# What is the first-order half-life for "0.50 M" of a substance whose radioactive decay has a rate constant of 0.69 xx 10^(-2) "min"^(-1)?

May 20, 2017

You don't need its concentration. (Why?)

t_("1/2") = "100.5 min"

The Integrated Rate Law for the first-order process is:

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

where:

• $\left[A\right]$ is the concentration of the reactant.
• ${\left[A\right]}_{0}$ is its initial concentration.
• $k$ is the rate constant in ${\text{time}}^{- 1}$.
• $t$ is of course the time in $\text{time}$ units.

For a half-life, we take $t = {t}_{\text{1/2}}$ such that $\left[A\right] = \frac{1}{2} {\left[A\right]}_{0}$. Therefore:

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

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

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

But since $\frac{1}{2} = {2}^{- 1}$ and $\ln \left({a}^{b}\right) = b \ln a$,

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

and:

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

See, the concentration doesn't matter for first-order...

$\textcolor{b l u e}{{t}_{\text{1/2") = (ln2)/(0.69 xx 10^(-2) "min}}^{- 1}}$

$=$ $\textcolor{b l u e}{\text{100.5 min}}$