# A first order reaction has a half life of 27.5 minutes. If the initial concentration of the reactant is 0.400 M, when will the concentration drop to 0.100 M?

Dec 3, 2015

We know that it is a first order reaction, so as a result, we know that there is only one participant, i.e. it is a unimolecular reaction.

We write this rate law as:

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

By separation of variables, we get (where ${t}_{0} = 0$ and $\left[A\right] = {\left[A\right]}_{0} / 2$):

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

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

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

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

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

$\setminus m a t h b f \left({t}_{\text{1/2}} = \frac{\ln 2}{k}\right)$

So the half-life is independent of the initial or current concentrations.

Therefore, we can assume that after the half-life passes a certain number of times, the concentration is exactly halved, that number of times.

So, we can figure this out by just saying that the concentration was quartered, meaning that two half-lives passed ($\text{1/"2^n = "1/4}$ when $n = 2$, where $n$ is the number of half-lives passed).

As a result, $2 {t}_{\text{1/2}}$ is the amount of time it took, i.e. $\textcolor{b l u e}{\text{55 min.}}$