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

`{:(,,N_(2),+,O_(2),hArr,2NO),(,"Initial",a,,(100-a),,0),(,"Final",(a-x),,(100-a-x),,2x):}`
Given `(2x)/(100)=(1.8)/(100)`
`:. X=0.9`
Also `K_(p)=K_(c)=([NO]^(2))/([N_(2)][O_(2)])`
`=((2x)^(2))/((a-x)(100-a-x))=2.1xx10^(-3)`
`:. a=79%`, `100-a=21%`
Note: Two values of `a` are obtained as `a=79%` and `a=21%`. Not that air contains more `N_(2)` than `O_(2)`, thus `a=79%`.