The reaction $2 NO + O_{2} \overset{k}{⟶} 2 {NO}_{2}$ follows the mechanism
$NO + NO \overset{}{\overset{k_{e q}}{⇌}} N_{2} O_{2}$ (in rapid equilibrium)
$N_{2} O_{2} + O_{2} \overset{k_{2}}{⇌} 2 {NO}_{2}$ (slow)
The rate law of the reaction will be

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$r = k {[{NO}_{2}]}^{2} [O_{2}]$ with $k = K_{eq} / k_{2}$

$r = k [NO]^{2} [O_{2}]$ with $k = k_{2} K_{eq}$

$r = k [NO]^{2} [O_{2}]$ with $k = 2 k_{2} K_{eq}$

$r = k {[{NO}_{2}]}^{2} [O_{2}]$ with $k = k_{2} / K_{eq}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Since the second step is slow, the rate law is $\frac{1}{2} \frac{d [{NO}_{2}]}{d t} = k_{2} [N_{2} O_{2}] [O_{2}]$ From the fast equilibrium, we have $K_{eq} = \frac{[N_{2} O_{2}]}{[NO]^{2}}$ This gives $[N_{2} O_{2}] = K_{eq} [NO]^{2}$ Hence, $\frac{1}{2} \frac{d [{NO}_{2}]}{d t} = k_{2} K_{eq} [{NO}^{2} [O_{2}]$