 Solution

Identify the transition-state separation with contact, $R=r_A+r_B$, take distinct atoms so $\sigma =1$, and let $E_A=N_A\Delta {\epsilon }_0^{‡}$. Then $\pi R^2=\rho$ and $\Delta {\epsilon }_0^{‡}/kT=E_A/RT$, yieldingThe velocity factor is the mean relative speed from the Maxwell-Boltzmann distribution. Thus this simple transition state theory model reproduces the stated collision theory expression. Molecular orientation constraints, quantum transmission and reaction-surface recrossing can invalidate the equality outside these simplifying assumptions.