https://doi.org/10.1351/goldbook.S05962

- In a kinetic analysis of a complex reaction involving unstable intermediates in low concentration, the rate of change of each such intermediate is set equal to zero, so that the rate equation can be expressed as a function of the concentrations of chemical species present in macroscopic amounts. For example, assume that
**X**is an unstable intermediate in the reaction sequence:S05962-1.pngS05962-2.pngConservation of mass requires that: \[[\text{A}] + [\text{X}] + [\text{D}] = [\text{A}]_{0}\] which, since [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 [X] is negligibly small, the rate of formation of**D**is essentially equal to the rate of disappearance of**A**, and the rate of change of [X] can be set equal to zero. Applying the steady state approximation (d[X]/dt = 0) allows the elimination of [X] from the kinetic equations, whereupon the rate of reaction 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 flow reactor a steady state implies a regime so that all concentrations are independent of time.

PAC, 1993,

PAC, 1994,