Problem 134 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following statements are correct about half-life period? (1) time required for \(99.9 \%\) completion of a reaction is 100 times the half-life period (2) time required for \(75 \%\) completion of a first-order reaction is double the half-life of the reaction (3) average life \(=1.44\) times the half-life for firstorder reaction

Step-by-Step Solution

Answer

Statements (2) and (3) are correct.

Step 1: Understand Half-life for a First-Order Reaction

The half-life ( 0) of a first-order reaction is the time required for the concentration of a reactant to decrease to half its initial value. For a first-order reaction, the half-life is independent of the initial concentration and is given by the formula: \[ t_{1/2} = \frac{0.693}{k} \] where \( k \) is the rate constant of the reaction.

Step 2: Analyze Statement (1)

We need to determine if the time required for 99.9% completion of a reaction is 100 times its half-life. For a first-order reaction, the relationship for completion is given by the formula: \[ t = \frac{2.303}{k} \log \left(\frac{[A]_0}{[A]}\right) \] For 99.9% completion, we have \([A] = 0.1\% \times [A]_0\). Substituting this gives a time : \[ t = \frac{2.303}{k} \log \left(\frac{1}{0.001}\right) = \frac{6.909}{k} = 10 \times \frac{0.693}{k} = 10 \times t_{1/2} \] Thus, statement (1) is false as the time is 10 times, not 100 times, the half-life.

Step 3: Analyze Statement (2)

For a 75% completion of a first-order reaction, \([A] = 25\% \times [A]_0\). Plug into the formula: \[ t = \frac{2.303}{k} \log \left(\frac{1}{0.25}\right) = \frac{2.303}{k} \log (4) = \frac{2.303 \times 0.602}{k} \approx \frac{1.386}{k} \] The half-life is \( \frac{0.693}{k} \), so the time required is indeed 2\( t_{1/2} \). Hence, statement (2) is true.

Step 4: Analyze Statement (3)

The average life (mean lifetime) for a first-order reaction is given by \( \frac{1}{k} \). The relationship with the half-life is: \[ \text{Average life} = \frac{1}{k} = \frac{t_{1/2}}{0.693} \approx 1.44 \times t_{1/2} \] Thus, statement (3) is correct.

Key Concepts

First-order reactions

In chemistry, first-order reactions are a class of reactions where the rate at which the reaction occurs is directly proportional to the concentration of one of the reactants. This means that as the concentration of the reactant decreases, the reaction rate also decreases. A defining characteristic is that the half-life of the reaction, which is the time it takes for half of the reactant to be converted into product, is constant over time and does not depend on the initial concentration of the reactant.

The formula used to describe 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 of the reaction. Because the half-life is independent of the initial concentration, first-order reactions are commonly observed in chemical processes.

• Characteristic constant half-life, regardless of reactant concentration
• Rate proportional to concentration of a single reactant

Rate constant

The rate constant, denoted as \( k \), is a crucial factor in the study of reaction kinetics. It is a measure of the speed of a chemical reaction. Specifically, in first-order reactions, the rate constant carries specific units of \( \text{s}^{-1} \), reflecting its time dependence.

Understanding the rate constant helps predict how quickly a reaction will proceed. For first-order reactions, it is used in equations that relate to both half-life and reaction time needed for a certain percentage of completion. The rate law for a first-order reaction can be expressed as:\[ R = k[A] \]where \( R \) is the reaction rate and \( [A] \) is the concentration of the reactant.

• Indicates reaction speed
• Specific to each reaction and determined experimentally

Reaction completion percentage

The percentage of reaction completion tells you how far along a reaction is, giving an idea of how much reactant has been converted to product. It's represented by comparing the remaining reactant concentration to the initial amount. For first-order reactions, there's a neat way to calculate the time required for any level of completion using integrated rate laws.

The general formula to determine the time \( t \) for a specific percentage completion \( x \% \) of a first-order reaction is:\[ t = \frac{2.303}{k} \log \left(\frac{[A]_0}{[A]}\right) \]To illustrate, for a reaction approaching 99.9% completion, the formula determines around 10 times the reaction's half-life is needed. This is calculated as:\[t \approx 10 \times t_{1/2}\]

• Reflects how much reactant remains or is consumed
• Calculated using the rate constant and logarithmic relations