Determine the standard enthalpy of formation of Fe2O3(s) given the thermochemical equations below. Fe(s) + 3 H2O(
) → Fe(OH) 3(s) + 3/2 H2(g)
ΔrH° = +160.9 kJ/mol-rxn
H2(g) + 1/2 O2(g) → H2O(
)
ΔrH° = -285.8 kJ/mol-rxn
Fe2O3(s) + 3 H2O(
) → 2 Fe(OH) 3(s)
ΔrH° = +288.6 kJ/mol-rxn
A) -252.6 kJ/mol-rxn
B) +163.7 kJ/mol-rxn
C) -824.2 kJ/mol-rxn
D) +33.2 kJ/mol-rxn
E) + 890.6 kJ/mol-rxn
Correct Answer:
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