Chemistry - Thermochemistry Question with Solution | TestHub

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