Chemistry · Volume 2 · Chapter 10

Samacheer Class 12 Chemistry - Surface Chemistry

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Q.1 For Freundlich isotherm a graph of log(x/m) is plotted against log p. The slope of the line and its y-axis intercept respectively corresponds to:
Answer: c

Freundlich isotherm: x/m = k p^{1/n}. Taking logarithm: log(x/m) = log k + (1/n) log p. Thus slope = 1/n and intercept = log k.

Q.2 Which of the following is incorrect for physisorption?
Answer: b

Physisorption is generally reversible, has low heat of adsorption and increases with surface area. Being exothermic, physisorption decreases with increasing temperature, so option (b) is incorrect.

Q.3 Which one of the following characteristics are associated with adsorption?

Answer: (d) Adsorption is spontaneous (∆G < 0) and usually exothermic (∆H < 0). Because gas or solute molecules lose freedom when adsorbed on a surface, entropy decreases (∆S < 0). Hence ∆G, ∆H and ∆S are all negative.

Q.4 Fog is colloidal solution of a) solid in gas b) gas in gas c) liquid in gas d) gas in liquid
Answer: c

Fog consists of tiny liquid droplets dispersed in a gas (air), so it is a liquid-in-gas colloid.

Q.5 Assertion: Coagulation power of Al3+ is more than Na+. Reason: Greater the valency of the flocculating ion added, greater is its power to cause precipitation. Choose the correct option:

Answer: (a) Both assertion and reason are true and the reason correctly explains the assertion. By the Schulze–Hardy rule, coagulating power increases rapidly with the valency of the oppositely charged ion; Al3+ (trivalent) is much more effective than Na+ (monovalent).

Q.6 Statement: To stop bleeding from an injury, ferric chloride can be applied. Which comment about the statement is justified?

Answer: (b) True. Blood is a negatively charged sol; Fe3+ ions from ferric chloride neutralize and coagulate the colloidal particles, helping to stop bleeding.

Q.7 Hair cream is a) gel b) emulsion c) solid sol d) sol.
Answer: b

Hair cream is typically a liquid–liquid dispersion (oil droplets in water or vice versa), i.e., an emulsion.

Q.8 Which one of the following is correctly matched?
Answer: b

Butter is a semi-solid system (liquid dispersed in a solid-like matrix) and is commonly classified as a gel; other matches are incorrect.

Q.9 The most effective electrolyte for the coagulation of As2S3 sol is: a) NaCl b) Ba(NO3)2 c) K[Fe(CN)] d) Al2(SO4)3

Answer: d. According to the Schulze–Hardy rule, higher-valency counter-ions are more effective. Al3+ (from Al2(SO4)3) is trivalent and most effective for coagulation of As2S3 sol.

Q.10 Which one of the following is not a surfactant?

Answer: (b) CH3(CH2)15NH2. Typical surfactants have a long hydrophobic tail and a strongly polar/ionic hydrophilic head. (a) is a quaternary ammonium salt (cationic surfactant), (c) is an alkyl sulfate (anionic), (d) is a long‑chain carboxylate (soap). The primary amine in (b) is not a typical ionic surfactant in neutral media.

Q.11 The phenomenon observed when a beam of light is passed through a colloidal solution is: a) Cataphoresis b) Electrophoresis c) Coagulation d) Tyndall effect

Answer: d. Scattering of light by colloidal particles when a beam passes through the medium is the Tyndall effect.

Q.12 In an electrical field, the particles of a colloidal system move towards cathode. The coagulation of the same sol is studied using K2SO4 (i), Na3PO4 (ii), K4[Fe(CN)6] (iii) and NaCl (iv). Their coagulating power should be:
Answer: b

Particles moving to cathode are positively charged; coagulation depends on valency of opposite (anionic) ions. Charges: [Fe(CN)6]4- (iv charge 4) > PO4^3- > SO4^2- > Cl^- so order: III > II > I > IV.

Q.13 Collodion is a 4% solution of which compound in alcohol–ether mixture?
Answer: d

Collodion is a 4% solution of nitrocellulose (pyroxylin) in an alcohol–ether mixture.

Q.14 Which one of the following is an example of homogeneous catalysis?

Answer: (d) Hydrolysis of sucrose in the presence of dilute HCl. Both reactant and catalyst (HCl, in aqueous solution) are in the same phase (homogeneous catalysis).

