[N2] = 3.0 mol L-1
[H2] = 2.0 mol L-1
[NH3] = 0.50 mol L-1
N2(g) + 3H2(g) ⇌ 2NH3
QC = \(\frac{[NH_3]^2}{[N_2][H_2]^3}\) = \(\frac{(0.50)^2}{3 \times (2)^3}\) = 0.0104
Thus at the stage of the reaction QC < KC
Therefore, the reaction is not at equilibrium. The reaction proceeds to form more products.