NCERT Solutions for Class 12 Chemistry Chapter 5 - Surface Chemistry

Intext Exercise

1. Write two characteristics of chemisorption.

Ans:

1. Chemisorption is a specific process that only occurs if there is a chemical bond between adsorbent and adsorbate.

2. Chemisorption is an endothermic reaction, so it is inversely proportional to temperature.
   
   

2. Why does the physisorption decrease with increase of temperature?

Ans: In Physisorption heat is released, this means it is an exothermic reaction. Exothermic reaction is inversely proportional to temperature according to Le Chateliere’s principle. Hence, physisorption decreases with increase of temperature.

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

Ans: Powdered substances have more surface area as compared to when they are in crystalline form. Physisorption is directly proportional to the surface area of the adsorbent that is why powdered substances are more effective adsorbents than their crystalline forms.

4. Why is it necessary to remove $\text{CO}$ when ammonia is obtained by Haber’s process?

Ans: In Haber’s process iron ($\text{Fe}$) is used as catalyst, $\text{CO}$ combines with $\text{Fe}$ to form $\text{Fe(CO}{{\text{)}}_{\text{5}}}.$ Hence, $\text{Fe}$ cannot act as catalyst hence slows the reaction.

5. Why is the ester hydrolysis slow in the beginning and becomes faster after some time?

Ans:

\[\]\[\text{Ester + Water}\to \text{Acid + Alcohol}\]

As seen above acid and alcohol is released in ester hydrolysis. Acid contains ${{\text{H}}^{\text{+}}}$ ions which act as catalyst (auto catalyst), Reaction is slow in the beginning and when ${{\text{H}}^{\text{+}}}$ is formed reaction speeds up.\[\]

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

Ans: The purpose of desorption in the catalysis process is to make the surface of the solid catalyst available for new adsorption of the reactants.

7. What modification can you suggest in the Hardy-Schulze law?

Ans: According to the Hardy-Schulze law, the higher the valence of the flocculating ion supplied, the greater its ability to precipitate.

Only the charge carried by an ion is taken into account by this law, not its size. The polarizing power of an ion increases as its size decreases. As a result, the Hardy-Schulze law can be tweaked in terms of the flocculating ion's polarizing strength. As a result, the modified Hardy-Schulze law can be stated as follows: ‘The greater the polarizing strength of the flocculating ion added, the greater its ability to generate precipitation.

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

Ans: Some ions that combine to produce the precipitate are adsorbed on the surface of the precipitate as it precipitates. As a result, it is necessary to wash the precipitate before quantifying it in order to eliminate any adsorbed ions or other contaminants.

NCERT Excercise

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

Ans:

2. What is the difference between physisorption and chemisorption?

Ans: The difference between the physisorption and Chemisorption are as following:

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

Ans: Adsorption is a phenomenon that occurs on the surface of a substance. As a result, the surface area has a direct relationship with adsorption. The surface area of a finely split material is big. With an increase in surface area, both physisorption and chemisorption rise. As a result, a finely divided substance works well as an adsorbent.

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

Ans: The rate of a gas's adsorption on a solid surface is influenced by a number of factors.

1. The gas's nature:

Gases that are easily liquefiable, such as $\text{N}{{\text{H}}_{\text{3}}}\text{,HCl}$ and others, are adsorbed to a greater extent than gases like ${{\text{H}}_{\text{2}}}\text{,}{{\text{O}}_{\text{2}}}\text{.}$ Because Van der Waals forces are stronger in easily liquefiable gases, this is the case.

2. The solid's surface area

The adsorption of a gas on a solid surface is proportional to the surface area of the adsorbent.

3. The impact of pressure

Adsorption is a reversible process that results in a drop in pressure. As a result, as pressure rises, adsorption rises as well.

4. The influence of temperature

The process of adsorption is exothermic. As a result, according to Le-principle, Chatelier's the amplitude of adsorption reduces as temperature rises.

5. What is an adsorption isotherm? Describe Freundlich adsorption isotherm.

Ans: The adsorption isotherm is a graph that shows the extent of adsorption against gas pressure $\left( P \right)$ at a constant temperature $\left( T \right).$

Freundlich adsorption isotherm: The Freundlich adsorption isotherm describes the empirical relationship between the amount of gas adsorbed by a unit mass of solid adsorbent and pressure at a given temperature.

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Case 1: At low pressure

The plot is straight and sloping, demonstrating that pressure is proportional to $x.$

Case 2: High-pressure situation:

When pressure exceeds saturated pressure, $x$ is no longer affected by $P$ values.

