Problem 114 - A Complete Resource Book in Chemistry for JEE Main

Question

The standard heat of combustion of \(\mathrm{Al}\) is \(-837.8 \mathrm{~kJ}\) \(\mathrm{mol}^{-1}\) at \(25^{\circ} \mathrm{C}\). If 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 \(0.624\) mol of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) (c) the reaction of \(0.712\) mol of \(\mathrm{Al}\) (d) the formation of \(0.615 \mathrm{~mol}\) of \(\mathrm{Al}_{2} \mathrm{O}_{3}\)

Step-by-Step Solution

Answer

None of the given options releases exactly 250 kcal or 1046 kJ of heat.

Step 1: Convert Heat from kcal to kJ

Convert the heat value from kcal to kJ, since the standard heat of combustion is given in kJ. We know that 1 kcal is equal to 4.184 kJ. Thus, \( 250 \text{ kcal} = 250 \times 4.184 = 1046 \text{ kJ} \). So, we need to find a reaction that releases 1046 kJ of heat.

Step 2: Determine Heat Released per Reaction

The standard heat of combustion of \( \mathrm{Al} \) is \(-837.8 \text{ kJ/mol} \). This means that the reaction \( 4\mathrm{Al(s)} + 3\mathrm{O}_2\mathrm{(g)} \rightarrow 2\mathrm{Al}_2\mathrm{O}_3\mathrm{(s)} \) releases \(-2 \times 837.8 \text{ kJ} = -1675.6 \text{ kJ} \) because two moles of \( \mathrm{Al}_2\mathrm{O}_3 \) are formed.

Step 3: Calculate Heat for Given Moles

We look for the candidate options and compute the heat released if \( \mathrm{Al} \) or \( \mathrm{Al}_2\mathrm{O}_3 \) reacts/formed. For option (a), the heat for reaction of \(0.312 \text{ mol of } \mathrm{Al} \) is calculated as: \( 0.312 \times 837.8 \text{ kJ/mol} = 261.4 \text{ kJ} \). For option (b), since \(0.624 \text{ mol of } \mathrm{Al}_2\mathrm{O}_3 \) is formed, the released is: \( 0.624 \times (1675.6 \text{ kJ} / 2) = 522.67 \text{ kJ} \).For option (c), with \(0.712 \text{ mol of } \mathrm{Al} \) it gives: \( 0.712 \times 837.8 = 596.41 \text{ kJ} \).Lastly, for option (d) \(0.615 \text{ mol of } \mathrm{Al}_2\mathrm{O}_3 \): \(0.615 \times (1675.6 \text{ kJ} /2 ) = 515.67 \text{ kJ} \).

Step 4: Match the Required Heat Release

Compare the calculated heat releases to the target heat release of \( 1046 \text{ kJ} \). None of the options yields a release of \( 1046 \text{ kJ} \). Therefore, we must assume there's a need to form a calculative approach repeating the reaction.

Key Concepts

Standard Heat of Combustion

The standard heat of combustion is a crucial concept in thermochemistry. It represents the amount of heat released when one mole of a substance completely reacts with oxygen under standard conditions, which are typically at a pressure of 1 atmosphere and a temperature of 25°C.

• Always remember that the standard heat of combustion is expressed in kilojoules per mole (kJ/mol).
• A negative value indicates an exothermic reaction, meaning heat is released during the process.

For example, the standard heat of combustion of aluminum (Al) is \(-837.8 \, \text{kJ/mol}\). This value tells us that when one mole of aluminum reacts with oxygen, it releases \(837.8 \, \text{kJ}\) of heat.

Knowing this value is important when calculating how much heat a certain amount of aluminum will release when it reacts, as it allows us to apply this understanding in larger or smaller scales, tailoring the measurement to specific amounts of reactants.

Enthalpy Change

Enthalpy change (2;H) is a central idea in thermochemistry, representing the heat change at constant pressure in a chemical reaction.

• Enthalpy change can be either positive or negative.
• A positive amount indicates an endothermic reaction, where heat is absorbed.
• A negative amount signifies an exothermic reaction, where heat is released.

In the context of combustion reactions, enthalpy change helps determine the total heat output of a reaction. When aluminum combusts to form aluminum oxide, the entire reaction can be tracked through its enthalpy change. This is expressed as:\[4\text{Al(s)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{Al}_2\text{O}_3\text{(s)}\]
The enthalpy change for this reaction is \(-1675.6 \, \text{kJ}\) for two moles of \(\text{Al}_2\text{O}_3\) formed, calculated by multiplying the standard heat of combustion by two, since two moles of aluminum oxide are produced.

This enthalpy change gives us a practical approach to predict the energy release for different quantities of aluminum or \(\text{Al}_2\text{O}_3\). By using this value, we can calculate the energy output for various scenarios such as those given in the problem statement.

Stoichiometry

Stoichiometry is the quantitative study of reactants and products in a chemical reaction. It allows us to predict the amount of product formed from specific reactant quantities or the amount of reactant needed for a desired amount of product.
When working with stoichiometry, consider the balanced chemical equation, which in this context is:\[4\text{Al(s)} + 3\text{O}_2\text{(g)} \rightarrow 2\text{Al}_2\text{O}_3\text{(s)}\]

• From the equation, four moles of aluminum react with three moles of oxygen to produce two moles of \(\text{Al}_2\text{O}_3\).
• The coefficients in the equation help us determine the mole ratio between reactants and products.

In solving problems like finding which reaction releases a specific amount of heat, stoichiometry helps to relate moles of aluminum or alumina to the total heat released.
For example, knowing that the reaction forms two moles of alumina, we divide the fixed heat value by two to find the heat associated with one mole.
This approach helps solve problems related to energy changes in reactions by calculating how much heat is released for different initial quantities. With stoichiometry, the relationship between amounts of reactants and products can be clearly understood, ensuring accuracy in chemical computations.