# Bihar Board Class 12 Chemistry Solutions Chapter 5 Surface Chemistry

Bihar Board Class 12 Chemistry Solutions Chapter 5 Surface Chemistry Textbook Questions and Answers, Additional Important Questions, Notes.

## BSEB Bihar Board Class 12 Chemistry Solutions Chapter 5 Surface Chemistry

### Bihar Board Class 12 Chemistry Surface Chemistry Intext Questions and Answers

Question 1.
Why are substances like platinum and palladium often used for carrying out electrolysis of aqueous solutions?
Metals like platinum and palladium are the least reactive among metals and so do not ordinarily react with the products on electrolysis Of aqueous solutions.

Question 2.
Why does physisorption decrease with the increase of temperature?
Physical adsorption/physisorption is an exothermic process and is reversible.
Solid + Gas⇌Adsorption + Heat .
Physisorption occurs at low temperature and increasing the temperature will decrease the adsorption in keeping with Le-Chatelier’s principle.

Question 3.
Why are powdered substances more effective adsorbents than their crystalline forms?
The extent of adsorption increases with the increase of surface area of the adsorbent. Thus finely divided metals and porous substances having large surface area are good adsorbents. The surface area of the crystalline substances is less than their powdered forms.

Question 4.
Why is it necessary to remove CO when ammonia is obtained by Haber’s process?
In Haber’s process for the production of NH3 from N2 and H2, iron is used as a solid catalyst. It is necessary to remove CO because it reacts with iron to form [Fe (CO)5) which is a liquid at room temperature and will interfere in the production of NH3 at higher temperature due to reaction of CO and H2.

Question 5.
Why is the ester hydrolysis slow in the beginning and becomes faster after sometime?
During the hydrolysis of the ester, one of the products formed is an organic acid which acts as a catalyst and accelerates the reaction after its formation.

Question 6.
What is the role of desorption in the process of catalysis?
After the gaseous reactants are adsorbed on the surface of a solid catalyst and the formation of intermediate with it, the process of desorption of the product molecules takes place. Thereby the surface of the solid catalyst is made available again for more reaction to occur.

Question 7.
What modification can you suggest in the Hardy Schulze law?
Hardy Schulze law can be modified as follows: As coagulation power is inversely proportional to coagulation/flocculation value, to compare the relative coagulating powers of two electrolytes for the same colloidal sol, we have
$$\frac{\text { Coagulating power of electrolyte } 1}{\text { Coagulating power of electrolyte } 2}$$ = $$\frac{\text { Coagulation value of electrolyte } 2}{\text { Coagulation value of electrolyte } 1}$$
For example, for coagulation of negative As2S3 sol
$$\frac{\text { Coagulation power of } \mathrm{AlCl}_{3}}{\text { Coagulating power of } \mathrm{NaCl}}$$ = $$\frac{\text { Coagulation value of } \mathrm{NaCl}}{\text { Coagulation value of } \mathrm{AlCl}_{3}}$$
= $$\frac{52}{0.093}$$ = 559
Thus AlCl3 has 559 times more coagulating power than NaCl.

Question 8.
Why is it essential to wash the precipitate with water before estimating it quantitatively?
It is necessary to wash the precipitates with water before estimating it quantitatively so as to remove any colloidal particles sticking to the precipitates. Colloids, as we know, can pass through ordinary filter paper, whereas precipitate cannot. Colloid particles size being, smaller to that of the precipitate can pass through ordinary filter paper.

### Bihar Board Class 12 Chemistry Surface Chemistry Text Book Questions and Answers

Question 1.
Distinguish’ between the meanings of the terms adsorption and absorption. Give one example of each.
Adsorption-The phenomenon of attracting and retaining the molecules of a substance on the surface of a liquid or a solid resulting into a higher concentration of the molecules on the surface is called adsorption. Adsorption is a Surface Phenomenon, i.e., it occurs only at the surface of the absorbent.
Example: Silica gel when placed in a vessel containing water vapours adsorbed the water vapours on its surface.

Absorption on the other hand, is a phenomenon in which the particles of a gas or liquid get uniformly distributed throughout the body of the solid.
Absorption is a Bulk Phenomenon, i.e., it occurs throughout the body of the material.
A piece of anhydrous CaCl2 kept in a beaker containing water absorbs water which gets uniformly distributed in CaCl2 to form hydrated CaCl2 (CaCl2.2H2O).

Question 2.
What is the difference between physisorption and chemisorption?
Physisorption-When a gas is held (adsorbed) on the surface of a solid by van der Waal’s forces (which are weak intermolecular forces of attraction) without resulting into the formation of any chemical bond between the adsorbate and the adsorbent, it is called ‘physical adsorption’ or ‘van der Waal’s adsorption’ or ‘Physisorption’.

It is characterised by (i) low heats of adsorption (20 – 40 kj mol-1, (ii) the process is reversible, (iii) no activation energy is required, (iv) proceeds at low temperature and decreases with increase in temperature, (v) not specific in nature, (vi) forms multimolecular layers.
Chemisorption, on the other hand, is a phenomenon where the gas is held on to the surface of a solid by forces similar to those of a chemical bond. This adsorption results in the formation of what is called ‘surface compound’.

It is characterised by:

• high heats of adsorption (40 – 400 kj mol-1),
• the process is, irreversible,
• it requires activation energy,
• This type of adsorption first increases with increase in temperature,
• It is specific in nature,
• It forms unimolecular layer.

Question 3.
Give reason why a finely divided substance is more effective as an adsorbent.
A finely divided substance has greater surface area. Greater the surface area of the adsorbent, greater is the volume of the adsorbed gases.

Question 4.
What are the factors which influence the adsorption of a it gas on a solid?
The extent of adsorption of a gas on a solid adsorbent is affected by the following factors:
(i) Nature of the gas-Different gases are adsorbed to different extents by the same adsorbent at the same temperature. A gas which has higher critical temperature will be adsorbed to a greater extent. In other words, a gas which is more easily liquefiable or is more soluble in water is more readily adsorbed.
Thus gases like NH3, HCl, CO2 etc. are adsorbed to a greater extent than the permanent gases such as H2, O2, N2 etc.

