The rate constant for the reaction below was determined to be 3.241x10-5 s-1 at 800 K. The activation energy of the reaction is 225 kJ/mol. What would be the value of the rate constant at 9.30×10² K? N₂O(g) → N₂(g) + 0(g)
The rate constant for the reaction below was determined to be 3.241x10-5 s-1 at 800 K. The activation energy of the reaction is 225 kJ/mol. What would be the value of the rate constant at 9.30×10² K? N₂O(g) → N₂(g) + 0(g)
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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![### Example Problem: Determining the Rate Constant for a Reaction
#### Problem Statement
The rate constant for the reaction below was determined to be \( 3.241 \times 10^{-5} \, \text{s}^{-1} \) at 800 K. The activation energy of the reaction is 225 kJ/mol. What would be the value of the rate constant at 930 K?
\[ \text{N}_2\text{O}(g) \rightarrow \text{N}_2(g) + \text{O}(g) \]
#### Steps to Solve
To determine the new rate constant at a different temperature, we can use the Arrhenius equation shown below:
\[ k = Ae^{-\frac{E_a}{RT}} \]
However, since we are comparing two temperatures, we can use the following form of the Arrhenius equation:
\[ \ln \left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
Where:
- \( k_1 \) and \( k_2 \) are the rate constants at temperatures \( T_1 \) and \( T_2 \) respectively,
- \( E_a \) is the activation energy,
- \( R \) is the gas constant (8.314 J/mol·K),
- \( T_1 \) and \( T_2 \) are the absolute temperatures (in Kelvin).
Using the given values:
- \( k_1 = 3.241 \times 10^{-5} \, \text{s}^{-1} \)
- \( T_1 = 800 \, \text{K} \)
- \( T_2 = 930 \, \text{K} \)
- \( E_a = 225 \, \text{kJ/mol} = 225000 \, \text{J/mol} \)
Calculate:
\[ \ln \left( \frac{k_2}{3.241 \times 10^{-5}} \right) = \frac{225000}{8.314} \left( \frac{1}{800} - \frac{1}{930} \right) \]
Calculate inside the parenthesis first:
\[ \frac{1}{800} - \frac{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc80d6140-4b5d-446e-8970-6dff282bf772%2F424b8587-f01a-4d8f-af49-9bb1597fb253%2Fr71y7zf_processed.png&w=3840&q=75)
Transcribed Image Text:### Example Problem: Determining the Rate Constant for a Reaction
#### Problem Statement
The rate constant for the reaction below was determined to be \( 3.241 \times 10^{-5} \, \text{s}^{-1} \) at 800 K. The activation energy of the reaction is 225 kJ/mol. What would be the value of the rate constant at 930 K?
\[ \text{N}_2\text{O}(g) \rightarrow \text{N}_2(g) + \text{O}(g) \]
#### Steps to Solve
To determine the new rate constant at a different temperature, we can use the Arrhenius equation shown below:
\[ k = Ae^{-\frac{E_a}{RT}} \]
However, since we are comparing two temperatures, we can use the following form of the Arrhenius equation:
\[ \ln \left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
Where:
- \( k_1 \) and \( k_2 \) are the rate constants at temperatures \( T_1 \) and \( T_2 \) respectively,
- \( E_a \) is the activation energy,
- \( R \) is the gas constant (8.314 J/mol·K),
- \( T_1 \) and \( T_2 \) are the absolute temperatures (in Kelvin).
Using the given values:
- \( k_1 = 3.241 \times 10^{-5} \, \text{s}^{-1} \)
- \( T_1 = 800 \, \text{K} \)
- \( T_2 = 930 \, \text{K} \)
- \( E_a = 225 \, \text{kJ/mol} = 225000 \, \text{J/mol} \)
Calculate:
\[ \ln \left( \frac{k_2}{3.241 \times 10^{-5}} \right) = \frac{225000}{8.314} \left( \frac{1}{800} - \frac{1}{930} \right) \]
Calculate inside the parenthesis first:
\[ \frac{1}{800} - \frac{
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