The oxidation of 2-butanone (CH3COC2H5) by the cerium(IV) ion in aqueous solution to form acetic acid...

Answer:

The given equation is

CH₃COCH₂CH₃ + 6Ce⁴⁺ + 3 H₂O 2 CH₃COOH + 6 Ce³⁺ + 6 H⁺

The rate equation for the above reaction is

[rate] = - d [2-butanone]/dt = [- 1/6] d[Ce⁴⁺] /dt = [1/2] d [acetic acid]/dt = [1/6] d[Ce³⁺]/dt = [1/6]d[H⁺]/dt ------(1)

given d[acetic acid]/dt = 5.2 x 10^-8 M/s

                         [1/2] d [acetic acid]/dt = [1/6]d[H⁺]/dt            [this relation is taken from (1)]

                          [1/2] * 5.2 x 10^-8 = [1/6]d[H⁺]/dt

therefore d[H⁺]/dt =[ 6/2] x 5.2 x 10^-8 M/s

                          = 15.6 x 10^-8 M/s

                          = 1.6 x 10^-7 M/s

Thus the rate of formation of [H⁺] = 1.6 x 10^-7 M/s

similarly the rate of consumption of Ce⁴⁺ is given by

              [- 1/6] d[Ce⁴⁺] /dt = [1/6]d[H⁺]/dt                     [ this relation is taken from (1) ]

                      d[Ce⁴⁺]/dt = - d[H⁺]/dt

                                     = - 1.6 x 10^-7 M/s

Thus the rate of consumption of Ce⁴⁺ = 1.6 x 10^-7 M/s