A) \[-\frac{d[{{N}_{2}}]}{dt}=\frac{1}{2}\frac{d[N{{H}_{3}}]}{dt}\]
B) \[3\frac{d[{{H}_{2}}]}{dt}=2\frac{d[N{{H}_{3}}]}{dt}\]
C) \[-\frac{1}{3}\frac{d[{{H}_{2}}]}{dt}=-\frac{1}{2}\frac{d[N{{H}_{3}}]}{dt}\]
D) \[-\frac{d[{{N}_{2}}]}{dt}=2\frac{d[N{{H}_{3}}]}{dt}\]
Correct Answer: A
Solution :
| [a] \[{{N}_{2}}+3{{H}_{2}}\rightleftharpoons 2N{{H}_{3}}\] |
| \[-\frac{d{{N}_{2}}}{dt}=-\frac{1}{3}\frac{d{{H}_{2}}}{dt}=+\frac{1}{2}-\frac{dN{{H}_{3}}}{dt}\] |
| \[-\frac{d{{N}_{2}}}{dt}=+\frac{1}{2}\frac{d[N{{H}_{3}}]}{dt}\] |
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