Chemistry and Chemical Reactivity (9th Edition)

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: 40

Answer

$E_a=269.3\ kJ/mol$

Work Step by Step

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