Problem 74 - A Complete Resource Book in Chemistry for JEE Main

Question

For the reaction \(\mathrm{H}_{2}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) the value of \(\Delta \mathrm{H}=-285.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(\Delta \mathrm{S}=0.163\) \(\mathrm{JK}^{-1} \mathrm{~mol}^{-1}\). The free energy change at \(300 \mathrm{~K}\). for the reaction, is (a) \(-289.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(437.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(-334.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(-291.6 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

The free energy change is approximately \\( -285.8 \\ ext{kJ/mol}\\), not matching any option suggest any option mismatch in given choices.

Step 1: Understand the concept

The free energy change, \( \Delta G \), for a reaction can be calculated using the equation \( \Delta G = \Delta H - T \Delta S \), where \( \Delta H \) is the change in enthalpy, T is the temperature in Kelvin, and \( \Delta S \) is the change in entropy.

Step 2: Convert Units

Since \( \Delta H \) is given in kJ/mol and \( \Delta S \) is in J/K mol, we need to convert \( \Delta S \) to kJ/K mol by dividing by 1000. This gives us \( \Delta S = 0.163 \, \rac{J}{K \, \text{mol}} = 0.000163 \, \rac{kJ}{K \, \text{mol}}\).

Step 3: Calculate \\( \\Delta G \\)

Substitute the given values into the equation: \( \Delta G = -285.8 \, \ ext{kJ/mol} - 300 \, \ ext{K} \cdot 0.000163 \, \ ext{kJ/K mol}\).

Step 4: Perform the Calculation

Calculate the term \( 300 \, \cdot 0.000163 \), which equals \( 0.0489 \, \ ext{kJ/mol}\). Subtract this from \( -285.8 \, \ ext{kJ/mol}\) to get \( \Delta G = -285.8489 \, \ ext{kJ/mol}\).

Step 5: Significant Figures and Final Answer

Based on the significant figures provided in \( \Delta H \), round off the final answer to three significant figures: \( \Delta G \approx -285.8 \ ext{kJ/mol}\).

Key Concepts

Thermodynamics

Thermodynamics is a fundamental branch of science that deals with the study of energy and its transformations. Within this framework, we focus on how energy and matter interact in a system, especially in processes involving heat and work. A key player in thermodynamics is the Gibbs Free Energy ( \( \Delta G \) ), which predicts whether a chemical reaction occurs spontaneously.

To understand Gibbs Free Energy, think of it as the balance struck between two main other concepts: enthalpy (\( \Delta H \)) and entropy (\( \Delta S \)). This provides valuable insights into reaction feasibility.

• If \( \Delta G \) is negative, the reaction tends to be spontaneous.
• If \( \Delta G \) is positive, the reaction is non-spontaneous under the given conditions.

The equation \( \Delta G = \Delta H - T \Delta S \) is fundamental as it incorporates both enthalpy and entropy changes along with temperature to determine the free energy change.

Enthalpy

Enthalpy, denoted as \( \Delta H \), is a measure of the total heat content in a system at constant pressure. It tells us how much heat is absorbed or released during a chemical reaction. If you've ever noticed a reaction being labeled as exothermic or endothermic, what's being referred to is its enthalpy change.

• An exothermic reaction releases heat, so \( \Delta H \) is negative.
• An endothermic reaction absorbs heat, resulting in a positive \( \Delta H \).

In the original exercise, the reaction under consideration releases heat. This is evidenced by \( \Delta H = -285.8 \text{ kJ/mol} \). The negative sign indicates that it is an exothermic reaction, meaning it could potentially drive itself by releasing energy into the surroundings.

Entropy

Entropy, symbolized as \( \Delta S \), represents the degree of disorder or randomness in a system. It's a measure that helps explain the second law of thermodynamics: the entropy of the universe tends to increase. This means systems naturally progress towards a state of greater disorder.

• A positive \( \Delta S \) indicates an increase in disorder.
• A negative \( \Delta S \) indicates a decrease in disorder.

In our exercise, \( \Delta S = 0.163 \text{ J/K mol} \), converted to kJ gives \( 0.000163 \text{ kJ/K mol} \). This positive entropy change suggests that the products of the reaction may be more disordered than the reactants. Yet, it plays a smaller role relative to enthalpy in determining \( \Delta G \) due to its smaller magnitude when scaled with temperature.