IIT-JAM Surface Chemistry — Freundlich and Langmuir Adsorption Isotherms

By Aniket Bhardwaj · 11 September 2026 · IIT-JAM Chemistry

Surface chemistry is a short unit that rewards precision. In an IIT-JAM paper it appears in two shapes: a conceptual question that asks you to distinguish physisorption from chemisorption or to identify the right isotherm, and a numerical that hands you two data points and expects you to extract the constants of the Freundlich or Langmuir equation. Both are fully doable in a couple of minutes once you know which linear plot goes with which model.

Why adsorption happens at all

A molecule inside a solid is pulled equally in every direction, but an atom at the surface has unbalanced forces acting on it. That residual attraction is what holds adsorbate molecules on the surface. The thermodynamic argument is worth memorising because JAM asks it directly:

ΔG = ΔH − TΔS. Adsorption is spontaneous, so ΔG < 0.
A gas losing translational freedom to sit on a surface has ΔS < 0, so −TΔS > 0.
Therefore ΔH must be negative — adsorption is always exothermic.

That is also why the extent of physisorption falls as temperature rises: heating an exothermic equilibrium shifts it backwards, exactly as Le Chatelier's principle predicts.

Physisorption versus chemisorption

Feature	Physisorption	Chemisorption
Force involved	van der Waals	Genuine chemical bond
Enthalpy of adsorption	Small — of the order of tens of kJ/mol	Large — of the order of hundreds of kJ/mol
Specificity	Not specific; any gas on any solid	Highly specific to the surface–adsorbate pair
Reversibility	Readily reversible	Usually irreversible
Layers formed	Multilayer possible	Monolayer only
Activation energy	Essentially none	Often significant
Effect of raising temperature	Decreases	Increases first, then decreases

The last row catches people out. Chemisorption needs activation energy, so warming a cold surface speeds it up; once enough energy is available, the exothermic equilibrium takes over and the extent falls again. The plot therefore passes through a maximum.

The Freundlich isotherm

x/m = k · p^1/n (n > 1, so 0 < 1/n < 1)
Linear form: log(x/m) = log k + (1/n) log p

x is the mass of gas adsorbed, m the mass of adsorbent, so x/m is the amount adsorbed per gram. Plot log(x/m) against log p: the slope is 1/n and the intercept is log k. The model is empirical — it fits mid-range pressures well but predicts no saturation, which is its known failure at high pressure.

Example 1 — extracting k and n from two data points. At 20 kPa, x/m = 0.60 g per gram of charcoal; at 80 kPa, x/m = 1.20 g/g. Find n and k, then predict x/m at 45 kPa.

Divide one Freundlich equation by the other so k cancels:
1.20 ÷ 0.60 = (80 ÷ 20)^1/n → 2 = 4^1/n.
Take logs: log 2 = (1/n) log 4 → 0.30103 = (1/n)(0.60206) → 1/n = 0.500, so n = 2.
Now find k from the first point: 0.60 = k × 20^0.5 = k × 4.47214
→ k = 0.60 ÷ 4.47214 = 0.1342 (in g g⁻¹ kPa^−1/2).
Check against the second point: 0.1342 × 80^0.5 = 0.1342 × 8.94427 = 1.200 ✓.
Prediction at 45 kPa: x/m = 0.1342 × 45^0.5 = 0.1342 × 6.70820 = 0.900 g/g.

Quick mental check: 45 = 20 × 2.25 and √2.25 = 1.5, so x/m should be 0.60 × 1.5 = 0.90. The two routes agree.

The Langmuir isotherm

Langmuir's model assumes a fixed number of identical sites, one molecule per site (monolayer only), and no interaction between adsorbed molecules. Writing θ for the fraction of sites occupied:

θ = Kp ÷ (1 + Kp) · equivalently x/m = a·p ÷ (1 + b·p)
Linear form: p ÷ (x/m) = 1/a + (b/a)·p → plot p/(x/m) against p
Monolayer capacity (p → ∞) = a/b

The two limits are the examinable part. At low pressure Kp ≪ 1, so θ ≈ Kp — the amount adsorbed is first order in pressure. At high pressure Kp ≫ 1, so θ → 1 — the amount adsorbed becomes zero order, independent of pressure, because the monolayer is full. Freundlich, with its fractional power, sits between these two extremes, which is why it works in the middle range and fails at both ends.

Example 2 — reading the Langmuir constants off a plot. A plot of p/(x/m) against p gives a straight line of intercept 2.00 and slope 0.500 in consistent units. Find a, b and the monolayer capacity.

