View source for MRD:sjl1218

{{fontcolor1|green|Good report showing understanding of the task at hand. I have written some comments which is hopefully beneficial. I would advise adding references to your answers.[[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 03:09, 23 May 2020 (BST)}} 

==Exercise 1: H<sub>A</sub> + H<sub>B</sub>-H<sub>C</sub> -> H<sub>A</sub>-H<sub>B</sub> + H<sub>C</sub>==

[[File:sjl1218-reactionscheme.png|500px|center|thumb|Initial Conditions of Reaction Involving 3 Atoms: H<sub>A</sub> + H<sub>B</sub>-H<sub>C</sub> -> H<sub>A</sub>-H<sub>B</sub> + H<sub>C</sub>]]

===Dynamics of the Transition State Region===
====The Transition State====
[[File:Sjl1218-pesurface.PNG|300px|right|thumb|center|Potential Energy Surface Diagram of H<sub>A</sub> + H<sub>B</sub>-H<sub>C</sub> -> H<sub>A</sub>-H<sub>B</sub> + H<sub>C</sub>]]
In a potential energy surface diagram, the reaction pathway (black line) proceeds along the minimum energy path linked by the reactants and products. The transition state is the PES diagram saddle point defined mathematically by ∂V(ri)/∂ri=0.

The transition state location can be identified if V(r<sub>AB</sub>)/∂r<sub>AB</sub>=0 and V(r<sub>BC</sub>)/∂r<sub>BC</sub>=0. This point can be distinguished from the local minima if the outcome of taking the partial second derivative of potential energy with respect to either r<sub>AB</sub> or r<sub>BC</sub> is 0.

{{fontcolor1|green|Okay, but would the second derivative to a axis be negative whilst to another be positive if it were a TS. Consider what a saddle point is...also may be worth reading up on 'Hessian'.  [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 02:39, 23 May 2020 (BST)}}

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

Since the H<sub>2</sub> + H<sub>1</sub> surface is symmetric, the transition state must have r<sub>AB</sub>=r<sub>BC</sub>. The location of the transition state position r<sub>ts</sub> can be estimated when ∂V(ri)/∂ri=0 and there will be minimal oscillation among the 3 atoms as there is no force acting along r<sub>AB</sub> and r<sub>BC</sub>.

[[File:Sjl1218-4.PNG|400px|center|thumb|Model for when r<sub>AB</sub> = r<sub>BC</sub> and p<sub>AB</sub>=p<sub>BC</sub>=0]]

The transition state position r<sub>ts</sub> is estimated as 91 pm. The internuclear distances vs time graph for r<sub>AB</sub> = r<sub>BC</sub>=91pm shows a 'flat' line,  meaning no forces are acting at this position. 

[[File:Sjl1218-ts.PNG|300px|center|thumb|Internuclear Distances vs Time Graph for r<sub>AB</sub> = r<sub>BC</sub>=91pm and p<sub>AB</sub>=p<sub>BC</sub>=0]]

{{fontcolor1|green| How do you think you could better accurately measure the TS?  [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 02:48, 23 May 2020 (BST)}}

====Calculating the reaction path====
 
The reaction path or minimum energy path is an infinitely slow trajectory (momenta and velocities resets to zero in each time step) obtained after locating the transition state position. 

The minimum energy path initially shows a short horizontal line of the atoms entering the transition state position followed by the decreasing rate of atoms H<sub>A</sub>-H<sub>C</sub> and H<sub>B</sub>-H<sub>C</sub> increasing in distance. Atoms H<sub>A</sub>-H<sub>B</sub> forms a bond and remains constant in distance.

Even though the MEP outlines a reaction, it does not take into account the mass and inertia of the atoms, therefore does not provide the realistic picture of the reaction. 
The MEP and dynamic trajectory differs in that the realistic trajectory shows oscillations (potential energy interactions) between atoms. At first the 3 atoms are in the transition state positions, showed by the horiontal lines. The trajectory then shows a linear increase in H<sub>A</sub>-H<sub>C</sub> and H<sub>B</sub>-H<sub>C</sub> distances and H<sub>A</sub>-H<sub>B</sub> oscillates about the same position.
[[File:Sjl1218-mep.PNG|300px|center|thumb|Minimum Energy Path (r<sub>BC</sub> = r<sub>ts</sub>+1 pm, r<sub>AB</sub> = r<sub>ts</sub> and p<sub>BC</sub> = p<sub>AB</sub> = 0 g.mol-1.pm.fs-1)]]
[[File:Sjl1218-traj.PNG|300px|center|thumb|Dynamic Trajectory (r<sub>BC</sub> = r<sub>ts</sub>+1 pm, r<sub>AB</sub> = r<sub>ts</sub> and p<sub>BC</sub> = p<sub>AB</sub> = 0 g.mol-1.pm.fs-1)]]

