For the reaction,$N_2 + 3H_2 \rightarrow 2NH_3,$ if $\frac{d[NH_3]}{dt} = 2 \times 10^{-4} \ mol \ L^{-1} \ s^{-1},$ the value of $-\frac{d[H_2]}{dt}$ would be .................. $mol \ L^{-1} \ s^{-1}$.

  • A
    $4 \times 10^{-4}$
  • B
    $6 \times 10^{-4}$
  • C
    $1 \times 10^{-4}$
  • D
    $3 \times 10^{-4}$

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