Answer
$2Al(s) + \frac{3}{2}O_2(g) -- \gt Al_2O_3(s)$ ; $\Delta_fH^{\circ}\{Al_2O_3\}$ = $-1675.7 kJ/mol$
Work Step by Step
1. Write the separated atoms, and their coefficients, to form a single $Al_2O_3$:
$2Al + 3O -- \gt Al_2O_3(s)$
2. Analyze the appendix J, and find the standard state for the elements.
- Divide the coefficient by the number of atoms in each compound in standard state:
$\frac{2}{1} Al(s) + \frac{3}{2}O_2(g) -- \gt Al_2O_3(s)$
$2Al(s) + \frac{3}{2}O_2(g) -- \gt Al_2O_3(s)$ $\Delta_fH^{\circ}\{Al_2O_3\}$ = $-1675.7 kJ/mol$
