Given $$ 2 \mathrm{CuO}(s) \longrightarrow 2 \mathrm{Cu}(s)+\mathrm{O}_{2}(g) \quad \Delta H^{\circ}=314.6 \mathrm{~kJ} $$ (a) Determine the heat of formation of \(\mathrm{CuO}\). (b) Calculate \(\Delta H^{\circ}\) for the formation of \(13.58 \mathrm{~g}\) of \(\mathrm{CuO}\).

To find the heat of formation of CuO, we need to use the given standard enthalpy change value. Given the balanced chemical equation, we see that 2 moles of CuO decompose to form 2 moles of Cu and 1 mole of O2 gas. The standard enthalpy change of this reaction is 314.6 kJ. If we divide the enthalpy change by the number of moles of CuO in the reaction, we can find the heat of formation for one mole of CuO: $$ \Delta H_{f}^{\circ}(\mathrm{CuO}) = \frac{\Delta H^{\circ}}{\text{moles of CuO decomposed}} = \frac{314.6 \mathrm{k~J}}{2 \mathrm{~moles~of~CuO}} = 157.3 \mathrm{~kJ/mol} $$ Therefore, the heat of formation of one mole of CuO is 157.3 kJ/mol.

To find the enthalpy change for the formation of 13.58 g of CuO, we first need to convert grams to moles using the molar mass of CuO. Then, we'll multiply the number of moles with the heat of formation value we calculated in part (a). 1. Calculate the number of moles of CuO: Molar mass of Cu = 63.55 g/mol Molar mass of O = 16.00 g/mol Molar mass of CuO = 63.55 + 16.00 = 79.55 g/mol $$ \text{moles of CuO} = \frac{\text{mass of CuO}}{\text{molar mass of CuO}} = \frac{13.58 \mathrm{~g}}{79.55 \mathrm{~g/mol}} = 0.1707 \mathrm{~mol} $$ 2. Calculate ΔH° for the formation of 13.58 g of CuO: $$ \Delta H = \Delta H_{f}^{\circ}(\mathrm{CuO}) \times \text{moles of CuO} = 157.3 \mathrm{~kJ/mol} \times 0.1707 \mathrm{~mol} = 26.84 \mathrm{~kJ} $$ Thus, the enthalpy change for the formation of 13.58 g of CuO is 26.84 kJ.