Answer
(a) 178.0 kJ/mol
(b) -595 kJ/mol
(c) 44.8 kJ/mol
Work Step by Step
$\Delta H^{\circ}_{rxn}=\Sigma n_{p}\Delta H^{\circ}_{f}(products)-\Sigma n_{r}\Delta H_{f}^{\circ}(reactants)$
We find:
(a) $\Delta H^{\circ}_{rxn}=[\Delta H_{f}^{\circ}(CaO,s)+\Delta H_{f}^{\circ}(CO_{2},g)]-[\Delta H_{f}^{\circ}(CaCO_{3},s)]$
$=[(-635.5\,kJ/mol)+(-393.5\,kJ/mol)]-(-1207\,kJ/mol)$
$=178.0\,kJ/mol$
(b) $\Delta H^{\circ}_{rxn}=[2\Delta H_{f}^{\circ}(HF,g)+\Delta H_{f}^{\circ}(I_{2},s)]-[2\Delta H_{f}^{\circ}(HI,g)+\Delta H_{f}^{\circ}(F_{2},g)]$
$=[2(-271\,kJ/mol)+(0\,kJ/mol)]-[2(26.5\,kJ/mol)+(0)]$
$=-595\,kJ/mol$
(c) $\Delta H^{\circ}_{rxn}=[6\Delta H_{f}^{\circ}(HF,g)+\Delta H_{f}^{\circ}(SO_{3},g)]-[\Delta H_{f}^{\circ}(SF_{6},g)+3\Delta H_{f}^{\circ}(H_{2}O,l)]$
$=[6(-271\,kJ/mol)+(-395.6\,kJ/mol)]-[(-1209\,kJ/mol)+3(-285.8\,kJ/mol)]$
$=44.8\,kJ/mol$
