Question: 82: 3\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2\mathrm{O}_{3}(\mathrm{g})
For the above reaction at 298\,\mathrm{K}, \mathrm{K}_{\mathrm{c}} is found to be 3.0\times10^{-59}.
If the concentration of \mathrm{O}_{2} at equilibrium is 0.040\,\mathrm{M}
then concentration of \mathrm{O}_{3} in M is
(1) 2.4\times10^{31}
(2) 1.2\times10^{21}
(3) 4.38\times10^{-32}
(4) 1.9\times10^{-63}
Answer: Option (3)
Explanation:
For the reaction 3\mathrm{O}_{2} \rightleftharpoons 2\mathrm{O}_{3},
the equilibrium constant expression is
\mathrm{K}_{\mathrm{c}}=\frac{[\mathrm{O}_{3}]^{2}}{[\mathrm{O}_{2}]^{3}}.
Given \mathrm{K}_{\mathrm{c}}=3.0\times10^{-59} and
[\mathrm{O}_{2}]=0.040\,\mathrm{M}.
Substituting the values,
3.0\times10^{-59}=\frac{[\mathrm{O}_{3}]^{2}}{(0.040)^{3}}.
Rearranging,
[\mathrm{O}_{3}]^{2}=3.0\times10^{-59}\times(0.040)^{3}.
(0.040)^{3}=6.4\times10^{-5}.
[\mathrm{O}_{3}]^{2}=3.0\times10^{-59}\times6.4\times10^{-5}=1.92\times10^{-63}.
Taking square root,
[\mathrm{O}_{3}]=\sqrt{1.92\times10^{-63}}=4.38\times10^{-32}\,\mathrm{M}.
Hence, the correct answer is Option (3).