Question

The gaseous decomposition of ozone, $2O_3⟶3O_2$, obeys the rate law $r=-\frac{d\left[O_3\right]}{dt}=\frac{k{\left[O_3\right]}^2}{\left[O_2\right]}$.
Show that the following mechanism is consistent with the above rate law:
$O_3\stackrel{K_{eq}}{\rightleftharpoons }{O}_2+O$ (fast)
$O+O_3\underset{}{\overset{k_t}{\to }}2O_2$ (slow)

Answer 1

From the slow rate determining step
$r=-\frac{d\left[O_3\right]}{dt}=k_t\left[O\right]\left[O_3\right]$
From the fast reaction,
$K_{eq}=\frac{\left[O_2\right]\left[O\right]}{\left[O_3\right]}$
or $\left[O\right]=\frac{K_{eq}\left[O_3\right]}{\left[O_2\right]}$
Substituting the value of $\left[O\right]$ in the above expression
$r=-\frac{d\left[O_3\right]}{dt}=\frac{k_1{K}_{eq}{\left[O_3\right]}^2}{\left[O_2\right]}=\frac{k{\left[O_3\right]}^2}{\left[O_2\right]}$