CBSE Class 12 Chemistry Surface Chemistry Questions

Unit - 5

SURFACE CHEMISTRY

VSA QUESTIONS (1 - MARK QUESTIONS)

Question 1. Why does a gas mixed with another gas not form a colloidal system?
Answer: A gas mixed with another gas forms a homogeneous mixture (true solution) because gases are completely miscible in all proportions and do not form a heterogeneous system consisting of dispersed phase and dispersion medium, which is a fundamental requirement for a colloidal system.
In simple words: Gases mix completely to form a clear, single-phase solution, whereas colloids require two separate phases.

Exam Tip: Remember that colloids are strictly heterogeneous systems. Gas-gas mixtures are always homogeneous.

Question 2. Why are adsorbate particles attracted and retained on the surface of adsorbent?
Answer: Adsorbate particles are attracted and retained due to the unbalanced residual attractive forces (residual valencies) present on the surface atoms of the adsorbent, which arise because surface molecules are not surrounded by other molecules from all sides like those in the bulk.
In simple words: Surface atoms have unsatisfied forces that pull and hold foreign molecules onto the solid surface.

Exam Tip: Use the exact term residual forces or residual valencies to score full marks.

Question 3. Explain the terms sorption and desorption.
Answer: Sorption is the process in which both adsorption (accumulation on the surface) and absorption (penetration into the bulk) take place simultaneously. Desorption is the reverse of adsorption, which is the process of removing an adsorbed substance from the surface of an adsorbent.
In simple words: Sorption means a substance sticks to the surface and goes inside too, while desorption means it comes off the surface.

Exam Tip: Sorption covers both phenomena; clearly mention both adsorption and absorption in your definition.

Question 4. "Chemisorption is highly specific." Illustrate with an example.
Answer: Chemisorption involves chemical bond formation between the adsorbate and adsorbent; therefore, it only occurs if chemical interaction is possible between them. For example, oxygen is chemisorbed on transition metals like iron due to oxide formation, but it does not react with all metals under the same conditions.
In simple words: Since chemical bonds are formed, chemisorption only happens between specific substances that can react chemically.

Exam Tip: Emphasize the formation of chemical bonds as the core reason for specificity.

Question 5. "Adsorbents in finely divided form are more effective." Why?
Answer: Finely divided forms of adsorbents provide a much larger surface area per unit mass compared to a solid block. Since adsorption is a surface phenomenon, greater surface area leads to a greater extent of adsorption.
In simple words: Breaking a solid into tiny pieces increases its exposed surface area, giving more room for adsorption to happen.

Exam Tip: Always link surface area directly to the extent of adsorption.

Question 6. Name two compounds used as adsorbent for controlling humidity.
Answer: Silica gel and alumina gel.
In simple words: Silica gel and alumina gel are commonly used to absorb moisture and keep things dry.

Exam Tip: Memorize these two common drying and dehumidifying agents.

Question 7. Mention one shape selective catalyst used to convert alcohol directly into gasoline.
Answer: ZSM-5 (a zeolite catalyst).
In simple words: ZSM-5 is a special sieve-like catalyst that converts alcohols straight into gasoline.

Exam Tip: ZSM-5 is a frequently asked example of a zeolite shape-selective catalyst.

Question 8. 'Generally high temperature is favourable for chemisorption.' Why?
Answer: Chemisorption has a significant activation energy barrier (similar to chemical reactions). Therefore, supplying heat (high temperature) provides the required activation energy to initiate and accelerate the bond-forming process.
In simple words: Chemical adsorption needs activation energy to form bonds, so higher temperatures help start the reaction.

Exam Tip: Do not confuse this with physisorption, which is exothermic and decreases at higher temperatures.

Question 9. Name the catalyst used in the following process : (a) Haber's process for the manufacture of \(\text{NH}_3\) gas. (b) Ostwald process for the manufacture of nitric acid.
Answer:
(a) Iron (Fe) with molybdenum (Mo) as promoter.
(b) Platinised asbestos or platinum gauze.
In simple words: Iron is used for ammonia production, and platinum is used for nitric acid production.