$\frac{x}{m}\propto {{P}^{{}^{1}/{}_{n}}}$

$\frac{x}{m}=k{{P}^{{}^{1}/{}_{n}}}$

Taking log both sides:

\[\log \frac{x}{m}=\log k+\frac{1}{n}\log P\]

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6. What do you understand about the activation of adsorbent? How is it achieved?

Ans: The adsorbing power of an adsorbent is boosted up by activating it. The following are some examples of how to activate an adsorbent:

(i) By expanding the absorbant's surface area. This can be accomplished by chopping it up or powdering it.

(ii) The adsorbent can also be activated by some specific treatments. Wood charcoal, for example, is activated by heating it between $650-1300{}^\circ \text{C}.$ Celsius in vacuum or air. It expels all absorbed or adsorbed gases, creating a place for gas adsorption.

7. What role does adsorption play in heterogeneous catalysis?

Ans: The catalyst is used in a catalytic process. Heterogeneous catalysis happens when the reactants are present in various stages. The adsorption hypothesis can be used to explain the heterogeneous catalytic action. The following steps make up the catalytic mechanism.

1. Reactant molecule adsorption on the catalyst surface.

2. A chemical reaction that occurs as a result of the creation of an intermediate.

3. Product desorption from the catalyst surface.

4. Product diffusion away from the catalyst surface.

The reactants are normally present in a gaseous state in this process, whereas the catalyst is present in a solid state. The gaseous molecules are then adsorbed on the catalyst's surface. The rate of reaction increases as the concentration of reactants on the catalyst's surface rises. In such reactions, the products have a low affinity for the catalyst and are quickly desorbed, allowing new reactants to access the surface.

8. Why is adsorption always exothermic?

Ans: The process of adsorption is always exothermic. There are two ways to understand this sentence.

(I) Adsorption causes the residual forces on the adsorbent's surface to diminish. The adsorbent's surface energy decreases as a result of this. Adsorption is usually exothermic as a result.

(II) $\Delta H$ 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., $\Delta S$ is negative. Now for a process to be spontaneous, $\Delta G$ should be negative.

$\Delta G=\Delta H-T\Delta S$

Since, $\Delta S$ is negative, $\Delta H$ has to be negative to make $\Delta G$ negative. Hence, adsorption is always exothermic.

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

Ans: The physical state of the dispersed phase and dispersion medium is one criterion for identifying colloids. There are eight different types of colloidal systems depending on the dispersed phase and dispersion medium (solid, liquid, or gas).

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

Ans: The impact of pressure: Adsorption is a reversible process that results in a drop in pressure. As a result, as pressure rises, adsorption rises as well.

Temperature's Impact: The process of adsorption is exothermic. As a result, according to Le-principle, Chatelier's the amplitude of adsorption reduces as temperature rises.

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

Ans:

1. Lyophilic sols: Lyophilic sols are colloidal sols made by mixing ingredients like gum, gelatin, and starch with a suitable liquid (dispersion medium). These sols are reversible in nature, which means that if two of the sol's parts are separated by any means, the sol will reverse (such as evaporation). The sol can then be made by mixing the dispersion medium with the dispersion phase and stirring the mixture again.

2. Lyophobic sols: When metals and their sulphides, for example, are combined with the dispersion medium, colloidal sols are not formed. Only particular processes can be used to make these colloidal sols. Lyophobic sols are those that are lyophobic. In nature, these sols are irreversible. As an example. Metals are divided into sols. The presence of a charge and the survival of colloidal particles now determine the stability of hydrophilic sols. Hydrophobic sols, on the other hand, are only stable because of the existence of a charge. As a result, the latter are considerably less stable than the former.
   
   

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?

Ans:

Associated colloid: At low quantities, some compounds behave like regular electrolytes. These chemicals, however, behave as colloidal solutions at larger concentrations due to the production of aggregated particles.

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

Ans: Enzymes are large protein molecules with a high molecular weight. When they are dissolved in water, they produce colloidal solutions. These are intricate. Living plants and animals produce nitrogenous organic molecules. Biochemical catalysts are another name for enzymes.

Various cavities with distinct shapes can be seen on the surface of enzymes. These have active groups like $\text{N}{{\text{H}}_{\text{3}}}\text{,COOH}$ and so on. The complementary-shaped reactant molecules fit into the cavities like a key fits into a lock. An active complex is formed as a result of this. The product is formed when the complex decomposes.

