CBSE Class–12 Subject Chemistry
Chapter-5 : Surface Chemistry
NCERT Solutions For Class 12 Chemistry Chapter 5 Surface Chemistry
NCERT TEXTBOOK QUESTIONS SOLVED
5.1. Write any two characteristics of Chemisorption.
Ans: Pt and Pd form inert electrodes, i.e., they are not attacked by the ions of the electrolyte or the products of electrolysis. Hence, they are used as electrodes for carrying out electrolysis.
5.2.Why does physisorption decrease with the increase of temperature?
Ans: Physisorption is an exothermic process :
According to Le-Chatelier’s principle, if T is increased, equilibrium shifts in the backward direction i.e., gas is released from the surface of solid.
5.3. Why are powdered substances more effective . adsorbents than their crystalline forms?
Ans:Powdered substances have greater surface area as compared to their crystalline forms. Greater the surface area, greater is the adsorption.
5.4.In Haber’s process, hydrogen is obtained by reacting methane with steam in presence of NiO as catalyst. The process is known as steam reforming. Why is it necessary to remove CO when ammonia is obtained by Haber’s process?
Ans: CO acts as a poison for the catalyst used in the manufacture of NH3 by Haber’s process. Hence, it is necessary to remove it.
5.5.Why is the ester-hydrolysis slow in the beginning and becomes faster after sometime?
Ans: The ester hydrolysis takes place as follows :
The acid produced in the reaction acts as an auto catalyst for the reaction. Hence, the reaction becomes faster after some time.
5.6. What is the role of desorption in the process of catalysis.
Ans: Desorption makes the surface of the solid- catalyst free for fresh adsorption of the reactants on the surface.
5.7. What modification can you suggest in the Hardy Schulze, law?
Ans: According to Hardy Schulze law, the coagulating ion has charge opposite to that on the colloidal particles. Hence, the charge on colloidal particles is neutralized and coagulation occurs.
The modification to this law is :
When oppositely charged sols are mixed in proper proportions to neutralize the charges of each other, coagulation of both the sol occurs.
5.8. Why is it essential to wash the precipitate with water before estimating it quantitatively?
Ans: Some amount of the electrolyte are mixed to form the ppt. Some of these electrolytes remains adsorbed on the surface of the particles of the ppt. Hence, it is essential to wash the ppt with water to remove the sticking electrolytes (or any other impurities) before estimating it quantitatively.
5.9. Distinguish between the meaning of the terms adsorption and absorption. Give one example of each.
Ans: This phenomenon of attracting and retaining the molecules of a substance by a solid (or a liquid) on its surface resulting into a higher concentration of the molecules on the surface is known as adsorption.
Absorption is different from adsorption. In absorption, the substance is uniformly distributed throughout the body of a solid or a liquid.
NH3 gets adsorbed on the charcoal where as NH3 when comes in contact with H20 gets absorbed by forming NH4OH solution of uniform concentration.
5.10.What is the difference between physisorption and chemisorption?
5.3. Give reason why a finely divided substance is more effective as an adsorbent?
Solution: Finely divided substance has large surface area and hence greater adsorption.
5.4. What are the factors which influence the adsorption of a gas on a solid?
Solution: The extent of adsorption of a gas on a solid depends upon the following factors:
(a) Nature of the adsorbate,
(b) Nature of the adsorbent,
(c) Temperature, and (d) Pressure.
5.5. What is an adsorption isotherm? Describe Freundlich adsorption isotherm.
Solution: Adsorption isotherm represents the variation of the mass of the gas adsorbed per gram of the adsorbent with pressure at constant temperature. Freundlich Adsorption isotherm:
Freundlich, in 1909, gave an empirical relationship between the quantity, of gas adsorbed by unit mass of solid adsorbent and pressure at a particular temperature. The relationship can be expressed by the following equation:
where x is the mass of the gas adsorbed by mass ‘m’ of the adsorbent at pressure P, k and n are constants which depend on the nature of the adsorbent and the gas at a particular temperature. The relationship is generally represented in the form of a curve where mass of the gas adsorbed per gram by the adsorbent is plotted against pressure. These curves indicate that at a fixed pressure, there is a decrease in physical adsorption with increase in temperature. These curves always seem to approach saturation at high pressure.
