Practice question
Question
For N₂O₄(g) <=> 2NO₂(g) , Kp = 0.16 at 400 K. If 0.5 moles of N₂O₄ are placed in a 1 L vessel, what is PNO₂ at equilibrium ( R = 0.0831 bar L/mol K )?
Explanation
Initial: PN₂O₄ = (0.5 × 0.0831 × 400/1) = 16.62 bar . Let 2x = PNO₂ , PN₂O₄ = 16.62 - x , total pressure = 16.62 + x . Kp = ((PNO₂)²/PN₂O₄) = ((2x)²/16.62 - x) = 0.16 , 4x² = 0.16 (16.62 - x) , x ≈ 0.8 , PNO₂ = 1.6 bar .
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