The \(K_H\) values of \(HCOOH\), \(CH_4\), \(Ar\) and \(CO_2\) at 298 K are \(1.83 × 10^{-5}\)kbar, \(0.413\) kbar, \(40\) kbr and \(1.67\) kbr respectively. Arrange them in the order of their decreasing solubility at constant pressure:
Answer & explanation
Correct answer: option 3
The correct answer is option 3. \(HCOOH > CH_4 > CO_2 > Ar \).
The order of decreasing solubility at constant pressure is 3. HCOOH > CH₄ > CO₂ > Ar.
Henry's law states that the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid at a constant temperature. However, the question specifies constant pressure. In this case, we can rely on the Henry's law constant \((K_H)\), which is a measure of a gas's tendency to dissolve in a liquid. A higher \(K_H\) value indicates greater solubility.
Looking at the provided \(K_H\) values:
\(HCOOH\) (Formic acid): \(1.83 × 10^{-5}\) kbar (lowest \(K_H\))
\(CH_4\) (Methane): \(0.413\) kbar
\(CO_2\) (Carbon dioxide): \(1.67\) kbar
\(Ar\) (Argon): \(40\) kbar (highest \(K_H\))
Based on the \(K_H\) values, we can arrange them in decreasing order of solubility:
Argon \((Ar)\) - Highest \(K_H\) (most soluble)
Carbon dioxide \((CO_2)\)
Methane \((CH_4)\)
Formic acid \((HCOOH)\) - Lowest \(K_H\) (least soluble)
Therefore, the correct answer is option 3. \(HCOOH > CH_4 > CO_2 > Ar\).