A hypothetical reaction A₂ + B₂ + 2AB follows the mechanism as given below:

${\mathrm{A}}_2\rightleftharpoons \mathrm{A}+\mathrm{A}$              (fast)

A + B₂ $\to$ AB + B    (slow)

A + B $\to$ AB            (fast)

The order of the overall reaction is

Since (A) is the intermediate reactive species whose concentration is determined from equilibrium step. 

slow step is:

$\mathrm{A}+{\mathrm{B}}_2⟶\mathrm{AB}+\mathrm{B}\text{ (slow) }$

r = k[A][B₂] . . . (i)

From equilibrium step

${\mathrm{A}}_2\rightleftharpoons \mathrm{A}+\mathrm{A}\text{ (fast) }$

${\mathrm{k}}_{\mathrm{eq}}=\frac{[\mathrm{A}{]}^2}{\left[{\mathrm{A}}_2\right]}$

$\therefore \left[\mathrm{A}\right]={\left(k_{\mathrm{eq}}\left[{\mathrm{A}}_2\right]\right)}^{1/2}$

substitute the value of [A] in Eq. (i)

$r=k\cdot k_{\mathrm{eq}}^{1/2}{\left[{\mathrm{A}}_2\right]}^{1/2}\left[{\mathrm{B}}_2\right]$

Thus, order of reaction 

$=\frac{1}{2}+1=1\frac{1}{2}$