MRD:JJ1516
Transition State
Along the minimum energy trajectory from reactants to products on a potential energy surface, the transition state is defined as the saddle point at which the potential energy of the particle system along the trajectory is a maximum. In a direction orthogonal to the reaction trajectory, the transition state is a minimum of potential energy. Thus, the gradient of all components of the potential energy is 0 at the transition state while the second derivative of the potential energy is negative with respect to a path along the trajectory and positive in a direction orthogonal to it.
Good--Sw2711 (talk) 11:12, 16 May 2018 (BST)
Energy minima are found at the reactant and product regions; here, the gradient of the potential energy along the reaction trajectory path is 0 while the second derivative of the potential energy along the reaction trajectory is positive.
While energy minima are points of stable equilibrium, the transition state is a point of unstable equilibrium. These two points may be distinguished by the different behaviours of reaction trajectories beginning at them. If a reaction trajectory begins with 0 momentum exactly at either a transition state or energy minimum, the trajectory will not move from the point in question. However, starting the reaction trajectory with 0 momentum at a slight deviation δri from the transition state in the direction of products/reactants will result in the trajectory falling towards the product/reactant region. Conversely, starting the trajectory at a slight deviation from an energy minimum will result in the trajectory collapsing back down to that energy minimum.
If the reaction trajectory is begun at a deviation from the transition state in a direction orthogonal to the minimum energy path between products/reactants, an oscillation of the trajectory about the transition state will occur with neither products nor reactants being formed.
Yeah, so the oscillation means there is a deviation from the TS--Sw2711 (talk) 11:12, 16 May 2018 (BST)
For the triatomic collision between atomic hydrogen (C) and molecular hydrogen (A-B), a transition state was located at rAB = rBC = 0.9102Å. A reaction trajectory begun at this point with 0 initial momentum was found to remain at said point.
So here at the saddle (not said) point. You should see a straight line, i.e. No movement at all. rather than a small oscillation--Sw2711 (talk) 11:12, 16 May 2018 (BST)
The Internuclear Distances vs Time (Figure 1) and trajectory Surface plots (Figure 2) are presented (below-right):
MEP/Dynamics Comparison
For the initial conditions rAB = rts+δ, r2 = rts where rts = 0.9102Å and δ = 0.01Å, both calculation types resulted in reactive trajectories (evident by the increase of the AB distance from bonding to non-bonding length). The trajectory calculated with the Dynamics calculation type resulted in products with energy in the vibrational mode, evident by the oscillating trajectory (Figure 3) and oscillations of the small BC distance (Figure 4). Conversely, the trajectory calculated by the MEP calculation type does not oscillate. Thus the MEP calculation type does indeed provide the minimum energy pathway to the products of the reactive trajectory as there is no excess energy to supply to the vibrational mode of the product. Yes, MEP provides the minimum energy pathway and you see no oscillation. So you think because there is no excess energy? Have you tried to give some initial momentum to your system and compare Dynamics and MEP? --Sw2711 (talk) 11:23, 16 May 2018 (BST) If the initial position conditions of the trajectory are changed to r1 = rts and r2 = rts + δ , it is expected that the trajectory changes from a reactive to an unreactive trajectory, that is, the reactants are re-obtained. This was indeed found after swapping the initial positions (Figure 5).
By setting the initial conditions of the trajectory to the position and negative momentum values at large t (Table 1) for the previous trajectory (where r1 = rts + 0.01Å and r2 = rts ) , the AB distance was found to reverse. That is, the H atom produced in the previous reaction returned along its exit path, colliding with the previously formed H2 molecule and reforming the transition state (Figures 6 & 7).
