What is the steady-state approximation and when is this approximation employed?

Answer 1

What it is:

The steady state approximation is a method employed to make the calculation of rate expressions easier. What is done is to assume that the concentration of any intermediate in the reaction is unchanged (the rate of intermediate produced is equal to the rate at which it is used). This will allow for the determination of an expression for the concentration of the intermediate that can then be substituted in to solve for the rate.

An Example:

X(OH)2 = X + H20 (disregard H20 as it is a solvent, and let the forward rate constant be k1 and the reverse be k-1)

X + I -> XI (with a rate constant of K2)

Now the rates of production and use of X must be equal so,

k-1[X]+ K2[I][X] = k1[X(OH)2]

solving for [X] gives,

[X] = k1[X(OH)2]/(k-1+ K2[I])

The reaction rate for the above reaction is:

v = k2[I][X]

We can not have the intermediate in the final answer so it is an easy thing to sub in the value of [X] we found via the steady state approximation.

Validity:

This approximation is known to be vaild only for systems in which the concentration of the intermediate is constant. This means the second step (the one using up the intermediate) must be faster than the first (the one producing intermediate) or k1 << k2.

Answer 2

The steady-state approximation assumes that the rate of change of an intermediate species is zero, simplifying the rate expression for a reaction. This approximation is employed when the concentration of the intermediate remains relatively constant over the course of the reaction, allowing for a simpler analysis of reaction kinetics.