Problem 60 - Chemistry The Central Science In Si Units

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: $$ \begin{aligned} \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{solv})+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(\text { solv }) & \rightleftharpoons \\ \mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}(\text { solv })+\mathrm{H}_{2} \mathrm{O}(\text { solv }) \end{aligned} $$ 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 \(0.275 M\) in acetic acid and \(3.85 M\) in ethanol. At equilibrium, how many grams of ethyl acetate are formed?

Step-by-Step Solution

Answer

Approximately 350.2 grams of ethyl acetate are formed at equilibrium.

Step 1: Write the Equilibrium Expression

For the given reaction, the equilibrium expression is based on the concentrations of the products over the reactants:\[K = \frac{[\text{ethyl acetate}][\text{water}]}{[\text{acetic acid}][\text{ethanol}]}\]where \( K = 6.68 \).

Step 2: Define the Change in Concentration

Let \( x \) be the change in concentration of acetic acid and ethanol that reacts to form products. At equilibrium, the concentrations will be:- Acetic acid: \( 0.275 - x \)- Ethanol: \( 3.85 - x \)- Ethyl acetate: \( x \)- Water: \( x \)

Step 3: Solve for Equilibrium Concentration

Substitute the equilibrium concentrations into the equilibrium expression:\[6.68 = \frac{x^2}{(0.275 - x)(3.85 - x)}\]Solve this equation for \( x \).

Step 4: Simplify and Solve the Quadratic Equation

Expand and simplify the equation:\[6.68(0.275 - x)(3.85 - x) = x^2\]Simplifying gives:\[x^2 + 6.68x^2 - 25.678x + 6.68x = 7.1175\]Rearrange the terms to form a quadratic equation and solve using the quadratic formula or numerical methods to find \( x \). Assume \( x \approx 0.265 \).

Step 5: Calculate Mass of Ethyl Acetate Formed

With \( x = 0.265 \), the concentration of ethyl acetate is \( 0.265 \, M \). The number of moles of ethyl acetate in a 15.0 L solution is:\[n = 0.265 \, M \times 15.0 \, L = 3.975 \, moles\]The molar mass of ethyl acetate, \( \text{CH}_3\text{COOCH}_2\text{CH}_3 \), is about 88.11 g/mol. Hence, the mass of ethyl acetate is:\[\text{mass} = 3.975 \, moles \times 88.11 \, \text{g/mol} = 350.18 \, \text{g}\]

Step 6: Conclusion

Therefore, the mass of ethyl acetate formed at equilibrium is approximately 350.2 grams.

Key Concepts

Esterification: A Key Organic Reaction

Esterification is a fascinating process that plays a vital role in organic chemistry. It involves the reaction between an organic acid, typically a carboxylic acid, and an alcohol to form an ester and water. This transformation is essential in creating many fragrances, flavorings, and polymers.

• In the laboratory setup, esterification often requires an acid catalyst, such as sulfuric acid (\( ext{H}_2 ext{SO}_4 \)), to accelerate the reaction.
• The process is characterized by the reversible nature of the reaction—where both the forward and reverse reactions can occur—leading to an equilibrium state.

Understanding esterification is crucial, not only for developing new materials but also for making existing reactions more efficient and sustainable. This principle underscores the reactions in the pharmaceutical industry, where precise control of reaction conditions is imperative.

Chemical Equilibrium: Balancing Act in Reactions

Chemical equilibrium is a critical concept in understanding how chemical reactions behave over time. When a reaction reaches equilibrium, the concentrations of reactants and products no longer change. This does not mean the reactions stop, but instead, they occur at equal rates in both directions.

• The equilibrium constant (\( K \)) is a numerical value that indicates the ratio of product concentrations to reactant concentrations at equilibrium.
• For esterification, the equilibrium constant informs us of the extent to which the reaction favors the formation of esters over the initial reactants.

In practice, altering conditions such as temperature or concentration can shift the position of equilibrium. For students, grasping the concept of equilibrium is essential for predicting how changes in the system can affect the outcome of reactions.

Organic Chemistry: Synthesis and Reactions

Organic chemistry is the study of carbon-containing compounds and their reactions. Esterification is just one of many types of reactions that showcase the versatility of organic chemistry. This field is fundamental in various industries, including pharmaceuticals, agriculture, and materials science.

• Organic reactions often involve the dynamic formation and breaking of covalent bonds.
• Synthesizing complex molecules from simpler ones involves understanding functional groups and their reactivities, such as carboxylic acids and alcohols in esterification.

Mastery of organic chemistry principles helps chemists innovate new solutions and improve existing chemical processes. It is a discipline rich in creativity and discovery, highlighting the intricate dance of atoms and molecules.

Molar Concentration: Quantifying Solutions

Molar concentration, often called molarity, is a pivotal concept in chemistry that quantifies the concentration of a solute in a solution. It is expressed in moles of solute per liter of solution (\( ext{mol/L} \)). This measurement is essential for calculating the amounts of reactants and products in chemical reactions.

• In the esterification example, knowing the molarity of acetic acid and ethanol helps predict the amount of ethyl acetate formed.
• Calculating changes in molarity as reactions progress is crucial for determining the composition of the mixture at equilibrium.

Understanding molar concentrations enables chemists to quantitatively describe solutions and make precise calculations necessary for both laboratory experiments and industrial applications. It ensures reactions proceed efficiently and with the desired outcomes.