===Reactive and Unreactive Trajectories===
Determining whether trajectories are reactive from their momenta when r<sub>BC</sub>=74 pm and r<sub>AB</sub>=200 pm.
{| class="wikitable"
!p<sub>1</sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>
!p<sub>2</sub>/ g.mol<sup>-1</sup>.pm.fs<sup>-1</sup>
!E<sub>tot</sub>
!Reactive?
!Description of the dynamics
!Illustration of the trajectory
|-
|<nowiki>-2.56</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-414.280</nowiki>
|Yes
|H<sub>B</sub>-H<sub>C</sub> collides with H<sub>A</sub> with sufficient kinetic energy to overcome the reaction barrier and forms H<sub>A</sub>H<sub>B</sub> and H<sub>C</sub>.
|[[File:Sjl1218-ani1.png|200px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-4.1</nowiki>
|<nowiki>-420.077</nowiki>
|No
|H<sub>B</sub>-H<sub>C</sub> approaches H<sub>A</sub> but does not react due to lack of kinetic energy/momenta.
|[[File:Sjl1218-ani2.png|200px]]
|-
|<nowiki>-3.1</nowiki>
|<nowiki>-5.1</nowiki>
|<nowiki>-413.977</nowiki>
|Yes
|H<sub>B</sub>-H<sub>C</sub> collides with H<sub>A</sub> with sufficient kinetic energy to overcome the reaction barrier and forms H<sub>A</sub>H<sub>B</sub> and H<sub>C</sub>.
|[[File:Sjl1218-ani3.png|200px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.1</nowiki>
|<nowiki>-357.277</nowiki>
|No
|H<sub>B</sub>-H<sub>C</sub> collides with H<sub>A</sub> and enters a temporary transition state but insufficient kinetic energy does not allow a reaction to occur. Atoms return to their starting positions.
|[[File:Sjl1218-ani4.png|200px]]
|-
|<nowiki>-5.1</nowiki>
|<nowiki>-10.6</nowiki>
|<nowiki>-349.477</nowiki>
|Yes
|H<sub>B</sub>-H<sub>C</sub> collides with H<sub>A</sub> with just enough kinetic energy to overcome the potential energy between H<sub>B</sub>-H<sub>C</sub> for the reaction to occur.
|[[File:Sjl1218-ani5.png|200px]]
|}
The hypothesis that higher values of momenta would be reactive is not precisely correct, this table shows that p<sub>BC</sub> must be greater  than p<sub>AB</sub> by a specific factor to be able reactive.

{{fontcolor1|green|But does that mean it is always reactive, Row 4 suggests you can add to your conclusions (crossing of the TS?)[[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 02:51, 23 May 2020 (BST)}}

===Transition State Theory===

The transition state theory attempts to provide a grearter understanding of the reaction by predicting the rate as well as the theromodynamic parameters of the reaction. However, it has its limitations. The transition state theory neglects the effects of quantum tunneling such that particles can bypass the activation energy of the transition state. Additionally, systems that have already crossed the transition state and formed products can recross the barrier to reform the reactants. Consequently, the rate calculated by the transition state theory is greater than the experimental value for the reasons that quantum effects are not taken into account.

{{fontcolor1|green|Perfect! Do you know of some other TST assumptions? Additionally, you could reference these assumptions you have learnt.[[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 02:54, 23 May 2020 (BST)}}

==Exercise 2: F + H-H and H + HF==

===PES Inspection===

====1. F + H-H -> F-H + H====

Conditions: r<sub>AB</sub> = 150 pm, r<sub>BC</sub> = 80 pm, p<sub>AB</sub> = 0 g.mol-1.pm.fs-1, p<sub>BC</sub> = 0.5 g.mol-1.pm.fs-1.
[[File:Sjl1218-sur3.png|center|thumb|PES of F + H-H -> F-H + H]]

Products are lower in potential energy than reactants, therefore reaction is exothermic.The H-F bond formed releases more energy than the energy taken to break the H-H bond, therefore H-F is greater in bond strength than H<sub>2</sub>.

{{fontcolor1|green|Good job. We can supplement your findings with actual experimental H-F and H-H bond energies (quick literature search and referencing should do the trick!). [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 02:55, 23 May 2020 (BST)}}

====2. F-H + H -> F + H-H====

Conditions: r<sub>AB</sub> = 80 pm, r<sub>BC</sub> = 150 pm, p<sub>AB</sub> = 15 g.mol-1.pm.fs-1, p<sub>BC</sub> = 0 g.mol-1.pm.fs-1.
[[File:Sjl1218-sur4.PNG|center|thumb|PES of F-H + H -> F + H-H]]

Products are higher in energy than reactants, therefore reaction is endothermic. The energy used to break the H-F bond is greater than the energy released for forming the H-H bond. The H-F bond is greater in bond strength.

