The following concentrations were obtained for the formation of ${\mathrm{NH}}_3$ from ${\mathrm{N}}_2$ and ${\mathrm{H}}_2$ at equilibrium at 500 K. $\left[{\mathrm{N}}_2\right]=1.5\times {10}^{-2}\mathrm{M}, \left[{\mathrm{H}}_2\right]=3.0\times {10}^{-2}\mathrm{M}$ and $\left[{\mathrm{NH}}_3\right] = 1.2 \times 10{ }^{-2}\mathrm{M}. \mathrm{Calculate} \mathrm{the} \mathrm{equilibrium} \mathrm{constant}.$