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) hArr 2NO (g)`
was established with an equilibrium constant `K _(c) = 2.1 xx 10 ^(-3).` At equilibrium the mole % of NO was 1.8. Estimate the initial composition of air in mole fraction of `N_(2) and O_(2).`

Options

A. 0.68, 0.32
B. 0.46, 0.54
C. 0.79, 0.21
D. 0.83, 017

Answer

C

Explanation

Let the total number of moles of `N_(2)` and `O_(2)` be 100 containig at moles of `N_(2)` initially
`{:(N_(2),+,CO_(2),hArr,2NO,),(a,,100-a,,0,"initial moles"),((a-x),,100-(a-x),,2x,"moles at equm."):}`
Mole % of NO at equilibrium `= (2x)/((a-x)-[100-(a-x)]-2x) prop 100 = 188 therefore x = 0.9`
Thus at equilibrium , moles of of `N_(2) = (a-0.9)`, moles of `O_(2) = (100-a-0.9) = (99.1=a)`
`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)) therefore a = 79.46%`
(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`