Class 12 Chemistry - Chapter Surface Chemistry NCERT Solutions | What role does adsorption play in hetero

Welcome to the NCERT Solutions for Class 12th Chemistry - Chapter Surface Chemistry. This page offers a step-by-step solution to the specific question from Exercise 2, Question 7: what role does adsorption play in heterogeneous ca....
Question 7

What role does adsorption play in heterogeneous catalysis?

Answer

The phenomenon of concentration of molecules of a gas or liquid at a solid surface is called adsorption.

Heterogenous catalysis is a catalysis in which the catalyst is in different physical phase from the reactant. The most important of such reactions are those in which the reactants are in the gas phase while the catalyst is a solid.

This heterogeneous catalytic action can be explained in terms of the adsorption theory. The mechanism of catalysis involves the following steps:

(i) Adsorption of reactant molecules on the catalyst surface.

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

(iii) De-sorption of products from the catalyst surface.

(iv) Diffusion of products away from the catalyst surface.

(v) Diffusion of reactants to the surface of the catalyst.

The reaction occurs by the contact of reactants with the catalyst surface.The molecule of reactant (gas) are adsorbed at the catalyst surface where they form an ‘adsorption complex’. The catalyst surface behave like the surface molecules of a liquid and are not surrounded by atoms or molecules of their kind. Therefore, they have unbalanced  or residual attractive forces (Vander waal’s forces) on the surface which can hold adsorbate particles.As the concentration of reactants on the surface of the catalyst increases, the rate of reaction also increases. In such reactions, the products have very less affinity for the catalyst and are quickly desorbed, thereby making the surface free for other reactants.

Popular Questions of Class 12 Chemistry

Recently Viewed Questions of Class 12 Chemistry

5 Comment(s) on this Question

Write a Comment: