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 moles of `N_2 and O_2` be 100 containing .a. mole of `N_2` initially.
`"a (100-a) 0 Initial moles"`
`N_(2)+O_(2) Leftrightarrow 2NO`
`"(a-x) (100-a-x) 2x Moles at eqb."`
Mole % of NO at equilibrium
`=(2x)/((a-x) +(100-a-x) +2x) xx 100=1.8`
x=0.9
Thus, at equilibrium
moles of `N_2=(a-0.9)`
moles of `O_2=(100-a-0.9)=(99.1-a)`
moles of `NO=2x=2 xx 0.9 =1.8`
`K_c=(2x//V)^2/((a-0.9)/(V)) ((99.1-a)/V)=((2x))^2/((a-0.9) (99.1-a)`
`2.1 xx 10^(-3) =((1.8))^2/((a-0.9) (99.1-a))`
a=79.6%
(we know that percentage of `N_2` in the air is more than that of `O_2)` mole fraction of `N_2=0.7946`
mole fraction of `O_2=0.2054`