Answer
$k_2=0.314\ s^{-1}$
Work Step by Step
From the Arrhenius equation:
$ln(k_2/k_1)=-E_a/R(1/T_2-1/T_1)$
$ln(k_2/0.0315\ s^{-1})=-260\ kJ/mol/8.314\ J/mol.K(1/850\ K-1/800\ K)$
$k_2=0.314\ s^{-1}$
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: 39
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