Practicing Success
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: |
\(Ar > CO_2 > CH_4 > HCCOH\) \(HCOOH > Ar > CH_4 > CO_2\) \(HCOOH > CH_4 > CO_2 > Ar \) \(Ar > CO_2 > HCOOH > CO_2\) |
\(HCOOH > CH_4 > CO_2 > Ar \) |
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. \(HCOOH > CH_4 > CO_2 > Ar\). |