CBSE Class 12 Chemistry Chapter 5 Surface Chemistry NCERT Solutions

Reviewed by NCERT Help Editorial Team · Updated on 29 Jul 2026

NCERT Solutions PDF Class 12 PDF

CBSE Class 12 Chemistry, Chapter 5: Surface Chemistry introduces students to the fascinating world of interactions occurring at surfaces. This chapter explores key concepts such as adsorption, differentiating between chemisorption and physisorption, and highlights the critical role of surface area. It also delves into the mechanisms of catalysis, including homogeneous and heterogeneous catalysis, and examines specific industrial applications like the Haber's process for ammonia synthesis. The Hardy-Schulze law, which explains the coagulation of colloidal particles, and the concept of autocatalysis are also discussed. Understanding these principles is fundamental for comprehending various chemical phenomena and their applications in diverse fields. These solutions aim to provide clear and concise explanations for intext questions, reinforcing learning and aiding in exam preparation.

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Board	CBSE
Class	Class 12
Subject	Chemiry
Session	2026
Language	English
Type	NCERT Solutions
Chapter	Chapter 5: Surface Chemistry - Intext Questions Solutions

Chapter summary

This chapter's NCERT Solutions focus on intext questions related to Surface Chemistry. It explains the characteristics of chemisorption and physisorption, the impact of temperature and surface area on adsorption, and the significance of catalyst poisoning in processes like Haber's. The solutions also cover autocatalysis and the Hardy-Schulze law, offering a modified perspective. These explanations are vital for understanding the fundamental concepts of adsorption, catalysis, and colloidal chemistry.

Learning outcomes

Understand the characteristics of chemisorption and physisorption.
Explain the effect of temperature and surface area on adsorption.
Identify the role of catalysts and catalyst poisons in chemical reactions.
Explain the concept of autocatalysis with an example.
Understand and apply the Hardy-Schulze law and its modification.

Topics covered

Paper topics

Adsorption
Physisorption
Chemisorption
Factors affecting adsorption (Surface area, Temperature)
Catalysis
Catalyst poisons
Haber's Process
Autocatalysis
Hardy-Schulze Law
Desorption

Important topics

Physisorption vs. Chemisorption
Factors affecting Adsorption
Role of Catalysts
Hardy-Schulze Law
Autocatalysis

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Questions and Solutions

Question 5.1

Write any two characteristics of Chemisorption.

Solution: Two key characteristics of chemisorption are:

High Specificity: Chemisorption is highly specific. It occurs only when there is a strong possibility of forming chemical bonds between the surface of the adsorbent and the molecules of the adsorbate. For example, oxygen is adsorbed on metals, but hydrogen is not, due to the difference in chemical bonding possibilities.
Irreversibility: Chemisorption is often irreversible or only partially reversible because it involves the formation of strong chemical bonds, similar to chemical reactions.

Question 5.2

Why does physisorption decrease with the increase of temperature?

Solution: Physisorption is an exothermic process, meaning it releases heat. According to Le-Chatelier's principle, if a system at equilibrium is subjected to a change in temperature, the position of equilibrium will shift in a direction that tends to counteract the change. In the case of physisorption, increasing the temperature provides energy to the adsorbed molecules, facilitating their desorption from the surface. Therefore, the extent of physisorption decreases as the temperature rises. It is generally more effective at lower temperatures.

Question 5.3

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

Solution: Powdered substances are more effective adsorbents because powdering a substance significantly increases its surface area compared to its crystalline form. Adsorption, particularly physisorption, is a surface phenomenon. The extent of adsorption is directly proportional to the surface area of the adsorbent. A larger surface area provides more active sites for the adsorbate molecules to adhere to, leading to more efficient adsorption.

Question 5.4

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

Solution: In Haber's process, which is used for the synthesis of ammonia from nitrogen and hydrogen, an iron catalyst is employed. Carbon monoxide (CO) is a common impurity that can be present in the reactant gases. CO acts as a catalyst poison; it strongly adsorbs onto the active sites of the iron catalyst, blocking them and significantly reducing the catalyst's efficiency and activity. Therefore, it is essential to remove CO from the synthesis gas to maintain the catalyst's performance and ensure a high yield of ammonia.

