A sample of air consisting of `N_(2)` and `O_(2)` was heated to `2500 K` until the equilibrium
`N_(2)(g)+O_(2)(g)hArr2NO(g)`
was established with an equlibrium constant, `K_(c )=2.1xx10^(-3)`. At equilibrium, the `"mole"%` of `NO` was `1.8`. Eatimate the initial composition of air in mole fraction of `N_(2)` and `O_(2)`.

Explanation

Let the total number of mole of `N_(2)` and `O_(2)` be `100` containing a mole of `N_(2)` initially.
`N_(2)+O_(2)hArr2NO`
`{:(a,(100-a),,0,"Initial mol"),((a-x),(100-a-x),,2x,"mole at equilibrium"):}`
mole `%` of NO at equilibrium
`=(2x)/((a-x)+(100-a-x)+2x)xx100=1.8`
`:. x=0.9`
Thus, at equilibrium,
mole of `N_(2)=(a-0.9)`
mole of `O_(2)=(100-a-0.9)=(99.1-a)`
"mole" of `NO=2x=2xx0.9=1.8`
`K_(c )=((2x//V)^(2))/(((a-0.9)/V)((99.1-a)/V))=((2x)^(2))/((a-0.9)(99.1-a))`
`2.1xx10^(-3)=((1.8)^(2))/((a-0.9)(99.1-a))`
`a=74.46%` and `20.54%`
As we know that `%` of `N_(2)` in the air is more than that of `O_(2)`.