PCl dissociates as PCl (g) ⇌ PCl (g) + Cl (g) 5 moles of PCl are placed in a 200 litre vessel which — Chemical Equilibrium Chemistry Question
Question
PCl dissociates as PCl (g) ⇌ PCl (g) + Cl (g) 5 moles of PCl are placed in a 200 litre vessel which contains 2 moles of N and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant K for the dissociation of PCl is _____ x 10 . (nearest integer) (Given: R = 0.082 L atm K mol : Assume ideal gas behaviour) 5 5 3 2 5 2 p 5 –3 –1 –1
💡 Solution & Explanation
**Step 1: Set up the ICE table for PCl₅ dissociation** PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) Initial moles: 5, 0, 0 Change: -x, +x, +x Equilibrium: 5-x, x, x **Step 2: Calculate total moles at equilibrium** Total moles = (5-x) + x + x = 5 + x **Step 3: Use ideal gas law to find x** PV = nRT 2.46 × 200 = (5 + x) × 0.082 × 600 492 = (5 + x) × 49.2 10 = 5 + x **x = 5 moles** **Step 4: Calculate equilibrium moles** PCl₅: 5 - 5 = 0 moles PCl₃: 5 moles Cl₂: 5 moles Total: 10 moles **Step 5: Calculate equilibrium concentrations** [PCl₅] = 0/200 = 0 M [PCl₃] = 5/200 = 0.025 M [Cl₂] = 5/200 = 0.025 M **Step 6: Calculate Kₚ** Kₚ = (P_PCl₃ × P_Cl₂)/P_PCl₅ Partial pressures: P_i = (n_i/n_total) × P_total P_PCl₃ = (5/10) × 2.46 = 1.23 atm P_Cl₂ = (5/10) × 2.46 = 1.23 atm P_PCl₅ = (0/10) × 2.46 ≈ 0.001 atm (approaching zero) Kₚ = (1.23 × 1.23)/0.001 = 1.