Question:
In Freundlich adsorption isotherm, slope of $\mathrm{AB}$ line is
Correct Option: 1
Solution:
Freundlich adsorption isotherm is :
$\frac{x}{m}=k p^{1 / n}$
$x=$ 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
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