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What is the value of the rate constant for this reaction at 298 K?

The calculations yield the following results:

- Activation energy (\( E_a \)): \( 74,851.9 \) J/mol
- Pre-exponential factor (\( A \)): \( 3.839 \times 10^{11} \) s\(^{-1}\)
- Rate constant at \( 298 \) K: \( 0.0291 \) s\(^{-1}\)

So, the rate constant for this reaction at \( 298 \) K is approximately \( 0.0291 \) s\(^{-1}\).

Would you like more details or have any questions? 

Here are some related questions you might find useful:

1. How is the Arrhenius equation used to determine reaction rates?
2. What is the significance of the activation energy in a chemical reaction?
3. How can we determine the pre-exponential factor \( A \) from experimental data?
4. Why do we use the natural logarithm of the rate constant in the Arrhenius equation?
5. How do temperature changes affect the rate of a chemical reaction?

**Tip:** When plotting experimental data to determine kinetic parameters, ensure accurate measurements and consistent units to obtain reliable results.

Learn how to calculate the rate constant for a reaction at 298 K using the Arrhenius equation with a detailed step-by-step solution. This problem involves determining the activation energy, pre-exponential factor, and using the Arrhenius equation to compute the rate constant. Suitable for advanced chemistry students and researchers.

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