Answer
$-108.7\,kJ$
Work Step by Step
We find:
$\Delta H^{\circ}=\Sigma n_{p}\Delta H_{f}^{\circ}(products)-\Sigma n_{r}\Delta H_{f}^{\circ}(reactants)$
$=[4\,mol\times\Delta H_{f}^{\circ}(HNO_{3},aq)+2\,mol\times\Delta H_{f}^{\circ}(NO,g)]-[3\,mol\times\Delta H_{f}^{\circ}(N_{2}O_{4},g)+2\Delta H_{f}^{\circ}(H_{2}O,l)]$
$=[4\,mol(-207.4\,kJ/mol)+2\,mol(91.3\,kJ/mol)]-[3\,mol(11.1\,kJ/mol)+2(-285.8\,kJ/mol)]$
$=-108.7\,kJ$