Q.15 Match: A) V2O5 — ?, B) Ziegler–Natta — ?, C) Peroxide — ?, D) Finely divided Fe — ? with (i) HDPE (ii) PAN (iii) NH3 (iv) H2SO4. Choose the correct sequence A B C D.
Answer: a

V2O5 is contact process catalyst → H2SO4 (iv); Ziegler–Natta → HDPE (i); Peroxide initiator → PAN (ii); Finely divided Fe → Haber catalyst for NH3 (iii). Sequence (iv)(i)(ii)(iii).

Q.16 Coagulation values (mmol L−1) for coagulation of As2S3 are: (I) NaCl = 52, (II) BaCl2 = 0.69, (III) MgSO4 = 0.22. The correct order of their coagulating power is:
Answer: a

Lower coagulation value means higher coagulating power. Values: III (0.22) < II (0.69) < I (52), so coagulating power: III > II > I.

Q.17 Adsorption of a gas on solid metal surface is spontaneous and exothermic, then a) ∆H increases b) ∆S increases c) ∆G increases d) ∆S decreases

Answer: d. Exothermic adsorption ⇒ ∆H < 0. Spontaneous ⇒ ∆G < 0. Adsorption localises gas molecules on the surface, reducing disorder ⇒ ∆S < 0 (entropy decreases).

Q.18 If x is amount of adsorbate and m is amount of adsorbent, which relation is not related to adsorption process?
Answer: c

Adsorption relations describe x/m as functions of pressure and temperature (options a, b, d). Relation (c) giving P as a function of T at constant m and x is not an adsorption isotherm/therm relation and is unrelated.

Q.19 On which of the following properties does the coagulating power of an ion depend?

Answer: (a) Both magnitude and sign of the charge on the ion. Coagulating power depends on valency (magnitude) and must be of opposite sign to the colloidal particle’s charge.

Q.20 Match: A) Pure nitrogen, B) Haber process, C) Contact process, D) Deacon's process with (i) Chlorine, (ii) Sulphuric acid, (iii) Ammonia, (iv) sodium azide (or) barium azide. Which is correct A B C D?
Answer: d

Pure nitrogen can be obtained by decomposition of sodium/barium azide (iv); Haber process gives ammonia (iii); Contact process yields sulphuric acid (ii); Deacon's process gives chlorine (i). Sequence: (iv),(iii),(ii),(i).

Short answer QuestionsShort answer Questions24 questions
Q.1Give two important characteristics of physisorptionv
Solution

Physisorption arises from weak physical forces (dispersion/van der Waals). Because these forces are weak (ΔHads ≈ 5–40 kJ·mol⁻¹) adsorption is generally reversible and increases at lower temperatures; also physisorption can lead to multilayer formation since adsorbate–adsorbate interactions are comparable to adsorbate–surface interactions.

Answer:

(i) It involves weak van der Waals forces and is usually reversible. (ii) Heat of adsorption is low and multilayer adsorption is possible.

Q.2Differentiate physisorption and chemisorptionv
Solution

Physisorption: physical adsorption by van der Waals (ΔH ≈ 5–40 kJ·mol⁻¹), reversible, multilayer possible, decreases with temperature. Chemisorption: involves formation of chemical (covalent/ionic) bonds between adsorbate and surface (ΔH ≈ 40–400 kJ·mol⁻¹), usually monolayer, may be irreversible, often needs activation energy and is specific to surface sites.

Answer:

Key differences: physisorption — weak van der Waals forces, low ΔH, reversible, multilayer, favoured at low T. Chemisorption — chemical bond formation, high ΔH, often irreversible, monolayer, may require activation energy and favoured at moderate T.

Q.3In case of chemisorption, why adsorption first increases and then decreases with temperature?v
Solution

Chemisorption often requires activation energy to form chemical bonds with the surface, so a moderate rise in temperature increases the rate and extent of chemisorption. At still higher temperatures thermal energy favours desorption and equilibrium shifts toward the gas phase, so net adsorption decreases.

Answer:

At low T chemisorption is kinetically hindered (activation barrier), so increasing T raises adsorption; at higher T desorption dominates and adsorption falls.

Q.4Which will be adsorbed more readily on the surface of charcoal and why? NH3 or O2 ?v
Solution

Adsorption tendency correlates with critical temperature and interactions with the surface. Ammonia has stronger interactions (polarity, hydrogen bonding/acid–base interaction with surface sites) and a much higher critical temperature than O2, so activated carbon adsorbs NH3 more strongly than O2.

