Chapter 4 - Energy and Chemical Reactions - Questions for Review and Thought - General Questions - Page 189f: 98a

$-69.14\,kJ/mol$

$\Delta H^{\circ}=\Sigma n_{p}\Delta H_{f}^{\circ}(products)-\Sigma n_{r}\Delta H_{f}^{\circ}(reactants)$ $=[\Delta H_{f}^{\circ}(CaCO_{3},s)+\Delta H_{f}^{\circ}(H_{2}O,g)]-[\Delta H_{f}^{\circ}(Ca(OH)_{2},s)+\Delta H_{f}^{\circ}(CO_{2},g)]$ $=[(-1206.92\,kJ/mol)+(-241.818\,kJ/mol)]-[(-986.09\,kJ/mol)+(-393.509\,kJ/mol)]$ $=-69.14\,kJ/mol$