Density of equilibrium mixture of `N_(2)O_(4)` and `NO_(2)` at `1 atm` and `384 K` is `1.84 g dm^(-3)`. Calculate the equilibrium constant of the reaction.
`N_(2)O_(4)hArr2NO_(2)`

Explanation

We know
`Pm=dRT`
`1xxm=1.84xx0.0821xx384`
`m=29xx2`
Vapour density (d) at equilibrium `=29`
Initial vapour density`=M//2=92//2=46`
therefore, degree of dissociation is:
`x=(D-d)/((n-1)d)=(46-29)/(29)=0.586`
For reaction
`N_(2)O_(4)hArr2NO_(2)`
`{:(t=0,1,,0,),(t_(eq),1-x,,2x,"Total moles"=1+x):}`
`p_(N_(2)O_(4))=(1-x)/(1+x)xxP, p_(NO_(2))=(2x)/(1+x)xxP`
`K_(p)=(4x^(2)P)/(1-x^(2))=(4xx(0.586)^(2)xx1)/(1-(0.586)^(2))=2.09 "atm"`