For the reaction, $N_2+3H_2\to 2NH_3$, $\frac {d[NH_3]}{dt}=2\times10^{-4} mol L^{-1} S^{-1}$, The value of $\frac{-d[H_2]}{dt}$ would be

Question

( A ) $3\times 10^{–4} mol L^{–1} s^{–1}$
( B ) $4 × 10^{–4} mol L^{–1} s^{–1}$
( C ) $1 × 10^{–4} mol L^{–1} s^{–1}$
( D ) $6 × 10^{–4} mol L^{–1} s^{–1}$