4 . 0 moles of argon and 5 . 0 moles of PCl 5 are introduced into an evacuated flask of 100 litre capacity at 610 K . The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6 . 0 atm . The K p for the reaction is [Given : R = 0 . 082 L atm K - 1 mol - 1 ]

Step-by-step solution

PCl 5 = 5 mole Ar = 4 mole P Total = 9 × 0 . 082 × 610 100 = 4 . 5   atm P PCl 5 = 5 × 4 . 5 9 = 2 . 5 ;   P Ar = 4 × 4 . 5 9 = 2 PCl 5 ⇌ PCl 3 + Cl 2 2 . 5 - P   P           P P total = 2 . 5 - P + P + P + P Ar = 6 P = 1 . 5 K p = 1 . 5 × 1 . 5 1 = 2 . 25