Question

In a Freundlich adsorption isotherm, if the slope is unity and k is 0.1, the extent of adsorption at 2 atm is (log 2 = 0.30)

• A. 0.6
• B. 0.4
• C. 0.2
• D. 0.8

Hint:

Freundlich isotherm: $\frac{x}{m} = kp^{1/n}$.

Answer 1

The Correct Option is C

The Freundlich adsorption isotherm is given by: $$ \frac{x}{m} = kp^{1/n} $$ where $x$ is the mass of adsorbate, $m$ is the mass of adsorbent, $p$ is the pressure, $k$ is a constant, and $n$ is another constant. We are given that the slope (1/n) is unity (1), and $k = 0.1$. We want to find $x/m$ at $p=2$ atm. $$ \frac{x}{m} = (0.1)(2^1) = 0.2 $$