(ii) Nature of the adsorbent-It is observed that the same gas is adsorbed to different extents by different solids at the same temperature.

(iii) Surface area of the adsorbent-The greater the surface area of the adsorbent, greater is the volume of the gas adsorbed. It is for this reason that substances like charcoal, silica gel are excellent adsorbents because they have highly porous structures and hence large surface areas.

(iv) Effect of pressure of the gas-Adsorption is accompanied with a decrease in pressure. Therefore, it is expected that extent of adsorption increases with increase in pressure. The extent of adsorption is generally expressed as $$\frac{\mathrm{x}}{\mathrm{m}}$$ where m is the mass of the adsorbent and x is that of the adsorbate when equilibrium is attained.
At equilibrium pressure ps, $$\frac{\mathrm{x}}{\mathrm{m}}$$ reaches its maximum value and no more adsorption takes place even if pressure is increased.

(v) Temperature-Adsorption decreases with the increase in temperature.

The amount of heat evolved when one mole of the gas is adsorbed on the adsorbent is called its heat of adsorption.

This is generally done by increasing the surface area (specific area) of the adsorbent:

• By making the surface of the adsorbent rough by mechanical rubbing or by chemical action. For example, depositing finely ‘ dispersed metals on the surface of the adsorbent by electroplating.
• By subdividing the adsorbent into smaller pieces pr grains,
• By removing the gases already adsorbed, e.g., charcoal is activated by heating in superheated steam or in vacuum at a temperature between 623 and 1273 K.

Question 5.
Tire extent of adsorption of a gas on a solid is generally expressed as $$\frac{\mathrm{x}}{\mathrm{m}}$$ where m is the mass of the adsorbent and x is the m number of moles of the adsorbate when dynamic equilibrium is attained between the free gas and the adsorbed gas. A related between $$\frac{\mathrm{x}}{\mathrm{m}}$$against the pressure at constant temperature is called ADSORPTION ISOTHERM. FREUNDLICH ADSORPTION ISOTHERM- $$\frac{x}{m}$$ = kP1/n (where n > 1). k and n are the parameters which depend upon the nature of the gas and the solid.

Applying logs.
log $$\frac{x}{m}$$ = log k + $$\frac{1}{\mathrm{n}}$$ log P
From the slope of the st. line obtained by plotting log P vs log $$\frac{x}{m}$$ the value of n can be found. Similarly from the intercept, the value of k can be determined.

Question 6.
What do you understand by activation of adsorbent? How is it achieved?
Activation of the adsorbent means increasing the adsorption power or capabilities of the adsorbent. It is necessary so as to increase the rate of adsorption. This can be achieved as follows :

• Metallic adsorbents are activated by mechanical rubbing or by subjecting it to some chemical reactions. ‘
• Sub-dividing the adsorbent into smaller pieces increases the adsorbing power of the adsorbent. As a result of sub-division: the surface area of the adsorbent increases and therefore the adsorbing power increases.
• Some adsorbents are activated by strong heating in contact with superheated steam. For example charcoal is activated by subjecting it to the action of superheated steam.

Question 7.
What role does adsorption play in heterogeneous catalysis?
Adsorbents such as Fe, VM2O5, finally divided Ni play a very; important role in the catalytical formation of NH3 (Haber’s Process), H2SO4 (contact process), hydrogenation of oils as heterogeneous catalysts respectively. In heterogeneous catalysis, the catalyst used is in different 1 phase that the reactants. Basically, it involves the following steps.

• Diffusion of the reactants at the surface of the catalyst.
• Adsorption of the molecules of the reactant at the active sites.
• Occurrence of the chemical reactions on the surface of the catalyst.
• Desorption of product molecules from the surface.
• Diffusion of products away from the surface of the catalyst.

Thus catalysts facilitate the reactions in many ways.

1. Adsorption increases the concentration of the reactants on the surface of the catalyst. Due to increased concentration of the reactants, the reactions proceed rapidly.
2. Adsorbed molecules get dissociated to form active species like free radicals which react faster than molecules.
3. The adsorbed molecules are not free to move about and, therefore, they collide with other molecules on the surface.
4. The heat of adsorption evolved acts as energy of activation for the reaction (CHEMISORPTION).

Question 8.
Adsorption is accompanied by decrease of randomness, i.e., this factor opposes the process, i.e., ΔS is +ve. For the process to be spontaneous factor ΔG must be – ve. Hence, according to the equation ΔG = ΔH – TΔS, ΔG can be – ve only if ΔH is – ve, i.e., process is exothermic. Hence adsorption is always an exothermic process.

Question 9.
How are colloidal solutions classified on the basis of physical states of the dispersed phase and dispersion medium?
CLASSIFICATION BASED ON PHYSICAL STATE OF DISPERSED PHASE AND DISPERSION MEDIUM—Depending upon whether the dispersed phase and the dispersion medium are solids, liquids or gases, light types of collodial systems are possible as detailed below.

Question 10.
Discuss the effect of pressure and temperature on the adsorption of gases on solids?
(a) Effect of pressure on adsorption of gases on solids can be explained by FREUNDLICH ADSORPTION ISOTHERM. Adsorption is a reversible process and is accompanied by decrease in pressure. Therefore, it is expected that extent of adsorption of adsorption increases with increase in pressure.
At equilibrium pressure ps, $$\frac{x}{m}$$ reaches its maximum value and no more adsorption takes place even if pressure is increased.

(i) At Low Pressure (The curve from A to B in the graph below)
$$\frac{x}{m}$$ ∝ p1
$$\frac{x}{m}$$ = kp1 where k is constant.

(ii) At High Pressure-
$$\frac{x}{m}$$ ∝ p0
or $$\frac{x}{m}$$ is constant
i.e., $$\frac{x}{m}$$ = kp0

combining the two
$$\frac{x}{m}$$ ∝ p0-1
or $$\frac{x}{m}$$ = kp1/n where n>1
Thus, in the intermediate range of pressure
$$\frac{x}{m}$$ = kp1/n where n> 1 ……………………. (1)
or log $$\frac{x}{m}$$ = log k + $$\frac{1}{\mathrm{n}}$$ log p. ……………………………………… (2)
Equations (1) and (2) are called Freundlich Adsorption Isotherm. It gives an expression for studying the change in adsorption of a gas on the surface of a solid with change in pressure at constant temperature.

