The value of $K_{c}$ for the reaction
$3 \mathrm{O}_{2}(\mathrm{~g}) \longleftrightarrow 2 \mathrm{O}_{3}(\mathrm{~g})$
is $2.0 \times 10^{-50}$ at $25^{\circ} \mathrm{C}$. If the equilibrium concentration of $\mathrm{O}_{2}$ in air at $25^{\circ} \mathrm{C}$ is $1.6 \times 10^{-2}$, what is the concentration of $\mathrm{O}_{3} ?$
The given reaction is:
$3 \mathrm{O}_{2(\mathrm{~g})} \longleftrightarrow 2 \mathrm{O}_{3(\mathrm{~g})}$
Then, $K_{C}=\frac{\left[\mathrm{O}_{3(\mathrm{~g})}\right]^{2}}{\left[\mathrm{O}_{2(\mathrm{~g})}\right]^{3}}$
It is given that $K_{C}=2.0 \times 10^{-50}$ and $\left[O_{2(\mathrm{~g})}\right]=1.6 \times 10^{-2}$.
Then, we have,
$2.0 \times 10^{-50}=\frac{\left[\mathrm{O}_{3(\mathrm{~g})}\right]^{2}}{\left[1.6 \times 10^{-2}\right]^{3}}$
$\Rightarrow\left[\mathrm{O}_{3(\mathrm{~g})}\right]^{2}=2.0 \times 10^{-50} \times\left(1.6 \times 10^{-2}\right)^{3}$
$\Rightarrow\left[\mathrm{O}_{3(\mathrm{~g})}\right]^{2}=8.192 \times 10^{-56}$
$\Rightarrow\left[\mathrm{O}_{3(\mathrm{~g})}\right]=2.86 \times 10^{-28} \mathrm{M}$
Hence, the concentration of $\mathrm{O}_{3}$ is $2.86 \times 10^{-28} \mathrm{M}$.