Question

At 293 K, the Henry law constant in water for N$_2$ and O$_2$ are 76.48 k bar and 34.86 k bar respectively. What is the ratio of mole fractions of N$_2$ and O$_2$ in water? (Assume partial pressures of N$_2$ and O$_2$ same at 293 K)

• A. 2.19
• B. 0.95
• C. 0.6
• D. 0.45

Hint:

Henry's law describes the solubility of gases in liquids. A higher Henry's law constant indicates lower solubility of the gas at a given partial pressure. In this problem, since nitrogen has a higher Henry's law constant than oxygen, its mole fraction in water will be lower for the same partial pressure.

Answer 1

The Correct Option is D

Step 1: State Henry's Law.
Henry's law is given by $P = K_H \cdot x$, where $P$ is the partial pressure, $K_H$ is Henry's law constant, and $x$ is the mole fraction.
Step 2: Apply Henry's Law to N$_2$ and O$_2$.
For N$_2$: $P_{N_2} = K_{H, N_2} \cdot x_{N_2} \implies x_{N_2} = \frac{P_{N_2}}{K_{H, N_2}}$ For O$_2$: $P_{O_2} = K_{H, O_2} \cdot x_{O_2} \implies x_{O_2} = \frac{P_{O_2}}{K_{H, O_2}}$
Step 3: Use the given values and the assumption $P_{N_2 = P_{O_2} = P$.}
$K_{H, N_2} = 76.48$ k bar
$K_{H, O_2} = 34.86$ k bar

Step 4: Calculate the ratio of mole fractions.
$$\frac{x_{N_2}}{x_{O_2}} = \frac{P/K_{H, N_2}}{P/K_{H, O_2}} = \frac{K_{H, O_2}}{K_{H, N_2}} = \frac{34.86 \text{ k bar}}{76.48 \text{ k bar}}$$ $$\frac{x_{N_2}}{x_{O_2}} \approx 0.4557$$ Rounding to two decimal places, the ratio is approximately 0.46, which is closest to option 4 (0.45). Final Answer: The final answer is $\boxed{0.45}$