Answer
$E_a=102.1\ kJ/mol$
Work Step by Step
From the Arrhenius equation:
$ln(k_2/k_1)=-E_a/R(1/T_2-1/T_1)$
$ln(1.5\times10^{-3}\ s^{-1}/3.46\times10^{-5}\ s^{-1})=-E_a/8.314\ J/mol.K(1/328\ K-1/298\ K)$
$E_a=102.1\ kJ/mol$
Chemistry and Chemical Reactivity (9th Edition)
by Kotz, John C.; Treichel, Paul M.; Townsend, John R.; Treichel, David A.
Published by
Cengage Learning
ISBN 10:
1133949649
ISBN 13:
978-1-13394-964-0
Chapter 14 Chemical Kinetics: The Rates of Chemical Reactions - Study Questions - Page 553c: 37
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