NCERT Solutions | Class 12 Chemistry Chapter 5

NCERT Solutions | Class 12 Chemistry Chapter 5 | Surface Chemistry 

CBSE Solutions | Chemistry Class 12

Check the below NCERT Solutions for Class 12 Chemistry Chapter 5 Surface Chemistry Pdf free download. NCERT Solutions Class 12 Chemistry  were prepared based on the latest exam pattern. We have Provided Surface Chemistry Class 12 Chemistry NCERT Solutions to help students understand the concept very well.

NCERT | Class 12 Chemistry

NCERT Solutions Class 12 Chemistry

Book:	National Council of Educational Research and Training (NCERT)
Board:	Central Board of Secondary Education (CBSE)
Class:	12th
Subject:	Chemistry
Chapter:	5
Chapters Name:	Surface Chemistry
Medium:	English

Surface Chemistry | Class 12 Chemistry | NCERT Books Solutions

You can refer to MCQ Questions for Class 12 Chemistry Chapter 5 Surface Chemistry to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

NCERT Exercises

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 1.

Why are substances like platinum and palladium often used for carrying out electrolysis of aqueous solutions?

Solution.

Due to their inert nature, these metals do not affect the products of electrolysis. They have good adsorbing capacity for hydrogen.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 2.

Why does physisorption decrease with the increase of temperature?

Solution.

Physical adsorption of a gas by a solid is generally reversible. Thus,
Solid + Gas \(\rightleftharpoons \) Gas / Solid + Heat
Since the adsorption process is exothermic, the physical adsorption occurs readily at low temperature and decreases with increasing temperature (Le Chatelier’s principle).

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 3.

Why are powdered substances more effective adsorbents than their crystalline forms?

Solution.

The extent of adsorption increases with increase in surface area of the adsorbent. Finely powdered substances have large, porous areas and act as good adsorbents.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 4.

Why is it necessary to remove CO when ammonia is obtained by Haber’s process?

Solution.

CO is a catalytic poison. It reacts with iron to form iron carbonyl thus inhibiting the activity of catalyst.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 5.

Why is the ester hydrolysis slow in the beginning and becomes faster after sometime?

Solution.

The acid formed during the reaction provides hydrogen ions which act as catalyst for the reaction and it becomes faster.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 6.

What is the role of desorption in the process of catalysis?

Solution.

In catalysis the products formed are desorbed and detached from the surface so that, more reactants can get adsorbed on the surface of catalyst.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 7.

What modification can you suggest in the Hardy Schulze law?

Solution.

The Hardy Schulze law considers the coagulation of sols because of neutralisation of their charges. Since coagulation can also occur by mixing two oppositely charged sols, it should also include “when oppositely charged sols are mixed in proper proportions to neutralize the charges of each other, the coagulation of both the sols occurs.”

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 8.

Why is it essential to wash the precipitate with water before estimating it quantitatively?

Solution.

Few impurities which are soluble in water and are adsorbed on the surface of the precipitate are removed by washing them with water.

NCERT Exercises

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 1.

Distinguish between the meaning of the terms adsorption and absorption. Give one example of each.

Solution.

The phenomenon of accumulation of the molecules of a substance on a solid or liquid surface resulting in the increased concentration of the molecules on the surface is called adsorption. In absorption, the substance is uniformly distributed throughout the bulk of the solution. A distinction can be made by taking an example of water vapours. Water vapours are absorbed by anhydrous calcium chloride but adsorbed by silica gel.

	Adsorption	Absorption
1.	It is a surface phenomenon, i.e., it occurs only on the adsorbent surface.	It occurs throughout the body of the material. It is called bulk phenomenon.
2.	The concentration on the adsorbent’s surface is different from that in the bulk.	The concentration is same throughout.
3.	The rate varies throughout the process.	The rate remains the same.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 2.

What is the difference between physisorption and chemisorption?

Solution.

