Class 12 Chemistry Chapter 5: Surface Chemistry - NCERT Solutions

This page focuses on the complete NCERT solutions for Class 12 Chemistry Chapter 5: Surface Chemistry. This section provides detailed, easy-to-understand solutions for all the questions from this chapter. These Surface Chemistry question answers will offer you valuable insights and explanations.

Surface chemistry is that branch of chemistry which deals with the study of phenomena occurring at the surfaces or interfaces. The surface or interface is represented by pulling a hyphen or slash between the two bulk phases involved i.e. liquid-liquid, solid-solid or liquid/solid, solid/liquid etc. Definition and mechanism of adsorption alongwith its classification into physical and chemical adsorption. Enumeration of the nature of colloidal state along with the preparation, properties, purification and uses of colloids are also given.

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Exercise 1 ( Page No. : 145 )

  • Q1 Write any two characteristics of Chemisorption.
    Ans:
    1. In chemisorption, the gas molecules or atoms are held to the solid surface by chemical bonds which can be a covalent bond or Ionic bond in nature.
    2. Chemisorptions is not reversible in nature because a surface compound is formed due to chemical interaction between the gas molecule & solid surface.

    Q2 Why does physisorption decrease with the increase of temperature
    Ans:

    Physisorption is exothermic in nature. Therefore, in accordance with Le-Chateliers’s principle, it decreases with an increase in temperature. This means that physisorption occurs more readily at a lower temperature.


    Q3 Why are powdered substances more effective adsorbents than their crystalline forms?
    Ans:

    Adsorption is a surface phenomenon.The extent of adsorption depends on the surface area. Therefore, adsorption is directly proportional to the surface area.Increase in the surface area of the adsorbent, increases the total amount of gas adsorbed. A finely divided substance like nickel, platinum & porous substances like charcoal, silica gel provides large surface area but crystalline substances has less surface as compared to finely divided substances, as a result powered substances are more effective adsorbents than their crystalline forms. 


    Q4 Why is it necessary to remove CO when ammonia is obtained by Haber's process?
    Ans:

    It is important to remove CO in the synthesis of ammonia as CO adversely affects the activity of the iron catalyst, used in Haber's process.


    Q5 Why is the ester hydrolysis slow in the beginning and becomes faster after sometime?
    Ans:

    Ester hydrolysis can be represented as:

    Ester + Water → Acid + Alcohol

    The acid produced in the reaction acts as a catalyst and makes the reaction faster. Substances that act as catalysts in the same reaction in which they are obtained as products are known as autocatalysts.


    Q6 What is the role of desorption in the process of catalysis?
    Ans:

    Desorption is a phenomenon where a substance is released from or through the surface on which it is adsorbed. It is opposite of adsorption process. Catalyst can be any chemical entity which helps in increasing the rate of reaction, but they itself do not take part in reaction i.e. they just support the chemical reaction to occur effectively. For catylytic action, the reactants get adsorbed on the surface of the catalyst and gets converted into the product after the specified procedure. The product or the unreacted reactant has to be removed from the surface of catalyst, desorption helps in removal of the reactant and product form the surface of catalyst and makes it free for the next reaction.


    Q7 What modification can you suggest in the Hardy-Schulze law?
    Ans:

    Hardy-Schulze law states that ‘the greater the valence of the flocculating ion added, the greater is its power to cause precipitation’.

    This law takes into consideration only the charge carried by an ion, not its size. The smaller the size of an ion, the more will be its polarising power. Thus, Hardy-Schulze law can be modified in terms of the polarising power of the flocculating ion. Thus, the modified Hardy-Schulze law can be stated as ‘the greater the polarising power of the flocculating ion added, the greater is its power to cause precipitation.’


    Q8 Why is it essential to wash the precipitate with water before estimating it quantitatively?
    Ans:

    The precipitate (product) formed out from any chemical reaction always contains some traces of unwanted materials (unreacted reactants or catalyst and certain impurities). These unwanted materials adsorbed onto the surface of the required product and settle down at the bottom of the container. For quantitative estimation of the product, these unwanted materials has to be removed. For removal of unwanted materials, different methods are used, one of them is washing with water. Because most of the impurities are water soluble, so they can be easily removed from the product. Water easily makes hydrogen bond with most of the elements (polar). After washing with water, the unwanted materials gets dissolved in water and removed out from the final product.


