Surface Chemistry Comprehensive Notes For Competitive Exams Download PDF

Surface Chemistry — Comprehensive Notes

Examination-oriented study notes for NEET, JEE (Main & Advanced), IIT-JAM, BITSAT, GATE, CSIR-NET, TGT/PGT

Table of Contents

1. Introduction to Surface Chemistry
2. Adsorption
   1. Adsorption, Absorption and Sorption
   2. Adsorption Mechanism and Thermodynamics
   3. Physisorption and Chemisorption
3. Factors Affecting Adsorption
   1. Adsorption of Gases on Solids
   2. Adsorption of Solids from Solutions
4. Adsorption Isotherms
   1. Freundlich Adsorption Isotherm
   2. Langmuir Adsorption Isotherm
   3. BET Theory of Multilayer Adsorption
   4. Determination of Surface Area of Solids
5. Applications of Adsorption
6. Catalysis
   1. Types of Catalysts and Promoters, Poisons
   2. Characteristics of Catalysts
7. Types of Catalytic Reactions
8. Theory of Heterogeneous Catalysis
9. Shape-Selective Catalysis by Zeolites
10. Enzyme Catalysis
11. Catalysts in Industry
12. Colloids
    1. Types of Colloids
    2. Mechanism of Micelle Formation
13. Preparation of Colloids
14. Purification of Colloidal Solutions
15. Properties of Colloidal Solutions
16. Coagulation and Hardy–Schulze Rule
17. Protection of Colloids
18. Emulsions
19. Role of Colloids in Natural Phenomena and Industry
20. Solved Examples
21. Practice MCQs (with Answers)

1. Introduction to Surface Chemistry

The surface of a material is an interface between a solid and a gas or between a solid and the solvent/solute in a solution. It is written by separating the two bulk phases with a hyphen — for example, solid–gas or solid–liquid. This interface is only a few molecules thick (about 0.5–2 nm in the solid bulk phase), and its area is governed by the particle size of the bulk phase. For gases no such interface exists because gases are completely miscible.

Surface chemistry is the study of chemical processes that occur at this interface. Surface phenomena are responsible for many natural and industrial changes — catalysis, electrode processes, crystallization, and the cleansing action of soaps and detergents. Rusting of iron (corrosion) is an undesirable surface change that is prevented by coating the surface with paint so as to deactivate it toward oxidation.

For an accurate study of surface phenomena the surface must be clean and must have the same composition as the bulk phase. Clean surfaces are prepared by removing impurities through heating, chemical reactions, or bombardment with electrons or ions. For absolutely clean surfaces, metals are subjected to an ultra-high vacuum of the order of 10⁻¹⁰ to 10⁻¹⁴ mm of Hg and stored under vacuum to prevent contamination by atmospheric gases.

When a molecule strikes a solid surface it may rebound, undergo a reaction, or get adsorbed. The adsorbed molecule may then diffuse on the surface, react, or dissolve into the bulk. Our focus here is on the equilibrium and dynamics of processes that occur at the interface — not when the molecule dissolves in the bulk phase.

2. Adsorption

Adsorption is the phenomenon of the assimilation of a higher concentration of any species at the surface of a solid or liquid compared to that present in the bulk of the material. Familiar examples include the accumulation of dust or soot on the skin when travelling in heavy traffic, and the protection from poisonous gases afforded by gas masks.

In adsorption, substances leave the gaseous or liquid phase and form a new layer on the surface of a solid substrate. The solid or liquid on whose surface the gas or liquid is adsorbed is called the adsorbent (or substrate); it must be a condensed phase. Solids in a finely divided state possess large surface areas and act as good adsorbents — silica gel, metals, clay, and activated charcoal are common examples. The gas or solution held on the surface is the adsorbate; it is atomic, ionic, or molecular, and is generally a liquid or gas.

Adsorption can be positive or negative. When the concentration of the adsorbate is greater on the surface than in the bulk, it is positive adsorption. When the concentration of the adsorbate is less on the surface than in the bulk, it is negative adsorption (for example, adsorption of water by charcoal from a concentrated KCl solution, where the solvent itself is taken up preferentially).

2.1 Adsorption, Absorption and Sorption

Adsorption is a surface phenomenon, whereas absorption is a bulk phenomenon in which the concentration of the molecular species is uniformly distributed throughout the body of the solid or liquid. In adsorption the concentration increases only at the surface; in absorption it increases uniformly throughout the bulk.

Absorption	Adsorption
Bulk phenomenon; concentration uniformly distributed throughout the solid or liquid.	Assimilation of higher concentration at the surface compared to the bulk.
Slower process; equilibrium takes longer time.	Faster process; equilibrium is attained in a shorter time.
Example: ammonia is absorbed by water.	Example: charcoal in a gas mask adsorbs poisonous gases.

The term sorption describes a process in which both absorption and adsorption occur simultaneously, and desorption is the reverse — the release of a substance from a surface. When a chalk stick is dipped in ink, the colour is adsorbed on the surface while the solvent is absorbed throughout the thickness of the chalk. During cooking of a curry, the yellow colouring compounds of turmeric are absorbed into the curry but are adsorbed on the surface of the pan, producing yellow stains.

Desorption occurs when the concentration (or pressure) of the substance in the bulk phase is lowered; it is brought about by increasing the temperature or reducing the pressure. Desorption is important in catalysis, where it frees the catalyst surface for fresh adsorption of reactants.

2.2 Adsorption Mechanism and Thermodynamics

The driving force for adsorption is the presence of unbalanced or residual forces at the solid or liquid surface. Surface molecules are attracted only from below and from the sides, unlike molecules in the bulk that are attracted uniformly from all directions. The surface molecules therefore tend to move inwards, where forces are uniform. Because of these residual forces, the surface is in a state of strain and tends to attract and retain molecules of another species, creating an excess concentration at the surface.

Adsorption reduces these residual forces, which causes a decrease in surface energy that appears as heat. This is the heat of adsorption, and adsorption is therefore exothermic.

The incoming adsorbate molecule faces two interactions from the substrate: electronic repulsion and steric restriction. If the substrate is aromatic or unsaturated, charge donated during adsorption-bond formation is conducted away from the site, reducing electronic repulsion. In a non-conducting adsorbent, the donated charge stays near the site and restricts the approach of further adsorbate molecules. The adsorbate molecule must approach the vacant site from within a certain cone-shaped region; otherwise steric restrictions prevent successful adsorption.

Since adsorption is spontaneous, the Gibbs free energy change must be negative:

ΔG = ΔH − TΔS   (1)

Because adsorption restricts the movement of gas molecules, it is always accompanied by a decrease in entropy (ΔS < 0). Since it is exothermic, ΔH is also negative. For ΔG to be negative, ΔH must be sufficiently more negative than TΔS so that (ΔH − TΔS) < 0. As adsorption proceeds, ΔH becomes less and less negative; at ΔH = TΔS, ΔG becomes zero and equilibrium is reached, with adsorption and desorption occurring at equal rates.

Conceptual point: Adsorption is exothermic because it is spontaneous with a negative ΔS. If ΔS is negative, then −TΔS is positive, so ΔH must be negative (and large enough) to keep ΔG negative. A common error is to think a process with negative ΔS cannot be spontaneous — it can, provided it is sufficiently exothermic.

2.3 Physisorption and Chemisorption

Based on the forces of attraction between adsorbent and adsorbate, adsorption is classified into two types:

1. Physisorption (physical adsorption or van der Waals adsorption) — e.g. adsorption of gases by charcoal.
2. Chemisorption (chemical adsorption or activated adsorption) — e.g. adsorption of oxygen on tungsten.

The two forms may occur simultaneously or may interchange with temperature. In the hydrogenation of alkenes over nickel, hydrogen first physisorbs on the nickel surface and, at higher temperature, dissociates into hydrogen atoms bonded to nickel by hydride formation (chemisorption). If silica gel and elemental bromine are placed at opposite ends of a sealed container, bromine vapours travel to the silica gel and are chemisorbed; once the chemisorbed monolayer is complete, successive layers are held by physisorption.

The enthalpy of chemisorption is greater than that of physisorption because strong chemical bonds are involved, whereas physisorption involves only weak van der Waals forces. Since bond formation in chemisorption requires a higher activation energy, it is also called activated adsorption.

The extent of both types increases with the surface area per unit mass of the adsorbent, because subdividing a solid creates more residual forces. In physisorption, the nature of the adsorbate also matters: easily liquefiable gases (gases with high critical temperatures) are more readily adsorbed because van der Waals forces are stronger near the critical temperature.

Physisorption	Chemisorption
Weak, long-range van der Waals forces.	Strong chemical bonds.
No surface complex formed.	A surface complex forms between adsorbate and adsorbent.
Reversible; gas recovered by lowering pressure or raising temperature.	Irreversible; involves chemical bond formation.
Not specific; can bond similarly to all gases.	Highly specific; occurs only where chemical bond formation is possible.
Occurs at low temperatures (low activation energy, ~5 kJ).	Occurs at high temperatures (high activation energy).
Decreases with increase in temperature.	Increases with increase in temperature.
Equilibrium attained rapidly.	Equilibrium attained relatively slowly.
Heat of adsorption is low (20–40 kJ).	Heat of adsorption is high (80–240 kJ mol−1).
Generally multilayered.	Monolayered.

3. Factors Affecting Adsorption

3.1 Adsorption of Gases on Solids

Adsorption of gases on a solid surface (sometimes called occlusion) depends on the following factors.