Hence, Step 1: $E+S\to E{{S}^{*}}$

Step 2: $E{{S}^{*}}\to E+P$

14. How are colloids classified on the basis of

(a) Physical states of components

Ans: Colloids are classified in a variety of ways:

On the basis of the physical state of components' (by components we mean the dispersed phase and dispersion medium). There are eight different forms of colloids depending on whether the components are solids, liquids, or gases.

(b) Nature of dispersion medium

Ans: On the basis of nature of dispersion medium, colloids are classified as,

(c) Interaction between dispersed phase and dispersion medium

Ans: On the nature of the interaction between the dispersed phase and the dispersion medium (solvent repelling) colloids are characterized as lyophilic (solvent attractive) or lyophobic (solvent repellent).

15. Explain what is observed

(a) When a beam of light is passed through a colloidal sol.

Ans: Light scattering is detected when a beam of light passes through a colloidal solution. The Tyndall effect is the name for this phenomenon. The path of the beam in the colloidal solution is illuminated by this scattering of light.

(b) An electrolyte, is added to hydrated ferric oxide sol.

Ans:  dissociates into  and  ions when introduced to ferric oxide sol. The ferric oxide sol particles are positively charged. In the presence of negatively charged  ions, they coagulate.

(c) Electric current is passed through a colloidal sol.

Ans: Colloidal particles are charged, and they might be positive or negative. The charge in the dispersion medium is equal and opposing. As a result, the entire system becomes neutral. The colloidal particles travel towards the oppositely charged electrode when an electric current is applied.

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

Ans: An emulsion is a colloidal solution in which both the dispersed phase and the dispersion medium are liquids. Emulsions are divided into two categories: liquid emulsions and solid emulsions.

1. Oil in water type: The dispersed phase is oil, and the dispersion medium is water. For instance, milk disappearing cream.

2. Water in an oil type: The dispersed phase is water, and the dispersion medium is oil. For example, cold cream.
   
   

17. What is demulsification? Name two demulsifiers.

Ans: Demulsification is the process of breaking down an emulsion into its constituent liquids. Surfactants, ethylene oxide, and other demulsifiers are examples.

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

Ans: Emulsification and micelle production are responsible for soap's cleaning activity. Soaps are sodium and post-sodium salts of long-chain fatty acids $\left( \text{R}-\text{CO}{{\text{O}}^{-}}\text{N}{{\text{a}}^{+}} \right).$ The sodium-attached end of the molecule is polar in nature, but the alkyl-end is non-polar. As a result, a soap molecule has both a hydrophilic (polar) and a hydrophobic (non-polar) component. When soap is applied to water containing dirt, the soap molecules surround the dirt particles in such a way that the hydrophobic parts of the soap molecules adhere to the dirt molecule while the hydrophilic parts point away. Micelle formation is the term for this. As a result, the polar group dissolves in water, whereas the non-polar group dissolves in dirt particles.

19. Give four examples of heterogeneous catalysis.

Ans:

(i) Sulfur dioxide is oxidized to generate sulfur trioxide. $\text{Pt}$ serves as a catalyst in this process.

\[\text{2S}{{\text{O}}_{\text{2}}}\to \text{2S}{{\text{O}}_{\text{3}}}\]

(ii) Ammonia is formed when dinitrogen and dihydrogen react in the presence of finely split iron.

\[{{\text{N}}_{\text{2}}}+\text{3}{{\text{H}}_{2}}\to \text{2N}{{\text{H}}_{\text{3}}}\]

(iii) Oswald's process: In the presence of platinum, ammonia is converted to nitric oxide.

\[\text{N}{{\text{H}}_{3}}+{{\text{O}}_{2}}\to \text{NO}\]

(iv) In the presence of $\text{Ni,}$ vegetable hydrogenation occurs.

$\text{Vegetable oil}+{{\text{H}}_{\text{2}}}\to \text{Vegetable ghee}$

20. What do you mean by activity and selectivity of catalysts?

Ans:

1. Catalyst activity: A catalyst's activity is defined as its capacity to accelerate the rate of a specific reaction. Chemisorptions are the most important component in determining a catalyst's activity. Reactant adsorption on the catalyst surface should not be excessively strong or too weak. It only needs to be powerful enough to activate the catalyst.

2. Selectivity of the catalyst: The selectivity of the catalyst refers to its capacity to steer a reaction to produce a certain product. For example, we can acquire different products for the reaction between ${{\text{H}}_{\text{2}}}$ and $\text{CO}$ by employing different catalysts.
   
