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 KH 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 = kH x

  • p is the partial pressure, and x is the mole fraction.
  • kH 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 kH = 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.

Was this answer helpful?

  
   

4 (14)

Upvote (23)

Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*

BOOK

Free Class

Ask
Question