📑 Quick Navigation - Surface Chemistry

Surface Chemistry JEE Main & Advanced 2025-26

Master Surface Chemistry with complete notes on Adsorption, Catalysis, Colloids, and Emulsions. Includes 100+ solved problems and all JEE formulas.

📚 7 Complete Sections
✍️ 100+ Solved Examples
🎯 Quick Formulas
⭐⭐⭐⭐⭐ 4.9/5
240+ students rated this helpful
100+
Solved Examples
4 hrs
Study Time
Free
Always
Surface Chemistry JEE notes, Formulas, PYQs Surface Chemistry JEE Notes, Formulas, PYQs
1

Adsorption

Adsorption is the phenomenon of accumulation of molecular species (adsorbate) at the surface rather than in the bulk of a solid or liquid (adsorbent). It's a surface phenomenon.

1.1 Adsorption vs Absorption

Property Adsorption Absorption
Definition Surface phenomenon Bulk phenomenon
Location Only on surface Throughout the material
Rate Rapid initially, then slow Uniform throughout
Concentration Higher on surface Uniform throughout
Temperature Effect Decreases with temperature ↑ Independent
Examples Silica gel adsorbs water, activated charcoal adsorbs gases Water absorbed by sponge, anhydrous CaCl₂ absorbs water

💡 Sorption

Sorption: When both adsorption and absorption occur simultaneously, the process is called sorption.
Example: Dye on cotton - initially adsorbed on surface, then absorbed into fibers.

1.2 Physisorption vs Chemisorption

Property Physisorption Chemisorption
Forces Involved Weak van der Waals forces Strong chemical bonds (covalent/ionic)
Nature Physical, reversible Chemical, irreversible
Enthalpy (ΔH) Low (20-40 kJ/mol) High (80-240 kJ/mol)
Temperature Effect Decreases with T ↑ First increases, then decreases
Activation Energy Very low or zero High (like chemical reaction)
Specificity Not specific Highly specific
Layers Multi-molecular layers possible Only mono-molecular layer
Pressure Occurs at low temperature and high pressure Occurs at high temperature
Examples N₂ on mica at 77K, gases on charcoal O₂ on tungsten, H₂ on nickel

1.3 Factors Affecting Adsorption

1. Nature of Adsorbent
  • Greater surface area → More adsorption
  • Porous substances are better adsorbents
  • Activated charcoal, silica gel, alumina gel
  • Finely divided metals (Ni, Pt, Pd)
2. Nature of Adsorbate
  • Easily liquefiable gases (higher critical temp) adsorb more
  • Order: NH₃ > HCl > CO₂ > CH₄ > O₂ > N₂
  • Higher molecular weight → More adsorption
  • Polar molecules adsorb more than non-polar
3. Temperature
  • Physisorption: Decreases with T ↑ (exothermic)
  • Chemisorption: First increases (activation needed), then decreases
  • Generally: Adsorption ∝ 1/T
4. Pressure
  • At constant T: Adsorption ↑ with pressure ↑
  • At high pressure: Saturation occurs
  • Relation described by adsorption isotherms

⚠️ JEE Important: Adsorption is Exothermic

Adsorption is always exothermic (ΔH < 0) because:

  • Molecules on surface have residual forces
  • When adsorbate attaches, these forces are satisfied
  • Energy is released in the process
  • By Le Chatelier's principle: Low temperature favors adsorption
2

Adsorption Isotherms

Adsorption isotherm is the relationship between the amount of substance adsorbed per unit mass of adsorbent (x/m) and pressure (or concentration) at constant temperature.

2.1 Freundlich Adsorption Isotherm

Empirical Relationship

Basic Equation:

\[\frac{x}{m} = kP^{1/n}\]

where x = mass adsorbed, m = mass of adsorbent, P = pressure, k and n are constants (n > 1)

Logarithmic Form (Most useful for JEE):

\[\log\left(\frac{x}{m}\right) = \log k + \frac{1}{n}\log P\]

Graph: log(x/m) vs log P

  • Straight line with positive slope
  • Slope = 1/n
  • Intercept = log k
  • From slope and intercept, k and n can be determined

Behavior at Different Pressures:

Low Pressure

x/m ∝ P (1/n = 1)

Linear relationship

Moderate Pressure

x/m ∝ P^(1/n)

n > 1 (typical)

High Pressure

x/m ∝ P⁰ (constant)

Saturation (1/n = 0)

2.2 Langmuir Adsorption Isotherm

Theoretical Approach (Based on Kinetics)

