91Ó°ÊÓ

First, let's reverse the second given equation and adjust the coefficients to get CuO on the product side: 2. $$\mathrm{CuO}_{(s)} + \mathrm{Cu}_{(s)} \longrightarrow \mathrm{Cu}_{2}\mathrm{O}_{(s)} \qquad\qquad\Delta H^{\circ}= -11 \text{ kJ}$$ We have reversed the second equation and multiplied by 2 because if we add equations 1 and 2, \(\mathrm{Cu}_{2}\mathrm{O}_{(s)}\) in the reactants will be canceled, and we would get the formation reaction of \(\mathrm{CuO}(s)\) from the reactants \(\mathrm{O_{2}}(g)\) and \(\mathrm{Cu}(s)\), which we want.

Now, let's add the first and second equations and their enthalpy changes to create a new combined equation: 1. $$2 \mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CuO}(s) \qquad\qquad\Delta H^{\circ}=-288 \text{ kJ}$$ + 2. $$\mathrm{CuO}_{(s)} + \mathrm{Cu}_{(s)} \longrightarrow \mathrm{Cu}_{2}\mathrm{O}_{(s)} \qquad\qquad\Delta H^{\circ}= -11 \text{ kJ}$$ \(4 \mathrm{CuO}(s) + \mathrm{Cu}(s) \longrightarrow 2 \mathrm{Cu}_{2}\mathrm{O}(s)+ 2 \mathrm{Cu}_{2}\mathrm{O}(s)+\mathrm{O}_{2}(g)\) Simplifying the above equation gives: $$\mathrm{Cu}(s) + \dfrac{1}{2}\mathrm{O}_{2}(g) \longrightarrow \mathrm{CuO}(s)$$ Adding the enthalpy changes of the two equations according to Hess's Law: $$\Delta H^{\circ}_{\text{combined}}= -288 \mathrm{kJ} - 11 \mathrm{kJ} = -299 \mathrm{kJ}$$

We have obtained the new combined equation with the enthalpy change \(\Delta H^{\circ}_{\text{combined}}\). Since the enthalpy change of this combined equation represents the formation of one mole of solid copper oxide (CuO) from its elements, it is also the standard enthalpy of formation of CuO: $$\Delta H_{f}^{\circ}(\text{CuO}) = \Delta H^{\circ}_{\text{combined}} = -299 \text{ kJ/mol}$$ So, the standard enthalpy of formation of solid copper oxide (CuO) is -299 kJ/mol.