BotanyIf PCl5 dissociates to 50%, the total pressure exerted by the reaction mixture at equilibrium when Kp = 3 atm is

If PCl₅ dissociates to 50%, the total pressure exerted by the reaction mixture at equilibrium when K_p = 3 atm is

Let initial mole of PCl₅ = 1 mole and degree of dissociation = x

${PCl}_{5} ⇌ {PCl}_{3} + {Cl}_{2} 1 0 0 - - - initial moles 1 - 0.5 0.5 0.5 - - moles at equilibrium$

partial pressure of each gas at equilibrium

PCl₅ = PCl₃ = Cl₂ = $\frac{0.5}{1.5} p$

3 = $\frac{\frac{P}{3} \times \frac{P}{3}}{\frac{P}{3}}$

P = 9 atm ( P = total pressure )