JEE Chemistry Surface Chemistry MCQs Set 19

Answer: (c) \( \ce{NH3} > \ce{CO2} > \ce{H2} \)

Explanation:
1. Physisorption increases with the ease of liquefaction of the gas, that is with stronger van der Waals forces and higher critical temperature.
2. Critical temperatures: \( \ce{NH3} \) 406 K, \( \ce{CO2} \) 304 K, \( \ce{H2} \) 33 K.
3. So the adsorbed volume order is \( \ce{NH3} > \ce{CO2} > \ce{H2} \).

Teacher's Note:
1. Easily liquefiable gases are adsorbed more because they have stronger intermolecular attraction.
2. Remember: higher critical temperature means greater physisorption.

Answer: (a) \( 1/n \)

Explanation:
1. Freundlich isotherm: \( x/m = K P^{1/n} \).
2. Taking logs: \( \log (x/m) = \log K + \tfrac{1}{n} \log P \).
3. Comparing with \( y = c + m x \), the slope is \( 1/n \) and the intercept is \( \log K \).

Teacher's Note:
1. The intercept gives \( \log K \) and the slope gives \( 1/n \), where \( 1/n \) lies between 0 and 1.
2. This log-log form is the standard way to test Freundlich behaviour.

Answer: (a) 80 to 240

Explanation:
1. Chemisorption involves formation of chemical bonds, so the heat evolved is large.
2. Typical values are 80 to 240 kJ/mol, whereas physisorption gives only 20 to 40 kJ/mol.

Teacher's Note:
1. Heat of chemisorption is of the order of chemical reaction enthalpies.
2. Physisorption values are much smaller because only weak van der Waals forces act.

Answer: (d) Enzymes are specific biological catalysts that possess well-defined active sites

Explanation:
1. Enzymes are biological catalysts, mostly proteins, with a specific active site that binds only particular substrates.
2. They can be poisoned, they are usually homogeneous (colloidal) catalysts and work best near body temperature (about 298 to 310 K).
3. Hence only the last statement is correct.

Teacher's Note:
1. Enzyme action follows the lock and key idea: the substrate fits the active site.
2. High temperature denatures enzymes and destroys their activity.

Answer: (d) preferential adsorption on the catalyst surface

Explanation:
1. A catalytic poison is adsorbed strongly on the active sites of the catalyst.
2. The sites become unavailable to the reactants, so the catalyst loses activity.

Teacher's Note:
1. Examples: arsenic compounds poison platinum in contact process, CO poisons iron/haemoglobin sites.
2. Poisoning is a surface phenomenon based on adsorption.

Answer: (a) suspension > colloidal > true solution

Explanation:
1. True solution: particle size below 1 nm.
2. Colloidal solution: 1 nm to 1000 nm.
3. Suspension: above 1000 nm.
4. So suspension > colloidal > true solution.

Teacher's Note:
1. Size decides the properties: suspensions settle, colloids scatter light, true solutions do neither.
2. Memorise the limits 1 nm and 1000 nm.

Answer: (c) the lower its value, the greater is the protecting power

Explanation:
1. Gold number is the minimum mass (in mg) of a protective colloid that prevents coagulation of 10 mL of standard gold sol on adding 1 mL of 10% NaCl.
2. A smaller mass needed means a better protective colloid.
3. So lower gold number means greater protecting power.

Teacher's Note:
1. Protecting power is inversely proportional to gold number.
2. Gelatin (0.005 to 0.01) is a better protector than starch (25).

Answer: (a) \( \ce{NaCl} > \ce{Na2SO4} > \ce{Na3PO4} \)

Explanation:
1. Particles moving to the cathode are positively charged, so the anions cause coagulation.
2. By the Hardy-Schulze rule, higher the charge on the effective ion, the smaller the amount needed.
3. Charges: \( \ce{PO4^3-} \) > \( \ce{SO4^2-} \) > \( \ce{Cl-} \), so coagulating values are NaCl > \( \ce{Na2SO4} \) > \( \ce{Na3PO4} \).

