Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Using the information provided, determine the activation energy for the reaction. Please explain why the hilighted answer is correct.. thank you!

The rate constant was determined at four temperatures for the following reaction:

\[2 \text{ClO} (g) \rightarrow \text{Cl}_2 (g) + \text{O}_2 (g)\]

The data were then plotted in the form \(\ln k\) versus \(1/T\) in the plot on the right, which was found to have a best-fit line:

\[y = -1590x + 28.0\]

**Graph Explanation:**

- The graph plots \(\ln k\) on the y-axis against \(1/T\) (in K\(^{-1}\)) on the x-axis.
- The graph is a straight line with a negative slope, indicating an exponential decrease in the reaction rate constant with increasing inverse temperature.
- The slope of the line \(-1590\) is related to the activation energy of the reaction.

**Question 23:**

Using the information provided, determine the activation energy for the reaction.

Options:
- A. 28.0 J/mol
- B. 1590 J/mol
- C. 1.32 × 10\(^4\) J/mol
- *D. 191 J/mol*
- E. 1.32 × 10\(^{-3}\) J/mol

The correct answer is highlighted: **C. 1.32 × 10\(^4\) J/mol**.
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Transcribed Image Text:The rate constant was determined at four temperatures for the following reaction: \[2 \text{ClO} (g) \rightarrow \text{Cl}_2 (g) + \text{O}_2 (g)\] The data were then plotted in the form \(\ln k\) versus \(1/T\) in the plot on the right, which was found to have a best-fit line: \[y = -1590x + 28.0\] **Graph Explanation:** - The graph plots \(\ln k\) on the y-axis against \(1/T\) (in K\(^{-1}\)) on the x-axis. - The graph is a straight line with a negative slope, indicating an exponential decrease in the reaction rate constant with increasing inverse temperature. - The slope of the line \(-1590\) is related to the activation energy of the reaction. **Question 23:** Using the information provided, determine the activation energy for the reaction. Options: - A. 28.0 J/mol - B. 1590 J/mol - C. 1.32 × 10\(^4\) J/mol - *D. 191 J/mol* - E. 1.32 × 10\(^{-3}\) J/mol The correct answer is highlighted: **C. 1.32 × 10\(^4\) J/mol**.
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