Some Basic Concepts of Chemistry 1 Question 22
43. $20 %$ surface sites have adsorbed $\mathrm{N}{2}$. On heating $\mathrm{N}{2}$ gas evolved from sites and were collected at $0.001 \mathrm{~atm}$ and 298 $\mathrm{K}$ in a container of volume is $2.46 \mathrm{~cm}^{3}$. Density of surface sites is $6.023 \times 10^{14} / \mathrm{cm}^{2}$ and surface area is $1000 \mathrm{~cm}^{2}$, find out the number of surface sites occupied per molecule of $\mathrm{N}_{2}$.
$(2005,3 \mathrm{M})$
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Solution:
- Partial pressure of $\mathrm{N}_{2}=0.001 \mathrm{~atm}$,
$$ T=298 \mathrm{~K}, V=2.46 \mathrm{dm}^{3} . $$
From ideal gas law : $p V=n R T$
$$ n\left(\mathrm{~N}_{2}\right)=\frac{p V}{R T}=\frac{0.001 \times 2.46}{0.082 \times 298}=10^{-7} $$
$\Rightarrow$ Number of molecules of $\mathrm{N}_{2}=6.023 \times 10^{23} \times 10^{-7}$
$$ =6.023 \times 10^{16} $$
Now, total surface sites available
$$ =6.023 \times 10^{14} \times 1000=6.023 \times 10^{17} $$
Surface sites used in adsorption $=\frac{20}{100} \times 6.023 \times 10^{17}$
$$ =2 \times 6.023 \times 10^{16} $$
$\Rightarrow$ Sites occupied per molecules
$$ =\frac{\text { Number of sites }}{\text { Number of molecules }}=\frac{2 \times 6.023 \times 10^{16}}{6.023 \times 10^{16}}=2 $$
