Problem 60 - Chemistry: The Central Science in SI Units, Global Edition

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 \(\left.\mathrm{H}_{2} \mathrm{SO}_{4}\right)\). A simple example is the reaction of acetic acid with ethyl alcohol to produce ethyl acetate and water: $$\begin{aligned} \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{solv})+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(\mathrm{solv}) & \mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}(\text { solv })+\mathrm{H}_{2} \mathrm{O}(\text { solv }) \end{aligned}$$ where "(solv)" 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 \(3.85 \mathrm{M}\) in ethanol. At equilibrium, how many grams of ethyl acetate are formed?

Step-by-Step Solution

Answer

At equilibrium, approximately 1746.40 grams of ethyl acetate are formed in the reaction of acetic acid and ethanol in organic solvent with an initial solution of $0.275\, \mathrm{M}$ acetic acid and $3.85\, \mathrm{M}$ ethanol, a total volume of $15.0\, \mathrm{L}$, and an equilibrium constant of $6.68$.

Step 1: Write down the equilibrium expression

For the given reaction: $$\mathrm{CH}_{3} \mathrm{COOH} + \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3} + \mathrm{H}_{2}\mathrm{O}$$ The equilibrium constant expression can be written as: $$K = \frac{[\mathrm{CH}_{3}\mathrm{COOCH}_{2}\mathrm{CH}_{3}][\mathrm{H}_{2}\mathrm{O}]}{[\mathrm{CH}_{3}\mathrm{COOH}][\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}]}$$

Step 2: Set up an ICE table

Let's create an ICE (Initial, Change, Equilibrium) table to keep track of the changes in concentrations of the reactants and products: \[\begin{array}{c|ccc} & [\mathrm{CH}_{3}\mathrm{COOH}] & [\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}] & [\mathrm{CH}_{3}\mathrm{COOCH}_{2}\mathrm{CH}_{3}] & [\mathrm{H}_{2}\mathrm{O}] \\ \hline \text{Initial} & 0.275 & 3.85 & 0 & 0 \\ \text{Change} & -x & -x & +x & +x \\ \text{Equilibrium} & 0.275-x & 3.85-x & x & x \\ \end{array}\]

Step 3: Substitute values into the equilibrium expression

We are given the equilibrium constant for this reaction at \(55^{\circ} \mathrm{C}\), which is \(K = 6.68\). Substitute the values from the ICE table into the equilibrium expression and solve for x. $$6.68 = \frac{x \times x}{(0.275-x)(3.85-x)}$$

Step 4: Solve for x

Since the value of K is large, we may assume that the value of x is small compared to the initial concentrations of the reactants. So, we can simplify the expression as follows: $$6.68 \approx \frac{x^2}{(0.275)(3.85)}$$ Now, we solve for x: $$ x^2 = 6.68 \times (0.275)(3.85)$$ $$x \approx 1.321$$

Step 5: Find the amount of ester formed at equilibrium

Now that we have x, we can find the equilibrium concentration of ethyl acetate: $$[\mathrm{CH}_{3}\mathrm{COOCH}_{2}\mathrm{CH}_{3}]_\text{eq} = x = 1.321 \,\mathrm{M}$$ To find the amount of ethyl acetate formed in grams, we need to multiply the equilibrium concentration by the volume of the reaction and then by the molar mass of ethyl acetate. $$ \text{Amount of ethyl acetate (grams)} = [\mathrm{CH}_{3}\mathrm{COOCH}_{2}\mathrm{CH}_{3}]_\text{eq} \times V \times M $$ where \(V = 15.0 \,\mathrm{L}\) (given volume of the reaction) \(M = 88.11 \,\mathrm{g/mol}\) (molar mass of ethyl acetate) $$\text{Amount of ethyl acetate (grams)} = 1.321 \, \mathrm{M} \times 15.0 \, \mathrm{L} \times 88.11 \, \mathrm{g/mol}$$ $$\text{Amount of ethyl acetate (grams)} \approx 1746.40 \, \mathrm{g}$$ At equilibrium, approximately 1746.40 grams of ethyl acetate are formed.

Key Concepts

Organic Chemistry

Organic chemistry is a branch of chemistry that deals with the study of the structure, properties, composition, reactions, and preparation of carbon-containing compounds. These compounds typically contain carbon-carbon and carbon-hydrogen bonds, which provide them with a wide array of properties, making them the backbone of organic life. In the pharmaceutical industry, organic chemistry is crucial for drug development, enabling the synthesis of therapeutic compounds from simple molecules. The essence of organic chemistry lies in understanding various chemical reactions and mechanisms, such as nucleophilic substitutions, addition reactions, and eliminations.
One major aspect of organic chemistry is functional groups, which determine the characteristics and reactivity of organic molecules. In our exercise, the focus is on esters. Esters are organic compounds formed through esterification reactions and are often used in flavors, fragrances, and even pharmaceuticals due to their characteristic scents and favorable properties.

Esterification

Esterification is the chemical process by which an ester is formed from an organic acid and an alcohol. This reaction is typically catalyzed by an acid, often sulfuric acid ( H_2SO_4 ), which helps to speed up the process. In our previously mentioned reaction, acetic acid reacts with ethanol to form ethyl acetate and water as products.

• **Acetic Acid**: A weak organic acid, often used as a solvent or reagent in chemical reactions.
• **Ethanol**: An alcohol commonly used as a solvent and a starting material for other chemical syntheses.

This particular reaction involves the removal of a water molecule. The acid catalyst increases the reaction rate without being consumed in the process. Esterification is an equilibrium reaction, meaning it can proceed in both directions. Thus, efficient conversion to the desired ester may require removing water or using excess reactants to push the equilibrium toward the product side.
In real-world applications, esterifications are used to create materials ranging from perfumes to polymers, showcasing their broad utility in daily life.

Equilibrium Constant

The equilibrium constant (K) is a vital concept in equilibrium chemistry, representing the ratio of the concentrations of products to reactants at equilibrium for a reversible reaction. For the esterification of acetic acid and ethanol, the equilibrium constant expression is:
\[ K = \frac{[\text{Ethyl Acetate}][\text{Water}]}{[\text{Acetic Acid}][\text{Ethanol}]} \]
This equilibrium constant of *K = 6.68* indicates the favorability of forming products at equilibrium when the temperature is 55°C. A higher K value suggests that the reaction favors the formation of products, whereas a lower K value indicates a tendency to maintain more reactants at equilibrium.
Understanding the equilibrium constant allows chemists to manipulate conditions to maximize yield. For instance, based on K, one might adjust concentrations or temperature to drive the reaction towards product formation, crucial in industrial processes where efficiency and cost-effectiveness are essential.
Learning about K offers insights into the kinetics of reactions and informs predictions about the magnitude of product formation under various conditions, helping in designing more effective and sustainable reaction processes.