Question

Calculate the enthalpy of reaction for the following reaction: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) ΔHfo(Fe2O3(s)) = -824.2 kJ/mol ΔHfo(CO(g)) = -110.5 ΔHfo(Fe(s)) = ? ΔHfo(CO2(g)) = -393.5 kJ/mol

Answer 1

a) C(s) + O2(g) → CO2(g) ΔH = -393.5 kJ/mol
b) 2 Fe(s) + 3/2 O2(g) → Fe2O3(s) ΔH = -824.2 kJ/mol
c) C(s) + 1/2 O2(g) → CO(g) ΔH = -110.5 kJ/mol

Steps:
1. Multiply (a) by 3
2. Reverse (b) and change sign of ΔH
3. Reverse (c), change sign of ΔH and multiply by 3

3C(s) + 3O2(g) → 3CO2(g) ΔH = (3) -393.5 kJ/mol = -1180.5 kJ        ------------     (1)
Fe2O3(s) ========⇒ 2 Fe(s) + 3/2 O2(g) ΔH = +824.2 kJ             ------------     (2)
3CO(g) ==========⇒ 3C(s) + 3/2O2 3(+110.5) = +331.5 kJ           ------------     (3)

now (2) + (3) - (1)
Fe2O3(s) + 3CO(g) ====⇒ 2Fe(s) + 3CO2(g) ΔH = -24.8 kJ