1. Temperature. Adsorption is exothermic, so by Le Chatelier's principle its magnitude increases with a decrease in temperature. At lower temperature the thermal energy is smaller, so more molecules are held to the surface by residual forces. A plot of adsorption against temperature at constant pressure is an adsorption isobar. For physisorption the isobar shows a decrease in x/m with increasing temperature; for chemisorption, x/m first increases (because activation energy must be supplied) and then decreases. A plot of temperature versus pressure for a fixed amount of adsorption is an adsorption isostere, which should be linear.

2. Pressure. At constant temperature the extent of adsorption (expressed as x/m, where x is the mass of adsorbate and m the mass of adsorbent) increases with pressure at low pressure, in agreement with Le Chatelier's principle — the system shifts toward fewer molecules when pressure is increased. Physical adsorption increases with pressure and eventually reaches a saturation value.

3. Nature of adsorbent. At a given temperature and pressure, the greater the surface area per unit mass, the greater the adsorption. Finely divided, porous solids therefore show a much larger extent of adsorption because subdivision generates more residual surface forces.

4. Nature of adsorbate. The molecular forces involved are more predominant in easily liquefiable and readily soluble gases (e.g. HCl, NH₃, Cl₂) than in permanent gases (e.g. H₂, N₂, O₂). Hence the former are adsorbed to a greater extent.

5. Surface area. The extent of adsorption increases with the surface area of the adsorbent; porous or finely divided forms therefore adsorb larger quantities.

3.2 Adsorption of Solids from Solutions

Solids also adsorb solutes from solutions — acetic acid is adsorbed from an acetic acid–water mixture when shaken with charcoal, and litmus solution becomes colourless when shaken with charcoal. The factors influencing adsorption from solution are:

1. An adsorbent adsorbs certain solutes in preference to others; the extent is greater for solutes of high molecular mass.
2. Adsorption decreases with rise in temperature (increased kinetic energy lets solute particles leave the surface).
3. Adsorption increases with the concentration of the solution. When the solute is taken up it is positive adsorption (e.g. KCl by charcoal from a concentrated solution); when the solvent is taken up it is negative adsorption (e.g. water by charcoal from a concentrated KCl solution).
4. The extent increases with the surface area of the adsorbent.
5. Adsorption increases as the solubility of the solute in the solvent decreases, and adsorption strength increases with the polarity of the solute.

4. Adsorption Isotherms

An adsorption isotherm describes the variation in the amount of gas adsorbed by an adsorbent at a particular temperature as pressure is varied. Different gases on different adsorbents show five types of isotherms: Type I corresponds to unimolecular adsorption, while Types II–V correspond to multimolecular adsorption.

4.1 Freundlich Adsorption Isotherm

Freundlich observed the variation of x/m with pressure p at constant temperature. The curve flattens at higher pressures, indicating a saturation point. Three cases arise:

Case 1 — Low pressure: the plot is straight and sloping, so x/m is directly proportional to p:

x/m = k p   (direct proportionality)

Case 2 — High pressure: when pressure exceeds the saturation pressure p_s, x/m becomes independent of p:

x/m = k p⁰ = constant

Case 3 — Intermediate pressure: x/m depends on p raised to a fractional power between 1 and 0. This is the Freundlich adsorption isotherm:

x/m = k p^1/n   (n > 1)

Taking logarithms:

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

A plot of log(x/m) versus log p is a straight line with slope 1/n and intercept log k. The factor 1/n varies between 0 and 1: when 1/n = 0, x/m is constant; when 1/n = 1, x/m varies directly with pressure. The relation is approximate and holds only over a limited range of pressure; it fails at high pressures. Other limitations: it is applicable only to physical adsorption and does not account for multilayer adsorption, and it ignores the role of the surface area of the adsorbent.

For adsorption from solutions, pressure is replaced by concentration C:

x/m = k C^1/n   (2)

log(x/m) = log k + (1/n) log C   (3)

A graph of log(x/m) versus log C is a straight line over small concentration ranges.

Solved Example 1. A graph of log(x/m) vs. log p is linear with slope 45° and intercept 0.3010 on the log(x/m) axis. Find the amount of gas adsorbed per gram of charcoal at p = 0.5 atm.
Solution: slope = 1/n = tan 45° = 1, so n = 1. log k = 0.3010 gives k = 2. Then x/m = k p^1/n = 2 × (0.5)¹ = 1.0.

Solved Example 2. Check the applicability of the Freundlich isotherm and find k and n for adsorption of a solute on a solid at 25 °C, given the data below.

Concentration (mol L−1)	0.05	0.10	0.50	1.00	1.50
x/m (g)	0.048	0.072	0.144	0.192	0.228

Solution: A plot of log C (x-axis) versus log(x/m) (y-axis) gives slope 1/n and intercept log k. The values obtained are k = 0.19 and n = 2.29.

Solved Example 3. Different concentrations of aqueous solutions were placed with charcoal. At equilibrium: C × 10⁻² = 2.0 and 4.0; x/m = 0.185 and 0.290. Calculate log k and n.
Solution: Using log(x/m) = log k + (1/n) log C with the two data pairs and solving, we get 1/n = 0.64, k = 83.5, and log k = 1.92.

4.2 Langmuir Adsorption Isotherm

In 1916, Langmuir gave an expression for the extent of adsorption based on these assumptions:

1. The solid surface contains only a fixed number of adsorption sites.
2. All sites are equivalent and the surface is homogeneous.
3. Each site adsorbs only one gas molecule; adsorption is monolayered.
4. The heat of adsorption is constant and identical for all sites.
5. Adsorbed molecules do not interact with one another.
6. Molecules tend to be adsorbed, and adsorbed molecules tend to evaporate (desorb) back into the gas phase.

Let θ be the fraction of surface covered. Then:

Rate of desorption = k_d θ   ;   Rate of adsorption = k_a p (1 − θ)

At equilibrium, the two rates are equal:

k_a p (1 − θ) = k_d θ   (4)

Solving for θ:

θ = (k_a/k_d) p / [1 + (k_a/k_d) p]   (5)

Substituting the distribution constant k₁ = k_a/k_d:

θ = k₁ p / (1 + k₁ p)   (6)

Since θ is proportional to x/m, we write x/m = k₂ θ, giving the Langmuir equation:

x/m = k₁ k₂ p / (1 + k₁ p)   (8)

Taking reciprocals:

p / (x/m) = (1/k₁k₂) + (1/k₂) p   (10)

A plot of p/(x/m) versus p is linear with slope 1/k₂ and intercept 1/(k₁k₂), from which k₁ and k₂ are obtained.

Langmuir's isotherm explains the Freundlich curve under limiting conditions:

• Low pressure: θ is small, so (1 − θ) ≈ 1 and k_ap = k_dθ ⇒ θ = k₁p. Hence x/m = k₁k₂p   (12) — directly proportional to p.
• High pressure: θ ≈ 1, so (1 − θ) = k_d/(k_ap) and θ = 1 − 1/(k₁p). At very high pressure, 1/(k₁p) becomes negligible, so x/m ≈ k₂   (15) — a constant (saturation).
• Intermediate pressure: the denominator (1 + k₁p) grows faster than the numerator, so x/m does not increase as fast as p.

Solved Example 4. Data for adsorption of CO on charcoal at 273 K are given below. Confirm the Langmuir isotherm and find (a) k₁ and V_max, (b) the fraction of surface covered at p = 600 mm Hg, and (c) ΔG° of adsorption at STP. Mass of charcoal = 3.022 g.

p (mm Hg)	100	200	300	400	500	600	700
V (cm3)	10.2	18.6	25.5	31.4	36.9	41.6	46.1

Solution: A plot of p/V versus p gives slope = 1/V_max = 0.009 and intercept = 1/(k₁V_max) = 9. So V_max = 111.11 cm³ and k₁ = 0.001 mm Hg⁻¹.
(b) At 600 mm Hg, θ = V/V_max = 41.6/111.11 = 0.374.
(c) ΔG_ads = −2.303 RT log k₁. With k₁ = 0.001 mm Hg⁻¹ = 0.76 atm⁻¹, ΔG_ads = −2.303 × 8.314 × 273 × log(0.76) = +0.62 kJ mol⁻¹.

Solved Example 5. A gas follows Langmuir adsorption with k₁ = 0.9 kPa⁻¹ at 27 °C. Find the pressure for fractional coverage (a) 0.1, (b) 0.5, (c) 0.95.
Solution: From θ = k₁p/(1 + k₁p), rearrange: p = [θ/(1 − θ)] × (1/k₁).
(a) p = (0.1/0.9) × (1/0.9) = 0.123 kPa; (b) p = (0.5/0.5) × (1/0.9) = 1.11 kPa; (c) p = (0.95/0.05) × (1/0.9) = 21.11 kPa.

Solved Example 6 (IIT-JEE 2003). 1 g of activated charcoal is added to 100 mL of 0.5 M acetic acid. The concentration becomes 0.49 M. If the surface area of charcoal is 3.01 × 10² m², find the area occupied by a single acetic acid molecule.
Solution: Moles adsorbed = 0.05 − 0.049 = 0.001 mol. Molecules adsorbed = 0.001 × 6.02 × 10²³ = 6.02 × 10²⁰. Area per molecule = (3.01 × 10²)/(6.02 × 10²⁰) = 5 × 10⁻¹⁹ m².

4.3 BET Theory of Multilayer Adsorption

The Langmuir isotherm is valid only at low pressure and high temperature, where monolayer adsorption and non-interacting adsorbate molecules are reasonable assumptions. At high pressure and low temperature, additional layers form and the Langmuir model fails. Brunauer, Emmett, and Teller proposed the BET theory to explain multilayer adsorption.

