https://doi.org/10.1351/goldbook.S05962
- In a kinetic analysis of a @[email protected] involving @[email protected] intermediates in low concentration, the rate of change of each such intermediate is set equal to zero, so that the @[email protected] can be expressed as a function of the concentrations of @[email protected] present in macroscopic amounts. For example, assume that X is an @[email protected] intermediate in the reaction @[email protected]: S05962-1.pngS05962-2.pngConservation of mass requires that: \[[\text{A}] + [\text{X}] + [\text{D}] = [\text{A}]_{0}\] which, since \([\text{A}]_{0}\) is constant, implies: \[-\frac{\text{d}[\text{X}]}{\text{d}t} = \frac{\text{d}[\text{A}]}{\text{d}t}+\frac{\text{d}[\text{D}]}{\text{d}t} .\] Since \([\text{X}]\) is negligibly small, the rate of formation of D is essentially equal to the @[email protected] of A, and the rate of change of \([\text{X}]\) can be set equal to zero. Applying the steady state approximation (\(\frac{\text{d}[\text{X}]}{\text{d}t} = 0\)) allows the @[email protected] of \([\text{X}]\) from the kinetic equations, whereupon the @[email protected] is expressed: \[\frac{\text{d}[\text{D}]}{\text{d}t} = -\frac{\text{d}[\text{A}]}{\text{d}t} = \frac{k_{1}\,k_{2}[\text{A}]\,[\text{C}]}{k_{-1}\,+k_{2}\,[\text{C}]}\]Note:
The steady-state approximation does not imply that \([\text{X}]\) is even approximately constant, only that its absolute rate of change is very much smaller than that of \([\text{A}]\) and \([\text{D}]\). Since according to the reaction scheme \(\frac{\text{d}[\text{X}]}{\text{d}t} = k_{2}\,[\text{X}]\,[\text{C}]\), the assumption that \([\text{X}]\) is constant would lead, for the case in which C is in large excess, to the absurd conclusion that formation of the product D will continue at a constant rate even after the reactant A has been consumed. - In a stirred @[email protected] a steady state implies a regime so that all concentrations are independent of time.