Question: 73. The Henry’s law constant \left(\mathrm{K}_{\mathrm{H}}\right) values of three gases (A, B, C) in water are 145, 2 \times 10^{-5} and 35 kbar, respectively. The solubility of these gases in water follow the order :
(1) \mathrm{B}>\mathrm{A}>\mathrm{C}
(2) \mathrm{B}>\mathrm{C}>\mathrm{A}
(3) \mathrm{A}>\mathrm{C}>\mathrm{B}
(4) \mathrm{A}>\mathrm{B}>\mathrm{C}
Answer: Option (2)
Explanation:
According to Henry’s law, the solubility of a gas in a liquid is inversely proportional to its Henry’s law constant.
\text{Solubility} \propto \frac{1}{\mathrm{K}_{\mathrm{H}}}This means that a gas with a smaller value of \mathrm{K}_{\mathrm{H}} will be more soluble in water.
The given Henry’s law constants are:
Gas A: \mathrm{K}_{\mathrm{H}} = 145 \ \text{kbar}
Gas B: \mathrm{K}_{\mathrm{H}} = 2 \times 10^{-5} \ \text{kbar}
Gas C: \mathrm{K}_{\mathrm{H}} = 35 \ \text{kbar}
Comparing the values, gas B has the smallest \mathrm{K}_{\mathrm{H}},
so it is the most soluble. Gas C has a lower \mathrm{K}_{\mathrm{H}} than gas A,
so it is more soluble than gas A.
Therefore, the order of solubility is:
\mathrm{B}>\mathrm{C}>\mathrm{A}Hence, the correct answer is Option (2).