MRD:hz7718

Exercise 1: H + H₂ system

Grade: 3/5 - Some good features to this report. Some of your numerical estimates were a little off, and you failed to complete the final excercise. Other individual comments are provided throughout.

Transition state

Mathematically, transition state is a saddle point which is on a graph of a function where slopes in orthogonal directions are zero.

For finding transition state on graph, the slopes of both r₁ and r₂ directions should be zero, which means δV/δr₁ = 0 and δV/δr₂ = 0. Also, if it is the maximum point in one direction, it should be minimum point in the orthogonal direction. In math, (δ²V/δr₁²)*(δ²V/δr₂²) - δ²V/(δr₁*δr₂) < 0.

A minimum stationary point means in both directions, it is the minimum point with zero gradients (δ²V/δx² > 0 and (δ²V/δr₁²)*(δ²V/δr₂²) -δ²V/(δr₁*δr₂) > 0). But transition state is the minimum point in one direction but a maximum point in the other orthogonal direction.

Where do these directions lie on your PES though?

saddle point on a graph	Minimum stationary point plot	Maximum stationary point plot

locating the transition state

When the system is at transition state, the potential energies of both p₁ and p₂ are zero. And because all the three atoms are in equilibrium, the change of the energy is zero, too. It means that the gradients of potential energy surface is zero and the force acting on atoms is zero, which indicates there is no oscillation of the three atoms and r₁ and r₂ will keep constant. On the graph, r₁ and r₂ equal to 90.7pm, and two lines are almost perfectly straight without oscillating, which means it is(or close to) the transition state. So r_ts=90.7pm.

A good estimate for the TS, well done!

Trajectories from r₁=r_ts+δ, r₁=r_ts

In MEP type graph, the trajectory follows the valley floor to H₁ + H₂-H₃. But in the dynamic type graph, the shape of trajectory is wavy. This is because on mep, all atoms have zero kinetic energy(velocities and momenta are zero), which causes that B-C bond has no vibration. However, in dynamic type, kinetic energy is included in the system, so B-C bond will vibrate so the values of r₂ will fluctuate.

What causes these momenta to be zero though? How does the code enforce this property?

MEP graph, r₁=91.7pm, r₂=90.7pm, p₁=p₂=0	Dynamic graph, r₁=91.7pm, r₂=90.7pm, p₁=p₂=0

Reactive and unreactive trajectories

p₁/ g.mol^-1.pm.fs^-1	p₂/ g.mol^-1.pm.fs^-1	E_tot/ kJ.mol^-1	Reactive?	Description of the dynamics
-2.56	-5.1	-414.28	YES	H_C atom and H_A-H_B molecule approach to each other at the beginning. When the system passes the transition state region, new bond H_B- H_C will form and H_A-H_B bond will break. Then H_A atom and H_B-H_C molecule move away to each other. Before H_B-H_C forming, the trajectory is a line with very little oscillation. This is because most kinetic energy of H_A-H_B is in translational energy. After H_B-H_C bond forming, the trajectory is wavy because most kinetic energy of H_B-H_C is in vibrational energy.
-3.1	-4.1	-420.08	NO	H_C atom and H_A-H_B molecule approach to each other at the beginning. But they do not pass the transition state region, H_A-H_B bond does not break and H_B-H_C bond does not form. Then H_C atom and H_A-H_B molecule move away to each other. H_A-H_B bond keeps vibrating due to the kinetic energy.
-3.1	-5.1	-413.98	YES	H_C atom and H_A-H_B molecule approach to each other very slowly because most of kinetic energy is in vibrational energy, which means the translational energy is very little. When H_A-H_B bond starts to break, H_B-H_C bond starts to form. After that, H_A atom and H_B-H_C molecule move away to each other. And H_B-H_C bond keeps vibrating.
-5.1	-10.1	-357.28	NO	H_C atom and H_A-H_B molecule approach to each other, when they passes the transition state region, H_A-H_B bond starts to break and H_B-H_C bond starts to form. After that, H_A atom moves away from H_B-H_C molecule but H_B-H_C molecule vibrates strongly. Then H_B-H_C bond breaks and H_A and H_B atoms approach to each other to form bond, which means this system recrosses the transition region and reverts to the reactants. In the end, vibrating H_A-H_B molecule and H_C atom move away to each other.
-5.1	-10.6	-349.48	YES	The system passes the transition region for three times: At first, H_C atom and H_A-H_B molecule get closer, H_B-H_C bond forms and H_A-H_B bond breaks. Then H_A atom returns to H_B-H_C molecule immediately and the system reverts to reactants. After that, H_B-H_C bond forms again and H_A-H_B bond breaks again. In the end, vibrating H_B-H_C molecule and H_A atom move away to each.

From the table above, we can know that even though the system has enough energy to react, it may still be unreactive.

