Class 12 Surface Chemistry — Adsorption and Colloids
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
| Adsorption | Absorption | |
|---|---|---|
| Where the substance goes | Sticks on the surface only | Goes uniformly right through the bulk |
| Concentration | Higher at the surface than inside | Same everywhere |
| Rate | Fast at first, then reaches equilibrium | Steady throughout |
| Example | Water vapour on silica gel | Water 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
| Feature | Physisorption | Chemisorption |
|---|---|---|
| Force involved | Weak van der Waals forces | A chemical bond with the surface |
| Enthalpy released | Small, of the order of 20–40 kJ/mol | Large, of the order of 80–240 kJ/mol |
| Specificity | Not specific — any gas on any solid | Highly specific — only if a bond can form |
| Reversibility | Readily reversible | Usually irreversible |
| Layers formed | Multimolecular layers possible | Monomolecular layer only |
| Effect of temperature | Decreases as temperature rises | First 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
- Nature of the gas. Gases that are easily liquefied — that is, gases with a higher critical temperature, such as SO2, NH3 and CO2 — are adsorbed more than H2, N2 or O2, because stronger intermolecular attraction means stronger attraction to the surface too.
- Surface area of the adsorbent. This is why activated charcoal, silica gel and finely divided metals are used — the same mass, far more surface.
- Pressure. Adsorption of a gas rises with pressure and then levels off as the surface saturates.
- Temperature. Since adsorption is exothermic, lowering the temperature increases physisorption.
The Freundlich adsorption isotherm — the chapter's one formula
Taking logarithms: log(x/m) = log k + (1/n) log p
What each symbol means:
- x — mass of gas adsorbed
- m — mass of adsorbent, so x/m is the mass adsorbed per gram of adsorbent
- p — equilibrium pressure of the gas (for adsorption from solution, replace p with the equilibrium concentration C)
- k and n — constants for a given adsorbent, adsorbate and temperature
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.
- Promoter — increases the catalyst's activity (molybdenum with iron in the Haber process).
- Poison — destroys it (arsenic on the platinum catalyst in the older contact process).
- Shape-selective catalysis — zeolites have pores of a particular size, so only molecules that fit can react. ZSM-5 is the standard example.
- Enzymes are biological catalysts of remarkable specificity, working within a narrow range of temperature and pH.
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 phase | Dispersion medium | Name | Example |
|---|---|---|---|
| Solid | Liquid | Sol | Paints, gold sol |
| Liquid | Solid | Gel | Cheese, butter, jellies |
| Liquid | Liquid | Emulsion | Milk |
| Solid | Gas | Aerosol | Smoke |
| Liquid | Gas | Aerosol | Fog, mist, clouds |
| Gas | Liquid | Foam | Whipped 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
- Bredig's arc method — an electric arc struck between two metal electrodes under water disperses the metal; used for platinum, silver, gold and copper sols.
- Peptisation — converting a fresh precipitate back into a sol by adding a small amount of a suitable electrolyte, which supplies the ions the particles adsorb.
- Dialysis — removing dissolved electrolyte impurities through a parchment or cellophane membrane, which passes ions but holds colloidal particles back. Applying an electric field speeds it up: electrodialysis.
- Tyndall effect — the scattering of light by colloidal particles, seen as a visible beam. It works because the particle size is comparable to the wavelength of light, which is why a true solution shows no Tyndall cone.
- Brownian movement — the ceaseless zig-zag motion of colloidal particles caused by uneven bombardment by molecules of the medium. It also helps keep a sol stable.
- Electrophoresis — colloidal particles carry a charge, so they migrate towards one electrode; the direction tells you the sign of the charge.
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.
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.
- Purification of drinking water — alum coagulates the negatively charged clay particles suspended in the water.
- Cottrell smoke precipitator — charged plates discharge and bring down the charged carbon particles in factory smoke.
- Delta formation — river water's colloidal clay is coagulated by the electrolytes in sea water where the two meet.
- Cleansing action of soap — micelles trap oil and grease in their hydrocarbon interior and carry it away in water.
- Medicine — colloidal forms of some drugs are absorbed more readily because of their very large surface area.
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
| Exam | How it appears |
|---|---|
| CBSE Class 12 | Distinguish-between pairs, Hardy–Schulze applications, colloid classification, short Freundlich numericals |
| ICSE / state boards | Similar coverage; check whether the Freundlich isotherm is in your own syllabus, as it is not universal |
| NEET / JEE | Assertion-reason and matching questions on adsorption types and colloid examples |
| IIT-JAM / GATE / CSIR-NET | Langmuir 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.