The half-life method for analysing rate data is useful when the rate law involves a single reactant, i.e.
To determine the values of k and i, we let the reaction proceed, measure the appropriate physical property and plot the data in a concentration versus time graph.
If the curve is a straight line (see graph above), the gradient or the change in concentration versus the change in time (i.e. rate of reaction) is a constant. This implies that i = 0 in eq22, i.e. a zero order reaction. We can further confirm this by finding the first three successive half-lives from the graph (there is no need to consider the subsequent half-lives) and substitute them, along with the initial concentration of the reactant, in eq20 from the article on half-life: . The three substitutions should give three different values of k that may be close to one another. The average of these values of k is then taken to give the rate equation: rate = kave.
If the curve is concave (see diagram above), it represents a higher order reaction. Similarly, the first three successive half-lives is determined from the graph. If the values are relatively constant, the graph corresponds to a first order reaction, with i = 1. The value of the rate constant is then found using eq19 from the article on half-life: , resulting in the rate equation: rate = k[A].
If the half-lives are very different from one another, they could be those of a second order reaction, i = 2. We can verify this by substituting the three half-life values, along with the initial concentration of the reactant, in eq21 from the article on half-life: , which should give three different values of k that may be close to one another. The average value of the rate constant can be found and hence, the rate equation: rate = kave[A]2.