The BET theory assumes a uniform surface of localized sites where adsorption at one site does not affect neighbouring sites. After the first layer forms, molecules continue to adsorb in subsequent layers (second, third, … nth). The surface area available for the nth layer equals that for the (n − 1)th layer. The ratio of the volume V of gas adsorbed to the monolayer volume V_m is:

V / V_m = C x / [(1 − x)(1 + (C − 1)x)]   (16)

where x = p/p₀ (p is the gas pressure at temperature T and p₀ is the saturated vapour pressure at the same temperature). In linear form:

x / [V(1 − x)] = 1/(C V_m) + [(C − 1)/(C V_m)] x   (17)

A plot of x/[V(1 − x)] versus x yields V_m and C, giving the monolayer volume.

4.4 Determination of Surface Area of Solids

If a monolayer forms, the fractional surface coverage is V/V_m. The Langmuir form V = V_m/(1 + 1/k₁p) allows V_m and k₁ to be obtained from a plot of 1/V versus 1/p. At higher pressures where Langmuir deviates, the BET equation is used. The area occupied by one adsorbate molecule is estimated from the density of the liquefied adsorbate. Nitrogen is commonly used at its boiling point (−195 °C); assuming spherical, closely packed molecules, the area of one nitrogen molecule is taken as 16.2 × 10⁻²⁰ m².

5. Applications of Adsorption

1. Pollution control and clarification of sugar: Activated carbon (extremely porous, very large surface area) is used in gas masks, for removal of volatile organic compounds, for removal of organic matter from drinking water, decolouration of vinegar and sugar solution, and control of odours in the food industry.
2. Filtration (molecular sieves): Molecular sieves contain tiny uniform pores that act as filters on the molecular level; small molecules pass through and are absorbed, large ones are not. They can absorb water up to about 22% of their own weight and function as desiccants.
3. Controlling humidity and drying air: Silica gel dries processed air (oxygen, natural gas) and adsorbs hydrocarbons from natural gas; silica and alumina gels control humidity in rooms.
4. Metallurgy: Fuller's earth is used in refining petroleum and vegetable oil by adsorbing unwanted materials.
5. Dyeing industry: Mordants such as alum adsorb dye particles.
6. Paint industry: Dissolved gases are removed from paint during manufacture to increase adherence; wetting agents (e.g. spirit in furniture painting) adsorb gaseous/liquid films on surfaces before painting.
7. Chromatography: Adsorption chromatography separates compounds by selective adsorption on a suitable adsorbent. The most readily adsorbed solute appears in the initial layers. Used in column, gas, and HPLC techniques.
8. Catalysis: The interaction of reactants/products with the surface should be neither too strong nor too weak. If too weak, the catalyst has little effect; if too strong, reactants are immobilized and fail to react. Activity therefore increases initially and then decreases with the strength of adsorption. Examples: Haber's process (N₂ + 3H₂ → 2NH₃, Fe catalyst) and hydrogenation of vegetable oils (Ni catalyst).
9. Indicators in titrations: KBr is titrated with AgNO₃ using eosin as an adsorption indicator.
10. Softening hard water: Ion-exchange resins (insoluble, cross-linked, long-chain polymers with functional groups; also zeolites, clay, soil humus) release one ion and adsorb another; used for water softening, demineralization, desalination, removal of heavy metals and ammonia, and radioactive waste treatment.
11. Removal of coloured impurities: Sugar solution is passed over a charcoal bed; coloured impurities are adsorbed, giving white sugar crystals.
12. Production of vacuum: After evacuation by a vacuum pump, residual traces of air are removed with adsorbents such as charcoal; adsorbents may also be placed between layers of Dewar flasks.
13. Separation of inert gases: Noble gases are separated based on differences in extent of adsorption on charcoal.
14. Biological applications: Adsorption removes toxins (germs) from the body and is used for recovery and concentration of vitamins and other biological substances.

6. Catalysis

Many reactions are accelerated by small amounts of substances that themselves remain chemically unaltered; these are catalysts. The action of a catalyst is specific to a reaction. Although the catalyst's amount and composition are unchanged at the end, it enters the reaction, forms an intermediate complex with a reactant, and this complex decomposes to give products with regeneration of the catalyst. A very small quantity of catalyst can cause a large increase in rate — for example, a small amount of MnO₂ greatly accelerates the decomposition of KClO₃. Since the catalyst is regenerated, it does not contribute energy and has no influence on the position of equilibrium.

6.1 Types of Catalysts, Promoters and Poisons

Positive catalysts (accelerators) open an alternative reaction path that allows lower-energy molecules to react, increasing the rate. Examples:

2SO₂(g) + O₂(g) —^Pt→ 2SO₃

N₂(g) + 3H₂(g) —^Fe→ 2NH₃(g)

Negative catalysts (inhibitors) retard the reaction rate. Examples include glycerol (which retards rusting of machinery), lead tetraethyl or nickel carbonyl (antiknock agents in engines), and a small quantity of alcohol (which retards oxidation of chloroform).

Promoters are substances that enhance the activity of a catalyst; for example, molybdenum promotes iron in ammonia synthesis. Anticatalysts deactivate a catalyst. Poisons decrease the catalyst's efficiency so that the reaction stops at an intermediate stage — for example, when BaSO₄ and quinoline are added with Pd to acetylene (Lindlar-type conditions), hydrogenation stops at ethene and ethane is not formed:

CH≡CH + H₂ —^Pd→ CH₂=CH₂ (+ H₂) —^{Pd + BaSO₄ + quinoline}→ CH₃CH₃ (does not form)

6.2 Characteristics of Catalysts

1. A catalyst remains unchanged in composition and amount at the end, though it may change physically (e.g. a solid recovered as a powder).
2. A small amount is usually sufficient; in some homogeneous reactions, however, the rate depends on the catalyst concentration.
3. A catalyst does not affect thermodynamic parameters (ΔH, ΔS, ΔG).
4. It does not shift the equilibrium, because it affects the rates of the forward and backward reactions equally.
5. Activity: an effective catalyst adsorbs reactants strongly enough for them to react, but not so strongly that products adhere permanently. Chemisorption is the main factor deciding activity. Pt catalyzes the formation of water: 2H₂(g) + O₂(g) —^Pt→ 2H₂O(l). Note: the metals of Groups 5 to 11 show an increasing order of catalytic activity in hydrogenation.
6. Selectivity (specificity): a catalyst directs the reaction to desired products. Different catalysts give different products from H₂ + CO:
   CO(g) + 3H₂(g) —^Ni→ CH₄(g) + H₂O(g)
   CO(g) + 2H₂(g) —^{Cu/ZnO–CrO₃}→ CH₃OH(g)
   CO(g) + H₂(g) —^Cu→ HCHO(g)

Solved Example 7. 20% of surface sites have adsorbed N₂. On heating, N₂ evolved was collected at 0.001 atm and 298 K in a 2.46 cm³ container. Density of surface sites = 6.023 × 10¹⁴ cm⁻², surface area = 1000 cm². Find the number of sites occupied per molecule of N₂. (IIT-JEE 2005)
Solution: By pV = nRT, n = (0.001 × 2.46 × 10⁻³)/(0.0821 × 298) = 1.0 × 10⁻⁷ mol. Molecules of N₂ = 6.023 × 10²³ × 10⁻⁷ = 6.023 × 10¹⁶. Total sites = 6.023 × 10¹⁴ × 1000 = 6.023 × 10¹⁷. Sites used = 20% of total = 12.04 × 10¹⁶. Sites per N₂ molecule = 12.04 × 10¹⁶ / 6.023 × 10¹⁶ = 2.

Solved Example 8. Name two physical quantities that remain unaffected when a reversible reaction is carried out in the presence of a catalyst.
Solution: Enthalpy change (ΔH) and the equilibrium constant (K_eq). The catalyst only speeds up the attainment of equilibrium.

7. Types of Catalytic Reactions

1. Homogeneous catalysis: the catalyst is in the same phase as the reactants. Gaseous-phase example — NO catalyzes the oxidation of CO:
   2CO(g) + O₂(g) —^NO(g)→ 2CO₂(g)
   Solution-phase example — acid hydrolysis of ethyl acetate:
   CH₃CH₂COOCH₃(l) + H₂O(l) —^HCl(l)→ CH₃CH₂COOH(aq) + CH₃OH(aq)
2. Heterogeneous catalysis: the catalyst is in a different phase from the reactants. Ethylene hydrogenation over nickel:
   CH₂=CH₂ + H₂ —^Ni→ CH₃CH₃
   Solid Pt or V₂O₅ catalyzes SO₂ → SO₃ in the Contact process; solid Fe in Haber's process. Gold catalyzes decomposition of N₂O:
   2N₂O —^Au→ 2N₂ + O₂
   N₂O is chemisorbed on Au; a bond forms between the O of N₂O and an Au atom, weakening the N–O bond and making it easier to break.
3. Autocatalysis: a product of the reaction speeds up the reaction. The hydrolysis of an ester by water is autocatalytic because the acid liberated catalyzes further hydrolysis.
4. Induced catalysis: one reaction influences the rate of another that does not ordinarily occur. Reduction of HgCl₂ by oxalic acid proceeds faster in the presence of KMnO₄, because reduction of KMnO₄ by oxalic acid induces the reduction of HgCl₂.
5. Acid–base catalysis: catalyzed by acids or bases. Examples — ester hydrolysis:
   CH₃COOC₂H₅ + H₂O —^[H+]→ CH₃COOH + C₂H₅OH
   and inversion of cane sugar:
   C₁₂H₂₂O₁₁ + H₂O —^[H+]→ C₆H₁₂O₆ + C₆H₁₂O₆ (glucose + fructose)
6. Enzyme catalysis: reactions catalyzed by enzymes (discussed in Section 10).

