Chapter 4 - Energy and Chemical Reactions - Questions for Review and Thought - Topical Questions - Page 189e: 81a

$178.32\,kJ/mol$

$\Delta H^{\circ}=\Sigma n_{p}\Delta H_{f}^{\circ}(products)-\Sigma n_{r}\Delta H_{f}^{\circ}(reactants)$ $=[\Delta H_{f}^{\circ}(CaO,s)+\Delta H_{f}^{\circ}(CO_{2},g)]-[\Delta H_{f}^{\circ}(CaCO_{3},s)]$ $=[(-635.09\,kJ/mol)+(-393.509\,kJ/mol)]-(-1206.92\,kJ/mol)$ $=178.32\,kJ/mol$