Chemistry 12th Edition

Published by McGraw-Hill Education
ISBN 10: 0078021510
ISBN 13: 978-0-07802-151-0

Chapter 6 - Thermochemistry - Questions & Problems - Page 264: 6.56

Answer

218.2 kJ/mol

Work Step by Step

$\Delta H_{rxn}^{\circ}=\Sigma n\Delta H_{f}^{\circ}(\text{products})-\Sigma m\Delta H_{f}^{\circ}(\text{reactants})$ where $m$ and $n$ denote the stoichiometric coefficients for the reactants and products. Then, $\Delta H_{rxn}^{\circ}=436.4\,kJ/mol=$ $[\Delta H_{f}^{\circ}(H)+\Delta H_{f}^{\circ}(H)]-\Delta H_{f}^{\circ}(H_{2})$ $=2\Delta H_{f}^{\circ}(H)-0=2\Delta H_{f}^{\circ}(H)$ $\implies \Delta H_{f}^{\circ}(H)=\frac{436.4\,kJ/mol}{2}=218.2\,kJ/mol$
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