NCERT Class 12 Chemistry Solutions Surface Chemistry with answers available in Pdf for free download. The NCERT Solutions for Class 12 Chemistry with answers have been prepared as per the latest syllabus, NCERT books and examination pattern suggested in Standard 12 by CBSE, NCERT and KVS. Solutions to questions given in NCERT book for Class 12 Chemistry are an important part of exams for Grade 12 Chemistry and if practiced properly can help you to get higher marks. Refer to more Chapter-wise Solutions for NCERT Class 12 Chemistry and also download more latest study material for all subjects
Surface Chemistry Class 12 NCERT Solutions
Class 12 Chemistry students should refer to the following NCERT questions with answers for Surface Chemistry in standard 12. These NCERT Solutions with answers for Grade 12 Chemistry will come in exams and help you to score good marks
Surface Chemistry NCERT Solutions Class 12
NCERT Class 12 Chemistry Solutions Surface Chemistry - NCERT Solutions prepared for CBSE students by the best teachers in Delhi.
Class XII Chemistry
Chapter 5 – Surface Chemistry
Write any two characteristics of Chemisorption.
1. Chemisorption is highly specific in nature. It occurs only if there is a possibility of chemical bonding between the adsorbent and the adsorbate.
2. Like physisorption, chemisorption also increases with an increase in the surface area of the adsorbent.
Why does physisorption decrease with the increase of temperature?
Physisorption is exothermic in nature. Therefore, in accordance with Le-Chateliere’s principle, it decreases with an increase in temperature. This means that physisorption occurs more readily at a lower temperature.
Why are powdered substances more effective adsorbents than their crystalline forms?
Powdered substances are more effective adsorbents than their crystalline forms because when a substance is powdered, its surface area increases and physisorption is directly proportional to the surface area of the adsorbent.
Why is it necessary to remove CO when ammonia is obtained by Haber’s process?
It is important to remove CO in the synthesis of ammonia as CO adversely affects the activity of the iron catalyst, used in Haber’s process.
Why is the ester hydrolysis slow in the beginning and becomes faster after sometime?
Ester hydrolysis can be represented as:
The acid produced in the reaction acts as a catalyst and makes the reaction faster. Substances that act as catalysts in the same reaction in which they are obtained as products are known as autocatalysts.
What is the role of desorption in the process of catalysis?
The role of desorption in the process of catalysis is to make the surface of the solid catalyst free for the fresh adsorption of the reactants on the surface.
What modification can you suggest in the Hardy-Schulze law?
Hardy-Schulze law states that ‘the greater the valence of the flocculating ion added, the greater is its power to cause precipitation.’
This law takes into consideration only the charge carried by an ion, not its size. The smaller the size of an ion, the more will be its polarising power. Thus, Hardy-Schulze law can be modified in terms of the polarising power of the flocculating ion. Thus, the modified Hardy-Schulze law can be stated as ‘the greater the polarising power of the flocculating ion added, the greater is its power to cause precipitation.’
Why is it essential to wash the precipitate with water before estimating it quantitatively?
When a substance gets precipitated, some ions that combine to form the precipitate get adsorbed on the surface of the precipitate. Therefore, it becomes important to wash the precipitate before estimating it quantitatively in order to remove these adsorbed ions or other such impurities.
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