Chemistry: Molecular Approach (4th Edition)

Published by Pearson
ISBN 10: 0134112830
ISBN 13: 978-0-13411-283-1

Chapter 20 - Exercises - Page 973: 45

Answer

$2.34\times10^{9}\,y$

Work Step by Step

Rate constant $k= \frac{0.693}{t_{1/2}}=\frac{0.693}{703\times10^{6}\,y}$ $=9.85775\times10^{-10}\,y^{-1}$ We can obtain the time taken using integrated rate law which is $\ln \frac{N_{t}}{N_{0}}=-kt$ $\ln \frac{10}{100}=-9.85775\times10^{-10}\,y^{-1}\times t$ $\implies -2.302585=-9.85775\times10^{-10}\,y^{-1}\times t$ $\implies t=\frac{-2.302585}{-9.85775\times10^{-10}\,y^{-1}}$ $=2.34\times10^{9}\,y$
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