Question

The standard heat of combustion of \(\mathrm{Al}\) is \(-837.8 \mathrm{~kJ}\) \(\mathrm{mol}^{-1}\) at \(25^{\circ} \mathrm{C}\). If \(\mathrm{Al}\) reacts with \(\mathrm{O}_{2}\) at \(25^{\circ} \mathrm{C}\), which of the following releases \(250 \mathrm{kcal}\) of heat? (a) the reaction of \(0.312 \mathrm{~mol}\) of \(\mathrm{Al}\) (b) the formation of \(0.624 \mathrm{~mol}\) of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) (c) the reaction of \(0.712 \mathrm{~mol}\) of \(\mathrm{Al}\) (d) the formation of \(0.615 \mathrm{~mol}\) of \(\mathrm{A} 1 \mathrm{O}_{3}\)

Short answer

Option (b) releases 250 kcal of heat.

Step-by-step solution

Convert kilojoules to kilocalories

To solve the problem, we first need to convert the heat energy given in kilojoules to kilocalories, because the energy released is mentioned in kilocalories. Using the conversion factor: \[ 1 ext{ kcal} = 4.184 ext{ kJ} \]Convert the given heat of combustion: \[ 837.8 ext{ kJ/mol} = \frac{837.8}{4.184} ext{ kcal/mol} \approx 200.3 ext{ kcal/mol} \]

Calculate moles needed for 250 kcal

We need to calculate how many moles of \( \text{Al} \) would theoretically release 250 kcal of heat. Set up the equation to solve for moles: \[ \text{Heat released} = \text{Moles of Al} \times 200.3 \text{ kcal/mol} \]\[ \text{Moles of Al} = \frac{250 \text{ kcal}}{200.3 \text{ kcal/mol}} \approx 1.248 \text{ mol} \]

Analyze each option

Now we will analyze each option given in the exercise:- Option (a): 0.312 mol of \( \text{Al} \) would release \[0.312 \times 200.3 \approx 62.5 \text{ kcal} \] (less than 250 kcal).- Option (b): Formation of \( 0.624 \text{ mol of Al}_2 \text{O}_3 \) involves 2 moles of \( \text{Al} \) for 1 mole of \( \text{Al}_2 \text{O}_3 \). Since the molar requirement for the reaction is doubly related due to \( \text{Al}_2 \text{O}_3 \) formation:\[2 \times 0.624 = 1.248 \text{ mol of Al}\] which releases exactly 250 kcal.- Option (c): 0.712 mol of \( \text{Al} \) would release \[0.712 \times 200.3 \approx 142.7 \text{ kcal} \] (less than 250 kcal).- Option (d): Formation of \( 0.615 \text{ mol of AlO}_3 \) is not a recognized chemical formula, so it is invalid for our purposes.

Choose the correct answer

Based on the analysis, option (b) matches the required condition exactly, as it yields 250 kcal by reacting the corresponding amount of aluminum with oxygen.

Key concepts

Heat of Combustion

When a chemical reaction occurs, such as the combustion of a metal or hydrocarbon, it releases energy. This release of energy is quantified as the "heat of combustion." For instance, when aluminum ocombines with oxygen, producing aluminum oxide, it releases heat. The heat of combustion serves as a crucial parameter in determining how much energy you can expect from such a reaction.
In this scenario, the heat of combustion for aluminum is given as \(-837.8 \ \text{kJ/mol}\). This means that when one mole of aluminum undergoes combustion, \(837.8\) kilojoules of energy are released as heat. This negative sign symbolizes an exothermic reaction, where energy exits the system. Understanding these values is vital for applications involving energy production, such as in rockets or power plants.
By converting this to kilocalories, using the conversion factor \(1 \ \text{kcal} = 4.184 \ \text{kJ}\), we find that the combustion releases about \(200.3\) kcal/mol.

Energy Conversion

Energy conversion is the process of changing one form of energy into another, such as converting heat into mechanical energy. In thermochemistry, we often need to express energy in different units. This helps in consistent calculations across various natural processes or industrial applications.
Converting the energy involved in chemical reactions into the same units makes it easier to compare and calculate the energy needs. For aluminum's combustion, initially given in kilojoules, it was essential to convert it to kilocalories for precise comparison and analysis. Hence, knowing conversion factors, like \(1 \ \text{kcal} = 4.184 \ \text{kJ}\), is crucial.

• 1 kcal = 4.184 kJ
• Energy from reactions can be expressed in different units depending on the context.
• Familiarizing with conversion factors enables better understanding and problem-solving in chemistry.

This capability to effortlessly switch between units can greatly aid in solving complex problems in the realm of thermodynamics.

Stoichiometry

Stoichiometry is the method in chemistry used to determine the proportions of elements in chemical reactions. It involves calculations based on balanced chemical equations. With stoichiometry, chemists predict the quantities of reactants needed and products formed in given chemical reactions.
In the example of forming aluminum oxide from aluminum and oxygen, stoichiometry tells us we need two moles of aluminum to form one mole of aluminum oxide (\(\text{Al}_2\text{O}_3\)). This relationship ensures that all reactants are used efficiently to form products without waste.
Applying the stoichiometric relationship in this case, we conclude that for each mole of \(\text{Al}_2\text{O}_3\) formed, we consume two moles of \(\text{Al}\). This fundamental principle of stoichiometry is what permits the precise calculation of energy released during specific reactions.

Chemical Reactions

Chemical reactions involve transforming substances through the breaking and forming of chemical bonds, leading to the production of different substances. A common example is the combustion of metals like aluminum. When aluminum reacts with oxygen, it forms aluminum oxide, a stable compound. This reaction, apart from yielding a product, also releases energy in the form of heat.
Chemical reactions can be classified broadly as exothermic or endothermic:

• Exothermic reactions: Release heat (e.g., combustion of aluminum).
• Endothermic reactions: Absorb heat during the process.

In practical applications, understanding these conditions helps in determining whether a reaction can be harnessed for generating or consuming energy.