Exam Tip: Learn industrial preparation catalysts thoroughly as they frequently appear in board exams.

Question 10. Explain the relationship given by Freundlich in adsorption isotherm.
Answer: Freundlich adsorption isotherm gives the quantitative relationship between the extent of adsorption (\(\frac{x}{m}\)) and equilibrium pressure (\(p\)) or concentration (\(C\)) at constant temperature. The mathematical equation is expressed as:
\(\frac{x}{m} = k p^{\frac{1}{n}}\) (where \(n > 1\))
Taking logarithm on both sides: \(\log\left(\frac{x}{m}\right) = \log k + \frac{1}{n}\log p\)
In simple words: It is an equation that relates how much gas is adsorbed on a solid based on the gas pressure.

Exam Tip: Be sure to state the logarithmic linear form of the equation and mention what \(x\), \(m\), \(k\), and \(n\) represent.

Question 11. Which group elements show maximum catalytic activity for hydrogenation reactions?
Answer: Group 7 to 9 elements (transition metals such as Fe, Co, Ni, Ru, Rh, Pd, Pt, etc.).
In simple words: Transition metals from groups 7 to 9 are the best catalysts for adding hydrogen to molecules.

Exam Tip: Mention transition metals of groups 7-9 explicitly.

Question 12. Why gas masks are used by miners in coal mines while working?
Answer: Gas masks contain suitable adsorbents (like activated charcoal) that adsorb poisonous and harmful gases present in coal mines, protecting miners from inhaling toxic fumes.
In simple words: Gas masks use activated charcoal to trap dangerous poisonous gases so miners can breathe safely.

Exam Tip: Connect the principle of adsorption to activated charcoal in gas masks.

Question 13. Write the chemical reaction involved in the preparation of sulphur sol.
Answer: Sulphur sol is prepared by oxidation of hydrogen sulphide with sulphur dioxide:
\(2\text{H}_2\text{S} + \text{SO}_2 \xrightarrow{\text{oxidation}} 2\text{H}_2\text{O} + 3\text{S (sol)}\)
In simple words: Reacting hydrogen sulphide with sulphur dioxide produces colloidal sulphur.

Question 14. Name the enzyme which converts milk into curd.
Answer: Lactobacilli (Lactase enzyme present in lactic acid bacteria).
In simple words: Lactobacilli bacteria produce enzymes that turn milk into curd.

Exam Tip: Write lactobacilli clearly as the standard answer.

Question 15. What are the optimum temperature and pH at which enzymes are highly active?
Answer: Optimum temperature is \(298-310\text{ K}\) and optimum pH is \(5\) to \(7\).
In simple words: Enzymes work best at body temperature and neutral to slightly acidic pH ranges.

Exam Tip: Mention both temperature range and pH range clearly.

Question 16. What are the physical states of dispersed phase and dispersion medium in foam rubber.
Answer: Dispersed phase is Gas (\(\text{G}\)) and dispersion medium is Solid (\(\text{S}\)).
In simple words: Foam rubber is made of gas trapped inside a solid structure.

Exam Tip: Memorize the colloidal system classification table (gel, foam, emulsion, aerosol, etc.).

Question 18. What is the composition of collodion solution?
Answer: A 4% solution of cellulose nitrate in a mixture of alcohol and ether.
In simple words: Collodion is a solution of cellulose nitrate dissolved in alcohol and ether.

Exam Tip: Note down the exact solvent mixture (alcohol and ether) and solute (cellulose nitrate).

Question 19. Why do colloidal particles show Brownian movement?
Answer: Due to the unbalanced bombardment of colloidal particles by the molecules of the dispersion medium from all directions.
In simple words: Liquid molecules constantly hit colloidal particles unevenly from all sides, making them zig-zag around.

