In figure, a straight line is given for Freundrich Adsorption (y=3 x+2.505). The value of (1/ n ) and log K are respectively.

Question

In figure, a straight line is given for Freundrich Adsorption (y=3 x+2.505). The value of (1/ n ) and log K are respectively. (A) 3 and 2.505 (B) 3 and 0.7033 (C) 0.3 and log 2.505 (D) 0.3 and 0.7033

Tardigrade Admin · Accepted Answer

( x / m )= Kp 1 / n log ( x / m )= log k +(1/ n ) log P . Y =3 x +2.505, (1/ n )=3, log K =2.505)

Q. In figure, a straight line is given for Freundrich Adsorption $(y = 3 x + 2.505)$ . The value of $\frac{1}{n}$ and $lo g K$ are respectively.

$3$ and $2.505$

$3$ and $0.7033$

$0.3$ and $lo g 2.505$

$0.3$ and $0.7033$

$\frac{x}{m} = K p^{1/ n}$
$lo g \frac{x}{m} = lo g k + \frac{1}{n} lo g P$
$Y = 3 x + 2.505, \frac{1}{n} = 3, lo g K = 2.505)$

Q. In figure, a straight line is given for Freundrich Adsorption (y=3x+2.505). The value of n1​ and logK are respectively.

3 and 2.505

3 and 0.7033

0.3 and log2.505

0.3 and 0.7033