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
`1 xx m =1.84 xx0.0821 xx 384`
`m=29 xx2 =58`
Vapour density (d) at equilibrium =29
initial vapour density = M/2=92/2 =46
`x=(D-d)/((n-1)d)=(46-29)/(29)=0.586`
`N_2O_4 hArr 2NO_2`
`{:(t=0,1,0),(t_(eq),1-x,2x):}`
Total moles =1 +x
`P_(N_2O_4)=(1-x)/(1+x)xxP`
`P_(NO_2)=(2x)/(1+x)xxP`
`K_p=(4x^2P)/(1-x^2)=(4xx(0.586)^2xx1)/(1-(0.586)^2)=2.09 atm`