Question

The chemical reaction ${2O}_3⟶{3O}_2$ proceeds as follows
$O_3\rightleftharpoons O_2+O$ ......(fast)
$O+O_3\rightleftharpoons {2O}_2$ ......(slow)
The rate of expression should be

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

Answer 1

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

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

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

Therefore $\left[O\right]=\frac{k_c\left[O_3\right]}{O_2}$-----------(iii)

By equation (iii) and (i) the intermediate $\left[O\right]$ eliminated as rate now

$k.k_c\frac{\left[O_3\right]\left[O_3\right]}{O_2}$

$=K^{\prime }\frac{[O_3{]}^2}{O_2}$

So $r=K^{\prime }\left[O_3{]}^2\right[O_2{]}^{-1}$

Option B is answer.