Answer
179.2 kJ
Work Step by Step
We find:
$\Delta H^{\circ}=\Sigma n_{p}\Delta H_{f}^{\circ}(products)-\Sigma n_{r}\Delta H_{f}^{\circ}(reactants)$
$=[1\,mol\times\Delta H_{f}^{\circ}(CaO,s)+1\,mol\times\Delta H_{f}^{\circ}(CO_{2},g)]-[1\,mol\times\Delta H_{f}^{\circ}(CaCO_{3},s)]$
$=[1\,mol(-634.9\,kJ/mol)+1\,mol(-393.5\,kJ/mol)]-[1\,mol(-1207.6\,kJ/mol)]$
$=179.2\,kJ$
