Question

20% surface sites have adsorbed $N_2$. On heating, $N_2$ gas evolved from sites and were collected at 0.001 $\text{atm}$ and 298 $\text{K}$ in a container of volume 2.46 ${\text{cm}}^3$. Density of surface sites is $6.023\times {10}^{14}{\text{cm}}^2$ and surface area is 1000 ${\text{cm}}^2$, find out the number of surface sites occupied per molecule of $N_2$.

• A. 4
• B. 2
• C. 10
• D. 5

Answer 1

The correct option is B 2

Partial pressure of $N_2$ = 0.001 $\text{atm}$
$T=298\text{K}$
$V=2.46{\text{dm}}^3$
From ideal gas law
$pV=nRT$
$n_{\left(N_2\right)}=\frac{pV}{RT}$
$n_{\left(N_2\right)}=\frac{0.001\times 2.46}{0.082\times 298}={10}^{-7}$
Number of molecules of $N_2$ = $6.023\times {10}^{23}\times {10}^{-7}=2\times 6.023\times {10}^{16}$
Now, total sites available$=6.023\times {10}^{14}\times 1000=6.023\times {10}^{17}$
Surface sites used in adsorption $=\frac{20}{100}\times 6.023\times {10}^{17}=2\times 6.023\times {10}^{16}$
Sites occupied per molecules $=\frac{\text{Number of sites}}{\text{Number of molecules}}=\frac{2\times 6.023\times {10}^{16}}{6.023\times {10}^{16}}=2$