Henry’s constant KH at a given pressure

The correct answer is Option (2) → increases with increase in temperature

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

    \( \displaystyle p=K_H\,x \)

At a given pressure, x = p/K_H. So a larger K_H means a lower solubility of the gas.

Dissolving a gas in a liquid is an exothermic process. When the temperature rises, the equilibrium shifts back (Le Chatelier’s principle) and the gas becomes less soluble.

Lower solubility at a higher temperature means a higher K_H. The data for oxygen in water show it: K_H is about 34.86 kbar at 293 K and 46.82 kbar at 303 K.

So K_H increases with increase in temperature.

Why the other options are incorrect

• Option (1) decreases with increase in temperature — Incorrect. This would mean gases dissolve better in hot water, which is the reverse of what is observed.

• Option (3) increases initially and then decreases with increase in temperature — Incorrect. The solubility of a gas falls steadily as the temperature rises, so K_H rises steadily; there is no maximum.

• Option (4) is independent of temperature — Incorrect. K_H is constant only at a fixed temperature; its value depends on the gas and on the temperature.

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     (1.11)

… 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. It can be seen from Table 1.2 that K_H values for both N₂ and O₂ increase with increase of temperature indicating that the solubility of gases increases with decrease of temperature.”

Table 1.2 (part):