The reaction
(r)
$2NO(g)$ $+$ $O_2(g)$ $\longrightarrow$ $ 2NO_2(g)$
is second order with respect to $NO$ and first order with respect to $O_2$.
Justify considering the following elementary processes:
(a)
$NO$ $+$ $O_2$ $\longrightarrow$ $NO_3$
(fast)
(b)
$NO_3$ $\longrightarrow$ $NO$ $+$ $O_2$
(fast)
(c)
$NO_3$ $+$ $NO$ $\longrightarrow$ $2NO_2$
(slow)
(c) imposes the rate, so:
$v_r=v_c=k_c[NO_3][NO]$
We have to substitute $[NO_3]$ supposing that its molarity finishes rapidly to be constant (equilibrium!):
Disappearance:
$v$ $=$ $k_c[NO_3][NO]$ $+$ $k_b[NO_3]$
(1)$v$ $\approx$ $k_b[NO_3]$, because $k_c\lt \lt k_b$
Appearance:
(2)$v_a$ $=$ $k_a[NO][O_2]$
Equilibrium:
$v=v_a$
$[NO_3]$ $=$ $\frac{k_a}{k_b}[NO][O_2]$
Introducing into (c):
$v_r$ $=$ $k_c\frac{k_a}{k_b}[NO]^2[O_2]$
