Answer:
ΔH = 46.2 kJ
Explanation:
Hello there!
In this case, according to the given set of reactions and the required one, it is possible to apply the Hess' rule in order to realize the following manipulations to obtain the required reaction:
1. Keep this reaction as it is.
2Fe(s) + O2(g) => 2FeO(s) ΔH = -545.6 kJ
2. Keep this reaction as it is.
4Fe(s) + 3O2(g) => 2Fe2O3(s) ΔH = -1,643.6 kJ
3. Reverse double this reaction so that we obtain:
2Fe3O4(s) => 6Fe(s) + 4O2(g) ΔH = -(-1,117.7 kJ)*2=2235.4kJ
Add them up to obtain:
2Fe(s) + O2(g) + 4Fe(s) + 3O2(g) + 2Fe3O4(s) => 6Fe(s) + 4O2(g) +2Fe2O3(s) + 2FeO(s)
Simplify common terms to obtain:
2 Fe3O4(s) => 2 Fe2O3(s) + 2 FeO(s)
Therefore, the enthalpy of such reaction is:
ΔH = 2235.4kJ -545.6 kJ -1,643.6 kJ
ΔH = 46.2 kJ
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