Let the given equations be: (i) $\frac{1}{2}\text{A} \rightarrow \text{B}; \Delta H_1 = +150\text{ kJ/mol}$ Multiplying equation (i) by 2, we get: $\text{A} \rightarrow 2\text{B}; \Delta H_4 = +300\text{ kJ/mol}$ ... (iv) (ii) $3\text{B} \rightarrow 2\text{C} + \text{D}; \Delta H_2 = -125\text{ kJ/mol}$ (iii) $\text{E} + \text{A} \rightarrow 2\text{D}; \Delta H_3 = +350\text{ kJ/mol}$ Reversing equation (iii), we get: $2\text{D} \rightarrow \text{E} + \text{A}; \Delta H_5 = -350\text{ kJ/mol}$ ... (v) Now, adding equations (iv), (ii), and (v): $(\text{A}) + (3\text{B}) + (2\text{D}) \rightarrow (2\text{B}) + (2\text{C} + \text{D}) + (\text{E} + \text{A})$ $\Rightarrow \text{B} + \text{D} \rightarrow \text{E} + 2\text{C}$ According to Hess's Law, the total enthalpy change for the reaction will be the sum of the enthalpy changes of these steps: $\Delta H = \Delta H_4 + \Delta H_2 + \Delta H_5$ $\Delta H = (+300) + (-125) + (-350) = -175\text{ kJ/mol}$