# Is the half-life of a first-order decay dependent on temperature?

Dec 13, 2016

Of course. As rate constants depend on temperature, so does the first-order half-life.

For a first-order process, let us consider simply a reactant $A$, whose rate law is:

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

where $\left[A\right]$ is the concentration of $A$, and $- \frac{d \left[A\right]}{\mathrm{dt}}$ is the rate of disappearance of $A$ as time passes.

To determine the change in concentration over time, we consider an interval of time from $0$ seconds to some general time $t$.

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

Integrate from $t = 0$ to $t = t$ on the left side, and from the initial concentration ${\left[A\right]}_{0}$ to the final concentration $\left[A\right]$ on the right side.

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

(the integral of $\frac{1}{x}$ is $\ln | x |$.)

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

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

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

If we let ${t}_{\text{1/2}}$ be the half-life time duration, then at $t = {t}_{\text{1/2}}$, we have that $\left[A\right] = \frac{1}{2} {\left[A\right]}_{0}$. Thus, we get:

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

$= - \ln \left(\frac{1}{2}\right) = \ln 2$

So...

$\textcolor{b l u e}{{t}_{\text{1/2}}} = \frac{\ln 2}{k} \approx \textcolor{b l u e}{\frac{0.693}{k}}$

Although the first-order half-life depends on the rate constant $k$ and not the initial concentration ${\left[A\right]}_{0}$, $\boldsymbol{k}$ is a function of temperature, and increases with increasing temperature.

Note the Arrhenius equation:

$\ln \left({k}_{2} / {k}_{1}\right) = - \frac{{E}_{a}}{R} \left[\frac{1}{T} _ 2 - \frac{1}{T} _ 1\right]$

where clearly, $k$ is related to $T$.

For instance, if ${T}_{2} > {T}_{1}$, then $\frac{1}{T} _ 2 - \frac{1}{T} _ 1 < 0$, and thus, $\ln \left({k}_{2} / {k}_{1}\right) > 0$, meaning that ${k}_{2} > {k}_{1}$. Therefore, we've shown that $k$ increases with increasing temperature.