A) \[3.0\times {{10}^{-4}}mol\,{{L}^{-1}}{{s}^{-1}}\]
B) \[6.0\times {{10}^{-4}}mol\,{{L}^{-1}}{{s}^{-1}}\]
C) \[1.5\times {{10}^{-4}}mol\,{{L}^{-1}}{{s}^{-1}}\]
D) \[4.5\times {{10}^{-4}}mol\,{{L}^{-1}}{{s}^{-1}}\]
Correct Answer: B
Solution :
[b] \[-\frac{d[{{O}_{2}}]}{dt}=+\frac{1}{2}\frac{d[S{{O}_{3}}]}{dt}\] \[3\times {{10}^{-4}}=\frac{1}{2}\frac{d[S{{O}_{3}}]}{dt}\] \[\frac{d[S{{O}_{3}}]}{dt}=6\times {{10}^{-4}}M/sec.\]You need to login to perform this action.
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