Question

In Freundlich adsorption isotherm, slope of AB line is :

(1) log n with (n > 1)

(2) n with (n, 0.1 to 0.5)

(3) log \(\frac{1}{n}\) with (n<1)

(4)  \(\frac{1}{n}\) with \(\Big(\frac{1}{n}=0\,to\,1\Big)\)

Answer 1

\(\frac{x}{m}=K(P)^{1/n}\)

log\(\Big(\frac{x}{m}\Big)\) = log K + \(\frac{1}{n}\)log P

y = c + mx

m = 1/n so slope will be equal to 1/n.