How is the half-life of a first-order reaction affected by change in concentration?

Jan 28, 2017

It's not. The half-life of a reactant is intrinsic to the identity of the reactant, which has its own rate constant.

A first-order reaction rate law, for the reaction

$A \to B$,

would be written as:

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

where $r \left(t\right)$ is the rate of reaction in $\text{M/s}$, $k$ is the rate constant in ${\text{s}}^{- 1}$, and $\frac{d \left[A\right]}{\mathrm{dt}}$ is the change in the concentration $\left[A\right]$ of $A$ over the change in time $t$ (otherwise known as the rate of disappearance).

When you separate variables by multiplying by $- \frac{\mathrm{dt}}{\left[A\right]}$ on both sides, you'd get:

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

To observe the process over time, we will be integrating from $0$ to $t$ on the left and ${\left[A\right]}_{0}$ (initial concentration) to $\left[A\right]$ (current concentration) on the right.

We get:

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

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

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

which is just a rearrangement of the integrated rate law for a first-order reaction.

Now, if we say that we want to know how much time it takes for $\left[A\right]$ to be $\frac{{\left[A\right]}_{0}}{2}$, we are asking for the half-life, ${t}_{\text{1/2}}$.

Therefore:

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

Since $\ln \left(\frac{1}{2}\right) = \ln \left({2}^{- 1}\right) = - \ln 2$, we have:

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

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

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

So, the half-life depends only on the rate constant for the first-order reaction that is occurring.