Answer
$-11.3\,kJ/mol$
Work Step by Step
We find:
$\Delta H^{\circ}_{rxn}=\Sigma n_{p}\Delta H_{f}^{\circ}(products)-\Sigma n_{r}\Delta H_{f}^{\circ}(reactants)$
$=[\Delta H_{f}^{\circ}(Cu_{2}O,s)]-[\Delta H_{f}^{\circ}(CuO,s)+\Delta H_{f}^{\circ}(Cu,s)]$
$=-168.6\,kJ/mol-[-157.3\,kJ/mol+0]$
$=-11.3\,kJ/mol$
