The Half-life of a Reaction

The Half-Life of a Reaction

The half-life of a reaction (t_1/2) is the time required for one-half of a given amount of reactant to be consumed. In each succeeding half-life, half of the remaining concentration of the reactant is consumed. Using the decomposition of hydrogen peroxide (see this lesson) as an example, we find that during the first half-life (from 0.00 hours to 6.00 hours), the concentration of H₂O₂ decreases from 1.000 M to 0.500 M.

During the second half-life (from 6.00 hours to 12.00 hours), it decreases from 0.500 M to 0.250 M; during the third half-life, it decreases from 0.250 M to 0.125 M. The concentration of H₂O₂ decreases by half during each successive period of 6.00 hours. The decomposition of hydrogen peroxide is a first-order reaction, and, as can be shown, the half-life of a first-order reaction is independent of the concentration of the reactant. However, half-lives of reactions with other orders depend on the concentrations of the reactants.

First-Order Reactions

We can derive an equation for determining the half-life of a first-order reaction from the alternate form of the integrated rate law as follows:

\(\begin{array}{}\\ \hfill \text{ln}\phantom{\rule{0.2em}{0ex}}\cfrac{{\left[A\right]}_{0}}{\left[A\right]}& =& kt\hfill \\ \hfill t& =& \text{ln}\phantom{\rule{0.2em}{0ex}}\cfrac{{\left[A\right]}_{0}}{\left[A\right]}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{1}{k}\hfill \end{array}\)

If we set the time t equal to the half-life, \({t}_{1\text{/}2},\) the corresponding concentration of A at this time is equal to one-half of its initial concentration. Hence, when \(t={t}_{1\text{/}2},\)\(\left[A\right]=\phantom{\rule{0.1em}{0ex}}\cfrac{1}{2}{\left[A\right]}_{0}.\)

Therefore:

\(\begin{array}{cc}\hfill {t}_{1\text{/}2}& =\text{ln}\phantom{\rule{0.2em}{0ex}}\cfrac{{\left[A\right]}_{0}}{\cfrac{1}{2}{\left[A\right]}_{0}}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{1}{k}\hfill \\ & =\text{ln}\phantom{\rule{0.2em}{0ex}}2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{1}{k}\phantom{\rule{0.1em}{0ex}}=0.693\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{1}{k}\hfill \end{array}\)

Thus:

\({t}_{1\text{/}2}=\phantom{\rule{0.1em}{0ex}}\cfrac{0.693}{k}\)

We can see that the half-life of a first-order reaction is inversely proportional to the rate constant k. A fast reaction (shorter half-life) will have a larger k; a slow reaction (longer half-life) will have a smaller k.

Example

Calculation of a First-order Rate Constant using Half-Life

Calculate the rate constant for the first-order decomposition of hydrogen peroxide in water at 40 °C, using the data given in the figure below.

The decomposition of H₂O₂\(\left({\text{2H}}_{2}{\text{O}}_{2}\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{2H}}_{2}\text{O}+{\text{O}}_{2}\right)\) at 40 °C is illustrated. The intensity of the color symbolizes the concentration of H₂O₂ at the indicated times; H₂O₂ is actually colorless.

Solution

The half-life for the decomposition of H₂O₂ is 2.16 \(×\) 10⁴ s:

\(\begin{array}{ccc}\hfill {t}_{1\text{/}2}& =& \cfrac{0.693}{k}\hfill \\ \hfill k& =& \cfrac{0.693}{{t}_{1\text{/}2}}\phantom{\rule{0.1em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\cfrac{0.693}{2.16\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{s}}\phantom{\rule{0.1em}{0ex}}=3.21\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\phantom{\rule{0.2em}{0ex}}{\text{s}}^{-1}\hfill \end{array}\)

Second-Order Reactions

We can derive the equation for calculating the half-life of a second order as follows:

\(\cfrac{1}{\left[A\right]}\phantom{\rule{0.1em}{0ex}}=kt+\phantom{\rule{0.2em}{0ex}}\cfrac{1}{{\left[A\right]}_{0}}\)

or

\(\cfrac{1}{\left[A\right]}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\cfrac{1}{{\left[A\right]}_{0}}\phantom{\rule{0.1em}{0ex}}=kt\)

If

\(t={t}_{1\text{/}2}\)

then

\(\left[A\right]=\phantom{\rule{0.1em}{0ex}}\cfrac{1}{2}{\left[A\right]}_{0}\)

and we can write:

\(\begin{array}{ccc}\hfill \cfrac{1}{\cfrac{1}{2}{\left[A\right]}_{0}}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\cfrac{1}{{\left[A\right]}_{0}}& =& k{t}_{1\text{/}2}\hfill \\ \hfill 2{\left[A\right]}_{0}-\phantom{\rule{0.2em}{0ex}}\cfrac{1}{{\left[A\right]}_{0}}& =& k{t}_{1\text{/}2}\hfill \\ \hfill \cfrac{1}{{\left[A\right]}_{0}}& =& k{t}_{1\text{/}2}\hfill \end{array}\)

Thus:

\({t}_{1\text{/}2}=\phantom{\rule{0.1em}{0ex}}\cfrac{1}{k{\left[A\right]}_{0}}\)

For a second-order reaction, \({t}_{1\text{/}2}\) is inversely proportional to the concentration of the reactant, and the half-life increases as the reaction proceeds because the concentration of reactant decreases. Consequently, we find the use of the half-life concept to be more complex for second-order reactions than for first-order reactions. Unlike with first-order reactions, the rate constant of a second-order reaction cannot be calculated directly from the half-life unless the initial concentration is known.

Zero-Order Reactions

We can derive an equation for calculating the half-life of a zero order reaction as follows:

\(\left[A\right]=\text{−}kt+{\left[A\right]}_{0}\)

When half of the initial amount of reactant has been consumed \(t={t}_{1\text{/}2}\) and \(\left[A\right]=\phantom{\rule{0.1em}{0ex}}\cfrac{{\left[A\right]}_{0}}{2}.\) Thus:

\(\begin{array}{ccc}\hfill \cfrac{{\left[\text{A}\right]}_{0}}{2}& =& \text{−}k{t}_{1\text{/}2}+{\left[\text{A}\right]}_{0}\hfill \\ \hfill k{t}_{1\text{/}2}& =& \cfrac{{\left[\text{A}\right]}_{0}}{2}\hfill \end{array}\)

and

\({t}_{1\text{/}2}=\phantom{\rule{0.2em}{0ex}}\text{}\cfrac{{\left[A\right]}_{0}}{2k}\)

The half-life of a zero-order reaction increases as the initial concentration increases.

Equations for both differential and integrated rate laws and the corresponding half-lives for zero-, first-, and second-order reactions are summarized in the table below.