A) \[rate=-\frac{d[{{N}_{2}}]}{dt}=-\frac{1}{3}\frac{d[{{H}_{2}}]}{dt}=\frac{1}{2}\frac{d[N{{H}_{3}}]}{dt}\]
B) \[rate=-\frac{d[{{N}_{2}}]}{dt}=-3\frac{d[{{H}_{2}}]}{dt}=2\frac{d[N{{H}_{3}}]}{dt}\]
C) \[rate=\frac{d[{{N}_{2}}]}{dt}=\frac{1}{3}\frac{d[{{H}_{2}}]}{dt}=\frac{1}{2}\frac{d[N{{H}_{3}}]}{dt}\]
D) \[rate=-\frac{d[{{N}_{2}}]}{dt}=-\frac{d[{{H}_{2}}]}{dt}=\frac{d[N{{H}_{3}}]}{dt}\]
Correct Answer: A
Solution :
\[{{N}_{2}}(g)+3{{H}_{2}}(g)\xrightarrow{{}}2N{{H}_{3}}(g)\] Instantaneous rate \[=-\frac{d[{{N}_{2}}]}{dt}=-\frac{1}{3}\frac{d[{{H}_{2}}]}{dt}=\frac{1}{2}\frac{d[N{{H}_{3}}]}{dt}\]You need to login to perform this action.
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