The reaction of ${NO}_{2} (g)$ and $O_{3} (g)$ is first-order in ${NO}_{2} (g)$ and $O_{3} (g)$
$2 {NO}_{2} (g) + O_{3} (g) ⟶ N_{2} O_{5} (g) + O_{2} (g)$
The reaction can take place by mechanism :
P) ${NO}_{2} + O_{3} \overset{slow}{⟶} {NO}_{3} + O_{2}$
${NO}_{3} + {NO}_{2} \overset{fast}{⟶} N_{2} O_{5}$
Q) $O_{3} \underset{k_{b}}{\overset{k_{o} fust}{⇌}} O_{2} + [O]$
${NO}_{2} + O \overset{slow}{⟶} {NO}_{3}$
${NO}_{3} + {NO}_{2} \overset{fast}{⟶} N_{2} O_{5}$
Select the correct mechanism.

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P only

Both P and Q

Q only

None of these

answer is C.

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Detailed Solution

P) $r = k [{NO}_{2}] [O_{3}]$

Q) $(Q) r = k [{NO}_{2}] [O]$

and, $[O] = k_{eq} \cdot \frac{[O_{3}]}{[O_{2}]}$

$r = \frac{k^{'} [{NO}_{2}] [O_{3}]}{[O_{2}]}$

Hence, the correct option is c.

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