In Statistical Transition State Theory, why is the concentration of the transition state for the forward and reverse reaction identical?

Question

For example on Wikipedia, and also in my lecture notes, it is assumed that for a reaction $$\ce{A + B<=> AB^{\ddagger} <=> P}$$ at equilibrium, $$[\ce{AB^{\ddagger}_f}] = [\ce{AB^{\ddagger}_r}],$$ but I don't see how the equation follows from the assumption, since there is no direct relation between transition state concentration and total reaction rate. And if it is, how can they simply be equal instead of depending on some constants?

Answer 1

In transition state theory, it is assumed that the forward and the reverse reaction occur via the same transition state. This is implied by assuming a quasi-equilibrium between the reactants and the products, which adheres to a common Boltzman statistics. Therefore the labelling $\ce{AB^\ddagger_r}$, $\ce{AB^\ddagger_f}$ is artificial, because it is the same transition state: $$\ce{AB^\ddagger_r} = \ce{AB^\ddagger} = \ce{AB^\ddagger_f},$$ and therefore the concentration is identical, too: $$[\ce{AB^\ddagger_r}] = [\ce{AB^\ddagger}] = [\ce{AB^\ddagger_f}].$$