For the given reaction $\text{A}_2(g) + \text{B}_2(g) \rightleftharpoons 2\text{AB}(g)$, the equilibrium constant expression ($K_c$) is written as 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 their stoichiometric coefficients .

$K_c = \frac{[\text{AB}]^2}{[\text{A}_2][\text{B}_2]}$

Given equilibrium concentrations are: $[\text{A}_2] = 3.0 \times 10^{-3} \text{ M}$ $[\text{B}_2] = 4.2 \times 10^{-3} \text{ M}$ $[\text{AB}] = 2.8 \times 10^{-3} \text{ M}$

Substituting these values into the equilibrium constant expression: $K_c = \frac{(2.8 \times 10^{-3})^2}{(3.0 \times 10^{-3})(4.2 \times 10^{-3})}$ $K_c = \frac{7.84 \times 10^{-6}}{12.6 \times 10^{-6}}$ $K_c = 0.622 \approx 0.6$