Teacher's Note:
1. Coagulating value is the minimum concentration needed, so it is inversely related to coagulating power.
2. The ion with charge opposite to the sol is the effective one.

Answer: (c) 2.1 L/g

Explanation:
1. Mass of \( \ce{O2} \) per gram of adsorbent = 3.6/1.2 = 3 g.
2. Moles = 3/32 = 0.094 mol.
3. Volume = \( 0.094 \times 22.4 \) (1 atm, 273 K gives 22.4 L/mol) is about 2.1 L per gram.

Teacher's Note:
1. First convert mass per gram of adsorbent, then use molar volume 22.4 L at STP.
2. Check units: the answer is per gram of adsorbent, not per gram of gas.

Answer: (b) 0.3 g

Explanation:
1. Moles adsorbed = \( 0.1 \text{ L} \times (0.6 - 0.5) \) = 0.01 mol.
2. Mass = 0.01 x 60 = 0.6 g on 2 g charcoal.
3. Per gram: 0.6/2 = 0.3 g.

Teacher's Note:
1. Work in three steps: moles lost from solution, then mass, then divide by adsorbent mass.
2. Molar mass of acetic acid is 60 g/mol.

Answer: (d) lower the activation energy of the reaction

Explanation:
1. Every catalyst works by providing an alternative path with lower activation energy.
2. The lower the activation energy, the faster the reaction; enzymes do this very effectively.

Teacher's Note:
1. A catalyst does not change \( \Delta H \) or the equilibrium position.
2. It only changes how fast equilibrium is reached.

Answer: (d) does not affect the equilibrium constant of the reaction

Explanation:
1. A catalyst speeds up the forward and backward reactions equally.
2. It lowers activation energy and raises the rate constant but leaves \( \Delta H \) and \( \Delta G^\circ \) unchanged.
3. So the equilibrium constant is not changed.

Teacher's Note:
1. Catalysts change kinetics, not thermodynamics.
2. They cannot make a non-spontaneous reaction spontaneous.

Answer: (b) B > A > C > D

Explanation:
1. Protective power is inversely proportional to gold number.
2. Ascending gold number: B (0.004) < A (0.04) < C (10) < D (40).
3. So protective power is B > A > C > D.

Teacher's Note:
1. Smaller gold number means a smaller mass of colloid protects the gold sol.
2. Always invert the order of gold numbers.

Answer: (d) \( 2.4 \times 10^{6} \)

Explanation:
1. Volume of gold = \( 1.9\times10^{-4}/19 = 10^{-5} \) cm\(^3\).
2. Volume of one particle = \( \tfrac{4}{3}\pi (10^{-6})^3 = 4.19\times10^{-18} \) cm\(^3\).
3. Particles per litre = \( 10^{-5}/4.19\times10^{-18} = 2.4\times10^{12} \).
4. 1 litre = \( 10^{6} \) mm\(^3\), so about \( 2.4\times10^{6} \) per mm\(^3\).

Teacher's Note:
1. Convert the radius carefully: 10 nm = \( 10^{-6} \) cm.
2. Remember 1 L = 1000 cm\(^3\) = \( 10^6 \) mm\(^3\).

Answer: (c) reversible transformation of gel into sol by stirring

Explanation:
1. Thixotropy is the property of a gel to become a sol on mechanical agitation.
2. On standing, the sol sets back to the gel, so the change is reversible.

Teacher's Note:
1. Examples: ferric hydroxide gel and some paints and ketchup.
2. Do not confuse it with imbibition, which is swelling of a gel.

Answer: (c) \( \ce{Ba^2+} \), 1.0

Explanation:
1. \( \ce{As2S3} \) sol is negatively charged, so the cation \( \ce{Ba^2+} \) is effective.
2. Amount of \( \ce{Ba^2+} \) = 4.0 mL x 0.005 M = 0.02 mmol.
3. Total volume = 20 mL = 0.020 L, so flocculation value = 0.02/0.020 = 1.0 mmol/L.