Exam Tip: The key phrase to include is "unbalanced bombardment by dispersion medium molecules".

Question 21. State the sign of entropy change involved when the molecules of a substances get adsorbed on a solid surface.
Answer: Negative (\(\Delta S = -\text{ve}\)).
In simple words: Adsorption restricts the movement of molecules, leading to a decrease in randomness (entropy).

Exam Tip: Since molecules lose degrees of freedom upon adsorption, randomness decreases and entropy becomes negative.

Question 22. Why does sky appear blue to us?
Answer: Due to the scattering of light by colloidal dust and gas molecules in the atmosphere (Tyndall effect). Shorter wavelengths of light (blue) are scattered much more strongly than longer wavelengths.
In simple words: Tiny particles in the atmosphere scatter blue light in all directions more than other colors.

Exam Tip: Connect the blue color of the sky to Rayleigh scattering / Tyndall effect of shorter wavelengths.

Question 23. What happens when hydrated ferric oxide and arsenious sulphide sols are mixed in almost equal proportions?
Answer: Mutual coagulation (precipitation) takes place because hydrated ferric oxide is positively charged while arsenious sulphide sol is negatively charged, resulting in neutralization of charges.
In simple words: Mixing oppositely charged colloids neutralizes them and causes them to clump together and precipitate.

Exam Tip: Mention mutual coagulation and state the charges of both sols.

Question 24. Gelatin is generally added to ice-cream. Why?
Answer: Gelatin acts as a protective colloid (emulsifier) that stabilizes the ice-cream emulsion and prevents ice crystals from growing too large, ensuring a smooth texture.
In simple words: Gelatin acts as a stabilizer to keep ice-cream smooth and prevent it from separating.

Exam Tip: Use the term "protective colloid" or "emulsifier".

Question 25. How is lake test for aluminium ion based upon adsorption?
Answer: Aluminium hydroxide precipitate (\(\text{Al}_2\text{O}_3 \cdot x\text{H}_2\text{O}\)) has a strong capacity to adsorb blue litmus dye from the solution, forming a blue colored floating lake.
In simple words: Aluminium precipitate traps blue dye on its surface, creating a colored lake-like spot.

Question 26. What is saturation pressure in Freundlich's isotherm?
Answer: Saturation pressure (\(P_s\)) is the pressure above which the extent of adsorption (\(\frac{x}{m}\)) becomes independent of pressure and reaches its maximum limit.
In simple words: It is the pressure where the solid surface is completely covered and adsorption can no longer increase.

Question 27. Mention the two conditions for the formation of micelles.
Answer:
1. Critical Micelle Concentration (CMC)
2. Kraft temperature (\(T_k\))
In simple words: Micelles only form above a specific concentration and above a specific temperature.

Exam Tip: Always state both CMC and Kraft temperature for micelle formation.

Question 28. How is Brownian movement responsible for the stability of sols?
Answer: The continuous zig-zag Brownian movement creates a stirring effect that counteracts the force of gravity, preventing colloidal particles from settling down at the bottom.
In simple words: Constant motion keeps colloidal particles suspended so they do not settle down due to gravity.

Exam Tip: Emphasize the counteracting effect against gravity.

Question 29. Which of the following is more effective in coagulating positively charged hydrated ferric oxide sol : (i) KCl (ii) \(\text{CaSO}_4\) (iii) \(\text{K}_3[\text{Fe}(\text{CN})_6]\).
Answer: \(\text{K}_3[\text{Fe}(\text{CN})_6]\).
In simple words: According to Hardy-Schulze rule, the ion with the highest negative charge is most effective at coagulating a positive sol.

Exam Tip: Cite the Hardy-Schulze rule: higher valency of the coagulating ion means greater coagulating power.

