JEE PYQ: Surface Chemistry Question 6
Question 6 - 2021 (24 Feb Shift 1)
Type: MCQ
In Freundlich adsorption isotherm, slope of $AB$ line is:
[Graph shows $\log \frac{x}{m}$ vs $\log P$ with points A and B on a straight line]
(a) $\frac{1}{n}$ with $\left(\frac{1}{n} = 0 \text{ to } 1\right)$
(b) $\log \frac{1}{n}$ with $(n < 1)$
(c) $\log n$ with $(n > 1)$
(d) $n$ with $(n, 0.1 \text{ to } 0.5)$
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Answer: 0
Solution
Freundlich adsorption isotherm: $\frac{x}{m} = kp^{1/n}$
$\log \frac{x}{m} = \frac{1}{n} \log p + \log k$
Comparing with $y = mx + c$: slope $= \frac{1}{n}$ (where $\frac{1}{n} = 0$ to $1$) and $n > 1$.
BETA
