Practice question
Question
The equilibrium constant Kp for N₂O₄(g) <=> 2NO₂(g) is 0.98 at 298 K. What is Kc at this temperature ( R = 0.0831 bar L/mol K )?
Explanation
Kp = Kc (RT)Δ n , Δ n = 2 - 1 = 1 , RT = 0.0831 × 298 ≈ 24.76 . Thus, Kc = (Kp/RT) = (0.98/24.76) ≈ 0.04 .
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