Physisorption		Chemisorption
1.	The adsorbate and adsorbent are held by weak van der Waals forces.	The adsorbate and adsorbent are held by forces similar to a chemical bond.
2.	Heat of adsorption is of the order of 20 kj/mol.	Heat of adsorption is of the order of 200 kj/mol.
3.	It is reversible.	It is irreversible.
4.	It decreases with increase in temperature and occurs at lower temperatures.	It increases with temperature and occurs at high temperature.
5.	It is not specific in nature, i.e., all gases are adsorbed on all solids to some extent.	It is specific in nature and occurs only when a chemical bond is formed between the adsorbate and adsorbent.
6.	Multimolecular layers may be formed on the adsorbent.	Usually unimolecular layer is formed on the adsorbent.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 3.

Give reason why a finely divided substance is more effective as an adsorbent.

Solution.

The extent of adsorption increases with increase in surface area of the adsorbent. Thus, finely divided metals and porous substances having large surface areas are good adsorbents.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 4.

What are the factors which influence the adsorption of a gas on a solid?

Solution.

Factors affecting adsorption of a gas on solids are :

Nature of the adsorbent : The same gas is adsorbed to different extents by different solids at the same temperature. Also, greater the surface area of the adsorbent, more is the gas adsorbed.

Nature of the adsorbate : Different gases are adsorbed to different extents by different solids at the same temperature. Higher the critical temperature of the gas, greater is its amount adsorbed.

Temperature : Since adsorption is an exothermic process, applying Le Chatelier’s principle, we can find out that adsorption decreases with an increase in temperature.

Specific area of the adsorbent : Surface area available for adsorption per gram of the adsorbent increases the extent of adsorption. Greater the surface area, higher would be the adsorption therefore, porous or powdered adsorbents are used.

Pressure : At constant temperature, the adsorption of gas increases with pressure.

Activation of adsorbent : It means increasing the adsorbing power of an adsorbent by increasing its surface area. It is done by :

1. making the adsorbent’s surface rough
2. removing gases already adsorbed
3. subdividing the adsorbent into smaller pieces.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 5.

What is an adsorption isotherm? Describe Freundlich adsorption isotherm.

Solution.

Adsorption isotherm is a graph between the amount of the gas adsorbed by an adsorbent and equilibrium pressure of the adsorbate at constant temperature. Freundlich obtained an empirical relationship between the quantity of gas adsorbed by unit mass of solid adsorbent and pressure at a particular temperature. It is mathematically represented in the following way :

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 6.

What do you understand by activation of adsorbent? How is it achieved?

Solution.

Activation of an adsorbent means increasing its adsorbing power by increasing the surface area of the adsorbent by making its surface rough, and removing adsorbed gases from it. With an increase in surface area the adsorption increases.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 7.

What role does adsorption play in heterogenous catalysis?

Solution.

Adsorption of reactants on solid surface of the catalysts increases the rate of reaction. There are many gaseous reactions of industrial importance involving solid catalysts. Manufacture of ammonia using iron as a catalyst, manufacture of H₂SO₄ by Contact process and use of finely divided nickel in the hydrogenation of oils are excellent examples of heterogeneous catalysis.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 8.

Why is adsorption always exothermic?

Solution.

During adsorption, there is always a decrease in residual forces of the surface, i.e., there is decrease in surface energy which appears as heat. Adsorption therefore, is invariably an exothermic process. In other words, ∆H of adsorption is always negative to keep the value of ∆G negative for the reaction to be spontaneous as ∆S decreases during adsorption.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 9.

How are the colloidal solutions classified on the basis of physical states of the dispersed phase and dispersion medium?

Solution.

Colloids can be classified into eight types depending upon the physical state of the dispersed phase and the dispersion medium.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 10.

Discuss the effect of pressure and temperature on the adsorption of gases on solids.

Solution.

1. Adsorption decreases with an increase in, temperature because it is an exothermic process and according to Le Chatelier’s principle the reaction will proceed in backward direction with increase in temperature.
2. At a constant temperature, adsorption increases with pressure.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 11.

What are lyophilic and lyophobic sols? Give one example of each type. Why are hydrophobic sols easily coagulated?

Solution.

There are two types of colloidal sols :

(i) Lyophilic sols : The word lyophilic means solvent loving. They are obtained by directly mixing the dispersed phase and the dispersion medium, e.g., sols of gum, gelatin, starch, etc. They are solvent attracting hence quite stable and cannot be coagulated easily.

(ii) Lyophobic sols : They cannot be prepared by directly mixing the dispersed phase and dispersion medium but are prepared by special methods, e.g., sols of metals. They are solvent repelling. Hydrophobic sols are easily coagulated due to repulsion between water and dispersed phase.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 12.

