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Law Of Chemical Equilibrium

Easychemistry

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?

Select the correct option:

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About This Question

Subject
chemistry
Chapter
equilibrium
Topic
law of chemical equilibrium
Difficulty
Easy
Year
2025
Tags
equilibrium constantK_c expressionlaw of mass actionstoichiometric coefficientsHaber process

Solution

Correct 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.

This easy difficulty chemistry question is from the chapter equilibrium, covering the topic of law of chemical equilibrium. It appeared in the 2025 exam.

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