Chapter 14: Problem 27
What is the half-life of a compound if 75 percent of a given sample of the compound decomposes in \(60 \mathrm{~min}\) ? Assume first-order kinetics.
Short Answer
Expert verified
The half-life of the compound is approximately 30 minutes.
Step by step solution
01
Understand First-order Kinetics
In first-order kinetics, the rate of reaction depends only on the concentration of one reactant. The rate law is expressed as \( \text{Rate} = k[A] \), where \( k \) is the rate constant and \( [A] \) is the concentration of the reactant.
02
Recognize the Percentage Decomposition
Given that 75% of the compound has decomposed, it means that 25% of the original concentration remains. This is used to determine the rate constant \( k \).
03
Use the Integrated Rate Law
For first-order reactions, the integrated rate law is \( \ln \left(\frac{[A]_t}{[A]_0}\right) = -kt \). Here, \( [A]_t \) is the final concentration (25% of \([A]_0\)), \( [A]_0 \) is the initial concentration, and \( t = 60 \) minutes.
04
Calculate the Rate Constant
Substitute into the equation: \( \ln \left( \frac{0.25[A]_0}{[A]_0} \right) = -k(60) \). Simplify to get \( \ln(0.25) = -k(60) \). Then solve for \( k \) to find \( k = -\frac{\ln(0.25)}{60} \approx 0.0231 \, ext{min}^{-1} \).
05
Find the Half-life Formula
For first-order reactions, the half-life \( t_{1/2} \) is given by \( t_{1/2} = \frac{0.693}{k} \).
06
Calculate the Half-life
Substitute \( k = 0.0231 \, ext{min}^{-1} \) into the half-life formula: \( t_{1/2} = \frac{0.693}{0.0231} \approx 30 \, \text{minutes} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-order kinetics
First-order kinetics is an essential concept in chemical kinetics, explaining how reaction rates depend on the concentration of a single reactant. In these reactions, the rate is directly proportional to the concentration of the reactant. Mathematically, this is shown as:
- Rate = k[A]
Rate constant
The rate constant, often symbolized as "k", is a pivotal parameter in the study of reaction kinetics. It provides the proportionality factor between the reaction rate and the concentration of reactants, especially in first-order reactions where the relationship is simple and direct. A higher value of "k" indicates a faster reaction. The units of the rate constant depend on the order of reaction.
- For first-order reactions, the unit is usually min-1 or s-1, as it involves time after simplifying the rate expression.
Integrated rate law
The integrated rate law for first-order reactions is a powerful tool enabling the transformation of rates into concentrations over time. This helps in calculating the remaining concentration of a reactant after a given period. For a first-order reaction, it takes the form:
- \( \ln \left(\frac{[A]_t}{[A]_0}\right) = -kt \)
Percentage decomposition
Percentage decomposition measures how much of the original material has reacted over a given period, usually shown as a percentage. It tells you how far a reaction has progressed and is particularly useful in kinetics to interpret the extent of reactions. For instance, saying that 75% has decomposed means only 25% of the original reactant remains. This information can then be used in the integrated rate law.In first-order kinetics:
- Calculate the remaining concentration, knowing that \(%\text{decomposed} = 100 - \%[A]_t\).
- \( \ln \left(\frac{[A]_t}{[A]_0}\right) = -kt \)