Question

The saponification (hydrolysis) of ethyl acetate occurs according to the stoichiometric relation CH 3 COOC 2 H 5 + OH - S CH 3 COO - + C 2 H 5 OH. The reaction can be followed by monitoring the disappearance of OH - . The following experimental results were obtained at 25 °C: Initial concentration of OH " (M) Initial concentration of CH 3 COOC 2 H 5 (M) Half-life, t ½ , s 0.0050 0.0050 2000 0.0100 0.0100 1000 There is no significant dependence on the concentrations of products of the reaction. a. What is the overall kinetic order of the reaction? b. Calculate a value for the rate coefficient, including appropriate units. c. How long would it take for the concentration of OH - to reach 0.0025 M for each experiment? d. Based on your answer to part (a), what are possible rate laws for this reaction? e. Carefully describe a single experiment that would enable you to decide which of the possibilities of part (d) is most nearly correct.

Answer 1

Complete Question in order

The saponification (hydrolysis) of ethyl acetate occurs according to the stoichiometric relation CH₃COOC₂H₅ + OH⁻ S CH 3 COO - + C 2 H 5 OH. The reaction can be followed by monitoring the disappearance of OH - . The following experimental results were obtained at 25 °C:

Initial Concentration Initial Concentration Half-life, tVi, s

of OH- (M) of CH3COOC2H5 (M)

0.0050 0.0050 2000

0.0100 0.0100 1000

There is no significant dependence on the concentrations of products of the reaction.

a. What is the overall kinetic order of the reaction?

b. Calculate a value for the rate coefficient, including appropriate units.

c. How long would it take for the concentration of OH - to reach 0.0025 M for each experiment?

d. Based on your answer to part (a), what are possible rate laws for this reaction?

e. Carefully describe a single experiment that would enable you to decide which of the possibilities of part (d) is most nearly correct.

Answer:

a. Second order

b. 0.10M⁻¹S⁻¹

c. 2000s, 3000s

d. 2

e. [OH⁻]²

Step-by-step explanation:

a. The reaction for the saponification of ethyl acetate is as follows:

CH3COOC2H5 + OH-

the following data were obtained:

Experiment [OH-] (M) [CH3COOC2H5] (M) Half-life t (s)

1 0.0050 0.0050 2000

2. 0.0100 0.0100 1000

The concentrations shown in the data are the initial concentration

The first-order half-life is given as follows:

`t_(1/2) = (0.693)/(k_1) ....... (1)`

Here, k1 is

the rate constant

The second-order half-life is given as follows:

`t = (1)/(k_2 [A]_0) .........(2)`

Here, k2 is the rate constant and [A]0 is the initial concentration of the reactant.

From the data, the half-life is halved when the initial concentrations of the reactants are doubled. Therefore the reaction is second order overall

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