10 g of a gas is adsorbed on 500 g of solid | vedclass

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Question

(D) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = kp^{1/n}$ For the first condition: $\frac{10}{500} = k(10)^{1/n} \quad

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

(D) The Freundlich adsorption isotherm is given by the equation: $\frac{x}{m} = kp^{1/n}$ For the first condition: $\frac{10}{500} = k(10)^{1/n} \quad ...(i)$ For the second condition: $\frac{14}{500} = k(20)^{1/n} \quad ...(ii)$ Dividing equation $(ii)$ by equation $(i)$: $\frac{14/500}{10/500} = \frac{k(20)^{1/n}}{k(10)^{1/n}}$ $\frac{14}{10} = (\frac{20}{10})^{1/n}$ $1.4 = 2^{1/n}$ Taking logarithm on both sides: $\log(1.4) = \frac{1}{n} \log(2)$ $0.1461 = \frac{1}{n} 	imes 0.3010$ $\frac{1}{n} = \frac{0.1461}{0.3010} \approx 0.485$ Rounding to the nearest option,the slope $\frac{1}{n}$ is approximately $0.5$.

List-$I$	List-$II$
$A$. Physisorption	$I$. Single layer adsorption
$B$. Chemisorption	$II$. $20-40 \, kJ \, mol^{-1}$
$C$. $N_{2(g)} + 3H_{2(g)} \xrightarrow{Fe(s)} 2NH_{3(g)}$	$III$. Chromatography
$D$. Analytical Application of Adsorption	$IV$. Heterogeneous catalysis

$10 \ g$ of a gas is adsorbed on $500 \ g$ of solid at $10 \ bar$. If the pressure is increased to $20 \ bar$,$14 \ g$ of the gas is adsorbed by the same solid at the same temperature. What is the slope of Freundlich adsorption isotherm?

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