Question 30. State the purpose of impregnating the filter paper with collodion solution.
Answer: Impregnating filter paper with collodion solution plugs the pores of ordinary filter paper, reducing pore size so that colloidal particles cannot pass through it (forming an ultra-filter paper).
In simple words: It shrinks the pores of regular filter paper so it can filter out tiny colloidal particles.

Exam Tip: Mention the creation of ultra-filter paper to separate colloids from true solutions.

Question 31. Mention one use of ZSM-5 catalyst.
Answer: It is used in the petroleum industry to convert alcohols directly into gasoline.
In simple words: It turns alcohol into petrol.

Exam Tip: Keep this answer concise and directly tied to petroleum refining.

SA (I) TYPE QUESTIONS (2 - MARK QUESTIONS)

Question 35. Explain the effect of temperature on the extent of physical and chemical adsorption.
Answer:
- Physisorption (Physical adsorption): It is exothermic in nature. According to Le Chatelier's principle, lowering temperature favors physisorption, so it decreases continuously with an increase in temperature.
- Chemisorption (Chemical adsorption): It requires activation energy. Initially, adsorption increases with temperature as activation energy is supplied, but at higher temperatures, it decreases due to desorption caused by thermal energy.
In simple words: Physical adsorption drops as temperature rises, while chemical adsorption first increases and then decreases at higher temperatures.

Exam Tip: Draw or mentally picture the adsorption isobar curve to explain both behaviors clearly.

Question 36. Define the term peptization and mention its cause.
Answer: Peptization is the process of converting a freshly prepared precipitate into colloidal particles by adding a small amount of a suitable electrolyte (peptizing agent).
Cause: The precipitate adsorbs common ions from the electrolyte onto its surface, developing a net electrical charge which causes particles to repel each other and disperse into a sol.
In simple words: Peptization breaks down a solid precipitate into colloid particles by adding an electrolyte that gives them like charges so they repel.

Question 37. What will be the charge on colloidal solutions in the following cases. Give reasons for the origin of charge.
(1) \(\text{AgNO}_3\) solution added to \(\text{KI}\) solution.
(2) \(\text{KI}\) solution added to \(\text{AgNO}_3\) solution.
Answer:
(1) When \(\text{AgNO}_3\) is added to excess \(\text{KI}\), \(\text{I}^-\) ions are preferentially adsorbed from the dispersion medium, giving the colloidal particles a negative charge (\([\text{AgI}]\text{I}^-/\text{K}^+\)).
(2) When \(\text{KI}\) is added to excess \(\text{AgNO}_3\), \(\text{Ag}^+\) ions are preferentially adsorbed, giving the colloidal particles a positive charge (\([\text{AgI}]\text{Ag}^+/\text{NO}_3^-\)).
Reason for origin of charge: Preferential adsorption of common ions from the dispersion medium onto the surface of the colloidal precipitate.
In simple words: The charge depends on which reactant is in excess, as its ions get adsorbed onto the precipitate surface.

Exam Tip: Always specify which ion is in excess and gets preferentially adsorbed.

Question 38. Write the factors upon which the catalytic reaction of shape-selective catalyst depends?
Answer:
1. The pore structure of the catalyst (zeolites).
2. The size and shape of reactant and product molecules.
In simple words: It depends on the size of the catalyst pores and whether the reactant and product molecules can fit through them.

Exam Tip: Mention pore structure and molecular dimensions as the two primary factors.

Question 39. Mention two examples of emulsifying agents for o/w emulsions and w/o emulsions.
Answer:
- For oil-in-water (o/w) emulsions: Proteins, gums, soaps.
- For water-in-oil (w/o) emulsions: Heavy metal salts of fatty acids, lampblack, organic solvents/cetyl alcohol.
In simple words: Soaps and proteins help oil mix in water, while heavy metal soaps help water mix in oil.

Exam Tip: Give standard and clear examples for both emulsion types.

