Question

The rate constants for the forward and backward reactions of hydrolysis of ester are \(1.1 \times 10^{-2}\) and \(1.5 \times\) \(10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) respectively. The equilibrium constant of the reaction, \(\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}+\mathrm{H}^{+} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOH}+\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) is: (a) \(6.53\) (b) \(7.34\) (c) \(7.75\) (d) \(8.33\)

Short answer

The equilibrium constant is 7.34 (option b).

Step-by-step solution

Understand the Reaction

The problem involves the equilibrium of the hydrolysis of an ester. Specifically, the reaction is given by \( \text{CH}_3 \text{COOC}_2 \text{H}_5 + \text{H}^+ \rightleftharpoons \text{CH}_3 \text{COOH} + \text{C}_2 \text{H}_5 \text{OH} \). We need the equilibrium constant for this reaction.

Formulate the Equilibrium Constant

The equilibrium constant \( K \) for the reaction can be expressed as the ratio between the forward rate constant \( k_f \) and the backward rate constant \( k_b \). Therefore, \( K = \frac{k_f}{k_b} \).

Insert Given Values

Insert the values of the forward rate constant \( k_f = 1.1 \times 10^{-2} \) and the backward rate constant \( k_b = 1.5 \times 10^{-3} \) into the equation for \( K \). This gives us \( K = \frac{1.1 \times 10^{-2}}{1.5 \times 10^{-3}} \).

Calculate Equilibrium Constant

Calculate \( K \) by dividing the two constants: \( K = \frac{1.1 \times 10^{-2}}{1.5 \times 10^{-3}} = 7.33 \).

Interpret the Result

The calculated equilibrium constant \( K \approx 7.33 \) is closest to the option \( 7.34 \). Thus, the answer is option (b) \( 7.34 \).

Key concepts

Hydrolysis of Ester

Hydrolysis of ester is a chemical reaction where an ester is broken down into its component alcohol and carboxylic acid. This reaction typically requires the presence of a catalyst, often an acid, to proceed effectively. In the given exercise, the hydrolysis involves the ester ethyl acetate, denoted as \( \text{CH}_3 \text{COOC}_2 \text{H}_5 \), which reacts with a proton \( \text{H}^+ \) to form acetic acid \( \text{CH}_3 \text{COOH} \) and ethanol \( \text{C}_2 \text{H}_5 \text{OH} \).
The reaction can proceed in both directions until a balance is reached and dynamic equilibrium is established. The concentration of reactants and products remains constant at this point, although they continuously interconvert. This equilibrium is crucial for calculating several constants related to the reaction, including the equilibrium constant, which tells us about the position of the equilibrium.

Forward Rate Constant

The forward rate constant, often denoted as \( k_f \), refers to the rate at which the forward reaction occurs, converting reactants into products. It is a fundamental component in kinetics that helps determine how fast a reaction proceeds towards equilibrium.
For the hydrolysis of an ester, the forward rate constant identifies the speed at which the ester and the proton form the carboxylic acid and alcohol. In this exercise, \( k_f \) is given as \( 1.1 \times 10^{-2} \ ext{mol} \ ext{L}^{-1} \ ext{s}^{-1} \). A higher forward rate constant would indicate a faster conversion of the ester to its hydrolysis products.
To understand the effect of \( k_f \), consider factors such as concentration of reactants and temperature, which can influence it. A higher concentration or temperature generally increases \( k_f \), pushing the reaction forward more rapidly.

Backward Rate Constant

The backward rate constant, denoted as \( k_b \), measures the rate at which the reverse reaction occurs, where products revert to reactants. In the hydrolysis of esters, \( k_b \) signifies the rate at which the formed carboxylic acid and alcohol combine to revert back to the ester.
For this particular problem, the backward rate constant is \( 1.5 \times 10^{-3} \ ext{mol} \ ext{L}^{-1} \ ext{s}^{-1} \). This value tells us how quickly the products can reform the original ester under equilibrium conditions. This constant is significant in determining the equilibrium constant \( K \). A higher \( k_b \) signifies a stronger tendency of products to form reactants, while a lower \( k_b \) suggests that the equilibrium favors the formation of products.

Equilibrium Calculation

Equilibrium calculation is an essential step to understand where a reaction stands when the rates of the forward and backward reactions are equal. The equilibrium constant \( K \) is a dimensionless number calculated from the ratio of the forward rate constant \( k_f \) to the backward rate constant \( k_b \).
From the step-by-step solution, \( K = \frac{1.1 \times 10^{-2}}{1.5 \times 10^{-3}} = 7.33 \). This value is very close to the choice \( 7.34 \) among the provided options.
A high equilibrium constant indicates that, at equilibrium, the reaction heavily favors the formation of products over reactants. The closer \( K \) is to 1, the more balanced the concentrations of reactants and products at equilibrium. Calculating \( K \) helps in predicting and quantifying the extent of a reaction, crucial for various applications in chemical engineering and synthesis.