Question

The rate constants for the decomposition of a molecule in the presence of oxygen are \(0.237 \times 10^{-4}\) L mol\(^{-1}\) s\(^{-1}\) at 0 °C and \(2.64 \times 10^{-4}\) L mol\(^{-1}\) s\(^{-1}\) at 25 °C. The activation energy for this reaction (rounded off to one decimal place) is \(\underline{\hspace{2cm}}\) kJ mol\(^{-1}\).
Given: \( R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \).

Hint:

The activation energy can be calculated using the Arrhenius equation, which relates the rate constant to the temperature and activation energy.

Answer 1

Correct Answer: 65

The activation energy \( E_a \) can be found using the Arrhenius equation:
\[ \ln \left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \] where \( k_1 \) and \( k_2 \) are the rate constants at temperatures \( T_1 \) and \( T_2 \), respectively. Substituting the given values:
\[ \ln \left( \frac{2.64 \times 10^{-4}}{0.237 \times 10^{-4}} \right) = \frac{E_a}{8.314} \left( \frac{1}{273 + 25} - \frac{1}{273} \right) \] Solving for \( E_a \), we get:
\[ E_a \approx 65.0 \, \text{kJ mol}^{-1}. \] Thus, the activation energy is \( 65.0 \, \text{kJ mol}^{-1} \).