Half-life, t1/2, 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.:


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 8.96×10-2 M is presented in the diagram above. The 1st half-life of H2O2 is about 480s. The 2nd and 3rd 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


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


In summary,


Simple rate equation Integral rate equation


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}

Next, we shall explore the various methods in monitoring the progress of a reaction and analysing the data obtained.


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