Use the logarithmic Arrhenius equation and the following data showing the dependence of temperature and reaction rate constant k to predict the activation energy barrier (E₂) for the reaction shown. Hint: try graphing In(k) against 1/T from the data in the table and estimate the intercept, which is In(A), and the slope, which is E₂/R. In case you get stuck, you can use the Arrhenius factor for this reaction (A = 8.1 x 10° mol¹Ls ¹) and substitute it into the equation. 2NO2 (9) T(K) 673 683 693 703 713 O E,- 1.2 x 105 kJmol-1 O E,-8.314 JK ¹mol-1 O E,=-119601 Jmol.1 O E, -7.7 x 106 J O E,-1.2 x 10² kJmol-1 O E, = 0.12 kJ k (mol ¹Ls¹) 7.8 10 14 2NO (9) 18 24 + O2(g)
Use the logarithmic Arrhenius equation and the following data showing the dependence of temperature and reaction rate constant k to predict the activation energy barrier (E₂) for the reaction shown. Hint: try graphing In(k) against 1/T from the data in the table and estimate the intercept, which is In(A), and the slope, which is E₂/R. In case you get stuck, you can use the Arrhenius factor for this reaction (A = 8.1 x 10° mol¹Ls ¹) and substitute it into the equation. 2NO2 (9) T(K) 673 683 693 703 713 O E,- 1.2 x 105 kJmol-1 O E,-8.314 JK ¹mol-1 O E,=-119601 Jmol.1 O E, -7.7 x 106 J O E,-1.2 x 10² kJmol-1 O E, = 0.12 kJ k (mol ¹Ls¹) 7.8 10 14 2NO (9) 18 24 + O2(g)
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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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![Use the logarithmic Arrhenius equation and the following data showing the dependence of temperature and reaction rate constant \( k \) to predict the activation energy barrier (\( E_a \)) for the reaction shown. Hint: try graphing ln(k) against 1/T from the data in the table and estimate the intercept, which is ln(A), and the slope, which is \( E_a/R \). In case you get stuck, you can use the Arrhenius factor for this reaction (A = \( 8.1 \times 10^9 \, \text{mol}^{-1}\text{L}\text{s}^{-1} \)) and substitute it into the equation.
\[ \text{2NO}_2 \, (g) \rightarrow \text{2NO} \, (g) + \text{O}_2 \, (g) \]
| \( T \, (K) \) | \( k \, (\text{mol}^{-1}\text{L}\text{s}^{-1}) \) |
|---------------|-----------------------------------|
| 673 | 7.8 |
| 683 | 10 |
| 693 | 14 |
| 703 | 18 |
| 713 | 24 |
Options:
- \( E_a = 1.2 \times 10^5 \, \text{kJmol}^{-1} \)
- \( E_a = 8.314 \, \text{JK}^{-1}\text{mol}^{-1} \)
- \( E_a = -119601 \, \text{Jmol}^{-1} \)
- \( E_a = 7.7 \times 10^{-6} \, \text{J} \)
- \( E_a = 1.2 \times 10^2 \, \text{kJmol}^{-1} \)
- \( E_a = 0.12 \, \text{kJ} \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa5cf8e31-f3df-44d1-8e16-edf3053310db%2Ff6f32bb5-d63f-4153-9418-e07ae959a801%2F6a50118_processed.png&w=3840&q=75)
Transcribed Image Text:Use the logarithmic Arrhenius equation and the following data showing the dependence of temperature and reaction rate constant \( k \) to predict the activation energy barrier (\( E_a \)) for the reaction shown. Hint: try graphing ln(k) against 1/T from the data in the table and estimate the intercept, which is ln(A), and the slope, which is \( E_a/R \). In case you get stuck, you can use the Arrhenius factor for this reaction (A = \( 8.1 \times 10^9 \, \text{mol}^{-1}\text{L}\text{s}^{-1} \)) and substitute it into the equation.
\[ \text{2NO}_2 \, (g) \rightarrow \text{2NO} \, (g) + \text{O}_2 \, (g) \]
| \( T \, (K) \) | \( k \, (\text{mol}^{-1}\text{L}\text{s}^{-1}) \) |
|---------------|-----------------------------------|
| 673 | 7.8 |
| 683 | 10 |
| 693 | 14 |
| 703 | 18 |
| 713 | 24 |
Options:
- \( E_a = 1.2 \times 10^5 \, \text{kJmol}^{-1} \)
- \( E_a = 8.314 \, \text{JK}^{-1}\text{mol}^{-1} \)
- \( E_a = -119601 \, \text{Jmol}^{-1} \)
- \( E_a = 7.7 \times 10^{-6} \, \text{J} \)
- \( E_a = 1.2 \times 10^2 \, \text{kJmol}^{-1} \)
- \( E_a = 0.12 \, \text{kJ} \)
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