Question

In Freundlich adsorption isotherm, slope of $AB$ line is :

• A. $log\text{n with}(n>1)$
• B. $\frac{1}{n}\text{with}(\frac{1}{n}=0to1)$
• C. $\text{n with}(n=0.1to0.5)$
• D. $log\frac{1}{n}\text{with}(n<1)$

Answer 1

The correct option is B $\frac{1}{n}\text{with}(\frac{1}{n}=0to1)$

According to Freundlich adsorption isotherm

$\frac{x}{m}-k.P^{1/n}(n>1)$

$log\frac{x}{m}=logk+\frac{1}{n}logP$

$\text{So, in the plot of}log\frac{x}{m}\text{vs log P, the slope is}\frac{1}{n},$

$where\frac{1}{n}\text{varies from 0 to 1 (since, n >1)}$

Hence, option (b) is correct.