A classic experiment in equilibrium studies dating from 1862 involved the reaction in solution of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) to produce ethyl acetate and water. $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}+\mathrm{H}_{2} \mathrm{O}$$ The reaction can be followed by analyzing the equilibrium mixture for its acetic acid content. $$\begin{array}{r} 2 \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq}) \rightleftharpoons \\ \mathrm{Ba}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{array}$$ In one experiment, a mixture of 1.000 mol acetic acid and 0.5000 mol ethanol is brought to equilibrium. A sample containing exactly one-hundredth of the equilibrium mixture requires \(28.85 \mathrm{mL} 0.1000 \mathrm{M}\) \(\mathrm{Ba}(\mathrm{OH})_{2}\) for its titration. Calculate the equilibrium constant, \(K_{c}\), for the ethanol-acetic acid reaction based on this experiment.

Step by step solution

01

Analyze the information given by the problem

The reaction is given, as well as the initial amounts of acetic acid and ethanol. From the titration data, we can infer the concentration of acetic acid in the equilibrium mixture. Since the titration involves two moles of acetic acid reacting with one mole of barium hydroxide, the concentration of barium hydroxide needed to react with all the acetic acid in the sample is half of the concentration of acetic acid. It follows that the concentration of acetic acid at equilibrium is 2 * 0.1000 M = 0.2000 M.

02

Calculate the change in moles for reactants and products

Given that the total amount of acetic acid at the start is 1.000 mol and its concentration at equilibrium is 0.2000 M, and assuming the total volume is 1 L, the change in moles of acetic acid (and hence ethanol, given its stoichiometry in the reaction) is 1.000 - 0.2000 = 0.800 mol. As a result, the change in moles of ethyl acetate (and water) is also 0.800 mol.

03

Determine the equilibrium concentrations

The initial concentration of ethanol was 0.5000 M, and it lost 0.800 mol throughout the reaction. Therefore, its equilibrium concentration is 0.5000 - 0.800 = -0.3000 M. However, a negative concentration does not make sense in physical terms. It seems that such an amount of ethanol was not available at the start. Therefore, we must adjust the changes in moles for all substances, which in practical terms means assuming all the initial ethanol is used up in the reaction. In that case, the changes in moles are -0.5000 mol for ethanol and acetic acid, and +0.5000 mol for ethyl acetate and water. The equilibrium concentration of ethanol is then 0, and those of acetic acid and ethyl acetate are 0.5000 M and 1.000 - 0.5000 = 0.5000 M, respectively.

04

Calculate the equilibrium constant

Applying the equilibrium constant expression \(K_c = [CH_3COOC_2H_5][H_2O]/[C_2H_5OH][CH_3COOH] = [0.5000][1] / [0][0.5000] = (\inf)\). Technically, the equilibrium constant is not defined for reactions where one of the reactant concentrations is zero, but in practice it implies that the reaction has gone to completion, i.e., all ethanol was converted to ethyl acetate.

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