Answer
(c) 23 kJ/mol-rxn
Work Step by Step
$\Delta G^{\circ}=-RT\ln K_{p}$
$=-(8.314\,Jmol^{-1}K^{-1})(425+273.15)K(\ln 0.018)$
$=23000\,J/mol=23\,kJ/mol$
Chemistry and Chemical Reactivity (9th Edition)
by Kotz, John C.; Treichel, Paul M.; Townsend, John R.; Treichel, David A.
Published by
Cengage Learning
ISBN 10:
1133949649
ISBN 13:
978-1-13394-964-0
Chapter 18 Principles of Chemical Reactivity: Entropy and Free Energy - 18-7 Calculating and Using Free Energy - Review & Check for Section 18-7 - Page 708: 3
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