]) 14:22, 8 May 2018 (BST)