For N₂ + 3H₂ ⇌ 2NH₃, Kc equals:

To determine the equilibrium constant expression ($K_c$) for the reversible reaction: $$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$$ We apply the **Law of Mass Action**, which states that the equilibrium constant is the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants, each raised to the power of its stoichiometric coefficient. ### Step-by-Step Breakdown: 1. **Identify Products and Reactants**: - Products: $NH_3$ - Reactants: $N_2$ and $H_2$ 2. **Identify Stoichiometric Coefficients**: - Coefficient for $NH_3$ is **2**. - Coefficient for $N_2$ is **1**. - Coefficient for $H_2$ is **3**. 3. **Construct the Expression**: - Numerator (Products): $[NH_3]^2$ - Denominator (Reactants): $[N_2]^1 \times [H_2]^3$ Putting it all together: $$K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}$$ This matches option A.