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Class 12 Surface Chemistry — Adsorption and Colloids

By Aniket Bhardwaj · 24 September 2026 · CBSE / ICSE

Surface chemistry is a chapter of definitions with one formula in the middle. Most of the marks come from being able to distinguish pairs — adsorption from absorption, physisorption from chemisorption, lyophilic from lyophobic — and from stating the Hardy–Schulze rule correctly. This guide covers the NCERT Class 12 chapter at board level, with the Freundlich isotherm worked in numbers. If you need the Langmuir and BET treatments, those belong to the postgraduate-entrance level and are linked at the end.

Adsorption is a surface effect — absorption is a bulk effect

AdsorptionAbsorption
Where the substance goesSticks on the surface onlyGoes uniformly right through the bulk
ConcentrationHigher at the surface than insideSame everywhere
RateFast at first, then reaches equilibriumSteady throughout
ExampleWater vapour on silica gelWater vapour in anhydrous calcium chloride

When both happen together the word is sorption. The substance that sticks is the adsorbate; the surface it sticks to is the adsorbent.

Why adsorption happens at all

An atom inside a solid is pulled equally in every direction. An atom at the surface has unbalanced forces acting on it, so the surface carries residual attraction. Adsorbing something satisfies part of that attraction, and energy is released.

So adsorption is always exothermic — ΔH is negative. It also reduces the freedom of the adsorbed molecules, so ΔS is negative too. Because ΔG = ΔH − TΔS, adsorption only continues while the negative ΔH term outweighs the −TΔS term. That single argument explains why adsorption equilibrium exists and why raising the temperature drives physisorption backwards. It is a favourite three-mark question.

Physisorption versus chemisorption

FeaturePhysisorptionChemisorption
Force involvedWeak van der Waals forcesA chemical bond with the surface
Enthalpy releasedSmall, of the order of 20–40 kJ/molLarge, of the order of 80–240 kJ/mol
SpecificityNot specific — any gas on any solidHighly specific — only if a bond can form
ReversibilityReadily reversibleUsually irreversible
Layers formedMultimolecular layers possibleMonomolecular layer only
Effect of temperatureDecreases as temperature risesFirst increases (it needs activation energy), then decreases

Different editions print slightly different enthalpy ranges; quote the range your own textbook gives, and describe the difference in words as well — "small" versus "large" earns the mark even if your remembered numbers are shaky.

What increases the adsorption of a gas

The Freundlich adsorption isotherm — the chapter's one formula

x/m = k · p1/n   (n > 1)

Taking logarithms: log(x/m) = log k + (1/n) log p

What each symbol means:

The logarithmic form is the useful one, because plotting log(x/m) against log p gives a straight line of slope 1/n and intercept log k. Where a textbook would print a graph, describe it in words in your answer: a straight line rising from left to right, its gradient equal to 1/n.

Its honest limitation: the Freundlich equation is purely empirical — it was fitted to data, not derived from a model — and it fails at high pressure, where the real curve flattens off while k·p1/n keeps rising. Say so if the question asks for limitations.

Worked example 1 — finding k and 1/n from two readings

Q. For a gas on charcoal at constant temperature, x/m = 2.0 g per gram of charcoal at an equilibrium pressure of 1 bar, and x/m = 8.0 g per gram at 16 bar. Find k and 1/n, then predict x/m at 4 bar.

Step 1 — find k. At p = 1 bar, p1/n = 1 whatever 1/n is, so x/m = k. Therefore k = 2.0.

Step 2 — find 1/n. Divide the two equations:
8.0 ÷ 2.0 = (16 ÷ 1)1/n
4 = 161/n
Taking logs: log 4 = (1/n) log 16 → 0.60206 = (1/n) × 1.20412
1/n = 0.60206 ÷ 1.20412 = 0.500

Step 3 — predict. At p = 4 bar:
x/m = 2.0 × 40.5 = 2.0 × 2 = 4.0 g per gram of charcoal

Check with the log form: log(x/m) = log 2.0 + 0.5 × log 4 = 0.30103 + 0.5 × 0.60206 = 0.30103 + 0.30103 = 0.60206, and the antilog of 0.60206 is 4.0. ✓

Worked example 2 — mass actually adsorbed

Q. 1.2 g of activated charcoal adsorbs 0.30 g of a gas at 1 bar. If 1/n = 0.5 for this system, what mass of the gas will the same charcoal adsorb at 4 bar, at the same temperature?

At 1 bar: x/m = 0.30 ÷ 1.2 = 0.25 g per gram, and since p = 1 bar this also equals k.

At 4 bar: x/m = 0.25 × 40.5 = 0.25 × 2 = 0.50 g per gram

Mass adsorbed, x = 0.50 × 1.2 = 0.60 g

Doubling the mass adsorbed needed a fourfold pressure increase — exactly what a square root relationship should give.

