A first order process has an experimentally determined rate constant of 6.0 x104s! at 500°C. If the initial concentration of the species is 0.0226 mol/L, calculate its concentration after 955 s.
A first order process has an experimentally determined rate constant of 6.0 x104s! at 500°C. If the initial concentration of the species is 0.0226 mol/L, calculate its concentration after 955 s.
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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![**Educational Exercise: Calculating Concentration in a First-Order Reaction**
**Problem Statement:**
A first-order process has an experimentally determined rate constant of \(6.0 \times 10^{-4} \, \text{s}^{-1}\) at \(500^\circ\text{C}\). If the initial concentration of the species is \(0.0226 \, \text{mol/L}\), calculate its concentration after \(955 \, \text{s}\).
**Solution:**
To solve for the concentration of a species in a first-order reaction, we use the integrated rate law for a first-order reaction, given by:
\[ [A] = [A]_0 e^{-kt} \]
Where:
- \([A]\) is the concentration of the species at time \(t\).
- \([A]_0\) is the initial concentration of the species.
- \(k\) is the rate constant.
- \(t\) is the time.
Given:
- \(k = 6.0 \times 10^{-4} \, \text{s}^{-1}\)
- \([A]_0 = 0.0226 \, \text{mol/L}\)
- \(t = 955 \, \text{s}\)
Substitute the given values into the equation:
\[ [A] = 0.0226 \, \text{mol/L} \times e^{-(6.0 \times 10^{-4} \, \text{s}^{-1}) \times 955 \, \text{s}} \]
First, calculate the exponent:
\[ -(6.0 \times 10^{-4} \, \text{s}^{-1}) \times 955 \, \text{s} = -0.573 \]
Then, calculate \(e^{-0.573}\):
\[ e^{-0.573} \approx 0.564 \]
Now, substitute back into the equation:
\[ [A] = 0.0226 \, \text{mol/L} \times 0.564 \approx 0.01276 \, \text{mol/L} \]
Therefore, the concentration of the species after \(955 \, \text{s}\) is approximately \(0.01276 \, \text{mol/L}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F24666b77-75da-49bd-bf84-8c4674dedd2c%2F19a708d7-a52f-4e8b-b47d-a04f5e3f4dd4%2Fb6k6qar.png&w=3840&q=75)
Transcribed Image Text:**Educational Exercise: Calculating Concentration in a First-Order Reaction**
**Problem Statement:**
A first-order process has an experimentally determined rate constant of \(6.0 \times 10^{-4} \, \text{s}^{-1}\) at \(500^\circ\text{C}\). If the initial concentration of the species is \(0.0226 \, \text{mol/L}\), calculate its concentration after \(955 \, \text{s}\).
**Solution:**
To solve for the concentration of a species in a first-order reaction, we use the integrated rate law for a first-order reaction, given by:
\[ [A] = [A]_0 e^{-kt} \]
Where:
- \([A]\) is the concentration of the species at time \(t\).
- \([A]_0\) is the initial concentration of the species.
- \(k\) is the rate constant.
- \(t\) is the time.
Given:
- \(k = 6.0 \times 10^{-4} \, \text{s}^{-1}\)
- \([A]_0 = 0.0226 \, \text{mol/L}\)
- \(t = 955 \, \text{s}\)
Substitute the given values into the equation:
\[ [A] = 0.0226 \, \text{mol/L} \times e^{-(6.0 \times 10^{-4} \, \text{s}^{-1}) \times 955 \, \text{s}} \]
First, calculate the exponent:
\[ -(6.0 \times 10^{-4} \, \text{s}^{-1}) \times 955 \, \text{s} = -0.573 \]
Then, calculate \(e^{-0.573}\):
\[ e^{-0.573} \approx 0.564 \]
Now, substitute back into the equation:
\[ [A] = 0.0226 \, \text{mol/L} \times 0.564 \approx 0.01276 \, \text{mol/L} \]
Therefore, the concentration of the species after \(955 \, \text{s}\) is approximately \(0.01276 \, \text{mol/L}\).
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