Question 40. Suggest a mechanism of enzyme catalysed reaction.
Answer: Enzyme-catalyzed reactions proceed via the lock-and-key mechanism involving two main steps:
Step 1: Binding of enzyme (\(E\)) with substrate (\(S\)) to form an activated enzyme-substrate complex (\(ES\)).
\(E + S \rightleftharpoons [ES]\)
Step 2: Decomposition of the complex to yield the product (\(P\)) and regenerate the enzyme (\(E\)).
\([ES] \rightarrow E + P\)
In simple words: Enzymes first lock with reactants to form a temporary complex, which then breaks apart into products and frees the enzyme.

Exam Tip: Write both chemical steps clearly using \(E\), \(S\), \(ES\), and \(P\).

Question 41. A small amount of silica gel and a small amount of anhydrous calcium chloride are placed separately in two beakers containing water vapour. Name of phenomenon that takes place in both the beakers.
Answer:
- Silica gel: Adsorption (water molecules are held only on the surface).
- Anhydrous calcium chloride: Absorption (water molecules enter the bulk and form a hydrated salt, \(\text{CaCl}_2 \cdot 2\text{H}_2\text{O}\)).
In simple words: Silica gel causes adsorption on its surface, while calcium chloride absorbs moisture uniformly throughout its mass.

Exam Tip: Clearly differentiate between surface adsorption (silica gel) and bulk absorption (anhydrous \(\text{CaCl}_2\)).

Question 42. Write the differences between adsorption and absorption?
Answer:
- Adsorption: It is a surface phenomenon where concentration of adsorbate is higher only at the surface. It occurs at a uniform rate and is reversible.
- Absorption: It is a bulk phenomenon where the substance is uniformly distributed throughout the bulk of the solid or liquid. It occurs at a constant rate.
In simple words: Adsorption happens only on the outside surface, whereas absorption goes all the way inside.

Exam Tip: Contrast surface vs. bulk in your comparison.

Question 43. How can physisorption be distinguished from chemisorption?
Answer:
- Physisorption is reversible, caused by weak van der Waals forces, has low heat of adsorption (\(20-40\text{ kJ/mol}\)), and does not require activation energy.
- Chemisorption is irreversible, caused by chemical bond formation, has high heat of adsorption (\(80-240\text{ kJ/mol}\)), and requires activation energy.
In simple words: Physisorption is weak and reversible without bonds, while chemisorption involves strong chemical bonds and is irreversible.

Question 44. Classify the following reactions as homogeneous and heterogeneous catalysis :
(a) \(\text{Vegetable oil (l)} + \text{H}_2\text{(g)} \xrightarrow{\text{Ni(s)}} \text{Vegetable ghee (s)}\)
(b) \(\text{C}_{12}\text{H}_{22}\text{O}_{11}\text{(aq)} + \text{H}_2\text{O(l)} \xrightarrow{\text{H}_2\text{SO}_4\text{(aq)}} \text{C}_6\text{H}_{12}\text{O}_6\text{(aq)} + \text{C}_6\text{H}_{12}\text{O}_6\text{(aq)}\)
Answer:
(a) Heterogeneous catalysis (Reactants are in liquid/gas phase, while the nickel catalyst is a solid).
(b) Homogeneous catalysis (Reactants and the sulphuric acid catalyst are all in the same liquid aqueous phase).
In simple words: Reaction (a) is heterogeneous because the catalyst is solid while reactants are liquid/gas. Reaction (b) is homogeneous because everything is in the liquid phase.

Exam Tip: Check the physical states of reactants and catalysts to determine homogeneity.

Question 45. In what way these are different : (a) a sol and a gel (b) a gel and an emulsion.
Answer:
(a) Sol is a colloidal system with a solid dispersed phase in a liquid dispersion medium where the liquid flows freely. Gel is a colloidal system with a liquid dispersed phase in a solid dispersion medium which is semi-rigid and jelly-like.
(b) Gel has liquid dispersed in solid, whereas an emulsion has liquid dispersed in another immiscible liquid.
In simple words: A sol flows like liquid, a gel is semi-solid, and an emulsion consists of two liquids mixed together.