Exercise 2 ( Page No. : 146 )

  • Q1 Distinguish between the meaning of the terms adsorption and absorption. Give one example of each.
    Ans:

    Adsorption is a surface phenomenon of accumulation of molecules of a substance at the surface rather than in the bulk of a solid or liquid. The substance that gets adsorbed is called the 'adsorbate' and the substance on whose surface the adsorption takes place is called the 'adsorbent'. Here, the concentration of the adsorbate on the surface of the adsorbent increases. In adsorption, the substance gets concentrated at the surface only. It does not penetrate through the surface to the bulk of the solid or liquid. For example, when we dip a chalk stick into an ink solution, only its surface becomes coloured. If we break the chalk stick, it will be found to be white from inside. On the other hand, the process of absorption is a bulk phenomenon. In absorption, the substance gets uniformly distributed throughout the bulk of the solid or liquid.


    Q2 What is the difference between physisorption and chemisorption?
    Ans:
    Physisorption Chemisorption

    1.

    In this type of adsorption, the adsorbate is attached to the surface of the adsorbent with weak van der Waal's forces of attraction.

    In this type of adsorption, strong chemical bonds are formed between the adsorbate and the surface of the adsorbent.

    2.

    No new compound is formed in the process.

    New compounds are formed at the surface of the adsorbent.

    3.

    It is generally found to be reversible in nature.

    It is usually irreversible in nature.

    4.

    Enthalpy of adsorption is low as weak van der Waal's forces of attraction are involved. The values lie in the range of 20-40 kJ mol - 1.

    Enthalpy of adsorption is high as chemical bonds are formed. The values lie in the range of 40-400 kJ mol - 1.

    5.

    It is favoured by low temperature conditions.

    It is favoured by high temperature conditions.

    6.

    It is an example of multi-layer adsorption

    It is an example of mono-layer adsorption.

     


    Q3 Give reason why a finely divided substance is more effective as an adsorbent.
    Ans:

    Adsorption is a surface phenomenon. The extent of adsorption depends on the surface area. Therefore, adsorption is directly proportional to the surface area.Increase in the surface area of the adsorbent, increases the total amount of gas adsorbed. A finely divided substance like nickel, platinum and porous substances like charcoal, silica gel provides large surface area. Both physisorption and chemisorption increase with an increase in the surface area. Hence, a finely divided substance behaves as a good adsorbent.


    Q4 What are the factors which influence the adsorption of a gas on a solid?
    Ans:

    There are various factors that affect the rate of adsorption of a gas on a solid surface.

    (1) Nature of the gas:

    Easily liquefiable gases such as NH3, HCl etc. are adsorbed to a great extent in comparison to gases such as H2, O2 etc. This is because Vander Waal's forces are stronger in easily liquefiable gases.

    (2) Surface area of the solid

    The greater the surface area of the adsorbent, the greater is the adsorption of a gas on the solid surface.

    (3) Effect of pressure

    Adsorption is a reversible process and is accompanied by a decrease in pressure. Therefore, adsorption increases with an increase in pressure.

    (4) Effect of temperature

    Adsorption is an exothermic process. Thus, in accordance with Le-Chatelier's principle, the magnitude of adsorption decreases with an increase in temperature.


    Q5 What is an adsorption isotherm? Describe Freundlich adsorption isotherm.
    Ans:

     

    The plot between the extent of adsorption (x/m) against the pressure of gas (P) at constant temperature (T) is called the adsorption isotherm.

    Freundlich adsorption isotherm:

    Freundlich adsorption isotherm gives an empirical relationship between the quantity of gas adsorbed by the unit mass of solid adsorbent and pressure at a specific temperature.

    From the given plot it is clear that at pressure PS,(x/m) reaches the maximum valve. Ps is called the saturation pressure. Three cases arise from the graph now.

