Question: 47: \mathrm{K}_{\mathrm{H}} value for some gases at the same temperature ‘ T ‘ are given
| Gas | \mathbf{K}_{\mathbf{H}} / \mathbf{k} bar |
|---|---|
| Ar | 40.3 |
| \mathrm{CO}_{2} | 1.67 |
| HCHO | 1.83 \times 10^{-5} |
| \mathrm{CH}_{4} | 0.413 |
where \mathrm{K}_{\mathrm{H}} is Henry’s Law constant in water. The order of their solubility in water is :
(1) \mathrm{Ar}<\mathrm{CO}_{2}<\mathrm{CH}_{4}<\mathrm{HCHO}
(2) \mathrm{Ar}<\mathrm{CH}_{4}<\mathrm{CO}_{2}<\mathrm{HCHO}
(3) \mathrm{HCHO}<\mathrm{CO}_{2}<\mathrm{CH}_{4}<\mathrm{Ar}
(4) \mathrm{HCHO}<\mathrm{CH}_{4}<\mathrm{CO}_{2}<\mathrm{Ar}
Answer: Option (1)
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 smaller value of \mathrm{K}_{\mathrm{H}} corresponds to higher solubility, and a larger value corresponds to lower solubility.
The given \mathrm{K}_{\mathrm{H}} values are:
\mathrm{Ar} = 40.3 \mathrm{CO}_{2} = 1.67 \mathrm{CH}_{4} = 0.413 \mathrm{HCHO} = 1.83 \times 10^{-5}Arranging the gases in decreasing order of \mathrm{K}_{\mathrm{H}}
(i.e., increasing solubility):
\mathrm{Ar}>\mathrm{CO}_{2}>\mathrm{CH}_{4}>\mathrm{HCHO}Hence, the increasing order of solubility in water is:
\mathrm{Ar}<\mathrm{CO}_{2}<\mathrm{CH}_{4}<\mathrm{HCHO}