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Question

For the reaction PCl₅(g) <=> PCl₃(g) + Cl₂(g) , if Kp = 1.8 at 500 K and R = 0.0831 bar L/mol K , what is Kc ?

Options

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Explanation

Kp = Kc (RT)Δ n , where Δ n = 2 - 1 = 1 . Thus, Kc = (Kp/RT) = (1.8/0.0831 × 500) = (1.8/41.55) ≈ 0.043 .