Two reaction schemes are kinetically equivalent if they imply the same rate law. For example, consider the two schemes (i) and (ii) for the formation of $\ce{C}$ from $\ce{A}$: \[(\rm{i}) \qquad \ce{A <-->[k_{1},\ce{OH^{-}}][k_{-1},\ce{OH^{-}}] B ->[k_{2}] C}\] Providing that $\ce{B}$ does not accumulate as a reaction intermediate. \[\frac{\rm{d}[\ce{C}]}{\rm{d}t} = \frac{k_{1} k_{2}[\ce{A}][\ce{OH^{-}}]}{k_{2} + k_{-1} [\ce{OH^{-}}]}\] \[(\rm{ii}) \qquad \ce{A <-->[k_{1}][k_{-1}] B ->[k_{2}][\ce{OH^{-}}] C}\] Providing that $\ce{B}$ does not accumulate as a reaction intermediate: \[\frac{\rm{d}[\ce{C}]}{\rm{d}t} = \frac{k_{1} k_{2}[\ce{A}][\ce{OH^{-}}]}{k_{-1} + k_{2} [\ce{OH^{-}}]}\] Both equations for \(\frac{\rm{d}[\ce{C}]}{\rm{d}t}\) are of the form: \[\frac{\rm{d}[\ce{C}]}{\rm{d}t} = \frac{r [\ce{A}][\ce{OH^{-}}]}{1 + s [\ce{OH^{-}}]}\] 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 $\ce{OH^{−}}$ catalyses the formation of $\ce{B}$, and necessarily also its reversion to $\ce{A}$, or is involved in its further transformation to $\ce{C}$. The two schemes are therefore kinetically equivalent under conditions to which the stated provisos apply.