Question

Given $$2 \mathrm{Al}_{2} \mathrm{O}_{3}(s) \longrightarrow 4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H^{\circ}=3351.4 \mathrm{~kJ}$$ (a) What is the heat of formation of aluminum oxide? (b) What is \(\Delta H^{\circ}\) for the formation of \(12.50 \mathrm{~g}\) of aluminum oxide?

Short answer

Answer: The heat of formation of aluminum oxide is -1675.7 kJ/mol. The enthalpy change for the formation of 12.50 g of aluminum oxide is -205.38 kJ.

Step-by-step solution

(a) Heat of Formation of Aluminum Oxide#

Since we are given the equation for the decomposition of aluminum oxide, we can reverse the equation and find the heat of formation of aluminum oxide. To do this, we will reverse the ΔH° value: $$4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s) \quad \Delta H^{\circ}=-3351.4 \mathrm{~kJ}$$ Now, we need to adjust the stoichiometry of the equation to obtain the heat of formation per mol of aluminum oxide. We'll divide the enthalpy change by 2 to get the value per mol: $$\Delta H_{f}^{\circ}(\mathrm{Al}_{2} \mathrm{O}_{3})=\frac{-3351.4 \mathrm{~kJ}}{2}=-1675.7 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}$$

(b) Enthalpy Change for Formation of 12.50 g of Aluminum Oxide#

To find the ΔH° for the formation of 12.50 g of aluminum oxide, we first need to convert grams to moles using the molar mass of aluminum oxide: $$\mathrm{Molar \ Mass \ of \ Al}_{2} \mathrm{O}_{3}=(2 \times 26.98)+(3 \times 16.00)=101.96 \mathrm{~g} \cdot \mathrm{mol}^{-1}$$ Now, convert the mass of aluminum oxide to moles: $$\mathrm{Moles \ of \ Al}_{2} \mathrm{O}_{3}=\frac{12.50 \mathrm{~g}}{101.96 \mathrm{~g} \cdot \mathrm{mol}^{-1}} = 0.1226 \mathrm{~mol}$$ Finally, multiply the moles of aluminum oxide by its heat of formation to obtain the total enthalpy change for the formation of 12.50 g of aluminum oxide: $$\Delta H^{\circ}=-1675.7 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \times 0.1226 \mathrm{~mol}=-205.38 \mathrm{~kJ}$$ The ΔH° for the formation of approximately 12.50 g of aluminum oxide is -205.38 kJ.