    On plotting the graph between log  (x/m)  and log P, a straight line is obtained with the slope equal to (1 / n) and the intercept equal to log k.

     

     


    Q6 What do you understand by activation of adsorbent? How is it achieved?
    Ans:

    By activating an adsorbent, we tend to increase the adsorbing power of the adsorbent. Some ways to activate an adsorbent are:

    (i) By increasing the surface area of the adsorbent. This can be done by breaking it into smaller pieces or powdering it.

    (ii) Some specific treatments can also lead to the activation of the adsorbent. For example, wood charcoal is activated by heating it between 650 K and 1330 K in vacuum or air. It expels all the gases absorbed or adsorbed, thus creates a space for adsorption of gases.


    Q7 What role does adsorption play in heterogeneous catalysis?
    Ans:

    The phenomenon of concentration of molecules of a gas or liquid at a solid surface is called adsorption.

    Heterogenous catalysis is a catalysis in which the catalyst is in different physical phase from the reactant. The most important of such reactions are those in which the reactants are in the gas phase while the catalyst is a solid.

    This heterogeneous catalytic action can be explained in terms of the adsorption theory. The mechanism of catalysis involves the following steps:

    (i) Adsorption of reactant molecules on the catalyst surface.

    (ii) Occurrence of a chemical reaction through the formation of an intermediate.

    (iii) De-sorption of products from the catalyst surface.

    (iv) Diffusion of products away from the catalyst surface.

    (v) Diffusion of reactants to the surface of the catalyst.

    The reaction occurs by the contact of reactants with the catalyst surface.The molecule of reactant (gas) are adsorbed at the catalyst surface where they form an ‘adsorption complex’. The catalyst surface behave like the surface molecules of a liquid and are not surrounded by atoms or molecules of their kind. Therefore, they have unbalanced  or residual attractive forces (Vander waal’s forces) on the surface which can hold adsorbate particles.As the concentration of reactants on the surface of the catalyst increases, the rate of reaction also increases. In such reactions, the products have very less affinity for the catalyst and are quickly desorbed, thereby making the surface free for other reactants.


    Q8 Why is adsorption always exothermic?
    Ans:

    Adsorption is always exothermic. This statement can be explained in two ways.

    (i) Adsorption leads to a decrease in the residual forces on the surface of the adsorbent. This causes a decrease in the surface energy of the adsorbent. Therefore, adsorption is always exothermic.

    (ii) ΔH of adsorption is always negative. When a gas is adsorbed on a solid surface, its movement is restricted leading to a decrease in the entropy of the gas i.e., ΔS is negative. Now for a process to be spontaneous, ΔG should be negative.

    ΔG = ΔH - TΔS

    Since ΔS is negative, ΔH has to be negative to make ΔG negative. Hence, adsorption is always exothermic.


    Q9 How are the colloidal solutions classified on the basis of physical states of the dispersed phase and dispersion medium?
    Ans:

    One criterion for classifying colloids is the physical state of the dispersed phase and dispersion medium. Depending upon the type of the dispersed phase and dispersion medium (solid, liquid, or gas), there can be eight types of colloidal systems.

    Dispersed phase Dispersion medium Type of colloid Example

    1.

    Solid

    Solid Solid Sol Gemstone

    2.

    Solid

    Liquid Sol Paint

    3.

    Solid

    Gas Aerosol Smoke

    4.

    Liquid

    Solid Gel Cheese

    5.

    Liquid

    Liquid Emulsion Milk

    6.

    Liquid

    Gas Aerosol Fog

    7.

    Gas

    Solid Solid foam Pumice stone

    8.

    Gas

    Liquid Foam Froth

     


    Q10 Discuss the effect of pressure and temperature on the adsorption of gases on solids.
    Ans:

    Effect of pressure & temperature-

    According to Le Chatelier’s principle, the magnitude of adsorption should increase with fall in temperature & since adsorption of a gas leads to decrease of pressure, the magnitude of adsorption increases with increase in pressure. If a equilibrium is subjected to a stress, the equilibrium shifts in such a way as to reduce the stress. Graphically, the variation of adsorption with pressure at given constant temperature is given by adsorption isotherm.