Exam Tip: Clearly state the dispersed phase and dispersion medium for each.

Question 46. State "Hardy Schulze Rule" with one example.
Answer: Hardy-Schulze rule states that higher is the valency of the flocculating (coagulating) ion, greater is its power to coagulate the oppositely charged sol.
Example: For coagulating a negatively charged arsenic sulphide sol, the flocculating power of cations follows the order: \(\text{Al}^{3+} > \text{Ba}^{2+} > \text{Na}^+\).
In simple words: Ions with higher electrical charges are much more effective at causing colloids to clump and precipitate.

Exam Tip: Always include the valency order in your example.

Question 47. What is an emulsifying agent? What role does it play in forming an emulsion?
Answer: An emulsifying agent (emulsifier) is a substance added to an emulsion to stabilize it. Role: It forms an interfacial film between the suspended droplets and the dispersion medium, reducing interfacial tension and preventing the droplets from coalescing.
In simple words: An emulsifier is a stabilizer that coats liquid droplets so they do not separate out.

Exam Tip: Mention interfacial tension reduction and stabilization.

Question 48. Define the terms : (a) Helmholtz electrical double layer. (b) Zeta potential.
Answer:
(a) Helmholtz electrical double layer: The combination of the two layers of opposite charges around a colloidal particle (the fixed layer of preferentially adsorbed ions and the mobile diffused layer of counter ions).
(b) Zeta potential (Electrokinetic potential): The potential difference between the fixed charged layer and the diffused mobile layer of ions surrounding a colloidal particle.
In simple words: The double layer is the dual layer of charges around a colloid, and zeta potential is the voltage difference between these layers.

Exam Tip: Use the alternative name electrokinetic potential for zeta potential.

Question 49. A graph between \(\frac{x}{m}\) and \(\log p\) is a straight line at an angle of \(45^\circ\) with intercept on the y-axis i.e. (\(\log k\)) equal to \(0.3010\). Calculate the amount of the gas absorbed per gram of the adsorbent under a pressure of \(0.5\) atmosphere.
Answer:
Given equation of adsorption isotherm:
\(\log\left(\frac{x}{m}\right) = \log k + \frac{1}{n}\log p\)
Slope of the line \(= \tan(45^\circ) = 1\), so \(\frac{1}{n} = 1\)
Intercept \(\log k = 0.3010\), so \(k = \text{antilog}(0.3010) = 2\)
Pressure \(p = 0.5\text{ atm}\)
Substituting the values into the equation:
\(\frac{x}{m} = k p^{\frac{1}{n}}\)
\(\frac{x}{m} = 2 \times (0.5)^1 = 2 \times 0.5 = 1\)
In simple words: Using the slope and intercept from the graph, we calculate \(k\) and \(n\), then substitute the pressure to find that 1 gram of gas is adsorbed.

Exam Tip: Remember that \(\tan(45^\circ) = 1\) gives \(\frac{1}{n} = 1\) and antilog of \(0.3010\) is \(2\).

Question 50. Mention the two necessary conditions for the observation of Tyndall Effect.
Answer:
1. The diameter of the dispersed particles is not much smaller than the wavelength of the light used.
2. The refractive indices of the dispersed phase and dispersion medium differ markedly in magnitude.
In simple words: Particles must be comparable in size to light wavelengths, and the two phases must have different refractive indices.

Exam Tip: State both particle size and refractive index conditions clearly.

Question 51. Account for the following : (a) Artificial rain can be caused by spraying electrified sand on the clouds. (b) Electrical precipitation of smoke.
Answer:
(a) Clouds are colloidal dispersions of water droplets carrying a charge. Spraying oppositely charged sand neutralizes these charges, causing the water droplets to coagulate and fall as artificial rain.
(b) Smoke is a colloidal suspension of carbon particles in air carrying an electrical charge. When passed through an Cottrell precipitator at high voltage, the particles get neutralized and precipitate out, leaving clean air.
In simple words: Both processes rely on neutralizing the electrical charge of colloidal particles so they clump together and settle down.