8. Theory of Heterogeneous Catalysis

In heterogeneous catalysis the catalyst and reactants are in different phases. In Haber's ammonia synthesis, the reaction occurs on an iron catalyst containing traces of aluminium and potassium oxides; H₂ and N₂ dissociate on the surface, combine to form NH₃, and the product desorbs. It is important to remove CO in ammonia synthesis because CO poisons the iron catalyst.

The theories explaining catalytic action are:

1. Intermediate compound theory. The catalyst forms a reactive, unstable intermediate with a reactant; this intermediate then reacts with the other reactant to give the product and regenerate the catalyst:

Step 1: A + X → C + AX (intermediate)
Step 2: AX + B → D + X

2. Adsorption theory. The catalyst surface has active sites where reactants adsorb as a unimolecular layer due to unsatisfied valence forces. If both reactants are adsorbed, they react because of close proximity; if only one is adsorbed, the other (gas/liquid) strikes it. Products desorb, freeing the active sites. The steps are:

1. Diffusion of reactant molecules to the surface.
2. Adsorption of reactants at active sites.
3. Formation of an intermediate complex and its reaction to give products.
4. Desorption of products and regeneration of the catalyst.
5. Diffusion of products away from the surface.

3. Combination of the two theories.

• The catalyst provides an alternative path by forming an adsorbed activated complex, lowering the activation energy.
• Active centres govern the amount of adsorption. Their number is increased by powdering the catalyst (larger surface area), by increasing surface roughness, or by preparing the catalyst in colloidal form. Rough surfaces are more effective because cracks, corners, and breaks contain more active sites.
• Many transition metals and their compounds act as catalysts because of their ability to exhibit multiple oxidation states and to provide large surface area. They form unstable intermediates and provide a new path with lower activation energy — e.g. V₂O₅ in the Contact process.

This combined mechanism explains most observed characteristics of catalysis and why the catalyst's amount and composition remain unchanged, but it cannot fully explain the action of promoters and poisons.

Exam tip: The activity–adsorption strength relationship is a classic exam concept. Activity rises initially with adsorption strength (more reactant held) but then falls (reactants become immobilized). The optimum is "strong enough to activate, weak enough to release."

9. Shape-Selective Catalysis by Zeolites

Shape-selective catalysts are those whose reactions depend on the pore structure of the catalyst and the size of the reactant and product molecules. Zeolites are naturally occurring hydrated aluminosilicates of the general formula:

M_x/n [(AlO₂)_x · (SiO₂)_y] · m H₂O

where n is the charge of the metal cation Mⁿ⁺ (usually Na⁺, K⁺, or Ca²⁺) and m is the (highly variable) number of moles of water of hydration. The characteristic feature is the openness of the honeycomb-like (AlO₂)_n network — a three-dimensional silicate network in which some Si sites are replaced by Al, giving an Al–O–Si network.

Zeolites have pores of various sizes and act as molecular sieves. They are heated in vacuum to lose water of hydration, making the structure porous. The pore size is generally 260–740 pm, which controls which molecules can be adsorbed. Zeolites are used for water purification and as selective adsorbents for gases or liquids; synthetic zeolites are now made to meet specific catalyst needs.

A synthetic zeolite, ZSM-5 (Na_nAl_nSi_96−nO₁₉₂ · 16 H₂O), is used in the petroleum industry to dehydrate alcohols directly into a mixture of hydrocarbons usable as high-octane petrol.

Why zeolites separate straight-chain from branched hydrocarbons: the pore size (260–740 pm) allows only small, straight-chain molecules to enter; larger branched or aromatic molecules cannot fit. Thus zeolites act as selective adsorbents (molecular sieves) in the petrochemical industry.

10. Enzyme Catalysis

Enzymes are complex nitrogenous biological macromolecules (most are proteins) produced by living plants and animals; they catalyze numerous reactions to maintain life processes and are therefore called biochemical catalysts. Examples of enzyme-catalyzed reactions:

1. Inversion of cane sugar (enzyme invertase):
   C₁₂H₂₂O₁₁ + H₂O —^invertase→ C₆H₁₂O₆ + C₆H₁₂O₆ (glucose + fructose)
2. Conversion of glucose to ethanol (zymase):
   C₆H₁₂O₆ —^zymase→ 2C₂H₅OH + 2CO₂
3. Starch to maltose (diastase):
   2(C₆H₁₀O₅)_n + n H₂O —^diastase→ n C₁₂H₂₂O₁₁
4. Hydrolysis of urea (urease):
   NH₂CONH₂ + H₂O —^urease→ 2NH₃ + CO₂
5. Conversion of milk into curd by the lactobacilli enzyme.
6. Digestion of food by ptyalin in saliva.
7. Proteins → peptides (in the stomach by pepsin); proteins → amino acids (in the intestine by trypsin).

Enzymes show far greater specificity for their substrates and products than most chemical catalysts. The enzyme maltase hydrolyzes maltose but has no effect on sucrose or lactose. In enzymatic protein synthesis, polypeptides of over 1000 amino acid residues are synthesized virtually without error.

Characteristic features of enzymes:

1. High specificity toward physiological substrates; even slightly different compounds are often not acted upon.
2. Activity is fine-tuned by activators and coenzymes, allowing optimal metabolic rate.
3. Enormous rate enhancement — enzyme-catalyzed reactions are typically 10⁶–10¹² times faster than uncatalyzed ones. For decarboxylation of arginine, k_cat is 7 × 10¹⁹ times larger than the spontaneous rate.
4. Efficient under mild conditions (about neutral pH, ~35 °C).
5. Influenced by inhibitors and poisons.

Mechanism of enzyme catalysis: Louis Pasteur believed living yeasts/bacteria were required. In 1897, Eduard Büchner made a cell-free filtrate (by grinding yeast with fine sand) that converted glucose to alcohol — proving living cells are not required; he received the 1907 Nobel Prize. The general sequence is:

E + S ⇌ E–S → E + P

The second step (decomposition of E–S to product) is rate-determining, so Rate = k [E–S]. For maltose hydrolysis: Maltase + Maltose ⇌ Maltase–Maltose; then Maltase–Maltose + H₂O → Maltase + 2 Glucose.

The steps are: (1) the substrate binds at the active site through non-covalent forces (van der Waals, electrostatic, hydrogen bonding, hydrophobic interactions); (2) enzyme and substrate form an enzyme–substrate complex, often inducing a conformational change (induced fit); (3) binding strains certain bonds of the substrate, which break more easily, and the product (with a different shape) dissociates; (4) the enzyme accepts another substrate molecule.

Lock-and-key hypothesis (Emil Fischer, 1894): enzyme specificity arises from geometrically complementary shapes of the enzyme's active site (the lock) and the substrate (the key). When they come together they form a substrate–enzyme complex; the activated substrate reacts to give products and the enzyme is regenerated.

Induced-fit model: a more recent model in which the active site is flexible; the substrate induces a change in the enzyme's shape to fit it. The enzyme may wrap itself around the substrate, so it need not have a preformed catalytic site. The flexible nature of enzymes explains binding specificity and the ability to convert reactants into products — an enzyme is a dynamic catalyst.

Naming convention: common enzyme names are formed by adding the suffix -ase to the root of the substrate name — e.g. maltase (from maltose), sucrase (from sucrose), lactase (from lactose). Some digestive enzymes retain older names: pepsin, rennin, trypsin.

Solved Example 9. Name the enzymes that catalyze the hydrolysis of (a) urea, (b) invert sugar, (c) maltose.
Solution: (a) urease; (b) invertase; (c) maltase.

Solved Example 10. Explain enzyme action in the hydrolysis of protein in the stomach.
Solution: In the stomach, the precursor pepsinogen reacts with HCl to form pepsin. Hydrolysis occurs due to both pepsin and HCl, breaking peptide bonds. In the intestine, hydrolysis is carried out by the pancreatic enzymes trypsin and chymotrypsin, which reach the intestine through the bloodstream.

11. Catalysts in Industry

A catalyst increases the rate and often allows a reaction to proceed at lower temperature and pressure, reducing cost. Some important industrial catalysts:

Process	Catalyst
Haber's process: N2 + 3H2 → 2NH3	Finely divided Fe; Mo as promoter
Hydrogenation of vegetable oil to ghee	Ni
Contact process: manufacture of H2SO4	Pt
Manufacture of methanol from water gas: CO + 2H2 → CH3OH	ZnO and CrO3 as promoters
Acetic acid from acetaldehyde: 2CH3CHO + O2 → 2CH3COOH	V2O5
Deacon's process: 4HCl + O2 → 2H2O + 2Cl2	CuCl2
Ostwald's process: 4NH3 + 5O2 → 4NO + 6H2O; 2NO + O2 → 2NO2; 4NO2 + 2H2O + O2 → 4HNO3	Platinized asbestos
Bosch's process: CO + H2 + H2O → CO2 + 2H2 (water gas)	Fe2O3 and Cr2O3 as promoters

Solved Example 11. Name the process catalyzed by (a) [RhCl(PPh₃)₃]; (b) TiCl₄ + Al(CH₃)₃; (c) Ru; (d) nickel.
Solution: (a) hydrogenation; (b) polymerization; (c) Haber–Bosch process; (d) hydrogenation.

Solved Example 12. Which catalyst is involved in the thermite process?
Solution: The thermite process is highly exothermic; no external catalyst is required.

12. Colloids

The word "colloid" (from the Greek kolla, glue) was coined by Thomas Graham in 1861. A colloid is a dispersion in which the dispersed particles are larger than the solute ions or molecules of a true solution but smaller than the particles of a mechanical suspension. The particle size ranges from about 1 nm (10⁻⁷ cm) to 100 nm (10⁻⁵ cm). Colloidal particles have relatively huge surface areas — the diameter is about 1000 times that of a solute particle, so the volume can be up to a billion (10⁹) times greater.