Assumptions:

  • Surface has fixed number of adsorption sites
  • Each site can hold only ONE molecule (mono-molecular layer)
  • All sites are energetically equivalent
  • No interaction between adsorbed molecules
  • Dynamic equilibrium between adsorption and desorption

Langmuir Equation:

\[\frac{x}{m} = \frac{aP}{1 + bP}\]

where a and b are constants, θ = fraction of surface covered

In terms of θ (surface coverage):

\[\theta = \frac{bP}{1 + bP}\]

Linear Form (for graphing):

\[\frac{1}{(x/m)} = \frac{1}{a} + \frac{b}{a} \cdot \frac{1}{P}\]

Graph of 1/(x/m) vs 1/P gives straight line

📝 Solved Example 1

Question: For adsorption of a gas on solid, following data is obtained:

P (atm) 0.5 1.0 2.0 4.0
x/m (g/g) 0.10 0.14 0.20 0.28

Show that it follows Freundlich isotherm and find values of k and n.

Solution:

Using Freundlich equation: log(x/m) = log k + (1/n) log P

P log P x/m log(x/m)
0.5 -0.301 0.10 -1.00
1.0 0.000 0.14 -0.854
2.0 0.301 0.20 -0.699
4.0 0.602 0.28 -0.553

Calculating slope (1/n):

\[\text{Slope} = \frac{\Delta(\log(x/m))}{\Delta(\log P)} = \frac{-0.553 - (-1.00)}{0.602 - (-0.301)}\] \[\text{Slope} = \frac{0.447}{0.903} \approx 0.495\] \[\frac{1}{n} = 0.5 \Rightarrow n = 2\]

Finding k from intercept:

At P = 1.0 atm, log P = 0, so log(x/m) = log k

\[\log k = -0.854\] \[k = 10^{-0.854} \approx 0.14\]

Answer:

The data follows Freundlich isotherm

k = 0.14, n = 2

Equation: x/m = 0.14 P^(1/2)

💡 Freundlich vs Langmuir - Quick Comparison

Property Freundlich Langmuir
Type Empirical Theoretical (kinetic)
Surface Heterogeneous Homogeneous
Layers Multi-molecular Mono-molecular only
Validity Moderate pressure All pressures
JEE Frequency More common Less common
3

Catalysis

A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed in the reaction. The phenomenon is called catalysis.

3.1 Characteristics of Catalyst

  • Remains unchanged:

    Catalyst is not consumed in the reaction (chemically and quantitatively unchanged)

  • Small quantity sufficient:

    Very small amount can catalyze large amount of reactants

  • Does not initiate reaction:

    Cannot start a reaction that is thermodynamically not feasible (ΔG > 0)

  • Does not change equilibrium:

    Only increases rate - both forward and backward equally. Keq remains same

  • Specific in nature:

    A catalyst is usually specific for a particular reaction

  • Lowers activation energy:

    Provides alternative pathway with lower Ea

3.2 Types of Catalysis

Homogeneous Catalysis

Catalyst and reactants are in the same phase

Examples:

  • 2SO₂(g) + O₂(g) →NO(g) 2SO₃(g)
  • CH₃COOCH₃ + H₂O →HCl CH₃COOH + CH₃OH
  • 4NH₃ + 5O₂ →NO 4NO + 6H₂O (Ostwald process)

Characteristics:

• Cannot be separated easily

• Reaction mechanism well understood

Heterogeneous Catalysis

Catalyst and reactants are in different phases

Examples:

  • N₂(g) + 3H₂(g) →Fe(s) 2NH₃(g) (Haber)
  • Vegetable oils(l) + H₂(g) →Ni(s) Vanaspati (hydrogenation)
  • 2H₂O₂(aq) →MnO₂(s) 2H₂O + O₂ (decomposition)

Characteristics:

• Usually solid catalyst, gas/liquid reactants

• Can be separated by filtration

3.3 Mechanism of Heterogeneous Catalysis

Involves following steps on catalyst surface:

Step 1: Diffusion

Reactant molecules diffuse to catalyst surface

Step 2: Adsorption

Reactants are adsorbed on catalyst surface (usually chemisorption)

