The value of Henry’s Law constant KH(in kbar) for He, H2, N2 and O2 i…

The correct answer is Option (4) → (D), (B), (C), (A)

According to Henry’s law, the partial pressure of a gas above a solution is proportional to its mole fraction in the solution:

    \( \displaystyle p=K_H\,x\quad\Rightarrow\quad x=\frac{p}{K_H} \)

At the same pressure and temperature, the solubility (x) of a gas is inversely proportional to K_H. The higher the value of K_H, the lower the solubility of the gas. Since the pressure is the same for all four gases, it does not change the order.

Arranging the gases by K_H:

    K_H: \( \ce{O2} \) (34.86) < \( \ce{H2} \) (69.16) < \( \ce{N2} \) (76.48) < He (144.97)
    Solubility: \( \ce{O2} \) > \( \ce{H2} \) > \( \ce{N2} \) > He

Matching the given gases:

• (D) \( \ce{O2} \): K_H = 34.86 kbar, most soluble

• (B) \( \ce{H2} \): K_H = 69.16 kbar

• (C) \( \ce{N2} \): K_H = 76.48 kbar

• (A) He: K_H = 144.97 kbar, least soluble

Shortcut: Options (2) and (4) both start with \( \ce{O2} \) (D) and end with He (A). \( \ce{H2} \) has the smaller K_H, so (B) comes before (C), which gives Option (4).

NCERT (2026-27 edition):
Solubility of a Gas in a Liquid – “… The most commonly used form of Henry’s law states that “the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution” and is expressed as:

    \( p = K_H\,x \qquad (1.11) \)

… Different gases have different K_H values at the same temperature (Table 1.2). This suggests that K_H is a function of the nature of the gas. It is obvious from equation (1.11) that higher the value of K_H at a given pressure, the lower is the solubility of the gas in the liquid.”