The Henry's law constant $K_H$ values of three gases (A, B, C) in water are 145, $2 imes 10^{-5}$, and 35 kb

Question

The Henry's law constant $K_H$ values of three gases (A, B, C) in water are 145, $2 	imes 10^{-5}$, and 35 kbar, respectively. Determine the order of solubility of these gases in water from highest to lowest:

Accepted Answer

According to Henry's law, the partial pressure of a gas in the vapour phase ($p$) is directly proportional to its mole fraction ($x$) in the solution, expressed as $p = K_H x$. This implies that at a constant pressure, the higher the value of Henry's law constant ($K_H$), the lower the solubility of the gas in the liquid .

Given the $K_H$ values:
$K_H(	ext{A}) = 145 	ext{ kbar}$
$K_H(	ext{B}) = 2 	imes 10^{-5} 	ext{ kbar}$
$K_H(	ext{C}) = 35 	ext{ kbar}$

The decreasing order of $K_H$ values is A > C > B. Since solubility is inversely proportional to $K_H$, the decreasing order of solubility will be exactly the reverse: B > C > A.

Step-by-Step Solution

Practice All CHEMISTRY Questions