Question

Consider the reaction : $N_2 (g) + 3H_2 (g) \rightarrow 2NH_3 (g)$
The equality relationship between $\frac{d[NH_3]}{dt}$ and $- \frac{d[H_2]}{dt}$ is
(a) $+ \frac{d[NH_3]}{dt} = -\frac{2}{3} \frac{d[H_2]}{dt}$
(b) $+ \frac{d[NH_3]}{dt} = -\frac{3}{2} \frac{d[H_2]}{dt}$
(c) $\frac{d[NH_3]}{dt} = -\frac{d[H_2]}{dt}$
(d) $\frac{d[NH_3]}{dt} = -\frac{1}{3} \frac{d[H_2]}{dt}$

Answer 1

a opption is the correct answer