Answer
240
Work Step by Step
We find:
$\Delta G^{\circ}=-RT\ln K_{p}$
$\implies \ln K_{p}=-\frac{\Delta G^{\circ}}{RT}$
$=-\frac{-13.6\times10^{3}\,J/mol}{(8.314\,Jmol^{-1}K^{-1})(298\,K)}=5.489245$
$K_{p}=e^{5.489245}=240$
Chemistry (7th Edition)
by McMurry, John E.; Fay, Robert C.; Robinson, Jill Kirsten
Published by
Pearson
ISBN 10:
0321943171
ISBN 13:
978-0-32194-317-0
Chapter 17 - Thermodynamics: Entropy, Free Energy and Equilibrium - Section Problems - Page 753: 113
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