The chemical reaction 2O₃+ 3O₂ proceeds as follows:

${\mathrm{O}}_3\rightleftharpoons {\mathrm{O}}_2+\mathrm{O}$            (fast)

$\mathrm{O}+{\mathrm{O}}_3\to 2{\mathrm{O}}_2$          (slow)

The rate law expression should be

$2{\mathrm{O}}_3\stackrel{k_{\mathrm{I}}}{⟶}3{\mathrm{O}}_2$

Mechanism:

${\mathrm{O}}_3\stackrel{k_{\mathrm{eq}}}{\rightleftharpoons }{\mathrm{O}}_2+\mathrm{O}$      (fast)

$\mathrm{O}+{\mathrm{O}}_3\stackrel{k}{⟶}2{\mathrm{O}}_2$(slow)

Rate of the reaction from the slow step is

$r=k\left[\mathrm{O}\right]\left[{\mathrm{O}}_3\right]$

Fast step is in equilibrium, the concentration of [O] is calculated from this step

$k_{\mathrm{eq}}=\frac{\left[{\mathrm{O}}_2\right]\left[\mathrm{O}\right]}{\left[{\mathrm{O}}_3\right]}\therefore \left[\mathrm{O}\right]=\frac{k_{\mathrm{u}}\cdot \left[{\mathrm{O}}_3\right]}{\left[{\mathrm{O}}_2\right]}$

Substitute the value of [O] in Eq. (i)

$r=\frac{\left.k\cdot k_{\mathrm{eq}}\cdot {\mathrm{O}}_3\right]\left[{\mathrm{O}}_3\right]}{{\mathrm{O}}_2}=\frac{k_1{\left[{\mathrm{O}}_3\right]}^2}{\left[{\mathrm{O}}_2\right]}=k_1{\left[{\mathrm{O}}_3\right]}^2{\left[{\mathrm{O}}_2\right]}^{-1}$