Question 24 of 36 Submit The activation energy for a particular reaction is 102 kJ/mol. If the rate constant is 1.35 x 10-4 s1 at 308 K, what is the rate constant at 273 K?
Question 24 of 36 Submit The activation energy for a particular reaction is 102 kJ/mol. If the rate constant is 1.35 x 10-4 s1 at 308 K, what is the rate constant at 273 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
Section: Chapter Questions
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![### Arrhenius Equation and Reaction Rates
#### Question 24 of 36
**Problem Statement:**
The activation energy for a particular reaction is 102 kJ/mol. If the rate constant is \(1.35 \times 10^{-4} \, \text{s}^{-1}\) at 308 K, what is the rate constant at 273 K?
#### Explanation:
To solve this problem, we need to use the Arrhenius Equation, which is given by:
\[ k = A e^{-E_a / (RT)} \]
Where:
- \( k \) is the rate constant
- \( A \) is the pre-exponential factor
- \( E_a \) is the activation energy
- \( R \) is the universal gas constant (8.314 J/mol·K)
- \( T \) is the temperature in Kelvin
Given:
- Activation Energy, \( E_a = 102 \, \text{kJ/mol} = 102 \times 10^3 \, \text{J/mol} \)
- Initial Rate Constant, \( k_1 = 1.35 \times 10^{-4} \, \text{s}^{-1} \) at \( T_1 = 308 \, \text{K} \)
- Final Temperature, \( T_2 = 273 \, \text{K} \)
We will use the modified form of the Arrhenius Equation to compare the rate constants at two different temperatures:
\[ \ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]
By substituting all known values into the equation, we can solve for the new rate constant \( k_2 \).
#### Steps to solve:
1. Calculate the difference in reciprocal temperatures:
\[
\frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{308} - \frac{1}{273}
\]
2. Calculate the exponential factor:
\[
\frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)
\]
3. Use the Arrhenius Equation to find \( k_2 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F17050110-9897-4bd6-b477-110f723d877a%2Fc7555a59-8858-430a-abd8-1e74a9d3a2dd%2Fabdinc_processed.png&w=3840&q=75)
Transcribed Image Text:### Arrhenius Equation and Reaction Rates
#### Question 24 of 36
**Problem Statement:**
The activation energy for a particular reaction is 102 kJ/mol. If the rate constant is \(1.35 \times 10^{-4} \, \text{s}^{-1}\) at 308 K, what is the rate constant at 273 K?
#### Explanation:
To solve this problem, we need to use the Arrhenius Equation, which is given by:
\[ k = A e^{-E_a / (RT)} \]
Where:
- \( k \) is the rate constant
- \( A \) is the pre-exponential factor
- \( E_a \) is the activation energy
- \( R \) is the universal gas constant (8.314 J/mol·K)
- \( T \) is the temperature in Kelvin
Given:
- Activation Energy, \( E_a = 102 \, \text{kJ/mol} = 102 \times 10^3 \, \text{J/mol} \)
- Initial Rate Constant, \( k_1 = 1.35 \times 10^{-4} \, \text{s}^{-1} \) at \( T_1 = 308 \, \text{K} \)
- Final Temperature, \( T_2 = 273 \, \text{K} \)
We will use the modified form of the Arrhenius Equation to compare the rate constants at two different temperatures:
\[ \ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]
By substituting all known values into the equation, we can solve for the new rate constant \( k_2 \).
#### Steps to solve:
1. Calculate the difference in reciprocal temperatures:
\[
\frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{308} - \frac{1}{273}
\]
2. Calculate the exponential factor:
\[
\frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)
\]
3. Use the Arrhenius Equation to find \( k_2 \
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