Q51 / 180 · JIPMER 2017 · Chemistry
For the chemical reaction, $2 \mathrm{O}_3 \rightleftharpoons 3 \mathrm{O}_2$
The reaction proceed as follows
$\begin{gathered}\mathrm{O}_3 \rightleftharpoons \mathrm{O}_2+\mathrm{O} \text { (fast) } \\ \mathrm{O}+\mathrm{O}_3 \longrightarrow 2 \mathrm{O}_2 \text { (slow) }\end{gathered}$
The rate law expression will be
- A$r=k^{\prime}\left[\mathrm{O}_3\right]^2$
- Correct answerB$r=k^{\prime}\left[\mathrm{O}_3\right]^2\left[\mathrm{O}_2\right]^{-1}$
- C$r=k^{\prime}\left[\mathrm{O}_3\right]\left[\mathrm{O}_2\right]$
- DUnpredictable
Solution
As slowest step is the rate determining step.
Hence, from slow reaction, $r=K[\mathrm{O}]\left[\mathrm{O}_3\right]...(i)$
From fast reaction, $K_C=\frac{\left[\mathrm{O}_2\right][\mathrm{O}]}{\left[\mathrm{O}_3\right]}$
$\begin{aligned} {[\mathrm{O}] } & =\frac{K_C \cdot\left[\mathrm{O}_3\right]}{\left[\mathrm{O}_2\right]}, \text { i.e. } \\ r & =\frac{K_C\left[\mathrm{O}_3\right]}{\left[\mathrm{O}_2\right]} \cdot\left[\mathrm{O}_3\right]\end{aligned}$
$\Rightarrow K_C \cdot\left[\mathrm{O}_3\right]^2\left[\mathrm{O}_2\right]^{-1}$
$\therefore$ Rate $(r)=K^{\prime}\left[\mathrm{O}_3\right]^2 \cdot\left[\mathrm{O}_2\right]^{-1}$ $\left(K^{\prime}=K_C \cdot K\right)$
Hence, the correct answer is option: 'B'.