Question 5.5

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

Solution: The hydrolysis of an ester in the presence of an acid catalyst is represented by the equation:

Ester + Water $\rightarrow$ Acid + Alcohol

Initially, the concentration of the acid catalyst is low (or zero if no external acid is added), so the reaction proceeds slowly. However, the reaction itself produces an acid (one of the products). This newly formed acid acts as a catalyst for the ester hydrolysis. As the reaction progresses, the concentration of this autocatalyst increases, leading to a progressive increase in the reaction rate. Thus, the ester hydrolysis starts slow and becomes faster over time due to autocatalysis.

Question 5.6

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

Solution: In heterogeneous catalysis, reactants are adsorbed onto the surface of the solid catalyst. After the reaction occurs on the surface, the products are formed. Desorption is the process by which these product molecules detach from the catalyst's surface. The role of desorption is crucial because it frees up the active sites on the catalyst's surface. This allows fresh reactant molecules to adsorb and react, ensuring the catalytic process can continue efficiently. If desorption is slow, the catalyst surface can become blocked by products, hindering further catalysis.

Question 5.7

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

Solution: The Hardy-Schulze law states that the flocculating power of an ion is directly proportional to the magnitude of its charge. For example, in the coagulation of a negative sol, the flocculating power increases in the order $Al^{3+} > Ba^{2+} > Na^{+}$. While this law is generally true, it primarily considers only the charge. A modification can be suggested by considering the polarising power of the flocculating ion, which is related to its size and charge distribution. Smaller ions with higher charge density tend to have greater polarising power and can more effectively destabilize colloidal particles. Therefore, a modified Hardy-Schulze law could state that the greater the polarising power of the flocculating ion, the greater is its power to cause precipitation (coagulation).

Question 5.8

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

Solution: When a precipitate is formed in a solution, it often carries down or adsorbs some impurities from the solution onto its surface. These impurities might be soluble ions from the original solution or other soluble salts formed during the reaction. If these impurities are not removed before quantitative estimation (e.g., by weighing the dried precipitate), they will contribute to the measured mass, leading to inaccurate results. Washing the precipitate with a suitable solvent, usually water, helps to remove these soluble impurities without dissolving the precipitate itself, thus ensuring a more accurate quantitative determination.

Common mistakes

Confusing chemisorption and physisorption characteristics.
Not understanding the effect of temperature on adsorption equilibrium.
Incorrectly applying the Hardy-Schulze law without considering ion size.
Overlooking the role of desorption in catalysis.

Revision tips

Focus on the key differences between physisorption and chemisorption.
Review the examples of catalysis and catalyst poisoning.
Practice applying the Hardy-Schulze law to different electrolyte combinations.
Understand the concept of autocatalysis and its mechanism.

Practice MCQs

Q1. Which of the following is a characteristic of chemisorption?

Q2. Why does physisorption decrease with increasing temperature?

Q3. Powdered substances are better adsorbents because:

Q4. In Haber's process for ammonia synthesis, why is CO removed?

Q5. What is an autocatalyst?

Frequently asked questions

What are the main differences between physisorption and chemisorption?

Physisorption is weak, reversible, non-specific, and occurs at low temperatures, while chemisorption is strong, often irreversible, highly specific, and can occur at higher temperatures due to chemical bond formation.

Why is a larger surface area important for adsorption?

Adsorption occurs on the surface of the adsorbent. A larger surface area provides more sites for the adsorbate molecules to attach, thus increasing the extent of adsorption.

What is the significance of desorption in catalysis?

Desorption is crucial in catalysis as it removes the product molecules from the catalyst's surface, freeing up active sites for fresh reactant molecules to adsorb and react.

How does the Hardy-Schulze law relate to flocculation?

The Hardy-Schulze law states that the flocculating power of an electrolyte is directly related to the magnitude of the charge on the ion that has the opposite sign to that of the colloid.

Can you give an example of autocatalysis?

Yes, the hydrolysis of esters is a common example where the acid produced acts as an autocatalyst, speeding up the reaction over time.

Why is CO removed in Haber's process?

CO is removed because it acts as a catalyst poison, deactivating the iron catalyst used in the synthesis of ammonia.

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