The reaction 2NO+O2⟶k2NO2 follows the mechanismNO+NO⇌ keq N2O2 (in rapid equilibrium)N2O2+O2⇌k22NO2 (slow)The rate law of the reaction will be

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

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, $\frac{1}{2} \frac{d [{NO}_{2}]}{d t} = k_{2} K_{eq} [{NO}^{2} [O_{2}]$

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

Solution:

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