For reaction : SO2(g)+ 1 O2(g) SO3(g) Kp = 2 1012 at 27ºC and 1 atm pressure. The Kc for the same re — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

**Step 1: Identify the relationship between Kp and Kc** Use the formula: **Kp = Kc(RT)^Δn** where Δn = moles of gaseous products − moles of gaseous reactants **Step 2: Calculate Δn** For the reaction: SO₂(g) + ½O₂(g) ⇌ SO₃(g) Δn = 1 − (1 + ½) = 1 − 1.5 = **−0.5** **Step 3: Rearrange to solve for Kc** Kc = Kp / (RT)^Δn **Step 4: Substitute values** - Kp = 2 × 10¹² - R = 0.082 L atm K⁻¹ mol⁻¹ - T = 27°C = 300 K - Δn = −0.5 RT = 0.082 × 300 = 24.6 (RT)^Δn = (24.6)^(−0.5) = 1/√24.6 = 1/4.96 ≈ **0.201** **Step 5: Calculate Kc** Kc = (2 × 10¹²) / 0.201 = (2 × 10¹²) × 4.97 Kc ≈ **9.94 × 10¹² ≈ 10 × 10¹² = 1 × 10¹³** Therefore, the answer is A.