Question

From the following data at 25°C, calculate the standard enthalpy of formation of FeO(s) and of Fe₂ O₃(s)

Reaction	∆rH° (kJ/mole)
(A) Fe2O3(s) +3C (g) →2Fe (s) +3CO (g)	492.6
(B) FeO(s) +C (g)→ Fe (s) +3CO(g)	155.8
(C) C(g) +O2(g) →CO2(g)	-393.51
(D) CO +11O2(g) →CO2 (g)	-282.98

Answer 1

 ∆H_f = -155.8 – 393.51 + 282.98 = –266.33

2Fe + 3/2 O₂ → Fe₂ O

2Fe + 3CO → Fe₂ O₃ + 3C ∆H = –492.6 

3C + 3O₂ → 3CO₂ ∆H =–3 × 393.51

3CO₂ → 3CO + 3/2 O₂ ∆x=+3 × 282.8

∴ Here 2Fe + 3/2 O₂ → Fe₂ O₃

∆H₂ = –492.6 – 3 × 393.51 + 3 × 282.98 = –824.2