What is the difference between multimolecular and macromoiecular colloids? Give one example of each. How are associated colloids different from these two types of colloids?

Solution.

Depending upon the type of particles of the dispersed phase, colloids are classified as : multimolecular, macromoiecular and associated colloids.

(i) Multimolecular colloids : On dissolution, a large number of atoms or smaller molecules of a substance aggregate together to form species having size in the colloidal range (diameter < 1 nm). The species thus formed are called multimolecular colloids. For example, a gold sol may contain particles of various sizes having many atoms. Sulphur sol consists of particles containing a thousand or more of S8 sulphur molecules.

(ii) Macromoiecular colloids : Macromolecules in suitable solvents form solutions in which the size of the macromolecules may be in the colloidal range. Such systems are called macromoiecular colloids. These colloids are quite stable and resemble true solutions in many respects. Examples of naturally occurring macromolecules are starch, cellulose, proteins and enzymes; and those of man-made macromolecules are polythene, nylon, polystyrene, synthetic rubber, etc.

(iii) Associated colloids (Micelles) : There are some substances which at low concentrations behave as normal strong electrolytes but, at higher concentrations exhibit colloidal behaviour due to the formation of aggregates. The aggregated particles thus formed are called micelles. These are also known as associated colloids. The formation of micelles takes place only above a particular temperature called Kraft temperature (Tk) and above a particular concentration called critical micelle concentration (CMC). On dilution, these colloids revert back to individual ions. Surface active agents such as soaps and synthetic detergents belong to this class. For soaps, the CMC is 10^-4 to 10^-3 mol L^-1. These colloids have both lyophobic and lyophilic parts. Micelles may contain as many as 100 molecules or more.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 13.

What are enzymes? Write in brief the mechanism of enzyme catalysis.

Solution.

Enzymes are complex nitrogenous organic compounds which act as biological catalysts and increase the rate of cellular processes. According to the lock and key model, like every lock has a specific key, similarly every enzyme acts at a specific substrate.

When the substrate fits the active site (lock) of the enzyme, the chemical change begins. But it has also been noticed that enzyme changes shape, when substrate lands at the active site. This induced-fit model of enzyme action pictures the substrate inducing the 1 active site to adopt a perfect fit, rather than a rigid shaped lock and key. Therefore, the new model for enzyme action is called induced fit model.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 14.

How are colloids classified on the basis of

1. physical states of components
2. nature of dispersion medium and
3. interaction between dispersed phase and dispersion medium?

Solution.

1. Refer answer number 9.
2. Depending upon the nature of dispersion medium colloids can be classified as sol if the dispersion medium is liquid, gel if the dispersion medium is solid. If the dispersion medium is water, the sol is called hydrosol and if the dispersion medium is alcohol, it is called alcosol. A colloid in which the dispersion medium as well as dispersed phase are liquids, is called emulsion.
3. Refer answer number 11.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 15.

Explain what is observed

1. when a beam of light is passed through a colloidal sol.
2. an electrolyte, NaCI is added to hydrated ferric oxide sol.
3. electric current is passed through a colloidal sol?

Solution.

(i) When a beam of light is passed through colloidal particles, its path becomes clearly visible and is known as Tyndall effect. It is due to scattering of light by colloidal particles. The bright cone of the light is called Tyndall cone.

(ii) When NaCI is added to hydrated ferric oxide sol coagulation takes place. Since ferric oxide is a positive sol, it is coagulated by the negative chloride ions.

(iii) When electric potential is applied across two platinum electrodes dipped in a colloidal solution, the colloidal particles move towards one or the other electrode. The movement of colloidal particles under an applied electric potential is called electrophoresis.

Positively charged particles move towards the cathode while negatively charged particles move towards the anode. Since all the colloidal particles in a given colloidal solution carry the same charge, the particles move to one or the other electrode depending on the charge.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 16.

What are emulsions? What are their different types? Give example of each type.

Solution.

