Problem 73 - A Complete Resource Book in Chemistry for JEE Main

Question

For a reaction at \(300 \mathrm{~K}\), enthalpy and entropy changes are \(-11.5 \times 10^{3} \mathrm{~J} \mathrm{~mol}^{-1}\) and \(-105 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) respectively. What is the change in Gibbs free energy? (a) \(25 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(30 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(15 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(20 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The change in Gibbs free energy is 20 kJ/mol, which is option (d).

Step 1: Understand Gibbs Free Energy Formula

The Gibbs free energy change (9G) can be calculated using the formula: \[\Delta G = \Delta H - T\Delta S\] where \(\Delta H\) is the change in enthalpy, \(T\) is the temperature in Kelvin, and \(\Delta S\) is the change in entropy.

Step 2: Substitute Given Values

Substitute the given values into the Gibbs free energy formula. We have:\(\Delta H = -11.5 \times 10^3 \mathrm{~J~mol}^{-1}\),\(T = 300 \mathrm{~K}\),\(\Delta S = -105 \mathrm{~J~K}^{-1} \mathrm{~mol}^{-1}\).Thus:\[\Delta G = (-11.5 \times 10^3) - (300 \times -105)\]

Step 3: Calculate the Temperature Times Entropy Term

Calculate the \(T\Delta S\) term:\[T\Delta S = 300 \times -105 = -31500 \mathrm{~J} \/ \mathrm{mol}^{-1}\]

Step 4: Solve for Gibbs Free Energy Change

Substitute \(T\Delta S\) back into the Gibbs free energy equation:\[\Delta G = (-11500) - (-31500) = 20000 \mathrm{~J} \/ \mathrm{mol}^{-1}\]Convert the result to kilojoules by dividing by 1000:\[\Delta G = 20 \mathrm{~kJ} \/ \mathrm{mol}^{-1}\]

Step 5: Determine the Correct Answer

The calculated Gibbs free energy change is \(20 \mathrm{~kJ} \/ \mathrm{mol}^{-1}\), which corresponds to option (d).

Key Concepts

Understanding Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), is a core concept in thermodynamics that expresses the total heat content transferred in a system during a chemical reaction at constant pressure. It's important to note that enthalpy is a state function, meaning it depends only on the initial and final states of the system, not the path taken to get there.

• A negative \( \Delta H \) indicates an exothermic reaction, where heat is released to the surroundings.
• A positive \( \Delta H \) signifies an endothermic reaction, where heat is absorbed from the surroundings.

In the given exercise, the enthalpy change is \(-11.5 \times 10^3 J/mol\), indicating that the reaction releases heat, hence it is exothermic. This gives us insight into how energy flows during the reaction, which is critical for predicting reaction behavior under different conditions.

The Role of Entropy Change

Entropy is a measure of the disorder or randomness in a system. The change in entropy, \( \Delta S \), is crucial in determining the feasibility and spontaneity of a reaction.

• When \( \Delta S \) is positive, there's an increase in disorder, which typically favors spontaneity.
• A negative \( \Delta S \) implies a decrease in disorder, often seen in reactions where order is increasing.

In our exercise, \( \Delta S = -105 J K^{-1} mol^{-1} \), suggesting a decrease in randomness. Despite the entropic penalty (loss of disorder), the dominance of enthalpic changes greatly influences the overall spontaneity of reactions, as seen in calculations of Gibbs Free Energy. This interplay between enthalpy and entropy is fundamental to understanding why certain reactions occur spontaneously.

Basics of Thermodynamics

Thermodynamics involves studying heat and energy transformations in chemical processes. Gibbs Free Energy, a pivotal thermodynamic function, is used to predict the spontaneity of reactions by combining enthalpy and entropy changes through the equation \( \Delta G = \Delta H - T\Delta S \).

• If \( \Delta G \) is negative, the reaction is spontaneous under constant temperature and pressure.
• A positive \( \Delta G \) means the reaction is non-spontaneous.
• If \( \Delta G \) equals zero, the system is at equilibrium.

This foundational concept explains why some reactions happen on their own while others require external intervention. In the given problem, temperature and the interplay between \( \Delta H \) and \( \Delta S \) are used to calculate \( \Delta G \), which ultimately tells whether the chemical process is favorable under the provided conditions.