Problem 25 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

For a dimerization reaction, \(2 \mathrm{~A}(\mathrm{~g}) \rightarrow \mathrm{A}_{2}(\mathrm{~g})\) at \(298 \mathrm{~K}, \Delta \mathrm{U}^{\Theta}=-20 \mathrm{~kJ} \mathrm{~mol}^{-1}, \Delta \mathrm{S}^{\Theta}=-30 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\), then the \(\Delta \mathrm{G}^{\Theta}\) will be \(\quad\) J.

Step-by-Step Solution

Answer

\( \Delta G^{\Theta} = -13537 \text{ J/mol} \).

Step 1: Identify Given Data

We are given the following data: \( \Delta U^{\Theta} = -20 \text{ kJ/mol} \), \( \Delta S^{\Theta} = -30 \text{ J K}^{-1} \text{mol}^{-1} \), and the temperature \( T = 298 \text{ K} \). Convert \( \Delta U^{\Theta} \) to Joules by noting that 1 kJ = 1000 J, so \( \Delta U^{\Theta} = -20000 \text{ J/mol} \).

Step 2: Understand Relation Between \( \Delta U \), \( \Delta H \) and \( \Delta G \)

The relation between the change in internal energy (\( \Delta U \)) and the enthalpy change (\( \Delta H \)) for a gaseous reaction is given by \( \Delta H = \Delta U + \Delta nRT \). Here, \( \Delta n \) is the change in moles of gas, which is \(-1\) for this reaction because 1 mole of gas is lost (2 mol of A form 1 mol of A2). \( R = 8.314 \text{ J K}^{-1} \text{mol}^{-1} \).

Step 3: Calculate Change in Enthalpy \( \Delta H \)

Calculate \( \Delta \text{n}RT \): \( \Delta n = -1 \), \( T = 298 \text{ K} \), \( R = 8.314 \text{ J K}^{-1} \text{mol}^{-1} \). Hence, \( \Delta nRT = -1 \times 8.314 \times 298 = -2477 \text{ J/mol} \). Now, \( \Delta H = \Delta U + \Delta nRT = -20000 - 2477 = -22477 \text{ J/mol} \).

Step 4: Calculate \( \Delta G \) Using Gibbs Free Energy Equation

Use the formula \( \Delta G^{\Theta} = \Delta H^{\Theta} - T\Delta S^{\Theta} \). Substitute the values: \( \Delta G^{\Theta} = -22477 - (298)(-30) \). Calculate \( T\Delta S^{\Theta} = 298 \times -30 = -8940 \text{ J/mol} \).

Step 5: Final Calculation of \( \Delta G \)

Substitute into the expression for \( \Delta G^{\Theta}: \Delta G^{\Theta} = -22477 + 8940 = -13537 \text{ J/mol} \). This gives us the change in Gibbs free energy for the reaction.

Key Concepts

Dimerization Reaction

A dimerization reaction involves the combination of two identical molecules to form a single dimer molecule. In the context of chemical reactions, this is particularly significant as it enables the formation of more complex structures from simpler elements or molecules. For example, in the dimerization of hydrogen cyanide (HCN) to form cyanoacetylene (HCCCN), each molecule of reactant acts as a monomer that comes together to form a dimer.
In the specific reaction we are discussing:

• The reaction is given by: \( 2 \ \mathrm{A}(\mathrm{g}) \rightarrow \mathrm{A}_{2}(\mathrm{g}) \).
• This indicates the transformation of two moles of the gaseous compound A into one mole of compound \( A_2 \).
• This is reflected in the change in the number of moles, namely \( \Delta n = -1 \).

This reduction in the number of gas molecules is important because it affects the associated changes in enthalpy and internal energy, providing deeper insights into the thermodynamic characteristics of the reaction.

Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), refers to the heat content change during a chemical reaction at constant pressure. It is a pivotal concept because it helps us understand how energy is absorbed or released during a process. For our example:

• Enthalpy change is calculated using the relation: \( \Delta H = \Delta U + \Delta nRT \).
• \( \Delta U \) is the internal energy change, and \( R \) is the ideal gas constant with a value of 8.314 J K\(^{-1}\) mol\(^{-1}\).
• \( T \) is the temperature in Kelvin, which is given as 298 K in this case.

For the dimerization reaction, the enthalpy change helps to determine how much heat is involved when two molecules come together to form a dimer. Negative enthalpy values, as calculated, suggest that the reaction is exothermic, releasing heat in the process, which is a common scenario for dimerization reactions.

Internal Energy Change

Internal energy change, represented by \( \Delta U \), relates to the total energy change within a system during a process that includes kinetic and potential energy across all molecules. It does not only represent the visible heat exchange but integrates the microscopic interactions within the molecules. This is fundamental for:

• Understanding how energy is conserved and transformed at a molecular level.
• Featuring in the calculation of other thermodynamic properties such as \( \Delta H \) (enthalpy change) and \( \Delta G \) (Gibbs free energy).
• In our calculation, \( \Delta U = -20000 \text{ J/mol} \).

Internal energy is an intrinsic property of the substances involved, indicating how energy transitions occur internally, contributing to our comprehensive understanding of energy transfer during chemical reactions.

Entropy Change

Entropy change, symbolized by \( \Delta S \), quantifies the degree of disorder or randomness within a system. In thermodynamics, it is crucial because it helps to determine the feasibility and spontaneity of a reaction. For the dimerization reaction, the entropy change is:

• Given as \( \Delta S^{\Theta} = -30 \text{ J K}^{-1} \text{mol}^{-1} \).
• The negative sign indicates that the disorder within the system decreases as the reaction progresses, typical of dimerization reactions where fewer but more orderly molecules are formed.
• Entropy change affects the Gibbs free energy \( \Delta G \), which further determines the spontaneity of the reaction.

By examining \( \Delta S \), we glean insight into how likely it is for a reaction to occur without external intervention, perpetuating a foundational understanding of thermodynamics as applied to chemical reactions.