Answer
177.8 kJ/mol
Work Step by Step
$\Delta H_{rxn}^{\circ}=\Sigma n\Delta H_{f}^{\circ}(\text{products})-\Sigma m\Delta H_{f}^{\circ}(\text{reactants})$
where $m$ and $n$ denote the stoichiometric coefficients for the reactants and products.
Then,
$\Delta H_{rxn}^{\circ}=$
$[\Delta H_{f}^{\circ}(CaO)+\Delta H_{f}^{\circ}(CO_{2})]-\Delta H_{f}^{\circ}(CaCO_{3})$
$=[(-635.6\,kJ/mol)+(-393.5\,kJ/mol)]-(-1206.9\,kJ/mol)$
$=177.8\,kJ/mol$
