3.0 moles of PCl is introduced in a 1 L vessel. The number of moles of PCl at equilibrium is _______ — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

**Step 1: Set up the equilibrium problem** The dissociation reaction is: PCl₅ ⇌ PCl₃ + Cl₂ Initial moles: PCl₅ = 3.0 mol; PCl₃ = 0; Cl₂ = 0 Volume = 1 L, so concentrations equal moles numerically. **Step 2: Create an ICE table** | | PCl₅ | PCl₃ | Cl₂ | |---|---|---|---| | I | 3.0 | 0 | 0 | | C | -x | +x | +x | | E | 3.0-x | x | x | **Step 3: Apply the equilibrium constant** For PCl₅ dissociation, Kc = 1.8 × 10⁻⁷ (standard value) Kc = [PCl₃][Cl₂]/[PCl₅] = x²/(3.0-x) Since Kc is very small, assume x << 3.0: 1.8 × 10⁻⁷ = x²/3.0 **Step 4: Solve for x** x² = 1.8 × 10⁻⁷ × 3.0 = 5.4 × 10⁻⁷ x = √(5.4 × 10⁻⁷) = 7.35 × 10⁻⁴ mol **Step 5: Calculate moles of PCl₅ at equilibrium** Moles of PCl₅ = 3.0 - x = 3.0 - 7.35 × 10⁻⁴ ≈ 2.99926 mol **Step 6: Express in required form** 2.99926 × 10³ = 2999.26 × 10⁰ When expressed as × 10⁻³: 2.99926 × 10³ mol