    The relationship between magnitude of adsorption & pressure can be expressed mathematically by Freundlich Adsorption isotherm i.e  a = Kpwhere a is the amount of gas adsorbed per unit mass of adsorbent at pressure p & K & n are constant depending upon the nature of gas & adsorbent.The value of n is less than 1 & therefore a does  not increase as rapidly as p.

    The effect of temperature on the extent of adsorption at a given pressure of adsorbate is given by adsorption isobar. 


    Q11 What are lyophilic and lyophobic sols? Give one example of each type. Why are hydrophobic sols easily coagulated?
    Ans:

    Sols are colloidal systems in which a solid is dispersed in a liquid. They are of two types:

    (i) Lyophilic sols (solvent loving):

    They are those in which the dispersed phase exhibits a definite affinity for the dispersion medium (liquid) or the solvent. For example: Dispersion of starch, gum & protein in water. The affinity of sol particles for the medium is due to hydrogen bonding with water (dispersion medium). They may have little or no charge at all & they donot exhibit Tyndall effect. These sols are reversible in nature i.e., if two constituents of the sol are separated by any means (such as evaporation), then the sol can be prepared again by simply mixing the dispersion medium with the dispersion phase and shaking the mixture.

    (ii) Lyophobic sols:

    They are those in which the dispersed phase has no attraction for the dispersion medium or the solvent. For example: Dispersion of gold, iron & sulphur in water. Their colloidal sols can be prepared only by special methods. The particles carry positive or negative charge& they do exhibit Tyndall effect. These sols are irreversible in nature. 

    Now, the stability of a lyophobic sol is due to the adsorption of positive or negative ions by the dispersed particles. The repulsive forces between the charged particles donot allow them to settle.If some how ,the charge is removed, there is nothing to keep the particles apart from each other. They aggregate or flocculate & settle down under the action of gravity. In lyophobic sols,they are not surrounded by adsorbed layer of dispersion medium, as a result they come together & coagulate. So whenever an excess of electrolyte is added or boiling is done, the electrolyte furnishes both positive & negative ions in the medium & the sol particles adsorb oppositely charged ions & gets discharged. The electrically neutral particles then aggregate & settle down as precipitate. The stability of hydrophilic sols depends on two things- the presence of a charge and the salvation of colloidal particles. On the other hand, the stability of hydrophobic sols is only because of the presence of a charge. Therefore, the latter are much less stable than the former.  


    Q12 What is the difference between multimolecular and macromolecular colloids? Give one example of each. How are associated colloids different from these two types of colloids?
    Ans:

    (i) In multi-molecular colloids, the colloidal particles are an aggregate of atoms or small molecules with a diameter of less than 1 nm. Examples of such colloids include gold sol and sulphur sol.

    (ii) In macro-molecular colloids, the colloidal particles are large molecules having colloidal dimensions. These particles have a high molecular mass. When these particles are dissolved in a liquid, sol is obtained. For example: starch, nylon, cellulose, proteins, enzymes etc.

    (iii) Certain substances tend to behave like normal electrolytes at lower concentrations. However at higher concentrations, these substances behave as colloidal solutions due to the formation of aggregated particles. Such colloids are called Associated colloids (micelles).


    Q13 What are enzymes? Write in brief the mechanism of enzyme catalysis.
    Ans:

    Enzymes are basically protein molecules of high molecular masses. These form colloidal solutions when dissolved in water. These are complex, nitrogenous organic compounds produced by living plants and animals. Enzymes are also called 'biochemical catalysts'.

    Mechanism of enzyme catalysis:

     

    On the surface of the enzymes, various cavities are present with characteristic shapes. These cavities possess active groups such as -NH2, -COOH, etc. The reactant molecules having a complementary shape fit into the cavities just like a key fits into a lock. This leads to the formation of an activated complex. This complex then decomposes to give the product.

