Answer
0.350
Work Step by Step
We find:
$\Delta G^{\circ}=-RT\ln K_{p}$
$2.60\,kJ/mol\times\frac{1000\,J}{1\,kJ}=-\frac{8.314\,J}{K\cdot mol}\times298\,K\times \ln K_{p}$
$\implies \ln K_{p}=-1.0494$
Or $K_{p}=e^{-1.0494}=0.350$
Chemistry 12th Edition
by Chang, Raymond; Goldsby, Kenneth
Published by
McGraw-Hill Education
ISBN 10:
0078021510
ISBN 13:
978-0-07802-151-0
Chapter 17 - Entropy, Free Energy, and Equilibrium - Questions & Problems - Page 805: 17.23
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