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Practice question

MCQ

Question

For N₂(g) + 3H₂(g) <=> 2NH₃(g) , Kp = 4.0 × 10⁻³ at 600 K. If 1 mole of N₂ and 3 moles of H₂ are in a 2 L vessel, what is PNH₃ at equilibrium ( R = 0.0831 bar L/mol K )?

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Explanation

Initial: PN₂ = (1 × 0.0831 × 600/2) = 24.93 bar , PH₂ = 74.79 bar . Let 2x be PNH₃ , PN₂ = 24.93 - x , PH₂ = 74.79 - 3x . Kp = ((PNH₃)²/PN₂ (PH₂)³) = ((2x)²/(24.93 - x)(74.79 - 3x)³) = 4.0 × 10⁻³ . Solving, x ≈ 0.8 , PNH₃ = 2 × 0.8 = 1.6 bar .

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