Be sure to answer all parts. -1 The rate constant of a reaction is 6.3 × 10s at 25°C, and the activation energy is 33.6 kJ/mol. What is k at 75°C? Enter your answer in scientific notation. 44.2 x 10 3 -1
Be sure to answer all parts. -1 The rate constant of a reaction is 6.3 × 10s at 25°C, and the activation energy is 33.6 kJ/mol. What is k at 75°C? Enter your answer in scientific notation. 44.2 x 10 3 -1
Chemistry
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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 answer I have is wrong. It’s not to the 3rd or -3rd power.
![**Chemistry: Reaction Rate Constant Calculation**
**Be sure to answer all parts.**
**Problem Statement:**
The rate constant of a reaction is \( 6.3 \times 10^{-3} \text{s}^{-1} \) at 25°C, and the activation energy is 33.6 kJ/mol. What is \( k \) at 75°C? Enter your answer in scientific notation.
**Solution:**
\( 44.2 \times 10^{3} \text{s}^{-1} \)
**Explanation:**
To solve this problem, we use the Arrhenius Equation:
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \( k \) is the rate constant
- \( A \) is the pre-exponential factor (frequency of collisions)
- \( E_a \) is the activation energy
- \( R \) is the gas constant (\(8.314 \text{ J/mol·K} \))
- \( T \) is the temperature in Kelvin
Given:
- Initial rate constant, \( k_1 = 6.3 \times 10^{-3} \text{s}^{-1} \)
- Initial temperature, \( T_1 = 25^\circ \text{C} = 298 \text{ K} \)
- Final temperature, \( T_2 = 75^\circ \text{C} = 348 \text{ K} \)
- Activation energy, \( E_a = 33.6 \text{ kJ/mol} = 33600 \text{ J/mol} \)
The relationship between the rate constants at two different temperatures can be expressed as:
\[ \ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]
Substituting the values into this equation will give the value of \( k_2 \) (the rate constant at 75°C).
Finally, the calculated rate constant at 75°C (\( k_2 \)) is:
\[ 44.2 \times 10^{3} \text{s}^{-1} \]
This solution includes intermediate steps and units to ensure a clear understanding of the calculation procedure.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe0ae19d5-3750-41df-884f-8ca3aa31e002%2Fc0bf30b7-de16-43a4-872a-605cad35c7bc%2Fhtzlsln_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Chemistry: Reaction Rate Constant Calculation**
**Be sure to answer all parts.**
**Problem Statement:**
The rate constant of a reaction is \( 6.3 \times 10^{-3} \text{s}^{-1} \) at 25°C, and the activation energy is 33.6 kJ/mol. What is \( k \) at 75°C? Enter your answer in scientific notation.
**Solution:**
\( 44.2 \times 10^{3} \text{s}^{-1} \)
**Explanation:**
To solve this problem, we use the Arrhenius Equation:
\[ k = A e^{-\frac{E_a}{RT}} \]
where:
- \( k \) is the rate constant
- \( A \) is the pre-exponential factor (frequency of collisions)
- \( E_a \) is the activation energy
- \( R \) is the gas constant (\(8.314 \text{ J/mol·K} \))
- \( T \) is the temperature in Kelvin
Given:
- Initial rate constant, \( k_1 = 6.3 \times 10^{-3} \text{s}^{-1} \)
- Initial temperature, \( T_1 = 25^\circ \text{C} = 298 \text{ K} \)
- Final temperature, \( T_2 = 75^\circ \text{C} = 348 \text{ K} \)
- Activation energy, \( E_a = 33.6 \text{ kJ/mol} = 33600 \text{ J/mol} \)
The relationship between the rate constants at two different temperatures can be expressed as:
\[ \ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]
Substituting the values into this equation will give the value of \( k_2 \) (the rate constant at 75°C).
Finally, the calculated rate constant at 75°C (\( k_2 \)) is:
\[ 44.2 \times 10^{3} \text{s}^{-1} \]
This solution includes intermediate steps and units to ensure a clear understanding of the calculation procedure.
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