1/2 Fe203 (s) + 3/2 CO (g) ® Fe(s) + 3/2 CO2 (g) DH = -11.5 kJ 3 Fe2O3 (s) + CO (g) ® 2 Fe3O4 (s) + CO2 (g) DH = -39 kJ 1/3 Fe304 (s) + 1/3 CÓ (g) ® FeO (s) + 1/3 CO2 (g) DH = 6 kJ Calculate DH for the reaction: FeO (s) + CO (g) ® Fe(s) + CO2 (g) DH = ? -56.5 kJ -11 kJ 21.5 kJ -44.5 kJ -24 kJ

Chemistry: The Molecular Science
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Author:John W. Moore, Conrad L. Stanitski
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Chapter4: Energy And Chemical Reactions
Section: Chapter Questions
Problem 97QRT
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I was able to do a few of these on my own, but I'm really stuck with this one. Should I be flipping the third reaction? If so, how does that help me cancel out the reactants and products on both sides? Which ones should I be multiplying?

1/2 Fe2O3 (s) + 3/2 CÓ (g) ® Fe(s) + 3/2 CO2 (g)
DH = -11.5 kJ
3 Fe2O3 (s) + CÓ (g) ® 2 Fe304 (s) + CO2 (g)
DH = -39 kJ
1/3 Fe304 (s) + 1/3 CÓ (g) ® FeO (s) + 1/3 CO2 (g)
DH = 6 kJ
Calculate DH for the reaction:
FeO (s) + CO (g) ® Fe(s) + CO2 (g)
DH = ?
R
-56.5 kJ
-11 kJ
O 21.5 kJ
-44.5 kJ
-24 kJ
Transcribed Image Text:1/2 Fe2O3 (s) + 3/2 CÓ (g) ® Fe(s) + 3/2 CO2 (g) DH = -11.5 kJ 3 Fe2O3 (s) + CÓ (g) ® 2 Fe304 (s) + CO2 (g) DH = -39 kJ 1/3 Fe304 (s) + 1/3 CÓ (g) ® FeO (s) + 1/3 CO2 (g) DH = 6 kJ Calculate DH for the reaction: FeO (s) + CO (g) ® Fe(s) + CO2 (g) DH = ? R -56.5 kJ -11 kJ O 21.5 kJ -44.5 kJ -24 kJ
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