MRD:reactiondynamics SM
Molecular reaction dynamics - Lab notebook
In this lab we use potential energy surfaces to monitor collisions between 3 isolated atoms in the gas phase. In the initial example we examine the collision between H-H and H, treating the system with classical mechanics. The system can be described using a potential energy surface as a function of r1 and r2:
On this surface the trajectory of the system i.e. it's change over time can be monitored (shown via a black line) given a set of initial conditions.
1.
Looking at the potential energy surface we can see the saddle point; the highest energy point on the lowest energy reaction route (i.e. the transition state). Given that the change in total potential energy of the system can be fully described using the variable relationships r1 (AB bond distance) and r2 (BC bond distance), the saddle point must lie when one of these is at a maximum (dv/dr = 0, dv2/dr2 < 0) and the other is at a minimum (dv/dr = 0, dv2/dr2 > 0). We can obtain a coordinate for the point by applying numerical methods to iteratively calculate values for the energy while varying the positions of the particles. The saddle point is found where both gradients converge to 0.
(Fv611 (talk) 16:44, 16 May 2017 (BST) Correct, but you didn't discuss minima at all, or how they differ from the Transition States)
2.
If we look at the line where r1=r2, the gradient perpendicular to the line is always 0 because the (H - H - H) system is symmetric. It follows that the minimum point on this line is the saddle point. If we set the system with no net momentum, i.e. leave it to oscillate naturally, we can vary r1=r2 until there is no movement - we can find the transition state like this because if we move along the r1=r2 line we will eventually find the minimum point of the line (saddle point overall); there is no movement at this point because the attractive and repulsive forces across the three particles are symmetrically balanced. At this point the potential energy is stationary, and therefore so is the kinetic energy as energy must be conserved overall.
By trial and error, using a and r1 = r2 bond distance of 0.910 A results in the atoms almost not oscillating. At (or very close to) this coordinate the attractive and repulsive forces are in equilibrium and no particle has a net momentum:
(Fv611 (talk) 16:44, 16 May 2017 (BST) Good)
3.
An MEP trajectory was run along with the equivalent dynamic trajectory (same initial conditions). Looking at the two reaction plots it can be seen that the MEP always follows the route where an infinitesimal change results in the greatest potential energy reduction to the system. During this change no energy is converted to vibration. On the other hand, the reaction dynamics plot shows that when there is a reduction in potential energy it is converted into vibrational kinetic energy; this causes the system to begin to oscillate.
(Fv611 (talk) 16:44, 16 May 2017 (BST) Infinitesimal change in what? Also, there is not much point in showing that the reaction system is symmetric, as we already know that. It would have been more interesting to compare the momenta vs time graph from the dynamic simulation and that from the mep.)
Focusing on the dynamic plot, we can look at how Internuclear distance varies with time and how internuclear momenta vary with time
These graphs clearly show Ha-Hb departing as a vibrating molecule from Hc.
Furthermore we can show the symmetry of the system by switching the initial positions. In this case Hb-Hc departs as a vibrating molecule from Ha;
4.
In order to observe some reactive an unreactive trajectories, the following conditions were set:
r1 = 0.74 and r2 = 2.0:
| p1 | p2 | Reactive? |
|---|---|---|
| -1.25 | -2.5 | yes |
| -1.5 | -2.0 | no |
| -1.5 | -2.5 | yes |
| -2.5 | -5.0 | no |
| -2.5 | -5.2 | yes |
The trajectories on the potential energy surface were observed as follows:
1)
This reaction proceeds normally after transversing the potential energy barrier.
2)
This reaction fails to overcome the barrier; the collision is unsuccessful and the incoming atom rebounds.
3)
Again, the reaction proceeds normally after transversing the potential energy barrier
4)
This system crosses the activation barrier, but the vibrations gained in the route taken cause the system to travel back over it as if no reaction had happened.
5)
In this case the system crosses the barrier and recrosses it twice resulting in a net reaction.
(Fv611 (talk) 16:44, 16 May 2017 (BST) Correct, but would have been nicer to include a discussion on the amount of energy involved in each reaction, and on the distribution between translational and vibrational)
5.