(b) Effect of temperature in adsorption of gases on solids- Adsorption is a process involving true equilibrium. The two opposing processes involved are Condensation (adsorption) of the gas molecules on the surface of the solid and evaporation (desorption) of the gas molecules from the surface of a solid into the gaseous phase.

As adsorption is accompanied by evolution of heat, so in accordance with Le-Chatelier’s principle, the magnitude of adsorption’ should decrease with rise in temperature and this is actually so.
A graph drazvn between extent of adsorption (x/m) and temperature (t) at constant pressure is called adsorption isobar. The adsorption isobars for physical adsorption and chemisorption are shown in Fig. below.

These isobars are different from each other. While a physical adsorption isobar shows a decrease in x/m as the temperature rises, the isobar of chemisorption shows an increase in the beginning and then a decrease as the temperature rises. This initial increase is due to the fact that, like chemical reactions, chemisorption also requires activation energy.

Question 11.
What are lyophilic and lyophobic sols? Give one example of each type. Why are hydrophobic sols easily coagulated?
LYOPHILIC SOLS [SOLVENT-ATTRACTING]-The word lyophilic means liquid-loving. Colloidal sols directly formed by substances like gum, gelatin, starch, rubber etc. on mixing with a suitable liquid (the dispersion medium) are called lyophilic sols. These sols are also called REVERSIBLE SOLS as on evaporating the dispersion medium, colloid can be obtained. Furthermore these SOLS are quite stable and cannot be easily coagulated, when dissolved in water, they are called hydrophilic sols.

LYOPHOBIC [SOLVENT-REPELLINGJ-SOLS-The word lyophobic means liquid-hating. Substances like metals, their sulphides etc. when simply mixed with the dispersion medium do not form the colloidal sol. Their colloidal sols can be prepared only by special methods. Such sols are called lyophobic sols. These sols are readily precipitated (or coagulated) on the addition of small amounts of electrolytes, by heating or by shaking and hence are not stable. Further, once precipitated or coagulated, they do not give back the colloidal sol . by simple addition of the dispersion medium. Hence, these sols are also called IRREVERSIBLE SOLS. Lyophobic sols need stabilizing agents for their preservation. When water is dispersion medium, they are called hydrophobic Sols. Hydrophobic sols get easily coagulated on the addition V of small amount of electrolytes or by heating or even shaking as they are not stable.

Question 12.
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?
Based upon the particle size and nature, these colloids are classified as :
(a) Multimolecular colloids,
(b) Macromolecular colloids and
(c) Associated colloids.
(a) Multimolecular colloids-In this type of colloids particles are aggregates of atoms or small molecules with molecular she less than 1 nm. For example, a gold sol consists of particles of various sizes which are clusters of several gold atoms. A sol of sulphur consists of colloidal particles which are aggregates of Sg molecules. These molecules in the aggregate are held by Van der Waal’s forces.

(b) Macromolecular colloids-In this type of colloids, colloidal particles are themselves large molecules having colloidal dimensions. They have very high molecular masses varying from thousands to millions. Naturally occurring macromolecules are such as starch, proteins and cellulose. Artificial macromolecules consist of polymers sTich as polyethene, nylon, plastics, polystyrene, etc. These colloidal solutions resemble true solutions in some respects.,

(c) Associated colloids-There are certain substances which behave as normal, strong electrolytes at low concentration but at higher concentration, they behave as colloidal solutions due to the formation of aggregated particles. Such colloids are called associated colloids and the aggregated particles are called micelles. Soaps and detergents are example of associated colloids.

Question 13.
What are enzymes? Write in brief the mechanism of enzyme catalysis.
Enzymes are complex nitrogenous organic compounds which are produced by living plants and animals. They are actually protein molecules of high molecular mass and form colloidal sols in water. They are very effective catalysts in numerous reactions especially those connected with natural processes. Several reactions occurring in the bodies of animals and plants to maintain life process as catalysed by enzymes. The enzymes are, thus, BIO-CHEMICAL CATALYSTS.
Inversion of cane sugar to glucose and fructose is brought about by the enzyme INVERTASE.

Mechanism of enzyme catalysis-There are a number of cavities present on the surface of colloidal particles of enzymes. These cavities are of characteristic shape and possess active groups. Such as -NH2-COOH-SH-OH etc. These are actually the active centres on the surface of enzyme particles. The molecules of the reactant (substrate) which have complementary shape, fit into these cavities just like a key fits into a lock. On account of the presence of active groups, an activated complex is formed which then decomposes to yield the products.
Thus, the enzyme-catalysed reactions may be considered to proceed in two steps.
Main reaction:

Question 14.
How are colloids classified on the basis of (a) physical states of components (b) nature of the dispersion medium and (c) interaction between dispersed phase and dispersion medium?
(a) Colloids are cLassified based on physical state of dispersed phase and dispersion medium as follows:

(b) Depending upon the nature of the dispersion medium colloidal sols are given specific names. For example.

(c) Depending upon the interaction between the dispersed phase and dispersion medium, the colloids are called LYOPHILIC and LYOPHOBIC SOLS. For details of Lyophilic and Lyophobic sols, please refer to the answer given under LYOPHILIC SOLS [SOLVENT-ATTRACTING]-The word lyophilic means liquid-loving. Colloidal sols directly formed by substances like gum, gelatin, starch, rubber etc. on mixing with a suitable liquid (the dispersion medium) are called lyophilic sols. These sols are also called REVERSIBLE SOLS as on evaporating the dispersion medium, colloid can be obtained. Furthermore, these SOLS are quite stable and cannot be easily coagulated, when dissolved in water, they are called hydrophilic sols.

