We were told in a first year introductory Thermodynamics and Kinetics lecture, and also from Wikipedia, that there are 3 main assumptions for the derivations used in Transition State Theory, but I'm not sure how or why they have been assumed?

  1. Reactants are in constant equilibrium with the transition state structure.
  2. The energy of the particles follow a Boltzmann distribution.
  3. Once reactants become the transition state, the transition state structure does not collapse back to the reactants.

I'm concerned with 1. and 3. particularly, as it seems as though they contradict each other, because how can the reactants be in equilibrium with the TS, if the TS structure cannot revert back into reactants? How do we then utilise these assumptions, is it so that we can write $$\ce{A + B <=> AB^‡ -> P}?$$


1 Answer 1


This is deeper discussed in Rate Constant Units and Eyring Equation, but I am going to post a very short take-home message here.

First, the three assumptions you have cited are not complete. They completely gloss over the most important one, i.e. nuclear and electron motion can be separated, energies (translation, vibration, rotation) can be treated additively.

Having gotten that out of the way, reactants are in constant thermal equilibrium with the transition state structure. This actually means that the reactant structure(s) is in equilibrium with all possible structures that can be reached from it. It is hence more correct to say that the reactants are in thermal equilibrium with all possible transition state structures.
This also basically ensures, that all these structures can be treated within a single, combined Boltzmann statistic, where the energy is distributed according to all the states of the assembly/molecule.
The next assumption is that the transition state will be crossed only once (point of no return), which means that when an assembly/molecule reaches this (high) energy state, it will form the product. This basically sets the limit to the Boltzmann statistic as all states following the transition state are not included anymore.
Putting both together we arrive at \begin{align} \ce{A + B <=> &AB^\ddagger -> P}\\ \ce{P <=> &AB^\ddagger -> A + B}. \end{align} However, you see that you can look at it from both channels, but that'll become to involved here. Please read the other post first.

All these assumption basically lead up to that the transition vector can be treated classical as translation.

In conclusion, the equilibrium part is necessary to write the combined Boltzmann statistic; the transition state collapses into products part is necessary to only include everything up to and including the TS, but nothing more into the statistic.

Also related: References for Transition State Theory and Quantum Transtion State Theory

  • $\begingroup$ Thank you for your reply and point in the right direction, I have my final exams right now, so I'll take some time to look into this a bit more and perhaps post some follow up. $\endgroup$
    – James Liu
    May 4, 2021 at 21:00

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