# How do you calculate half life decay?

Jun 16, 2017

It depends on the order of the reaction with respect to the reactant undergoing half-life decay.

The common half-life expressions for one reactant $A$ are:

${t}_{\text{1/2}} = \frac{{\left[A\right]}_{0}}{2 k}$ (zero order)

${t}_{\text{1/2}} = \frac{\ln 2}{k}$ (first order)

${t}_{\text{1/2}} = \frac{1}{k {\left[A\right]}_{0}}$ (second order)

${t}_{\text{1/2}} = \frac{3}{2 k {\left[A\right]}_{0}^{2}}$ (third order)

where ${\left[A\right]}_{0}$ is the initial concentration in $\text{M}$ and $k$ is the rate constant in the appropriate units to obtain time units of $\text{s}$.

If you wish to derive them, that requires some calculus. I can do first order as an example, as that is the hardest one. For the general rate law

$r \left(t\right) = k \left[A\right] = - \frac{d \left[A\right]}{\mathrm{dt}}$,

of the first-order reaction

$A \to B$,

separation of variables gives:

$- k \mathrm{dt} = \frac{1}{\left[A\right]} d \left[A\right]$

Now, integrate from $0$ to $t$ on the left-hand side and ${\left[A\right]}_{0}$ to $\left[A\right]$ for the right-hand side.

$- {\int}_{0}^{t} k \mathrm{dt} = {\int}_{{\left[A\right]}_{0}}^{\left[A\right]} \frac{1}{\left[A\right]} d \left[A\right]$

The integral of $\frac{1}{x}$ is $\ln | x |$, and $\left[A\right] \ge 0$, so:

$- k t = \ln \left[A\right] - \ln {\left[A\right]}_{0}$

For a half-life, $\left[A\right] = \frac{1}{2} {\left[A\right]}_{0}$, i.e. the concentration of $A$ has halved. So:

$- k {t}_{\text{1/2}} = \ln \left(\frac{1}{2} {\left[A\right]}_{0}\right) - \ln {\left[A\right]}_{0}$

$k {t}_{\text{1/2}} = - \ln \left(\frac{1}{2} \cancel{\frac{{\left[A\right]}_{0}}{{\left[A\right]}_{0}}}\right)$

=> color(blue)(t_"1/2 " ("1st order") = (ln2)/k)