Transition state theory assumes an equilibrium between reactants and the transition state. The rate of product formation is then dependent on the nature of the transition state and it's position in the equilibrium; the rate can be estimated using kinetic theory. These trajectories just described show some of the reactions go via a higher energy minimum point than the transition state (saddle point). This means if just the equilibrium position of the saddle point is being considered, not all of the potential reaction pathways are being considered. This mean experimental rate constants should be greater than those predicted using transition state theory.
(Fv611 (talk) 16:44, 16 May 2017 (BST) Correct, but you are missing an important point to be made about recrossing)
Flourine
1.
The H-H bond enthalpy is 436 kJ/mol, whereas the H-F bond enthalpy is 565 kJ/mol. This means the H2 + F ---> H-F + H reaction is exothermic with an enthalpy change of reaction of -129 kJ/mol. Conversely the H-F + H ---> H2 + F reaction is endothermic with the opposite enthalpy change of reaction.
(Fv611 (talk) 16:44, 16 May 2017 (BST) Correct, but you are not discussing the relation to the potential energy surface)
2. & 3.
Using trial and error, the transition state was approximated at r1 = 0.75 (A-B distance) and r2 = 1.81 (B-C distance). This as done by modifying r1 and r2 with no initial momentum and finding the shortest trajectory at a fixed number of steps i.e. starts on the flattest region of the surface - gradient closest to 0. This point is found closer to the reactants. This agrees with Hammond's postulate which states the energy of the transition state is closer to the reactants if they are higher in energy than the products.
A point was chosen slightly to the products side of the transition state to see the energy change when the almost transition state decayed back to the reactants. From the potential energy / time graph it can be seen that the activation energy in the forward direction is 0.27 kcal/mol:
Similarly the activation energy for the reverse reaction was found by changing the starting coordinate so it was on the other side of the transition state. The potential energy / time graph showed the activation energy in this direction to be about 30 kcal/mol. Note during this dynamic pathway significant vibration manifested. The potential energy level of the products can be considered the lowest point in this oscillation:
(Fv611 (talk) 16:44, 16 May 2017 (BST) Good)
4.
A dynamic reaction was set up:
The reaction above shows how energy is conserved during the reaction; the kinetic energy / time graph mirrors the potential energy / time graph, showing that there is always a constant energy distributed between the two forms. Experimentally, the increase in kinetic energy (in both translational and vibrational forms) causes an increase in temperature. The total random kinetic energy for the system (internal energy) is proportional to temperature where the heat capacity is the proportionality constant. This means the change in temperature can be measured and change in internal energy estimated as long as the heat capacity is known.
(Fv611 (talk) 16:44, 16 May 2017 (BST) Good)
5.
The H-H + F ---> H-F + H reaction was then studied using the following initial conditions:
r1 = 0.74 r2 = 2.4 p2 = -0.5
p1 was varied between -3 and 3
It was found that some of the reactions went and others didn't with no particular pattern. Interestingly the reactions with did not go could be made to go by slightly changing p2. Overall this show the importance of both components of momentum for the success of the reaction. The roles of the two different components of momentum are as follows:
For the forward reaction, before the transition state, the A-B (H-H) momentum is vibration across the 'gutter' in the energy surface and B-C momentum is translation down it. To get to the transition state, A-B vibration doesn't help, but B-C translation does. Once we're past the transition state, B-C momentum becomes vibration and A-B becomes translation because the gutter is now at right angles (representing the change in bonding). This means to move away from the transition state we now need A-B momentum while the B-C momentum doesn't contribute to this translation. Overall this shows both components of momentum are important for the reaction to go to completion.
If the transition state is in one 'gutter' like in this case, only momentum in one (the translational) direction is productive in overcoming it. Some vibrational momentum is needed however to get around the curve in the surface without rebounding. Conversely, if the transition state is in the middle of the curve, both components of momentum are equally important in overcoming the transition state.
(Fv611 (talk) 16:44, 16 May 2017 (BST) Nice discussion, but you could have mentioned Polany's rules)