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

(a) The heat of formation for 1 mole of aluminum oxide is 1675.7 kJ. (b) The heat change for the formation of 12.50 grams of aluminum oxide is approximately 205 kJ.

Step-by-step solution

(a) Heat of Formation of Aluminum Oxide

We are given the heat change for the decomposition of aluminum oxide: $$ 2 \mathrm{Al}_2 \mathrm{O}_3(s) \longrightarrow 4 \mathrm{Al}(s)+3 \mathrm{O}_2(g) \quad \Delta H^{\circ} = -3351.4 \mathrm{~kJ} $$ The above reaction is the decomposition of 2 moles of aluminum oxide into 4 moles of aluminum and 3 moles of oxygen. We can express the formation of aluminum oxide as: $$ 4 \mathrm{Al}(s) + 3 \mathrm{O}_2(g) \longrightarrow 2 \mathrm{Al}_2 \mathrm{O}_3(s) $$ From the given reaction, $$ \Delta H^{\circ}_\textrm{decomposition} = -3351.4 \mathrm{~kJ} $$ We know \(\Delta H^{\circ}_\textrm{formation} = -\Delta H^{\circ}_\textrm{decomposition}\), which gives us: $$ \Delta H^{\circ}_\textrm{formation} = -(-3351.4 \mathrm{~kJ}) = 3351.4 \mathrm{~kJ} $$ So, the heat of formation for 2 moles of aluminum oxide is 3351.4 kJ. For 1 mole of aluminum oxide, the heat of formation is: $$ \Delta H^{\circ}_\textrm{formation(1 mole)} = \frac{3351.4 \mathrm{~kJ}}{2} = 1675.7 \mathrm{~kJ} $$

(b) \(\Delta H^{\circ}\) for 12.50g of Aluminum Oxide

Now, we need to calculate how many moles of aluminum oxide are represented by 12.50 grams. The molar mass of aluminum oxide is: $$ \mathrm{M_{Al_2O_3}} = (2 \times \mathrm{M_{Al}}) + (3 \times \mathrm{M_{O}}) = 2 \times 26.98 + 3 \times 16 = 101.96 \mathrm{\frac{g}{mol}} $$ The number of moles of aluminum oxide in 12.50 grams can be calculated: $$ \mathrm{moles_{Al_2O_3}} = \frac{\mathrm{mass_{Al_2O_3}}}{\mathrm{M_{Al_2O_3}}} = \frac{12.50 \mathrm{g}}{101.96 \mathrm{\frac{g}{mol}}} = 0.1225 \mathrm{mol} $$ Now, we can find the heat change for the formation of 12.50 grams (0.1225 moles) of aluminum oxide using the heat of formation for 1 mole of aluminum oxide: $$ \Delta H^{\circ}_\textrm{formation(12.50g)} = 0.1225 \mathrm{mol} \times 1675.7 \mathrm{\frac{kJ}{mol}} = 205 \mathrm{~kJ} $$ So, the heat change for the formation of 12.50 grams of aluminum oxide is approximately 205 kJ.

Key concepts

Heat of formation

The heat of formation is a crucial concept in thermochemistry. It refers to the heat change that occurs when one mole of a compound is formed from its constituent elements in their standard states. For aluminum oxide (\( \mathrm{Al}_2 \mathrm{O}_3} \)), this involves combining elemental aluminum (\( \mathrm{Al}(s) \)) and oxygen gas (\( \mathrm{O}_2(g) \)). When you know the heat involved in the decomposition process, you can find the heat of formation by reversing the sign of the enthalpy change.

• In this exercise, the decomposition of 2 moles of aluminum oxide has an enthalpy change (\( \Delta H^{\circ} \)) of -3351.4 kJ.
• Consequently, the heat required to form 2 moles of aluminum oxide is +3351.4 kJ, because formation is the reverse of decomposition.
• To find the heat of formation for 1 mole, we divide this value by 2, resulting in +1675.7 kJ.

Understanding the heat of formation can help in predicting whether reactions are spontaneous and how much energy is required or released during chemical processes.

Enthalpy change

Enthalpy change is a key topic in thermochemistry, signifying the heat absorbed or released during a chemical reaction at constant pressure. It allows us to grasp how a process affects a system's energy. For reactions like the formation or decomposition of aluminum oxide, we use this concept to assess energy transformations.

• The enthalpy change for the decomposition of aluminum oxide shows how much heat is released when the compound breaks down into its elemental parts.
• Conversely, formation reactions require an equivalent amount of heat, but their signs are positive due to heat absorption.
• Essentially, the enthalpy change provides a numeric value representing the difference in energy between reactants and products.

By understanding and calculating enthalpy changes, you can predict how different substances will interact energetically during a reaction.

Aluminum oxide decomposition

The decomposition of aluminum oxide is an interesting reaction that demonstrates the conversion of a solid compound into its elements, aluminum and oxygen.

• This process is expressed in the reaction: \\[ 2 \mathrm{Al}_2 \mathrm{O}_3(s) \longrightarrow 4 \mathrm{Al}(s)+3 \mathrm{O}_2(g) \]
• In this decomposition, energy is released, denoted by a negative enthalpy change. This means that the reaction is exothermic.
• Conversely, forming aluminum oxide from its elements requires the same amount of energy, but it is absorbed, indicating an endothermic process.
• Understanding the energetics of aluminum oxide decomposition helps in designing and controlling industrial processes involving this compound.

These concepts show the dual nature of chemical reactions, where decomposition and formation balance energy among reactants and products.