JEE Chemistry Surface Chemistry MCQs Set 10

Answer: (c) accumulates on the surface of the other substance

Explanation:
1. Adsorption is the accumulation of a substance on the surface of another (absorption is uniform penetration throughout the bulk).

Teacher's Note:
1. Physical adsorption is weak (van der Waals), reversible, multilayer and decreases with a rise in temperature.
2. Chemisorption involves chemical bonds, is specific, irreversible and forms a single layer.

Answer: (b) temperature decreases

Explanation:
1. Adsorption is exothermic, so by Le Chatelier's principle lowering the temperature favours it.

Teacher's Note:
1. Physical adsorption is weak (van der Waals), reversible, multilayer and decreases with a rise in temperature.
2. Chemisorption involves chemical bonds, is specific, irreversible and forms a single layer.

Answer: (a) one

Explanation:
1. Chemisorption involves bond formation with surface atoms, so it stops at a unimolecular layer.

Teacher's Note:
1. Physical adsorption is weak (van der Waals), reversible, multilayer and decreases with a rise in temperature.
2. Chemisorption involves chemical bonds, is specific, irreversible and forms a single layer.

Answer: (d) It is independent of temperature

Explanation:
1. Chemisorption needs activation energy, so it first increases with temperature and then falls; it is not temperature independent.
2. It is slow, irreversible and specific.

Teacher's Note:
1. Physical adsorption is weak (van der Waals), reversible, multilayer and decreases with a rise in temperature.
2. Chemisorption involves chemical bonds, is specific, irreversible and forms a single layer.

Answer: (b) exothermic

Explanation:
1. Adsorption decreases the residual forces and the freedom of the adsorbate, so both enthalpy and entropy decrease; \( \Delta G<0 \) requires \( \Delta H<0 \).

Teacher's Note:
1. Physical adsorption is weak (van der Waals), reversible, multilayer and decreases with a rise in temperature.
2. Chemisorption involves chemical bonds, is specific, irreversible and forms a single layer.

Answer: (d) gel

Explanation:
1. Liquid in solid is a gel (e.g. cheese, jelly).
2. Aerosol is solid or liquid in gas, sol is solid in liquid and foam is gas in liquid or solid.

Teacher's Note:
1. Learn the standard catalysts: Fe for Haber, V2O5 for the contact process, Ni for hydrogenation, AlCl3 for Friedel-Crafts.
2. Colloid properties depend on particle size (1 to 1000 nm): the Tyndall effect, Brownian motion and electrophoresis are key tests.

Answer: (c) platinum black

Explanation:
1. Finely divided platinum (and palladium) are outstanding adsorbents for hydrogen.

Teacher's Note:
1. Physical adsorption is weak (van der Waals), reversible, multilayer and decreases with a rise in temperature.
2. Chemisorption involves chemical bonds, is specific, irreversible and forms a single layer.

Answer: (d) \( P^{1/n} \)

Explanation:
1. \( \frac{x}{m}=kP^{1/n} \), with n greater than 1.

Teacher's Note:
1. Physical adsorption is weak (van der Waals), reversible, multilayer and decreases with a rise in temperature.
2. Chemisorption involves chemical bonds, is specific, irreversible and forms a single layer.

Answer: (c) It may be multimolecular

Explanation:
1. Chemisorption is restricted to a monolayer.
2. Multilayer adsorption is a feature of physical adsorption.

Teacher's Note:
1. Physical adsorption is weak (van der Waals), reversible, multilayer and decreases with a rise in temperature.
2. Chemisorption involves chemical bonds, is specific, irreversible and forms a single layer.

Answer: (a) chemisorption

Explanation:
1. Strong chemical bonds between adsorbate and surface define chemisorption.

Teacher's Note:
1. Physical adsorption is weak (van der Waals), reversible, multilayer and decreases with a rise in temperature.
2. Chemisorption involves chemical bonds, is specific, irreversible and forms a single layer.

Answer: (d) activated coconut charcoal

Explanation:
1. Noble gases are adsorbed physically, and charcoal (especially cooled) is the usual adsorbent for them.

Teacher's Note:
1. Physical adsorption is weak (van der Waals), reversible, multilayer and decreases with a rise in temperature.
2. Chemisorption involves chemical bonds, is specific, irreversible and forms a single layer.

Answer: (a) It will decrease

Explanation:
1. Physical adsorption is exothermic and weak; higher temperature releases the adsorbed gas.

Teacher's Note:
1. Physical adsorption is weak (van der Waals), reversible, multilayer and decreases with a rise in temperature.
2. Chemisorption involves chemical bonds, is specific, irreversible and forms a single layer.

Answer: (c) The catalyst influences the direct and the reverse reactions to the same extent

Explanation:
1. A catalyst lowers the activation energy of both directions by the same amount, so K is unchanged and equilibrium is reached faster.

Teacher's Note:
1. A catalyst lowers the activation energy and speeds up the forward and reverse reactions equally.
2. It does not change the equilibrium constant, \( \Delta H \) or the extent of the reaction.

Answer: (a) finely divided iron is used as catalyst

Explanation:
1. Iron is the catalyst, with molybdenum (a promoter) increasing its activity.

