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 approximately 7.34.

Step 1: Understanding the Equilibrium Constant

The equilibrium constant \( K \) for a reaction at equilibrium can be determined using the relationship between the rate constants of the forward reaction (\( k_f \)) and the backward reaction (\( k_b \)). The formula is: \[ K = \frac{k_f}{k_b} \]

Step 2: Rate Constants Substitution

Substitute the given rate constants into the equilibrium constant formula. The forward rate constant \( k_f \) is \(1.1 \times 10^{-2}\) mol L\(^{-1}\) s\(^{-1}\) and the backward rate constant \( k_b \) is \(1.5 \times 10^{-3}\) mol L\(^{-1}\) s\(^{-1}\).

Step 3: Calculate the Equilibrium Constant

Using the relationship \( K = \frac{k_f}{k_b} \):\[ K = \frac{1.1 \times 10^{-2}}{1.5 \times 10^{-3}} \]Calculate \( K \) by performing the division.

Step 4: Solving the Calculation

Perform the division of the numbers:\[ K = \frac{1.1}{1.5} \times \frac{10^{-2}}{10^{-3}} = \frac{1.1}{1.5} \times 10^{1} \]Simplify further:\[ K = \frac{1.1}{1.5} \times 10 \approx 0.7333 \times 10 \approx 7.333 \]

Step 5: Match the Answer

Since \( K \approx 7.33 \), we can match this to the closest provided option, which is option (b) \( 7.34 \).

Key Concepts

Rate Constants

When discussing chemical reactions, rate constants are pivotal in understanding how quickly reactions proceed. The rate of a chemical reaction is often expressed as the change in concentration of reactants or products over time. The rate constant, denoted as either \( k_f \) for forward reactions or \( k_b \) for backward reactions, helps us understand this rate in quantitative terms.
The units of rate constants can vary depending on the order of the reaction. For a first-order reaction, the unit is s\(^{-1}\). However, for the reaction discussed here, which involves concentrations, the unit is typically mol L\(^{-1}\) s\(^{-1}\). Knowing the rate constant tells us how fast a reaction will occur at a given condition, such as concentration and temperature.

Forward Reaction

In the context of chemical reactions, the forward reaction refers to the process where reactants are converted to products. Considering the hydrolysis of ester, the forward reaction is the breaking of ester bonds in the presence of an acid to form carboxylic acid and an alcohol.
For the given problem, the forward rate constant \( k_f \) is \(1.1 \times 10^{-2}\) mol L\(^{-1}\) s\(^{-1}\). This numeric value indicates the speed at which the reactants, such as esters, will transform into products like carboxylic acids, under the influence of an acidic environment.
The forward reaction often depends on several factors, including the nature of the reactants, temperature, and presence of a catalyst such as an acid in this hydrolysis process, which speeds up the breakdown.

Backward Reaction

A backward reaction, also known as the reverse reaction, is the process where products convert back into reactants. This is crucial in reaching a state of equilibrium, where the rates of forward and backward reactions are equal, maintaining a constant concentration of reactants and products.
In our example of ester hydrolysis, the backward reaction would involve the recombination of carboxylic acid and alcohol to reform the ester and water. The rate constant for this backward reaction \( k_b \) is \(1.5 \times 10^{-3}\) mol L\(^{-1}\) s\(^{-1}\).
The value of \( k_b \) helps us understand how likely the molecules are to revert to their original species and how this impacts the position of equilibrium.

Hydrolysis of Ester

The hydrolysis of ester is a classic organic reaction where an ester molecule reacts with water to produce a carboxylic acid and an alcohol. In an acidic condition, this process is accelerated, making acids effective catalysts for the reaction.
This reaction is reversible, which means that the products can also react to form the original ester and water, contributing to the state of equilibrium. The balance between the forward reaction rate (ester turning into acid and alcohol) and backward reaction rate (acid and alcohol turning back into ester) establishes the equilibrium constant \( K \).
Understanding hydrolysis is important in many biological and chemical processes where esters are functional groups in fats and phospholipids, especially as metabolism and breakdown involve similar reactions.