Answer
B
Work Step by Step
For first-order processes:
$ln\left( \dfrac{[N_2O_5]_t}{[N_2O_5]_0}\right)=-kt$
$t_{1/2}=\dfrac{0.693}{k}$
For 99% decomposition:
$ln(0.01)=-0.693\times\frac{t}{t_{1/2}}$
$\frac{t}{t_{1/2}}=6.65$
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 - 14-4 Concentration-Time Relationships: Integrated Rate Laws - Review & Check for Section 14-4 - Page 528: 1
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