Teacher's Note:
1. Learn the standard catalysts: Fe for Haber, V2O5 for the contact process, Ni for hydrogenation, AlCl3 for Friedel-Crafts.
2. Colloid properties depend on particle size (1 to 1000 nm): the Tyndall effect, Brownian motion and electrophoresis are key tests.

Answer: (b) poison

Explanation:
1. Arsenic compounds bind to the platinum surface and reduce its activity: a catalytic poison.

Teacher's Note:
1. A catalyst lowers the activation energy and speeds up the forward and reverse reactions equally.
2. It does not change the equilibrium constant, \( \Delta H \) or the extent of the reaction.

Answer: (b) oxide of nitrogen

Explanation:
1. Oxides of nitrogen (NO) act as a homogeneous catalyst in the lead chamber process.

Teacher's Note:
1. Learn the standard catalysts: Fe for Haber, V2O5 for the contact process, Ni for hydrogenation, AlCl3 for Friedel-Crafts.
2. Colloid properties depend on particle size (1 to 1000 nm): the Tyndall effect, Brownian motion and electrophoresis are key tests.

Answer: (c) vanadium pentoxide

Explanation:
1. Vanadium pentoxide (\( \ce{V2O5} \)) is the modern catalyst (platinised asbestos was used earlier but is easily poisoned).

Teacher's Note:
1. Learn the standard catalysts: Fe for Haber, V2O5 for the contact process, Ni for hydrogenation, AlCl3 for Friedel-Crafts.
2. Colloid properties depend on particle size (1 to 1000 nm): the Tyndall effect, Brownian motion and electrophoresis are key tests.

Answer: (d) they always increase activation energy

Explanation:
1. Enzymes act as catalysts and therefore lower the activation energy.

Teacher's Note:
1. A catalyst lowers the activation energy and speeds up the forward and reverse reactions equally.
2. It does not change the equilibrium constant, \( \Delta H \) or the extent of the reaction.

Answer: (a) zymase

Explanation:
1. Zymase (in yeast) ferments glucose to ethanol and carbon dioxide.

Teacher's Note:
1. Learn the standard catalysts: Fe for Haber, V2O5 for the contact process, Ni for hydrogenation, AlCl3 for Friedel-Crafts.
2. Colloid properties depend on particle size (1 to 1000 nm): the Tyndall effect, Brownian motion and electrophoresis are key tests.

Answer: (b) anhydrous \( \ce{AlCl3} \)

Explanation:
1. This is Friedel-Crafts alkylation; anhydrous AlCl3 is the Lewis acid catalyst.

Teacher's Note:
1. Learn the standard catalysts: Fe for Haber, V2O5 for the contact process, Ni for hydrogenation, AlCl3 for Friedel-Crafts.
2. Colloid properties depend on particle size (1 to 1000 nm): the Tyndall effect, Brownian motion and electrophoresis are key tests.

Answer: (a) It changes the equilibrium constant

Explanation:
1. A catalyst does not change the equilibrium constant (K depends only on temperature).

Teacher's Note:
1. A catalyst lowers the activation energy and speeds up the forward and reverse reactions equally.
2. It does not change the equilibrium constant, \( \Delta H \) or the extent of the reaction.

Answer: (b) maltose → glucose

Explanation:
1. Maltase hydrolyses maltose into two molecules of glucose.
2. Fructose is formed from sucrose (invertase).

Teacher's Note:
1. Learn the standard catalysts: Fe for Haber, V2O5 for the contact process, Ni for hydrogenation, AlCl3 for Friedel-Crafts.
2. Colloid properties depend on particle size (1 to 1000 nm): the Tyndall effect, Brownian motion and electrophoresis are key tests.

Answer: (d) Ni

Explanation:
1. Finely divided nickel is used industrially (Sabatier-Senderens).

Teacher's Note:
1. Learn the standard catalysts: Fe for Haber, V2O5 for the contact process, Ni for hydrogenation, AlCl3 for Friedel-Crafts.
2. Colloid properties depend on particle size (1 to 1000 nm): the Tyndall effect, Brownian motion and electrophoresis are key tests.

Answer: (b) a catalyst that is a product of the reaction

Explanation:
1. In autocatalysis, the product formed catalyses the reaction, e.g. \( \ce{Mn^2+} \) in the oxalic acid and permanganate reaction.

Teacher's Note:
1. Learn the standard catalysts: Fe for Haber, V2O5 for the contact process, Ni for hydrogenation, AlCl3 for Friedel-Crafts.
2. Colloid properties depend on particle size (1 to 1000 nm): the Tyndall effect, Brownian motion and electrophoresis are key tests.

Answer: (a) oxidation of ammonia to form nitric acid

Explanation:
1. The Ostwald process uses platinum-rhodium gauze for the oxidation of ammonia.
2. Hardening of oils uses Ni, and methanol synthesis uses ZnO/\( \ce{Cr2O3} \).

Teacher's Note:
1. Learn the standard catalysts: Fe for Haber, V2O5 for the contact process, Ni for hydrogenation, AlCl3 for Friedel-Crafts.
2. Colloid properties depend on particle size (1 to 1000 nm): the Tyndall effect, Brownian motion and electrophoresis are key tests.