Surface Chemistry Question 9
Question 9 - 01 February - Shift 2
In figure, a straight line is given for Freundrich Adsorption $(y=3 x+2.505)$. The value of $\frac{1}{n}$ and $\log K$ are respectively.
(1) 0.3 and $\log 2.505$
(2) 0.3 and 0.7033
(3) 3 and 2.505
(4) 3 and 0.7033
Show Answer
Answer: (3)
Solution:
$\frac{X}{m}=Kp^{1 / n}$
$\log \frac{x}{m}=\log k+\frac{1}{n} \log P$
$.Y=3 x+2.505, \frac{1}{n}=3, \log K=2.505)$
