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 present as a saddle point along the product and reactant channel. The reactant channel has a potential energy minima where ∂V(r1)/∂r1=0, and the product channel has a minima where ∂V(r2)/∂r2=0. The transition state is a maximum along these channels, however the transition state is a minimum orthogonal to these channels (saddle point). Here ∂V2(ri)/∂(ri)2<0 at the saddle point and ∂V2(ri)/∂(ri)2 = 0 at the reaction channels.<gallery>
File:Ruman Surface Plot q1.png
</gallery>

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 point position is found when the A-C distance and B-C distance both 0.908Å. The potential energy surface point is evidence of this as it shows the bond to be single point at the saddle point. Furthermore, the Internuclear Distances vs Time shows the A-C and A-B plots to overlap, be constant throughout, indicating no oscillation to be present.<gallery>
File:Q2 surface plot ruman top 90888.png| Surface plot
File:Q2 suface plot side rumaaan 908.png| Surface plot (side view)
File:Q2 internuclear rumaan 908.png| Internuclear Distances vs Time
</gallery>'''Comment on how the ''mep'' and the trajectory you just calculated differ.'''

The MEP is the minimum energy path that corresponds to infinitely slow motion, where the velocity always reset to zero in each time step. The momenta is set to zero, hence no oscillation is present, this therefore displays a plot in which the trajectory is straight follows the valley floor to H1+ H2-H3.

The calculation type was reset to Dynamics, to show the effect mass has upon the trajectory. The plot obtained shows inertia to sway the motion of the atoms, hence leading to a more curved trajectory pattern.<gallery>
File:RUman MEPs plot surf.png| MEP Surface Plot (20,000 steps used)
File:RUman dynamics plot surf.png| Dynamics Surface Plot
</gallery>'''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"
!Experiment Number
!p1
!p2
!Total energy
!Reactive/Unreactive trajectory
!Plot
!Descriptoin
|-
|1
|<nowiki>-1.25</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-99.018</nowiki>
|Reactive
|[[File:Ruman no.1 surface.png|340x340px]]
|The trajectory shown reaches, then exceeds the transition state's saddle point, leading to the formation of products, indicated by the AB product channel.

|-
|2
|<nowiki>-1.5</nowiki>
|<nowiki>-2.0</nowiki>
|<nowiki>-100.456</nowiki>
|Unreactive
|[[File:Ruman no.2 surface.png|340x340px]]
|The trajectory does not reach the saddle point, hence the transition state is not reached, and BC product is not formed.

|-
|3
|<nowiki>-1.5</nowiki>
|<nowiki>-2.5</nowiki>
|<nowiki>-98.956</nowiki>
|Reactive
|[[File:Ruman no.3 surface.png|340x340px]]
|Here the trajectory surpasses the transition state, leading to the formation of the AB product. The energy is lower than that of experiment 1.

|-
|4
|<nowiki>-2.5</nowiki>
|<nowiki>-5.0</nowiki>
|<nowiki>-84.956</nowiki>
|Unreactive
|[[File:RUMAN CONTOUR EXP4.png|340x340px]]
|The trajectory shows a disordered pathway throughout as the system does not utilise the entire minimum energy reaction pathway. The transition state is reached, however the reaction does not fully proceed to the product pathway. Suggesting the trajectory is unreactive. Atom B oscillates between atoms A and C before reforming the initial AB reactant.

|-
|5
|<nowiki>-2.5</nowiki>
|<nowiki>-5.2</nowiki>
|<nowiki>-83.416</nowiki>
|Reactive
|[[File:RUMAN CONTOUR EXP5555.png|340x340px]]
|The trajectory shown here is similar to that of experiment 4. However this lower energy pathway proceeds down the product channel, as indicating a successful reactive pathway. Atom B oscillates between atoms A and C before reforming the initial BC reactant. The internuclear Vs time plot for experiment 5 is similar to that of experiment 4.
|}

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 two most basic assumptions of transition state theory 1:

1)Separation of electronic and nuclear motions, which is known as the Born-Oppenheimer approxiamtion in quantum mechanics

2)Reactant molecules are distributed among their states in accordance with the Maxwell-Boltzmann distribution

This theory makes three further assumptions:

3) Molecular systems that cross the transition state (forming products), cannot reform reactants

4) In the transition state, motion along the reaction coordinate can be treated as only transnational

5) When no equilibrium is present between products and reactants, transition states that are becoming products are distrusted among their states (follows the Maxwell-Boltzmann laws) Not necessary, follows from first assumption

<gallery>
File:Ruman no.4 surface try birds.png| Experiement 4 birds-eye view
File:Ruman no.4 interrr try.png | Experiement 4 Internuclear Distance Vs Time
File:Ruman no.5 surface.png |Experiment 5
File:Ruman no.5 surface topview.png | Experiment 5,Surface plot birds-eye view
</gallery>

The results obtained from experiments 4 and 5, does not follow assumption 3 of the transition state theory. Both experiments show the trajectory crossing the transition state, then returning to reactant energy channel. These simulations were run with higher momentum values than needed in order to react, thus leading to high energy trajectory's which do not act as expected.

