Question

The chemical reaction, ${2O}_2⟶{3O}_2$ proceeds as:
$O_3\rightleftharpoons O_2+\left[O\right]\left(fast\right)\left[O\right]+O_3⟶{2O}_2\left(slow\right)$
The rate law expression will be:

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

Answer 1

The correct option is A Rate$=K\left[O\right]\left[O_3\right]$

slow step

rate=$k\times \left[O\right]\left[O_3\right]$ equation (1)

also for equilibrium equating 1 and 2

$k_c=$$\frac{\left[O\right]\left[O_3\right]}{\left[O_2\right]}$ equation(2)

$\left[o\right]=\frac{k_c\left[O_3\right]}{\left[O_2\right]}$ equation(3)

by 3 and 1 the intermediate [O] is eliminated as rate

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

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

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

Where $k^1=k\times k_c$