Surface ChemistryHard
Question
At 70 K, the adsorption of N2(g) at iron surface obeys Freundlich adsorption isotherm. The following data is collected experimentally.
P (/bar) 4 25 64
(x/m) 0.2 0.5 0.8
Here, x/m is the mass (in g) of N2(g) adsorbed per g of iron at P bar pressure. The moles of N2(g) adsorbed per g iron at 36 bar and 70 K is
Options
A.6/10
B.3/140
C.3/70
D.3/280
Solution
$\frac{x}{m} = K.P^{\frac{1}{n}}$
$\Rightarrow 0.2 = K \times (4)^{\frac{1}{n}} (1) $$${0.5 = K \times (25)^{\frac{1}{n}} (2) }{0.8 = K \times (64)^{\frac{1}{n}} (3)}$$
From (1), (2) & (3): $K = \frac{1}{10}\text{and }n = 2$
$\therefore\left( \frac{x}{m} \right)_{\text{required}} = K \times (36)^{\frac{1}{n}} = 0.6$
Hence, moles of N2 adsorbed per gm of iron = $\frac{0.6}{28} = \frac{3}{140}$.
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