For the reaction shown, the equilibrium constant, $K_c$, is $2\,\text{mol}^{-2}\text{dm}^6$ at $600\,\text{K}$. $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)}$ Calculate the equilibrium concentration of $\text{NH}_3$ at $600\,\text{K}$ if the equilibrium concentrations of $\text{N}_2$ and $\text{H}_2$ are each $2\,\text{mol dm}^{-3}$?
- A$\sqrt{8}\,\text{mol dm}^{-3}$
- B$\sqrt{16}\,\text{mol dm}^{-3}$
- C$\sqrt{32}\,\text{mol dm}^{-3}$
- D$32\,\text{mol dm}^{-3}$