Answer
$19.0\,kJ/mol$
Work Step by Step
We find:
$\Delta H^{\circ}_{rxn}=\Sigma n_{p}\Delta H_{f}^{\circ}(products)-\Sigma n_{r}\Delta H_{f}^{\circ}(reactants)$
$=[3\Delta H_{f}^{\circ}(FeO,s)+\Delta H_{f}^{\circ}(CO_{2},g)]-[\Delta H_{f}^{\circ}(Fe_{3}O_{4},s)+\Delta H_{f}^{\circ}(CO,g)]$
$=[3(-272\,kJ/mol)+(-393.5\,kJ/mol)]-[(-1118\,kJ/mol)+(-110.5\,kJ/mol)]$
$=19.0\,kJ/mol$