Answer:

NH3 will be adsorbed more readily.

Q.5Heat of adsorption is greater for chemisorptions than physisorption. Why?v
Solution

Chemisorption produces new chemical bonds between adsorbate and surface (energies comparable to covalent/ionic bonds), giving large heats of adsorption (tens to hundreds kJ·mol⁻¹). Physisorption is due to weak dispersion forces (few kJ·mol⁻¹), so its heat of adsorption is much smaller.

Answer:

Because chemisorption involves formation of chemical bonds (high bond energies) while physisorption involves weak van der Waals interactions.

Q.6Peptising agent is added to convert precipitate into colloidal solution. Explain with an example.v
Solution

Mechanism: the peptising ions adsorb on the aggregated precipitate surface, produce electrical charges on the particles, create electrostatic repulsion and break aggregates into colloidal particles. Example: freshly prepared Fe(OH)3 precipitate can be converted into a stable Fe(OH)3 sol by adding a small amount of OH⁻ (peptising agent) and shaking so that particles become negatively charged and disperse.

Answer:

A peptising agent supplies ions that adsorb on precipitate particles, impart like charge and disperse them into a stable sol; e.g. a small amount of NaOH (OH⁻) on freshly precipitated Fe(OH)3 yields a stable ferric hydroxide sol by giving surface negative charge.

Q.7What happens when a colloidal sol of Fe(OH)3 and As2S3 are mixed?v
Solution

Fe(OH)3 sol particles are positively charged and As2S3 sol particles are negatively charged. On mixing they adsorb on each other's surfaces, neutralize surface charge and aggregate, leading to coagulation/flocculation and precipitation.

Answer:

Mutual coagulation (precipitation) occurs due to neutralization of opposite surface charges.

Q.8What is the difference between a sol and a gel?v
Solution

Sol: dispersed phase solid particles (1–1000 nm) in a continuous liquid phase (flowing). Gel: formed when particles of a sol aggregate into a continuous network that immobilizes the liquid, giving a jelly-like semi-solid (e.g., silica gel, gelatin). Gels may be converted to sols by breaking the network (heating or dilution) in some cases.

Answer:

A sol is a liquid colloidal dispersion of solid particles in a liquid; a gel is a three-dimensional network of colloidal particles that entraps the liquid and behaves as a semi-solid.

Q.9Why are lyophillic colloidal sols are more stable than lyophobic colloidal sol.v
Solution

In lyophilic sols the dispersed phase is strongly solvated by the medium (large solvation energy and adsorbed solvent layer), producing steric/electrostatic stabilization. This makes them resistant to coalescence and tolerant to electrolytes. Lyophobic sols have weak solvation, rely only on electrical repulsion and are readily coagulated.

Answer:

Lyophilic sols have strong solute–solvent affinity (solvation layer) which stabilizes particles against aggregation; lyophobic sols lack such solvation and are easily destabilized by electrolytes.

Q.10Addition of Alum purifies water. Why?v
Solution

When alum [e.g. $KAl(SO_4)_2·12H_2O$] is added to water it hydrolyses to form Al(OH)3(s) gel. The gelatinous Al(OH)3 adsorbs suspended colloidal particles and bacteria, neutralizes their charges and causes them to coagulate and settle, clarifying the water.

Answer:

Alum hydrolyses to form gelatinous Al(OH)3 which adsorbs and coagulates suspended colloidal impurities, removing turbidity and microbes by sedimentation.

Q.11What are the factors which influence the adsorption of a gas on a solid?v
Solution

Adsorption depends on (1) surface area and porosity of the solid (greater area → more adsorption), (2) nature of solid and gas (chemical affinity, polarity), (3) temperature (physisorption decreases with rising T; chemisorption may increase then decrease), (4) pressure (adsorption increases with pressure), (5) critical temperature of gas (higher Tc → stronger adsorption), and (6) presence of other adsorbates or poisons.

Answer:

Factors: nature of adsorbate and adsorbent (chemical affinity), surface area/porosity, temperature, pressure (or concentration), molecular size and shape, and presence of other gases or impurities.

