A) \[+\frac{d[N{{H}_{3}}]}{dt}=-\frac{2}{3}\frac{d[{{H}_{2}}]}{dt}\]
B) \[+\frac{d[N{{H}_{3}}]}{dt}=-\frac{3}{2}\frac{d[{{H}_{2}}]}{dt}\]
C) \[\frac{d[N{{H}_{3}}]}{dt}=-\frac{d[{{H}_{2}}]}{dt}\]
D) \[\frac{d[N{{H}_{3}}]}{dt}=-\frac{1}{3}\frac{d[{{H}_{2}}]}{dt}\]
Correct Answer: A
Solution :
[a] If we write rate of reaction in terms of concentration of \[N{{H}_{3}}\] and \[{{H}_{2}}\] then Rate of reaction \[=\frac{1}{2}\frac{d[N{{H}_{3}}]}{dt}=-\frac{1}{3}\frac{d[{{H}_{2}}]}{dt}\] So, \[=\frac{d[N{{H}_{3}}]}{dt}=-\frac{2}{3}\frac{d[{{H}_{2}}]}{dt}\]You need to login to perform this action.
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