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For the chemical reaction $N_2(g)+3H_2(g)\rightleftharpoons2NH_3(g)$ the correct option is:
A
$-\frac{1}{3}\frac{d[H_2]}{dt}=-\frac{1}{2}\frac{d[NH_3]}{dt}$
B
$-\frac{d[N_2]}{dt}=2\frac{d[NH_3]}{dt}$
C
$-\frac{d[N_2]}{dt}=\frac{1}{2}\frac{d[NH_3]}{dt}$
D
$3\frac{d[H_2]}{dt}=2\frac{d[NH_3]}{dt}$
Detailed Solution
$N_2+3H_2\rightleftharpoons2NH_3$
Rate of reaction $=-\frac{d[N_2]}{dt}=-\frac{1}{3}\frac{d[H_2]}{dt}=+\frac{1}{2}\frac{d[NH_3]}{dt}$
So $-\frac{d[N_2]}{dt}=\frac{1}{2}\frac{d[NH_3]}{dt}$ is correct.