Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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![The image depicts a problem related to the Arrhenius equation in the context of chemical kinetics.
### Problem Statement:
A certain reaction has an activation energy of 61.40 kJ/mol. At what Kelvin temperature will the reaction proceed 5.50 times faster than it did at 323 K?
### Solution Attempt:
The user has entered the following response:
\[ T = 300.59 \, \text{K} \]
However, this response has been marked as "Incorrect."
### Instructions:
To solve this problem, one would typically use the Arrhenius equation, which relates the rate constant \( k \) of a reaction to the temperature \( T \):
\[ k = A e^{\frac{-E_a}{RT}} \]
Where:
- \( k \) is the rate constant,
- \( A \) is the pre-exponential factor,
- \( E_a \) is the activation energy,
- \( R \) is the gas constant (8.314 J/(mol·K)),
- \( T \) is the temperature in Kelvin.
Given that the reaction rate increases by a factor of 5.50, we can set up a ratio of the rate constants at two different temperatures.
### Additional Information:
To find the correct temperature, you would typically rearrange and solve the logarithmic form of the Arrhenius equation.](https://content.bartleby.com/qna-images/question/b6e4df00-119f-450b-8f4c-37fa1d1e07f7/3428730e-878f-45ba-a063-87302c6247c8/6a5t70l_processed.png)
Transcribed Image Text:The image depicts a problem related to the Arrhenius equation in the context of chemical kinetics.
### Problem Statement:
A certain reaction has an activation energy of 61.40 kJ/mol. At what Kelvin temperature will the reaction proceed 5.50 times faster than it did at 323 K?
### Solution Attempt:
The user has entered the following response:
\[ T = 300.59 \, \text{K} \]
However, this response has been marked as "Incorrect."
### Instructions:
To solve this problem, one would typically use the Arrhenius equation, which relates the rate constant \( k \) of a reaction to the temperature \( T \):
\[ k = A e^{\frac{-E_a}{RT}} \]
Where:
- \( k \) is the rate constant,
- \( A \) is the pre-exponential factor,
- \( E_a \) is the activation energy,
- \( R \) is the gas constant (8.314 J/(mol·K)),
- \( T \) is the temperature in Kelvin.
Given that the reaction rate increases by a factor of 5.50, we can set up a ratio of the rate constants at two different temperatures.
### Additional Information:
To find the correct temperature, you would typically rearrange and solve the logarithmic form of the Arrhenius equation.
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