Surface Chemistry

Surface chemistry covers adsorption, catalysis, and colloids. Advanced keeps this conceptual but precise: distinguishing physisorption from chemisorption, the Freundlich isotherm, enzyme/heterogeneous catalysis, and colloid properties (Tyndall, Brownian motion, coagulation, Hardy-Schulze rule).

Key results

• Adsorption: physisorption (weak van der Waals, reversible, multilayer, decreases with T) vs chemisorption (chemical bond, specific, increases then decreases with T).
• Freundlich isotherm: x/m = k·p^(1/n), 0 < 1/n < 1; plateaus at high pressure.
• Catalysis: lowers Ea via an alternative path; heterogeneous acts at active sites (adsorption theory); enzymes are highly specific (lock-and-key).
• Colloids: dispersed phase in dispersion medium; lyophilic (stable) vs lyophobic (needs stabiliser). Tyndall effect, Brownian motion confirm colloid.
• Coagulation: oppositely charged ion of higher charge is more effective (Hardy–Schulze); Al³⁺ > Ba²⁺ > Na⁺ for a negative sol.

Feature	Physisorption	Chemisorption
Force	van der Waals	chemical bond
Enthalpy	low (20–40 kJ)	high (80–240 kJ)
Layers	multilayer	monolayer
Temperature	decreases	rises then falls

Exam Tricks & Tips

• 🎯 Physisorption decreases with temperature; chemisorption first increases then decreases — the temperature trend is the key discriminator.
• 🎯 Hardy–Schulze rule: coagulating power rises steeply with the charge of the oppositely charged ion (Al³⁺ » Na⁺ for a negative sol).
• 🎯 Tyndall effect and Brownian motion are the diagnostic tests for a colloid — true solutions show neither.
• 🎯 A catalyst lowers activation energy without being consumed and does not change ΔG or the equilibrium constant.
• 🎯 The Freundlich exponent 1/n lies between 0 and 1; adsorption becomes pressure-independent at saturation (multilayer/monolayer plateaus).
• ❌ Common mistake: assuming enzymes work like ordinary heterogeneous catalysts — they are far more specific and pH/temperature sensitive.

Expected exam pattern

Usually 1 conceptual question (often multiple-correct) on adsorption type, colloid properties, or coagulation ordering. Occasionally a Freundlich-isotherm interpretation.

Quick recap

Physisorption (van der Waals, multilayer, falls with T) vs chemisorption (bond, monolayer, rises then falls). Freundlich x/m = kp^(1/n). Catalyst lowers Ea only. Colloids show Tyndall/Brownian; Hardy–Schulze: higher opposite charge coagulates best.

Cover the answer, recall, then check. 11 cards on surface chemistry for JEE Advanced.

Q1. Key difference in forces between physisorption and chemisorption.
A1. Physisorption uses weak van der Waals forces; chemisorption forms chemical bonds.

Q2. Temperature dependence of physisorption vs chemisorption.
A2. Physisorption decreases with T; chemisorption increases then decreases.

Q3. Freundlich adsorption isotherm.
A3. x/m = k·p^(1/n), with 0 < 1/n < 1.

Q4. How does a catalyst speed a reaction?
A4. It provides an alternative path with lower activation energy; ΔG and K are unchanged.

Q5. Two diagnostic properties of a colloid.
A5. Tyndall effect (light scattering) and Brownian motion.

Q6. Hardy–Schulze rule.
A6. Coagulating power increases with the charge of the ion opposite to the sol's charge.

Q7. Order of coagulating power for a negatively charged sol.
A7. Al³⁺ > Ba²⁺ > Na⁺ (higher positive charge coagulates faster).

Q8. Lyophilic vs lyophobic sols.
A8. Lyophilic are solvent-loving and stable; lyophobic are solvent-hating and need a stabiliser.

Q9. Which adsorption is specific and forms a monolayer?
A9. Chemisorption (a surface chemical bond, hence one layer).

Q10. What theory explains heterogeneous catalysis?
A10. The adsorption theory — reactants adsorb on active sites, react, then desorb.

Q11. Why is enzyme catalysis highly specific?
A11. A precise active-site geometry (lock-and-key) fits only particular substrates.

Surface chemistry studies what happens at interfaces — adsorption, catalysis and colloids. It is a factual, high-return topic: JEE Advanced tests the adsorption isotherms, the distinction between adsorption types, and colloid properties and their preparation/purification.

Core concept: molecules accumulate at surfaces (adsorption), enabling catalysis; colloids are intermediate-sized dispersions with characteristic optical and electrical behaviour.

