Answer
80. kJ/mol
Work Step by Step
We find:
$\Delta G^{\circ}=-RT\ln K_{w}$
$=-(8.314\,Jmol^{-1}K^{-1})(25+273)K\times(\ln 1.0\times10^{-14})$
$=8.0\times10^{4}\,J/mol$
$=80.\,kJ/mol$
Chemistry (4th Edition)
by Burdge, Julia
Published by
McGraw-Hill Publishing Company
ISBN 10:
0078021529
ISBN 13:
978-0-07802-152-7
Chapter 18 - Questions and Problems - Page 869: 18.42
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