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

Answer

C

Explanation

`underset(0.1)(N_2O_4)hArrunderset(0)(2NO_2)`
`(0.1 -alpha),2alpha`
`:. P prop 0.1`
If V and T are constant `(P prop 0.1 + alpha)`
`P=(0.1 +alpha)//0.1`
`K_p=([2alpha]^2)/([0.1 -alpha])xx[(P)/(0.1xxalpha)]`
or `K_p = (40 alpha^2)/([0.1-alpha])=0.14`
`alpha = 0.017`
`NO_2 = 0.017 xx2 = 0.034 ` mole