Problem 12 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The reaction \(2 \mathrm{X} \square \mathrm{B}\) is a zeroth order reaction. If the initial concentration of \(X\) is \(0.2 \mathrm{M}\), the half-life is \(6 \mathrm{~h}\). When the initial concentration of \(X\) is \(0.5 \mathrm{M}\), the time required to reach its final concentration of \(0.2 \mathrm{M}\) will be : [Main Jan. 11, 2019 (II)](a) \(9.0 \mathrm{~h}\) (b) \(12.0 \mathrm{~h}\) (c) \(18.0 \mathrm{~h}\) (d) \(7.2 \mathrm{~h}\)

Step-by-Step Solution

Answer

(c) 18.0 h

Step 1: Understand Zeroth Order Reactions

In a zeroth order reaction, the rate of reaction is constant and does not depend on the concentration of the reactants. The rate law for a zeroth order reaction is given by \( \text{rate} = k \).

Step 2: Use Half-Life Formula for Zeroth Order Reaction

For zeroth order reactions, the half-life \( t_{1/2} \) is given by the formula \( t_{1/2} = \frac{[X]_0}{2k} \), where \([X]_0\) is the initial concentration and \(k\) is the rate constant. Here, \( t_{1/2} = 6 \text{ h} \) and \([X]_0 = 0.2 \text{ M} \).

Step 3: Solve for Rate Constant k

Using the half-life formula, substitute the known values: \( 6 = \frac{0.2}{2k} \). Solving for \(k\), we get \( k = \frac{0.2}{12} = 0.01667 \text{ M/h} \).

Step 4: Calculate Time for Concentration Change

The time \(t\) to decrease from an initial concentration \([X]_0 = 0.5 \text{ M}\) to a final concentration \([X] = 0.2 \text{ M}\) is given by the zeroth order relationship: \( [X]_0 - [X] = kt \). Substitute \([X]_0 = 0.5 \text{ M}\), \([X] = 0.2 \text{ M}\), and \(k = 0.01667 \text{ M/h}\): \[ 0.5 - 0.2 = 0.01667 \cdot t \].

Step 5: Solve for Time t

From the equation \(0.3 = 0.01667 \cdot t\), solve for \(t\): \( t = \frac{0.3}{0.01667} \approx 18 \text{ h} \).

Key Concepts

Rate Constant

The rate constant, often symbolized by the letter \(k\), is a crucial factor in the study of chemical kinetics. For a zeroth order reaction, the rate of reaction is constant and independent of the concentration of the reactants. This means that the rate constant \(k\) directly equals the rate of reaction.

For our zeroth order reaction, the formula is straightforward:

• The rate law is given by \( ext{rate} = k \).

In our problem, by using the half-life formula, we've determined that the rate constant \(k\) is \(0.01667 \, \mathrm{M/h}\). This insight helps calculate how fast the reactants turn into products over time, without depending on how much reactant is present.

This constancy simplifies calculations, making zeroth order reactions easier to analyze compared to other reaction orders.

Half-Life

In chemical kinetics, the half-life is the time required for the concentration of a reactant to reach half of its initial value. For zeroth order reactions, half-life can often seem confusing because it varies based on the initial concentration.

The half-life for a zeroth order reaction is calculated using the formula:

• \( t_{1/2} = \frac{[X]_0}{2k} \)
• \([X]_0\) is the initial concentration
• \(k\) is the rate constant

In the given problem, the initial concentration is \(0.2 \, \mathrm{M}\) with a half-life of \(6 \, \mathrm{h}\), allowing us to calculate \(k\).

It's crucial to understand that unlike first-order reactions (where half-life is constant), in zeroth order reactions, as the initial concentration increases, the half-life increases. This is because half-life is directly proportional to the initial concentration due to the rate's dependency only on \(k\).

Reaction Rate

The reaction rate for zeroth order reactions is uniquely independent of the concentration of the reactants.

This means that the velocity at which products are formed is constant throughout the reaction, regardless of how much of the reactant is left. For the reaction in question, we primarily used reaction rate concepts to calculate the time it takes for the concentration of \(X\) to decrease from \(0.5 \, \mathrm{M}\) to \(0.2 \, \mathrm{M}\).

• The relationship used was: \( [X]_0 - [X] = kt \)
• \( [X]_0 \) is the starting concentration \(0.5 \, \mathrm{M}\)
• \([X]\) is the final concentration \(0.2 \, \mathrm{M}\)

Using the reaction rate and constant \(k\), we computed the time to be \(18 \, \mathrm{h}\). Understanding the influence of reaction rate in such processes is integral in predicting how quickly a reaction proceeds at any given time.