LYOPHOBIC [SOLVENT-REPELLINGJ-SOLS-The word lyophobic means liquid-hating. Substances like metals, their sulphides etc. when simply mixed with the dispersion medium do not form the colloidal sol. Their colloidal sols can be prepared only by special methods. Such sols are called lyophobic sols. These sols are readily precipitated (or coagulated) on the addition of small amounts of electrolytes, by heating or by shaking and hence are not stable. Further, once precipitated or coagulated, they do not give back the colloidal sol . by simple addition of the dispersion medium. Hence, these sols are also called IRREVERSIBLE SOLS. Lyophobic sols need stabilizing agents for their preservation. When water is dispersion medium, they are called hydrophobic Sols. Hydrophobic sols get easily coagulated on the addition V of small amount of electrolytes or by heating or even shaking as they are not stable.

Question 15.
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.?
(i) When a beam of light is passed through a colloidal sol kept in dark, the path of the light is illuminated by a bluish-white light. This phenomenon of scattering of light by the colloidal particles in the SOL is called TYNDALL EFFECT.

(ii) When an electrolyte like NaCl is added to Fe(OH)3 sol the positive charge on the colloidal particles is neutralized by the Cl- ions released by the electrolyte. Thus rendered neutral, the colloidal particles grow in size and thus are coagulated or precipitated. This phenomenon of rendering the charged colloidal particles neutral in the sol on the addition of an excess of electrolyte resulting in their precipitation is called COAGULATION or FLOCCULATION.

(iii) When electric current is passed through some of the colloidal sols, the colloidal sol particles start moving towards the opposite charge electrodes. This phenomenon of the movement of colloidal particles under an applied electric field towards the oppositely charged electrode is called CATAPIIORESIS or ELECTROPHORESIS. If the particles accumulate near the negative electrode, the charge on the particles is positive. On the other hand, if the particles accumulate near the positive electrode, they have negative charge on them.

Question 16.
What are emulsions? What are their different types? Give example of each type.
Emulsion are colloids in which both the dispersed phase and the dispersion medium are liquids. Emulsions are of two types:

1. OIL-IN-WATER TYPE
2. WATER-IN-OIL TYPE,

1.OIL-IN-WATER TYPE of emulsions. Here dispersed phase is
oil and dispersion medium is water. Examples are milk in which fats are dispersed in water. Vanishing cream is another example of oil-in-water type of emulsion.

2. WATER-IN-OIL TYPE of Emulsions—In this emulsion, water acts as the dispersed phase while the oil behaves as the dispersion medium. For example, butter, cod-liver oil, cold cream etc.

Question 17.
What is demulsification? Name two demulsifiers.
The process of separating of an emulsion into its constituent liquids is called demulsification. The various techniques applied for demulsification are boiling and centrifugation which destroy the emulsifying agents.

Question 18.
Action of soap is due to emulsification and micelle formation. Comment.
The cleansing action of soap is due to its tendency to act as micelle and form emulsion. Soap acts as an emulsifier by lowering the interfacial tensions between oil and water and thus stabilises the emulsion. Soap, say, sodium stearate is composed of long-chain of alkyl group called tail and polar part of COO ion called head.

Washing action of soap is due to the emulsification of grease and taking it away with dirt or dust present in grease.

Explanation-The cleansing action of soap can be explained keeping in mind that a soap molecule contains a non-polar hydrophobic ^ group and a polar hydrophilic group. The dirt is held on the surface of ^ clothes by the oil or grease which is present there. Since oil or grease are not soluble in water, therefore, the dirt, particles cannot be removed by simply washing the cloth with water. When soap is applied, the nonpolar alkyl group dissolves in oil droplets while the polar – COO Na+ groups remain dissolved in water. In this way, each oil droplet is surrounded by negative charge. These negatively charged oil droplets cannot coalesce and a stable emulsion is formed. These oil droplets (containing dirt particles) can be washed away with water along with dirt particles.

Question 19.
Give four example of heterogeneous catalysis.
In heterogeneous catalysis, the catalyst is present in a different phase than that of the reactant. Here a catalyst is generally a solid and the reactants also used. It is also known as Surface catalysis.
Examples of heterogenous catalysis are:
(i) Production of Ammonia by Haber’s process uses Fe as a solid catalyst.

(ii) Synthesis of CH3OH from CO and H2 using a mixture of copper,
ZnO and Cr2O3 as catalyst.

Question 20.
What do you mean by activity and selectivity of catalysts?
The ability/effectiveness of the catalyst depends upon its two important aspects-activity and selectivity.
Two important aspects of the solid catalysts are activity and selectivity.

(a) Activity-The ability of catalysts to accelerate the rates of chemical reactions is called activity. Catalysts can accelerate the rates of reactions to a very large extent. In some cases, the rate of reaction have been found to increase by the extent of 1010 times by the use of catalyst. For example, a mixture of pure hydrogen and pure oxygen does not react at all in the absence of catalyst, however, in the presence of platinum as catalyst the mixture reacts with explosive violence to form water.

Similarly, the hydrogenation of ethylene to form ethane does not take place at all in the absence of a catalyst. However, in the presence of finely divided metal such as nickel, platinum or palladium, the reaction occurs at room temperature.

It may be noted that the granular forms of the solid catalyst have large surface area. Hence, they are more active than the bulk forms.
(b) Selectivity-It is the ability of the catalyst to direct a reaction to yield a particular product. For example, acetylene on reaction with H2 in the presence of Pt as catalyst gives ethane while in the presence of Lindlar’s catalyst (Palladium, partially inactivated by quinoline or heavy metal ions) it gives ethylene.

A heterogeneously catalysed reaction involves several steps. The initial step is usually chemisorption of reactants on the surface of the catalyst. The places at the surface where reacting molecules may becomes adsorbed are called active sites. The adsorbed reactants then undergo diffusion, dissociation and chemical reaction at the surface. Finally, the products formed undergo desorption.

Question 21.
Describe some features of catalysis by zeolites.
Zeolites are microporous aluminosilicates of the general formula
Mx/n[(AlO2)x(SiO2)y].mH2O
where n is the valency of the metal cation, Mn+. Zeolites possess high porosity due to the presence of one-, two- or three-dimensional networks of interconnected channels and cavities of molecular dimensions. Zeolites are widely used as catalyst in petrochemical industry for cracking and isomerization of hydrocarbons. The important feature of zeolite catalysis is shape selectivity. The catalytic activity of zeolites depends upon the size of the cavities (cages) and pores (apertures) present in them. The pore size in zeolites is generally between 260 pm and 740 pm.

