\begin{aligned} & \frac{1}{2} A \longrightarrow B ; \quad \Delta H=150 \mathrm{~kJ} / \mathrm{mol} \\ & 3 B \longrightarrow 2 C+D ; \Delta H=-125 \mathrm{~kJ} / \mathrm{mol} \\ & E+A \longrightarrow 2 D ; \Delta H=+350 \mathrm{~kJ} / \mathrm{mol} \\ & \hline \text { By }[2 \times(\mathrm{i})+(\mathrm{ii})]-(\mathrm{iii}), \text { we have } \\ & B+D \longrightarrow E+2 C \\ & \therefore \quad \Delta H=150 \times 2+(-125)-350 \\ & \quad=-175 \mathrm{~kJ} / \mathrm{mol} . \end{aligned}