An equilibrium mixture at 300 K contains N 2 O 4 and NO 2 at 0 . 28 and 1 . 1 atm pressure, respectively. If the volume of the container is doubled, the new equilibrium pressure of these two gases are respectively.

Step-by-step solution

N 2 O 4 g ⇌ 2 NO 2 g Pressure at equilibrium K p = p NO 2 2 p N 2 O 4 = 1 . 1 2 0 . 28 = 4 . 32   atm If volume of the container is doubled, the pressure will reduced to half N 2 O 4 ⇌ 2 N 2 O New pressure, 0 . 028 2 - p 1 . 1 2 + 2 p K p = 1 . 1 2 + 2 p 2 0 . 28 2 - p = 4 . 32 On solving, we get p = 0 . 045 ∴   p N 2 O 4 = 0 . 14 - 0 . 045 = 0 . 095   atm p NO 2 = 0 . 55 + 2 × 0 . 045 = 0 . 64   atm