**[B] = [A] _{0} [k_{1} /(k_{2} – k_{1}) (e^{-k1t} – e^{-k2t})**

By consideration of each of the terms in the equation show that if k_{2} is very much larger than k_{1} i.e. k_{2} » k_{1} then [B] and d[B]/dt will be both small and approximately constant.

**Now, **

When k_{2} >> k_{1} then the following approximations may be made. From below Figure we have as k_{2} is dominantly large then exp (- k_{2}t) ≈ 0 and as k_{1} is small and positive then exp (- k_{2}t) ≈ exp(0) ≈ 1. Finally (k_{2} — k_{1}) ≈ k_{2}.

As k_{1}/k_{2} is a small number thus [B] will be small compared to [A]_{0}. Using the same approximations as above with the equation for the rate of forming the intermediate d[B]/dt we have

**d[B]/dt = [A] _{0} [k_{1} /(k_{2} – k_{1}) (e^{-k1t} – e^{-k2t}) ≈ [A]_{0} [k_{1} /k_{2}] (e^{-k1t} – e^{-k2t})**

**d[B]/dt ≈ – [A] _{0} (k_{1}^{2} /k_{2}) **

As k_{1}/k_{2} is a small number then **k _{1}^{2} /k_{2 }**is an even smaller number and thus the rate of change of the [B] concentration is exceedingly small, i.e. [B] is approximately constant. Thus the intermediate concentration [B] is both small and approximately constant under the approximations which follow from where k

_{2 }>> k

_{1}and this is called the “Steady State Approximation” in Chemical Reactions kinetics, see below figure.

The steady state approximation does not assume the Chemical Reactions intermediate concentration to be constant and therefore its time derivative being zero. The steady state approximation assumes that the variation in the concentration of the intermediate is approximately zero and as the concentration of the intermediate is so low that even a large relative variation in its concentration is small if compared quantitatively against those of the reactant and product.