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Question

For the reaction 3A(g) + B(g) <=> 2C(g) , Kc = 8 at 500 K. If 1.5 moles of A and 0.5 moles of B are placed in a 1 L vessel, what is [C] at equilibrium?

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Explanation

Initial: [A] = 1.5 M , [B] = 0.5 M , [C] = 0 . Let 2x be moles of C formed, so A decreases by 3x , B by x . At equilibrium: [A] = 1.5 - 3x , [B] = 0.5 - x , [C] = 2x . Kc = ([C]²/[A]³[B]) = ((2x)²/(1.5 - 3x)³ (0.5 - x)) = 8 , (4x²/(1.5 - 3x)³ (0.5 - x)) = 8 . Solving iteratively, x ≈ 0.25 , (0.5)² / [(0.75)³ × 0.25] = 0.25 / 0.1055 ≈ 2.37 (adjust), x ≈ 0.4 , [C] = 2 × 0.4 = 0.8 M .

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