Answer
$$3 Fe_2O_3(s) \longrightarrow \frac{1}{2} O_2(g) + 2 Fe_3O_4(s)$$
$ \Delta H^o_{rxn} = 236 \ kJ/mol$
Work Step by Step
1. Reverse the first equation and multiply it by 3:
$$3 Fe_2O_3(s) \longrightarrow 6 Fe(s) + \frac{9}{2}O_2(g)$$
$\Delta H^o_{rxn} = (-3)(-824.2 \ kJ/mol) = 2473 \ kJ/mol$
2. Multiply the second equation by 2:
$$6 Fe(s) + 4 O_2(g) \longrightarrow 2 Fe_3O_4(s)$$
$\Delta H^o_{rxn} = (2)(-1118.4 \ kJ/mol) = -2236.8 \ kJ/mol$
3. Sum the equations:
$$3 Fe_2O_3(s) + 6 Fe(s) + 4 O_2(g) \longrightarrow 6 Fe(s) + \frac{9}{2} O_2(g) + 2 Fe_3O_4(s)$$
$$3 Fe_2O_3(s) \longrightarrow \frac{1}{2} O_2(g) + 2 Fe_3O_4(s)$$
$ \Delta H^o_{rxn} = 2473 \ kJ/mol - 2236.8 \ kJ/mol = 236 \ kJ/mol$
