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The rate constant of a first-order process that has a half-life of 3.50 min is Select one: O A. 0.693 В. 3.30 x10-3 C. 1.65 x 10-2 O D. 0.198 O E. 1.98

The rate constant of a first-order process that has a half-life of 3.50 min is Select one: O A. 0.693 В. 3.30 x10-3 C. 1.65 x 10-2 O D. 0.198 O E. 1.98

Chemistry
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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### Problem Statement: The rate constant of a first-order process that has a half-life of 3.50 min is ______ s⁻¹. ### Choose One: - **A.** \( 0.693 \) - **B.** \( 3.30 \times 10^{-3} \) - **C.** \( 1.65 \times 10^{-2} \) - **D.** \( 0.198 \) - **E.** \( 1.98 \) --- ### Explanation: For a first-order reaction, the rate constant \( k \) can be determined from the half-life \( t_{1/2} \) using the formula: \[ t_{1/2} = \frac{0.693}{k} \] Given \( t_{1/2} = 3.50 \) minutes, we first need to convert this into seconds: \[ 3.50 \text{ min} \times 60 \text{ s/min} = 210 \text{ s} \] Now, substituting the given half-life into the formula, we solve for \( k \): \[ 210 = \frac{0.693}{k} \] Rearranging for \( k \): \[ k = \frac{0.693}{210} \approx 3.30 \times 10^{-3} \text{ s}^{-1} \] The correct answer is **B**.
Transcribed Image Text:### Problem Statement: The rate constant of a first-order process that has a half-life of 3.50 min is ______ s⁻¹. ### Choose One: - **A.** \( 0.693 \) - **B.** \( 3.30 \times 10^{-3} \) - **C.** \( 1.65 \times 10^{-2} \) - **D.** \( 0.198 \) - **E.** \( 1.98 \) --- ### Explanation: For a first-order reaction, the rate constant \( k \) can be determined from the half-life \( t_{1/2} \) using the formula: \[ t_{1/2} = \frac{0.693}{k} \] Given \( t_{1/2} = 3.50 \) minutes, we first need to convert this into seconds: \[ 3.50 \text{ min} \times 60 \text{ s/min} = 210 \text{ s} \] Now, substituting the given half-life into the formula, we solve for \( k \): \[ 210 = \frac{0.693}{k} \] Rearranging for \( k \): \[ k = \frac{0.693}{210} \approx 3.30 \times 10^{-3} \text{ s}^{-1} \] The correct answer is **B**.
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