Question Detail
Question
Consider the following reversible reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). In a 3.00 liter container, the following amounts are found in equilibrium at 400 °C: 0.0420 mole N₂, 0.516 mole H₂, and 0.0357 mole NH₃. Evaluate Kc.
Correct Answer
Option B is the correct answer.
Detailed Explanation
[N₂] = 0.0420/3.00 = 0.0140 M, [H₂] = 0.516/3.00 = 0.172 M, [NH₃] = 0.0357/3.00 = 0.0119 M. Kc = [NH₃]²/([N₂][H₂]³) = (0.0119)²/(0.0140 × (0.172)³) = 0.0001416/(0.0140 × 0.00509) = 0.0001416/0.00007126 = 1.987 ≈ 1.99.
Hint
Convert moles to concentrations using volume, then apply Kc expression.