Question

question 15 (1 point) saved
what is the correct equilibrium constant expression for the following reaction?
2ncl₃(g) ↔ n₂(g) + 3cl₂(g)
k_eq = n₂cl₂³ / ncl₃²
k_eq = ncl₃² / n₂cl₂³
k_eq = n₂cl₂ / ncl₃
k_eq = ncl₃ / n₂cl₂
k_eq = n₂³cl₂³ / ncl₃²

Explanation:

Step1: Recall Equilibrium Constant Formula

For a reaction \( aA + bB
ightleftharpoons cC + dD \), the equilibrium constant \( K_{eq} \) is \( K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b} \), where concentrations of products are in the numerator and reactants in the denominator, each raised to their stoichiometric coefficients.

Step2: Apply to Given Reaction

The reaction is \( 2NCl_3(g)
ightleftharpoons N_2(g) + 3Cl_2(g) \). Here, reactant is \( NCl_3 \) with coefficient 2, products are \( N_2 \) (coefficient 1) and \( Cl_2 \) (coefficient 3). So \( K_{eq}=\frac{[N_2][Cl_2]^3}{[NCl_3]^2} \).

Answer:

\( K_{eq} = \frac{[N_2][Cl_2]^3}{[NCl_3]^2} \) (the first option: \( K_{\text{eq}} = \frac{[\text{N}_2][\text{Cl}_2]^3}{[\text{NCl}_3]^2} \))