A) \[{{k}_{1}}={{k}_{2}}={{k}_{3}}\]
B) \[3{{k}_{1}}=6{{k}_{2}}=2{{k}_{3}}\]
C) \[2{{k}_{1}}=3{{k}_{2}}=6{{k}_{3}}~~\]
D) \[6{{k}_{1}}=3{{k}_{2}}=2{{k}_{3}}\]
Correct Answer: B
Solution :
\[-\frac{1}{2}\frac{d[N{{H}_{3}}]}{dt}=\frac{d[{{N}_{2}}]}{dt}=\frac{1}{3}\frac{d[{{H}_{2}}]}{dt}\] \[\frac{1}{2}{{k}_{1}}(N{{H}_{3}})={{k}_{2}}(N{{H}_{3}})=\frac{1}{3}{{k}_{3}}(N{{H}_{3}})\] or \[\frac{{{k}_{1}}}{2}={{k}_{2}}=\frac{{{k}_{3}}}{3}\,or\,3\,{{k}_{1}}=6{{k}_{2}}=2{{k}_{3}}\]You need to login to perform this action.
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