Law Of Chemical Equilibrium

Question

For the reversible reaction N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g), the equilibrium constant expression K_c is correctly written as which of the following?

RankGuru · Accepted Answer

The law of chemical equilibrium states that for a reaction aA + bB \rightleftharpoons cC + dD at equilibrium, the equilibrium constant K_c = [C]^c[D]^d / ([A]^a[B]^b), where concentrations are raised to the power of their stoichiometric coefficients. For N_2 + 3H_2 \rightleftharpoons 2NH_3, products are in the numerator and reactants in the denominator with their respective stoichiometric coefficients as exponents: K_c = [NH_3]^2 / ([N_2][H_2]^3). This is the only correct form. Option B is wrong because it is the expression for 1/K_c, the reverse reaction's equilibrium constant. Option C is wrong because it does not account for the stoichiometric coefficients of 2 for NH_3 and 3 for H_2. Option D is incorrect because N_2 should have an exponent of 1, not 2, matching its coefficient in the balanced equation. This expression is directly derived from NCERT Equilibrium chapter treatment of Guldberg-Waage law and appears frequently in JEE Main as a conceptual question. The plausibility check: units of K_c here are (mol/L)^(2−1−3) = (mol/L)^{-2}, confirming the denominator has higher total power than numerator.

Law Of Chemical Equilibrium

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