Past exam of the chemistry course of the year 4 of the natural sciences course of the University of Cambridge / 2022 / Paper 3 / Section F / 46 / b / ii

Take $s=x-x^{‡}$ with $0<x^{‡}<a$. A trajectory crossing to the right at the barrier top goes downhill on exit and cannot return in this model. Its initial Boltzmann weight contains $e^{-\beta V_0}$; negative-velocity trajectories are on the reactant side at the commitment time. ThereforeDividing by the same $Q_r$ gives the result in the energy-integral calculation. No width appears in the classical rate because every classically allowed trajectory passes through the flat barrier.

Put the dividing surface at $x^{‡}<0$. The immediate positive flux includes all positive momenta and gives $C_{fs}(0^{+})=1/(2\pi \hslash \beta )$, an overestimate. Trajectories with $p^2/(2m)<V_0$ hit the barrier, return across this surface, and cease contributing to the later product-side indicator. After their recrossing transient, only $p>\sqrt{2m{V}_0}$ survives:The exact rate is again unchanged; the transmission correction to the reactant-side transition-state estimate is $e^{-\beta V_0}$. The return time depends on surface placement and velocity, so the plateau means a timescale-separated limiting flux, not the claim that every trajectory has one identical finite recrossing time.