Which of the following graph correctly represents the plots of $$K_H$$ at 1 bar gases in water versus temperature?

For a gas dissolved in water, Henry’s law is written as $$p = K_H \, x$$, where
$$p$$ = partial pressure of the gas above the solution,
$$x$$ = mole-fraction of the gas in the liquid phase,
$$K_H$$ = Henry’s law constant.

The solubility of the gas is inversely proportional to $$K_H$$: a lower $$K_H$$ means a larger mole-fraction (greater solubility), while a higher $$K_H$$ means a smaller mole-fraction (lower solubility).

Dissolution of almost every common gas in water is an exothermic process (heat is evolved). For an exothermic equilibrium the Le-Châtelier principle predicts that increasing temperature shifts the equilibrium in the direction that absorbs heat, i.e. towards the gaseous state. Hence the solubility of the gas decreases as temperature rises.

Mathematically, the temperature dependence of $$K_H$$ is obtained from the van’t Hoff relation:

$$\frac{d(\ln K_H)}{dT} = \frac{\Delta H}{R\,T^{2}} \qquad -(1)$$

Here $$\Delta H$$ is the enthalpy change for dissolution. Because $$\Delta H$$ is negative (exothermic), the right-hand side of $$(1)$$ is negative. Integrating, $$\ln K_H$$ increases with temperature, so $$K_H$$ itself increases monotonically with temperature.

Therefore a correct plot of $$K_H$$ versus temperature must show an increasing (upward-rising) curve.

Among the given alternatives, only Option D depicts $$K_H$$ steadily increasing as the temperature rises. All other graphs show either a decrease or a non-monotonic behaviour, contradicting the thermodynamic prediction.

Hence, the correct choice is:
Option D which is: $$K_H$$ increases with temperature.