Exam Tip: Mention charge neutralization and coagulation for both phenomena.

Question 52. Write chemical equations for the preparation of sols : (a) Gold sol by reduction. (b) hydrated ferric oxide sol by hydrolysis.
Answer:
(a) Gold sol by reduction:
\(2\text{AuCl}_3 + 3\text{HCHO} + 3\text{H}_2\text{O} \rightarrow 2\text{Au (sol)} + 3\text{HCOOH} + 6\text{HCl}\)
(b) Hydrated ferric oxide sol by hydrolysis:
\(\text{FeCl}_3 + 3\text{H}_2\text{O} \rightarrow \text{Fe(OH)}_3\text{ (sol)} + 3\text{HCl}\)
In simple words: Gold chloride is reduced to form gold sol, and ferric chloride is hydrolyzed with boiling water to form ferric hydroxide sol.

Exam Tip: Balance both preparation equations correctly with their respective byproducts.

Question 53. How can the two emulsions can be distinguished : (a) oil in water type (O/W) and (b) water in oil type (W/O)?
Answer:
1. Dye test: Adding an oil-soluble dye to the emulsion; if the background becomes colored, it is W/O, and if only droplets are colored, it is O/W.
2. Dilution test: Adding water; if it mixes easily, it is O/W (water is the external phase).
3. Conductivity test: O/W emulsions conduct electricity well, whereas W/O emulsions are poor conductors.
In simple words: Emulsions can be told apart using water dilution, dye tests, or electrical conductivity tests.

Exam Tip: Mention at least two standard tests (dilution and conductivity) for full marks.

SA (II) TYPE QUESTIONS (3 - MARK QUESTIONS)

Question 54. Write the difference between (a) catalysts and enzymes (b) promoters and poisons.
Answer:
(a) Catalysts vs Enzymes:
- Catalysts are usually inorganic chemical substances that accelerate reactions and can operate over wide temperature and pressure ranges.
- Enzymes are complex protein molecules produced by living organisms that act as biochemical catalysts, highly specific and active under optimum physiological conditions (\(298-310\text{ K}\)).
(b) Promoters vs Poisons:
- Promoters are substances that enhance the activity of a catalyst (e.g., Mo in Haber's process).
- Poisons are substances that decrease or destroy the activity of a catalyst (e.g., CO in Haber's process).
In simple words: Catalysts speed up reactions, enzymes are biological catalysts, promoters boost catalyst activity, and poisons destroy it.

Exam Tip: Define each term clearly with a brief example to secure all 3 marks.

Question 55. Write the steps of 'Modern Adsorption Theory of Heterogenous Catalysis.'
Answer:
The modern mechanism (adsorption theory) involves 5 steps:
1. Diffusion of reactants to the surface of the solid catalyst.
2. Adsorption of reactant molecules on the catalyst surface.
3. Occurrence of the chemical reaction on the catalyst surface through intermediate formation.
4. Desorption of reaction products from the catalyst surface.
5. Diffusion of products away from the catalyst surface.
In simple words: Reactants move to the surface, stick to it, react to form products, detach, and diffuse away.

Exam Tip: List all 5 steps in correct chronological order.

Question 56. Mention the two important features of solid catalysts and explain with the help of suitable examples.
Answer:
1. Activity: It is the ability of a catalyst to accelerate a chemical reaction. For example, a mixture of \(\text{H}_2\) and \(\text{O}_2\) does not react at room temperature, but in the presence of platinum catalyst, water is formed explosively.
2. Selectivity: It is the ability of a catalyst to direct a reaction to yield a specific product. For example, using different catalysts with carbon monoxide and hydrogen yields different products like methane, formaldehyde, or methanol:
\(\text{CO} + 3\text{H}_2 \xrightarrow{\text{Ni}} \text{CH}_4 + \text{H}_2\text{O}\)
\(\text{CO} + 2\text{H}_2 \xrightarrow{\text{Cu/ZnO-Cr}_2\text{O}_3} \text{CH}_3\text{OH}\)
In simple words: Activity means how fast a catalyst speeds up a reaction, and selectivity means its ability to choose a specific product.

