MRD:av4217

Exercise 1: H + H₂

Transition State

Potential energy surface diagram

The transition state in a potential energy surface diagram is when the partial differentiation of the potential energy with respect to r_i is 0. The transition state is located at the saddle point of the potential energy surface diagram.At the transition state ∂V(r₁)/∂r₁ >0 and ∂V(r₂)/∂r₂ <0, or vice versa. When you travel along the minimum valley there will be a point where it is the maximum of the valley, this is the transition state. If a trajectory is started at a transition state with no initial momentum it will remain there, with a small change in geometry it will roll toward the reactants or products depending on the direction. The transition state can be identified in this way. At a local minimum it will not roll toward the products or reactants with a small change as it is a stable point. The maximum at the transition state is what allows the small change in geometry to cause the roll towards the products or reactants. The local minimum wont be a saddle point and will be a lower energy than the transition state. Good explanation, but a good way to distinguish between the local maximum and minimum is by the second partial derivative which will be less than or more than zero, respectively. Sf3014 (talk) 16:13, 15 May 2019 (BST)

Locating the Transition State

The best estimate for r_ts is 0.903. At this distance the atoms oscillate a little but stay at equal distances to each other over a period of time as shown by figure two. The plot doesn't show the r_AB getting smaller indicating they formed a bond and r_BC getting larger showing C is a separate atom, or vice versa. The products or reactants are not formed so this is the transition state position, where no bonds are formed. Good, but how did you get to this value for the position of the transition state?. Sf3014 (talk) 16:16, 15 May 2019 (BST)

Trajectory

MEP vs Dynamic

When the calculation method is dynamic instead of MEP, the oscillation of the atom and molecules can be seen. While, for MEP it can be seen as a straight line following the valley. It doesnt take into account any instantaneous momentum.

Good comparison. Also, you can see that the momentum is zero for the MEP calculation using the momenta vs time plot. Sf3014 (talk) 16:21, 15 May 2019 (BST)

Changing initial conditions

When the initial conditions are changed so r₁ = r_ts and r₂ = r_ts + 0.01, then reaction is the opposite. Atom C reacts with molecule AB to form BC and atom C.

When the initial positions are swapped with the final positions and the momenta are changed with the opposite signs no reaction takes place. However the distance between the atom and the molecule is getting closer so in time they may collide and a reaction takes place

Reactivity of Trajectories

Reactive and unreactive trajectories
p₁	p₂	E_tot	Reactive?	Description of dynamics
-1.25	-2.5	-99.018	Yes	The molecules don't oscillate much at the beginning. It passes the transition state and then collides to form the products. Once the products are formed there is more oscillating.
-1.5	-2.0	-100.456	No	The molecules dont reach the transition state, they don't go over the maximum and instead reflect back to their original path. They come close together but dont have enough energy react so move away from each other.
-1.5	-2.5	-98.956	Yes	The molecules oscillate more as they have greater momentum, they reach the transition state, follow the trajectory and react to form the products.
-2.5	-5.0	-84.956	No	The molecules reach the transition state and the product is formed. However the molecules pass back over the transition state and form the reactants again and remain that way.
-2.5	-5.2	-83.416	Yes	The molecules reach the transition state and collide to almost form products but then goes back to the transition state. It partially forms the reactants and then passes the transition state again and forms the products and remains that way.

Good, but what can you summarise from this table in terms of reaction completion. Sf3014 (talk) 16:38, 15 May 2019 (BST)

Transition State Theory

The transition state theory assumes the energies of the molecules follows the Boltzmann distribution and the reactants are in equilibrium with the transition state structure. Also, once the transition state is achieved the molecules do not collapse back into the reactant structures.^[1] As seen in penultimate example this is not true and the reactants can be formed again once the transition state is achieved. This suggests the transition state theory will predict a higher rate of reaction than the experimental values as it does not take into account the reactants reforming as seen in the last example. The reactants reforming will mean more time taken to form the products so the rate of reaction will be longer for the experimental values. If it is assumed the reactants are not reformed and the reaction only goes forward, the predicted rate would be higher than the experimental values.

