(d) Half-life Method to Determine the Order of Irreversible Reaction

Consider the reaction \[\alpha A + \beta B + \cdots \rightarrow \text{ products}\] for which, \[-r_A = kC_A^{a}C_B^{b}\cdots\] where \(a, b, \ldots\) are the orders of reaction with respect to \(A,B,\ldots\) respectively. If we start with stoichiometric ratio of reactants, then they will be present in the reactor in the same proportion through out the reaction. Thus for reactants \(A\) and \(B\), at any time, \(C_B/C_A=\beta/\alpha\). Using this in above equation, we get \[\begin{align*} -\frac{dC_A}{dt} &= kC_A^{a}\left(\frac{\beta}{\alpha}C_A\right)^{b}\cdots =k\left(\frac{\beta}{\alpha}\right)^b\cdots C_A^{a+b+\cdots} \\ &= k'C_A^n \qquad (\text{where }n=a+b+\cdots)\end{align*}\] where \(n\) is the overall order of the reaction.

Integrating the above equation for \(n\ne1\) gives, \[\begin{align*} C_A^{1-n}-C_{A0}^{1-n} &= (n-1)kt \nonumber \\ \text{At $t=t_{1/2}$, $C_A=C_{A0}/2$. Substituting this in above, we get} \\ (C_{A0}/2)^{1-n} - C_{A0}^{1-n} &= (n-1)kt \nonumber \\ \Longrightarrow \quad t_{1/2} = \frac{(2^{n-1}-1)}{(n-1)k}C_{A0}^{1-n} \qquad (\text{for } n\ne 1) \tag*{(9)}\end{align*}\] Eqn.(9) is represented in Fig.(8).


The overall order of reaction is obtained graphically using the data from a series of half-life experiments. Note that all the runs are to be conducted at different \(C_{A0}\).

Fractional Life Method:

Let \(F=C_A/C_{A0}\) = fraction of initial concentration (\(C_{A0}\)) after a time \(t_F\). Then, \[t_F = \frac{(F^{1-n}-1)}{(n-1)k}C_{A0}^{1-n} \qquad (\text{for } n\ne 1)\]