For 2NOCl(g) 2NO(g) + Cl₂(g) , Kp = 1.8 × 10⁻² at 500 K. If 2 moles of NOCl are placed in a 2 L vessel, what is the tota

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Question

For 2NOCl(g) <=> 2NO(g) + Cl₂(g) , Kp = 1.8 × 10⁻² at 500 K. If 2 moles of NOCl are placed in a 2 L vessel, what is the total pressure at equilibrium ( R = 0.0831 bar L/mol K )?

Options

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43.42 bar Correct 41.55 bar 45.0 bar 39.0 bar

Explanation

Initial: PNOCl = (2 × 0.0831 × 500/2) = 41.55 bar . Let 2x mol dissociate, PNOCl = 41.55(1 - x) , PNO = 41.55x , PCl₂ = 41.55 × (x/2) , total pressure = 41.55(1 - x + x + (x/2)) = 41.55(1 + (x/2)) . Kp = ((PNO)² PCl₂/(PNOCl)²) = ((41.55x)² (41.55 (x/2))/[41.55(1 - x)]²) = 1.8 × 10⁻² , (41.55x³/2(1 - x)²) = 0.018 , x³ = 8.67 × 10⁻⁴ (1 - x)² , x ≈ 0.09 , total pressure = 41.55 × 1.045 = 43.42 bar .

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