Depending upon the size of the molecules of reactants and products and the size of the pores of the zeolites, reactions proceed in specific manner.
For example, zeolite catalyst known as ZSM-5 converts alcohols to gasoline. The alcohol is dehydrated in the cavities and hydrocarbons are formed. The shape selectivity in the reactions can be judged from the conversion of methanol and 1 -heptanol to hydrocarbon mixtures.

Question 22.
What is shape-selective catalysis?
Thaeafalytical reaction that depends upon the pore structure of the catalyst and size of the reactant and product molecules is called shape-selective catalysis. Zeolites are good shape-selective catalysts, because of their honeycomb-like structures. Zeolites are alumino-silicates, i.e., three-dimensional network silicates in which some silicon atoms are replaced by aluminium atoms. Zeolites are heated in vacuum before use when they are dehydrated and the cavities in cagelike structures become vacant.

The size of the pores is between 260-740 pm. Thus only those molecules can be adsorbed in these pores whose size is small enough to enter into the cavities and also leave easily. Reactions taking place in zeolites depend upon the size and shape of the reactant and product molecules as well as the pores and cavities of the zeolites. That is why these reactions are called Shape-selective catalysis reactions.

Question 23.
Explain the following terms:
(i) Electrophoresis,
(ii) Coagulation,
(iii) Dialysis,
(iv) Tyndall Effect.
(i) Electrophoresis/Cataphoresis-The particles of the colloidal solution are electrically charged and carry same type of charge; either positive or negative. The dispersion medium has an equal and opposite charge making the system neutral as a whole. Due to similar nature of the charge carried by the particles, they repel each other and do not combine to form bigger particles. That is why a sol is stable and particles do not settle down. Arsenious sulphide, gold, silver and platinum particles in their respective colloidal sols are negatively charged while particles of ferric hydroxide, aluminium hydroxide are positively charged.

The existence of the electric charge is shown by the phenomenon of electrophoresis. It involves the “movement of colloidal particles either towards the cathode or anode, under the influence of the electrical field”. The apparatus used for electrophoresis is shown in Fig.

The colloidal solution is placed in a U-tube fitted with platinum electrodes. On passing an electric current, the charged colloidal particles move towards the oppositely charged electrode. Thus, if arsenious sulphide sol is taken in the U-tube, its particles move towards the anode.

(ii) Coagulation-The presence of small amounts of electrolytes is necessary for the stability of the colloids. However, when an electrolyte is added in larger concentrations, the particles of the sol take up the ions which are oppositely charged and thus get neutralised. The neutral particles, then, start aggregating, gives particles of larger size which are then precipitated.

This process of aggregation of colloidal particles into an insoluble precipitated by the addition of same suitable electrolyte is known as coagulation. At lower concentration of electrolytes, the aggregation of particles is called FLOCCULATION that can be reversed on shaking while at higher concentration of the electrolyte coagulation takes place and the same cannot be reversed by simple shaking.

(iii) Dialysis-The process of separating a crystalloid and a colloid is called Dialysis. Particles of true solutions can pass through parchment paper or cellophane membrane. On the other hand, sol’ particles cannot pass through these membranes. A bag made up of such a membrane is tilled with the colloidal solution and is then suspended in freshwater. The electrolyte particles pass out leaving behind the colloidal solution. Movement of ions across the membrane can be expedited by applying electric current through two electrodes. This method is very fast and is known as Electrodialysis.

(iv) Tyndall Effect-If a strong beam of light is passed through a colloidal sol placed in dark, the path of the beam gets illuminated. The phenomenon of scattering a light by the colloidal particles in the sol when light of suitable frequency is incident upon it is called Tyndall Effect. The illuminated path of beam is called TYNDALL CONE. The path gets illuminated by bluish-white .light. The same phenomenon is observed due to scattering of light by dust particles when a beam of sunlight enters a dark room through a slit. Mist pierced by headlights at height also displays the same effect. True solutions do not exhibit Tyndall effect, because the particles in them are too small in size to scatter light.

Question 24.
Give four uses of emulsions.
(i) Concentration of ores in metallurgy-Several sulphides ores or metals like Zn, Cu are concentrated by Froth Flotation Process which involves the use of exclusions like that of pine oil and water. The ore particles get preferentially wet by oil and with the passage of air, these ore particles rise to the surface as foam and is skimmed off. The impurities sink to the bottom.

(ii) In Medicine-The various pharmaceuticals and cosmetics available in liquid form such as cod liver oil, B-complex, ointments are emulsions of water-in-oil type. Some of these are readily absorbed by the intestines.

(iii) Cleansing action of soap-The cleansing action of soap is based upon the formation of oil-in-water type of emulsion.
(iv) Milk-Milk is an important constituent of our daily diet is an emulsion of fat in water.

Question 25.
What are micelles? Give an example of a micellar system?
Micelles-Certain substances like soaps and detergents behave as electrolytes at low concentration. However, at high concentration, they constitute colloidal sols, which are known as associated colloids. The colloidal behaviour of such substances is due to the formation of aggregates or clusters in solutions. Such aggregated particles are known as micelles. Thus, micelle may be defined as the cluster or aggregated particles formed by association colloids in solutions. The formation of micelle takes place above certains concentration called critical micellization concentration (CMC). Every micelle system has a specific value of CMC. The formation of micelles takes place only above a particular temperature called KRAFT TEMPERATURE (Tk).

Micelle are generally formed by the specific type of molecules which have lyophilic as well as lyophobic ends. Such molecules are known as surface .active molecules or surfactant molecules. Sodium oleate C17H33COONa+ (one of the soaps) is a typical example of such typo of molecule. The long hydrocarbon part of oleate radical (C17H33) is lyophobic end while COONa+ is lyophilic end.

Question 26.
Explain the terms with suitable examples : (1) Alcohol, (2) Aerosol and (3) Hydrosol.

