The following concentrations were observed at $$500$$ K for the formation of $$NH_3$$ from $$N_2$$ and $$H_2$$. At equilibrium: $$[N_2] = 2 \times 10^{-2}$$ M, $$[H_2] = 3 \times 10^{-2}$$ M and $$[NH_3] = 1.5 \times 10^{-2}$$ M. Equilibrium constant for the reaction is ______.

We need to find the equilibrium constant for the formation of $$NH_3$$ from $$N_2$$ and $$H_2$$. The balanced equation for this reaction is $$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$$.

At equilibrium the concentrations are $$[N_2] = 2 \times 10^{-2}$$ M, $$[H_2] = 3 \times 10^{-2}$$ M, and $$[NH_3] = 1.5 \times 10^{-2}$$ M. The expression for the equilibrium constant $$K_c$$ is $$K_c = \frac{[NH_3]^2}{[N_2][H_2]^3}$$, so substituting these values gives $$K_c = \frac{(1.5 \times 10^{-2})^2}{(2 \times 10^{-2})(3 \times 10^{-2})^3}$$.

Since $$(1.5 \times 10^{-2})^2 = 2.25 \times 10^{-4}$$ and $$(3 \times 10^{-2})^3 = 27 \times 10^{-6} = 2.7 \times 10^{-5}$$, multiplying the denominator yields $$(2 \times 10^{-2})(2.7 \times 10^{-5}) = 5.4 \times 10^{-7}$$. Therefore, $$K_c = \frac{2.25 \times 10^{-4}}{5.4 \times 10^{-7}} = \frac{2.25}{5.4} \times 10^{3} = 0.4167 \times 10^{3} \approx 417$$.

Therefore, the equilibrium constant is 417.