    Hence,

    Step 1: E + S → ES+

    (Activated complex)

    Step 2: ES+ → E + P


    Q14 How are colloids classified on the basis of (i) Physical states of components (ii) Nature of dispersion medium and (iii) Interaction between dispersed phase and dispersion medium?
    Ans:

    Colloids can be classified on various bases:

    (i) On the basis of the physical state of the components (by components we mean the dispersed phase and dispersion medium). Depending on whether the components are solids, liquids or gases, we can have eight types of colloids.

    (ii) On the basis of the dispersion medium, sols can be divided as:

    Dispersion medium Name of sol
    Water Aquasol or hydrosol
    Alcohol Alcosol
    Benzene Benzosol
    Gases Aerosol

    (iii) On the basis of the nature of the interaction between the dispersed phase and dispersion medium, the colloids can be classified as lyophilic (solvent attracting) and lyophobic (solvent repelling).


    Q15 Explain what is observed (i) When a beam of light is passed through a colloidal sol. (ii) An electrolyte, NaCl is added to hydrated ferric oxide sol. (iii) Electric current is passed through a colloidal sol?
    Ans:

    (i) When a beam of light is passed through a colloidal solution, then scattering of light is observed. This is known as the Tyndall effect. This scattering of light illuminates the path of the beam in the colloidal solution.

    (ii) When NaCl is added to ferric oxide sol, it dissociates to give Na+ and Cl-­ ions. Particles of ferric oxide sol are positively charged. Thus, they get coagulated in the presence of negatively charged Cl- ions.

    (iii) The colloidal particles are charged and carry either a positive or negative charge. The dispersion medium carries an equal and opposite charge. This makes the whole system neutral. Under the influence of an electric current, the colloidal particles move towards the oppositely charged electrode. When they come in contact with the electrode, they lose their charge and coagulate.


    Q16 What are emulsions? What are their different types? Give example of each type.
    Ans:

    The colloidal solution in which both the dispersed phase and dispersion medium are liquids is called an emulsion.

    There are two types of emulsions:

    (a) Oil in water type:

    Here, oil is the dispersed phase while water is the dispersion medium. For example: milk, vanishing cream, etc.

    (b) Water in oil type:

    Here, water is the dispersed phase while oil is the dispersion medium. For example: cold cream, butter, etc.


    Q17 What is demulsification? Name two demulsifiers.
    Ans:

    The process of decomposition of an emulsion into its constituent liquids is called demulsification. Examples of demulsifiers are surfactants, ethylene oxide etc.


    Q18 Action of soap is due to emulsification and micelle formation. Comment.
    Ans:

    The cleansing action of soap is due to emulsification and micelle formation. Soaps are basically sodium and potassium salts of long chain fatty acids, R-COO-Na+. The end of the molecule to which the sodium is attached is polar in nature, while the alkyl-end is non-polar. Thus, a soap molecule contains a hydrophilic (polar) and a hydrophobic (non-polar) part.

    When soap is added to water containing dirt, the soap molecules surround the dirt particles in such a manner that their hydrophobic parts get attached to the dirt molecule and the hydrophilic parts point away from the dirt molecule. This is known as micelle formation. Thus, we can say that the polar group dissolves in water while the non-polar group dissolves in the dirt particle. Now, as these micelles are negatively charged, they do not coalesce and a stable emulsion is formed.


    Q19 Give four examples of heterogeneous catalysis.
    Ans:

    (i) Oxidation of sulphur dioxide to form sulphur trioxide. In this reaction, Pt acts as a catalyst.

    (ii) Formation of ammonia by the combination of dinitrogen and dihydrogen in the presence of finely divided iron.

    This process is called the Haber's process.

    (iii) Oswald's process: Oxidation of ammonia to nitric oxide in the presence of platinum.

    (iv) Hydrogenation of vegetable oils in the presence of Ni.


    Q20 What do you mean by activity and selectivity of catalysts?
    Ans:

    (a) Activity of a catalyst:

    The activity of a catalyst is its ability to increase the rate of a particular reaction. Chemisorption is the main factor in deciding the activity of a catalyst. The adsorption of reactants on the catalyst surface should be neither too strong nor too weak. It should just be strong enough to make the catalyst active.

