Question

For the hypothetical second order reaction: \(\mathrm{A} \rightarrow\) Products, the general rate law is: Rate \(=\mathrm{k}[\mathrm{A}]^{2}\). How long is the third half life of the reaction if \([\mathrm{A}]_{0}\) is \(0.080 \mathrm{M}\) and the first half life is 22 minutes. a. \(48 \mathrm{~min}\) b. \(66 \mathrm{~min}\) c. \(88 \mathrm{~min}\) d. \(78 \mathrm{~min}\)

Short answer

The third half-life is 88 minutes (option c).

Step-by-step solution

Understanding Second-Order Reactions

For a second-order reaction, the half-life formula is given by \( t_{1/2} = \frac{1}{k[A]_0} \), where \( t_{1/2} \) is the half-life, \( k \) is the rate constant, and \( [A]_0 \) is the initial concentration of the reactant.

Calculating the Rate Constant

Given \( t_{1/2} = 22 \) minutes for the first half-life and \( [A]_0 = 0.080 \) M. The rate constant \( k \) can be determined from the first half-life:\[22 = \frac{1}{k \cdot 0.080}\]Solving for \( k \), we get \( k = \frac{1}{22 \times 0.080} \approx 0.568 \text{ M}^{-1}\text{min}^{-1} \).

Calculating the Concentration After the First and Second Half-Lives

For a second-order reaction, the concentration after each half-life is related to the initial concentration. After the first half-life (22 minutes), \[[A]_{1} = \frac{[A]_0}{2} = \frac{0.080}{2} = 0.040 \text{ M}\]After the second half-life (another 22 minutes), \[[A]_{2} = \frac{0.040}{2} = 0.020 \text{ M}\]

Determining the Third Half-Life

The third half-life decreases the concentration to half of \( [A]_2 \): \[[A]_{3} = \frac{0.020}{2} = 0.010 \text{ M}\]Using the half-life formula: \[t_{1/2} = \frac{1}{k[A]_2} = \frac{1}{0.568 \times 0.020} \approx 88 \text{ minutes}\]

Conclusion

By comparing the calculated third half-life to the given options, we find that the third half-life is closest to 88 minutes.

Key concepts

Rate Law

In the study of chemical kinetics, the rate law is an equation that relates the rate of a chemical reaction to the concentration of its reactants. For the second-order reaction provided (\( \mathrm{A} \rightarrow \) Products), the rate law is expressed as \( \text{Rate} = k[\mathrm{A}]^2 \). Here, \( k \) is the rate constant, and \( [\mathrm{A}] \) is the concentration of the reactant. This means that the rate of the reaction is directly proportional to the square of the concentration of \( \mathrm{A} \).

This dependency indicates that small changes in the concentration of \( \mathrm{A} \) will have a larger impact on the reaction rate, making second-order reactions particularly sensitive to concentration changes. Therefore, doubling the concentration of \( \mathrm{A} \) will quadruple the reaction rate.

• Rate law helps predict how fast a reaction will proceed.
• It provides insight into the reaction mechanism.

Understanding the rate law is key to controlling reactions in industrial and laboratory settings, aiding in the optimization of chemical processes.

Half-Life

Half-life in chemistry refers to the time required for half of the reactant to be consumed in a reaction. For a second-order reaction like this one, the half-life is not constant and depends on the initial concentration. The formula for the half-life of a second-order reaction is \( t_{1/2} = \frac{1}{k[A]_0} \), where \( t_{1/2} \) is the half-life, \( k \) is the rate constant, and \( [A]_0 \) is the initial concentration.

This dependency on concentration means that the half-life increases as the reaction proceeds because the concentration of the reactant decreases. For the scenario given, the first half-life is 22 minutes at \( 0.080 \text{ M} \). As the concentration becomes less, the time taken to halve it again, or subsequent half-lives, becomes longer.

• The first half-life was 22 minutes with \( [A]_0 = 0.080 \, \text{M} \).
• After the first half-life, the concentration decreases, making the second or third half-life longer than the first.

Thus, in second-order reactions, understanding how concentration affects half-life is important for predicting how long a reaction will take to progress.

Rate Constant

The rate constant, \( k \), is a proportionality constant in the rate law expression and it provides crucial information about the speed of the reaction. In second-order reactions, the units of \( k \) are \( \text{M}^{-1}\text{min}^{-1} \), reflecting that the reaction rate depends on the concentration squared.

To determine \( k \) in this example, we used the first half-life. With \( t_{1/2} = 22 \) minutes and \( [A]_0 = 0.080 \text{ M} \), we calculated \( k = \frac{1}{22 \times 0.080} \approx 0.568 \text{ M}^{-1}\text{min}^{-1} \). This value of \( k \) indicates how fast the reaction proceeds at the beginning when the concentration is highest.

• The rate constant \( k \) remains the same throughout the reaction, even though the half-life increases.
• The calculated \( k \) helps to find subsequent half-lives and other kinetics information.

It's crucial to understand that the rate constant gives insight into the reaction's inherent speed, providing a benchmark to compare different reactions or the same reaction under different conditions.