For the reaction ${\mathrm{N}}_2+3{\mathrm{H}}_2\to 2{\mathrm{NH}}_3,$ the  $\frac{\mathrm{d}\left[{\mathrm{NH}}_3\right]}{\mathrm{dt}}=2\times {10}^{-4}{\mathrm{Ms}}^{-1}$ Therefore the - $\frac{\mathrm{d}\left[{\mathrm{N}}_2\right]}{\mathrm{dt}}$ is given as

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