Question

The reaction ; $2O_3$ → $3O_2$, is assigned the following mechanism.

(I) $O_3\leftrightharpoons O_2+O\left(ii\right)O_3+O\stackrel{slow}{\rightleftharpoons }2O_2$

The rate law of the reaction will therefore be

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

Answer 1

The correct option is B

$r\alpha \left[O_3{]}^2\right[O_2{]}^{-1}$

Clearly, step II is the Rate Determining Step. Hence the rate of overall reaction = Rate of Step II $r=K_{II}\left[O\right]\left[O_3\right]$ ...(i)

For the equilibrium of Step I, we get:

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)

Using (iii) in (i), we get:

$\Rightarrow r=\frac{K_{II}{K}_C\left[O_3\right]\left[O_3\right]}{\left[O_2\right]}=K\text{'}\left[O_3{]}^2\right[O_2{]}^{-1}$

In other words $r\propto \left[O_3{]}^2\right[O_2{]}^{-1}$