Problem 9 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

For a reaction, \(\mathrm{A}(\mathrm{g}) \rightarrow \mathrm{A}(\mathrm{l}) ; \Delta \mathrm{H}=-3 \mathrm{RT}\). The correct statement for the reaction is: (a) \(\Delta \mathrm{H}=\Delta \mathrm{U} \neq \mathrm{O}\) (b) \(\Delta \mathrm{H}=\Delta \mathrm{U}=\mathrm{O}\) (c) \(|\Delta \mathrm{H}|<|\Delta \mathrm{U}|\) (d) \(|\Delta \mathrm{H}|>|\Delta \mathrm{U}|\)

Step-by-Step Solution

Answer

The correct answer is (d) \(|\Delta \mathrm{H}| > |\Delta \mathrm{U}|\).

Step 1: Understanding Enthalpy Change

Given that for the reaction \( \mathrm{A}(\mathrm{g}) \rightarrow \mathrm{A}(\mathrm{l}) \), \( \Delta \mathrm{H} = -3 \mathrm{RT} \). This indicates that the enthalpy change (\( \Delta \mathrm{H} \)) for this reaction is negative, meaning it is an exothermic reaction where heat is released as the gas condenses into a liquid.

Step 2: Relating Enthalpy and Internal Energy

The relationship between the change in enthalpy (\( \Delta H \)) and the change in internal energy (\( \Delta U \)) is given by the equation: \( \Delta H = \Delta U + \Delta n_{gas}RT \). Here, \( \Delta n_{gas} \) changes as the moles of gaseous substances change from reactants to products.

Step 3: Calculate Change in Moles of Gas

In this reaction, \( \mathrm{A}(\mathrm{g}) \rightarrow \mathrm{A}(\mathrm{l}) \), the gaseous moles decrease from 1 to 0, thus \( \Delta n_{gas} = 0 - 1 = -1 \). Substitute \( \Delta n_{gas} = -1 \) into the relationship \( \Delta H = \Delta U + \Delta n_{gas}RT \).

Step 4: Substitute and Simplify Equation

Substitute \( \Delta n_{gas} = -1 \) into \( \Delta H = \Delta U - RT \). This implies \( \Delta H = \Delta U - RT \). Given that \( \Delta H = -3RT \), we can write \( -3RT = \Delta U - RT \), which simplifies to \( \Delta U = -3RT + RT = -2RT \).

Step 5: Compare Magnitudes of \( \Delta H \) and \( \Delta U \)

From \( \Delta H = -3RT \) and \( \Delta U = -2RT \), the absolute value \(|\Delta H| = 3RT\) is greater than \(|\Delta U| = 2RT\). Therefore, \(|\Delta H| > |\Delta U|\).

Key Concepts

Enthalpy Change

When we talk about enthalpy change (\( \Delta \mathrm{H} \)), we're discussing the heat energy change in a chemical reaction. This concept is an essential aspect of thermodynamics, helping us understand how energy is transferred in reactions. In our specific reaction, \( \mathrm{A}(\mathrm{g}) \to \mathrm{A}(\mathrm{l}) \), the enthalpy change is \( -3 \mathrm{RT} \). This negative value signals that energy is released during the process, a hallmark of an exothermic reaction.

In simpler terms:

• \( \Delta \mathrm{H} \) represents the heat exchange at constant pressure.
• A negative \( \Delta \mathrm{H} \) indicates an exothermic process where heat exits the system.

Understanding enthalpy helps us predict how much energy is needed or released in a reaction, crucial for both practical applications like engineering and theoretical studies like chemical kinetics.

Internal Energy

Internal energy (\( \Delta \mathrm{U} \)) deals with the total energy within a system, encompassing both kinetic and potential energy of molecules. For our reaction, understanding the change in internal energy clarifies how energy behaves when a gas becomes a liquid. The formula linking enthalpy change and internal energy is:\[\Delta \mathrm{H} = \Delta \mathrm{U} + \Delta n_{\text{gas}}RT\]This equation highlights how changes in gas moles affect internal energy.

For \( \mathrm{A}(\mathrm{g}) \to \mathrm{A}(\mathrm{l}) \):

• Moles of gas reduce, hence \( \Delta n_{\text{gas}} = -1 \).
• We adapt this into our equation, resulting in \( \Delta \mathrm{U} = -2RT \), indicating an internal change as energy condenses.

Comprehending internal energy is vital as it gives insight into the unseen molecular forces and helps understand broader thermodynamic principles.

Exothermic Reaction

An exothermic reaction is one where energy is emitted, often as heat. The change \( \Delta \mathrm{H} = -3RT \) in the reaction \( \mathrm{A}(\mathrm{g}) \to \mathrm{A}(\mathrm{l}) \) indicates it's exothermic. This release of energy is characteristic of processes that move from a higher-energy state to a lower-energy one.

In this case:

• The gas \( \mathrm{A} \) transitions to liquid \( \mathrm{A} \), losing energy.
• Heat is released to the surroundings, typically increasing the temperature of the environment.
• We found that \(|\Delta \mathrm{H}| > |\Delta \mathrm{U}|\), meaning the energy released as heat is greater compared to energy changes within the system itself.

Recognizing exothermic reactions is beneficial in various fields, providing safer and more efficient chemical process designs.