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 \(\mathrm{L}^{-1} \mathrm{~s}^{-1}\) 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) \(8.33\)

Step-by-Step Solution

Answer

The equilibrium constant is (b) 7.34.

Step 1: Identify the Given Data

We are given the forward reaction rate constant, \( k_f = 1.1 \times 10^{-2} \) mol L\(^{-1}\) s\(^{-1}\), and the backward reaction rate constant, \( k_b = 1.5 \times 10^{-3} \) mol L\(^{-1}\) s\(^{-1}\). The task is to find the equilibrium constant \( K \) for the reaction.

Step 2: Formula for Equilibrium Constant

Recognize that the equilibrium constant \( K \) for a reversible reaction can be found using the formula \( K = \frac{k_f}{k_b} \), where \( k_f \) is the rate constant for the forward reaction and \( k_b \) is the rate constant for the backward reaction.

Step 3: Substitute the Values into the Formula

Substitute the given rate constants into the equilibrium constant formula: \( K = \frac{1.1 \times 10^{-2}}{1.5 \times 10^{-3}} \).

Step 4: Calculate the Equilibrium Constant

Perform the division to calculate \( K \): \[ K = \frac{1.1 \times 10^{-2}}{1.5 \times 10^{-3}} = \frac{1.1}{1.5} \times 10^{1} \approx 7.33 \].

Step 5: Select the Closest Answer

Compare the calculated \( K = 7.33 \) with the provided options. The closest answer to 7.33 is (b) 7.34.

Key Concepts

Rate Constants

Rate constants are pivotal in understanding how fast or slow a reaction proceeds. Essentially, the rate constant (\(k\)) is a proportionate value that relates the reaction rate to the concentrations of the reactants. It provides a numeric insight into the speed and dynamics of chemical reactions. For instance:

• Forward Reaction Rate Constant (\(k_f\)): This refers to the speed at which the reactants are converted into products. In the given exercise, \(k_f = 1.1 \times 10^{-2}\) mol L\(^{-1}\) s\(^{-1}\).


• Backward Reaction Rate Constant (\(k_b\)): Conversely, this rate constant measures the speed at which products revert to reactants. The exercise specifies \(k_b = 1.5 \times 10^{-3}\) mol L\(^{-1}\) s\(^{-1}\).

These values shed light on the inherent speed of forward and backward processes, facilitating predictive modeling of the reaction kinetics.

Hydrolysis of Ester

The hydrolysis of ester is a common chemical reaction where an ester reacts with water, leading to the formation of an alcohol and an acid. This type of reaction is prevalent in both industrial applications and biological systems. In our given scenario, the ester undergoing hydrolysis is \(\mathrm{CH}_{3}\mathrm{COOC}_{2}\mathrm{H}_{5}\), and it involves the following parts:

• Ester: The starting material. Here, it is ethyl acetate, represented as \(\mathrm{CH}_{3}\mathrm{COOC}_{2}\mathrm{H}_{5}\).


• Acid Formation: The hydrolysis leads to the formation of acetic acid \(\mathrm{CH}_{3}\mathrm{COOH}\).


• Alcohol Formation: The second product is ethanol \(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}\).

Hydrolysis of esters is a reversible process, impacted by factors like temperature, catalysts, and concentration of reactants.

Reversible Reactions

Reversible reactions are chemical processes that can proceed in both directions—forward and backward. In the context of equilibrium, they are characterized by the formation of reactants and products proceeding simultaneously. Consider the given reaction:\[\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5} + \mathrm{H}^{+} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOH} + \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\]In this scenario, hydrolysis and its reverse process constantly occur, and the direction depends on conditions such as

• Concentration of reactants and products
• Temperature and pressure
• Presence of catalysts that can shift equilibrium forward or reverse

The point of balance, or equilibrium, is achieved when the rates of the forward and reverse reactions are equal, resulting in stable concentrations of products and reactants.

Reaction Kinetics

Reaction kinetics delves into the rate and mechanism of chemical reactions. It encompasses everything from the factors influencing reaction rates to the steps that make up a reaction pathway. Understanding kinetics is crucial for controlling reactions in laboratories and industries. Key components include:

• Rate Laws: These are mathematical relationships that describe how the rate depends on reactant concentrations. Rate laws help in deriving the rate constant ( \(k\)) for reactions.


• Order of Reaction: It defines the relationship between reactant concentration and reaction rate. The total order is the sum of the powers of concentration terms in the rate law.


• Temperature Effect: Generally, an increase in temperature raises reaction rates, as particles collide with greater energy.


• Catalysts: Substances that increase reaction rates without being consumed, often providing an alternative pathway with lower activation energy.

Through studying reaction kinetics, chemists and engineers can not only predict how a reaction progresses but also optimize conditions for maximum efficiency.