Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question
![**Problem Statement:**
After 40.0 minutes, 22.0% of a compound has decomposed. What is the half-life of this reaction assuming first-order kinetics?
**Calculation Area:**
\[ t_{1/2} = \boxed{\phantom{}} \; \text{min} \]
**Explanation:**
To solve this problem, we apply the concepts of first-order kinetics. In first-order reactions, the rate depends linearly on the concentration of one reactant. The half-life \( t_{1/2} \) is calculated by using the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
where \( k \) is the rate constant. To find \( k \), we use the first-order rate equation:
\[ \ln \left( \frac{[A]_0}{[A]} \right) = kt \]
Given:
- \( [A]_0 = \) initial concentration
- \( [A] = \) concentration after time \( t \)
After 40 minutes, 22.0% has decomposed, so 78.0% remains (100% - 22%).
\[ \ln \left( \frac{100}{78} \right) = k \times 40 \]
Calculate \( k \), and then plug the value into the half-life equation to find \( t_{1/2} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F04d91ec5-5dc8-475a-a219-706d13e893d1%2F9218ac4f-5762-4ca6-a850-150e47def998%2Fycpbczr.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
After 40.0 minutes, 22.0% of a compound has decomposed. What is the half-life of this reaction assuming first-order kinetics?
**Calculation Area:**
\[ t_{1/2} = \boxed{\phantom{}} \; \text{min} \]
**Explanation:**
To solve this problem, we apply the concepts of first-order kinetics. In first-order reactions, the rate depends linearly on the concentration of one reactant. The half-life \( t_{1/2} \) is calculated by using the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
where \( k \) is the rate constant. To find \( k \), we use the first-order rate equation:
\[ \ln \left( \frac{[A]_0}{[A]} \right) = kt \]
Given:
- \( [A]_0 = \) initial concentration
- \( [A] = \) concentration after time \( t \)
After 40 minutes, 22.0% has decomposed, so 78.0% remains (100% - 22%).
\[ \ln \left( \frac{100}{78} \right) = k \times 40 \]
Calculate \( k \), and then plug the value into the half-life equation to find \( t_{1/2} \).
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