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Question
For the reaction, \[\ce{3A + 2B -> C + D}\], the differential rate law can be written as ______.
Options
`1/3 ("d"("A"))/"dt" = ("d"["C"])/"dt" = k ["A"]^n ["B"]^m`
`- ("d"["A"])/"dt" = ("d"["C"])/"dt" = k ["A"]^n ["B"]^m`
`+ 1/3 ("d"["A"])/"dt" = ("d"["C"])/"dt" = k ["A"]^n ["B"]^m`
`- 1/3 ("d"["A"])/"dt" = ("d"["C"])/"dt" = k ["A"]^n ["B"]^m`
MCQ
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Solution
For the reaction, \[\ce{3A + 2B -> C + D}\], the differential rate law can be written as `underline(- 1/3 ("d"["A"])/"dt" = ("d"["C"])/"dt" = k ["A"]^n ["B"]^m)`.
Explanation:
For the reaction
\[\ce{3A + 2B -> C + D}\]
From the rate law,
`= - 1/3 ("d"["A"])/"dt" = ("d"["C"])/"dt" = k ["A"]^n ["B"]^m`
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