Answer
-204 kJ/mol
Work Step by Step
We find:
$\Delta G^{\circ}_{rxn}=\Delta G_{f}^{\circ}=-RT\ln K_{p}=-(8.314\,Jmol^{-1}K^{-1})(298\,K)(\ln 5.62\times10^{35})$
$=-204\times10^{3}\,J/mol=-204\,kJ/mol$
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: 17.30
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.