1. ALCOHOL-It is a colloidal sol in which the dispersion medium is alcohol. Example: certain polymeric sols.
2. AEROSOLS are the colloidal systems in which the dispersion medium is air. Examples: Smoke, dust storm, haze, mist fog, insecticide spray etc.
3. HYDROSOLS-The colloidal systems in which dispersion medium is water are called Hydrosols. Example: Starch sol, soap leather, whipped cream, soda-water etc.

Question 27.
Comment on the statement that “colloid is not a substance but a state of substance”.
The given statement that “colloid is not a substance but a state of a substance” is true. It is because the same substance can exist as a colloid under certain conditions and as a crystalloid under certain other conditions. For example, NaCl in water behaves as a crystalloid with which it forms a true solution, while in benzene it behaves as s colloid.

Similarly, dilute soap solution behaves like a crystalloid while its concentrated solution behaves as a colloid (called associated colloid) It is the size of the particles which matters, i.e, the state in which the substance exists. If the size of the particles lies in the range 1 nm to 100 nm, it is called the colloidal state. On the other hand, if the size of the particles is greater than 100 nm, it exists as suspension and if particle size is less than 1 nm, it exists as true solution and behaves like a crystalloid. The colloidal state is intermediate state between true solution and suspension.

### Bihar Board Class 12 Chemistry Surface Chemistry Additional Important Questions and Answers

Question 1.
What happens when a colloidal sol of Fe(OH)3 and As2S3 are mixed?
Their mutual precipitation takes place.

Question 2.
Which of the two: absorption or adsorption is a surface: phenomenon?

Question 3.
What is observed when sodium chloride is added to a colloidal solution of ferric hydroxide?
The sol gets coagulated.

Question 4.
Give one example each of
(i) Micelles system
(ii) Macromolecular colloid.
(i) Sodium stearate C17H35COONa+
(ii) Proteins.

Question 5.
What does reciprocal of gold number indicate?
Reciprocal of gold number is a measure of protective power of a colloid. Smaller the value of gold number, greater will be its protecting power.

Question 6.
Define Gold number.
The minimum amount of the protective power of a colloid in milligrams required to just percentage the coagulation of a 10 ml of a given red gold sol when 1 ml of 10% solution of sodium chloride is added to it.

Question 7.
Define HARDY-SCHULZE rule.
Greater the valency of the active ion or flocculating ion, greater will be its coagulating power.

Question 8.
The conductance of an emulsion increases on adding ’ common salt. What type of emulsion is this? 1
Oil-in-water type.

Question 9.
What do mean by Brownian movement in the case of a colloidal sol? j
The colloidal particles zig-zag or random motion is called Brownian movement. R. Brown was the first to study the zig-zag motion of pollen grains suspended in water.

Question 10.
What is the specific role of alum added to water?
The drinking water can be purified by precipitation of suspended colloidal particles. A small amount of Alum ;
[K2SO4.Al2(SO4)3.24H2O] is added. The Al3+ ions neutralize the charge on the particles and they get coagulated.

Question 11.
What happens when freshly precipitated ferric hydroxide is treated with ferric chloride? (P.S.B. 2005)
Peptisation takes place and a positively charged colloidal sol of ferric hydroxide is obtained.

Question 12.
What is the range of particle size in colloidal sol in nm? (H.S.B..2003) ,
1-100 nm.

Question 13.
What type of substances form lyophobic sols? (H.S.B. 2003)
Substances such as metals, their sulphides etc. which do not mix directly with the dispersion medium to form a colloidal sol.

Question 14.
How will you prepare colloidal solution of gold? (P.S.B. 2005)
By Bredig’s arc process or by reduction of AUCl3 with SnCl2.
2AUCl3 + 3SnCl2 → 2Au + 3 SnCl4

Question 15.
What is Occlusion? ,
The adsorption of gases on the surface of metals is called occlusion.

Question 16.
How does chemical adsorption of a gas on a solid vary with temperature? (D.S.B. 2003) ;
Chemical adsorption first increases and then decreases with increase of temperature.

Question 17.
Write down the heterogeneous catalyst involved in the polymerisation of ethylene. (l.I.T. 2003)
TiCl4 + trialkyl aluminium (Zeigler-Natta Catalyst).

Question 18.
What happens when gelatin is added to gold sol? (P.S.B.2001)
Gold sol which is lyophobic sol starts behaving like a lyophilic colloid when gelatin is added to it.

Question 19.
How will you obtain a colloidal sol of arsenious sulphide? (H.S.B. 2001)
Bypassing H2S through arsenious oxide solution.
Al2G3 + 3H2S → Ar2S3 + 3H2O

Question 20.
What happens when an electric field is applied to a colloidal dispersion? (A.I.S.B. 2002)
Colloidal particles move towards the oppositely charged electrode, get neutralised and coagulated (electrophoresis takes place).

Question 21.
What is ZSM-5? What is its formula?
ZSM-5 is a zeolite sieve of molecular porosity 5. Its formula is Hx(AlO2)x (SiO2)96-x]. 16H2O.

Question 22.
Which will absorb more gas, a lamp of charcoal or its powder and why? (D.S.B. 2005 C)
Powdered charcoal will adsorb more gas because it has more surface area than a lemp of charcoal.

Question 23.
How can we remove moisture from glass apparatus?
Anhydrous CaCl2 when placed in the glass apparatus will adsorb moisture from it.

Question 24.
What is the main cause of charge on a colloidal sol?
It is due to the adsprption of common ions in the electrolyte on the surface of the colloidal particles, e.g., Fe3+ ions from FeCl3 on the surface of Fe(OH)3 particles.

Question 25.
Give an expression of Freundlich isotherms.
$$\frac{x}{m}$$ = kp1/n or log x = log k + $$\frac{1}{\mathrm{n}}$$ log p
where m = mass of adsorbent; x = mass of adsorbate; p = gas pressure, k = constant and n is an integer.

Question 26.
What do x and m represent in the following expression? (C.B.S.E. Sample Paper 1997)
$$\frac{x}{m}$$ = kp1/n

Question 27.
Name two industrial processes in which heterogeneous catalysts are employed. (A.I.S.B. 1997)
Haber’s process for the manufacture of ammonia [Fe-catalyst] and contact process for manufacture of H2SO4 [V2O5-Catalyst].

