`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 `L mol^(-1)J^(-1)`
Temperature `(T)=400K`
`pV=nRT`
`pxx1L=0.15 mol xx0.083 "bar L mol"^(-1)K^(-1)xx400 K`
`p=4.98` bar
`N_(2)O hArr 2NO_(2)`
Initial pressure 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=498+x`
`x=9.15-4.98=4.17` bar
Partial pressure at equlibrium 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_(N_(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`