• The chemical reaction $2{\mathrm{O}}_3\to 3{\mathrm{O}}_2$ proceeds as follows :   
  ${\mathrm{O}}_3\stackrel{\text{ Fast }}{⟶}{\mathrm{O}}_2+\mathrm{O}\text{ and }\mathrm{O}+{\mathrm{O}}_3\stackrel{\text{ Slow }}{⟶}2{\mathrm{O}}_2$
  the rate law expression should be 

$$\begin{gathered} \text{ The slow step (step-2) is the rate-determining step } \\ \mathrm{r}=\mathrm{k}\left[{\mathrm{O}}_3\right]\left[\mathrm{O}\right]................\left(1\right) \\ \mathrm{From} \mathrm{step}-1, \\ \begin{array}{rl} & {\mathrm{K}}_{\mathrm{eq}}=\frac{\left[{\mathrm{O}}_2\right]\left[\mathrm{O}\right]}{\left[{\mathrm{O}}_3\right]}\\ & \text{or}\left[\mathrm{O}\right]=\frac{{\mathrm{K}}_{\mathrm{eq}}\left[{\mathrm{O}}_3\right]}{\left[{\mathrm{O}}_2\right]}\dots \dots \dots ..\text{(ii)}\end{array} \\ \text{From eq.(i) and (ii),} \\ \begin{array}{rl} & \mathrm{r}={\mathrm{kK}}_{\mathrm{eq}}\frac{{\left[{\mathrm{O}}_3\right]}^2}{\left[{\mathrm{O}}_2\right]}={\mathrm{k}}^{\mathrm{\prime }}{\left[{\mathrm{O}}_3\right]}^2{\left[{\mathrm{O}}_2\right]}^{-1}\\ & [\because {\mathrm{k}}^{\mathrm{\prime }}=\mathrm{kKeq}]\end{array} \end{gathered}$$

Therefore, the correct option is (B).