Answer
0.35
Work Step by Step
We find:
$\Delta G^{\circ}=-RT\ln K_{p}$
$\implies \ln K_{p}=-\frac{\Delta G^{\circ}}{RT}=-\frac{2.60\times10^{3}\,J/mol}{(8.314\,Jmol^{-1}K^{-1})(25+273)K}$
$=-1.0494$
$\implies K_{p}=e^{-1.0494}=0.35$
Chemistry (4th Edition)
by Burdge, Julia
Published by
McGraw-Hill Publishing Company
ISBN 10:
0078021529
ISBN 13:
978-0-07802-152-7
Chapter 18 - Questions and Problems - Page 869: 18.41
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