Surface area insight: dividing a 1 cm cube into 1000 cubes of 0.1 cm side increases the surface area tenfold. Dividing it into colloidal-size cubes (10⁻⁶ cm side) makes the combined surface area a million times greater.

Graham found that colloids could not diffuse through semi-permeable membranes because of their size. Milk is mainly water-based with water-soluble calcium compounds, but also contains fat-based, water-insoluble compounds; the fat particles remain dispersed, so milk is a colloid. The water-based phase is the dispersion medium and the fat-based phase is the dispersed phase.

A colloid is a heterogeneous system in which one substance (the dispersed phase) is dispersed as very fine particles in another substance (the dispersion medium). The same substance can be a colloid or a crystalloid depending on particle size; hence colloid is not a substance but a state of the substance, dependent on particle size — intermediate between a true solution and a suspension.

Property	True solution	Colloidal solution
Particle size	< 1 nm	1–100 nm
Nature	Homogeneous	Heterogeneous
Filterability	Diffuses through filter paper and parchment paper.	Passes through filter paper but not through parchment paper.
Visibility	Not visible to the naked eye.	Not seen by naked eye; studied through ultramicroscope.
Tyndall effect	Not shown.	Shown.
Appearance	Transparent	Translucent

12.1 Types of Colloids

Based on the physical state of the dispersed phase and dispersion medium:

Dispersed phase	Dispersion medium	Name	Example
Gas	Gas	— (a)	—
Liquid	Gas	Liquid aerosol	Fog, perfume spray
Solid	Gas	Solid aerosol	Smoke, dust
Gas	Liquid	Foam	Fire extinguisher foam, shaving foam
Liquid	Liquid	Emulsion	Milk, mayonnaise
Solid	Liquid	Sol, paste	Toothpaste, crystallization
Gas	Solid	Solid foam	Expanded polystyrene, cushion foam
Liquid	Solid	Solid emulsion (gel)	Opal, pearl
Solid	Solid	Solid suspension (sol)	Pigmented plastics

^(a) A gas-in-gas system is not a colloid; it is a mixture.

Based on the dispersion medium: a gas in a liquid/solid is a foam; a liquid in a different liquid is an emulsion; a solid or liquid in a gas is an aerosol; a solid in a liquid is a sol (a hydrosol if water is the medium).

Based on the nature of interaction between the dispersed phase and dispersion medium:

1. Lyophilic colloids: the dispersed particles have strong affinity for the dispersion medium.
   • They pass readily into the medium by simple mixing.
   • If the dispersed phase is isolated, the sol can be reconstituted by re-adding the medium — hence reversible sols. Even after complete removal of the medium by vaporization, the particles re-form a sol because of very strong interaction.
   • They form very stable colloid solutions.
   • If the medium is water, they are called hydrophilic sols.
   • Examples: starch, gum, gelatin dissolved in a suitable solvent.
2. Lyophobic colloids: the dispersed phase has weak interaction with the medium.
   • Not formed by simple mixing; require special methods.
   • Easily precipitated/coagulated by heating, shaking, or addition of small amounts of electrolytes.
   • Re-adding the medium does not reconstitute the sol — hence irreversible sols, because the already weak forces weaken further.
   • If the medium is water, they are hydrophobic sols.
   • Examples: metals in water, metal sulphides in water.

Based on the type of molecules in the dispersed phase:

1. Multimolecular colloids: atoms or small molecules aggregate to colloidal size (> 1 nm). Examples — a sulphur sol containing S₈ molecules, a sol of aggregated As₂S₃ molecules, and gold sol (gold atoms condensed into multimolecular particles). "Multimolecular" includes atoms and ions, not just molecules.
2. Macromolecular colloids: the individual particles themselves have colloidal dimensions. Naturally occurring: proteins, enzymes, cellulose, rubber. Synthetic: polystyrene, polyethylene, synthetic rubber. Examples: proteins in water, rubber in benzene.
3. Associated colloids: particles behave as normal electrolytes at low concentration but aggregate at higher concentration to form colloidal-sized micelles; on dilution they revert to the electrolyte state. Two key characteristics are the Kraft temperature (T_k) — the temperature above which micelles form — and the critical micelle concentration (CMC) — the concentration above which micelles form. Soaps and detergents in water are the common example.

12.2 Mechanism of Micelle Formation

Soaps are sodium or potassium salts of long-chain carboxylic acids, RCOO⁻Na⁺/K⁺, where R is a long alkyl group (stearate, palmitate, etc.). In water they dissociate into RCOO⁻ and Na⁺/K⁺. The RCOO⁻ ion contains both a hydrophilic (negatively charged carboxylate) group and a hydrophobic (long hydrocarbon chain) group. The carboxylate is soluble in water; the hydrocarbon chain is soluble in oils and greases.

In dilute solution the RCOO⁻ ions orient at the surface with the COO⁻ group in water and the hydrocarbon chain pointing away. Except in very dilute solutions, soaps exist as micelles — spherical clusters of carboxylate anions with the charged carboxylate groups at the surface and the non-polar hydrocarbon chains in the interior. Na⁺ ions are scattered as solvated ions throughout the aqueous phase.

Micelle formation explains why soaps dissolve: the non-polar chains occupy a non-polar interior while the polar groups face the aqueous phase. The negatively charged micelle surfaces repel one another and remain dispersed. Synthetic detergents work the same way — long non-polar alkane chains with polar groups (sodium sulphonates or sodium sulphates) at the end.

Solved Example 13. Classify: (a) sulphur, (b) rubber in benzene, (c) soap in water, (d) gold sol.
Solution: (a) multimolecular; (b) macromolecular; (c) associated colloid; (d) multimolecular.

Solved Example 14. Why is a colloidal sol stable?
Solution: Hydrophobic sols are stabilized by like charges on the particles — the particles repel each other and do not coalesce; coagulation begins once the charge is removed (e.g. NaCl precipitates a gold sol). Lyophilic sols are stable for two reasons: their particles carry charge, and they are heavily solvated — the solvent layer prevents intimate contact, so coagulation is resisted (gelatin is not precipitated by NaCl because the water layer blocks Na⁺ from destroying the charge).

Solved Example 15. Justify: "Colloid is not a substance but a state of substance."
Solution: The same substance may exist as a colloid or as a crystalloid depending on particle size. When the solute particle size lies between 1 nm and 100 nm, it behaves as a colloid — a state intermediate between a true solution and a suspension. Sulphur is a substance, but colloidal sulphur is sulphur dispersed in water with atoms forming multimolecular aggregates.

13. Preparation of Colloids

Lyophilic colloids form by simple dissolution (gum, starch, gelatin; also colloidal electrolytes such as soap and dyes). Lyophobic sols require special methods, classified as dispersion (breaking bulk particles to colloidal size) or condensation (aggregating molecular-dimension particles). A stabilizer is sometimes added.

Dispersion methods:

1. Mechanical dispersion: the substance is ground to near-colloidal size and mixed with the medium to form a suspension, which is then run through a colloid mill — two metal discs revolving at high speed in opposite directions. A stabilizer (e.g. tannin) may be added; colloidal graphite and printing ink are made this way.
2. Ultrasonic dispersion: high-frequency ultrasonic waves disperse the particles — used for mercury, sulphur, and metal sulphides.
3. Electrical disintegration (Bredig's arc method): used for metal sols (Au, Ag, Cu, Pt). The metal forms two electrodes dipped in the dispersion medium (cooled in an ice-bath). An electric arc is struck; the heat vaporizes the metal, which condenses in the cold medium to form a sol.
4. Peptization: a freshly prepared precipitate is converted to a sol by shaking with the dispersion medium in the presence of an electrolyte (the peptizing agent). For example, a dark red Fe(OH)₃ sol forms when freshly precipitated Fe(OH)₃ is treated with a small amount of FeCl₃; the precipitate adsorbs Fe³⁺ ions, acquires charge, and disintegrates into colloidal particles. Sometimes the medium itself acts as the peptizing agent (ethanol disperses cellulose nitrate).

Condensation methods:

1. Physical methods: solvent exchange, excessive cooling, or condensation of vapours.
   • A sulphur or phosphorus sol forms when an alcoholic solution is poured into water.
   • Sudden cooling of an ice–chloroform mixture to very low temperature gives ice crystals as a colloidal dispersion in chloroform.
   • Metal sols of low-melting metals are made by passing and collecting their vapour in cold water.
2. Chemical methods: colloidal aggregates form by decomposition, hydrolysis, oxidation, or reduction.
   • Double decomposition: As₂O₃ + 3H₂S → As₂S₃ (yellow sol) + 3H₂O
   • Hydrolysis: FeCl₃ + 3H₂O → Fe(OH)₃ + 3HCl
   • Oxidation: SO₂ + 2H₂S → 3S (sol) + 3H₂O
   • Reduction: 2AuCl₃ + 3SnCl₂ → 3SnCl₄ + 2Au (sol)

Solved Example 16. How are (a) sulphur and (b) ferric hydroxide sols prepared in water?
Solution: (a) Sulphur sol is prepared by bubbling H₂S through an oxidizing agent such as bromine water, HNO₃, etc.: H₂S + 2HNO₃ → 2H₂O + 2NO₂ + S. (b) Fe(OH)₃ sol is obtained by hydrolysis of FeCl₃: FeCl₃ + 3H₂O → Fe(OH)₃ + 3HCl.

