Half-life

Half-life, t_1/2, is the time for the amount of a reactant to reduce to half of its initial value. For the first order decomposition of hydrogen peroxide, the rate equation is given by eq13 of the previous article: $ln\frac{[H_2O_2]}{[H_2O_2]_0}=-kt$ .

Assuming that volume of the reacting mixture remains constant, the ratio of the amounts of peroxide, n, at t = t and t = 0 is the same as the ratio of the concentrations of peroxide, $\frac{n}{V}$ , at t = t and t = 0. Hence, hydrogen peroxide’s half-life is the time taken for its initial concentration to fall to half, i.e.:

$ln\frac{\frac{1}{2}[H_2O_2]_0}{[H_2O_2]_0}=-kt_{\frac{1}{2}}$

$t_{\frac{1}{2}}=\frac{ln2}{k}\approx \frac{0.693}{k}\; \; \; \; \; \; \; \; 19$

Eq19 shows that the half-life of a species in a first order reaction does not depend on its concentration and only depends on the rate constant, k, of the reaction.

Experimental data for the decomposition of H₂O₂ with an initial concentration of 8.96×10^-2M is presented in the diagram above. The 1^st half-life of H₂O₂ is about 480s. The 2^nd and 3^rd half-lives, which are the time taken for $\frac{8.96\times10^{-2}}{2}$ M to fall to $\frac{8.96\times10^{-2}}{4}$ M and $\frac{8.96\times10^{-2}}{8}$ M respectively, are also about 480s each. Hence, the experimental value of the successive half-lives of the first order decomposition of H₂O₂ is a constant and is consistent with the theoretically derived eq19.

Using eq15 and eq17 from the previous article, the half-lives of a zero order reaction and a second order reaction for a chemical species A,are

$t_{\frac{1}{2}}=\frac{[A]_0}{2k}\; \; \; \; \; \; \; \; 20$

and

$t_{\frac{1}{2}}=\frac{1}{k[A]_0}\; \; \; \; \; \; \; \; 21$

respectively.

In summary,

Order	Simple rate equation	Integral rate equation	Half-life
0	$rate=k$	$[A]=-kt+[A]_0$	$t_{\frac{1}{2}}=\frac{[A]_0}{2k}$
1	$rate=k[A]$	$ln[A]=-kt+[A]_0$	$t_{\frac{1}{2}}=\frac{ln2}{k}$
2	$rate=k[A]^2$	$\frac{1}{[A]}=kt+\frac{1}{[A]_0}$	$t_{\frac{1}{2}}=\frac{1}{k[A]_0}$