Answer
$1.2\times10^{-25}$
Work Step by Step
$\Delta _{r}G^{\circ}=-RT\ln K_{p}$
$\implies K_{p}=e^{-\frac{\Delta_{r}G^{\circ}}{RT}}=e^{-\frac{142.12\times10^{3}\,J/mol}{(8.314\,Jmol^{-1}K^{-1})(298\,K)}}$
$=1.2\times10^{-25}$
Chemistry: The Molecular Science (5th Edition)
by Moore, John W.; Stanitski, Conrad L.
Published by
Cengage Learning
ISBN 10:
1285199049
ISBN 13:
978-1-28519-904-7
Chapter 16 - Thermodynamics: Directionality of Chemical Reactions - Questions for Review and Thought - More Challenging Questions - Page 737h: 119b
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