Problem 145 - A Complete Resource Book in Chemistry for JEE Main

Question

For the reaction, \(\mathrm{Ag}_{2} \mathrm{O}(\mathrm{s}) \rightleftharpoons 2 \mathrm{Ag}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) \(\Delta \mathrm{H}, \Delta \mathrm{S}\) and \(\mathrm{T}\) are \(40.657 \mathrm{~kJ} \mathrm{~mol}^{-1}, 109 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) and \(373 \mathrm{~K}\) respectively. Find the free energy change \((\Delta \mathrm{G})\) of the reaction.

Step-by-Step Solution

Answer

The free energy change \( \Delta G \) of the reaction is 0 \( \text{kJ} \cdot \text{mol}^{-1} \).

Step 1: Understand the Given Variables

We are given \( \Delta H = 40.657 \, \text{kJ} \cdot \text{mol}^{-1} \), \( \Delta S = 109 \, \text{J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \), and \( T = 373 \text{ K} \). Remember to convert \( \Delta S \) to \( \text{kJ} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \) by dividing by 1000.

Step 2: Convert Entropy Change

Convert \( \Delta S \) from \( \text{J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \) to \( \text{kJ} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \) to match the units of \( \Delta H \). Thus, \( \Delta S = \frac{109}{1000} = 0.109 \, \text{kJ} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \).

Step 3: Apply the Gibbs Free Energy Equation

The Gibbs free energy change \( \Delta G \) is calculated using the equation: \( \Delta G = \Delta H - T\Delta S \). Substitute \( \Delta H = 40.657 \, \text{kJ} \cdot \text{mol}^{-1} \), \( T = 373 \text{ K} \), and \( \Delta S = 0.109 \, \text{kJ} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \) into the equation.

Step 4: Calculate \( T \times \Delta S \)

Compute the product \( T \times \Delta S = 373 \times 0.109 = 40.657 \, \text{kJ} \cdot \text{mol}^{-1} \).

Step 5: Calculate \( \Delta G \)

Substitute the calculated value of \( T \times \Delta S \) into the equation for \( \Delta G \): \( \Delta G = 40.657 - 40.657 = 0 \, \text{kJ} \cdot \text{mol}^{-1} \).

Step 6: Verify the Calculation

Ensure each step and calculation aligns without any oversight. The result seems valid as the \( T \times \Delta S \) term equals \( \Delta H \), resulting in \( \Delta G = 0 \).

Key Concepts

Thermodynamic equations

Thermodynamic equations are essential tools in chemistry that help us understand how energy is transferred in chemical reactions. One crucial equation is the Gibbs Free Energy equation, which marries various energetic aspects of a system. This equation is expressed as: \[ \Delta G = \Delta H - T\Delta S \] - \( \Delta G \) represents the change in Gibbs Free Energy. It tells us whether a reaction is spontaneous. If \( \Delta G < 0 \), the reaction is spontaneous; if \( \Delta G = 0 \), the system is at equilibrium.- \( \Delta H \) denotes the enthalpy change, representing the total energy absorbed or released.- \( T \Delta S \) involves temperature \( T \) and entropy change \( \Delta S \), contributing to energy dispersion within the system.This equation emphasizes the balance between enthalpy and entropy in determining the spontaneity of reactions. By examining these thermodynamic parameters, we can predict reaction behaviors under various conditions.

Enthalpy change

Enthalpy change, or \( \Delta H \), is a measure of the heat energy change in a reaction at constant pressure. It reflects the difference in energy between reactants and products. If \( \Delta H \) is positive, the reaction absorbs energy (endothermic), and if negative, it releases energy (exothermic).In our context, with \( \Delta H = 40.657 \, \text{kJ} \cdot \text{mol}^{-1} \), it indicates that the reaction involving silver oxide decomposition requires an input of energy. The positive \( \Delta H \) signifies the breaking of bonds necessitates more energy than what is released during product formation.Enthalpy is critical in assessing the energy level transitions in chemical reactions. It's as if we are accounting for every bit of energy going in and out, ensuring conservation and understanding reaction energetics better.

Entropy change

Entropy, symbolized as \( \Delta S \), is the degree of randomness or disorder within a system. An increase in entropy indicates a move towards more disorder. The universe tends towards higher entropy states, reflecting in various natural processes.In our reaction, \( \Delta S = 109 \, \text{J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \) signals a greater degree of randomness among the products compared to the reactants. This is because going from solid silver oxide to a mixture of solid silver and gaseous oxygen increases possible microstates, hence higher entropy.Understanding entropy changes provides insight into why some reactions occur without external input, as they may contribute positively (energetically favorably) to the universe's entropy. It balances the energy equation alongside enthalpy for determining reaction feasibility.