Chemistry (4th Edition)

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: 5.6.4


e) -110.5 kJ/mol

Work Step by Step

From the first reaction $\Delta H_{rxn}^{\circ}=\Delta H_{f}^{\circ}(CO_{2}(g))-[\Delta H_{f}^{\circ}(C(graphite))+\Delta H_{f}^{\circ}(O_{2}(g))]$ As $O_{2}$ is the most stable form of oxygen and graphite is the most stable allotrope of carbon $\Delta H_{f}^{\circ}(C(graphite))=\Delta H_{f}^{\circ}(O_{2}(g))=0$ $\implies -393.5\,kJ/mol=\Delta H_{f}^{\circ}(CO_{2}(g))$ From the second reaction, $\Delta H_{rxn}^{\circ}=\Delta H_{f}^{\circ}(CO_{2}(g))-[\Delta H_{f}^{\circ}(CO(g))+\frac{1}{2}\times\Delta H_{f}^{\circ}(O_{2}(g))]$ $\implies -283.0\,kJ/mol=-393.5\,kJ/mol-[\Delta H_{f}^{\circ}(CO(g))+ \frac{1}{2}\times0]$ $\implies \Delta H_{f}^{\circ}(CO(g))=-393.5\,kJ/mol+283.0\,kJ/mol$ $=-110.5\,kJ/mol$
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