| Trajectory Characteristic | Value at large t |
|---|---|
| AB Distance | 9.994820817384259Å |
| BC Distance | 0.7579773465903802Å |
| AB Momentum | 2.477984140672245 |
| BC Momentum | 1.290848677272897 |
(Table 1)
Reactive and Unreactive Trajectories
This part is good--Sw2711 (talk) 11:27, 16 May 2018 (BST)
| Trajectory Name | p1 | p2 | Total Energy | Reactivity | Comment |
|---|---|---|---|---|---|
| Alpha | -1.25 | -2.5 | -99.018 kCal•mol-1 | Reactive | Atom C approaches molecule AB and reacts forming atom A and molecule BC containing energy in its vibrational mode. (Figure 9) |
| Beta | -1.5 | -2.0 | -100.456 kCal•mol-1 | Unreactive | Atom C approaches weakly oscillating AB before rebounding away. Closest approach of C to molecule AB = 1.89Å (Figure 9) |
| Gamma | -1.5 | -2.5 | -98.956 kCal•mol-1 | Reactive | Atom C approaches molecule AB forming oscillating BC and atom A. (Figure 10) |
| Delta | -2.5 | -5.0 | -84.956 kCal•mol-1 | Unreactive (Reactants reformed after reaction) | Atom C approaches molecule AB initially forming products CB + A. Atom A reacts with molecule BC to reform atom C and molecule AB with large amplitude oscillations. (Figure 11) |
| Epsilon | -2.5 | -5.2 | -83.416 kCal•mol-1 | Reactive | Atom C approaches oscillating molecule AB reacting to form atom A and molecule BC with large amplitude oscillations. (Figure 12) |
Transition State Theory
I think you forgot to answer this question--Sw2711 (talk) 11:28, 16 May 2018 (BST)
F - H - H System
How did you get these numbers? I think you understand. But I still need some evidence.--Sw2711 (talk) 11:35, 16 May 2018 (BST)
| Trajectory Characteristic | Energetics | ΔEa | Comment |
|---|---|---|---|
| F + H2 > HF + H | Exothermic | ≈ 0 kCal•mol-1 | Energy required for H - H bond cleavage < Energy released by H - F bond formation. H - F bond is stronger than H - H bond. |
| HF + H > H2 + F | Endothermic | + 29.961 kCal•mol-1 | Energy required for H - F bond cleavage > Energy released by H - H bond formation. H - H bond formed is weaker than H - F bond broken. |
Transition State
Hammond's Postulate states that the progression from one state within a chemical reaction to consecutive one will only involve small a small reorganisation of chemical structure if the two states are similar in energy. Assuming that H2 + F is kinetically as well as thermodynamically unstable with respect to HF + H, the activation energy for this reaction will be very small. Thus, by invoking Hammond's Postulate, the transition state will be expected to lie very near to the reactant region of the potential energy surface (by assigning H2 + F as AB + C, this corresponds to a transition state in the region where the AB distance is small and the BC distance is large). Using the bonding distance of H2 (0.74 Å), an approximate location for the transition state was found at rAB = 0.74Å, rBC = 1.815 Å (Figure 13). By checking the H2 bond length is a good approach to start. But I need some evidence to confirm this is a TS. Qualitative descriptions is not really enough--Sw2711 (talk) 11:41, 16 May 2018 (BST)
Performing an MEP calculation from a trajectory starting at a slight deviation (+ 0.02Å) in the reactant direction (H2 + F) from the approximate transition state shows a constant value of potential energy (Figure 14) from 0 to 20000 time steps. Thus the activation energy for the forward reaction of H2 + F > HF + H has been approximated as negligible. I think your attempt is very very good. Have you tried to zoom in to the figure to check the energy change? --Sw2711 (talk) 11:48, 16 May 2018 (BST)
Conversely, the activation energy for the reaction HF + H > H2 + F has been calculated as +29.961 kCal•mol-1 (MEP plot of potential energy vs time is shown in Fig 15).
Reaction Dynamics
The Animation and Internuclear Momenta vs Time plots for a reactive trajectory of the F + H2 > HF + F reaction (initial conditions listed below) indicate that the product molecule is formed with energy in its vibrational mode. As energy is conserved in the reaction system considered, IR spectroscopy could be used to quantify the reaction (translational) energy transferred to vibrational energy. I probably need a bit more explanation on how this is done. --Sw2711 (talk) 11:49, 16 May 2018 (BST)
I think you have missed another question => Discuss how the distribution of energy between different modes (translation and vibration) affect the efficienty of the reaction. and how this is influenced by the position of the transition state--Sw2711 (talk) 11:55, 16 May 2018 (BST)