Question

For the reaction, \[\ce{3A + 2B -> C + D}\], the differential rate law can be written as ______.

विकल्प

• `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`

Answer 1

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`