The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy E¸=69.0 kJ/mol. If the -1 rate constant of this reaction is 5.7 × 10² M ''s -1 at 304.0 °C, what will the rate constant be at 256.0 °C? Round your answer to 2 significant digits. -1 k =
The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy E¸=69.0 kJ/mol. If the -1 rate constant of this reaction is 5.7 × 10² M ''s -1 at 304.0 °C, what will the rate constant be at 256.0 °C? Round your answer to 2 significant digits. -1 k =
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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![### Rate Constant of a Reaction
The rate constant of a certain reaction is known to obey the Arrhenius equation and to have an activation energy \(E_a = 69.0 \, \text{kJ/mol}\). If the rate constant of this reaction is \(5.7 \times 10^2 \, \text{M}^{-1} \cdot \text{s}^{-1}\) at \(304.0 \, ^\circ\text{C}\), what will the rate constant be at \(256.0 \, ^\circ\text{C}\)?
**Round your answer to 2 significant digits.**
The Arrhenius equation is given by:
\[ k = A \cdot e^{\frac{-E_a}{RT}} \]
Where:
- \( k \) is the rate constant.
- \( A \) is the pre-exponential factor.
- \( E_a \) is the activation energy.
- \( R \) is the universal gas constant (\(R \approx 8.314 \, \text{J/mol} \cdot \text{K}\)).
- \( T \) is the temperature in Kelvin (K).
To find the rate constant at a different temperature, you can use the modified Arrhenius equation:
\[ \ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]
Given:
- \( k_1 = 5.7 \times 10^2 \, \text{M}^{-1} \cdot \text{s}^{-1} \)
- \( T_1 = 304.0 \, ^\circ\text{C} = 577.15 \, \text{K} \)
- \( T_2 = 256.0 \, ^\circ\text{C} = 529.15 \, \text{K} \)
- \( E_a = 69.0 \, \text{kJ/mol} = 69000 \, \text{J/mol} \)
### Solution Steps:
1. Convert temperatures to Kelvin:
- \( T_1 = 304.0 + 273.15 = 577.15 \, \text{K} \)
- \(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf1fd5ea-80f6-47fe-ac10-02a0625e13d5%2Fec8799d9-0c59-4aac-b7b9-4865c919d771%2F6r0s6ljh_processed.png&w=3840&q=75)
Transcribed Image Text:### Rate Constant of a Reaction
The rate constant of a certain reaction is known to obey the Arrhenius equation and to have an activation energy \(E_a = 69.0 \, \text{kJ/mol}\). If the rate constant of this reaction is \(5.7 \times 10^2 \, \text{M}^{-1} \cdot \text{s}^{-1}\) at \(304.0 \, ^\circ\text{C}\), what will the rate constant be at \(256.0 \, ^\circ\text{C}\)?
**Round your answer to 2 significant digits.**
The Arrhenius equation is given by:
\[ k = A \cdot e^{\frac{-E_a}{RT}} \]
Where:
- \( k \) is the rate constant.
- \( A \) is the pre-exponential factor.
- \( E_a \) is the activation energy.
- \( R \) is the universal gas constant (\(R \approx 8.314 \, \text{J/mol} \cdot \text{K}\)).
- \( T \) is the temperature in Kelvin (K).
To find the rate constant at a different temperature, you can use the modified Arrhenius equation:
\[ \ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]
Given:
- \( k_1 = 5.7 \times 10^2 \, \text{M}^{-1} \cdot \text{s}^{-1} \)
- \( T_1 = 304.0 \, ^\circ\text{C} = 577.15 \, \text{K} \)
- \( T_2 = 256.0 \, ^\circ\text{C} = 529.15 \, \text{K} \)
- \( E_a = 69.0 \, \text{kJ/mol} = 69000 \, \text{J/mol} \)
### Solution Steps:
1. Convert temperatures to Kelvin:
- \( T_1 = 304.0 + 273.15 = 577.15 \, \text{K} \)
- \(
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