    (b) Selectivity of the catalyst:

    The ability of the catalyst to direct a reaction to yield a particular product is referred to as the selectivity of the catalyst. For example, by using different catalysts, we can get different products for the reaction between H2 and CO.


    Q21 Describe some features of catalysis by zeolites.
    Ans:

    Zeolites are alumino-silicates that are micro-porous in nature. Zeolites have a honeycomb-like structure, which makes them shape-selective catalysts. They have an extended 3D-network of silicates in which some silicon atoms are replaced by aluminium atoms giving Al-O-Si framework. The reactions taking place in zeolites are very sensitive to the pores and cavity size of the zeolites. Zeolites are commonly used in the petrochemical industry.


    Q22 What is shape selective catalysis?
    Ans:

    A catalytic reaction which depends upon the pore structure of the catalyst and on the size of the reactant and the product molecules is called shape-selective catalysis. For example, catalysis by zeolites is a shape-selective catalysis. The pore size present in the zeolites ranges from 260-740 pm. Thus, molecules having a pore size more than this cannot enter the zeolite and undergo the reaction.


    Q23 Explain the following terms: (i) Electrophoresis (ii) Coagulation (iii) Dialysis (iv) Tyndall effect.
    Ans:

    (i) Electrophoresis:

    The movement of colloidal particles under the influence of an applied electric field is known as electrophoresis. Positively charged particles move towards the cathode, while negatively charged particles move towards the anode. As the particles reach oppositely charged electrodes, they become neutral and get coagulated.

     (ii) Coagulation:

    The process of settling down of colloidal particles i.e., conversion of a colloid into a precipitate is called coagulation.

    (iii) Dialysis:

    The process of removing a dissolved substance from a colloidal solution by the means of diffusion through a suitable membrane is known as dialysis. This process is based on the principle that ions and small molecules can pass through animal membranes unlike colloidal particles.

    (iv) Tyndall effect:

    When a beam of light is allowed to pass through a colloidal solution, it becomes visible like a column of light. This is known as the Tyndall effect. This phenomenon takes place as particles of colloidal dimensions scatter light in all directions.


    Q24 Give four uses of emulsions.
    Ans:

    Four uses of emulsions:

    (i) Cleansing action of soaps is based on the formation of emulsions.

    (ii) Digestion of fats in intestines takes place by the process of emulsification.

    (iii) Antiseptics and disinfectants when added to water form emulsions.

    (iv) The process of emulsification is used to make medicines.


    Q25 What are micelles? Give an example of a micellers system.
    Ans:

    Micelles formation is done by substances soaps and detergents when dissolved in water.The Molecules of such substances contain a hydrophobic and a hydrophilic part. when present in water, these substances arrange themselves in spherical structure in such a manner that their hydrophobic parts are present towards the centre, while the hydrophilic parts are pointing towards outside (as shown in the given figure). This is known as micelles formation.


    Q26 Explain the terms with suitable examples: (i) Alcosol (ii) Aerosol (iii) Hydrosol
    Ans:

    (i) Alcosol: A colloidal solution having alcohol as the dispersion medium and a solid substance as the dispersed phase is called an alcosol. For example: colloidal sol of cellulose nitrate in ethyl alcohol is an alcosol.

    (ii) Aerosol: A colloidal solution having a collection of solid particles or liquid droplets dispersed in air is called an aerosol. For example: fog, smoke, etc.

    (iii) Hydrosol: A colloidal solution having water as the dispersion medium and a solid as the dispersed phase is called a hydrosol. For example: starch sol or gold sol.


    Q27 Comment on the statement that "colloid is not a substance but a state of substance".
    Ans:

    Common salt (a typical crystalloid in an aqueous medium) behaves as a colloid in a benzene medium. Hence, we can say that a colloidal substance does not represent a separate class of substances. When the size of the solute particle lies between 10 nm and 100 nm, it behaves as a colloid.

    Hence, we can say that colloid is not a substance but a state of the substance which is dependent on the size of the particle. A colloidal state is intermediate between a true solution and a suspension.


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