Problem 148 - A Complete Resource Book in Chemistry for JEE Main

Question

For a reaction, \(\mathrm{A} \rightarrow \mathrm{B}+\mathrm{C}\), it was found that at the end of 10 min from the start the total optical, rotation of the system was \(50^{\circ} \mathrm{C}\) and when the reaction is complete it was \(100^{\circ}\). Assuming that only \(\mathrm{B}\) and \(\mathrm{C}\) are optically active and dextro rotator, the rate constant of this first order reaction would be (a) \(6.9 \mathrm{~min}^{-1}\) (b) \(0.069 \mathrm{~min}^{-1}\) (c) \(6.9 \times 10^{-2} \mathrm{~min}^{-1}\) (d) \(0.69 \mathrm{~min}^{-1}\)

Step-by-Step Solution

Answer

The rate constant is closest to option (b), 0.069 min⁻¹, or (c) 6.9 × 10⁻² min⁻¹.

Step 1: Understand the Problem

We are given a first-order reaction \(A \rightarrow B+C\), where only product \(B\) and \(C\) contribute to the optical activity. We need to find the rate constant \(k\) using the initial and final optical rotations.

Step 2: Identify Key Values

The optical rotation at 10 minutes is \(50^\circ\) and when the reaction is complete, the optical rotation is \(100^\circ\). This implies that when the reaction is complete, the optical rotation is twice that at 10 minutes.

Step 3: Use First-Order Kinetics Equation

The formula for first-order reactions relating initial concentration \([A]_0\), concentration at time \(t\) \([A]\), and the rate constant \(k\) is:\[[A] = [A]_0 \cdot e^{-kt}\]Since optical rotation correlates linearly with concentration, we use similar reasoning with rotations.

Step 4: Set Up the Relation for Optical Rotation

Using the given optical rotations as equivalents of concentrations, we have:- Initial equivalent concentration = 0 (since reaction has not started)- Rotation at \(t = 10\) min is half of the final, so \(\theta(10) = \frac{1}{2} \times 100^\circ = 50^\circ\)Using first-order reaction relations:\[50 = 100 (1 - e^{-kt})\]

Step 5: Solve for the Rate Constant

Rearrange the relation:\[1 - e^{-10k} = \frac{1}{2}\]Substitute and solve for \(k\):\[e^{-10k} = \frac{1}{2}\]Take the natural logarithm on both sides:\[-10k = \ln\left(\frac{1}{2}\right)\]Finally, solve for \(k\):\[k = -\frac{1}{10} \ln\left(\frac{1}{2}\right)\]

Step 6: Calculate and Finalize

Calculate the value:\[\ln\left(\frac{1}{2}\right) = -0.693\]Substitute:\[k = -\frac{-0.693}{10} = 0.0693 \text{ min}^{-1}\]Option (b) and (c) both match this value.

Key Concepts

First-order Reaction

In chemical kinetics, a **first-order reaction** is one where the rate depends linearly on the concentration of one reactant. This means the rate at which the reaction occurs is directly proportional to the concentration of the reacting species. For our exercise, the reaction is \( A \rightarrow B + C \), where only products \( B \) and \( C \) exhibit optical activity. Understanding how the concentration of reactants changes over time is crucial for determining the behavior of a first-order reaction.

The general formula used to describe the kinetics of a first-order reaction is:

• \([A] = [A]_0 \, e^{-kt}\)

Where:

• \([A]_0\) is the initial concentration of the reactant \( A \).
• \([A]\) is the concentration of \( A \) at time \( t \).
• \( k \) is the rate constant.
• \( t \) is the time elapsed.

Since this is a first-order reaction, the relationship shows how the concentration diminishes exponentially over time as the reactants are converted into products.

Rate Constant

The **rate constant**, \( k \), is a fundamental aspect of the rate equation in kinetics. It provides a measure of the speed of a reaction at a given temperature. For a first-order reaction, the units of the rate constant are \( \text{time}^{-1} \), which in this case is \( \text{min}^{-1} \).

To find the rate constant in our problem, we used the optical rotations as analogs to concentrations. The calculation involved equating the optical rotations at specific times to the concentrations, demonstrating how the reactions proceeds.

• We began with the optical rotation at 10 minutes being half the final rotation (indicating half the reaction is complete).
• The relationship used for the rate constant was derived from first-order reaction equations: \( 1 - e^{-10k} = \frac{1}{2} \).
• Solving this equation provided \( k = 0.069 \, \text{min}^{-1} \).

This interpretation allows us to see why the rate constant is fundamental in understanding the kinetics of any chemical reaction.

Optical Rotation

**Optical rotation** is the angle by which plane-polarized light is rotated when it passes through an optically active substance. In this exercise, we measured the change in optical rotation to determine the concentration of optically active products.

Both \( B \) and \( C \) from the reaction \( A \rightarrow B + C \) were assumed to be optically active and dextrorotatory (meaning they rotate light to the right). By knowing the initial and final optical rotations, you can infer the extent of reaction completion.

• The rotation observed after 10 minutes was \( 50^{\circ} \).
• The complete reaction had an optical rotation of \( 100^{\circ} \).
• The change in optical rotation formed the basis for calculating the rate constant, as this change was directly related to the extent of the reaction.

Optical rotation becomes a powerful tool to track reaction kinetics, especially in cases where concentration changes relate to changes in an optical property.

Reaction Kinetics

**Reaction kinetics** is the study of the rates of chemical processes and the underlying factors affecting these rates. For the given problem, reaction kinetics related the progression of the chemical reaction to observable changes in optical properties.

By using reaction kinetics principles, we were able to:

• Determine the influence of reactant concentration on reaction rate for a first-order reaction.
• Use the changes in optical rotation over time as a direct indicator of how much reaction has occurred.
• Derive the rate constant, \( k \), from measurable properties of the system (optical rotation and time).

Understanding these concepts supports comprehensive insights into how chemical processes unfold over time, equipping you to predict and control these reactions effectively.