Question

The chemical reaction, $2O_3\to 3O_2$ proceeds as follows;
$O_3\rightleftharpoons O_2+O$ ....(Fast)
$O+O_3\to 2O_2$ ....(Slow)
The rate law expression should be:

• A. $r=K[O_3{]}^2$
• B. $r=K\left[O_3{]}^2\right[O_2{]}^{-1}$
• C. $r=K\left[O_3{]}^{[}{O}_2\right]$
• D. unpredictable

Answer 1

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

For slow step, $rate=K\left[O\right]\left[O_3\right]$ .....(i)

Also for equilibrium $K_c=\frac{\left[O\right]\left[O_2\right]}{\left[O_3\right]}$ ....(ii)

$\therefore \left[O\right]=\frac{K_c\left[O_3\right]}{\left[O_2\right]}$ ....(iii)

By Eqs. (iii) and (i), the intermediate [O] is eliminated as:

$rate=\frac{K\cdot K_c\left[O_3\right]\left[O_3\right]}{\left[O_2\right]}=\frac{K[O_3{]}^2}{\left[O_2\right]}$.