Chemistry (4th Edition)

Published by McGraw-Hill Publishing Company
ISBN 10: 0078021529
ISBN 13: 978-0-07802-152-7

Chapter 5 - Questions and Problems - Page 221: 5.29


480.3 kJ

Work Step by Step

Given that $\hspace{2cm}\Delta H= +483.6\,kJ/mol$ $\hspace{2cm}P= 1.00\,atm$ $\hspace{2cm}\Delta V= 32.7\,L$ Recall that $\Delta U= \Delta H-P\Delta V$ $=483.6\,kJ-1.00\,atm\times32.7\,L$ $=483.6\,kJ-32.7\times0.1013\,kJ$ (As $1\,L\cdot atm=101.3\,J=0.1013\,kJ$) $=483.6\,kJ-3.31\,kJ=480.3\,kJ$
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