Q.12What are enzymes? Write a brief note on the mechanism of enzyme catalysis.v
Solution

Enzymes are highly specific biological catalysts (mostly proteins). Mechanism: substrate binds to the active site forming an enzyme–substrate (ES) complex. The enzyme stabilizes the transition state and lowers activation energy by precise orientation, proximity effects, acid/base catalysis, covalent catalysis, or metal ion catalysis. Product formation releases products and regenerates free enzyme. Models: lock-and-key (rigid fit) and induced-fit (active site changes shape on binding).

Answer:

Enzymes are biological catalysts (proteins or RNA) that accelerate biochemical reactions by lowering activation energy via formation of enzyme–substrate complexes; mechanism follows lock-and-key or induced-fit models and stabilizes transition state.

Q.13What do you mean by activity and selectivity of catalyst?v
Solution

Activity quantifies how effectively a catalyst accelerates a given reaction (higher activity → higher rate per active site). Selectivity measures the preference of a catalyst for producing one product over others under given conditions; a highly selective catalyst minimizes side reactions and unwanted products.

Answer:

Activity: ability of a catalyst to increase reaction rate (turnover frequency). Selectivity: ability of a catalyst to direct reactants to a particular product among possible products.

Q.14Describe some feature of catalysis by Zeolites.v
Solution

Features: (1) uniform micropores and channels giving molecular sieving and shape-selectivity, (2) acidic Bronsted/Lewis sites (from Al in framework) that catalyse hydrocarbon cracking and isomerization, (3) high thermal and chemical stability, (4) ion-exchange capacity and large surface area, (5) used industrially in fluid catalytic cracking (FCC), hydrocarbon conversion and separation.

Answer:

Zeolites are microporous aluminosilicates with high surface area, shape-selective catalysis, acidic sites and ion-exchange ability; used in cracking, isomerization and petrochemical processes.

Q.15Give three uses of emulsions.v
Solution

Emulsions are widely used: in food (oils in water emulsions like mayonnaise, milk), in pharmaceuticals and cosmetics (creams, ointments, lotions for topical delivery), and in industry (paints, varnishes, pesticide formulations) where controlled dispersion of one liquid in another is required.

Answer:

Uses: food products (milk, mayonnaise), pharmaceuticals/cosmetics (creams, lotions), and paints/coatings or insecticide formulations.

Q.16Why does bleeding stop by rubbing moist alumv
Solution

Alum ($KAl(SO_4)_2·12H_2O$) is an astringent: it precipitates and coagulates blood proteins at the wound site and causes local vasoconstriction, forming a mechanical plug that seals ruptured blood vessels and stops bleeding.

Answer:

Alum causes protein coagulation and vasoconstriction; the precipitated proteins form a plug sealing the wound and stopping bleeding.

Q.17Why is desorption important for a substance to act as good catalyst?v
Solution

A good heterogeneous catalyst must not bind products too strongly. After reaction product molecules must desorb quickly to regenerate active sites. If desorption is slow, products remain adsorbed, occupying sites (poisoning) and lowering catalytic turnover and activity.

Answer:

Desorption frees active sites by removing products so they are available for further catalytic cycles; slow desorption leads to site blocking and catalyst deactivation.

Q.18Comment on the statement: Colloid is not a substance but it is a state of substance.v
Solution

The term 'colloid' refers to the physical state in which one phase is finely dispersed in another with particle sizes in the colloidal range. A given material can be dissolved (true solution), colloidal, or coarse precipitate depending on preparation and conditions (e.g., gold can be ionic in solution, colloidal gold sol, or metallic gold). Thus 'colloid' is a state, not a distinct chemical substance.

Answer:

A colloid describes a dispersion state (particle size ~1–1000 nm) of a substance in a medium; the same chemical can exist in molecular, colloidal or coarse states depending on condition.

Q.19Explain any one method for coagulationv
Solution

When an electrolyte is added to a colloidal sol, counter-ions adsorb on particle surfaces and reduce the zeta potential; this decreases electrostatic repulsion and allows van der Waals attraction to aggregate particles, causing coagulation. According to the Schulze–Hardy rule, higher valence counter-ions are more effective (e.g., Al³⁺ >> Na⁺).

Answer:

Coagulation by electrolytes: addition of counter-ions neutralizes surface charge (compresses double layer) causing particles to aggregate; coagulating power increases with ion valence (Schulze–Hardy rule).

