# How do you derive first order half life?

Feb 5, 2017

You can always start from the first-order rate law:

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

for the reaction:

$A \to B$

Separate variables:

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

Integrate from the initial to the final state. It is convenient to set ${t}_{0} = 0$.

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

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

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

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

For the half-life, at time $t = {t}_{\text{1/2}}$, the concentration of $A$ dropped to $\frac{1}{2} {\left[A\right]}_{0}$. Therefore:

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

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

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

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

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

Therefore:

$\textcolor{b l u e}{{t}_{\text{1/2}} = \ln \frac{2}{k}}$