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 553b: 25


a) $1.0\times10^{4} s$ b) $3.4\times10^{4} s$

Work Step by Step

a) $t_{1/2}=\frac{0.693}{k}=\frac{0.693}{6.7\times10^{-5}s^{-1}}=1.0\times10^{4} s$ b) $k= \frac{2.303}{t}\log \frac{[R]_{0}}{[R]}$ Given that $[R]= \frac{[R]_{0}}{10}$ Therefore, $t= \frac{2.303}{k}\log\frac{[R]_{0}}{[R]_{0}/10}$ $=\frac{2.303}{k}\log 10= \frac{2.303}{k}$ $=\frac{2.303}{6.7\times10^{-5}s^{-1}}=3.4\times10^{4}s$
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