# What is the order of the reaction if doubling the concentration of A divides the half-life in 4?

Sep 20, 2017

This is third order. See the derivation below for why... but note that third-order decompositions are quite rare...

The rate law for a third-order reaction

$A \to B$

is:

$r \left(t\right) = k {\left[A\right]}^{3} = - \frac{\Delta \left[A\right]}{\Delta t}$

For an infinitesimally small change, we write:

$k {\left[A\right]}^{3} = - \frac{d \left[A\right]}{\mathrm{dt}}$

And we rearrange to get:

$- k \mathrm{dt} = \frac{1}{{\left[A\right]}^{3}} d \left[A\right]$

The integral then gives:

$- {\int}_{0}^{t} k \mathrm{dt} = {\int}_{{\left[A\right]}_{0}}^{\left[A\right]} \frac{1}{{\left[A\right]}^{3}} d \left[A\right]$

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

$k t = \frac{1}{2 {\left[A\right]}^{2}} - \frac{1}{2 {\left[A\right]}_{0}^{2}}$

Thus, the third-order integrated rate law is:

$\overline{\underline{| \stackrel{\text{ ")(" "1/(2[A]^2) = kt + 1/(2[A]_0^2)" }}{|}}}$

A half-life is when $\left[A\right] = \frac{1}{2} {\left[A\right]}_{0}$, i.e. the concentration of the reactant halves. Thus, we define the half-life ${t}_{\text{1/2}}$ as:

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

$\frac{2}{{\left[A\right]}_{0}^{2}} = k {t}_{\text{1/2}} + \frac{\frac{1}{2}}{{\left[A\right]}_{0}^{2}}$

$\frac{3}{2 {\left[A\right]}_{0}^{2}} = k {t}_{\text{1/2}}$

Therefore, the third-order half-life is given by:

$\textcolor{b l u e}{{t}_{\text{1/2}} = \frac{3}{2 k {\left[A\right]}_{0}^{2}}}$

Given that the half-life changed by a factor of $\frac{1}{4}$, you can convince yourself that doubling the concentration of $A$ quarters the half-life.