Question:

In Freundlich adsorption isotherm, slope of $\mathrm{AB}$ line is

1. $\frac{1}{n}$ with $\left(\frac{1}{n}=0\right.$ to 1$)$

2. $\log \frac{1}{n}$ with $(n<1)$

3. $\log n$ with $(n>1)$

4. $\mathrm{n}$ with $(\mathrm{n}, 0.1$ to $0.5)$

Correct Option: 1

Solution:

$\frac{x}{m}=k p^{1 / n}$

$x=$ mass of adsorbate

$\mathrm{m}=$ mass of adsorbent

$\mathrm{P}=$ eq. pressure

$\mathrm{k}_{1} \mathrm{n}=\frac{1}{\mathrm{n}} \log \mathrm{p}+\log \mathrm{k}$

$y=m x+c$

compairing

$m=\frac{1}{n}=$ slope $\left[\frac{1}{n}=0\right.$ to 1$]$

n > 1