Question

Manufacturing Ethyl acetate \(\left(\mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}\right), \mathrm{a}\) solvent used in making varnishes and lacquers, can be produced by the reaction between ethanol and acetic acid. The equilibrium system is described by the equation \(\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \rightleftharpoons\) $$ \mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}+\mathrm{H}_{2} \mathrm{O} $$ Calculate \(K_{\text { eq using these equilibrium concentrations: }}\) \(\left[\mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}\right]=2.90 M,\left[\mathrm{CH}_{3} \mathrm{COOH}\right]=0.316 M,\) \(\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right]=0.313 M,\) and \(\left[\mathrm{H}_{2} \mathrm{O}\right]=0.114 M\)

Short answer

The equilibrium constant \(K_\text{eq}\) for the given reaction can be calculated using the expression: $$ K_\text{eq} = \frac{[\mathrm{CH}_{3}\mathrm{COOCH}_{2}\mathrm{CH}_{3}][\mathrm{H}_{2}\mathrm{O}]}{[\mathrm{CH}_{3}\mathrm{COOH}][\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}]} $$ Substituting the given equilibrium concentrations: $$ K_\text{eq} = \frac{(2.90 \mathrm{M})(0.114 \mathrm{M})}{(0.316 \mathrm{M})(0.313 \mathrm{M})} $$ Calculating the value of \(K_\text{eq}\): $$ K_\text{eq} \approx 2.916 $$

Step-by-step solution

Write the expression for the equilibrium constant

The general expression for the equilibrium constant, \(K_\text{eq}\), can be written as follows: $$ K_\text{eq} = \frac{[\text{Products}]}{[\text{Reactants}]} $$ For the given reaction, the equilibrium constant expression becomes: $$ K_\text{eq} = \frac{[\mathrm{CH}_{3}\mathrm{COOCH}_{2}\mathrm{CH}_{3}][\mathrm{H}_{2}\mathrm{O}]}{[\mathrm{CH}_{3}\mathrm{COOH}][\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}]} $$

Substitute the given equilibrium concentrations

Substitute the given equilibrium concentrations into the equilibrium constant expression: $$ K_\text{eq} = \frac{(2.90 \mathrm{M})(0.114 \mathrm{M})}{(0.316 \mathrm{M})(0.313 \mathrm{M})} $$

Calculate the value of \(K_\text{eq}\)

Perform the calculation to determine the equilibrium constant: $$ K_\text{eq} = \frac{(2.90)(0.114)}{(0.316)(0.313)} \approx 2.916 $$ So, the equilibrium constant \(K_\text{eq}\) for the given reaction is approximately 2.916.

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is crucial in the study of reactions like the synthesis of Ethyl acetate—a common solvent. It's a dynamic state where the rates of the forward and reverse reactions are equal, leading to constant concentrations of reactants and products in a closed system over time.

However, 'equilibrium' doesn't imply that the reactants and products are present in equal amounts, but rather that their quantities no longer change with time. In the given exercise, reaching equilibrium means the continuous conversion of ethanol and acetic acid into Ethyl acetate and water happens at the same rate as their reformation from the products. This balance can be quantified by the equilibrium constant, denoted as \(K_\text{eq}\).

The \(K_\text{eq}\) value gives us the ratio of the concentration of products to the concentration of reactants at equilibrium, raised to the power of their stoichiometric coefficients from the balanced chemical equation. In cases like the production of Ethyl acetate, where the equation involves one mole each of reactants and products, the stoichiometric coefficients are understood to be one.

Reaction Quotient

The reaction quotient, \(Q\), plays a pivotal role in predicting the direction in which a reaction mixture will shift to reach equilibrium. It's akin to an 'instant snapshot' of a reaction's progress, using the same formula as \(K_\text{eq}\) but with the current concentrations rather than those at equilibrium.

To calculate \(Q\), one would use the given or measured concentrations of the reactants and products at any point in time during the reaction. For the example exercise, if we were to calculate \(Q\) before the reaction had reached equilibrium, we'd plug those concentrations into the equation we used for \(K_\text{eq}\).

If \(QK_\text{eq}\), the reaction will favor the reverse path. When \(Q=K_\text{eq}\), the system is at equilibrium. The reaction quotient is a dynamic figure that changes as a reaction proceeds, in contrast to the constant value of the equilibrium constant once equilibrium is established.

Le Chatelier's Principle

Digging deeper into the behavior of reactions under stress, we consider Le Chatelier's principle. It states that if an external change is applied to a system at equilibrium, the system will adjust itself in a way that counteracts the change and a new equilibrium is established.

For instance, if we were to change the concentration of either reactants or products in our Ethyl acetate reaction, Le Chatelier's principle suggests the equilibrium will shift to reduce that change. Increasing the concentration of a reactant or product drives the equilibrium in the direction that consumes the added substance. Conversely, removal of a substance shifts the equilibrium toward producing it.

Temperature and pressure are other factors that can affect equilibrium. For endothermic reactions, heat is a reactant, and increasing temperature causes the equilibrium to shift toward products, effectively raising \(K_\text{eq}\). In contrast, lowering the temperature favors the reactants. Changes in pressure, while only affecting reactions involving gases, manipulate the equilibrium similarly, shifting it towards the side with fewer gas molecules when pressure is increased.