`13.8 g` of `N_(2)O_(4)` was placed in a `1L` reaction vessel at `400 K` and allowed to attain equilibrium
`N_(2)O_(4) (g) hArr 2NO_(2)(g)`
The total pressuers at equilibrium was found to be `9.15` bar. Calculate `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`
Numner of moles `= 13.8g//92 g = 0.15` of the gas (n)
Gas constant (R) `=0.083 "bar L mol"^(-1) K^(-1) xx 400 K`
`p = 4.98 "bar"`
`N_(2)O_(4) 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 = 4.98 + x`
`x = 9.15 - 4.98 = 4.17` bar
Particle pressures at equlibrium are,
`p_(N_(2)O_(4)) = 4.98 - 4.17 = 0.81"bar"`
`p_(NO_(2)) = 2x = 2 xx 4.17 = 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)(0.083 xx 400)^(1)`
`K_(c) = 2.586 = 2.6`