Question

The chemical reaction $2O_3\to 3O_2$ involves two elementary steps :
$O_3\stackrel{fast}{\to }{O}_2+O;O+O_3\stackrel{Slow}{\to }2O_2$
The differential rate law expression will be :

• A. $rate=k[O_3{]}^2$
• B. $rate=k\left[O_3{]}^2\right[O_2]$
• C. $rate=k\left[O_3{]}^2\right[O_2{]}^{-1}$
  
• D. $rate=k\left[O_3{]}^2\right[O_2{]}^{-3}$

Answer 1

Answer: (C)

$rate=k\left[O_3{]}^2\right[O_2{]}^{-1}$

From the slow step, $rate=K\left[O\right]\left[O_3\right]....\left(i\right)$
From the fast reaction, $rate=k\left[O_3{]}^2\right[O_2{]}^{-1}$
Or $\left[O\right]=\frac{K\left[O_3\right]}{\left[O_2\right]}.....\left(ii\right)$
From (i) & (ii), rate = $k^{\prime }K=\frac{[O_3{]}^2}{\left[O_2\right]}=k\frac{[O_3{]}^2}{\left[O_2\right]}$