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Chemistry · JEE Advanced only

Freundlich Adsorption Isotherm (JEE Advanced)

Not in the JEE Main syllabus, but it is in JEE Advanced. Study time: about 2 to 3 hours for understanding the concept, practicing graph analysis, and solving numerical problems.

Last updated: 26 October 2026

Direct answer

The Freundlich Adsorption Isotherm is an empirical equation that describes how the amount of gas adsorbed on a solid surface varies with pressure at constant temperature.

The Freundlich Adsorption Isotherm is an empirical equation that describes how the amount of gas adsorbed on a solid surface varies with pressure at constant temperature. It works well for physical adsorption at moderate pressures but fails at very high pressures. The equation relates the mass of adsorbate per unit mass of adsorbent to the equilibrium pressure of the gas.

Official syllabus

What the JEE Advanced syllabus says

“Freundlich adsorption isotherm”

Source: JEE Advanced 2026 Information Bulletin, Annexure-I. Closest JEE Main topic: Preparation of lyophilic and lyophobic sols.

JEE Main covers basic concepts of colloids and sols, but Advanced specifically tests the mathematical treatment of adsorption through isotherms. You need to understand the Freundlich equation, its graphical analysis, and how to extract constants from experimental data.

Key ideas

What you must understand

  • Adsorption is a surface phenomenon where gas or liquid molecules stick to a solid surface, different from absorption which is a bulk phenomenon
  • The Freundlich isotherm is empirical (based on experiments) and works for heterogeneous surfaces with non-uniform adsorption sites
  • At low pressure, adsorption is directly proportional to pressure (n = 1), at high pressure it becomes independent of pressure (n approaches infinity)
  • The constants k and n depend on the nature of adsorbent, adsorbate, and temperature
  • Physical adsorption decreases with increasing temperature while chemical adsorption may initially increase
  • The isotherm fails at very high pressures because it does not predict a saturation limit
Formulas

Formulas and key facts

  • x/m = k × p^(1/n) where x = mass of adsorbate, m = mass of adsorbent, p = pressure, k and n are constants
  • log(x/m) = log k + (1/n) × log p (linear form for graph plotting)
  • 1/n typically ranges from 0 to 1 for most systems
  • At low pressure: x/m ∝ p (when 1/n = 1)
  • At high pressure: x/m = constant (when 1/n → 0)
  • Slope of log(x/m) vs log p graph = 1/n
  • Intercept of log(x/m) vs log p graph = log k
  • k indicates adsorption capacity, 1/n indicates adsorption intensity
Avoid

Common mistakes

  • Confusing x/m (mass ratio) with x alone, the equation uses mass of adsorbate per unit mass of adsorbent
  • Forgetting that the slope gives 1/n not n, so you must take reciprocal to find n
  • Assuming Freundlich isotherm works at all pressures, it actually fails at very high pressures where saturation occurs
  • Mixing up log and ln in calculations, JEE problems typically use log base 10 for this isotherm
Worked example

A JEE Advanced-level example

Question. For adsorption of a gas on charcoal, the following data was obtained at 300 K: At pressure 0.5 atm, x/m = 0.4 g/g. At pressure 2.0 atm, x/m = 0.8 g/g. Find the values of k and 1/n in the Freundlich isotherm. Also calculate x/m at pressure 8.0 atm.

Show solution
  1. Write the linear form: log(x/m) = log k + (1/n) × log p
  2. For point 1: log(0.4) = log k + (1/n) × log(0.5), which gives -0.398 = log k + (1/n) × (-0.301)
  3. For point 2: log(0.8) = log k + (1/n) × log(2.0), which gives -0.097 = log k + (1/n) × (0.301)
  4. Subtracting equation 1 from equation 2: -0.097 - (-0.398) = (1/n) × (0.301 - (-0.301)), so 0.301 = (1/n) × 0.602, giving 1/n = 0.5
  5. Substituting 1/n = 0.5 in equation 2: -0.097 = log k + 0.5 × 0.301 = log k + 0.1505, so log k = -0.2475, giving k = 0.566
  6. At p = 8.0 atm: x/m = k × p^(1/n) = 0.566 × (8.0)^0.5 = 0.566 × 2.828 = 1.6 g/g

Answer: k = 0.566, 1/n = 0.5, and x/m at 8.0 atm = 1.6 g/g

Solved twice independently; both solutions agree.

FAQs

Frequently asked questions

Is Freundlich Adsorption Isotherm in the JEE Main syllabus?

No. It is in the JEE Advanced 2026 syllabus only. The closest JEE Main topic is Preparation of lyophilic and lyophobic sols.

How long does Freundlich Adsorption Isotherm take to prepare for JEE Advanced?

About 2 to 3 hours for understanding the concept, practicing graph analysis, and solving numerical problems, if you already know the related JEE Main topic.

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