Question 57. How are the following colloids different from each other in respect of dispersion medium and dispersed phase? Give one example of each type. (a) An aerosol (b) A hydrosol (c) An emulsion.
Answer:
(a) Aerosol: Dispersed phase is Solid or Liquid, Dispersion medium is Gas. Example: Fog, smoke.
(b) Hydrosol: Dispersed phase is Solid, Dispersion medium is Water (Liquid). Example: Gold sol, starch sol.
(c) Emulsion: Dispersed phase is Liquid, Dispersion medium is Liquid. Example: Milk, butter.
In simple words: Aerosols have particles in gas, hydrosols have particles in water, and emulsions have liquid drops in another liquid.

Exam Tip: State both phases and an example for each of the three types.

Question 58. What happens : (a) by persistent dialysis of a sol. (b) when river water meets the sea water. (c) when alum is applied on cuts during bleeding.
Answer:
(a) Persistent dialysis removes traces of stabilizing electrolytes completely, making the sol unstable and causing it to coagulate (precipitate).
(b) River water contains colloidal clay particles, while sea water contains various dissolved electrolytes (\(\text{NaCl}\), etc.). When river water meets sea water, the electrolytes cause coagulation of clay, forming river deltas.
(c) Alum provides trivalent aluminium ions (\(\text{Al}^{3+}\)) which neutralize the negatively charged blood colloidal particles, causing rapid coagulation (clotting) of blood to stop bleeding.
In simple words: Dialysis removes stabilizers causing precipitation, sea salts cause river mud to form deltas, and alum clots blood.

Exam Tip: Explain each phenomenon using the concept of coagulation and electrolyte action.

Question 59. Distinguish between multimolecular, macromolecular and associated colloids with the help of one example of each.
Answer:
- Multimolecular colloids: Formed by the aggregation of a large number of atoms or small molecules (diameter \(< 1\text{ nm}\)). Example: Sulphur sol (\(\text{S}_8\)), Gold sol.
- Macromolecular colloids: Formed by giant molecules whose size itself is in the colloidal range. Example: Starch, cellulose, proteins.
- Associated colloids (Micelles): Substances that behave as normal electrolytes at low concentration but associate into colloidal aggregates at higher concentrations (above CMC). Example: Soaps and synthetic detergents.
In simple words: Multimolecular are small molecules clumping together, macromolecular are single giant molecules, and associated colloids form clusters above a certain concentration.

Exam Tip: Give precise definitions and examples for all three colloidal types.

Question 60. (a) Which property of colloids is responsible for the sun to look red at the time of setting? (b) \(\text{C}_2\text{H}_2\) on addition with \(\text{H}_2\) forms ethane in presence of palladium catalyst but if reaction is carried in the presence of barium sulphate and quinoline, the product is ethene and not ethane. Why?
Answer:
(a) Scattering of light (Tyndall effect). At sunset, sunlight travels a longer distance through the atmosphere where blue light is scattered away by dust and air molecules, leaving red light to reach our eyes.
(b) Barium sulphate (\(\text{BaSO}_4\)) in the presence of quinoline acts as a catalyst poison (Lindlar's catalyst), which lowers the activity of the palladium catalyst so that the hydrogenation stops at ethene and does not proceed further to ethane.
In simple words: Red sun is due to light scattering, and adding barium sulphate poisons the catalyst to stop the reaction halfway at ethene.

Exam Tip: Mention light scattering for part (a) and catalyst poisoning for part (b).