Answer
e) 205 kJ/mol.
Work Step by Step
We find:
$\Delta G^{\circ}=-RT\ln K_{sp}$
$=-(8.314\,Jmol^{-1}K^{-1})(25+273)K\times(\ln 1.1\times10^{-36})$
$=205\times10^{3}\,J/mol$
$=205\,kJ/mol$
Chemistry (4th Edition)
by Burdge, Julia
Published by
McGraw-Hill Publishing Company
ISBN 10:
0078021529
ISBN 13:
978-0-07802-152-7
Chapter 18 - Section 18.6 - Free Energy and Chemical Equilibrium - Checkpoint - Page 861: 18.6.4
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