Problem 47 - A Complete Resource Book in Chemistry for JEE Main

Question

For the reaction, \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 3 \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}\) (I) at constant temperature, \(\Delta \mathrm{H}-\Delta \mathrm{E}\) is (a) \(+3 \mathrm{RT}\) (b) \(-\mathrm{RT}\) (c) \(+\mathrm{RT}\) (d) \(-3 \mathrm{RT}\)

Step-by-Step Solution

Answer

The correct answer is (d) \(-3 \mathrm{RT}\).

Step 1: Understand the Reaction

The chemical equation given is for the combustion of propane: \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 3 \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(l)\). In this reaction, gases are the only reactants and products.

Step 2: Calculate Moles of Gases

Determine the number of gas moles in the reactants and products. Initially, we have 1 mole of \(\mathrm{C}_{3}\mathrm{H}_{8}\) and 5 moles of \(\mathrm{O}_{2}\), totaling 6 moles of gas in the reactants. In the products, we have 3 moles of \(\mathrm{CO}_{2}\) gas. Hence, the change in moles of gas (\(\Delta n_g\)) is given by: \(\Delta n_g = 3 - 6 = -3\).

Step 3: Relate Enthalpy and Internal Energy Changes

The relationship between the change in enthalpy (\(\Delta H\)) and the change in internal energy (\(\Delta E\)) at constant temperature and pressure is given by: \(\Delta H = \Delta E + \Delta n_gRT\). Here, \(\Delta n_g\) is the change in moles of gas, \(R\) is the gas constant, and \(T\) is temperature.

Step 4: Calculate \(\Delta H - \Delta E\)

In the formula \(\Delta H = \Delta E + \Delta n_gRT\), solving for \(\Delta H - \Delta E\) gives: \(\Delta H - \Delta E = \Delta n_gRT\). Substitute \(\Delta n_g = -3\) into the equation: \(\Delta H - \Delta E = -3RT\).

Step 5: Select the Correct Answer

The calculated value \(\Delta H - \Delta E = -3RT\) corresponds to option (d).

Key Concepts

Combustion Reaction

When we talk about a combustion reaction, we're referring to a type of chemical reaction where a substance combines with oxygen, releasing energy in the form of light or heat. Combustion reactions are exothermic, which means they give off energy. In the case of the reaction we have here, propane (\[ \text{C}_3\text{H}_8 \]) is burned in the presence of oxygen (\[ \text{O}_2 \]), to form carbon dioxide (\[ \text{CO}_2 \]) and water (\[ \text{H}_2\text{O} \]). This is a typical combustion reaction for hydrocarbons and is vital in powering engines and generating energy both in industrial and household settings.

• Propane (\[ \text{C}_3\text{H}_8 \]) acts as the fuel.
• Oxygen (\[ \text{O}_2 \]) is the oxidizer.
• The products are carbon dioxide and water, which are both stable, lower-energy forms.

Understanding combustion reactions is crucial, as it applies to several practical fields such as energy generation and environmental management.

Moles of Gases

In chemical reactions, especially those involving gases, it is important to account for the moles of gases involved on both the reactant and product sides of the equation. In the given reaction, we start with:

• 1 mole of propane (\[ \text{C}_3\text{H}_8 \])
• 5 moles of oxygen (\[ \text{O}_2 \])

This sums up to 6 moles of gas as reactants.
On the products side, we end up with:

• 3 moles of carbon dioxide (\[ \text{CO}_2 \])
• Note that water (\[ \text{H}_2\text{O} \]) is in liquid form here, so it does not contribute to the gaseous moles.

Therefore, the change in moles of gases (\[ \Delta n_g \]) is calculated as follows: \[ \Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 3 - 6 = -3 \]. Understanding this change is crucial for determining how the pressure and volume might be affected in a closed system as the reaction proceeds.

Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), represents the total heat content change during a reaction, at constant pressure. When a chemical reaction occurs, bonds break and new bonds form, leading to an exchange of energy with the surroundings.
For combustion reactions, this is typically a negative value, indicating that energy is being released. In our case, we're dealing with a special relationship where the enthalpy change needs to factor in not just the energy aspect but also the physical changes in gas moles:

• The formula connecting enthalpy change to internal energy and gas moles change is: \[ \Delta H = \Delta E + \Delta n_gRT \].
• \( R \) is the ideal gas constant, and \( T \) is the temperature.

Therefore, the changes in gaseous moles result in additional enthalpy considerations, which are necessary for a complete energy analysis of the reaction.

Internal Energy Change

The change in internal energy, represented as \( \Delta E \), refers to the energy changes within the system itself, excluding the effects of pressure-volume work. It encompasses the energy required to break bonds in the reactants and form new bonds in the products. In thermodynamics, understanding the distinction between \( \Delta E \) and \( \Delta H \) is essential.

• While \( \Delta H \) considers both the internal energy and the work of expanding against atmospheric pressure, \( \Delta E \) focuses solely on the internal IP transformations.
• The relationship is given by: \[ \Delta H = \Delta E + \Delta n_gRT \]

When the moles of gases change, as in our exercise with \( \Delta n_g = -3 \), it means the expansion work done by the gases also changes. This work, expressed as \( -RT \times \Delta n_g \), shifts the perceived energy content of the reaction, making it crucial to account for both types of energy changes to fully capture the reaction's dynamics.