Question

The chemical reaction, $2O_3\to 3O_2$ proceeds as
$O_3\rightleftharpoons O_2+\left[O\right]\left(fast\right)$
$\left[O\right]+O_3\to 2O_2\left(slow\right)$
The rate law expression will be :

• A. $Rate=l\left[O\right]\left[O_3\right]$
• B. $Rate=k\left[O_3{]}^2\right[O_2{]}^{-1}$
• C. $Rate=k[O_3{]}^2$
• D. $Rate=k\left[O_2\right]\left[O\right]$

Answer 1

The correct option is B $Rate=k\left[O_3{]}^2\right[O_2{]}^{-1}$

$O_3\underset{k_1}{\stackrel{k_1}{\rightleftharpoons }}{O}_2+\left[O\right]\left(fast\right)$
$\left[O\right]+O_3\stackrel{k_2}{\to }=2O_2\left(slow\right)$
Rate of reaction is determined by slow step hence,
$Rate=k_2\left[O\right]\left[O_3\right]$
$\left[O\right]$ is unstable intermediate so substitute the value of $\left[O\right]$ in above equation.
Rate of forward reaction $=k_1\left[O_3\right]$
Rate of backward reaction $=k_{-1}\left[O_2\right]\left[O\right]$
At equilibrium,
Rate of forward reaction $=$ Rate of backward reaction
$k_1\left[O_3\right]=k_{-1}\left[O_2\right]\left[O\right]$
$\left[O\right]=\frac{k_1\left[O_3\right]}{k_{-1}\left[O_2\right]}$
$Rate=k_1\left(\frac{k_1\left[O_3\right]}{k_{-1}\left[O_2\right]}\right)\left[O_3\right]=\frac{k[O_3{]}^2}{\left[O_2\right]}$.