Question

The oxidation of 2-butanone (CH3COC2H5) by the cerium(IV) ion in aqueous solution to form acetic acid (CH3CO2H) occurs according to the following balanced equation: CH3COC2H5(aq)+6Ce4+(aq)+3H2O(l)→ 2CH3CO2H(aq)+6Ce3+(aq)+6H+(aq)

Q\If acetic acid appears at an average rate of 5.2×10⁻⁸ M/sM/s what is Δ[H+]ΔtΔ[H+]Δt during the same time interval?

Express your answer to two significant figures and include the appropriate units.

Q\What is the average rate of consumption of Ce4+Ce4+ during the same time interval?

Express your answer to two significant figures and include the appropriate units.

Answer 1

Answer:

The given equation is

CH₃COCH₂CH₃ + 6Ce⁴⁺ + 3 H₂O $\rightarrow$ 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