A sample of air consisting mainly of `N_2` and `O_2` is heated at 2500K till the given equilibrium attains- `N_2(g) + O_2 (g) hArr 2NO(g)`. If the value of equilibrium constant `(K_c)` at this temperature is `2.1xx10^(-3)` and the percentage of number of moles of NO(g) in the above equilibrium mixture is 1.8 %, then what will be the mole fractions of `N_2` and `O_2` in the sample of air used'?

Explanation

Suppose, initial total number of moles of `N_2` and `O_2` in the sample of air = 1 mol. In the sample, if the initial number of moles of `N_2` is n, then the same for `O_2` is (1-n) mol.
Let, the number of moles of `N_2` reacts at equilibrium=`prop`
Initial number of moles : `N_2+O_2 `n" "(1-n)" "0`
Number of moles at equilibrium: `(n-prop)(1-n-prop)2prop`
Total number of moles at equilibrium `=n-prop+1-n-prop+2prop=1`
so, mole percentage of NO`=(2prop)/(1_xx100=1.8`
`:.prop=9xx10^(-3)`
`K_c=([NO]^2)/([N_2][O_2])=(n_(NO)^2)/(n_(N_2)xxn_(O_2))`

`=((2xx9xx10^(-3))^2)/((n-9xx10^(-3)(n-9xx10^(-3))=0.154`
`:.` n=0.21
so, initial number of moles of `O_2`=0.21 and that of `N_2=1-0.21=0.79`. Hence, in the sample of air, mole fraction of `O_2`=0.21 and that of `N_2`=0.79