In Freundlich adsorption isotherm, slope of AB line is :

Question

In Freundlich adsorption isotherm, slope of AB line is : (A)  log n with ( n >1) (B) n with ( n , 0.1 to 0.5) (C)  log (1/n) with (n<1) (D) (1/ n ) with ((1/ n )=0. to .1)

Tardigrade Admin · Accepted Answer

(x/m)=K(P)1 / n log (( x / m ))= log K +(1/ n ) log P y = c + mx m =1 / n so slope will be equal to 1 / n <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/c45a05422c0ddf0c80febac305c7eb61-.png" /> Hence 0 le (1/n) le 1

Q. In Freundlich adsorption isotherm, slope of AB line is :

$lo g n$ with $(n > 1)$

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

$lo g \frac{1}{n}$ with $(n < 1)$

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

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

$lo g (\frac{x}{m}) = lo g K + \frac{1}{n} lo g P$

$y = c + m x$

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

Hence $0 \leq \frac{1}{n} \leq 1$

Q. In Freundlich adsorption isotherm, slope of AB line is :

logn with (n>1)

n with (n,0.1 to 0.5)

logn1​ with (n<1)

n1​ with (n1​=0 to 1)

mx​=K(P)1/n log(mx​)=logK+n1​logP y=c+mx m=1/n so slope will be equal to 1/n Hence 0≤n1​≤1

$lo g n$ with $(n > 1)$

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

$lo g \frac{1}{n}$ with $(n < 1)$

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

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

$lo g (\frac{x}{m}) = lo g K + \frac{1}{n} lo g P$

$y = c + m x$

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

Hence $0 \leq \frac{1}{n} \leq 1$