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Practice question

MCQ

Question

For 2NO₂(g) <=> N₂O₄(g) , Kc = 200 at 298 K. If 0.1 mol NO₂ is placed in a 1 L vessel, what is [N₂O₄] at equilibrium?

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Explanation

Initial: [NO₂] = 0.1 M , [N₂O₄] = 0 . Let x = [N₂O₄] , [NO₂] = 0.1 - 2x . Kc = ([N₂O₄]/[NO₂]²) = (x/(0.1 - 2x)²) = 200 , x = 200 (0.1 - 2x)² , sqrtx = 14.14 (0.1 - 2x) , x ≈ 0.045 M (solving iteratively).

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