MRD：01508610

EXERCISE 1: H + H2 system

We will be looking at collision and reaction between diatomic molecule of hydrogen and an atom of H in a linear configuration in the gas phase. In simple terms, the atom A collides with the molecule B-C and forms a new molecule with B, while C is detached as a separate atom. The interatomic distances between A and B, B and C are labelled as r_AB and r_BC respectively.

Dynamics from the transition state region

On a potential energy surface diagram, how is the transition state mathematically defined? How can the transition state be identified, and how can it be distinguished from a local minimum of the potential energy surface?

• The molecular geometry and chemical reaction dynamics are analysed with a potential energy surface (PES) where the necessary points can be evaluated and classified according to first and second derivatives of the energy with respect to position, V(r), which respectively are the gradient and the curvature. Stationary points are those with zero gradient (∂V(r_i)/∂r_i=0) and have physical meaning: energy minima (or local minima of the PES) correspond to physically stable chemical species (such as reactant and product) and saddle points representing transition states, the maximum on the lowest energy pathway (or minimum energy path) that linking reactant and product.

• Mathematically, a saddle point is a point on the surface of a function where the slopes in orthogonal directions are all zero, but which is not a local extremum of the function. The transition state is a first-order saddle point. In this H + H2 system, the surface is symmetrical, which means that the transition state must be exactly half way between the reactants and products. In another words, the transition state lies somewhere on the diagonal line where the distances r_AB and r_BC are equal.

• The hydrogen atoms will have no movement at the saddle point since the force on atoms (i.e. the negative derivative of PES with respect to the coordinate r_i where the force acting on) will be zero at transition state. Consequently, if one starts a trajectory exactly at the transition state, with no initial momentum, it will remain there forever. Therefore, the method to identify the transition state position from a PES is to change the values of r_AB and r_BC until the forces along AB and BC are both zero. In fact, forces are negative on one side of the PES and positive one the other side. The transition state is located where forces across zero.

Trajectories from r_AB = r_BC: locating the transition state

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

• By testing different initial conditions with r_AB = r_BC, and p₁ = p₂ = 0.0 g.mol^-1.pm.fs^-1, the transition state position (r_ts) is found to be the point where distances r_AB and r_BC are equal to 90.78 pm. At that position, forces are both -0.001 KJ.mol^-1.pm^-1 (close to zero).

• Since the H + H2 surface is symmetrical, the transition state must have r_AB = r_BC. If we start a trajectory on the ridge r_AB = r_BC there is no gradient in the direction at right angles to the ridge, thus the trajectory will oscillate on the ridge and never fall off.

• To verify the transition state position, “Internuclear Distances vs Time” plots for some relevant r_AB = r_BC trajectories are shown below. In these graphical display the values of r_AB(t), r_BC(t) and r_AC(t) are given (Y axis) against time t (X axis). At transition state, distance r_AB and r_BC (and thus r_AC) are constant since all three atoms are stationary at the saddle point. The potential energy is maximized while the kinetic energy is zero. This geometry is transition structure.

A “Internuclear Distances vs Time” plot at transition state.

• In contrast, when initial conditions increased to r_AB = r_BC =100 pm, it shows an oscillatory behaviour for the whole trajectory corresponding to the vibration of atom A and atom C, since there are both potential and kinetic energy. The distance r_AB and r_BC are overlapped, since they have the same amplitude of vibration and the vibrations are symmetrical around atom B. The distance r_AC) has doubled the amplitude of vibration (i.e. the sum or superposition of the amplitude). The oscillatory behaviour is changed after the transition state, at t≈270 fs, to a dissociation pattern where r_AB (and thus r_AC) keeps increasing with a slight oscillatory behaviour while r_BC remained at the same level with a higher vibrational frequency. This means that the atom A leaves away from the molecule B-C with the translational energy while molecule B-C is left with vibrational energy, which resembles the reactant.

