Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 17 - Spontaneity, Entropy, and Free Energy - Exercises: 57

Answer

C$H_{4}$ + 2$O_{2}$ < C$O_{2}$ + 2$H_{2}$O $\Delta$G$^{\circ}$= -817 kJ/$mol_{rxn}$

Work Step by Step

Given: 2$H_{2}$ + C < C$H_{4}$/ $\Delta$G$^{\circ}$= -51 kJ 2$H_{2}$ + $O_{2}$ < 2 $H_{2}$O/ $\Delta$G$^{\circ}$= -474 kJ C + $O_{2}$ < C$O_{2}$/ $\Delta$G$^{\circ}$= -394 kJ Algebraically Fixed Equations so Undesired Components Cancel Out (Add all $\Delta$G$^{\circ}$ values to get the final $\Delta$G$^{\circ}$) : C$H_{4}$ < 2$H_{2}$ + C/ $\Delta$G$^{\circ}$= +51 kJ 2$H_{2}$ + $O_{2}$ < 2 $H_{2}$O/ $\Delta$G$^{\circ}$= -474 kJ C + $O_{2}$ < C$O_{2}$/ $\Delta$G$^{\circ}$= -394 kJ C$H_{4}$ + 2$O_{2}$ < C$O_{2}$ + 2$H_{2}$O $\Delta$G$^{\circ}$= -817 kJ/$mol_{rxn}$
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