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For the reaction, $N_2 + 3H_2 \rightarrow 2NH_3$, if $\frac{d[NH_3]}{dt} = 2\times10^{-4}$ mol $L^{-1}s^{-1}$, the value of $\frac{-d[H_2]}{dt}$ would be
A
$1\times10^{-4}$ mol $L^{-1}s^{-1}$
B
$3\times10^{-4}$ mol $L^{-1}s^{-1}$
C
$4\times10^{-4}$ mol $L^{-1}s^{-1}$
D
$6\times10^{-4}$ mol $L^{-1}s^{-1}$
Detailed Solution
For $N_2 + 3H_2 \rightarrow 2NH_3$: Rate $= -\frac{d[N_2]}{dt} = -\frac{1}{3}\frac{d[H_2]}{dt} = \frac{1}{2}\frac{d[NH_3]}{dt}$
$-\frac{1}{3}\frac{d[H_2]}{dt} = \frac{1}{2}\frac{d[NH_3]}{dt}$
$-\frac{d[H_2]}{dt} = \frac{3}{2}\frac{d[NH_3]}{dt} = \frac{3}{2}\times2\times10^{-4}$
$-\frac{d[H_2]}{dt} = 3\times10^{-4}$ mol $L^{-1}s^{-1}$
$-\frac{1}{3}\frac{d[H_2]}{dt} = \frac{1}{2}\frac{d[NH_3]}{dt}$
$-\frac{d[H_2]}{dt} = \frac{3}{2}\frac{d[NH_3]}{dt} = \frac{3}{2}\times2\times10^{-4}$
$-\frac{d[H_2]}{dt} = 3\times10^{-4}$ mol $L^{-1}s^{-1}$