In Freundlich adsorption isotherm,


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


Freundlich adsorption isotherm is :

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

$x=$ mass of adsorbate

$\mathrm{m}=$ mass of adsorbent

$\mathrm{P}=$ eq. pressure

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

$y=m x+c$


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

n > 1

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