Question:
Observe the following data given in the table. (\(K_H\) = Henry's law constant)
Gas CO₂ Ar HCHO CH₄ \(K_H\) (k bar at 298 K) 1.67 40.3 \(1.83 \times 10^{-5}\) 0.413
The correct order of their solubility in water is
Observe the following data given in the table. (\(K_H\) = Henry's law constant)
| Gas | CO₂ | Ar | HCHO | CH₄ |
|---|---|---|---|---|
| \(K_H\) (k bar at 298 K) | 1.67 | 40.3 | \(1.83 \times 10^{-5}\) | 0.413 |
The correct order of their solubility in water is
Show Hint
Remember the inverse relationship: \textbf{High \(K_H\)} = \textbf{Low Solubility}. A simple way to think about it is that a high \(K_H\) means a high pressure is needed to dissolve even a small amount of gas. Formaldehyde (HCHO) is extremely soluble in water (forming formalin), so it must have a very low \(K_H\), which is consistent with the data.
Updated On: Oct 17, 2025
- CO₂>CH₄>HCHO>Ar
- Ar>HCHO>CH₄>CO₂
- HCHO>CH₄>CO₂>Ar
- CO₂>HCHO>CH₄>Ar
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
Henry's Law describes the solubility of a gas in a liquid. It states that the partial pressure (\(p\)) of a gas in the vapor phase is directly proportional to the mole fraction (\(x\)) of the gas in the solution. The proportionality constant is the Henry's Law constant, \(K_H\).
Step 2: Key Formula or Approach:
The mathematical form of Henry's Law is:
\[ p = K_H \cdot x \] Here, \(x\) (mole fraction) is a measure of the gas's solubility. We can rearrange the formula to express solubility:
\[ x = \frac{p}{K_H} \] This relationship shows that for a given partial pressure of a gas, its solubility (\(x\)) is inversely proportional to the Henry's Law constant (\(K_H\)). Therefore, a higher value of \(K_H\) corresponds to lower solubility.
Step 3: Detailed Explanation:
To find the correct order of solubility, we need to arrange the gases in order of increasing \(K_H\) values. The order of decreasing solubility will be the reverse.
Let's list the given \(K_H\) values:
- \(K_H(\text{HCHO}) = 1.83 \times 10^{-5}\) kbar (smallest value) - \(K_H(\text{CH}_4) = 0.413\) kbar - \(K_H(\text{CO}_2) = 1.67\) kbar - \(K_H(\text{Ar}) = 40.3\) kbar (largest value) The order of increasing \(K_H\) is:
\[ \text{HCHO}<\text{CH}_4<\text{CO}_2<\text{Ar} \] Since solubility is inversely proportional to \(K_H\), the order of decreasing solubility will be the reverse of the order above:
\[ \text{HCHO}>\text{CH}_4>\text{CO}_2>\text{Ar} \] Step 4: Final Answer:
The correct order of solubility is HCHO>CH₄>CO₂>Ar. This matches option (C).
Henry's Law describes the solubility of a gas in a liquid. It states that the partial pressure (\(p\)) of a gas in the vapor phase is directly proportional to the mole fraction (\(x\)) of the gas in the solution. The proportionality constant is the Henry's Law constant, \(K_H\).
Step 2: Key Formula or Approach:
The mathematical form of Henry's Law is:
\[ p = K_H \cdot x \] Here, \(x\) (mole fraction) is a measure of the gas's solubility. We can rearrange the formula to express solubility:
\[ x = \frac{p}{K_H} \] This relationship shows that for a given partial pressure of a gas, its solubility (\(x\)) is inversely proportional to the Henry's Law constant (\(K_H\)). Therefore, a higher value of \(K_H\) corresponds to lower solubility.
Step 3: Detailed Explanation:
To find the correct order of solubility, we need to arrange the gases in order of increasing \(K_H\) values. The order of decreasing solubility will be the reverse.
Let's list the given \(K_H\) values:
- \(K_H(\text{HCHO}) = 1.83 \times 10^{-5}\) kbar (smallest value) - \(K_H(\text{CH}_4) = 0.413\) kbar - \(K_H(\text{CO}_2) = 1.67\) kbar - \(K_H(\text{Ar}) = 40.3\) kbar (largest value) The order of increasing \(K_H\) is:
\[ \text{HCHO}<\text{CH}_4<\text{CO}_2<\text{Ar} \] Since solubility is inversely proportional to \(K_H\), the order of decreasing solubility will be the reverse of the order above:
\[ \text{HCHO}>\text{CH}_4>\text{CO}_2>\text{Ar} \] Step 4: Final Answer:
The correct order of solubility is HCHO>CH₄>CO₂>Ar. This matches option (C).
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