These are liquid-liquid colloidal systems, i.e., the dispersion of finely divided droplets in another liquid. If a mixture of two immiscible or partially miscible liquids is shaken, a coarse dispersion of one liquid in the other is obtained which is called emulsion. Generally, one of the two liquids is water. There are two types of emulsions :

1. Oil dispersed in water (O/W type) and
2. Water dispersed in oil (W/O type).

In the first system, water acts as dispersion medium. Examples of this type of emulsion are milk and vanishing cream. In milk, liquid fat is dispersed in water. In the second system, oil acts as dispersion medium. Common examples of this type are butter and cream.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 17.

What is demulsification? Name two demulsifiers.

Solution.

The process of converting the emulsion back into two distinct components, oil and water is called demulsification. This can be done by

1. boiling
2. freezing
3. changing pH
4. electrostatic precipitation.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 18.

Action of soap is due to emulsification and micelle formation. Comment.

Solution.

Soap is sodium or potassium salt of a higher fatty acid and may be represented as RCOO^– Na⁺ (e.g., sodium stearate CH₃(CH₂)]₁₆ COO^– Na⁺, which is a major component of many bar soaps). When dissolved in water, it dissociates into RCOO^– and Na+ ions. The RCOO^– ions, however, consist of two parts – a long hydrocarbon chain R (also called non-polar ‘tail’) which is hydrophobic (water repelling), and a polar group COO^– (also called polar-ionic ‘head’), which is hydrophilic (water loving).

The RCOO^– ions are, therefore, present on the surface with their COO^– groups in water and the hydrocarbon chains R staying away from it and remain at the surface. But at critical micelle concentration, the anions are pulled into the bulk of the solution and aggregate to form a spherical shape with their hydrocarbon chains pointing towards the centre of the sphere with COO^– part remaining outward on the surface of the sphere. An aggregate thus formed is known as ‘ionic micelle’.

The cleansing action of soap is due to the fact that soap molecules form micelle around the oil droplet in such a way that hydrophobic part of the stearate ions is in the oil droplet and hydrophilic part projects out of the grease droplet like the bristles

(a) Grease on cloth
(b) Stearate ions (from soap) arranging around the grease droplets
(c) Micelle formed

Since the polar groups can interact with water, the oil droplet surrounded by stearate ions is now pulled in water and removed from the dirty surface. Thus soap helps in emulsification and washing away of oils and fats. Tne negatively charged sheath around the globules prevents them from coming together and forming aggregates.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 19.

Give four examples of heterogeneous catalysis.

Solution.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 20.

What do you mean by activity and selectivity of catalysts?

Solution.

(a) Activity : The activity of a catalyst depends upon the strength of chemisorption to a large extent. The reactants must get adsorbed reasonably strongly on to the catalyst to become active. But adsorption must not be so strong that they are immobilised. It is observed that maximum activity is shown by elements of groups 7 – 9 of the periodic table
2H₂ + O₂ \(\underrightarrow { Pt }\) 2H₂O

(b) Selectivity : The selectivity of a calatyst is its ability to yield a particular product in the reaction e.g.,

Thus, a selective catalyst can act as a catalyst in one reaction and may fail to catalyse another reaction.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 21.

Describe some features of catalysis by zeolites.

Solution.

(a) Zeolites are hydrated aluminosilicates which have a three dimensional network structure containing water molecules in their pores.
(b) The pores are made vacant by heating before catalysis.
(c) The reactions taking place in zeolites depend upon the size and shape of reactant and product molecules and also on the pores and cavities in them, e.g., ZSM-5 converts alcohols to hydrocarbons by dehydrating them.

Alcohols \(\underrightarrow { ZSM-5 }\) Hydrocarbons

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 22.

What is shape selective catalysis?

Solution.

The catalytic reaction that depends upon the pore structure of the catalyst and the size of the reactant and product molecules is called shape-selective catalysis. Zeolites are good shape-selective catalysts because of their honeycomb-like structures. They are microporous aluminosilicates with three dimensional network of silicates in which some silicon atoms are replaced by aluminium atoms giving Al-O-Si framework. The reactions taking place in zeolites depend upon the size and shape of reactant and product molecules as well as upon the pores and cavities of the zeolites. They are found in nature as well as synthesised for catalytic selectivity.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 23.

Explain the following terms :

1. Electrophoresis
2. Coagulation
3. Dialysis
4. Tyndall effect

Solution.

