Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company
ISBN 10: 0078021529
ISBN 13: 978-0-07802-152-7

Chapter 18 - Questions and Problems - Page 869: 18.43

Answer

$79\,kJ/mol$

Work Step by Step

We find: $\Delta G^{\circ}=-RT\ln K_{sp}$ $=-(8.314\,Jmol^{-1}K^{-1})(25+273)K\times(\ln 1.6\times10^{-14})$ $=79000\,J/mol$ $=79\,kJ/mol$
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