Easily liquefiable gases such asNH₃, HCl etc. are adsorbed to a great extent in comparison to gasessuch as H₂, O₂ etc. This is because Van der Waal’sforces are stronger in easily liquefiable gases.

(2) Surface area of thesolid

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

(3) Effect of pressure

Adsorption is a reversibleprocess and is accompanied by a decrease in pressure. Therefore, adsorptionincreases with an increase in pressure.

(4) Effect of temperature

Adsorption is an exothermicprocess. Thus, in accordance with Le-Chatelier’s principle, the magnitude ofadsorption decreases with an increase in temperature.

Question - 12 : -
What is an adsorption isotherm?Describe Freundlich adsorption isotherm.

Answer - 12 : -

The plot between the extent of adsorption

against the pressure of gas(P) at constant temperature (T) is called the adsorption isotherm.

Freundlich adsorptionisotherm:

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

Fromthe given plot it is clear that at pressure P_S,

reaches the maximumvalve. Ps is called the saturation pressure. Three cases arisefrom the graph now.

Case I- At low pressure:

Theplot is straight and sloping, indicating that the pressure in directlyproportional to

i.e.

Case II- At high pressure:

Whenpressure exceeds the saturated pressure,

becomes independent of P values.

Case III- At intermediatepressure:

Atintermediate pressure,

depends on P raised to thepowers between 0 and 1. This relationship is known as the Freundlich adsorptionisotherm.

On plotting the graph between log

and log P, a straight line is obtainedwith the slope

equal to

and the intercept equal to log k.

Question - 13 : -
What do you understand byactivation of adsorbent? How is it achieved?

Answer - 13 : -

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

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

(ii) Somespecific treatments can also lead to the activation of the adsorbent. Forexample, wood charcoal is activated by heating it between 650 K and 1330 K invacuum or air. It expels all the gases absorbed or adsorbed and thus, creates aspace for adsorption of gases.

Question - 14 : -
What role does adsorption play inheterogeneous catalysis?

Answer - 14 : -

Heterogeneous catalysis:

A catalytic process in which thecatalyst and the reactants are present in different phases is known as aheterogeneous catalysis. This heterogeneous catalytic action can be explainedin terms of the adsorption theory. The mechanism of catalysis involves thefollowing steps:

(i) Adsorptionof reactant molecules on the catalyst surface.

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

(iii) De-sorptionof products from the catalyst surface

(iv) Diffusionof products away from the catalyst surface.

In this process, the reactantsare usually present in the gaseous state and the catalyst is present in thesolid state. Gaseous molecules are then adsorbed on the surface of thecatalyst. As the concentration of reactants on the surface of the catalyst increases,the rate of reaction also increases. In such reactions, the products have veryless affinity for the catalyst and are quickly desorbed, thereby making thesurface free for other reactants.

Question - 15 : -
Why is adsorption always exothermic?

Answer - 15 : -

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

(i) Adsorptionleads 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, adsorptionis always exothermic.

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

ΔG = ΔH − TΔS

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

Question - 16 : -
How are the colloidal solutionsclassified on the basis of physical states of the dispersed phase anddispersion medium?

Answer - 16 : -

One criterion for classifyingcolloids 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

Question - 17 : -
Discuss the effect of pressureand temperature on the adsorption of gases on solids.

Answer - 17 : -

Adsorption is a reversible processand is accompanied by a decrease in pressure. Therefore, adsorption increaseswith an increase in pressure.

Effect of temperature

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

Question - 18 : -
What are lyophilic and lyophobicsols? Give one example of each type. Why are hydrophobic sols easilycoagulated?

Answer - 18 : -

(i) Lyophilic sols:

Colloidal sols that are formed bymixing substances such as gum, gelatin, starch, etc. with a suitable liquid(dispersion medium) are called lyophilic sols. These sols are reversible innature i.e., if two constituents of the sol are separated by any means (such asevaporation), then the sol can be prepared again by simply mixing thedispersion medium with the dispersion phase and shaking the mixture.

(ii) Lyophobic sols:

When substances such as metalsand their sulphides etc. are mixed with the dispersion medium, they do not formcolloidal sols. Their colloidal sols can be prepared only by special methods.Such sols are called lyophobic sols. These sols are irreversible in nature. Forexample: sols of metals.

Now, the stability of hydrophilicsols depends on two things- the presence of a charge and the salvation ofcolloidal particles. On the other hand, the stability of hydrophobic sols isonly because of the presence of a charge. Therefore, the latter are much lessstable than the former. If the charge of hydrophobic sols is removed (byaddition of electrolytes), then the particles present in them come closer andform aggregates, leading to precipitation.

Question - 19 : -
What is the difference betweenmultimolecular and macromolecular colloids? Give one example of each. How areassociated colloids different from these two types of colloids?

Answer - 19 : -

(i) Inmulti-molecular colloids, the colloidal particles are an aggregate of atoms orsmall molecules with a diameter of less than 1 nm. The molecules in theaggregate are held together by van der Waal’s forces of attraction. Examples ofsuch colloids include gold sol and sulphur sol.

(ii) Inmacro-molecular colloids, the colloidal particles are large molecules havingcolloidal dimensions. These particles have a high molecular mass. When theseparticles are dissolved in a liquid, sol is obtained. For example: starch,nylon, cellulose, etc.

(iii) Certainsubstances tend to behave like normal electrolytes at lower concentrations.However, at higher concentrations, these substances behave as colloidalsolutions due to the formation of aggregated particles. Such colloids arecalled aggregated colloids.

Question - 20 : -
What are enzymes? Write in briefthe mechanism of enzyme catalysis.

Answer - 20 : -

Enzymes are basically proteinmolecules of high molecular masses. These form colloidal solutions whendissolved in water. These are complex, nitrogenous organic compounds producedby living plants and animals. Enzymes are also called ‘biochemical catalysts’.

Mechanismof enzyme catalysis:

the surface of the enzymes,various cavities are present with characteristic shapes. These cavities possessactive groups such as −NH₂, −COOH, etc. The reactant moleculeshaving a complementary shape fit into the cavities just like a key fits into alock. This leads to the formation of an activated complex. This complex thendecomposes to give the product.

Hence,

Step 1: E +S → ES⁺

(Activated complex)

Step 2: ES⁺ →E + P

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