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:
x/m = k p1/n, where n > 1.

Step 2: Analyzing the Options:

(A) Equation:
This is the correct representation of the Freundlich adsorption isotherm.

(B) Temperature Effect:
Physisorption involves weak van der Waals forces and is an exothermic process.
So, ΔH < 0.

According to Le Chatelier's principle, increasing temperature favors the reverse process, that is, desorption.
Therefore, adsorption decreases as temperature increases.

The statement says adsorption is more at high temperature, which is incorrect.

(C) Slope:
Taking logarithm on both sides:
log(x/m) = log k + (1/n) log p

So, a plot of log(x/m) versus log p is a straight line with slope 1/n.
Hence, this statement is correct.

(D) Limitation:
The Freundlich isotherm is empirical and fails at high pressure because adsorption reaches saturation and becomes independent of pressure.
So, it is valid only over a limited intermediate pressure range.
Hence, this statement is correct.

Step 3: Final Answer:

Statement (B) is not correct.