Problem 98 - A Complete Resource Book in Chemistry for JEE Main

Question

The rate constant of first-order reaction is \(10^{-2} \mathrm{~min}^{-1}\). The half-life period of reaction is (a) \(693 \mathrm{~min}\) (b) \(69.3 \mathrm{~min}\) (c) \(6.93 \mathrm{~min}\) (d) \(0.693 \mathrm{~min}\)

Step-by-Step Solution

Answer

The half-life period is 69.3 minutes, option (b).

Step 1: Understand the Concept of Half-life for First-order Reactions

For a first-order reaction, the half-life is the time required for the concentration of a reactant to decrease to half its initial concentration. The formula for the half-life of a first-order reaction is given by \( t_{1/2} = \frac{0.693}{k} \), where \( t_{1/2} \) is the half-life and \( k \) is the rate constant.

Step 2: Insert Given Values into the Formula

From the problem, the rate constant \( k \) is given as \( 10^{-2} \mathrm{~min}^{-1} \). Substitute this value into the half-life formula: \( t_{1/2} = \frac{0.693}{10^{-2} \mathrm{~min}^{-1}} \).

Step 3: Perform the Calculation

Calculate the half-life using the provided values: \( t_{1/2} = \frac{0.693}{10^{-2}} = 69.3 \mathrm{~min} \). This calculation simplifies to multiplying 0.693 by 100, resulting in a half-life of 69.3 minutes.

Step 4: Conclusion with Correct Option

Based on the calculation, the correct answer for the half-life period of the reaction is \(69.3 \mathrm{~min}\). Therefore, the answer corresponds to option (b).

Key Concepts

Half-life Calculation

In first-order reactions, determining the half-life is a straightforward process, and it plays a crucial role in understanding how quickly a reactant is consumed. The half-life, denoted as \( t_{1/2} \), is the time required for the concentration of a reactant to reduce to half its initial amount. For first-order reactions, this value is constant, irrespective of the initial concentration. The formula used to calculate the half-life for a first-order reaction is:

• \( t_{1/2} = \frac{0.693}{k} \)

where \( k \) represents the rate constant of the reaction.
In our example, the rate constant \( k \) is given as \( 10^{-2} \mathrm{~min}^{-1} \). By plugging this value into the formula, we find:

• \( t_{1/2} = \frac{0.693}{10^{-2}} = 69.3 \mathrm{~min} \)

This calculation shows that the half-life is 69.3 minutes, meaning every 69.3 minutes, the reactant's concentration halves. Understanding this can help predict how long a reaction will take to proceed through various stages.

Rate Constant

The rate constant, \( k \), is a fundamental parameter in the study of reaction kinetics. It provides a measure of the speed of a reaction. In a first-order reaction, the rate constant is directly proportional to the rate of reaction, and it is expressed in units of \( \mathrm{time}^{-1} \) (e.g., \( \mathrm{min}^{-1} \), \( \mathrm{sec}^{-1} \)).
For the exercise in question, the rate constant is given as \( 10^{-2} \mathrm{~min}^{-1} \). This means that every unit of time (in this case, a minute), the reaction progresses at a consistent rate, consuming the reactant in proportion to its current concentration.
The rate constant can be affected by several factors including temperature, pressure, and the presence of catalysts. As the rate constant is crucial for calculating other parameters, like half-life, it's essential for students to understand how it impacts the reaction kinetics.

Reaction Kinetics

Reaction kinetics involves studying the rates of chemical reactions and the factors that influence them. This field provides insight into how quickly reactants are transformed into products. For first-order reactions, the rate of reaction is proportional to the concentration of one reactant. The general rate equation is expressed as:

• \( ext{rate} = k[A] \)

where \([A]\) is the concentration of the reactant.
In the context of our exercise, we are dealing with a first-order reaction where the chosen reactant is constantly being reduced to half its concentration over uniform time intervals due to a consistent rate constant, \( k \).
Factors like temperature and pressure play significant roles in influencing reaction kinetics. With an increase in temperature, for instance, the rate increases, leading to faster conversions.
Understanding the interplay between these elements helps in predicting how a reaction will proceed under different conditions. Such insights are valuable not only in academics but also in industrial applications where chemical reactions need to be optimized for efficiency. By grasping these kinetics principles, students can better predict outcomes and adjust conditions for desired results.