Question 28.
To which colloidal system does milk belong? (P.S.B. 1999)
Liquid in liquid or oil-in-water type (emulsion).

Question 29.
Why do colloidal solutions exhibit Tyndall effect?
It is due to the size of colloidal particles (10° – 1000 A°) that they can scatter light.

Question 30.
Indicate a chemical reaction involving a homogeneous catalyst.

Question 31.
What happens when a beam of light is passed through AS2S3 sol? (P.S.B. 2001)
The path of light becomes visible because of Tyndall effect.

Question 32.
How can we make dialysis fast? (P.S.B. 1997)
By applying electric field across the parchment bag.

Question 1.
Name the factors on which the adsorption of a gas on a solid depends.
The extent of adsorption of a gas on a solid is affected by

• nature of the gas,
• nature of the solid,
• surface area of the solid,
• pressure of the gas,
• temperature,

Question 2.
What happens when persistent dialysis of a colloidal solution is carried out?
The stability of a colloidal sol is due to the presence of a small amount of electrolyte in it. On persistent dialysis, the electrolyte is completely removed. As a result, the colloidal sol becomes unstable and gets coagulated.

Question 3.
What is the difference in the nature of a dilute soap solution and a concentrated soap solution?
Dilute soap solution behaves like a true solution and a concentrated soap solution behaves like a colloidal sol.

Question 4.
Why are lyophilic sols more stable than Lyophobic colloidal sols?
Lyophilic sols are highly hydrated due to adsorption of water molecules as they are solvent-attracting. Lyophobic sols, on the other hand, hate the dispersion medium and are thus less stable.

Question 5.
How will you distinguish between an oil-in-water type and ‘ water-in-oil type of emulsion?
A small amount of an oil-soluble dye is added to the emulsion:
If it is water-in-oil type of emulsion, it becomes deeply coloured, otherwise, it remains colourless.

Question 6.
For the coagulation of 100 ml of arsenious sulphide sol, 5 ml of 1M NaCl is required. What is the flocculation value of NaCl?
(H.S.B. 2004)
5 ml of 1M NaCl contains NaCl = $$\frac{1}{1000}$$ x 5 moles .
= 5 millimoles
Thus 100 ml of As2S3 sol requires NaCl for complete coagulation = 5 millimoles
∴ 1L (1000 ml) of the sol will require for complete coagulation =50 millimoles of NaCl
∴ By definition, flocculation value of NaCl = 50.

Question 7.
The coagulation of 100 ml of a colloidal sol. of gold is completely prevented by addition 1 of 0.25 g of starch to it before adding 1 ml of 10% NaCl solution. Find out the gold number of starch.
Starch added to 100 ml of gold sol to completely prevent coagulation by 1 ml of 10% NaCl sol = 0.25 g = 250 mg starch required to be added to 10 ml of gold sol to completely prevent coagulation by 1 ml of 10% NaCl solution = 25 mt, By definition, gold number of starch = 25.

Question 8.
What do you mean by coagulation or flocculation value?
Trie minimum amount of an electrolyte (millimoles) that must be added to one litre of a colloidal sol so as to bring about complete coagulation or flocculation is called the coagulation or flocculation value of an electrolyte.

Question 9.
Which will be adsorbed more readily on the surface of charcoal and why – NH3 or CO2? (A.I.S.B. 2004 C)
NH3 has higher critical temperature than that of CO2, i.e., NH3 gas is more easily liquefiable than CO2 Hence NH3 has greater intermolecular forces of attraction and hence will be adsorbed more readily.

Question 10.
How is adsorption of a gas related to its critical temperature?
Higher is the critical temperature of a gas, greater is the case of liquefaction, i.e., greater are the Van der Waal’s forces of attraction and hence greater is the adsorption.

Question 11.
In case of chemisorption, why adsorption first increases and then decreases?
In the case of chemisorption, the head supplied in the beginning acts as activation energy required for chemisorption. The decrease of adsorption afterwards is due to exothermic nature of adsorptions equilibrium.

Question 12.
How can the constants k and n of the Freundlich adsorption equation be calculated? „
$$\frac{\mathrm{x}}{\mathrm{m}}$$ = kp1/n, ∴ log $$\frac{\mathrm{x}}{\mathrm{m}}$$ = log k + $$\frac{1}{\mathrm{n}}$$ log p.
Thus on plotting log $$\frac{\mathrm{x}}{\mathrm{m}}$$ Vs. log p, a straight line with slope = $$\frac{1}{\mathrm{n}}$$ and intercept on y-axis = log k is obtained.
Hence measuring the slope and intercept both k and n can be calculated.

Question 13.
Why silica gel is used as a dehumidifier?
Silica gel has strong adsorbing power for humidity, i.e.; moisture present in air. Hence it is used as a dehumidifier.

Question 14.
What type of colloidal sols are formed in the following?
(i) Sulphur vapour are passed through cooled water,
(ii) White of an egg is mixed with water,
(iii) Soap solution.
(i) Multimolecular because sulphur molecules associate together to form a colloidal sol.
(ii) Macromolecular because protein molecules present in the white of the egg are macromolecules soluble in water.
(iii) Associated because RCOO ions associate together to form micelles.

Question 15.
Which one of the following electrolytes is most effective for the coagulation of Fe(OH)3 sol and why? (A.I.S.B. 2004)
Fe(OH)3 is a positively charged sol. According to Hardy Schulze rule, greater the charge on the oppositely charged ion of the electrolyte added, more effective it is in bringing about coagulation. Hence Na3PO4 (containing PO4-3 ions) is most effective.

Question 16.
How do size of the particles of the adsorbant pressure of gas and prevailing temperature influence the extent of adsorption of a gas on a solid?

• Smaller the size of the particles of the adsorbent, greater is the surface area, greater is the adsorption.
• At constant temperature, adsorption first increases with increase in pressure and then attains equilibrium at high pressure.
• In physical adsorption, it decreases with increase of temperature, but in chemisorption, first it increases and then decreases.

