The equilibrium constant for the reaction:
CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g) is 3.0 at 500 K. In a 2.0 L vessel, 60 g of water gas [equimolar mixture of CO(g) and H₂(g)] and 90 g steam is initially t

Question

The equilibrium constant for the reaction:
CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g) is 3.0 at 500 K. In a 2.0 L vessel, 60 g of water gas [equimolar mixture of CO(g) and H₂(g)] and 90 g steam is initially taken. What is the equilibrium concentration of H₂(g)?

Prashant Kotian · Accepted Answer

Correct answer: A. Equimolar water gas (CO + H₂) has avg molar mass = (28+2)/2 = 15 g/mol. 60 g water gas = 4 moles total, meaning 2 moles CO and 2 moles H₂. 90 g steam = 5 moles H₂O. Initial: CO=2, H₂O=5, CO₂=0, H₂=2. Change: CO=2-x, H₂O=5-x, CO₂=x, H₂=2+x. K_c = [x(2+x)] / [(2-x)(5-x)] = 3.0. 2x + x² = 3(10 - 7x + x²) => 2x² - 23x + 30 = 0. Solving yields x = 1.5. Eq moles H₂ = 2 + 1.5 = 3.5 mol. Concentration of H₂ = 3.5 / 2.0 = 1.75 M. Therefore, correct answer is A.

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