For the chemical reaction 2 O _ 3 ⟶ 3 O _ 2 , the reaction proceeds...

For the chemical reaction

$2{O}_{3} ⟶ 3{O}_{2}$ , the reaction proceeds as follows

${O}_{3} ⇌ {O}_{2} + O$ (Fast)

$O + {O}_{3} ⟶ 2{O}_{2}$ ,& (Slow)

the rate law expression should be given as

Options

A $r = k\left[{O}_{3}\right]2\left[{O}_{2}\right]- 1$
B $r = k\left[{O}_{3}\right]2$
C $r = k \left[{O}_{3}\right] \left[{O}_{2}\right]$
Dunpredictable

Correct answer

A. $r = k\left[{O}_{3}\right]2\left[{O}_{2}\right]- 1$

Step-by-step solution

Let $k$ be the rate constant of given reaction

$2{O}_{3}\overset{k}{⟶}3{O}_{2}$

From the slowest step $r = k^{'} \left[O\right]\left[{O}_{3}\right]...& \left(i\right)To eliminate \left[O\right], (From fast step)$ {k}_{eq} = \left[{O}_{2}\right]\left[O\right]\left[{O}_{3}\right]

\left[O\right] = {k}_{eq.} \left[{O}_{3}\right]\left[{O}_{2}\right] $Now, substituting value of \left[O\right]$ in Eq \left(i\right) $$ r = k^{'}{k}_{eq} \left[{O}_{3}\right]\left[{O}_{3}\right] \left[{O}_{2}\right]

Let $k^{'}{k}_{eq} = k$

Thus, $r = k \left[{O}_{3}\rright]2\left[{O}_{2}\right]-$ 1 $$

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