The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy E = 50.0 kJ/mol. If the rate constant reaction is 4.6 × 10² M-¹-1 "S at 83.0 °C, what will the rate constant be at 42.0 °C? Round your answer to 2 significant digits. -1 -1 k = MS S X
The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy E = 50.0 kJ/mol. If the rate constant reaction is 4.6 × 10² M-¹-1 "S at 83.0 °C, what will the rate constant be at 42.0 °C? Round your answer to 2 significant digits. -1 -1 k = MS S X
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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![**Title: Calculating the Rate Constant of a Reaction Using the Arrhenius Equation**
**Introduction:**
The rate constant \((k)\) of a chemical reaction is a crucial factor in predicting how fast a reaction occurs. Understanding its dependence on temperature and activation energy allows chemists to control reaction rates effectively.
**Problem Statement:**
The rate constant of a certain reaction is known to obey the Arrhenius equation and has an activation energy \(E_a = 50.0\) kJ/mol. If the rate constant of this reaction is \(4.6 \times 10^2\) M\(^{-1}\)s\(^{-1}\) at \(83.0\)°C, what will the rate constant be at \(42.0\)°C?
**Solution Steps:**
1. **Restate Given Data:**
- Activation Energy (\(E_a\)): 50.0 kJ/mol
- Initial Rate Constant (\(k_1\)): \(4.6 \times 10^2\) M\(^{-1}\)s\(^{-1}\)
- Initial Temperature (\(T_1\)): 83.0°C
- Final Temperature (\(T_2\)): 42.0°C
2. **Convert Temperatures to Kelvin:**
\[ T_1 = 83.0 + 273.15 = 356.15 \text{ K} \]
\[ T_2 = 42.0 + 273.15 = 315.15 \text{ K} \]
3. **Apply the Arrhenius Equation:**
The Arrhenius equation links the rate constant \(k\) with temperature \(T\) and activation energy \(E_a\):
\[ k = A \exp\left(-\frac{E_a}{RT}\right) \]
Where:
- \(A\) is the pre-exponential factor
- \(R\) is the universal gas constant (8.314 J/mol·K)
- \(E_a\) is the activation energy (J/mol)
4. **Use the Two-Point Form of the Arrhenius Equation:**
Relate the rate constants at two temperatures:
\[ \ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F550116fe-4392-44b3-9444-167bfb5405dc%2F3601d88a-3c17-4a27-aa4b-9c5ee10f8d96%2Fdsaq3e_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Calculating the Rate Constant of a Reaction Using the Arrhenius Equation**
**Introduction:**
The rate constant \((k)\) of a chemical reaction is a crucial factor in predicting how fast a reaction occurs. Understanding its dependence on temperature and activation energy allows chemists to control reaction rates effectively.
**Problem Statement:**
The rate constant of a certain reaction is known to obey the Arrhenius equation and has an activation energy \(E_a = 50.0\) kJ/mol. If the rate constant of this reaction is \(4.6 \times 10^2\) M\(^{-1}\)s\(^{-1}\) at \(83.0\)°C, what will the rate constant be at \(42.0\)°C?
**Solution Steps:**
1. **Restate Given Data:**
- Activation Energy (\(E_a\)): 50.0 kJ/mol
- Initial Rate Constant (\(k_1\)): \(4.6 \times 10^2\) M\(^{-1}\)s\(^{-1}\)
- Initial Temperature (\(T_1\)): 83.0°C
- Final Temperature (\(T_2\)): 42.0°C
2. **Convert Temperatures to Kelvin:**
\[ T_1 = 83.0 + 273.15 = 356.15 \text{ K} \]
\[ T_2 = 42.0 + 273.15 = 315.15 \text{ K} \]
3. **Apply the Arrhenius Equation:**
The Arrhenius equation links the rate constant \(k\) with temperature \(T\) and activation energy \(E_a\):
\[ k = A \exp\left(-\frac{E_a}{RT}\right) \]
Where:
- \(A\) is the pre-exponential factor
- \(R\) is the universal gas constant (8.314 J/mol·K)
- \(E_a\) is the activation energy (J/mol)
4. **Use the Two-Point Form of the Arrhenius Equation:**
Relate the rate constants at two temperatures:
\[ \ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(
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