Question

The Henry's law constant ($K_H$) values for four gases $W, X, Y,$ and $Z$ in water at $298\text{ K}$ are $4.0 \times 10^4$, $2.5 \times 10^{-2}$, $1.5 \times 10^3$, and $8.0 \times 10^1\text{ bar}$, respectively. What is the correct order of their solubility in water under the same partial pressure?

• A. $W > Y > Z > X$
• B. $X > Z > Y > W$
• C. $X > Y > Z > W$
• D. $W > Z > Y > X$

Hint:

For gas solubility problems, always remember: Higher $K_H \Rightarrow$ Lower solubility at a given pressure.

Answer 1

The Correct Option is B

Concept: According to Henry's Law, the solubility of a gas in a liquid is directly proportional to the partial pressure of that gas above the surface of the liquid. Mathematically, this relation is expressed as: \[ p = K_H \cdot x \] Where:
• $p$ is the partial pressure of the gas in the vapor phase.
• $K_H$ is the Henry's law constant for the specific gas-solvent system.
• $x$ is the mole fraction of the dissolved gas in the solution (representing its solubility). Rearranging the fundamental equation to isolate the solubility term ($x$) yields: \[ x = \frac{p}{K_H} \] This expression mathematically demonstrates that at a fixed or constant partial pressure ($p$), the solubility ($x$) of a gas is inversely proportional to its Henry's law constant ($K_H$). Consequently, a higher value of $K_H$ implies a lower mole fraction of the gas in the solution phase.

Step 1: Comparing the given values of Henry's law constant ($K_H$) for each gas.
The specific numerical values given for the constants at $298\text{ K}$ are: K_H(W) &= 4.0 \times 10^4 bar
K_H(X) &= 2.5 \times 10^{-2} bar
K_H(Y) &= 1.5 \times 10^3 bar
K_H(Z) &= 8.0 \times 10^1 bar Arranging these numerical constants in a strictly increasing sequence based on their exponents and magnitudes gives: \[ 2.5 \times 10^{-2} & Lt; 8.0 \times 10^1 & Lt; 1.5 \times 10^3 & Lt; 4.0 \times 10^4 \] \[ \text{Therefore: } K_H(X) & Lt; K_H(Z) & Lt; K_H(Y) & Lt; K_H(W) \]

Step 2: Inverting the order to find the decreasing sequence of gas solubility.
Because solubility ($x$) shares an inverse physical relationship with $K_H$ ($x \propto 1/K_H$), the gas with the absolute lowest constant will exhibit the highest capacity to dissolve. Inverting the inequality chain established in Step 1 establishes the final relative solubility order: \[ x_X > x_Z > x_Y > x_W \] Hence, the correct decreasing order of solubility under identical pressure conditions is $X > Z > Y > W$.