=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 mean bond enthalpy for the H-H is 436 KJmol-1, whilst the H-F enthalpy is 565 KJmol-1. This therefore shows the F + H2 reaction to be exothermic as an excess of 129 KJmol-1 of energy is given out to the surroundings due to the formation of the stronger H-F bond. This therefore can be used to predict that the H + HF reaction will be endothermic as energy is needed to break the strong H-F bond and form the weak H-H bond 2.

'''Locate the position of the transition state'''

{| class="wikitable"
!System
!r1
!r2
!Transition state (kcal/mol)
|-
|F-H-H
|1.8106
|0.745
|<nowiki>-103.752</nowiki>
|}
<gallery>
File:F-h-h ruman transstate.png| Transition state

</gallery>

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

Calculation type was set to MEP, with 200,000 steps.

F + H2 System, the activation energy was found to be EA= 30.2533 kcal/mol. Where the r1 = 1.8306 and r2 = 0.745<gallery>
File:RUmaan Surface Plot activ h2.png| Surface plot of the activation energy
File:Ruman energy h2.png | Energy vs Time (potential energy)

</gallery>

H + F-H System, the activation energy was found to be EA= 0.246777. Where r1 - 1.8206 and r2=0.750 <gallery>
File:Ruman Surface Plot hf.png| Surface plot of activation energy
File:RUman energy hf.png|Energy vs Time (potential energy)

</gallery>

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

{| class="wikitable"
!System
!r1
!r2
!p1
!p2
!Total energy (kcal/mol)
|-
|F + H2
|2.5
|0.745
|<nowiki>-0.8</nowiki>
|0
|<nowiki>-103.752</nowiki>
|}
These conditions result in a reactive trajectory, allowing the formation of H-F product.

The first law of thermodynamic states "That the total energy of an isolated system is constant", This means that energy cannot be created or destroyed, but can be transformed from one form to another. The internuclear momenta vs time graph shows the momentum of A to be constant until B-C is in a reasonable range to react. This shows a large amount of potential energy being converted into vibration energy. This vibrational energy causes the bond to oscillate at a high rate. Furthermore this large transfer of energy from potential to vibrational can be confirmed experimentally by measuring the change in electromagnetic radiation emitted in the IR region as the reaction proceeds.<gallery>
File:Momtena vs time ruman h2 2.png| Internuclear Momenta vs Time graph
</gallery>

'''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 most important factor in determining the distribution of products is the location of the early barrier (in the reactants channel) and late barrier (in the products channel) on the reaction coordinate1.

Polanyi's empirical rules show that translational energy most effective to cross early barrier (early transition state, resembles reactants), whereas if the reactants vibrational energy is too high, the reaction may not proceed. This is the opposite for late energy barrier (late transition state, resembles product), where vibrational energy is preferred over translational energy.

'''F + H2 System (Earlier barrier)'''

{| class="wikitable"
!Situation
!rFH
!rHH
!pFH
!pHH
!Reactive/unreactive
!
|-
|High translational energy
|2.4
|0.7453
|<nowiki>-0.7</nowiki>
|<nowiki>-0.2</nowiki>
|Reactive
|[[File:Ruman early trans case.png|340x340px]]
|-
|High vibrational energy
|2.4
|0.7453
|<nowiki>-0.7</nowiki>
|<nowiki>-2</nowiki>
|Unreactive
|[[File:Ruman early vib case.png|340x340px]]
|}

'''H + H-F System (Late barrier)'''

{| class="wikitable"
!Situation
!rFH
!RHH
!pFH
!pHH
!Reactive/unreactive
!
|-
|High translational energy
|0.91
|2.4
|<nowiki>-7</nowiki>
|<nowiki>-7</nowiki>
|Unreactive
|[[File:Ruman late trans case.png|340x340px]]
|-
|High vibrational energy
|0.91
|2.4
|<nowiki>-0.1</nowiki>
|<nowiki>-9</nowiki>
|Reactive
|[[File:Ruman late vib case.png|340x340px]]
|}

The location of energy barrier also plays a key role in the formation of products, as it controls which distribution of energy modes in the reactants is most likely to result in reaction. For a given transition state, the favored reactant energy distribution is the inverse of the favored product energy distribution. Vibrational energy is most effective for passage over late barriers - high translational energy may not lead to reaction even if provided in exccess of the barrier height. An excess of translational energy may simply lead to mollecules "coliding" with the replusive walls of the potential surface, and "rolling" back down into the reactant well. Conversely, an early barrier is best overcome by translational energy. Excess vibrational energy will lead to molecules oscillating from side to side yet never reaching the top of the barrier. [CDK] These rules for understanding simple reaction dynamics are known as the Polanyi's rules. As F + H2 has an early transition state, high translatoinal energy is required for efficient reaction. The reverse is true for H + HF.

= References =

1) J. I. Steinfeld, J. S. Francisco, W. L. Hase Chemical Kinetic and Dynamics, Chapter 10, 297-310

2) P.Atkins and J. de Paula, Physical Chemistry, 10th Edition:2014

2018-05-18T15:41:38Z

Aa516:

File:Ruman early vib case.png

2018-05-18T15:41:38Z

Aa516:

File:Ruman late trans case.png

2018-05-18T15:41:37Z

Aa516:

File:Ruman late vib case.png

2018-05-18T15:41:37Z

Aa516: Aa516 uploaded a new version of File:Ruman late vib case.png

2018-05-17T17:01:14Z

Aa516:

File:Ruman energy h2.png

2018-05-17T17:01:13Z

Aa516: