When does Raoult's law become a special case of Henry's law?

Raoult’s law becomes a special case of Henry’s law when K_H becomes equal to p₁⁰ in Henry’s law.

At given temperature liquids vaporize. At equilibrium, the pressure exerted by the vapour of the liquid over the liquid phase is referred to as vapour pressure.

According to Raoult’s law:

p = p⁰x

• p is the partial pressure
• x is the mole fraction

p⁰ is the vapour pressure of the pure component.

According to Henry’s law:

p = k_H x

• p is the partial pressure, and x is the mole fraction.
• k_H is the proportionality constant (Henry’s constant)

Comparing both the equations we get that partial pressure of the volatile component or gas is directly proportional to its mole fraction in solution. Hence, Raoult’s law is a special case of Henry’s law when k_H = p⁰.

The Henry’s constant is approximately equal to the vapour pressure only for ‘ideal’ mixtures (mixtures obeying the Raoult’s law over the entire range of composition), which doesn’t exist. Still, closely related substances like benzene and toluene can form a close to an ideal mixture where this case applies.

The Raoult’s law and Henry’s law are obeyed in opposite composition ranges.

Hence, Raoult’s law is a special case of Henry’s law.