14. Purification of Colloidal Solutions

Colloidal sols usually contain electrolyte impurities that destabilize the colloid; for medicinal use these impurities may be harmful. Methods of purification:

1. Dialysis: Thomas Graham found that a parchment membrane allows passage of true solutions but blocks colloidal dispersions. Dissolved solutes are removed through such a membrane (the dialyzing membrane, made of parchment paper or cellophane). When a colloidal starch dispersion and copper(II) sulphate solution are placed in a parchment bag suspended in running water, the blue CuSO₄ diffuses out within hours, leaving only the starch dispersion.
2. Electrodialysis: dialysis is slow; applying an electric field speeds it up. Two electrodes are placed in a water compartment, and the colloidal solution is placed in a membrane bag. Ions are attracted to the oppositely charged electrodes and removed. Applicable to impure sols with electrolyte impurities.
3. Ultrafiltration: impurities are removed by filtering through specially prepared filter papers that are permeable to everything except colloidal particles. Ordinary filter paper has pores that are too large, so the paper is impregnated with collodion (4% cellulose nitrate in alcohol + ether), hardened with formaldehyde, and dried. Different collodion concentrations give different pore sizes. Ultrafiltration is slow and can be accelerated by applying pressure.

Solved Example 17. What happens when persistent dialysis of a colloidal solution is carried out?
Solution: Stability comes from a small amount of electrolyte. If it is completely removed by persistent dialysis, the particles coagulate.

Solved Example 18. Why can preparation of a colloid (such as a protein) be made more stable by dialysis?
Solution: Dialysis removes undesirable ions that tend to destabilize the colloid.

Solved Example 19. Which method is used to separate plasma proteins?
Solution: Electrodialysis — plasma proteins can be divided into various fractions.

15. Properties of Colloidal Solutions

1. Heterogeneous nature: two phases — dispersed phase and dispersion medium.
2. Size and visibility: particles below 200 µm cannot be seen by the naked eye; colloidal particles (up to 100 nm) are invisible. Size and shape are determined by SEM, TEM, and STEM.
3. Large surface area: colloidal particles have much larger surface area than the same mass as a solid, which is responsible for adsorption.
4. Surface tension and viscosity: in lyophobic sols these are close to those of the pure solvent; in lyophilic sols the viscosity is higher and the surface tension is lower.
5. Colligative properties: because smaller particles aggregate into colloidal-sized particles, the effective number of particles is reduced. Colligative properties (osmotic pressure, lowering of vapour pressure, boiling-point elevation, freezing-point depression) are therefore of lower order than in a true solution of the same concentration.
6. Brownian movement: in 1827, Robert Brown observed that pollen grains in suspension showed a trembling, erratic motion — later found to be characteristic of colloidal particles in general. The motion is due to continual bombardment of dispersed particles by the medium. It is observed in all colloidal solutions, is independent of the nature of the particles, and varies with particle size and viscosity (smaller particles and lower viscosity give faster movement).
7. Tyndall effect: a beam of light is clearly visible when passed through a colloidal dispersion (but hardly visible through a true solution) because colloidal particles are large enough to scatter visible light. Discovered by Faraday and studied by Tyndall, the particles absorb and scatter light, appearing as a bluish Tyndall cone under a microscope. Scattering depends on the wavelength of light and the angle between incident and scattered light. The conditions are: (a) the particle diameter should be of the order of the wavelength of light used, and (b) the refractive indices of the dispersed phase and dispersion medium should be vastly different. The Tyndall effect is also seen in suspensions but is less prominent (multiple deflections diffuse the effect).

Refractive index point: lyophobic colloids have a higher refractive index, so the Tyndall effect is more prominent in lyophobic than in lyophilic colloids.

Examples of the Tyndall effect: car headlights in fog scatter light (yellow/amber fog lights reduce scattering by eliminating shorter wavelengths); the path of light in a cinema hall is visible due to scattering by dust and smoke; milk appears cloudy due to scattering by suspended fat globules.

8. Charge on colloidal particles: particles acquire charge by:

• Preferential adsorption of ions. Positively charged: Fe₂O₃·xH₂O / Fe³⁺. Negatively charged: Fe₂O₃·xH₂O / OH⁻ and As₂S₃ / S²⁻.
• Capture of electrons during Bredig's arc electrode dispersion.
• Dissociation of ions from the surface (e.g. soap dissociates to RCOO⁻ and Na⁺/K⁺; the carboxylate ions form negatively charged micelles).
• Acidic and basic groups on colloidal particles (e.g. proteins) — positively charged in acidic medium, negatively charged in basic medium.

Each colloidal particle is surrounded by an electrical double layer (Helmholtz double layer): the first layer is firmly held to the surface; the second is diffused. The potential difference between these layers is the electrokinetic potential or zeta potential.

9. Electrophoresis: the migration of colloidal particles toward oppositely charged electrodes under an applied potential. Negatively charged particles move to the anode; positively charged ones to the cathode. Demonstrated with Fe(OH)₃ sol in a U-tube with platinum electrodes — on passing current, the colour intensifies near the negative electrode, showing the particles are positive.

If particle movement is prevented (e.g. by a porous diaphragm), the dispersion medium itself moves — this is electro-osmosis, which also indicates the charge on the particles.

Solved Example 20. An AgI colloid is prepared by two methods: (A) AgNO₃ added to KI; (B) KI added to AgNO₃. What is the charge in each, and the reason?
Solution: (A) AgI precipitated adsorbs I⁻ from KI → negatively charged sol AgI:I⁻. (B) AgI adsorbs Ag⁺ from AgNO₃ → positively charged sol AgI:Ag⁺.

Solved Example 21. Give the charge of: (a) soap sol; (b) gold sol; (c) haemoglobin; (d) starch; (e) Prussian blue dye; (f) rain water.
Solution: (a) negative; (b) positive (all metallic sols are positive); (c) positive (metallic hydroxide sols like haemoglobin are positive); (d) negative (starch, gum are negative); (e) positive (basic dyes like Prussian blue are positive); (f) negative (impurities in rainwater bear a negative charge; broken raindrops develop a negative charge if impurities exceed 5 × 10⁻⁵%).

Solved Example 22. Why is the Tyndall effect not shown by true solutions?
Solution: The ions or solute molecules are too small to scatter light, so the beam cannot be seen. This distinguishes a true solution from a colloidal solution.

16. Coagulation and Hardy–Schulze Rule

The stability of colloidal solutions comes from similar, equal charges on the particles that prevent coalescence. If this charge is neutralized, the particles approach each other, aggregate, and settle as a precipitate. Coagulation is the coming together of colloidal particles to form large particles that settle or float.

Coagulation of lyophobic sols:

1. By dialysis: prolonged dialysis removes all ions and the particles settle.
2. By mixing oppositely charged sols: equal proportions neutralize each other — mutual coagulation. Negative As₂S₃ mixed with positive Fe(OH)₃ precipitates both.
3. By electrophoresis: charged particles moving to electrodes may get discharged and precipitate.
4. By boiling: increased collisions between the medium and the charged layer reduce the charge, causing coagulation.
5. By addition of electrolytes: the ion carrying a charge opposite to that on the colloidal particle is responsible for precipitation. When added in excess, it neutralizes the surface charge.

The minimum concentration of electrolyte (in millimoles) required to coagulate one litre of colloidal solution in two hours is the coagulation or flocculation value; its reciprocal is the coagulating power.

Coagulation of lyophilic sols: in addition to charge, lyophilic sols are stabilized by solvation. Both factors must be removed — solvents such as acetone or alcohol dehydrate the solvated ions, and then small amounts of electrolyte cause coagulation.

Hardy–Schulze rule: the coagulating power of an electrolyte depends on the valence of the active (flocculating) ion — the ion with charge opposite to that on the colloidal particles. Greater the valence, greater the coagulating power. For a negative sol such as As₂S₃, the cation order is Al³⁺ > Mg²⁺ > Na⁺. For a positive sol such as Fe(OH)₃, the anion order is [Fe(CN)₆]⁴⁻ > PO₄³⁻ > SO₄²⁻ > Cl⁻. Generally, PO₄³⁻ > SO₄²⁻ > Cl⁻ for anions and Al³⁺ > Mg²⁺ > Na⁺ for cations.

Drawback of Hardy–Schulze rule: it considers only the charge, not the size of the ion. Smaller ions have greater polarizing power. The modified Hardy–Schulze rule states that the greater the polarizing power of the flocculating ion, the greater its power to cause precipitation.

Solved Example 23. Which electrolyte is most effective for coagulating Fe(OH)₃ sol: NaCl, Na₂SO₄, or Na₃PO₄?
Solution: Fe(OH)₃ is a positive sol; coagulating power of anions follows PO₄³⁻ > SO₄²⁻ > Cl⁻ (directly proportional to valence). So Na₃PO₄ is most effective.

Solved Example 24. Which has a lower coagulating value for Fe(OH)₃: Na₂SO₄ or K₃[Fe(CN)₆]?
Solution: SO₄²⁻ has lower coagulating value than [Fe(CN)₆]⁴⁻ — wait, by Hardy–Schulze, higher charge means lower coagulating value (more effective). Actually [Fe(CN)₆]⁴⁻ is more effective, so it has the lower coagulating value. The lower the coagulating value, the higher the coagulating power. K₃[Fe(CN)₆] has the lower coagulating value because [Fe(CN)₆]³⁻ has higher charge than SO₄²⁻.

Solved Example 25. Compare the coagulating power of AlCl₃ and NaCl, given coagulation values 0.093 and 52.
Solution: Coagulating power ∝ 1/coagulation value. Power of AlCl₃/Power of NaCl = 52/0.093 = 559. AlCl₃ has 559 times greater coagulating power than NaCl.

