Rrp17

EXERCISE 1: H + H₂ system

If H collides with a H₂ molecule, there system can be defined as an AB and C. {{ This is a slightly pointless statement as "an AB and C" doesn't really mean anything! Make sure you are always clear and spell everything out very explicitly, i.e. saying the atoms can be labelled... etc.Rk2918 (talk) 13:25, 3 June 2019 (BST)

How is the transition state defined?

The transition state is the maximum on the minimum energy path that links reactants to products; this point is where, ∂V(r_i)/∂r_i=0, the gradient of the potential energy is zero and the distance r_i = r_AB = r_BC .

Plot 1:The surface plot of the reaction between H₁-H₂ and H₃. The energies are represented by red and blue colours (high and low potential energy) and the black line is the trajectory of the reaction.

It can be seen that the transition state can be distinguished from a local minimum by plotting a tangent at the inflection point of the minimum energy path. The first derivative, ∂V²/∂q1_q1²>0and the second derivative ∂V²/∂q2_q2²<0 This would give you a maximum however, a tangent perpendicular to the q_BC would derive the minimum; this location is known as the saddle point and thus the transition state of a reaction. The transition state is a saddle point, it is where the minimum , q₁ and max ,q₂ intersect.

Transition state position

H + H₂ is symmetric and hence, the transition state must have r₁ = r₂, the momentum are zero and there is no gradient at the ridge. The location of the transition state can can be seen in plot 2 and the transition state position (r_ts) was found to be 0.9077 Å. This was determined by choosing a random value of r₁ and changing the inter-nuclear distance until the force equalled zero.

Plot 2 A plot of inter-nuclear distance vs time for the bond distance of 0.907 Å.

Minimum energy path and trajectory

The trajectory is a reaction path of minimum energy where the velocities/momenta are set to zero in each time step. This is where we displayed the transition state position by r₁=r_ts + 0.01 and the momenta is zero. This is shown in plot 3.

However, MEP cannot characterise a reaction in terms of the motion of atoms. The reaction calculation under 'Dynamics' that is shown in plot 4.

It can be seen that the MEP plot is a plateaued line as the kinetic energy is reset to zero at every interval and, there molecules are not vibrating however, with the dynamic plot, the line is seen as oscillating as the molecules have momentum.

MEP	Dynamic
Plot 3 The MEP surface plot.	Plot 4 The trajectory surface plot

Reactive and nonreactive trajectories

caption

p1	p2	Total Energy (Kcal/mol)	Reactive?	Trajectory contour plot	Dynamic description
-1.25	-2.5	-99.018	yes		The trajectory passes through the transition state as the reaction has enough energy. The B-C bond breaks and the A-B bond forms.
-1.5	-2	-100.456	no		The trajectory does not pass through the activation barrier as there is not enough energy
-1.5	-2.5	-98.956	yes		The trajectory goes through the transition state as the reaction has sufficient energy to overcome the activation barrier and the B-C bond breaks whilst the A-B bond forms.
-2.5	-5	-84.956	no		The B-A bond formed however, the trajectory reverts back to the reactants and the B-C bond reforms.
-2.5	-5.2	-83.416	yes		the B-A bond forms, only to revert back and re-form the B-C bond. Subsequently, the B-A bond forms again.

It is clear that the higher the total energy the more likely the reaction will go to completion as the energy is greater than the activation energy. However, it can be seen that with the last two contour plots, the products do not form-another variable affects the outcome of the reaction other than activation energy!

Assumptions of Transition State Theory

The Transition State Theory explains how the rate of a chemical reaction by which the reaction passes through a transition state and it has three assumptions:

-The energy of the particles follow a Boltzmann distribution -The reactants and transition state structure is in equilibrium -The transition state structure does not revert back to the reactants once the reactants form the transition state

Exercise 2: F-H-H system

Using the energy surface, the F + H₂ is an exothermic reaction as the products are lower in energy than the reactants.This bond breaking process must mean that the H-F bond is stronger than the H-H and this relates t the ionic nature of the bond due to the large difference in electronegativity between the atoms. Thus, the reverse reaction must be endothermic.

Surface plot of the F-H-H system with AB=HF and BC=H₂

The approximate position of the transition state

Using a MEP contour plot, the saddle point (where there is no momentum) was determined. AB=1.908 Å , BC=0.745 Å. The use of an internuclear distance vs time also highlighted this.

Contour Plot	Zoom of Contour
Contour plot of trajectory	Trajectory zoomed in

Activation energy

The activation energy for both reactions was determined by calculating the difference between the reactants/products and the transition state using an energy vs time plot .

-For the H + HF reaction the activation energy was found to be 30 kcal/mol.

-For the H₂ + F reaction the activation energy was found to be 0.026 kcal/mol. This was determined by zooming in on the energy vs time plot: number = 5000 and the BC distance was deviated to 1.82 angstroms and the activation energy is the difference between the total and potential energies.

Finding the activation energy for the H₂ + F reaction

Finding the activation energy for the H + HF reaction

Mechanism for the release of the reaction energy

Given that this reaction is exothermic, the thermal energy released would result in an overall increase in kinetic energy and a fall of the potential. Moreover, this exothermic reaction can be determined with calorimetry with sufficient insulation to prevent heat loss.

The initial conditions that result in a reactive trajectory: HH distance of 0.74 HF distance of 2.3 HF momentum of -2.2 HH momentum of -1.9

The results are shown below and show that the potential energy is a mirror of the kinetic energy and the potential energy converts to kinetic- this is how energy is conserved.

Energy vs Time

Momentum vs Time

Distribution of energy between different modes

Polanyi's rules¹ explain that the efficiency of the reaction is affected by the distribution of energy between different modes; an exothermic reaction has an early transition state and translational energy plays a more fundamental role than vibrational energy in the efficiency of the reaction and vice versa. The r_HH momentum corresponds to the vibrational energy and the r_HF corresponds to the translational energy.

-For the H₂ + F reaction, it can be seen using a dynamic contour plot, when translational energy is greater than vibrational energy, the reaction is efficient and is complete. This exothermic reaction follows Polanyi's rules.

Incomplete	Complete
High translational energy leads to a complete reaction	Low translational energy does not lead to a complete reaction

-For the HF + F reaction, the transition state is late and vibrational energy, according to Polyani's rules, is more important than translational and allows a more efficient reaction for this endothermic reaction.

Complete	inComplete
High vibrational energy, reaction complete	Low vibrational energy, reaction incomplete

References

[1]Theoretical Study of the Validity of the Polanyi Rules for the Late-Barrier Cl + CHD3 Reaction, Zhaojun Zhang, Yong Zhou,† and Dong H. Zhang, dx.doi.org/10.1021/jz301649w | J. Phys. Chem. Lett. 2012, 3, 3416−3419