Question 1.
(b) How adsorption proceeds in solutions.
(a) Adsorption-A substance which has different concentration at the surface than in the bulk phases is said to be adsorbed and the existence of a substance at a surface in a different, concentration than in the adjoining bulk phase is called adsorption.

ADSORBENT, ADSORBATE AND DESORPTION-The material upon whose surface the adsorption takes place is called an adsorbent while the molecular species that gets adsorbed are called adsorbate. The process of removal of an adsorbed substance from the surface on which it is adsorbed is called desorption, if is reverse of adsorption and can be brought about by heating or by reducing the pressure.

ABSORPTION-It is the phenomenon where there is uniform distribution of a substance in the bulk of another substance. For example, Ammonia is absorbed by water whereas it is adsorbed by charcoal. Similarly, water vapours are absorbed by anhydrous CaCl2 but are adsorbed by silica gel.

Enthalpy of Adsorption-The enthalpy change taking place for adsorption of one mole of adsorbate on an adsorbent smface is called enthalpy or heat of adsorption. In general heat of adsorption for 4 chemisorptions is more than that of physisorption. Heats of chemisorption are of the order of 40 to 400 kj mol-1 while those of physisorption are generally less than 40 kj mol-1. It is also observed that during adsorption there is a decrease in entropy i.e., TΔS is negative. Thus, for a process of adsorption,
Δ𝜏H = – ve; Δ𝜏S = – ve; Δ𝜏G = – ve.

COMPETING ADSORPTION-There is always a competition, between different adsorbates to adsorb on the absorbent. A strongly ‘ adsorbable substance can displace a weakly absorbed substance. For 1 example, on the charcoal taken in gas masks, gases such as O2, N2 etc. are already adsorbed. However, in the presence of poisonous gases such as Cl2 or CH4, which are very strongly adsorbable, O2 and N2, get displaced whereas CH4 and Cl2 get adsorbed.

If a mixture of gases is allowed to adsorb on a particular adsorbent, the more strongly adsorbable adsorbate adsorbs to greater extent than its partial pressure indicates. For example, moisture though present in, small proportion in air is strongly adsorbed by silica. Similarly, charcoal. also adsorbs polluting gases present in air in small concentration.

(b) ADSORPTION FROM SOLUTIONS—Solids can adsorb dissolved substances from solutions also. Common examples are the adsorption of colour impurities by activated Charcoal (carbon) in decolouring of solutions.
The chromatographic separation is possible due to different adsorption tendencies of the solutes of a solution. Many inorganic precipitates also act as adsorbents.

Freundlich adsorption and Langmuir adsorption isotherms have been found to be applicable in the adsorptions from solutions. Temperature dependence here also is similar to that for adsorption of gases. However, in place of equilibrium pressures, we use equilibrium concentrations of the adsorbates in the solutions and the isotherms take the form
$$\frac{\mathrm{x}}{\mathrm{m}}$$ = Kc1/n (n> 1), ………………. (1)
$$\frac{\mathrm{x}}{\mathrm{m}}$$ = $$\frac{a c}{1+b c}$$ ………………………. (2)
It is called LANGMUIR ADSORPTION ISOTHERM-The parameters of equations (1) and (2) can be determined as in the cases of adsorption of gases on solids.

Question 2.
(a) How will you distinguish between true solutions, colloidal solutions and suspensions?
(b) What are the applications of colloids in daily life?
(a) Points of difference between true solutions, colloidal COLLOIDS including emulsions find a number of uses in our daily life and industry. Some of the uses are given below :

COLLOIDS including emulsion find a number of uses in our daily life and industry. some of the uses are given below:
1. Rubber plating-The negatively charged rubber particles from the rubber sol are deposited on wares, handles of different tools, rubber gloves, etc., by electroplating.

2. Sewage disposal-Sewage water contains particle of dirt, rubbish etc., which are of colloidal size, carry charge and, therefore do not settle down easily. The particles can be removed by electrophoresis. Dirty water is passed through a tunnel fitted with metallic electrode which is maintained at high potential difference. The particles migrate to the oppositely charged electrode, lose their charge and are coagulated. The deposited matter is used as a manure and the water left behind is used for irrigation.

3. Cottrell smoke precipitator-Smoke is made free of colloidal particles by passing it through Cottrell precipitator installed in the chimney of an industrial plant. It consists of two metal discs charged to a high potential. The charge on the colloidal dust and carbon particles of smoke is neutralized and they are precipitated down, while free gases come out through the chimney.

4. Formation of delta-River water is a sort of colloidal solution of clay in water. Sea water, on other hand, contains a large number of dissolved salts and- is a sort of electrolyte. When river water comes in contact with sea water coagulation of colloidal particles occurs. These coagulated clay particles settle dpwn at the point of contact and gradually river bed starts rising. This leads river water to .adopt different course and ultimately result in the formation of delta.

5. Artificial rain-Artificial rain is caused by spraying oppositely charged colloidal dust or sand particles over the cloud by aeroplane. The colloidal water particles of the clouds get neutralised and coagulate to bigger water drops which cause artificial rain.

6. Purification of water-The colloidal particles present in water can be precipitated by addition of small amount of alum (K2SO4. Al2 (SO4)3 24H2O. The Al3+ ions furnished by alum help in coagulation of impurities which are present in the colloidal form.

7. Preparation of nano-materials-These materials are prepared for use as catalyst by using reverse micelles.
8. In Medicines-A wide variety of medicinal and Pharmaceutical preparations are emulsions. It is believed that in this form they can be more effective and are easily assimilated.
9. In disinfectants-The disinfectants such as Dettol arid Ivsol give emulsions of the oil-in-water type when mixed with water.
10. In Metallurgical preparations-Froth Floatation process uses the concentrated sulphide ore in emulsion of pine oil in water, when sulphide ore forms a froth and can be separated,
11. Building roads-Asphalt emulsified in water is used for building roads without the necessity of melting the asphalt.

Question 3.
Write brief notes on:
(i) Homogeneous catalysis,
(ii) Heterogeneous catalysis,
(iii) Enzyme catalysis.