Answer
$-53.9\,kJ$
The reaction is spontaneous.
Work Step by Step
$\Delta H^{\circ}=\Sigma n_{p}\Delta H_{f}^{\circ}(products)-\Sigma n_{r}\Delta H_{f}^{\circ}(reactants)$
$=[1\,mol\times\Delta H_{f}^{\circ}(CaO,s)+1\,mol\times\Delta H_{f}^{\circ}(CO_{2},g)]-[1\,mol\times\Delta H_{f}^{\circ}(CaCO_{3},s)]$
$=[1\,mol(-634.9\,kJ/mol)+1\,mol(-393.5\,kJ/mol)]-[1\,mol(-1207.6\,kJ/mol)]$
$=179.2\,kJ$
$\Delta S^{\circ}=\Sigma n_{p}S^{\circ}(products)-\Sigma n_{r}S^{\circ}(reactants)$
$=[S^{\circ}(CaO,s)+S^{\circ}(CO_{2},g)]-[S^{\circ}(CaCO_{3},s)]$
$=[1\,mol(38.1\,Jmol^{-1}K^{-1})+1\,mol(213.8\,Jmol^{-1}K^{-1})]-[1\,mol(91.7\,Jmol^{-1}K^{-1})]$
$=160.2\,JK^{-1}=0.1602\,kJK^{-1}$
$\Delta G=\Delta H-T\Delta S$
$=(179.2\,kJ)-(1455\,K)(0.1602\,kJK^{-1})$
$=-53.9\,kJ$
Because $\Delta G$ is negative, the reaction is spontaneous.