A “Internuclear Distances vs Time” plot at r_AB = r_BC =100 pm.

Trajectories from r_BC = r_ts+δ, r_AB = r_ts

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

The reaction path (minimum energy path or mep) is calculated with these initial conditions involving: • r_BC = r_ts+1 pm = 91.78 pm; • r_AB = r_ts = 90.78 pm; • p_BC = p_AB = 0 g.mol^-1.pm.fs^-1; • then change the calculation type from dynamics to MEP; • increase the number/size of steps to 5000/0.1(fs).

Then repeat the calculation with the same initial conditions used to calculate the reaction path, but change the calculation type back to Dynamics.

The calculated trajectories are shown in contour plots below, which both start from the transition state and simply follows the valley floor to H_C + H_A-H_B. The mep is a smooth curve with no oscillatory behaviour, while the dynamic trajectory is an oscillatory curve.

The contour plot of the reaction path calculated with mep.

The contour plot of the reaction path calculated with Dynamics.

The mep has the property that any point on the path is at an energy minimum in all directions perpendicular to the path. The mep can also be described as the union of steepest descent paths from the saddle points to the minima (that is why the mep passes through at least one first-order saddle point.)

From the “Internuclear Distances vs Time” diagrams below, it shows a constant increasing rate of r_BC (and thus r_AC) after the transition state in Dynamic trajectory, whereas it shows a reducing rate of the increase of the r_BC (and thus r_AC) in the mep. In addition, the molecule A-B has slight vibrational (or oscillatory) behaviour but has no vibration in the mep.

The “Internuclear Distances vs Time” plot of the mep with r_BC = r_ts+1 pm = 91.78 pm.

The “Internuclear Distances vs Time” plot of Dynamic trajectory with r_BC = r_ts+1 pm = 91.78 pm.

As mentioned before, the variation of momentum p (the forces acting on a given interatomic coordinate r_i) will depend on the derivative of the potential energy surface with respect to that coordinate. Simply, the momenta over time, p(t), is the gradient of V(r).

From the “Momenta vs Time” diagrams below, the momenta of all atoms are zero in the mep since the momenta/velocities are always reset to zero in each time step when calculating the mep. while it have various values in Dynamic trajectory: the momentum of A-B decreases to slightly negative values and then rises to ≈2.5 g.mol^-1.pm.fs^-1 and keeps oscillating; whereas the momentum of B-C, similarly, decreases to negative values and then increases dramatically to and remains at ≈5 g.mol^-1.pm.fs^-1.

The “Momenta vs Time” plot of the mep with r_BC = r_ts+1 pm = 91.78 pm.

The “Momenta vs Time” plot of Dynamic trajectory with r_BC = r_ts+1 pm = 91.78 pm.

If we change the initial conditions to: • r_BC = r_ts = 90.78 pm; • r_AB = r_ts + 1 pm = 91.78 pm. The “Internuclear Distances vs Time” and “Momenta vs Time” plots will look like this:

The “Internuclear Distances vs Time” plot of the mep when r_AB = r_ts + 1 pm.

The “Internuclear Distances vs Time” plot of Dynamic trajectory when r_AB = r_ts + 1 pm.

The “Internuclear Distances vs Time”and “Momenta vs Time” plots of both the mep and Dynamic trajectory have exactly the same pattern as those plots before the initial conditions are changed, except that line A-B (representing r_AB) and line B-C (representing r_BC) are interconverted. This means that the reaction proceeds in the same way but opposite direction (towards the valley floor to H_A + H_B-H_C).

The “Momenta vs Time” plot of the mep when r_AB = r_ts + 1 pm.

The “Momenta vs Time” plot of Dynamic trajectory when r_AB = r_ts + 1 pm.

Take note of the final values of the positions r1(t) r2(t) and p1(t) p2(t) for your trajectory for large enough t.