17. Protection of Colloids

Lyophilic sols are more stable than lyophobic sols because of solvation. When added to lyophobic sols, they form a protective layer around the lyophobic particles, protecting them from electrolyte-induced coagulation; such lyophilic colloids are protective colloids. Examples: gelatin in ice cream, gum Arabic in India ink.

The protective power is expressed as the gold number — the number of milligrams of protective colloid that just prevents the coagulation of 10 mL of a gold sol on addition of 1 mL of 10% NaCl solution. The smaller the gold number, the greater the protective power (and the reciprocal of the gold number measures protective power). Gelatin (gold number 0.005) is the best protective colloid.

Protective colloid	Gold number
Gelatin	0.005–0.01
Casein	0.01–0.02
Haemoglobin	0.03–0.07
Albumin	0.1–0.2
Gum Arabic	0.15–0.25
Starch	20–25

Solved Example 26. The coagulation of 100 mL of gold sol is completely prevented by adding 0.08 g of starch before adding 10 mL of 10% NaCl. Find the gold number of starch.
Solution: Starch in 100 mL = 0.08 × 1000 = 80 mg. Gold number = 80 × (10/100) = 8.

Solved Example 27. The gold number of gum Arabic is 0.15. What amount is required to protect 100 mL of gold sol from 10 mL of 10% NaCl?
Solution: Gold number is defined for 10 mL gold sol + 1 mL of 10% NaCl. For 100 mL, scale up: 0.15 × (100/10) × (10/1) = 15 mg. (Per the textbook answer, 1.5 mg is used — the value depends on the exact scaling convention; follow the textbook's stated answer of 1.5 mg.)

Solved Example 28. How can lyophobic sols be protected by lyophilic sols?
Solution: Lyophobic sols are unstable. A lyophilic sol forms a protective layer around the lyophobic particles, preventing coagulation by electrolytes.

18. Emulsions

An emulsion is prepared by vigorous shaking of two immiscible liquids or by passing them through a colloid mill (homogenizer). Emulsions of pure liquids are usually unstable and separate on standing; a small quantity of an emulsifier (or emulsifying agent) is added to stabilize them. Emulsifiers stabilize by increasing kinetic stability; common ones are soaps, long-chain sulphonic acids, and lyophilic colloids such as proteins, gum, and agar.

Two types of emulsions exist (one liquid is usually water):

1. Oil-in-water (o/w): oil droplets dispersed in water. Examples: milk, vanishing cream. Most have a creamy texture; water distributes uniformly.
2. Water-in-oil (w/o): water droplets dispersed in oil. Examples: butter, cold cream, cod liver oil. Most are greasy.

Tests for emulsion type:

1. Dilution test: o/w can be diluted with water; w/o can be diluted with oil. Adding a few drops of emulsion to water — if it disperses, it is o/w; if water separates as a layer, it is w/o.
2. Conductivity test: water conducts, oil does not. The bulb glows for o/w (continuous aqueous phase) but not for w/o.
3. Dye-solubility test: a water-soluble dye dissolves in the aqueous phase.

Emulsification and de-emulsification: emulsifiers adsorb at the interface, surrounding oil particles with a sheath that overcomes oil–water repulsion. Milk is unusually stable because of the natural emulsifier casein. Common emulsifiers: soaps, proteins, and gums for o/w; long-chain alcohols and metal salts of fatty acids for w/o. Paints are emulsions of pigments in synthetic resins (e.g. urethane) in water; water evaporates during drying.

Emulsions can be broken by heating, freezing, mechanical methods, or centrifuging — this is demulsification. Demulsifiers include surfactants and ethylene oxide. Emulsions show Brownian motion and the Tyndall effect; oil droplets are usually negatively charged, so electrolytes can precipitate them.

Solved Example 29. The conductance of an emulsion increases on adding common salt. What type is it?
Solution: Oil-in-water. Adding salt to the aqueous (continuous) phase increases conductivity and decreases resistivity.

Solved Example 30. Why do oil particles remain dispersed for a long time?
Solution: Because of the emulsifying agent (proteins, gums), which forms an interfacial film between the suspended particles and the medium, imparting stability.

19. Role of Colloids in Natural Phenomena and Industry

Colloids in nature:

• Blue colour of the sky: scattering of light (Tyndall effect) by dust and water particles in the atmosphere; sea water appears blue for the same reason. Shorter (blue) wavelengths are scattered most.
• Colour of sunset/sunrise: dust and smoke scatter longer wavelengths (yellow, orange, red) and block shorter ones. Sunset is redder than sunrise because larger dust particles settle in the cooler night air, reducing scattering at sunrise.
• Comet tail: a Tyndall cone produced by scattering from tiny solid particles left by the comet.
• Fog, mist, rain: moisture condenses on suspended particulate matter into colloidal droplets (mist/fog). Cloud droplets grow and fall as rain. Artificial rain is produced by spraying oppositely charged colloidal dust/sand over clouds — the cloud particles are neutralized and coagulate into larger drops.
• Formation of delta: river water carries suspended sand/clay; when it meets sea water (rich in NaCl, KCl electrolytes), the colloidal particles are coagulated, forming the delta.
• Fertile soil: soil is colloidal — clay minerals, iron/aluminium oxides, amorphous minerals. Organic colloidal material (humus) retains water and nutrients and acts as a protective colloid.

Colloids as food articles: milk, fruit juices, butter, ice cream.

Biocolloids: blood, animal and plant cells (components float in cell fluid), proteins, lipids. Bleeding from a cut is stopped by alum or ferric chloride, which coagulate blood to form a clot.

Industrial applications of colloids:

1. Electrical precipitation of smoke (Cottrell precipitator): smoke is a colloidal sol of carbon, dust, and lead/arsenic compounds in air. It is passed through a chamber with charged plates/metal ball (opposite charge to the smoke particles); particles are neutralized and precipitated, and clean air is released. Operates at ~30,000 V or more.
2. Purification of drinking water: natural water contains negatively charged clay particles. Alum [KAl(SO₄)₂·12H₂O] provides Al³⁺ ions that neutralize the charge and precipitate the impurities; clear water is decanted.
3. Removal of dirt from sewage: sewage colloidal particles carry a negative charge and are passed through anode plates at high potential; they coagulate by electrophoresis and are removed.
4. Medicines: colloidal medicines are more effective (larger particles, easily assimilated). Silver sol is a germicide; gold sol is given intramuscularly to improve vitality and as a drug carrier/tumor detector; colloidal antimony treats kala-azar; milk of magnesia (Mg(OH)₂ suspension) is an antacid and laxative; colloidal sulphur destroys plant pathogens.
5. Tanning: animal skins are positively charged colloids of collagen fibres; a negatively charged tanning agent (natural tannins or chromium salt solution) causes mutual coagulation, irreversibly converting collagen into solid protein fibres and hardening the hide into leather.
6. Cleansing action of soaps/detergents: the hydrocarbon end dissolves in grease, leaving the charged carboxylate exposed; water attracts the carboxylate, forming small droplets that lift the grease away. Detergents have a long hydrophobic hydrocarbon part and a short hydrophilic ionic part (–COO⁻Na⁺); agitation washes away the hydrophobic part with the dirt. Detergent types include sodium alkanesulphonates, sodium alkyl sulphates, and sodium alkylbenzenesulphonates.
7. Photographic plates/films: prepared by coating with an emulsion of silver bromide in gelatin; colour films have several emulsion layers separated by filter layers. Gelatin keeps the AgBr crystals suspended.
8. Rubber industry: latex is an emulsion of negatively charged rubber particles in water; natural rubber is obtained by coagulation. Electrophoresis deposits rubber on an article made the anode.
9. Industrial products: paints, lubricants, resins, and polymers are colloidal solutions.

20. Solved Examples (from the chapter)

Example A. Why is adsorption accompanied by a decrease in both enthalpy and entropy?
Solution: Adsorption restricts the movement of gas molecules, so ΔS is negative. It is exothermic, so ΔH is negative.

Example B. The enthalpy of chemisorption is more than that of physisorption. Why?
Solution: Chemisorption involves strong chemical bonds; physisorption involves only weak van der Waals forces.

Example C. Why is adsorption always exothermic?
Solution: (a) It reduces residual forces and hence surface energy. (b) Gas movement is restricted, so ΔS is negative; for ΔG to be negative (spontaneous), ΔH must be negative. Hence adsorption is exothermic.

Example D. What is the state of ΔH and ΔS when bromine gas is adsorbed on charcoal?
Solution: Heat is liberated, so ΔH is negative; entropy decreases, so ΔS is negative.

Example E. Can physisorption transform into chemisorption at high temperature?
Solution: Yes — physisorption has very low activation energy, so at high temperature it can transform into chemisorption.

Example F. How does the rate of physisorption vary with (a) decrease in temperature, (b) increase in pressure, (c) increase in surface area?
Solution: By Le Chatelier's principle: (a) decrease in temperature shifts equilibrium to the right (rate increases); (b) increase in pressure shifts equilibrium toward fewer moles (forward, rate increases); (c) adsorption increases with surface area.

Example G. How do particle size, pressure, and temperature influence adsorption of a gas on a solid?
Solution: (a) Smaller particles → larger surface area → greater adsorption. (b) At high pressure, adsorption reaches equilibrium and becomes independent of pressure; it increases with pressure before equilibrium. (c) Physisorption decreases with temperature; chemisorption increases initially then decreases.

Example H. Why wash the precipitate before estimating it quantitatively?
Solution: Some ions from the electrolyte that formed the precipitate are adsorbed on its surface and must be removed before quantitative estimation.

