Answer
$-1134.4\,kJ/mol$
Work Step by Step
$\Delta _{r}G^{\circ}=\Sigma n_{p}\Delta _{f}G^{\circ}(products)-\Sigma n_{r}\Delta _{f}G^{\circ}(reactants)$
$=[\Delta_{f}G^{\circ}(BaO,s)+\Delta_{f}G^{\circ}(CO_{2},g)]-[\Delta_{f}G^{\circ}(BaCO_{3},s)]$
$\implies 219.7\,kJ/mol=[(-520.38\,kJ/mol)+(-394.359\,kJ/mol)]-[\Delta_{f}G^{\circ}(BaCO_{3},s)]$
$\implies \Delta_{f}G^{\circ}(BaCO_{3},s)=[(-520.38\,kJ/mol)+(-394.359\,kJ/mol)]-(219.7\,kJ/mol)$
$=-1134.4\,kJ/mol$