• step number = 5000

• step size = 0.1 fs

• r_BC = 74.77117297476231 pm

• r_AB = 3777.013307800053 pm

• p_BC = 1.9461593911826354 g.mol^-1.pm.fs^-1

• p_AB = 5.073830709923566 g.mol^-1.pm.fs^-1

The atom A is extremely far from the molecule B-C and keeps moving away. The energies are: • kinetic energy = +19.657 KJ.mol^-1

• potential energy = -434.999 KJ.mol^-1

• total energy = -415.342 KJ.mol^-1

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 energy terms are the same as above, but in this case, the reverse sign of momenta results in the reserve of the moving direction of the free atom (atom A in this case). The atom A is very far from the molecule B-C and it keeps approaching the molecule B-C.

Reactive and unreactive trajectories

For the initial positions r_BC = 74 pm and r_AB = 200 pm, run trajectories with the following momenta combination and complete the table. Complete the table below by adding the total energy, whether the trajectory is reactive or unreactive, and provide a plot of the trajectory and a small description for what happens along the trajectory. What can you conclude from the table?

caption

pBC/ g.mol-1.pm.fs-1	pAB/ g.mol-1.pm.fs-1	Etot	Reactive?	Description of the dynamics	Illustration of the trajectory
-2.56	-5.1	-414.280	Yes	The reaction can occurred. The atom A approaches the molecule B-C and forms a new molecule A-B which leaves away with most vibrational energy and a small amount of kinetic energy (KE), while atom C is detached as a separate atom which moves away with most KE.
-3.1	-4.1	-420.077	No	The reaction doesn't occur. The atom A approaches the molecule B-C and return back without colliding with the molecule.
-3.1	-5.1	-413.977	Yes	The reaction can occur. The atom A approaches the molecule B-C and forms a new molecule A-B which leaves away with most vibrational energy and a small amount of kinetic energy (KE), while atom C is detached as a separate atom which moves away with most KE.
-5.1	-10.1	-357.277	No	A case of barrier recrossing.. The atom A approaches the molecule B-C and forms a temporary new molecule A-B with strong vibrational motion and the molecule A-B move back and collide with atom C to form the B-C again.
-5.1	-10.6	-349.477	Yes	The reaction can occur. The atom A approaches the molecule B-C and break B-C bond and makes atom B bouncing in-between and eventually form a new molecule A-B.

From this table, you can observe that some trajectories have a total energy greater than the activation energy and still do not react. Therefore, for the reaction to proceed, there are two conditions should be met:(1) the total energy is lager than the activation energy; (2)the interaction between atoms should be appropriate for the collision to happen with correct energy and direction. For instance, if the approaching atom A move opposite to the molecule B-C, there will be no collision and the reaction will definitely not occur.

• (The transition state is found to be at r_AB= 230.0 pm; r_BC=74.0 pm; forces along AB and AC = +0.015 and +0.012 respectively; KE = 0; PE = total E = -434.471.)

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) explains the reaction rates of elementary chemical reactions. The theory assumes a special type of chemical equilibrium (quasi-equilibrium) between reactants and activated transition state complexes. These are three properties of TST:

1.Rates of reaction can be studied by examining activated complexes near the saddle point of a potential energy surface. The details of how these complexes are formed are not important. The saddle point itself is called the transition state. 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.

TST is used primarily to understand qualitatively how chemical reactions take place. Simply, the Transition State Theory (TST) tells whether a reaction will happen if cross the TS barrier.

So the difference here is that our program is working with a single molecule, the TST deals with a bulk system with a Boltzmann speed distribution. The main differences between experimental and predicted rate constant values will be due to (1) ignoring barrier recrossing and (2) ignoring QM behaviour.

EXERCISE 2: F - H - H system

F + H2 --> HF + H

A F atom (A) approaches a hydrogen molecule (B-C) to forms a new HF molecule (A-B) and a separate H atom (C).

PES inspection

By inspecting the potential energy surfaces, 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.

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