Answer
The standart formation enthalpy of aluminium oxide.
$Δ_fH° = -1675.7 kJ/mol$
Work Step by Step
$2 Al_2O_3(s) --> 4 Al(s) + 3 O_2(g) Δ_rH° = 3351.4 kJ/mol$
Invert the reaction, and do the same for the formation enthalpy.
$4 Al(s) + 3 O_2(g) --> 2 Al_2O_3(s) Δ_rH° = -3351.4 kJ/mol$
Divide all the coefficients by 2, and do the same for the formation enthalpy.
$2 Al(s) + \frac{3}{2} O_2(g) --> 1 Al_2O_3(s) Δ_fH° = \frac{(-3351.4)}{2} kJ/mol$
$2 Al(s) + \frac{3}{2} O_2(g) --> Al_2O_3(s) Δ_fH° = -1675.7 kJ/mol$
