Question

Around `20%` surface sites have adsorbed `N_(2)`. On heating `N_(2)` gas evolved form sites and were collected at 0.001 atm and 298 K in a container of volume `2.46cm^(3)` the density of surface sites is `6.023xx10^(14)cm^(-2)` and surface area is `1000cm^(2)` find out the number of surface sites occupied per molecule of `N_(2)`.

Answer 1

Given `P_(N_(2))=0.001 atm, T=298 K`,
`V=2.46 cm^(3)=2.46xx10^(-3)"litre"`
`:. n_(N_(2))=(PV)/(RT)=(0.001xx2.46xx10^(-3))/(0.0821xx298)= 1.0xx10^(-7)`
`:.`Molecule of absorbed `N_(2)`
`=6.023xx10^(23)xx1.0xx10^(-7)=6.023xx10^(16)`
Total surface sites availiable
`="Number of sites"//cm^(2)xx"Area"`
`=6.023xx10^(14)xx1000=6.025xx10^(17)`
Surface sites at which `N_(2)` was absorbed
`=(20)/(100)xx6.023xx10^(17)=12.046xx10^(16)`
`:.` No. of sites absorbed per molecule per molecule of `N_(2)`
`=(12.046xx10^(16))/(6.023xx10^(6))=2`