Question

In the Freundlich adsorption isotherm, log $\left(\frac{x}{m}\right)=logk+\frac{1}{n}logp,$ the value of n is

• A. any value from 0 to 1
• B. a positive integer
• C. a negative integer
• D. a positive or a negative fractional number

Answer 1

The correct option is B a positive integer

By Freundlich Adsorption Isotherm,
log $\left(\frac{x}{m}\right)=logk+\frac{1}{n}logP$

$\frac{x}{m}$ is the extent of adsorption which means the amount of gas adsorbed by the adsorbent with pressure at constant temperature.

k and n are constants that depends on the nature of gas and adsorbent.

In Freundlich Adsorption Isotherm,
At low temperature, 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, $\frac{1}{n}=0$
$\Rightarrow \frac{x}{m}=k$

Hence,
The value of $\frac{1}{n}$ lies between 0 to 1 under different pressure ranges.
So the value of 'n' cannot be a negative integer or a fractional number.
It can only be a positive integer.