Taking log of equation (i), we get
5.6. What do you understand by activation of adsorbent? How is it achieved?
Solution: Activation of an adsorbent means increasing it’s adsorbing power by increasing the surface area of the adsorbent by making it’s surface rough, by removing already adsorbed gases from it and by subdividing the adsorbent into smaller pieces or grains..
5.7. What role does adsorption play in heterogeneous catalysis?
Solution: In heterogenous catalysis, generally the reactants are gaseous whereas catalyst is a solid. The reactant molecules are adsorbed on the surface of the solid catalyst by physical adsorption or chemisorption. As a result, the concentration of the reactant molecules on the surface increases further leading to increase in rate of reaction. Alternatively, one of the reactant molecules undergoes fragmentation on the surface of the solid catalyst producing active species which react faster. The product molecules in either case have no affinity for the solid catalyst and are deadsorbed making the surface free for fresh adsorption.
5.8. Why is adsorption always exothermic?
Solution: When a gas is adsorbed on the surface of a solid, its entropy decreases, i.e., ΔS is negative.
Now, ΔG = ΔH-TΔS For a process to be spontaneous, ΔG must be negative. As here, ΔS is negative, therefore, TΔS is positive ΔG can be negative only if ΔH is negative Hence, adsorption is always exothermic.
5.9. 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.
5.10. Discuss the effect of pressure and temperature on the adsorption of gases on solids.
Solution: (i) Adsorption decreases with an increase in temperature since it is an exothermic process by applying Le Chatelier’s principle the reaction will proceed in backward direction with increase in temperature.
(ii) At a constant temperature, adsorption increases with pressure.
5.11. What are lyophilic and lyophobic sols? Give one example of each type. Why are hydrophobic sols easily coagulated?
Solution: Lyophillic colloids (solvent loving) are those substances that directly pass into the colloidal state when brought- in contact with the solvent, e.g., proteins, starch, rubber, etc.
These sols are quite stable because of the strong attractive forces between the particles of disperse phase and the dispersion medium.
Lyophobic colloids (solvent hating) are those substances that do not form the colloidal sol readily when mixed with the dispersion medium.These sols are less stable than the lyophilic sols. Examples of lyophobic sols include sols of metals and their insoluble compounds like sulphides and hydroxides.
The stability of hydrophobic sol is only due to the presence of charge on the colloidal parties. If charge is removed, e.g., by addition of suitable electrolytes, the particles will come nearer to each other to form aggregate, i.e., they will coagulate and settle down. On the other hand, the stability of hydrophilic sol is due to charge as well as solvation of the colloidal particles. Thuf, for coagulation to occur easily both the mentioned factors have to be removed.
5.12. What is the difference between multimolecular and macromolecular colloids? Give one example of each.
How a re associated colloids different from these two types of colloids?
Solution: Comparison of some important characteristic of multimolecular, macro molecular and associated colloids.
5.13. What are enzymes? Write in brief the mechanism of enzyme catalysis.
Solution: Enzymes are complex nitrogenous organic compounds which act as a 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 as 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 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.
5.14. How are colloids classified on the basis of
(i) physical states of components
(ii) nature of dispersed phase and
(iii) interaction between dispersed phase and dispersion medium?
Solution: (i) Colloids can be classified into eight types depending upon the physical state of the dispersed phase and the dispersion medium.
(ii) Depending upon the type of the particles of the dispersed phase, colloids are classified as : Multimolecular, macro- molecular and associated colloids
(a) Multimolecular colloids: The colloids
in which the colloidal particles consist of aggregates of atoms or small molecules are called multimolecular colloids .
