Answer
e) -58.04 kJ/mol
Work Step by Step
$\Delta H_{rxn}^{\circ}=$$\Delta H_{f}^{\circ}(N_{2}O_{4}(g))-[2\times\Delta H_{f}^{\circ}(NO_{2}(g))]$
$=9.66\,kJ/mol-(2\times33.85\,kJ/mol)$
(values are substituted from appendix 2)
$=-58.04\,kJ/mol$
