Question

$20\%$ of surface sites are occupied by $N_2$ molecules. The density of surface site is $6.023\times {10}^{14}c{m}^{-2}$ and total surface area is $1000$ $c{m}^2$. The catalyst is heated to $300$ $K$ while $N_2$ is completely desorbed into a pressure of $0.001$ atm and volume $2.46$ $c{m}^3$. Find the number of active sites occupied by each $N_2$ molecule:

Answer 1

Number of surface sites occupied by $N_2$ molecules $=20\%$.

Density of surface sites $=6.023\times {10}^{14}c{m}^{-2}$.

Total surface area $=1000c{m}^2$.

Total number of surface sites $=$ Density $\times$ Total surface area $=6.023\times {10}^{14}\times 1000$ $=6.023\times {10}^{17}$.

Sites occupied by nitrogen molecule $=\left(\frac{20}{100}\right)\times 6.023\times {10}^{17}=1.2046\times {10}^{17}$

Number of nitrogen molecules $=\frac{PV{N}_A}{RT}$

GIven, $T=300K,P=0.001atm,V=2.46c{m}^3$

Number of nitrogen molecules $=\frac{0.001\times 2.46\times {10}^{-3}}{(0.0821\times 300)}\times 6.023\times {10}^{23}=6.0\times {10}^{16}$.

Therefore, number of sites occupied by each nitrogen molecule $=\frac{\text{Sites occupied by nitrogen molecules}}{\text{Number of nitrogen molecules}}$

$=\frac{1.2046\times {10}^{17}}{6.0\times {10}^{16}}=2$.