Problem 21 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

For the following esterification reaction \\[ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\left(\mathrm{p}+\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(\mathrm{D}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}\left(\mathrm{f}+\mathrm{H}_{2} \mathrm{O}(0)\right.\right. \\] the equilibrium constant at \(298 \mathrm{K}\) is \(3.8 .\) A mixture containing \(0.5 \mathrm{moldm}^{-3}\) each of ethanol and ethanoic acid was reacted in a sealed flask at \(298 \mathrm{K}\). After a certain time, the concentrations of each had changed to 0.39 moldm \(^{-3}\). (Section 15.3 ) (a) Had the reaction reached equilibrium? (b) If not, what would the concentration of \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{2} \mathrm{H}_{5}(\mathrm{aq})\) be at equilibrium? (c) In practice, the reaction is carried out so as to remove the water as it forms. Explain why this is done.

Step-by-Step Solution

Answer

(a) The reaction has not reached equilibrium. (b) At equilibrium, [CH₃CO₂C₂H₅] = 0.34 moldm⁻³. (c) Removing water shifts equilibrium toward ester production.

Step 1: Write the Equilibrium Expression

For the esterification reaction, the equilibrium expression in terms of the equilibrium constant \( K_c \) is given as: \[ K_c = \frac{[\text{CH}_3\text{CO}_2\text{C}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{C}_2\text{H}_5\text{OH}][\text{CH}_3\text{CO}_2\text{H}]} \]Given \( K_c = 3.8 \) at 298 K.

Step 2: Check Initial Reaction Conditions

Initially, the mixture contains 0.5 moldm\(^{-3}\) of ethanol and 0.5 moldm\(^{-3}\) of ethanoic acid. After a certain time, the concentrations are both 0.39 moldm\(^{-3}\). This indicates that each substance has changed by 0.11 moldm\(^{-3}\).

Step 3: Determine the Reaction Quotient (Q)

The reaction quotient, \( Q_c \), is evaluated using the changed concentrations. The concentration of ester formed is 0.11 moldm\(^{-3}\):\[ Q_c = \frac{0.11^2}{0.39^2} \]Calculate \( Q_c \) to compare with \( K_c \).

Step 4: Calculate \( Q_c \) and Compare with \( K_c \)

\[ Q_c = \frac{0.11^2}{0.39^2} = \frac{0.0121}{0.1521} \approx 0.0795 \]Since \( Q_c (0.0795) < K_c (3.8) \), the reaction has not reached equilibrium.

Step 6: Set Up an ICE Table

We need to find equilibrium concentrations when the reaction reaches equilibrium.Define the changes in concentrations:- Ethanol and ethanoic acid: \([0.5 - x]\), each- Ester: \([0 + x]\)- Water: \([0 + x]\)Using these, equilibrium concentrations are substituted in the \( K_c \) expression.

Step 7: Solve for Equilibrium Position

Substitute the expressions from the ICE table back into the equilibrium equation:\[ K_c = \frac{x^2}{(0.5-x)^2} = 3.8 \]Solve for \( x \) to find \( x = 0.34 \) moldm\(^{-3}\).

Step 8: Calculating Equilibrium Concentrations

From Step 6's substitution, we have:- The equilibrium concentration of ester (\( \text{CH}_3\text{CO}_2\text{C}_2\text{H}_5 \)) will be \([0.34]\)moldm\(^{-3}\).

Step 9: Discussion of Water Removal

Removing water as it forms drives the equilibrium towards products by Le Chatelier's Principle, increasing the yield of ester. This occurs because water is a product of the reaction.

Key Concepts

Esterification Reaction

Esterification is a chemical reaction that involves the formation of an ester from an acid and an alcohol. In our example, the reaction involves ethanol and ethanoic acid forming ethyl acetate and water. The general form of the esterification reaction is:

• Alcohol + Acid ⇌ Ester + Water

The reaction is reversible, meaning it can proceed in both directions. In practice, esterification often requires a catalyst, such as an acid, to increase the reaction rate. The production of esters is a key industrial process, vital for creating materials like soaps and synthetic flavors.

Equilibrium Constant

The equilibrium constant, denoted as \( K_c \), provides a quantitative measure of the position of equilibrium in a chemical reaction at a given temperature. For the esterification reaction in question, the equilibrium constant is expressed as:\[ K_c = \frac{[\text{CH}_3\text{CO}_2\text{C}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{C}_2\text{H}_5\text{OH}][\text{CH}_3\text{CO}_2\text{H}]} \]At equilibrium, this value is 3.8 at 298 K. A high \( K_c \) value suggests that the products are favored at equilibrium, while a low \( K_c \) value indicates that the reactants are favored. In this example, since \( K_c \) is greater than 1, products are relatively favored, indicating that the formation of ester and water is favored at equilibrium.

Reaction Quotient

The reaction quotient, \( Q_c \), is an expression that is similar to the equilibrium constant but is used to determine the reaction's direction at any point in time before equilibrium is reached. It is defined similarly:\[ Q_c = \frac{[\text{CH}_3\text{CO}_2\text{C}_2\text{H}_5][\text{H}_2\text{O}]}{[\text{C}_2\text{H}_5\text{OH}][\text{CH}_3\text{CO}_2\text{H}]} \]In our case, \( Q_c \) was calculated to be approximately 0.0795, whereas \( K_c \) is 3.8. Since \( Q_c < K_c \), this indicates that more products will form as the reaction moves towards equilibrium. Therefore, the reaction will proceed forward in order to increase the concentrations of the products until equilibrium is reached.

Le Chatelier's Principle

Le Chatelier's Principle helps to predict how a change in conditions can affect the position of equilibrium in a chemical reaction. It states that if a system at equilibrium experiences a change in concentration, temperature, or pressure, the equilibrium will shift to counteract that change.
In the esterification reaction, removing water from the system as it forms is a practical application of this principle. By decreasing the concentration of one of the products, the system will shift the equilibrium toward producing more ester and water, thus increasing the yield of ester. This approach is often used in industrial processes to maximize the production of desired products.