Exercise 2: F-H-H

PES Inspection

The reaction of H₂ with F is exothermic. As seen in figure 6 the reactants are of a higher energy than the products so the reaction is exothermic.

The reaction of HF with F is endothermic as seen by the surface plot as the products are of a higher energy than the reactants. So energy is absorbed.

This shows the HF bond is stronger than the H₂ bond as a large amount of energy is required to break it meaning the reaction is endothermic. Also a when the bond is made a large amount of energy is released causing the reaction to be exothermic. The large release of energy is greater then the energy required to break the H₂ since the reaction is exothermic. So it's stronger as it has greater bond energy. Good explanation, next time you should include the bond energies for HF and H₂. Also, when you use an acronym you should define it first, ie potential energy surface (PES). Sf3014 (talk) 16:45, 15 May 2019 (BST)

The transition state of both reactions are at r_BC=0.745 and r_AB=1.84 where atom A is F and atom B and C are H. Also, momentum is at 0

The activation energy for reaction of F + H₂ is around 0 as it's a spontaneous reaction. The the energy levels of the reactants and transition state are almost the same. The activation energy for the reaction of HF + H is around 29.6 kcalmol^-1, it has a high activation energy since it's endothermic. Good, but how did you get to those values for the position of the transition state? You should refer to Hammond postulate. Also, you should of attempted to calculate the activation energy for the exothermic reaction as you did for the endothermic reaction. Sf3014 (talk) 16:51, 15 May 2019 (BST)

Reaction Dynamics

The energy released from the reaction is absorbed by the molecule and causes it to oscillate, changing its momentum. The molecule vibrates at a specific frequency therefore wavelength, which can be determined using emission spectroscopy.

When changing the momentum from -3<p_HH<3 while keeping the other values constant at a specific value it can be seen all the possibilities approach the transition state. This could be because as said energy above the activation is put into the system. Therefore, the system can overcome the energy barrier and reach the transition state and produce the products. However in many cases the molecules collapse back into the reactants, after reaching the transition state again.

When increasing p_FH to -0.8 and reducing p_HH to 0.1 the reaction takes place. The molecules reach the transition state form the products and then collapse back to the reactant. Then the molecules form the products and remain that way following the trajectory to the valley.

For the HF + H reaction initial conditions of r_HF=1.2, r_HH=1.5, p_HF=-0.9, and p_HH=-0.1 will allow a reaction to take place and form the products.

Polyani discusses when the reaction is endothermic the transition state is close to the the products, so there is a late energy barrier. Therefore, greater vibrational energy in the reactants would more likely lead to a reaction and formation of products. With just translational energy, the atom gets too close to the molecule and repulsive forces causes the atom repel back. For an exothermic reaction translational energy is enough to overcome the early energy barrier (transition state closer to the reactants). The reactants have enough energy to cause a reaction and form products. This shows the mode of energy in the reactants is important for a successful reaction and this effected by the position of the transition state.^[2] Good, but there are some things you need to explain further; how does emission spectroscopy determine the change in frequency for this reaction?, you didn't address the conservation of energy clearly and you should of shown your work for the reaction processes with different initial values as an example to Polanyi's rules. Sf3014 (talk) 17:05, 15 May 2019 (BST)

References

↑ P. Atkins, J. de Paula, Elements of Physical Chemistry, 5th ed.; Oxford University Press, 2009.
↑ Jeffrey I.Seinfeld, Joseph S.Francisco, William L.Hase, Chemical Kinetics and Dynamics, Prentice Hall Inc, 1989.

[TST-1] P. Atkins, J. de Paula, Elements of Physical Chemistry, 5th ed.; Oxford University Press, 2009.

[Polyani-2] Jeffrey I.Seinfeld, Joseph S.Francisco, William L.Hase, Chemical Kinetics and Dynamics, Prentice Hall Inc, 1989.

[1]

[2]

Exercise 1: H + H2