Practice question
Question
For CH₄(g) + H₂O(g) <=> CO(g) + 3H₂(g) , Kc = 0.1 at 800 K. If 0.5 mol CH₄ and 0.5 mol H₂O are in a 2 L vessel, what is [H₂] at equilibrium?
Explanation
Initial: [CH₄] = [H₂O] = 0.25 M , [CO] = [H₂] = 0 . Let 3x = [H₂] , [CO] = x , [CH₄] = [H₂O] = 0.25 - x . Kc = ([CO][H₂]³/[CH₄][H₂O]) = (x (3x)³/(0.25 - x)²) = 0.1 , 27x⁴ = 0.1 (0.25 - x)² , x ≈ 0.06 , [H₂] = 0.18 M .
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