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Question

In A(g) + 2B(g) <=> C(g) , Kc = 8 and at equilibrium [A] = 0.1 M , [B] = 0.2 M , [C] = 0.16 M . What happens if the volume is doubled?

Options

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Explanation

Initial Kc = (0.16/(0.1)(0.2)²) = 40 (not 8, assume adjusted). Doubling volume halves concentrations: [A] = 0.05 , [B] = 0.1 , [C] = 0.08 , Q = (0.08/(0.05)(0.1)²) = 160 > Kc , shifts left.