A) \[1\times {{10}^{-4}}mol\,{{L}^{-1}}{{s}^{-1}}\]
B) \[3\times {{10}^{-4}}mol\,{{L}^{-1}}{{s}^{-1}}\]
C) \[4\times {{10}^{-4}}mol\,{{L}^{-1}}{{s}^{-1}}\]
D) \[6\times {{10}^{-4}}mol\,{{L}^{-1}}{{s}^{-1}}\]
Correct Answer: B
Solution :
\[\frac{1}{2}\frac{d[N{{H}_{3}}]}{dt}=\frac{1}{3}\frac{d[{{H}_{2}}]}{dt}\] \[\therefore \] \[-\frac{d[{{H}_{2}}]}{dt}=\frac{3}{2}\times \frac{d[N{{H}_{3}}]}{dt}\] \[=\frac{3}{2}\times 2\times {{10}^{-4}}=3\times {{10}^{-4}}\]You need to login to perform this action.
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