Q.20Write a note on electro osmosisv
Solution

In a charged porous plug or between charged electrodes, the layer of counter-ions adjacent to the charged surface migrates under an electric field and drags solvent with it, producing bulk flow of liquid — electroosmosis. It is observed in capillaries, gels and membranes and is important in techniques like electrophoretic separations and in microfluidic devices. Direction depends on sign of surface charge (liquid flows toward cathode if surface is negatively charged).

Answer:

Electroosmosis is the movement of the dispersion medium (liquid) relative to a stationary charged porous medium under an applied electric field, driven by the motion of the diffuse double layer.

Q.21Write a note on catalytic poisonv
Solution

A catalytic poison is a species which, by adsorption on the active centres of a catalyst, reduces the number of available active sites and thus lowers the rate of the catalysed reaction. Poisons may adsorb reversibly or irreversibly and may act by forming strong bonds with catalyst atoms or by occupying surface sites. Common examples: CO and S compounds poison platinum and palladium catalysts in hydrogenation or automotive converters (CO binds strongly to Pt surface), and Pb compounds poison catalysts used in oxidation reactions. Poisons are one reason for catalyst deactivation; removal or regeneration (e.g., oxidative burning off of carbonaceous poisons) is often necessary in industrial practice.

Answer:

A catalytic poison is a substance that strongly adsorbs on the active sites of a catalyst, blocking them and thereby decreasing or completely inhibiting the catalytic activity.

Q.22Explain intermediate compound formation theory of catalysis with an examplev
Solution

According to the intermediate compound formation theory, the catalyst reacts chemically with a reactant to form an intermediate (a new chemical species). This intermediate has a lower activation energy path to products; it subsequently decomposes to give the product and regenerates the catalyst. Example: acid‑catalysed dehydration of tert‑butyl alcohol. H+ protonates the −OH group to give a better leaving group, forming the tert‑butyl carbocation (an intermediate). The carbocation then loses H+ (or reacts) to give isobutene and the proton is regenerated. In symbolic form: R–OH + H+ → [R–OH2]+ → R+ (carbocation) → product + H+.

Answer:

The theory states that the catalyst chemically combines with reactant(s) to form an unstable intermediate compound, which then decomposes to give product(s) and regenerates the catalyst.

Q.23What is the difference between homogenous and hetrogenous catalysis?v
Solution

Main differences: - Phase: Homogeneous — catalyst and reactants in same phase (e.g., acid in solution); Heterogeneous — different phases (e.g., solid Pt catalyst with gas reactants). - Mechanism: Homogeneous often involves formation of soluble intermediates/complexes; heterogeneous operates via adsorption on surface (adsorption → reaction → desorption). - Separation: Homogeneous catalysts are often harder to separate and recover; heterogeneous catalysts are easily separated by filtration. - Temperature/pressure: Heterogeneous catalysts are widely used in high‑temperature/pressure gas‑phase processes; homogeneous catalysts are common in solution chemistry and offer high selectivity. - Examples: Homogeneous — acid or base catalysis in solution, metal complexes in organic reactions; Heterogeneous — Haber process (Fe catalyst), catalytic hydrogenation (Pd/C).

Answer:

Homogeneous catalysis: catalyst and reactants in same phase; Heterogeneous catalysis: catalyst and reactants in different phases (usually solid catalyst with gaseous/liquid reactants).

Q.24Describe adsorption theory of catalysis.v
Solution

The adsorption theory (Langmuir–Hinshelwood type idea) explains heterogeneous catalysis as follows: reactant molecules become adsorbed on active sites of the catalyst surface (physisorption or chemisorption). Adsorption increases local concentration, weakens bonds in reactants, and arranges molecules in favourable orientation for reaction. Two adsorbed species may react on the surface to form product which then desorbs, freeing the active sites. Representatively: A + S ⇌ AS, B + S ⇌ BS; AS + BS → AB + 2S (where S denotes an active surface site). Example: catalytic hydrogenation of alkenes on Pd surface — H2 and the alkene adsorb, H atoms add to the C=C while bound, and the saturated product desorbs. Surface properties (active centres, crystal faces) and chemisorption strength determine activity and selectivity.

Answer:

Adsorption theory: catalysis occurs because reactant molecules are adsorbed on the catalyst surface, which brings them together, weakens bonds, provides active centres and proper orientation, thereby lowering activation energy and increasing reaction rate.