Equilibrium concentrations for Kc in ammonia synthesis

Steps:

• Reaction is N₂ + 3H₂ ⇌ 2NH₃; initial: 2.0 mol N₂, 6.0 mol H₂, 2.4 mol NH₃; equilibrium N₂ = 2.32 mol.
• Reverse extent ξ = 2.32 - 2.0 = 0.32 mol; equilibrium H₂ = 6.0 + 3(0.32) = 6.96 ≈ 7 mol; NH₃ = 2.4 - 2(0.32) = 1.76 mol.
• With V = 1 dm³, Kc = [NH₃]² / ([N₂][H₂]³) = (1.76)² / (2.32 × 7³).
• Option A provides the approximated expression using 1.76 and 7 for direct Kc calculation.

Why A is correct:

• Matches Kc formula with equilibrium values and approximation of H₂ as 7 mol for simplicity in expression.

Why the others are wrong:

• B: Incomplete numerical value without relation to Kc formula.
• C: Uses incorrect numbers (3.208, 32) not matching equilibrium moles.
• D: Garbled and lacks valid equilibrium data.

Final answer: A