Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 6 - Sections 6.1-6.10 - Exercises - Problems by Topic - Page 290: 88d

Answer

-976.4 kJ

Work Step by Step

$\Delta H^{o}_{rxn}=\Sigma n_{p}\Delta H_{f}^{o}(products)-\Sigma n_{r}\Delta H_{f}^{o}(reactants)$ $=[1(\Delta H^{o}_{f,\,N_{2}(g)})+4(\Delta H^{o}_{f,\,H_{2}O(g)})]-[1(\Delta H^{o}_{f,\,N_{2}O_{4}(g)})+4(\Delta H^{o}_{f,\,H_{2}(g)})]$ $=[1(0)+4(-241.8\,kJ)]-[1(9.16\,kJ)+4(0)]$ $=-976.4\,kJ$
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