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For the reaction 2A + B ⇌ 3C + D, the equilibrium constant Kc is:
A$K_c = [A]^2[B]/([C]^3[D])$ (reactants over products)
B$K_c = [C]^3[D]/([A]^2[B])$ (products over reactants)
C$K_c = [C][D]/([A][B])$ (no stoichiometry)
D$K_c = ([C]+[D])/([A]+[B])$ (sums)
Answer & Solution
Correct answer: B. $K_c = [C]^3[D]/([A]^2[B])$ (products over reactants)
Kc = (products)^stoichiometric exponent / (reactants)^stoichiometric exponent. So Kc = [C]³[D] / ([A]²[B]).
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