A possible mechanism for the conversion of ozone to oxygen in the upper atmosphere is.
$O_{3} (g) ⇌ O_{2} (g) + O (g)$ (fast equilibrium)
$O (g) + O_{3} (g) ⟶ 2 O_{2} (g)$ (slow)
Which rate law is consistent with this mechanism?

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$Rate = k [O_{3}]$

$Rate = k {[O_{3}]}^{2}$

$Rate = k [O_{8}] [O]$

$Rate z k {[O_{3}]}^{2} {[O_{2}]}^{- 1}$

answer is D.

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Detailed Solution

Hence rate of reaction is given as:

$rate = k_{2} [O] [O_{3}]$

By chemical $e q^{n}$

$\begin{aligned} k_{1} = \frac{[O_{2}] [0]}{O_{3}} \\ [O] = \frac{k_{1} [O_{3}]}{[O_{2}]} \end{aligned}$

So,

$\begin{array}{c} rate = k_{1} k_{2} {[O_{3}]}^{2} {[O_{2}]}^{- 1} \\ r = k {[O_{3}]}^{2} {[O_{2}]}^{- 1} \end{array}$

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