===Locating the Transition State Position===

The approximate position of the transition state is F-H = 180 pm and H-H = 75 pm. Horizontal line shown on the internuclear distance vs time graph represents minimal oscillations among the atoms, ∂V(ri)/∂ri=0 no forces are acting on the F-H-H direction so transition state is reached.

[[File:Sjl1218-ts1.png|center|thumb|Internuclear Distance vs Time graph for r<sub>AB</sub> = 180 pm, r<sub>BC</sub> = 75 pm, p<sub>AB</sub> = 0 g.mol-1.pm.fs-1, p<sub>BC</sub> = 0 g.mol-1.pm.fs-1]]

===Calculating the Activation Energy===

The activation energy can be calculated by subtracting the potential energy of the reactants from the potential energy the TS. The potential energy of the transition state is -433.943 kJ.mol-1. The potential energy of the reactants can be obtained by performing an MEP of energy vs time for the reaction and observing the starting potential energy. 

Activation energies of:

1. F + H-H -> F-H + H: -433.943 - -434.288 = +0.345 kj.mol-1
[[File:Sjl1218-6.png|center|thumb|Energy vs Time MEP for F + H-H -> F-H + H with same reaction conditions]]

2. F-H + H -> F + H-H: -433.943 - -496.862 = +62.919 kj.mol-1
[[File:Sjl1218-7.png|center|thumb|Energy vs Time MEP for F-H + H -> F + H-H with same reaction conditions]]

{{fontcolor1|green|Good method. I would have expected  value in the range of 1. 1 - 1.5 kJmol-1 and 2. 125-127 kJmol-1. Nonetheless, your methodology is in the right direction. [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 03:01, 23 May 2020 (BST)}}

===Reaction Dynamics===
====1. F + H-H -> F-H + H====

Initial conditions that result in a reactive trajectory: r<sub>AB</sub> = 150 pm, r<sub>BC</sub> = 80 pm, p<sub>AB</sub> = 0 g.mol-1.pm.fs-1, p<sub>BC</sub> = 0.5 g.mol-1.pm.fs-1.
[[File:Sjl1218-8.png|300px|center|thumb|Illustration of reactants: F + H-H]] [[File:Sjl1218-9.png|center|thumb|Illustration of products: F-H + H]]

[[File:Sjl1218-10.png|center|thumb|Energy vs Time MEP of F + H-H -> F-H + H]]

This reaction is exothermic but the total energy is conserved in a reaction, the energy release mechanism is explained below:  

Originally, H-H is a molecule held together by strong potential energy. As it approaches atom F, potential energy between H-H is transferred to kinetic energy which releases energy in the reaction. The momenta vs time graph (orange line) shows the spikes in momentum/ kinetic energy for H-H (B-C) when H (B) is transferred to atom F (A). The momentum of this orange line then flattens, all kinetic energy is transferred to the formation of F-H (A-B) bond which shows oscillations in momentum by the blue line.

To confirm this experimentally, measure the change in temperature of the reaction with a thermometer which should show an increase in temperature as the reaction is exothermic.

{{fontcolor1|green|Sure we could use a technique to measure the temperature change. What about studying the vibrational energies?  [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 03:03, 23 May 2020 (BST)}}

====2. F-H + H -> F + H-H====

In this reaction, a late transition state is formed because of the high electronegativity of atom F. According to Hammond's postulate, the energy of the transition state is similar to the products because the structure of the late transition state resembles the products. 

Polanyi's rules states that vibrational energy is more efficient in promoting a reaction with a late barrier than translational energy. The two PES diagrams below shows the effectiveness of 1. high vibrational energy and 2. high kinetic energy, in overcoming the barrier. The first figure shows a reactive trajectory (A-B + C -> A + B-C) is achieved from a high distribution in vibrational energy. On the other hand, the trajectory was unreactive in the high kinetic energy case.

Conditions: r<sub>AB</sub> = 75 pm, r<sub>BC</sub> = 180 pm, p<sub>AB</sub> = 20 g.mol-1.pm.fs-1, p<sub>BC</sub> = 0 g.mol-1.pm.fs-1.
[[File:Sjl1218-11.png|center|thumb|1. High Vibrational Energy: PES of F-H + H -> F + H-H]]

Conditions: r<sub>AB</sub> = 75 pm, r<sub>BC</sub> = 180 pm, p<sub>AB</sub> = 20 g.mol-1.pm.fs-1, p<sub>BC</sub> = 10 g.mol-1.pm.fs-1.
[[File:Sjl1218-12.png|center|thumb|2. High Kinetic Energy: PES of F-H + H -> F + H-H]]

{{fontcolor1|green|Good. You could discuss the counter reaction of exothermic F + H2 with an early barrier. [[User:Mys18|Mys18]] ([[User talk:Mys18|talk]]) 03:07, 23 May 2020 (BST)}}