Chemistry: Atoms First (2nd Edition)

Published by Cengage Learning
ISBN 10: 1305079248
ISBN 13: 978-1-30507-924-3

Chapter 11 - Exercises - Page 487g: 69

Answer

$9.5\times10^{-5}L/m•s$

Work Step by Step

We know that log $k_{2}$-log $k_{1}$= $\frac{E_{a}}{2.303R}[\frac{1}{T_{1}}-\frac{1}{T_{2}}]$ log $k_{2}$=log $k_{1}+\frac{E_{a}}{2.303R}[\frac{1}{T_{1}}-\frac{1}{T_{2}}]$ = log$(3.52\times10^{-7})+ \frac{186000J/mol}{2.303\times8.314JK^{-1}mol^{-1}}[\frac{1}{555K}-\frac{1}{645K}]$ log $k_{2}$= -6.45+2.43= -4.02 $k_{2}$ = $9.5\times10^{-5}L/m•s$
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