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?

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
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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.
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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