Answer
e) -110.5 kJ/mol
Work Step by Step
From the first reaction
$\Delta H_{rxn}^{\circ}=\Delta H_{f}^{\circ}(CO_{2}(g))-[\Delta H_{f}^{\circ}(C(graphite))+\Delta H_{f}^{\circ}(O_{2}(g))]$
As $O_{2}$ is the most stable form of oxygen and graphite is the most stable allotrope of carbon $\Delta H_{f}^{\circ}(C(graphite))=\Delta H_{f}^{\circ}(O_{2}(g))=0$
$\implies -393.5\,kJ/mol=\Delta H_{f}^{\circ}(CO_{2}(g))$
From the second reaction,
$\Delta H_{rxn}^{\circ}=\Delta H_{f}^{\circ}(CO_{2}(g))-[\Delta H_{f}^{\circ}(CO(g))+\frac{1}{2}\times\Delta H_{f}^{\circ}(O_{2}(g))]$
$\implies -283.0\,kJ/mol=-393.5\,kJ/mol-[\Delta H_{f}^{\circ}(CO(g))+ \frac{1}{2}\times0]$
$\implies \Delta H_{f}^{\circ}(CO(g))=-393.5\,kJ/mol+283.0\,kJ/mol$
$=-110.5\,kJ/mol$