For Example: gold sol, sulphur sol etc.
(b) Macromolecular colloids : The colloids in which large particles of colloidal range having high molecular masses are dissolved in a suitable liquid are called macromolecular colloids.
Example: proteins, starch and cellulose form macromolecular colloids.
(c) Associated colloids (Micelles): Those colloids which behave as normal strong electrolyte at low concentration but show colloidal properties at high concentration due to the formation of aggregated particles of colloidal dimension’s. Such substances are also referred to as associated colloids.
(iii) (a) Lyophillic colloids (solvent loving) are those substances that directly pass into the colloidal state when brought in contact with the solvent, e.g., proteins, starch, rubber, etc.
These sols are quite stable because of the strong attractive forces between the particles of dispersed phase and dispersion medium.
(b) Lyophobic colloids (solvent hating) are those substances that do not form the colloidal sol readily when mixed with the dispersion medium.
These sols are less stable than the lyophilic sols.
5.15. Explain what is observed
(i) when a beam of light is passed through a colloidal sol.
(ii) an electrolyte, NaCI is added to hydrated ferric oxide sol.
(iii) electric current is passed through a colloidal sol.
Solution: (i) Scattering of light by the colloidal particles takes place and the path of light becomes visible (Tyndall effect).
(ii) The positively charged colloidal particles of Fe(OH)3 get coagulated by the oppositely charged Cl– ions provided by NaCl.
(iii) On passing electric current, colloidal particles move towards the oppositely charged electrode where they lose their charge and get coagulated.
5.16. What are emulsions? What are their different types? Give example of each type.
Solution: Emulsions: It is a colloidal system in which both the dispersed phase and the dispersion medium are liquids, e.g., milk consists of small drop’s of liquid fat dispersed in water.
Types of emulsions:
(a) Oil-in-water type in which small droplets of an oil are dispersed in water, e.g., milk, cod-liver oil, etc.
(b) Water-in-oil type in which water droplets are dispersed in an oil medium, e.g., butter.
5.17. How do emulsifires stabilise emulsion? Name two emulsifiers.
Solution: The role of an emulsifier in stabilising an emulsion can be explained in two ways:
(a) It is believed that an emulsifier gets concentrated at the oil-water interface i.e., the surface at which oil and water come in contact with each other. It forms a protective coating around each drop of oil and thus, prevents the oil drop from coming in contact with one another. The oil drops remain suspended in water and are not coagulated.
(b) According to an another view, the role of the emulsifier is the sjame as that of lubricant in a machine. Just as a lubricant reduces the friction in the various parts of machine, an emulsifier also tries to reduce the interfacial tension between oil and water by suitable means. Thus, oil and water remain in company of each other and do not get separated. The commonly used emulsifying agents are soaps, detergents, lyophilic colloids, proteins, gums, gelatin, caesin, agar etc.
5.18. 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. 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. The negatively charged sheath around the globules prevents them from coming together and forming aggregates.
(a) Grease on cloth
(b) Stearate ions (from soap) arranging around the grease droplets
(c) Micelle formed
5.19. Give four examples of heterogeneous catalysis.
Solution: In heterogeneous catalysis, the catalyst is present in a different phase than that of the reactants,e.g.,
5.20. What do you mean by activity and selectivity of catalysts?
Solution: Important features of solid catalyst:
(a) Activity: The activity of a catalyst is its ability to accelerate chemical reactions. It depends upon the strength of chemisorption to a large extent.
The catalytic activity of a metal for hydrogenation increases as we move from Group 5 metals to Group 11. The maximum activity is shown by metals of Groups 7, 8 and 9.
(b) Selectivity: The selectivity of a catalyst is its ability to direct a reaction to yield a particular product. It mean a substance which acts as a catalyst in one reaction may not act as a catalyst in other reaction e.g., we get different products when we use different catalysts in the reaction between H2 and CO.