Question

The chemical reaction, 2O₃ $\to$ 3O₂ proceeds as follows :
Step 1 : O₃ ${\rightleftharpoons}$ O₂ + O .......(fast)
Step 2 : O + O₃ $\to$ 2O₂ .......(slow)
The rate law expression should be

• A. $r\, = \,k'[O_3][O_2]$
• B. $r = k'[O_3]^2[O_2]^{-1}$
• C. $r = k'[O_3]^2$
• D. unpredictable

Answer 1

Answer: (B)

As the slowest step is the rate determining step, hence from step 2,
$r = k\left[O_{3}\right]\left[O\right]\quad\quad\quad\quad\quad\quad\quad\quad...\left(i\right)$
From step 1, $K_{eq} = \frac{\left[O_{2}\left[O\right]\right]}{\left[O_{3}\right]}$
or $\left[O\right]=\frac{K_{eq}\left[O_{3}\right]}{\left[O_{2}\right]}\quad \quad \quad \quad \quad \quad \quad \quad ...\left(ii\right)$
From eq. (i) and (ii),
r = k $K_{eq} \frac{\left[O_{3}\right]^{2}}{\left[O_{2}\right]}=k'\left[O_{3}\right]^{2}\left[O_{2}\right]^{-1}\quad\left[\because \,k' = kK_{eq}\right]$