   

21. Describe some features of catalysis by zeolites.

Ans: Zeolites are alumino – silicates with microporous surfaces. Because of their honeycomb-like structure, zeolites are shape-selective catalysts. They have an $\text{Al}-\text{O}-\text{Si}$ framework with an expanded 3D-network of silicates in which certain silicon atoms are substituted by aluminium atoms. The pores and cavity size of zeolites are extremely important in the reactions that take place within them. In the petrochemical sector, zeolites are widely used.

22. What is shape selective catalysis?

Ans: A catalytic reaction that is influenced by the catalyst's pore shape as well as the size of the reactant and product. Shape-selective catalysis is a term used to describe how molecules react with each other. Catalysis using zeolites, for example, is a shape-selective catalysis. The pore size of the zeolites varies between $260-74\text{0 pm}.$ Molecules with pore sizes larger than this are unable to penetrate the zeolite and complete the reaction.

23. (i) Electrophoresis

Ans: Electrophoresis is the movement of colloidal particles under the influence of an applied electric field. The cathode attracts positively charged particles, while the anode attracts negatively charged particles. The particles become neutral and coagulate as they reach oppositely charged electrodes.

(ii) Coagulation

Ans: Coagulation is the process of setting down colloidal particles, or the conversion of a colloid to a precipitate. Persistent dialysis, persistent boiling, electrophoresis, and mutual coagulation are all methods for achieving coagulation.

(iii) Dialysis

Ans: Dialysis is the process of extracting a dissolved component from a colloidal solution through diffusion through a membrane. This method is based on the fact that ions and tiny molecules, unlike colloidal particles, can flow through mammalian membranes.

(iv) Tyndall effect

Ans: The Tyndall effect: When a beam of light passes through a colloidal solution, it appears as a column of light. The Tyndall effect is the name for this phenomenon. Because colloidal particles scatter light in all directions, this effect occurs.

24. Give four uses of emulsions.

Ans: Emulsions are used in four ways: 

(i) Soaps' cleansing activity is based on the creation of emulsions.

(ii) Emulsification is the process by which lipids are digested in the intestines.

Antiseptics and disinfectants form emulsions when mixed with water.

(iv) Medicines are made via the emulsification technique.

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

Ans: Micelles are powerful electrolytes that act normally at low concentrations but become colloids at high concentrations due to the formation of aggregates. Associated colloids is another name for them. Micelle creation occurs above a certain concentration, known as the Critical Micelle Concentration, and at a specific temperature, known as the Kraft Temperature. They revert to the real solution when related colloids are diluted. For example, Soap $\left( {{\text{C}}_{\text{17}}}{{\text{H}}_{\text{35}}}\text{COONa} \right)$ sodium stearate.

26. Explain the terms with suitable examples:

(i) Alcosol

Ans: A colloidal solution of a solid (dispersed phase) in alcohol is known as alcosol (dispersion medium). For example, cellulose nitrate colloidal solution in ethyl alcohol.

(ii) Aerosol

Ans: A colloidal solution of a liquid (dispersed phase) in gas is called an aerosol (Dispersion medium). Fog is an example.

(iii) Hydrosol

Ans: Fresh flowers, leaves, fruits, and other plant materials are distilled into hydrosols, which are water-based products.

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

Ans: In a benzene medium, common salt (a typical crystalloid in an aqueous medium) acts as a colloid. As a result, we may argue that colloidal substances do not belong to a distinct class of substances. It acts as a colloid when the size of the solute particle is between 1 nm and 100 nm.

As a result, we may argue that colloid is not a substance but a condition of a material that is determined by particle size. Between a true solution and a suspension, a colloidal condition exists.

Surface Chemistry Class 12 NCERT Solution PDF Download

The NCERT Solutions for Class 12 Chemistry is available in PDF formats. Students can easily access the PDF of the solutions from e-learning websites like Vedantu. These PDF versions can act as a handbook and can prove extremely helpful for quick references and revision purposes.

Below-listed are some of the examples from Class 12 Chemistry Chapter 5 NCERT Solutions:

• NCERT Solutions of Surface Chemistry Class 12: Question 1 & 2

The solutions to these questions help the students understand the characteristics of Chemisorption and Physisorption. It illustrates the chemical bonds and the reaction to temperature in both these adsorption processes.

