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

\( \Delta G = 0 \text{ kJ mol}^{-1} \). The reaction is at equilibrium.

Step 1: Understanding the Formula

We need to calculate the Gibbs free energy change \( \Delta G \) using the formula: \( \Delta G = \Delta H - T \Delta S \), where \( \Delta H \) is the enthalpy change, \( T \) is the temperature in Kelvin, and \( \Delta S \) is the entropy change. This equation relates these thermodynamic quantities.

Step 2: Substituting Known Values

Now, substitute the known values into the formula: \( \Delta G = 40.657 \text{ kJ mol}^{-1} - 373 \text{ K} \times 109 \text{ J K}^{-1} \text{ mol}^{-1} \). Note that we need to convert \( \Delta S \) from J to kJ by dividing by 1000, as \( \Delta H \) is in kJ.

Step 3: Converting Units

Convert \( \Delta S \) from \( \text{J K}^{-1} \text{ mol}^{-1} \) to \( \text{kJ K}^{-1} \text{ mol}^{-1} \) as follows: \( 109 \text{ J K}^{-1} \text{ mol}^{-1} = 0.109 \text{ kJ K}^{-1} \text{ mol}^{-1} \). This ensures consistency in units.

Step 4: Calculating \( T \Delta S \)

Calculate \( T \Delta S \) by multiplying temperature \( T \) with \( \Delta S \) in the same units: \( T \Delta S = 373 \text{ K} \times 0.109 \text{ kJ K}^{-1} \text{ mol}^{-1} = 40.657 \text{ kJ mol}^{-1} \).

Step 5: Finding \( \Delta G \)

Substitute \( T \Delta S \) back into the Gibbs free energy equation: \( \Delta G = 40.657 \text{ kJ mol}^{-1} - 40.657 \text{ kJ mol}^{-1} \). This simplifies to \( \Delta G = 0 \text{ kJ mol}^{-1} \).

Key Concepts

Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), represents the heat absorbed or released during a reaction at constant pressure. It's an essential concept in thermodynamics and chemistry. In the context of the given exercise, \( \Delta H \) is given as 40.657 kJ/mol for the decomposition of \( \mathrm{Ag}_2\mathrm{O} \) into silver and oxygen. Understanding enthalpy involves recognizing whether a reaction is endothermic or exothermic.

- **Endothermic reactions:** Absorb heat, resulting in a positive \( \Delta H \).
- **Exothermic reactions:** Release heat, resulting in a negative \( \Delta H \).

In our example, a positive \( \Delta H \) indicates that the decomposition of \( \mathrm{Ag}_2\mathrm{O} \) occurs by absorbing heat from its surroundings, making it endothermic.

Entropy Change

Entropy change, symbolized as \( \Delta S \), measures the degree of disorder or randomness in a system. It's a crucial concept related to the second law of thermodynamics, which states that the entropy of an isolated system always increases over time. In the exercise, \( \Delta S \) is given as 109 J K\( ^{-1} \) mol\( ^{-1} \).

Here, the decomposition of \( \mathrm{Ag}_2\mathrm{O} \) increases entropy because a solid is converted into a more disordered gas phase. This change represents a natural tendency of systems to move towards higher entropy.

- **Positive \( \Delta S \):** Indicates an increase in disorder.
- **Negative \( \Delta S \):** Indicates a decrease in disorder.

Understanding entropy gives you insight into the feasibility of a reaction, as higher entropy often favors a reaction proceeding under given conditions.

Thermodynamics

Thermodynamics is the branch of physical science that deals with the relations between heat and other forms of energy. It helps us understand how energy is converted and transferred in chemical processes, like the reaction in our exercise. Three key aspects are considered:

- **First Law of Thermodynamics:** Energy is conserved; energy lost by a system is gained by its surroundings, and vice versa.
- **Second Law of Thermodynamics:** Entropy tends to increase; a system is more likely to favor states with higher entropy.

The third law establishes the impossibility of reaching absolute zero, where entropy of a pure crystalline substance is zero. With Gibbs free energy, which combines entropy and enthalpy into a single value (\( \Delta G \)), we can determine the spontaneity of a reaction:
- **If \( \Delta G < 0 \):** The reaction is spontaneous.
- **If \( \Delta G > 0 \):** The reaction is non-spontaneous.

Thus, a balance of enthalpy and entropy changes, along with temperature (373 K in our case), helps evaluate reaction feasibility using the formula: \( \Delta G = \Delta H - T\Delta S \).

Unit Conversion

Unit conversion is pivotal in solving problems involving measurements in different units. In chemistry, consistency in units is essential to make accurate calculations. For the given exercise, it's important to convert entropy change (\( \Delta S \)) from joules to kilojoules since enthalpy change (\( \Delta H \)) is given in kilojoules. This keeps the units uniform, preventing errors in calculations.

Here's how to perform the necessary conversions:
- **Converting J to kJ:** Divide by 1000.

By converting \( \Delta S = 109 \text{ J K}^{-1}\text{ mol}^{-1} \) to \( 0.109 \text{ kJ K}^{-1}\text{ mol}^{-1} \), we can accurately compute Gibbs free energy using \( \Delta G = \Delta H - T\Delta S \). The same principle applies whenever dealing with unit conversions: convert all units to desired or common units before computations.