Teacher's Note:
1. The effective ion has the charge opposite to that of the sol.
2. Divide by the total volume of the mixture, not by the sol volume.

Answer: (b) Latex is a negatively charged sol

Explanation:
1. Rubber latex is a colloidal dispersion of rubber particles (in a protein-stabilised medium) in water.
2. The particles carry a negative charge, as shown by their migration to the anode.
3. So latex is a negatively charged sol.

Teacher's Note:
1. Coagulation of latex by acids or electrolytes is used to obtain rubber.
2. Negative sols are coagulated most effectively by high-valent cations.

Answer: (b) II = IV < III < I

Explanation:
1. Ferric hydroxide sol is positive, so anions coagulate it.
2. Effective anions: \( \ce{Cl-} \) (II and IV), \( \ce{SO4^2-} \) (III), \( \ce{PO4^3-} \) (I).
3. Coagulating power rises with charge: II = IV < III < I.

Teacher's Note:
1. KCl and NaCl have the same effective ion, so equal power.
2. Always identify the sign of the sol first.

Answer: (d) At 83 K, multimolecular layers are formed

Explanation:
1. At 83 K, nitrogen is near its liquefaction temperature, so it is adsorbed by weak van der Waals forces (physisorption).
2. Physisorption at low temperature and high pressure is multilayered.

Teacher's Note:
1. Physisorption is favoured at low temperature.
2. Chemisorption needs activation and forms a monolayer.

Answer: (c) On adsorption, the residual forces on the surface are increased

Explanation:
1. Adsorption is exothermic, so \( \Delta H \) is negative and heat is released.
2. Gas molecules become restricted on the surface, so \( \Delta S \) is negative.
3. Adsorption satisfies the unbalanced residual forces on the surface, so they decrease, not increase.

Teacher's Note:
1. Spontaneity needs \( \Delta G < 0 \); with \( \Delta S < 0 \) it is \( \Delta H \) that must drive adsorption.
2. Hence adsorption is favoured at lower temperature.

Answer: (c) \( \log (x/m) \) vs \( \log p \)

Explanation:
1. \( \log (x/m) = \log K + \tfrac1n \log p \) is linear in \( \log p \).
2. So the straight line is \( \log (x/m) \) against \( \log p \).

Teacher's Note:
1. Plain \( x/m \) against \( p \) gives a curve.
2. The same form is used to find K and n experimentally.

Answer: (c) both are exothermic

Explanation:
1. Both processes release energy as the adsorbate settles on the surface.
2. Hence both have negative \( \Delta H \); only the magnitudes differ.

Teacher's Note:
1. Spontaneous adsorption always has \( \Delta G < 0 \) and \( \Delta S < 0 \), so \( \Delta H \) must be negative.
2. Physisorption has smaller heat than chemisorption.

Answer: (b) Gelatin

Explanation:
1. Protective colloids are lyophilic colloids that surround lyophobic particles and prevent coagulation.
2. Gelatin is a lyophilic colloid with very low gold number; the others are lyophobic.

Teacher's Note:
1. Other protective colloids: gum arabic, starch, albumin.
2. Used in ice-cream and photographic emulsions.

Answer: (c) \( \ce{Al2(SO4)3} \)

Explanation:
1. Gold sol is negative, so the effective ion is the cation.
2. Highest cation charge is \( \ce{Al^3+} \), so \( \ce{Al2(SO4)3} \) is most effective.

Teacher's Note:
1. Note that \( \ce{K4[Fe(CN)6]} \) has a highly charged anion, which is not the effective ion here.
2. Hardy-Schulze rule: power increases with valence of the oppositely charged ion.

Answer: (d) \( \ce{[Fe(CN)6]^4-} \)

Explanation:
1. For a positive sol the effective ions are anions.
2. The ion with the highest charge, \( \ce{[Fe(CN)6]^4-} \), has the greatest coagulating power and therefore the lowest flocculation value.

Teacher's Note:
1. Flocculation value is inversely related to coagulating power.
2. Charge order here: 4- > 3- > 2- > 1-.