Problem 65 - A Complete Resource Book in Chemistry for JEE Main

Question

Under the same reactions conditions, initial concentration of \(1.386 \mathrm{~mol} \mathrm{dm}^{-3}\) of a substance becomes half in \(40 \mathrm{sec}\) and 20 sec through \(1^{\text {st }}\) order and zero order kinetics respectively. Ratio \(\left[\frac{\mathrm{K}_{1}}{\mathrm{~K}_{0}}\right]\) of the rate constants for first order \(\left(\mathrm{k}_{1}\right)\) and zero order \(\left(\mathrm{k}_{0}\right)\) of the reactions is (a) \(0.5 \mathrm{~mol}^{-1} \mathrm{dm}^{-3}\) (b) \(1 \mathrm{~mol} \mathrm{dm}^{-3}\) (c) \(1.5 \mathrm{~mol} \mathrm{dm}^{-3}\) (d) \(2 \mathrm{~mol}^{-1} \mathrm{dm}^{3}\)

Step-by-Step Solution

Answer

The ratio \( \frac{k_1}{k_0} \) is 0.5 \( \mathrm{~mol}^{-1} \mathrm{dm}^{-3} \).

Step 1: Understand Kinetics Premise

First-order reactions have a rate that depends on the concentration of one reactant. With first-order kinetics, half-life \( t_{1/2} \) can be calculated using the formula \( t_{1/2} = \frac{0.693}{k_1} \). Zero-order reactions have a rate constant \( k_0 \) independent of reactant concentration, and the half-life is given by \( t_{1/2} = \frac{[A]_0}{2k_0} \), where \( [A]_0 \) is the initial concentration.

Step 2: Calculate First-Order Rate Constant \( k_1 \)

Given that the substance's half-life is 40 seconds for first-order kinetics, we use the formula for half-life of a first-order reaction: \( 40 = \frac{0.693}{k_1} \). Solving for \( k_1 \), we get \( k_1 = \frac{0.693}{40} \approx 0.017325 \text{ sec}^{-1} \).

Step 3: Calculate Zero-Order Rate Constant \( k_0 \)

For zero-order kinetics, with a half-life of 20 seconds and initial concentration \( [A]_0 = 1.386 \mathrm{~mol} \, ext{dm}^{-3} \), we use the formula: \( 20 = \frac{1.386}{2k_0} \). Solving for \( k_0 \), we find \( k_0 = \frac{1.386}{2 \times 20} = \frac{1.386}{40} \approx 0.03465 \mathrm{~mol} \, ext{dm}^{-3} \, ext{sec}^{-1} \).

Step 4: Calculate the Ratio of Rate Constants \( \frac{k_1}{k_0} \)

Now, to find the ratio \( \frac{k_1}{k_0} \), we divide \( k_1 \) by \( k_0 \): \( \frac{0.017325}{0.03465} \approx 0.5 \).

Step 5: Select the Correct Answer

Comparing our calculated ratio to the options provided, the ratio \( \frac{k_1}{k_0} \approx 0.5 \) corresponds to choice (a), which is \( 0.5 \mathrm{~mol}^{-1} \mathrm{dm}^{-3} \).

Key Concepts

First-Order Reaction

In chemical kinetics, a first-order reaction depends solely on the concentration of one reactant. Over time, this dependence dictates the speed at which the reaction proceeds. It is crucial for students to recognize that the rate of a first-order reaction is directly proportional to the concentration of the reacting substance. The formula for the rate of a first-order reaction can be written as: \( \text{Rate} = k_1 [A] \), where \( k_1 \) is the rate constant, and \( [A] \) is the concentration of the reactant.
A unique characteristic of first-order reactions is their half-life, which is constant and independent of the initial concentration. This means that the time required for half of the reactant to convert into product remains the same throughout the reaction. The half-life formula for a first-order reaction is \( t_{1/2} = \frac{0.693}{k_1} \). This concept helps in simplifying calculations and allows us to predict how long it will take for a reaction to proceed to a certain extent.
When working with these reactions in exercises, always remember that as the concentration drops, so does the rate of the reaction—but the half-life remains unaffected. This feature distinguishes first-order reactions from others in chemistry and is vital for understanding reaction dynamics.

Zero-Order Reaction

Zero-order reactions present a unique scenario in chemical kinetics. Unlike other reaction orders, the rate of a zero-order reaction is constant and independent of the concentration of the reactant. This means that, irrespective of how much reactant is present, the reaction process proceeds at a constant rate, determined solely by the rate constant \( k_0 \).
The rate law for zero-order reactions is expressed as \( \text{Rate} = k_0 \). Since the reaction rate is not influenced by the concentration of the reactant, it simplifies calculations significantly when you know the rate constant. However, the half-life of a zero-order reaction is dependent on the initial concentration of the reactant, unlike first-order reactions. The half-life formula is \( t_{1/2} = \frac{[A]_0}{2k_0} \), where \( [A]_0 \) is the starting concentration.
In practical terms, zero-order reactions often occur when a catalyst is saturated, or in photochemical reactions where light intensity rather than chemical concentration dictates the rate. Providing insights into these processes shows the intriguing variety of reaction dynamics students might encounter in their studies.

Rate Constant

The rate constant is a pivotal element in understanding chemical kinetics for both first and zero-order reactions. It is denoted by \( k \) and serves as a proportionality factor that relates the rate of a reaction to the concentrations of the reactants. Each reaction order has its rate constant with unique properties and units.
For a first-order reaction, the rate constant has units of \( \text{sec}^{-1} \). It is directly involved in calculating the half-life of the reaction through the equation \( t_{1/2} = \frac{0.693}{k_1} \). This makes predicting the reaction's progression straightforward, regardless of initial concentration.
In the case of zero-order reactions, the units for the rate constant are \( \text{mol} \, \text{dm}^{-3} \, \text{sec}^{-1} \). It is important here as it allows for the calculation of half-life with the formula \( t_{1/2} = \frac{[A]_0}{2k_0} \), showing a direct connection to the initial concentration. Ultimately, mastering the concept of the rate constant is crucial for students as it is the key to unlocking a deeper understanding of the speed at which reactions occur and how they can be quantitatively analyzed in various conditions.