Transition State Theory

In TST predictions, if the reactants with enough energy cross the transition state, it will never come back. However, the experimental results show the system can recross the transition state region and reform the reactants. SO, overall, the TST overestimates the reaction rate. And the TST has limitations: 1. High temperature, there are complicated vibrations which may lead the transition state far away from the saddle point of potential energy surface. 2. Quantum tunneling: Molecules and atoms can still tunnel across the barrier even with not enough energy. It will slightly underestimates the reaction rate. ¹

A sensible explanation of the shortcomings of TST. Some really good ideas here. Well done!

Exercise 2: F-H-H system

PES inspection

potential energy surface graph	Internuclear distance vs time plot

The potential energy surface graph show that the system H₂ and F has higher potential energy than HF and H. It means H₂ + F to HF + H reaction is exothermic and HF + H to H₂ + F reaction is endothermic, which indicates that the bond strength of H-F is stronger than H-H.

So, from H₂ + F to HF + H, the system needs to absorb energy from the environment, and from HF + H to H₂ + F, the system releases energy to the environment.

When it comes to the transition state location, it can be analyzed by Hammond's postulate. ² For example, in the reaction H₂ + F to HF + H which is an exothermic reaction. The transition state is close to the structure of the reactants(so it is an early transition state). It means the distance between F_A and H_B is very large and the distance between H_B and H_C is smaller, which matches what potential energy surface graph shows. Unlikely the H-H-H system, the transition state of F-H-H system is not symmetric.

In the Internuclear distance vs time plot, F_A-H_B is 180pm and H_B-H_C is 74pm. And both lines are almost straight, which means all atoms only slightly vibrate. So, the change of potential energy is zero.

Then we can know the black point in potential energy surface graph is the transition state(where F_A-H_B=180pm, H_B-H_C=74pm and pH₁=pH₂=0).

A sensible estimation. How did you confirm that this was a saddle point, and not a potnetial energy minimum though?

Activation EnergyFor finding the actvition energy, the steps are extended to 3500. When F_A-H_B is 180pm and H_B-H_C is 74pm, we can find the Contour plot 1 shows that the transition state finally forms HF + H. So, from Graph 1, we can know the activation energy is 121.6 kJ/mol for the reaction HF + H to H₂ + F.

Contour Plot 1	Graph 1

When F_A-H_B is 184pm and H_B-H_C is 74pm, the Contour plot 2 shows that the transition state finally forms H₂ + F. So, from Graph 2, we can know the activation energy is 0.03 kJ/mol for the reaction H₂ + F to HF + H.

A sensible estimate of the activation energy for FH+H. your estimate for F+H2 is a bit too low, likely because you've not let the reaction go to completion.

Contour Plot 2	Graph 2

Reaction dynamics

In reaction H₂ + F to HF + H, F_A-H_B is 184pm and H_B-H_C is 74pm, p1=-1 and p2=-2. From the contour plot and animation, the product HF keeps vibrating and H_C molecule moves away from HF. This is because the reaction is exothermic, the potential energy transfers to kinetic energy which includes vibrational energy and translational energy. However, from Momentum vs Time plot, it shows most of potential energy transfers to vibrational energy instead of translational energy because the H_B_A momentum fluctuates strongly but H_c-H_B momentum keeps a relatviely low value.

The energy released mechanism can be confirmed by IR Spectroscopy. Because H-F vibration has dipole moment and it is active in IR. We can just measure the peak absorbance of H-F vibration at different time after reaction, if the absorbance is large, the vibrational energy of H-F is more. We can also use calorimetry to measure the heat produced by this reaction, and the heat energy measured is the change of kinetic energy of the products including vibrational energy and translational energy.

No, this is incorrect. Simply looking at the peak of an IR spectrum is not enough? How would we detect vibrationally excited states in IR spectroscopy?

Contour Plot	Momentum vs Time

Translational energy VS Vibrational Energy

Graph 1 represents an exothermic reaction with an early transition state. And in exothermic reaction, translational energy can make the reaction more efficient because it can help the reactants pass the early transition state region. If most kinetic energy of reactants is vibrational energy, reactants cannot pass the early transition region in exothermic reaction.³

Graph 2 represent an endothermic reaction. Endothermic reaction has a late transition state. The reactants need more vibrational energy to cross the late transition state region. And translational energy cannot help reactants pass the late transition state.³

YOu need to provide some examples (i.e. trajectories for the F+h2 system) here to back up your assertion.

Graph 1	Graph 2

Reference

1.Veser, Götz. "Experimental and theoretical investigation of H2 oxidation in a high-temperature catalytic microreactor." Chemical Engineering Science 56.4 (2001): 1265-1273.

2.G.S. Hammond. A Correlation of Reaction Rates. J. Am. Chem. Soc. 1955, 77 (2): 334–338. doi:10.1021/ja01607a027.

3.J.C. Polanyi. Some Concepts in Reaction Dynamics. Science 1987, 236 (4802): 680-690. doi:10.1126/science.236.4802.680

Exercise 1: H + H2 system