Question

The reaction of an organic acid with an alcohol, in organic solvent, to produce an ester and water is commonly done in the pharmaceutical industry. This reaction is catalyzed by strong acid (usually \(\mathrm{H}_{2} \mathrm{SO}_{4} ) .\) A simple example is the reaction of acetic acid with ethyl alcohol to produce ethyl acetate and water: $$\begin{aligned} \mathrm{CH}_{3} \mathrm{COOH}(s o l v)+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(s o l v) & \rightleftharpoons \\ \mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}(\mathrm{solv}) &+\mathrm{H}_{2} \mathrm{O}(\text {solv}) \end{aligned}$$ where \(^{a}(s o l v)^{\prime \prime}\) indicates that all reactants and products are in solution but not an aqueous solution. The equilibrium constant for this reaction at \(55^{\circ} \mathrm{C}\) is 6.68 . A pharmaceutical chemist makes up 15.0 \(\mathrm{L}\) of a solution that is initially 0.275 \(\mathrm{M}\) in acetic acid and 3.85\(M\) in ethanol. At equilibrium, how many grams of ethyl acetate are formed?

Short answer

At equilibrium, 9290.14 grams of ethyl acetate are formed.

Step-by-step solution

Write the equilibrium expression

First, we will write the equilibrium expression in terms of the molar concentrations of reactants and products involved in the given equilibrium reaction. For the reaction: \(\mathrm{CH_3COOH} + \mathrm{CH_3CH_2OH} \rightleftharpoons \mathrm{CH_3COOCH_2CH_3} + \mathrm{H_2O}\), the equilibrium constant (K) expression is: \(K = \frac{[\mathrm{CH_3COOCH_2CH_3}][\mathrm{H_2O}]}{[\mathrm{CH_3COOH}][\mathrm{CH_3CH_2OH}]}\)

Set up an ICE table and express equilibrium concentrations in terms of change

Next, we set up an ICE table to represent the initial, change, and equilibrium concentrations of the reactants and products. Let x moles of acetic acid and ethanol react. | | CH3COOH | CH3CH2OH | CH3COOCH2CH3 | H2O | |-------|---------|----------|------------|-------| | I (M) | 0.275 | 3.85 | 0 | ? | C (M) | -x | -x | +x | +x | | E (M) | 0.275-x | 3.85-x | x | ? (Note: The initial concentration of water is not required in this calculation since it's not involved in the equilibrium constant).

Substitute equilibrium concentrations into the equilibrium expression and solve for x

Now, we substitute the equilibrium concentrations of the reactants and products from the ICE table into the K expression: \(6.68 = \frac{x(\text{Since x is small, water concentration remains almost constant})}{(0.275 - x)(3.85 - x)}\) We can assume that x is small compared to the initial concentrations (0.275 and 3.85), thus simplifying the equation: \(6.68 = \frac{x}{0.275 \times 3.85}\). Solve for x: \(x = 0.275 \times 3.85 \times 6.68 = 7.03 \: \mathrm{M}\)

Calculate the mole and mass of the product, ethyl acetate, at equilibrium

To find the grams of ethyl acetate formed at equilibrium, we first calculate the number of moles: No of moles of ethyl acetate = Molarity × volume = 7.03 × 15.0 = 105.45 \: \mathrm{moles} Now, we calculate the mass of ethyl acetate: Mass = No of moles × Molar mass = 105.45 \: \mathrm{moles} \times 88.1 \: \frac{\mathrm{g}}{\mathrm{mol}} = 9290.14 \: \mathrm{g}\) At equilibrium, 9290.14 grams of ethyl acetate are formed.

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is crucial when studying reactions such as esterification, where reagents like an organic acid and an alcohol react to form an ester and water. At equilibrium, the rate of the forward reaction, where the acid and alcohol form the ester, equals the rate of the reverse reaction, where the ester and water re-form the original substances. This balance does not imply that the reactants and products are present in equal amounts, but rather that their concentrations remain constant over time. In the context of the pharmaceutical industry, achieving the correct balance can be essential for the efficient production of drugs.

It's vital to note that while the reaction can reach a state where the concentration of reactants and products cease to change, they are not static. Molecules are continuously reacting, and equilibrium represents a dynamic balance between the forward and reverse reactions. Therefore, equilibrium can be disturbed by changes in concentration, pressure, or temperature, invoking Le Chatelier's principle, which predicts the direction of the reaction's shift in response to such changes.

ICE Table

An ICE table, which stands for Initial, Change, Equilibrium, is a powerful tool used to organize and calculate the concentrations of reactants and products at equilibrium. When setting up an ICE table for an esterification reaction, begin by listing the initial concentrations (I) of all species involved. Then, determine the change (C) in concentration for each substance as the system moves towards equilibrium. Usually, this involves introducing a variable, such as 'x', to represent the change.

For instance, if we start with acetic acid and ethanol, we assume that a certain amount, 'x', will convert to ethyl acetate and water at equilibrium. This 'x' is then subtracted from the initial concentrations of the reactants and added to the products. By applying the equilibrium constant, we use these expressions to solve for 'x', which allows us to determine the concentrations at equilibrium (E). Keep in mind, the ICE table is an essential simplification strategy which involves approximations that streamline calculations.

Equilibrium Constant

The equilibrium constant (K) quantitatively describes the chemical equilibrium of a reaction. It is calculated as the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants, with each concentration raised to the power of its stoichiometric coefficient in the balanced equation. For esterification, this would mean that K can help us to understand how acetic acid and ethyl alcohol create esters.

The magnitude of the equilibrium constant gives insight into the position of equilibrium. A high K value suggests that at equilibrium, the products are favored and present in larger concentrations. Conversely, a low K value indicates that the reactants are favored. It is important to note that K is temperature-dependent, so a change in temperature leads to a change in the equilibrium constant, influencing whether more product or reactant is favored.