At 600 K, 2 mol of NO are mixed with 1 mol of O₂.
2NO₍ₘ₎ + O₂₍ₘ₎ ⇌ 2NO₂₍ₘ₎
The reaction occurring as above comes to equilibrium under a total pressure of 1 atm. Analysis of the system shows that 0.6 mol of oxygen are present at equilibrium. The equilibrium constant for the reaction is _____. (Nearest integer).
2NO₍ₘ₎ + O₂₍ₘ₎ ⇌ 2NO₂₍ₘ₎
The reaction occurring as above comes to equilibrium under a total pressure of 1 atm. Analysis of the system shows that 0.6 mol of oxygen are present at equilibrium. The equilibrium constant for the reaction is _____. (Nearest integer).
Answer: 2
Solution
Initial: NO = 2 mol, O₂ = 1 mol. At equilibrium: O₂ = 0.6 mol, so x = 0.4 mol reacted. Therefore: NO = 2 - 2(0.4) = 1.2 mol, NO₂ = 2(0.4) = 0.8 mol. Total moles = 1.2 + 0.6 + 0.8 = 2.6 mol. Mole fractions: χ(NO) = 1.2/2.6, χ(O₂) = 0.6/2.6, χ(NO₂) = 0.8/2.6. Partial pressures (P_total = 1 atm): P(NO) = 1.2/2.6, P(O₂) = 0.6/2.6, P(NO₂) = 0.8/2.6. Kₚ = [P(NO₂)]²/([P(NO)]² × P(O₂)) = (0.8/2.6)²/((1.2/2.6)² × (0.6/2.6)) = 0.64/(1.44 × 0.6) = 0.64/0.864 ≈ 1.93 ≈ 2.
JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Integer answer