Find total pressure at equilibrium for N₂O₄ dissociation

Practice question

Question

For N₂O₄(g) <=> 2NO₂(g) , Kp = 0.16 at 298 K. If the initial pressure of N₂O₄ is 2 atm in a closed vessel, what is the total pressure at equilibrium ( R = 0.0831 bar L/mol K )?

Options

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2.0 atm 2.27 atm Correct 2.54 atm 3.0 atm

Explanation

Let PNO₂ = 2x , PN₂O₄ = 2 - x , total pressure = 2 - x + 2x = 2 + x . Kp = ((PNO₂)²/PN₂O₄) = ((2x)²/2 - x) = 0.16 , (4x²/2 - x) = 0.16 , 4x² = 0.32 - 0.16x , 4x² + 0.16x - 0.32 = 0 , x² + 0.04x - 0.08 = 0 , x = (-0.04 pm sqrt0.0016 + 0.32/2) , x ≈ 0.27 . Total pressure = 2 + 0.27 = 2.27 atm .

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