The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy E =31.0 kJ/mol. If the rate constant of this reaction is 8.5 × 10° M¯'s -1 at 254.0 °C, what will the rate constant be at 336.0 °C? Round your answer to 2 significant digits.

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### Determining the Rate Constant Using the Arrhenius Equation

#### Problem Statement:
The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy \( E_a = 31.0 \, \text{kJ/mol} \). If the rate constant of this reaction is \( 8.5 \times 10^5 \, M^{-1} \cdot s^{-1} \) at \( 254.0 \, ^\circ\text{C} \), what will the rate constant be at \( 336.0 \, ^\circ\text{C} \)?

Round your answer to 2 significant digits.

#### Solution Explanation:

The Arrhenius equation is given by:
\[ k = A e^{\left(-\frac{E_a}{RT}\right)} \]
where:
- \( k \) is the rate constant,
- \( A \) is the pre-exponential factor,
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant (\( 8.314 \, \text{J/mol} \cdot \text{K} \)),
- \( T \) is the temperature in Kelvin (K).

Given data:
- \( E_a = 31.0 \, \text{kJ/mol} = 31.0 \times 10^3 \, \text{J/mol} \) (conversion from kJ to J)
- Initial Rate Constant \( k_1 = 8.5 \times 10^5 \, M^{-1} \cdot s^{-1} \) at \( T_1 = 254.0 \, ^\circ\text{C} = 527.15 \, \text{K} \) (conversion from Celsius to Kelvin)
- Final temperature \( T_2 = 336.0 \, ^\circ\text{C} = 609.15 \, \text{K} \)

The Arrhenius equation can be used in a ratio form to find the new rate constant \( k_2 \) at temperature \( T_2 \):
\[ \ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \
Transcribed Image Text:### Determining the Rate Constant Using the Arrhenius Equation #### Problem Statement: The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy \( E_a = 31.0 \, \text{kJ/mol} \). If the rate constant of this reaction is \( 8.5 \times 10^5 \, M^{-1} \cdot s^{-1} \) at \( 254.0 \, ^\circ\text{C} \), what will the rate constant be at \( 336.0 \, ^\circ\text{C} \)? Round your answer to 2 significant digits. #### Solution Explanation: The Arrhenius equation is given by: \[ k = A e^{\left(-\frac{E_a}{RT}\right)} \] where: - \( k \) is the rate constant, - \( A \) is the pre-exponential factor, - \( E_a \) is the activation energy, - \( R \) is the universal gas constant (\( 8.314 \, \text{J/mol} \cdot \text{K} \)), - \( T \) is the temperature in Kelvin (K). Given data: - \( E_a = 31.0 \, \text{kJ/mol} = 31.0 \times 10^3 \, \text{J/mol} \) (conversion from kJ to J) - Initial Rate Constant \( k_1 = 8.5 \times 10^5 \, M^{-1} \cdot s^{-1} \) at \( T_1 = 254.0 \, ^\circ\text{C} = 527.15 \, \text{K} \) (conversion from Celsius to Kelvin) - Final temperature \( T_2 = 336.0 \, ^\circ\text{C} = 609.15 \, \text{K} \) The Arrhenius equation can be used in a ratio form to find the new rate constant \( k_2 \) at temperature \( T_2 \): \[ \ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \
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