Online Test: Surface Chemistry | Class 12 Chemistry

Question 1 of 20

Adsorption is best described as:

a surface phenomenon

a bulk phenomenon

an endothermic bulk process

distribution throughout the solid

Explanation: In adsorption the substance concentrates at the surface, unlike absorption which is a bulk process.

Question 2 of 20

When both adsorption and absorption occur together the process is called:

desorption

sorption

occlusion

condensation

Explanation: Sorption is the combined occurrence of adsorption and absorption.

Question 3 of 20

Physical adsorption is favoured by:

high temperature

low temperature

low pressure only

chemical bonding

Explanation: Physisorption is exothermic, so it increases as temperature is lowered.

Question 4 of 20

The enthalpy of chemisorption is typically of the order:

$20$–$40\ \text{kJ mol}^{-1}$

$2$–$4\ \text{kJ mol}^{-1}$

$80$–$240\ \text{kJ mol}^{-1}$

zero

Explanation: Chemisorption involves chemical bonds, so its enthalpy ($80$–$240\ \text{kJ mol}^{-1}$) is much higher than physisorption.

Question 5 of 20

In $\frac{x}{m}=k\,p^{1/n}$, the value of $\frac{1}{n}$ lies between:

$1$ and $2$

$0$ and $1$

$2$ and $3$

$-1$ and $0$

Explanation: Since $n\ge 1$, $\frac{1}{n}$ lies between $0$ and $1$.

Question 6 of 20

A plot of $\log\frac{x}{m}$ versus $\log p$ gives a straight line with intercept equal to:

$\frac{1}{n}$

$n$

$\log k$

$k$

Explanation: From $\log\frac{x}{m}=\log k+\frac{1}{n}\log p$, the intercept is $\log k$ and the slope is $\frac{1}{n}$.

Question 7 of 20

Which gas is adsorbed to the greatest extent on activated charcoal?

$\text{SO}_2$

$\text{N}_2$

$\text{H}_2$

$\text{O}_2$

Explanation: Easily liquefiable gases with higher critical temperature, such as $\text{SO}_2$, are adsorbed most.

Question 8 of 20

A catalyst speeds up a reaction by:

raising the temperature

increasing $\Delta H$

providing a path of lower activation energy

shifting the equilibrium right

Explanation: A catalyst offers an alternative path of lower activation energy without changing $\Delta G$ or equilibrium.

Question 9 of 20

The Haber process for ammonia uses the catalyst:

finely divided iron

$\text{V}_2\text{O}_5$

platinum gauze

nickel

Explanation: Finely divided iron (with Mo as promoter) catalyses $\text{N}_2+\text{H}_2\to\text{NH}_3$.

Question 10 of 20

Oxidation of $\text{SO}_2$ by $\text{NO}$ in the lead-chamber process is an example of:

heterogeneous catalysis

homogeneous catalysis

enzyme catalysis

no catalysis

Explanation: All species are gaseous, so the catalyst and reactants are in the same phase: homogeneous catalysis.

Question 11 of 20

The order of catalytic activity for hydrogenation is generally:

$\text{Fe}>\text{Ni}>\text{Pd}>\text{Pt}$

$\text{Ni}>\text{Pt}>\text{Pd}>\text{Fe}$

$\text{Pt}>\text{Pd}>\text{Ni}>\text{Fe}$

$\text{Pd}>\text{Fe}>\text{Pt}>\text{Ni}$

Explanation: Activity for hydrogenation decreases in the order $\text{Pt}>\text{Pd}>\text{Ni}>\text{Fe}$.

Question 12 of 20

Zeolites act as shape-selective catalysts because they:

are highly coloured

have pores of molecular dimensions

are liquids

have no surface area

Explanation: Their microporous structure admits only molecules of suitable size and shape, giving shape selectivity.

Question 13 of 20

Which statement about enzymes is correct?

They are inorganic salts

They are highly specific protein catalysts

They work best at very high temperatures

They are consumed in the reaction

Explanation: Enzymes are protein catalysts, highly specific and efficient, active near an optimum temperature and pH.

Question 14 of 20

The size of colloidal particles lies in the range:

$<1\ \text{nm}$

$1$–$1000\ \text{nm}$

$>1000\ \text{nm}$

$1$–$10\ \text{mm}$

Explanation: Colloidal particles are $1$–$1000\ \text{nm}$; below this is a true solution and above is a suspension.

Question 15 of 20

Gold sol and sulphur sol are examples of:

macromolecular colloids

associated colloids

multimolecular colloids

true solutions

Explanation: They are aggregates of a large number of small atoms/molecules, i.e. multimolecular colloids.

Question 16 of 20

Purification of a sol using a parchment membrane is called:

peptisation

electrophoresis

coagulation

dialysis

Explanation: Dialysis removes dissolved electrolyte by diffusion of ions through a parchment/semipermeable membrane.

Question 17 of 20

For a positively charged sol, the most effective coagulating ion is:

$[\text{Fe(CN)}_6]^{4-}$

$\text{Na}^+$

$\text{Ba}^{2+}$

$\text{Al}^{3+}$

Explanation: A positive sol is coagulated by anions, and by the Hardy-Schulze rule the highest-charge anion $[\text{Fe(CN)}_6]^{4-}$ is most effective.

Question 18 of 20

Butter is an emulsion of the type:

oil in water

gas in liquid

water in oil

solid in solid

Explanation: Butter is water dispersed in oil/fat, a water-in-oil (w/o) emulsion.

Question 19 of 20

The conversion of a fresh precipitate into a colloidal sol by adding a little electrolyte is called:

dialysis

coagulation

peptisation

electrodialysis

Explanation: Peptisation is a dispersion method that turns a fresh precipitate into a sol using a peptising electrolyte.

Question 20 of 20

Alum is used to purify muddy water because it:

dissolves the clay

coagulates the negatively charged clay sol

adds a pleasant colour

forms a true solution

Explanation: The $\text{Al}^{3+}$ ions coagulate the negatively charged colloidal clay particles, which then settle.

Online Test — Surface Chemistry