The reaction $2\mathrm{NO}+{\mathrm{O}}_2\stackrel{k}{⟶}2{\mathrm{NO}}_2$ follows the mechanism

$\mathrm{NO}+\mathrm{NO}\stackrel{ }{\stackrel{ k_{eq}}{\rightleftharpoons }}{\mathrm{N}}_2{\mathrm{O}}_2$ (in rapid equilibrium)

${\mathrm{N}}_2{\mathrm{O}}_2+{\mathrm{O}}_2\stackrel{k_2}{\rightleftharpoons }2{\mathrm{NO}}_2$ (slow)

The rate law of the reaction will be

Since the second step is slow, the rate law is $\frac{1}{2}\frac{\mathrm{d}\left[{\mathrm{NO}}_2\right]}{\mathrm{d}t}=k_2\left[{\mathrm{N}}_2{\mathrm{O}}_2\right]\left[{\mathrm{O}}_2\right]$ From the fast equilibrium, we have $K_{\mathrm{eq}}=\frac{\left[{\mathrm{N}}_2{\mathrm{O}}_2\right]}{[\mathrm{NO}{]}^2}$ This gives $\left[{\mathrm{N}}_2{\mathrm{O}}_2\right]=K_{\mathrm{eq}}[\mathrm{NO}{]}^2$Hence, $\frac{1}{2}\frac{\mathrm{d}\left[{\mathrm{NO}}_2\right]}{\mathrm{d}t}=k_2{K}_{\mathrm{eq}}\left[{\mathrm{NO}}^2\left[{\mathrm{O}}_2\right]\right.$