Two reaction schemes are kinetically equivalent if they imply the same @R05141@. 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 @R05171@. \[\frac{\text{d}[\text{C}]}{\text{d}t} = \frac{k_{1}\ k_{2}\ [\text{A}]\ [\text{OH}^{-}]}{k_{2}\,+\,k_{-1}\ [\text{OH}^{-}]}\] \[\text{(ii)}\qquad \text{A}\overset{k_{1}}{\underset{k_{-1}}\rightleftarrows }\text{B}\overset{k_{2}}{\underset{\text{OH}^{-}}\rightarrow }\text{C}\] Providing that B does not accumulate as a @R05171@: \[\frac{\text{d}[\text{C}]}{\text{d}t} = \frac{k_{1}\ k_{2}\ [\text{A}]\ [\text{OH}^{-}]}{k_{-1}\,+\,k_{2}\ [\text{OH}^{-}]}\] 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 @T06446@ to C. The two schemes are therefore kinetically equivalent under conditions to which the stated provisos apply.