Question

In Freundlich Adsorption isotherm, the value of $\frac{1}{n}$ is :

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

Answer 1

The correct option is C between 0 and 1 in all cases

By Freundlich Adsorption Isotherm,
$\frac{x}{m}\propto p^{\frac{1}{n}}$
n > 1
$\frac{x}{m}$ is the extent of adsorption which means the amount of gas adsorbed by the adsorbent with pressure at constant temperature.
$\frac{x}{m}=k{p}^{\frac{1}{n}}....\left(1\right)$
k and n are constants that depends on the nature of gas and adsorbent.
Taking log of equation (1) gives,

log $\left(\frac{x}{m}\right)=logk+\frac{1}{n}logP$

Here, the factor $\frac{1}{n}$ can have value between 0 to 1 based on the pressure range.

Freundlich isotherm holds good over a limited range of low pressure.
Hence, at low pressure, the isotherm is a straight line
$\therefore \frac{1}{n}=1$

$\Rightarrow \frac{x}{m}=kp$


At intermediate pressure,
$0<\frac{1}{n}<1$
Hence,
$\Rightarrow \frac{x}{m}=k{p}^{\frac{1}{n}}$


At high pressure, the extent of adsorption attains a constant value and becomes independent of pressure.
Thus, $\frac{1}{n}=0$
$\Rightarrow \frac{x}{m}=k{p}^0$
$\Rightarrow \frac{x}{m}=k$


$\therefore$
The value of $\frac{1}{n}$ lies between 0 to 1 under different pressure ranges.