# How do you find first order half life?

Aug 31, 2017

${t}_{\text{1/2}} = \frac{\ln 2}{k}$

Well, consider a general first-order rate law for a one-reactant reaction:

$A \to B$

$r \left(t\right) = k \left[A\right]$

where $k$ is the rate constant and $\left[A\right]$ is the concentration of $A$ in $\text{M}$.

This is numerically equal to:

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

where $\frac{d \left[A\right]}{\mathrm{dt}}$ denotes an instantaneous rate of change in concentration of $A$ over time.

By separation of variables:

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

Integrate the left-hand side from time zero to time $t$, and the right-hand side from initial to current concentration.

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

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

Thus, we obtain the first-order integrated rate law:

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

For a half-life, we have a current concentration of ${\left[A\right]}_{0} = \frac{1}{2} \left[A\right]$, so:

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

where ${t}_{\text{1/2}}$ is the half-life.

Using the properties of logarithms:

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

$\implies \ln \left(\frac{1}{2}\right) = - k {t}_{\text{1/2}}$

$\implies \ln 2 = k {t}_{\text{1/2}}$

$\implies \textcolor{b l u e}{\overline{\underline{| \stackrel{\text{ ")(" "t_"1/2" = (ln2)/k" }}{|}}}}$