Problem 70 - A Complete Resource Book in Chemistry for JEE Main

Question

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

Step-by-Step Solution

Answer

The equilibrium constant is (b) 7.34.

Step 1: Understand Reaction and Rate Constants

We are given the rate constants for the forward and backward reactions of the hydrolysis of ester. The forward rate constant is given as \( k_f = 1.1 \times 10^{-2} \) mol \( L^{-1} s^{-1} \), and the backward rate constant is given as \( k_b = 1.5 \times 10^{-3} \) mol \( L^{-1} s^{-1} \).

Step 2: Recall the Formula for Equilibrium Constant

For a reversible reaction reaching equilibrium, the equilibrium constant \( K_{eq} \) is the ratio of the forward reaction rate constant \( k_f \) to the backward reaction rate constant \( k_b \). Thus, we have: \[ K_{eq} = \frac{k_f}{k_b} \]

Step 3: Calculate the Equilibrium Constant

Substitute the given values of \( k_f \) and \( k_b \) into the equation for \( K_{eq} \): \[ K_{eq} = \frac{1.1 \times 10^{-2}}{1.5 \times 10^{-3}} \]

Step 4: Solve the Division

Perform the division: \[ K_{eq} = \frac{1.1}{1.5} \times 10^{-2 + 3} = \frac{1.1}{1.5} \times 10^{1} \]. Calculate \( \frac{1.1}{1.5} \approx 0.733 \). Thus, \( K_{eq} \approx 0.733 \times 10^{1} = 7.33 \).

Step 5: Select Closest Answer

Compare the calculated equilibrium constant \( K_{eq} = 7.33 \) with the options given. The closest value from the options is (b) 7.34.

Key Concepts

Equilibrium Constant

The equilibrium constant, often represented as \( K_{eq} \), is a crucial parameter in understanding chemical reactions. It provides insight into the balance between the reactants and products in a reversible chemical reaction at equilibrium. This constant is determined by the ratio of the concentrations of products to reactants, each raised to the power of their coefficients in the balanced chemical equation.

In the context of reversible reactions, the equilibrium constant remains fixed at a given temperature. It reflects the extent of the reaction; a large \( K_{eq} \) suggests that, at equilibrium, the reaction mixture contains more products than reactants. Conversely, a small \( K_{eq} \) indicates a predominance of reactants. For instance, in our exercise, the equilibrium constant is calculated as approximately 7.34, which means the products are favored slightly over the reactants at equilibrium.

• Useful for predicting the direction of the reaction.
• Helps in calculating concentrations at equilibrium.

Understanding \( K_{eq} \) enables chemists to anticipate how altering conditions like temperature or pressure could shift the equilibrium of a reaction, allowing better control over chemical processes.

Rate Constants

Rate constants are fundamental in describing the speed at which reactions proceed. They are denoted as \( k_f \) for the forward reaction and \( k_b \) for the backward reaction. Unlike the equilibrium constant, rate constants can vary with changes in temperature and are specific to the conditions of each reaction.

In our exercise, the forward rate constant \( k_f = 1.1 \times 10^{-2} \) mol \( L^{-1} s^{-1} \) and the backward rate constant \( k_b = 1.5 \times 10^{-3} \) mol \( L^{-1} s^{-1} \) tell us how quickly reactants are converted to products and vice versa. By comparing these constants, we can understand the balance of forces driving the reaction in either direction. The formula \( K_{eq} = \frac{k_f}{k_b} \) showcases the relationship between rate constants and equilibrium position.

• Indicates the reaction efficiency under certain conditions.
• Utilized in modeling the kinetics of reactions.

Understanding these constants helps in adjusting reaction conditions to favor faster reaction completion towards either side of the reaction.

Reversible Reactions

Reversible reactions are dynamic processes where products can re-form reactants. This characteristic leads reactions to reach an equilibrium state, where the rate of the forward reaction equals the rate of the reverse reaction.

The exercise on ester hydrolysis exemplifies a reversible chemical reaction. In this two-directional process, ester reacts to form acetic acid and ethanol, while acetic acid and ethanol can convert back to ester. The equilibrium constant, as calculated, informs us about the favorability of product formation versus reactant re-formation.

• Dynamic equilibrium signifies a balance with no net change in the concentration of reactants and products.
• Reactions are influenced by changes in concentration, pressure, and temperature following Le Chatelier’s principle.

Such a concept of reversibility is fundamental across chemistry, illustrating the flexible nature of chemical processes, fostering a deeper comprehension of chemical dynamics and predictability of reaction outcomes.