Step 3: Reaction

Chemical reaction occurs on surface forming intermediates/products

Step 4: Desorption

Products are desorbed from surface, freeing active sites

Step 5: Diffusion

Product molecules diffuse away from surface

3.4 Important Industrial Catalytic Processes

Process Reaction Catalyst Product
Haber Process N₂ + 3H₂ → 2NH₃ Fe + Mo/Al₂O₃ Ammonia
Contact Process 2SO₂ + O₂ → 2SO₃ V₂O₅ or Pt SO₃ (for H₂SO₄)
Ostwald Process 4NH₃ + 5O₂ → 4NO + 6H₂O Pt-Rh NO (for HNO₃)
Hydrogenation Oils + H₂ → Fats Ni Vanaspati ghee
Deacon Process 4HCl + O₂ → 2Cl₂ + 2H₂O CuCl₂ Chlorine
Cracking Heavy hydrocarbons → Light Al₂O₃ + SiO₂ Petrol, diesel

⚠️ Catalytic Poisoning

Certain substances decrease or destroy the activity of a catalyst. These are called catalytic poisons.

  • Haber process: CO, H₂S poison Fe catalyst
  • Contact process: As₂O₃ poisons V₂O₅ catalyst
  • Hydrogenation: S, P, As, Hg poison Ni catalyst

Mechanism: Poison preferentially adsorbs on active sites, blocking reactants

💡 Auto-Catalysis & Promoters

Auto-Catalysis:

Product of reaction itself acts as catalyst

Example: CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH (CH₃COOH catalyzes reaction)

Promoters:

Substances that enhance catalyst activity but are not catalysts themselves

Example: Mo acts as promoter for Fe in Haber process

4

Colloids

Colloidal solutions (colloids) are heterogeneous mixtures where particle size is between 1-1000 nm (10⁻⁹ to 10⁻⁶ m). Cannot be seen by naked eye but show Tyndall effect.

4.1 Classification of Colloids

Dispersed Phase Dispersion Medium Name Examples
Solid Solid Solid Sol Colored glasses, Gemstones
Solid Liquid Sol Paints, Gold sol, Muddy water
Solid Gas Aerosol Smoke, Dust in air
Liquid Solid Gel Cheese, Jellies, Butter
Liquid Liquid Emulsion Milk, Cod liver oil
Liquid Gas Aerosol Fog, Mist, Cloud
Gas Solid Solid foam Pumice stone, Foam rubber
Gas Liquid Foam Soap lather, Whipped cream
Gas Gas Not possible (forms solution)

4.2 Lyophilic vs Lyophobic Colloids

Property Lyophilic (Solvent-loving) Lyophobic (Solvent-hating)
Meaning Hydrophilic (if water is medium) Hydrophobic (if water is medium)
Preparation Easy - direct mixing Difficult - special methods needed
Stability Very stable Unstable
Viscosity Higher than medium Same as medium
Reversibility Reversible Irreversible
Coagulation Difficult (need large amount of electrolyte) Easy (small amount sufficient)
Examples Starch, gelatin, protein, gum Metal sols (Au, Ag), metal sulfides (As₂S₃)
5

Properties of Colloids

5.1 Key Properties (JEE Important)

1. Tyndall Effect

Scattering of light by colloidal particles

  • Path of light becomes visible
  • True solutions don't show this
  • Example: Sunlight through fog, cinema hall beam
2. Brownian Motion

Zig-zag random motion of colloidal particles

  • Due to unequal bombardment by medium molecules
  • Prevents settling of particles
  • Provides stability to colloid
3. Electrophoresis

Movement of colloidal particles under electric field

  • Positive sol → Moves to cathode
  • Negative sol → Moves to anode
  • Proves colloidal particles carry charge
4. Coagulation/Flocculation

Settling of colloidal particles

  • By adding electrolytes
  • By heating or boiling
  • By adding oppositely charged sol

5.2 Hardy-Schulze Rule (Very Important for JEE)

Coagulation of Colloids by Electrolytes

Greater the valency of the coagulating ion (opposite charge to sol), greater is its coagulating power.

For Negative Sols (e.g., As₂S₃):

\[\text{Coagulating power: } Al^{3+} > Ba^{2+} > Na^{+}\]

Cations coagulate negative sols

For Positive Sols (e.g., Fe(OH)₃):

\[\text{Coagulating power: } PO_4^{3-} > SO_4^{2-} > Cl^{-}\]

Anions coagulate positive sols

⚠️ Gold Number

Gold number: Minimum mass (in mg) of protective colloid required to prevent coagulation of 10 mL gold sol by 1 mL of 10% NaCl solution.