Intercept = 1/a = 2.00 → a = 0.500.
Slope = b/a = 0.500 → b = 0.500 × a = 0.500 × 0.500 = 0.250.
Monolayer capacity = a/b = 0.500 ÷ 0.250 = 2.00 g per gram of adsorbent.

Verify at a specific pressure, say p = 4.00: x/m = (0.500 × 4.00) ÷ (1 + 0.250 × 4.00) = 2.00 ÷ 2.00 = 1.00 g/g, which is half the monolayer capacity. And indeed θ = 0.5 occurs when bp = 1, i.e. p = 1/b = 4.00 ✓.

Example 3 — surface coverage directly. For a gas with K = 0.20 kPa⁻¹, find θ at 15 kPa, and the pressure at which the surface is half covered.

Kp = 0.20 × 15 = 3.00.
θ = 3.00 ÷ (1 + 3.00) = 3.00 ÷ 4.00 = 0.75 — the surface is 75 % covered.
Half coverage: θ = 0.5 needs Kp = 1, so p = 1 ÷ K = 1 ÷ 0.20 = 5.0 kPa.

The general result p(θ = ½) = 1/K is worth memorising; it is the fastest way to read K off any Langmuir-type data set.

Beyond the monolayer, and the rest of the unit

Multilayer adsorption is handled by the BET model, an extension of Langmuir in which layers stack on layers; its main use is measuring the surface area of a catalyst from the monolayer volume. In catalysis, the Langmuir picture supplies the standard mechanism: reactants adsorb on active sites, react on the surface, and desorb. Two ideas follow from it — a promoter increases the number or activity of sites, while a poison blocks them irreversibly. Zeolites act as shape-selective catalysts because only molecules that fit their pores can reach the active sites at all.

For colloids, keep four facts ready: the Tyndall effect proves particle sizes are in the colloidal range; Brownian motion opposes settling; electrophoresis shows the particles carry charge; and the Hardy–Schulze rule says the coagulating power of an ion rises sharply with the magnitude of its charge, and the effective ion is the one whose charge is opposite to that of the sol. Emulsions and micelles complete the unit — above the critical micelle concentration, surfactant molecules aggregate with their hydrophobic tails inward in an aqueous medium.

Common mistakes that cost marks

Confusing adsorption with absorption. Adsorption is a surface phenomenon; absorption is uniform throughout the bulk. "Sorption" means both are happening.
Letting 1/n exceed 1. The Freundlich exponent must lie between 0 and 1 for the equation to make physical sense. If your slope comes out above 1, recheck which axis is which.
Plotting the wrong linear form. Freundlich needs log–log; Langmuir needs p/(x/m) against p. Using a log–log plot for Langmuir data will not give a straight line.
Claiming chemisorption always increases with temperature. It increases first, because of activation energy, then decreases like every exothermic process.
Saying "adsorption increases with pressure" without a limit. Langmuir saturates; only the empirical Freundlich equation keeps rising, and that is its defect, not a property of real surfaces.
Forgetting that ΔS is negative for adsorption — this is the whole basis of proving the process is exothermic, and it is a favourite one-mark reasoning question.

How this unit is examined

Question shape	What it tests	Route to the answer
Two (p, x/m) pairs given, find k and n	Freundlich algebra	Divide the equations, take logs, back-substitute
Slope and intercept of a linear plot given	Recognising the linear form	1/a from the intercept, b/a from the slope
Find θ, or the pressure for a stated θ	Langmuir limits	θ = Kp/(1 + Kp); p(θ = ½) = 1/K
Distinguish physisorption and chemisorption	Concept table	Enthalpy magnitude, specificity, layers, temperature trend
Order of coagulating power of ions	Hardy–Schulze rule	Rank by charge of the oppositely charged ion
Why is adsorption exothermic?	Thermodynamic reasoning	ΔG < 0 with ΔS < 0 forces ΔH < 0

Confirm the exact syllabus scope and marking scheme from the current official IIT-JAM notification before you plan how much time to give this unit.

Fit your isotherm data properly. Both isotherms are solved by turning curved data into a straight line — log(x/m) against log p for Freundlich, p/(x/m) against p for Langmuir. The free Linear Regression Calculator returns the slope, intercept and correlation for any set of points, so you can extract k, n, a and b from real lab data instead of only from two-point exam questions.

Open the Linear Regression Calculator →

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