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 =

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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:
   - \( T_1 = 304.0 + 273.15 = 577.15 \, \text{K} \)
   - \(
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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