• NCERT Solution for Surface Chemistry Class 12: Question 3 & 4

The solution to question 3 helps the students understand how an increase in the surface area makes the powdered substance a better adsorbent than crystalline forms.

Similarly, the solution to Question 4 also explains the Necessity to remove CO to obtain ammonia (by Haber’s process) in detail. Students get a precise understanding from the detailed explanation of all the critical concepts.

• Surface Chemistry NCERT Solution Class 12: Question 5 & 6

The solution provides insights into the role and description of the process of catalysis. It also teaches the students how ester hydrolysis is represented as Ester + Water = Acid + Alcohol. Learning this solution will help the students easily answer the very short questions of 1 mark during the exam.

• Class 12 Chemistry Chapter 5 NCERT Solution: Question 7

The Surface Chemistry Class 12 NCERT solutions help the students to have an understanding of modifications that can be made in Hardy Schulze Law. The solution illustrates in detail how it can be modified in terms of the polarising power of flocculating ions. The student can clarify the concept of Hardy Schulze law through the explanation.

The NCERT Solution of Class 12 Chemistry chapter 5 surface Chemistry also provides an answer to in-text questions as given in the following.

Example of Class 12 Chemistry Chapter 5 NCERT Solutions for In - Text Question

Example 1: Question 1

Question one offers a detailed definition of terms like adsorption. It also provides the characteristic distinction between the two processes. The solution is essential for students to properly understand the basics of the chapter.

Example 2: Question 2

Through elaborate tables and charts, the solution presents the fundamental difference between the process of Physisorption and Chemisorption. The tables are easier to understand for the students and they can easily compare the difference between the two processes.

Example 3: Question 8

The solution to question 8 will help the students understand the concept of Enzyme and the mechanism of enzyme catalysis. The solution comes with a detailed illustrative diagram that will explain the process by providing concepts.

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The marks distribution for Chapter 5 is given in the table below.

Class 12 Surface Chemistry Marks Distribution

Benefits of Surface Chemistry Class 12 NCERT PDF

The PDF version of Surface Chemistry Class 12 NCERT Solutions are convenient for students due to the following factors.

• Free download option.

• Illustration through diagrams.

• Elaborate explanation of questions.

• Curated by expert teachers.

• Adequate examples for better understanding.

By thoroughly studying the textbooks and rigorously practising the exercises and sample papers, any student can garner the necessary expertise to ace their board exams as well as prepare themselves for any competitive exams that they might sit for in future.

NCERT Solutions for Class 12 Chemistry 

• Chapter 1 - The Solid State

• Chapter 2 - Solutions

• Chapter 3 - Electrochemistry

• Chapter 4 - Chemical Kinetics

• Chapter 5 - Surface Chemistry

• Chapter 6 - General Principles and Processes of Isolation of Elements

• Chapter 7 - The p-Block Elements

• Chapter 8 - The d and f Block Elements

• Chapter 9 - Coordination Compounds

• Chapter 10 - Haloalkanes and Haloarenes

• Chapter 11 - Alcohols, Phenols and Ethers

• Chapter 12 - Aldehydes, Ketones and Carboxylic Acids

• Chapter 13 - Amines

• Chapter 14 - Biomolecules

• Chapter 15 - Polymers

• Chapter 16 - Chemistry in Everyday life

NCERT Solutions Class 12 Chemistry - Related Links

Click to access the links listed below to refer to Vedantu’s CBSE Class 12 Chemistry syllabus, important questions, revision notes,  previous years’ questions papers and sample papers.

• Class 12 Chemistry Revision Notes

• CBSE Class 12 Chemistry Previous Year Question Papers 

• CBSE Sample Papers for Class 12 Chemistry with Solutions

• CBSE Syllabus for Class 12 Chemistry Syllabus (Term 1& 2)

• Important Questions for CBSE Class 12 Chemistry

• NCERT Exemplar for Class 12 Chemistry

• NCERT Solutions for Class 12 Maths

• NCERT Solutions for Class 12 Physics

• NCERT Solutions for Class 12 Biology

• Revision Notes for Class 12

• Important Questions for Class 12

Vedantu’s NCERT Solutions deal with all crucial topics concerning Chapter 5 Surface Chemistry. These have been carefully and meticulously developed to help students understand the definitions and classifications extensively. The NCERT solutions also prepare students to think analytically and apply their concepts to the questions asked in the exam. For all medical and JEE aspirants, NCERT Class 12 Chemistry Solutions are a must in order to kickstart their preparation and stay ahead of the curve.