0.1 mole of `N_(2O_(4)(g)` was sealed in a tude under one atmospheric conditions at `25^(@)C` Calculate the number of moles of `NO_(2)(g)` preesent , if the equilibrium `N_(2)O_(4)(g)hArr2NO_(2)(g)(K_(P)=0.14)` is reached after some time :

Options

A. `1.8xx10^(2)`
B. `2.8xx10^(2)`
C. `0.034`
D. `2.8xx10^(-2)`

Answer

C

Explanation

`underset(0.1)(N_(2)O_(4))harrunderset(0)2NO_(2)`
`(.1 - alpha)` `2alpha`
`.: P prop 0.1`
If V and T are constants `(P prop 0.1 + alpha)`
`P = (0.1 + alpha)//0.1`
`K_(p) = ([2alpha]^(2))/([0.1 - alpha])xx[(P)/(0.1+ alpha)]`
or `K_(p) = (40 alpha^(2))/([0.1-alpha]) = 0.14`
`alpha = 0.017`
`NO_(2) = 0.017xx2 = 0.034` mole