Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 5 / d / Solution

Eliminate cis using $[\mathrm{cis}]=C_{\mathrm{tot}}-[\mathrm{trans}]$:ThusBoth directions restore the perturbed distribution: an excess of trans both increases its reverse consumption and reduces cis available for forward formation. This explains why the chemical relaxation time is $1/(k_1^{\prime }+k_{-1}^{\prime })$, rather than the reciprocal of only one rate constant.