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A sample of air consisting of N(2) and O...

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)`.

Text Solution

Verified by Experts

`N_(2) + O_(2) rarr 2NO`
Given, 2x/100 = 1.8/100 implies x = 0.9
Also, `K_(p) = K_(c) = [NO]^(2)/([N_(2)][O_(2)]) = (2x)^(2)/((a-x)(100-a-x)) = 2.1 xx 10^(-3)`
a = 79% 100-a = 21%
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