Chapter 14 Chemical Kinetics: The Rates of Chemical Reactions - Study Questions - Page 553c: 40

$E_a=269.3\ kJ/mol$

Work Step by Step

From the Arrhenius equation: $ln(k_2/k_1)=-E_a/R(1/T_2-1/T_1)$ $ln(1.02\times10^{-3}\ s^{-1}/1.10\times10^{-4}\ s^{-1})=-E_a/8.314\ J/mol.K(1/783\ K-1/743\ K)$ $E_a=269.3\ kJ/mol$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.