Question

Which of the following is not correct about Freundlich adsorption isotherm?

• A. x/m = k p^1/n (n > 1)
• B. Extent of adsorption of gas is more at high temperature than at low temperature
• C. \( \frac{1}{n} \) represents the slope of the isotherm (log-log plot)
• D. \( \log \frac{x}{m} = \log k + \frac{1}{n} \log p \) holds good over a limited range of pressures

Hint:

Contrast with Chemisorption: Chemisorption first increases with temperature (needs activation energy) and then decreases. Physisorption always decreases with temperature.

Answer 1

The Correct Option is B

Step 1: Understanding Freundlich Isotherm:

It describes physical adsorption (Physisorption). Equation: \( \frac{x}{m} = k p^{1/n} \), where \( n \textgreater 1 \).
Step 2: Analyzing the Options:

- (A) Equation:
Correct representation. - (B) Temperature Effect:
Physisorption involves weak van der Waals forces and is an exothermic process (\( \Delta H \textless 0 \)). According to Le Chatelier's principle, increasing temperature favors the reverse process (desorption). Thus, adsorption decreases as temperature increases. The statement says adsorption is more at high temperature, which is Incorrect. - (C) Slope:
Taking logs, \( \log \frac{x}{m} = \log k + \frac{1}{n} \log p \). A plot of \( \log(x/m) \) vs \( \log p \) is linear with slope \( 1/n \). Correct. - (D) Limitation:
The empirical isotherm fails at high pressures where adsorption reaches saturation (independent of pressure). It is valid only for a limited intermediate pressure range. Correct.
Step 3: Final Answer:

Statement (B) is not correct.