Answer
b) -543.2 kJ/mol
Work Step by Step
$\Delta H_{rxn}^{\circ}=$$2\times \Delta H_{f}^{\circ}(HF)-[\Delta H_{f}^{\circ}(H_{2})+\Delta H_{f}^{\circ}(F_{2})]$
$=(2\times-271.6\,kJ/mol)-(0+0)$
$=-543.2\,kJ/mol$
Chemistry (4th Edition)
by Burdge, Julia
Published by
McGraw-Hill Publishing Company
ISBN 10:
0078021529
ISBN 13:
978-0-07802-152-7
Chapter 5 - Section 5.6 - Standard Enthalpies of Formation - Checkpoint - Page 215: 5.6.1
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