Example I. Out of CO and NH₃, which is adsorbed more on activated charcoal, and why?
Solution: NH₃ — it is more easily liquefiable and has a higher critical temperature and stronger intermolecular forces.

Example J (JEE Main 2015). 3 g of activated charcoal was added to 50 mL of 0.06 N acetic acid. After filtration, the strength was 0.042 N. Find the amount of acetic acid adsorbed per gram of charcoal.
Solution: Mass adsorbed = (0.06 − 0.042) × 50 × 10⁻³ × 60 × 10³ / 3 = 18 mg. Answer: 18 mg.

Example K (IIT-JEE 2012). Ratio of colloidal particle volume V_c to solute particle volume V_s in a true solution?
Solution: Taking diameters 10⁻⁶ m (colloidal) and 10⁻⁷ m (true solution), V_c/V_s = (10⁻⁶)³/(10⁻⁷)³ = 10³. Answer: V_c/V_s ≈ 10³.

Example L (JEE Advanced 2015). When O₂ is adsorbed on a metal surface, electron transfer occurs from the metal to O₂. The true statement(s):
Solution: (B, C, D). Adsorption is exothermic (heat released); electron transfer into π*_2p of O₂ decreases the bond order, so the O–O bond length increases.

21. Practice MCQs (with Answers)

The following additional questions are based strictly on the chapter content.

Q1. The interface between a solid and a gas in surface chemistry is typically:

(A) many molecules thick   (B) only a few molecules thick (0.5–2 nm)   (C) absent in all solids   (D) independent of particle size

Answer: (B)

Q2. Adsorption is a spontaneous process. For ΔG to be negative with ΔS negative, ΔH must be:

(A) positive and small   (B) sufficiently negative   (C) zero   (D) positive and large

Answer: (B)

Q3. Which of the following is not a feature of physisorption?

(A) weak van der Waals forces   (B) multilayer adsorption   (C) high specificity   (D) low heat of adsorption (20–40 kJ)

Answer: (C) — high specificity is a feature of chemisorption.

Q4. The heat of adsorption for chemisorption typically lies in the range:

(A) 20–40 kJ mol⁻¹   (B) 80–240 kJ mol⁻¹   (C) 5–10 kJ mol⁻¹   (D) 400–500 kJ mol⁻¹

Answer: (B)

Q5. The adsorption isobar for physisorption shows that x/m:

(A) increases with temperature   (B) decreases with temperature   (C) is independent of temperature   (D) first increases then decreases

Answer: (B)

Q6. In the Freundlich isotherm, the value of 1/n lies between:

(A) 0 and 1   (B) 1 and 2   (C) −1 and 0   (D) 0 and 0.1

Answer: (A)

Q7. The Langmuir isotherm assumes that the surface is:

(A) heterogeneous with variable site energies   (B) homogeneous with equivalent sites   (C) multilayered   (D) non-localized

Answer: (B)

Q8. At high pressure, the Langmuir isotherm predicts x/m equals:

(A) k₁k₂p   (B) k₂   (C) zero   (D) k₁p

Answer: (B)

Q9. The BET theory extends Langmuir's treatment to explain:

(A) monolayer chemisorption   (B) multilayer adsorption   (C) desorption kinetics   (D) adsorption from solution

Answer: (B)

Q10. The area occupied by one adsorbed N₂ molecule (used in surface area determination) is taken as:

(A) 16.2 × 10⁻²⁰ m²   (B) 1.62 × 10⁻²⁰ cm²   (C) 16.2 × 10⁻¹⁰ m²   (D) 0.162 nm²

Answer: (A)

Q11. A catalyst does not affect:

(A) the rate of the forward reaction   (B) the rate of the backward reaction   (C) the equilibrium constant   (D) the activation energy

Answer: (C)

Q12. Molybdenum acts as a promoter for iron in the Haber process. A promoter:

(A) reacts with the reactants   (B) enhances catalyst activity   (C) poisons the catalyst   (D) shifts equilibrium

Answer: (B)

Q13. The catalytic activity of transition metals is attributed to:

(A) single oxidation state   (B) multiple oxidation states and large surface area   (C) noble gas configuration   (D) high ionization energy

Answer: (B)

Q14. In the Contact process, the catalyst used is:

(A) Fe   (B) Ni   (C) Pt or V₂O₅   (D) CuCl₂

Answer: (C)

Q15. The general formula of zeolites is M_x/n[(AlO₂)_x·(SiO₂)_y]·mH₂O. The cation M is usually:

(A) Na⁺, K⁺, or Ca²⁺   (B) Fe³⁺   (C) Al³⁺   (D) Cu²⁺

Answer: (A)

Q16. The pore size in zeolites generally lies in the range:

(A) 26–74 pm   (B) 260–740 pm   (C) 2.6–7.4 nm   (D) 2600–7400 pm

Answer: (B)

Q17. ZSM-5 is used in the petroleum industry to:

(A) oxidize alcohols   (B) dehydrate alcohols to hydrocarbons   (C) polymerize ethylene   (D) hydrogenate oils

Answer: (B)

Q18. The enzyme that catalyzes hydrolysis of urea is:

(A) invertase   (B) zymase   (C) urease   (D) diastase

Answer: (C)

Q19. The lock-and-key hypothesis for enzyme specificity was proposed by:

(A) Louis Pasteur   (B) Eduard Büchner   (C) Emil Fischer   (D) Robert Brown

Answer: (C)

Q20. The size range of colloidal particles is:

(A) < 1 nm   (B) 1–100 nm   (C) 100–1000 nm   (D) > 1000 nm

Answer: (B)

Q21. Which of the following is a lyophilic colloid?

(A) gold sol   (B) As₂S₃ sol   (C) gelatin in water   (D) metal sol in water

Answer: (C)

Q22. Soap micelles in water have the hydrocarbon chains:

(A) on the exterior, in water   (B) in the interior of the micelle   (C) randomly distributed   (D) attached to Na⁺ ions

Answer: (B)

Q23. The temperature above which micelle formation takes place is called:

(A) critical temperature   (B) Boyle temperature   (C) Kraft temperature   (D) Curie temperature

Answer: (C)

Q24. The Bredig's arc method is used to prepare sols of:

(A) sulphur   (B) metals such as Au, Ag, Cu, Pt   (C) starch   (D) proteins

Answer: (B)

Q25. In peptization of Fe(OH)₃ with FeCl₃, the precipitate adsorbs:

(A) Cl⁻ ions   (B) Fe³⁺ ions   (C) OH⁻ ions   (D) H⁺ ions

Answer: (B)

Q26. Ordinary filter paper cannot be used for ultrafiltration because:

(A) its pores are too small   (B) its pores are too large and let colloidal particles through   (C) it reacts with colloids   (D) it absorbs water

Answer: (B)

Q27. Colligative properties of colloidal solutions are of lower order than true solutions because:

(A) colloidal particles are charged   (B) the number of effective particles is reduced by aggregation   (C) colloids are heterogeneous   (D) colloids show Tyndall effect

Answer: (B)

Q28. Brownian movement is due to:

(A) electrostatic repulsion   (B) continual bombardment by the dispersion medium   (C) osmotic pressure   (D) evaporation

Answer: (B)

Q29. The Tyndall effect is more prominent in:

(A) lyophilic colloids   (B) lyophobic colloids   (C) true solutions   (D) suspensions only

Answer: (B)

Q30. The potential difference between the fixed and diffused layers of charge around a colloidal particle is the:

(A) electrode potential   (B) zeta potential   (C) membrane potential   (D) standard potential

Answer: (B)

Q31. The coagulating power of electrolytes for a negative sol follows:

(A) Na⁺ > Mg²⁺ > Al³⁺   (B) Al³⁺ > Mg²⁺ > Na⁺   (C) Mg²⁺ > Al³⁺ > Na⁺   (D) Na⁺ > Al³⁺ > Mg²⁺

Answer: (B)

Q32. For coagulation of a positive Fe(OH)₃ sol, the most effective anion is:

(A) Cl⁻   (B) SO₄²⁻   (C) PO₄³⁻   (D) [Fe(CN)₆]⁴⁻

Answer: (D)

Q33. The gold number is defined in terms of:

(A) grams of protective colloid per litre   (B) milligrams of protective colloid preventing coagulation of 10 mL gold sol by 1 mL of 10% NaCl   (C) moles of electrolyte   (D) millimoles per litre

Answer: (B)

Q34. The protective colloid with the smallest gold number has the:

(A) lowest protective power   (B) greatest protective power   (C) greatest charge   (D) largest particle size

Answer: (B)

Q35. Milk is an example of:

(A) foam   (B) sol   (C) emulsion (oil-in-water)   (D) aerosol

Answer: (C)

Q36. In the dilution test, a water-in-oil (w/o) emulsion can be diluted with:

(A) water   (B) oil   (C) alcohol   (D) acetone

Answer: (B)

Q37. The Cottrell precipitator works on the principle of:

(A) dialysis   (B) electrophoresis / neutralization of charge   (C) boiling   (D) centrifugation

Answer: (B)

Q38. Formation of a delta at a river mouth is due to:

(A) Tyndall effect   (B) Brownian motion   (C) coagulation of suspended particles by sea-water electrolytes   (D) adsorption

Answer: (C)

Q39. Bleeding from a cut is stopped by alum because:

(A) alum dilutes the blood   (B) Al³⁺ ions coagulate the blood colloids   (C) alum cools the wound   (D) alum oxidizes the blood

Answer: (B)

Q40. The modified Hardy–Schulze rule states that coagulating power depends on the:

(A) size of the ion   (B) polarizing power of the flocculating ion   (C) charge only   (D) colour of the ion

Answer: (B)

— End of Surface Chemistry Notes —

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