Adsorption

Beginner — physisorption vs chemisorption

Adsorption is the accumulation of a substance at a surface (distinct from absorption, which is bulk uptake). Physisorption (van der Waals): weak, reversible, multilayer, low enthalpy (20–40 kJ/mol), favoured at low temperature, not specific. Chemisorption (chemical bonds): strong, often irreversible, monolayer, high enthalpy (80–240 kJ/mol), specific, and increases then decreases with temperature. Both are exothermic and increase with surface area (why powders and porous solids adsorb well).

Intermediate — adsorption isotherms

The Freundlich isotherm: x/m = k P^(1/n) (0 < 1/n < 1) — a log–log plot is linear but it fails at high pressure. The Langmuir isotherm assumes monolayer adsorption on identical sites and saturates at high pressure. Adsorption underlies heterogeneous catalysis: reactants adsorb on the catalyst surface, react, and desorb (the reason surface area and active sites matter).

Colloids

Types and preparation

Colloids have particle sizes 1–1000 nm (between true solutions and suspensions). Classified by phase (sol, gel, aerosol, emulsion, foam) and by affinity: lyophilic (solvent-loving, reversible, stable) vs lyophobic (solvent-hating, needs a stabiliser, easily coagulated). Prepared by dispersion (Bredig's arc, peptisation) or condensation methods; purified by dialysis (removing ions through a semipermeable membrane).

Advanced — colloid properties and coagulation

• Tyndall effect: scattering of light by colloidal particles (a visible beam path) — distinguishes colloids from true solutions.
• Brownian motion: random zig-zag from molecular bombardment, which keeps particles suspended.
• Electrophoresis: charged colloidal particles migrate in an electric field.
• Coagulation: adding an electrolyte neutralises the charge and precipitates the colloid; the Hardy–Schulze rule says the coagulating power of the ion increases sharply with its charge (Al³⁺ > Ca²⁺ > Na⁺ for a negative sol). Emulsions need emulsifiers (soaps stabilise oil-in-water). Applications: delta formation, Cottrell precipitator, water purification (alum).

Worked example

Which ion coagulates a negatively charged As₂S₃ sol most effectively: Na⁺, Ba²⁺ or Al³⁺? By the Hardy–Schulze rule, the coagulating power of the oppositely charged ion rises steeply with its charge. For a negative sol, the positive counter-ion matters, so Al³⁺ (charge +3) is far more effective than Ba²⁺ (+2), which beats Na⁺ (+1). The order Al³⁺ > Ba²⁺ > Na⁺ (roughly following the Schulze–Hardy charge dependence) is a guaranteed exam fact.

How JEE Advanced tests this

Distinguishing physisorption from chemisorption by given data (enthalpy, temperature dependence, reversibility); Freundlich-isotherm log-plot interpretation; identifying colloid type and preparation/purification method; Tyndall/Brownian/electrophoresis property matching; and Hardy–Schulze coagulation-power ordering.

Exam tricks & shortcuts

• Physisorption: weak, multilayer, low T; chemisorption: strong, monolayer, specific.
• Freundlich x/m = kP^(1/n); linear on a log–log plot.
• Hardy–Schulze: higher counter-ion charge ⇒ much stronger coagulation.
• Mnemonic: "Physical is fickle, chemical commits."

Key formulas / concepts

• Freundlich adsorption isotherm: x/m = k p^(1/n) (1/n between 0 and 1); log(x/m) = log k + (1/n)log p.
• Physisorption (weak van der Waals, reversible, low T) vs chemisorption (strong, specific, higher T).
• Colloids: Tyndall effect, Brownian motion; classified by dispersed phase/medium.
• Catalysis lowers activation energy; enzyme catalysis is highly specific.

Adsorption increases with pressure (Freundlich isotherm); colloids and catalysts are key surface phenomena.

Worked Example

Problem: Adsorption of a gas on charcoal follows the Freundlich isotherm x/m = k·P^{1/n}. Experiment gives x/m = 0.5 g/g at P = 2 atm and x/m = 1.0 g/g at P = 8 atm. Find the constants 1/n and k.

Solution:
Take logarithms of the Freundlich equation:
log(x/m) = log k + (1/n) log P.

Write it for both data points:
log 0.5 = log k + (1/n) log 2 … (i)
log 1.0 = log k + (1/n) log 8 … (ii)

Subtract (i) from (ii):
log(1.0/0.5) = (1/n)(log 8 − log 2) = (1/n) log 4
log 2 = (1/n) log 4
0.301 = (1/n)(0.602) → 1/n = 0.5.

Put 1/n = 0.5 into (i):
log 0.5 = log k + 0.5 × log 2
−0.301 = log k + 0.5(0.301) = log k + 0.150
log k = −0.451 → k = 10^(−0.451) ≈ 0.354.

Answer: 1/n = 0.5 and k ≈ 0.354, so x/m ≈ 0.354·P^{0.5}.