`13.8g` of `N_(2)O_(4)` was placed in a `1L` reaction vessel at `400K` and allowed to attain equilibrium
`N_(2)O_(4)(g)hArr2NO_(2)(g)`
The total pressure at equilibrium was found to be `9.15` bar . Calcualate `K_(c)`, `K_(p)` and partial pressure at equilibrium.

Explanation

We know `pV=nRT`
Total volume `(V)=1L`
Molecular mass of `N_(2)O_(4)=92g`
Number of moles `=13.8g//92g=0.15` of the gas `(n)`
Gas constant `(R)=0.083` bar `Lmol^(-1)R^(-1)`
Temperature `(T)=400K`
`pV=nRT`
`pxx1L=0.15molxx0.083` bar `Lmol^(-1)K^(-1)xx400K`
`p=4.98` bar
`{:(,N_(2)O_(4),,,hArr,2NO_(2)),("Intial",4.98"bar",,,,0),("At equilibrium",(4.98-x)"bar",,,,2x"bar"):}`
Hence,
`P_("total")` at equilibrium `=p_(N_(2)O_(4))+p_(NO_(2))`
`9.15=(4.98-x)+2x`
`9.15=4.98+x`
`x=9.15-4.98=4.17"bar"`
Partial pressures at equilibrium are,
`P_(N_(2)O_(4))=4.98-4.17=0.81` bar
`p_(NO_(2))=2x=2xx4.17=8.34` bar
`K_(p)=(p_(NO_(2)))^(2)//p_(N_(2)O_(4))`
`=(8.34)^(2)//0.81=85.87`
`K_(p)=K_(c)(RT)^(Deltan)`
`85.87-K_(c)(0.083xx400)^(1)`
`K_(c)=2.586=2.6`