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 \( K_{eq}=\frac{[N_2]^1[Cl_2]^3}{[NCl_3]^2} \), which matches the first option.

Answer:

A. \( K_{\text{eq}} = \frac{[\text{N}_2][\text{Cl}_2]^3}{[\text{NCl}_3]^2} \)