A) \[-\frac{d[N{{H}_{3}}]}{dt}=-\frac{3}{2}\left[ \frac{d[{{H}_{2}}]}{dt} \right]\]
B) \[\frac{d\left[ N{{H}_{3}} \right]}{dt}=-\frac{d\left[ {{H}_{2}} \right]}{dt}\]
C) \[\frac{d\left[ N{{H}_{3}} \right]}{dt}=-\frac{1}{3}\frac{d\left[ {{H}_{2}} \right]}{dt}\]
D) \[+\frac{d\left[ N{{H}_{3}} \right]}{dt}=-\frac{2}{3}\frac{d\left[ {{H}_{2}} \right]}{dt}\]
Correct Answer: D
Solution :
Consider the reaction \[{{N}_{2}}(g)+3{{H}_{2}}(g)\to 2N{{H}_{3}}(g)\] Rate of reaction \[=\frac{-1}{3}\frac{d[{{H}_{2}}]}{dt}=-\frac{d[{{N}_{2}}]}{dt}=\frac{1}{2}\frac{d[N{{H}_{3}}]}{dt}\] So, \[\frac{d[N{{H}_{3}}]}{dt}=-\frac{-2}{3}\frac{d[{{H}_{2}}]}{dt}\]
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