Question

In Freundlich adsorption isotherm, the value of \( \frac{1}{n} \) is:

• A. 1 in case of chemisorption
• B. 1 in case of physisorption
• C. between 0 and 1 in all cases
• D. between 2 and 4 in all cases

Hint:

In the Freundlich adsorption isotherm, the value of \( \frac{1}{n} \) lies between 0 and 1 in most cases, indicating the extent of adsorption, with physisorption typically having a value closer to 1 and chemisorption having a value between 0 and 1.

Answer 1

The Correct Option is C

The Freundlich adsorption isotherm is an empirical relationship that describes the adsorption of molecules onto a solid surface. The equation is given by: \[ x/m = k P^{1/n} \] Where: - \( x/m \) is the amount of adsorbate adsorbed per unit mass of adsorbent. - \( P \) is the pressure of the gas at equilibrium. - \( k \) and \( n \) are empirical constants, where \( n \) is a measure of the adsorption intensity. The value of \( \frac{1}{n} \) has a significant impact on the adsorption process: - When \( \frac{1}{n} = 1 \), the adsorption follows a linear isotherm, which corresponds to physisorption (physical adsorption). - When \( \frac{1}{n} \) is between 0 and 1, the adsorption process is exothermic and is typical for physisorption. - For chemisorption, the value of \( \frac{1}{n} \) is typically between 0 and 1. -
Therefore, in all cases, the value of \( \frac{1}{n} \) is between 0 and 1.
Thus, the correct answer is Option (3). Diagram for Freundlich Adsorption Isotherm: Here is the diagram showing the Freundlich adsorption isotherm: ```latex \begin{center} \includegraphics[width=0.4\textwidth]{q37_digrm.png} \end{center}