At 293 K,the Henry law constant in water for | vedclass

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(D) According to Henry's law: $P = x \cdot K_H$. For $N_2$: $P_{N_2} = x_{N_2} \cdot K_H(N_2)$. For $O_2$: $P_{O_2} = x_{O_2} \cdot K_H(O_2)$. Given t

Vedclass · Accepted Answer

$0.45$

Vedclass · Answer

(D) According to Henry's law: $P = x \cdot K_H$. For $N_2$: $P_{N_2} = x_{N_2} \cdot K_H(N_2)$. For $O_2$: $P_{O_2} = x_{O_2} \cdot K_H(O_2)$. Given that $P_{N_2} = P_{O_2}$,$K_H(N_2) = 76.48 \ kbar$,and $K_H(O_2) = 34.86 \ kbar$. The ratio of mole fractions is: $\frac{x_{N_2}}{x_{O_2}} = \frac{P_{N_2} / K_H(N_2)}{P_{O_2} / K_H(O_2)} = \frac{K_H(O_2)}{K_H(N_2)}$. Substituting the values: $\frac{x_{N_2}}{x_{O_2}} = \frac{34.86}{76.48} \approx 0.45$.

At $293 \ K$,the Henry law constant in water for $N_2$ and $O_2$ are $76.48 \ kbar$ and $34.86 \ kbar$ respectively. What is the ratio of mole fractions of $N_2$ and $O_2$ in water? (Assume partial pressures of $N_2$ and $O_2$ are the same at $293 \ K$)

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