kinetic equivalence

https://doi.org/10.1351/goldbook.K03403
Two reaction schemes are kinetically equivalent if they imply the same @[email protected] For example, consider the two schemes (i) and (ii) for the formation of C from A: \[\text{(i)}\qquad \text{A}\overset{k_{1},\text{OH}^{-}}{\underset{k_{-1},\text{OH}^{-}}\rightleftarrows }\text{B}\overset{k_{2}}{\rightarrow }\text{C}\] Providing that B does not accumulate as a @[email protected] \[\frac{\text{d}[\text{C}]}{\text{d}t} = \frac{k_{1}\ k_{2}\ [\text{A}]\ [\text{OH}^{-}]}{k_{2}\,+\,k_{-1}\ [\text{OH}^{-}]}\] \[\text{(ii)}\qquad \text{A}\overset{k_{1}}{\underset{k_{-1}}\rightleftarrows }\text{B}\overset{k_{2}}{\underset{\text{OH}^{-}}\rightarrow }\text{C}\] Providing that B does not accumulate as a @[email protected]: \[\frac{\text{d}[\text{C}]}{\text{d}t} = \frac{k_{1}\ k_{2}\ [\text{A}]\ [\text{OH}^{-}]}{k_{-1}\,+\,k_{2}\ [\text{OH}^{-}]}\] Both equations for \(\frac{\text{d}[\text{C}]}{\text{d}t}\) are of the form: \[\frac{\text{d}[\text{C}]}{\text{d}t} = \frac{r\ [\text{A}]\ [\text{OH}^{-}]}{1\,+\,s\ [\text{OH}^{-}]}\] where \(r\) and \(s\) are constants (sometimes called 'coefficients in the rate equation'). The equations are identical in their dependence on concentrations and do not distinguish whether OH catalyses the formation of B, and necessarily also its reversion to A, or is involved in its further @[email protected] to C. The two schemes are therefore kinetically equivalent under conditions to which the stated provisos apply.
Source:
PAC, 1994, 66, 1077. (Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994)) on page 1133 [Terms] [Paper]