2AL(s) + fe2O3(s) \(\longrightarrow\) AL2O3(s) + 2fe(s)
ΔfH(AL2O3) = - 1675.7 KJ/mol and ΔfHfe2O3 = - 828.4 KJ/mol
ΔrH = ΔfH product - ΔfH reactant
= (ΔfH(AL2O3) + ΔfHfe(s)) - (ΔfHAL(s) + ΔfH(fe2O3))
= (-1675.7 KJ/mol + 0) - 0 + (-828.4 KJ/mol)
ΔrH = -847.3 KJ/mol