Question

Consider the reaction,
$N_2(g)+3H_2(g) \to 2NH_3(g)$
The equality relationship between $\frac{d[NH_3]}{dt}$ and $-\frac{d [H_2]}{dt}$ is :-

• A. $\frac{d [NH_3]}{dt}=-\frac{1}{3} \frac{d [H_2]}{dt}$
• B. $+\frac{d [NH_3]}{dt}=-\frac{2}{3} \frac{d [H_2]}{dt}$
• C. $+\frac{d [NH_3]}{dt}=-\frac{3}{2} \frac{d [H_2]}{dt}$
• D. $\frac{d [NH_3]}{dt}=-\frac{d [H_2]}{dt}$

Answer 1

Answer: (B)

For the reaction $ N _{2}( g )+3 H _{2}( g ) \to 2 NH _{3}( g )$

$\frac{- d \left[ N _{2}\right]}{ dt }=-\frac{1}{3} \frac{ d }{ dt }\left(\left[ H _{2}\right]\right)=\frac{1}{2} \frac{ d }{ dt }\left(\left[ NH _{3}\right]\right)$

$\therefore \frac{ d }{ dt }\left(\left[ NH _{3}\right]\right)=-\frac{2}{3} \frac{ d }{ dt }\left(\left[ H _{2}\right]\right)$