Catalysis — the short board-level version

A catalyst changes the rate of a reaction by offering a path of lower activation energy, and is recovered chemically unchanged. In homogeneous catalysis the catalyst is in the same phase as the reactants; in heterogeneous catalysis it is in a different phase, usually a solid surface with gaseous or liquid reactants, and the mechanism is five steps: diffusion to the surface, adsorption, reaction on the surface, desorption of products, diffusion away.

Colloids — the three classifications you must know

A colloid is a two-phase system in which the particle size of the dispersed phase lies roughly between 1 nm and 1000 nm — bigger than a true solution, smaller than a suspension.

1. By the physical state of the two phases:

Dispersed phaseDispersion mediumNameExample
SolidLiquidSolPaints, gold sol
LiquidSolidGelCheese, butter, jellies
LiquidLiquidEmulsionMilk
SolidGasAerosolSmoke
LiquidGasAerosolFog, mist, clouds
GasLiquidFoamWhipped cream, froth

2. By affinity for the medium: lyophilic sols (starch, gum, gelatin) are solvent-loving, reversible and stable on their own; lyophobic sols (metal sols, As2S3) are solvent-hating, irreversible and need a stabiliser — a lyophilic sol added for that purpose is a protective colloid.

3. By particle type: multimolecular colloids, where many small atoms or molecules aggregate (gold sol, S8 sol); macromolecular colloids, where a single large molecule is already colloidal (starch, cellulose, proteins, polymers); and associated colloids or micelles, which behave as ordinary electrolytes at low concentration and form aggregates above the critical micelle concentration and above the Kraft temperature — soaps and detergents.

Preparing, purifying and testing colloids

Coagulation and the Hardy–Schulze rule

A lyophobic sol is stable because all its particles carry the same charge and repel one another. Neutralise that charge and the particles clump and settle. That is coagulation.

Hardy–Schulze rule: the coagulating power of an ion increases sharply with its charge, and the effective ion is the one carrying the charge opposite to that of the sol.

Worked example 3. Arsenious sulphide sol is negatively charged. Arrange NaCl, BaCl2 and AlCl3 in increasing order of coagulating power for it, and explain. Then say which of Na3PO4, Na2SO4 and NaCl coagulates a positively charged Fe(OH)3 sol best.

For the negative As2S3 sol, the coagulating ion is the cation. The cations are Na+, Ba2+ and Al3+, so the order of coagulating power is
NaCl < BaCl2 < AlCl3.

For the positive Fe(OH)3 sol, the coagulating ion is the anion. The anions are Cl, SO42− and PO43−, so Na3PO4 is the most effective.

Note that the chloride ion is common to all three electrolytes in the first list and the sodium ion is common to all three in the second — the examiner has arranged it that way so that only the charge of the active ion varies. Do not quote numerical coagulation values unless your textbook gives them; the order is what is being tested.

Emulsions and everyday applications

An emulsion is one liquid dispersed in another. The two kinds are oil in water (milk, vanishing cream) and water in oil (butter, cold cream). Because the two liquids separate on standing, an emulsifying agent such as soap or a protein is added to stabilise the interface.

Mistakes that cost marks

  • Writing "absorption" when the answer is "adsorption". If the substance concentrates on the surface, it is adsorption. Read the word in the question carefully too.
  • Saying chemisorption always decreases with temperature. It first increases, because forming a bond with the surface needs activation energy, and only then falls. Physisorption falls from the start.
  • Choosing the coagulating ion with the same charge as the sol. It is always the oppositely charged ion that does the work.
  • Treating the Freundlich equation as a derived law. It is empirical and breaks down at high pressure.
  • Forgetting that 1/n normally lies between 0 and 1, so x/m rises more slowly than pressure does.
  • Claiming a true solution shows the Tyndall effect. Its particles are far too small to scatter visible light appreciably.
  • Confusing a protective colloid with an emulsifying agent. A protective colloid stabilises a lyophobic sol; an emulsifier stabilises an emulsion of two liquids.

Where this chapter is examined

ExamHow it appears
CBSE Class 12Distinguish-between pairs, Hardy–Schulze applications, colloid classification, short Freundlich numericals
ICSE / state boardsSimilar coverage; check whether the Freundlich isotherm is in your own syllabus, as it is not universal
NEET / JEEAssertion-reason and matching questions on adsorption types and colloid examples
IIT-JAM / GATE / CSIR-NETLangmuir and BET isotherms, surface excess and catalysis kinetics — well beyond board level

Qualitatively this is a high-yield chapter for board revision because so much of it is definitional and can be secured in a single focused sitting. Do not skip the numericals, though — the Freundlich questions are easy marks that many students throw away.

Fit the Freundlich line the way a chemist does. Because log(x/m) = log k + (1/n) log p is a straight line, you can enter your log p values as x and your log(x/m) values as y in the linear regression tool: the slope it returns is 1/n and the intercept is log k. That is exactly how Worked Example 1 is done from real data instead of two neat points.

Open the Linear Regression (y = mx + c) Calculator →

Preparing for Class 12 boards? ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre plus online classes across India — details at abcchemistry.in. Families in Delhi-NCR who prefer one-to-one home tuition can see delhihometutor.com.