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Practice question

MCQ

Question

For X₂(g) <=> 2X(g) , Kc = 0.25 at 500 K. If 0.4 mol X₂ is in a 2 L vessel, what is the degree of dissociation at equilibrium?

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Explanation

Initial: [X₂] = (0.4/2) = 0.2 M . Let α be the degree of dissociation, [X₂] = 0.2(1 - α) , [X] = 0.2 × 2α = 0.4α . Kc = ([X]²/[X₂]) = ((0.4α)²/0.2(1 - α)) = (0.16α²/0.2(1 - α)) = 0.25 , 0.8α² = 0.25(1 - α) , 0.8α² + 0.25α - 0.25 = 0 , α ≈ 0.5 .

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