(i) Electrophoresis : Refer answer number 15 (iii)

(ii) Coagulation or precipitation : The stability of the lyophobic sols is due to the presence of charge on colloidal particles. If somehow, the charge is removed, the particles will come nearer to each other to form aggregates (or coagulate) and settle down under the force of gravity. The process of settling down of colloidal particles is called coagulation.

(iii) Dialysis : It is the process of removing dissolved substances from a colloidal solution by means of diffusion through a suitable membrane. Since particles (ions or smaller molecules) in a true solution can pass through animal membrane (bladder) or parchment paper or cellophane sheet but not the colloidal particles, the membrane can be used for dialysis. The apparatus used for this purpose is called dialyser. A bag of suitable membrane containing the colloidal solution is suspended in a vessel through which fresh water is continuously flowing. The molecules and ions diffuse through membrane into the outer water and pure colloidal solution is left behind.

(iv) Tyndall effect : Refer answer number 15 (i)

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 24.

Give four uses of emulsions.

Solution.

1. Some of the medicines are effective as emulsions.
2. Paints are emulsions which are used in our daily life.
3. Soaps and detergents act as cleansing agents, action of which is based on emulsification.
4. Photographic films are coated with emulsion of AgBr on its surface.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 25.

What are micelles? Give an example of a micelle system.

Solution.

Micelles are substances that behave as normal strong electrolytes at low concentration but at high concentrations behave as colloids due to formation of aggregates. They are also called associated colloids, e.g., soaps and detergents. They can form ions and may contain 100 or more molecules to form a micelle.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 26.

Explain the terms with suitable examples :

1. Alcosol
2. Aerosol
3. Hydrosol

Solution.

(i) Alcosol : The sol in which alcohol is used as dispersion medium is called alcosol e.g., sol of cellulose nitrate in ethyl alcohol.

(ii) Aerosol : The sol in which dispersion medium is gas and dispersed phase is either solid or liquid, the colloidal system is called aerosol e.g., fog, insecticides, sprays, etc.

(iii) Hydrosol : The sol in which dispersion medium is water is called hydrosol e.g., starch sol.

NCERT Solutions for Class 12 Chemistry Chapter 5, Question 27.

Comment on the statement that’colloid is not a substance but a state of substance’.

Solution.

Colloid is not a substance, but a state of substance because the same substance may exist as a colloid or crystalloid under different conditions e.g., sulphur. Colloidal solution of sulphur consists of sulphur molecules dispersed in water. In this state, sulphur atoms combine to form multimolecules whose size lies between 1 nm to 1000 nm and form colloidal state. Sulphur forms true solution in carbon disulphide. Similarly soap is a solution at low concentration but a colloid at higher concentration.

NCERT Class 12 Chemistry

Class 12 Chemistry Chapters | Chemistry Class 12 Chapter 5

Chapterwise NCERT Solutions for Class 12 Chemistry

NCERT Solutions For Class 12 Chemistry Chapter 1 The Solid State
NCERT Solutions For Class 12 Chemistry Chapter 2 Solutions
NCERT Solutions For Class 12 Chemistry Chapter 3 Electro chemistry
NCERT Solutions For Class 12 Chemistry Chapter 4 Chemical Kinetics
NCERT Solutions For Class 12 Chemistry Chapter 5 Surface Chemistry
NCERT Solutions For Class 12 Chemistry Chapter 6 General Principles and Processes of Isolation of Elements
NCERT Solutions For Class 12 Chemistry Chapter 7 The p Block Elements
NCERT Solutions For Class 12 Chemistry Chapter 8 The d and f Block Elements
NCERT Solutions For Class 12 Chemistry Chapter 9 Coordination Compounds
NCERT Solutions For Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes
NCERT Solutions For Class 12 Chemistry Chapter 11 Alcohols Phenols and Ethers
NCERT Solutions For Class 12 Chemistry Chapter 12 Aldehydes Ketones and Carboxylic Acids
NCERT Solutions For Class 12 Chemistry Chapter 13 Amines
NCERT Solutions For Class 12 Chemistry Chapter 14 Biomolecules
NCERT Solutions For Class 12 Chemistry Chapter 15 Polymers
NCERT Solutions For Class 12 Chemistry Chapter 16 Chemistry in Everyday Life

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