Rate laws – integral form

An integral rate law mathematically expresses the rate of a reaction in terms of the initial concentration and the measured concentration of one or more reactants over a particular time.

Such a rate law can be derived from its differential form via simple calculus. Consider the decomposition of hydrogen peroxide to oxygen:

$H_2O_2(aq)\rightleftharpoons H_2O(l)+\frac{1}{2}O_2(g)$

The rate law is experimentally determined to be first order: rate = k[H₂O₂]. If we are monitoring the progress of the reaction by measuring the change in concentration of the peroxide, the differential form of the rate law is:

$-\frac{d[H_2O_2]}{dt}=k[H_2O_2]\; \; \; \; \; \; \; \; 12$

Rearranging and integrating eq12,

$\int_{[H_2O_2]_0}^{[H_2O_2]}\frac{1}{[H_2O_2]}d[H_2O_2]=-k\int_{0}^{t}dt$

where [H₂O₂]₀ is the concentration of the peroxide at t = 0, i.e. its initial concentration. We then get:

$ln\frac{[H_2O_2]}{[H_2O_2]_0}=-kt\; \; \; or\; \; \; ln[H_2O_2]=-kt+ln[H_2O_2]_0\; \; \; \; \; \; \; \; 13$

Eq13 is the integral form of the first order rate law for the decomposition of hydrogen peroxide. The general equation for a species, A, that participates in a first order reaction of vA → B is:

$ln\frac{[A]}{[A]_0}=-vkt\; \; \; or\; \; \; ln[A]=-vkt+ln[A]_0\; \; \; \; \; \; \; \; 14$

For a zero order reaction, e.g. the decomposition of excess N₂O on hot platinum, $N_2O\rightarrow N_2+\frac{1}{2}O_2$ , the rate equation is $-\frac{d[N_2O]}{dt}=k$ , and by integrating both sides of the differential rate equation, its integral form is: