For the reaction, N_2+3 H_2 → 2 NH_3, if (d [ NH_3])/dt=2 × 10^-4 mol L^-1 s^-1, the value of (- d [ H_2])/dt would be:

For the reaction, $\displaystyle \mathrm{N}_2+3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3$, if $\displaystyle \frac{\mathrm{d}\left[\mathrm{NH}_3\right]}{\mathrm{dt}}=2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$, the value of $\displaystyle \frac{-\mathrm{d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}$ would be :