Answer
(a) -124.0 kJ/mol
(b) -1656 kJ/mol
(c) 624.6 kJ/mol
Work Step by Step
We find:
$\Delta H^{\circ}_{rxn}=\Sigma n_{p}\Delta H^{\circ}_{f}(products)-\Sigma n_{r}\Delta H_{f}^{\circ}(reactants)$
(a) $\Delta H^{\circ}_{rxn}=[\Delta H_{f}^{\circ}(N_{2}O,g)+2\Delta H_{f}^{\circ}(H_{2}O,l)]-[\Delta H_{f}^{\circ}(NH_{4}NO_{3},s)]$
$=[(82.05\,kJ/mol)+2(-285.8\,kJ/mol)]-(-365.6\,kJ/mol)$
$=-124.0\,kJ/mol$
(b) $\Delta H^{\circ}_{rxn}=[\Delta H_{f}^{\circ}(Fe_{2}O_{3},s)+4\Delta H_{f}^{\circ}(SO_{2},g)]-[2\Delta H_{f}^{\circ}(FeS_{2},s)+\frac{11}{2}\Delta H_{f}^{\circ}(O_{2},g)]$
$=[(-824.2\,kJ/mol)+4(-296.8\,kJ/mol)]-[2(-177.5\,kJ/mol)+\frac{11}{2}(0)]$
$=-1656\,kJ/mol$
(c) $\Delta H^{\circ}_{rxn}=[\Delta H_{f}^{\circ}(SiC,s)+2\Delta H_{f}^{\circ}(CO,g)]-[\Delta H_{f}^{\circ}(SiO_{2},s)+3\Delta H_{f}^{\circ}(C,graphite)]$
$=[(-65.3\,kJ/mol)+2(-110.5\,kJ/mol)]-[(-910.9\,kJ/mol)+3(0)]$
$=624.6\,kJ/mol$