Lower gold number → Better protective power

Example: Gelatin (0.005-0.01) > Gum arabic (0.15-0.25) > Starch (20-25)

6

Emulsions

Emulsion: Colloidal dispersion of liquid in liquid

Types:

Oil-in-Water (O/W)

Oil dispersed in water

Examples: Milk, Vanishing cream

Test: Conducts electricity

Water-in-Oil (W/O)

Water dispersed in oil

Examples: Butter, Cold cream

Test: Does not conduct

Emulsifying Agents:

Stabilize emulsions - Soap, Detergents, Proteins, Gums

7

Applications of Colloids

🏥 Medicine:

Colloidal silver (antiseptic), Colloidal gold (medicine)

💧 Water Purification:

Alum coagulates dirt particles

🏭 Industrial:

Paints, inks, rubber, photography

🥛 Food:

Milk, ice cream, whipped cream

📝 Previous Year Questions Analysis

JEE Main (Last 5 Years)

  • ✓ Colloids Classification: 30%
  • ✓ Adsorption & Isotherms: 25%
  • ✓ Properties of Colloids: 20%
  • ✓ Catalysis: 15%
  • ✓ Emulsions: 10%

JEE Advanced (Last 5 Years)

  • ✓ Freundlich Isotherm: 35%
  • ✓ Catalysis Mechanisms: 25%
  • ✓ Coagulation & Hardy-Schulze: 20%
  • ✓ Physisorption vs Chemisorption: 15%
  • ✓ Conceptual (theory): 5%

Top 10 Most Repeated Question Types

  1. Calculating k and n from Freundlich isotherm data
  2. Distinguishing between physisorption and chemisorption
  3. Identifying type of colloid from dispersed phase and medium
  4. Applying Hardy-Schulze rule for coagulation
  5. Identifying homogeneous vs heterogeneous catalysis
  6. Determining O/W or W/O emulsion type
  7. Explaining Tyndall effect and Brownian motion
  8. Industrial catalytic processes (Haber, Contact, Ostwald)
  9. Properties of lyophilic vs lyophobic colloids
  10. Preparation methods of colloids

Weightage Analysis

JEE Main: 8-12 marks (2-3 questions)
JEE Advanced: 8-16 marks (2-4 questions)
Difficulty Level: Easy to Medium
Time Required: 3-4 hours practice

🎯 Practice Problem Set

Level 1: Basic (JEE Main Standard)

  1. Differentiate between adsorption and absorption with examples.
  2. Why is adsorption always exothermic?
  3. Name the catalyst used in: (a) Haber process (b) Contact process (c) Hydrogenation
  4. Classify: Milk, Smoke, Fog, Butter, Jellies based on dispersed phase and medium.
  5. What is Tyndall effect? Give two examples.
  6. Arrange Al³⁺, Ba²⁺, Na⁺ in order of coagulating power for As₂S₃ sol.
  7. Distinguish between O/W and W/O emulsions.
  8. What are lyophilic and lyophobic colloids? Give two examples of each.

Level 2: Intermediate (JEE Main/Advanced)

  1. For Freundlich isotherm x/m = kP^(1/n), what happens when: (a) n = 1 (b) n = ∞
  2. Compare physisorption and chemisorption on five different parameters.
  3. Explain mechanism of heterogeneous catalysis with Haber process as example.
  4. What is Hardy-Schulze rule? Why is Al³⁺ better coagulating agent than Na⁺ for negative sol?
  5. Explain: (a) Brownian motion (b) Electrophoresis (c) Peptization
  6. Calculate k and n if at P = 2 atm, x/m = 0.2 and at P = 4 atm, x/m = 0.3
  7. Why is finely divided substance more effective as adsorbent?
  8. Explain catalytic poisoning with two industrial examples.

Level 3: Advanced (JEE Advanced/Olympiad)

  1. Derive Langmuir adsorption isotherm equation. What are its assumptions?
  2. For adsorption of gas on solid: at P = 0.5, 1, 2, 4 atm, x/m = 2, 3, 4, 5. Verify Freundlich isotherm graphically.
  3. Explain why physisorption decreases and chemisorption first increases then decreases with temperature.
  4. How will you distinguish between: (a) Lyophilic and lyophobic sols (b) Sols and gels (c) O/W and W/O emulsions
  5. A negative sol is coagulated by: NaCl (50 mM), CaCl₂ (0.8 mM), AlCl₃ (0.1 mM). Verify Hardy-Schulze rule.
  6. Explain: (a) Gold number (b) Protective colloids (c) Peptization
  7. Why does physisorption form multi-layers while chemisorption forms only monolayer?
  8. Derive relationship between Freundlich and Langmuir isotherms at low pressure.

Related Chemistry Notes