Problem 61 - 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 \(-105 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) respec- tively. 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 \( 20 \text{ kJ/mol} \), matching option (d).

Step 1: Understand the Gibbs Free Energy Formula

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

Step 2: Plug in Given Values into the Formula

From the problem, the given values are \( \Delta H = -11.5 \times 10^{3} \text{ J/mol} \), \( T = 300 \text{ K} \), and \( \Delta S = -105 \text{ J/K/mol} \). Substitute these values into the Gibbs free energy equation: \( \Delta G = (-11.5 \times 10^{3}) - (300)(-105) \).

Step 3: Perform the Calculations

First, calculate the entropy component: \( 300 \times -105 = -31500 \text{ J/mol} \). Now substitute this back into the formula: \( \Delta G = -11500 - (-31500) \). Simplifying, \( \Delta G = -11500 + 31500 = 20000 \text{ J/mol} \).

Step 4: Convert Units and Match Answer

Since the options are given in \( \text{kJ/mol} \), convert \( 20000 \text{ J/mol} \) to \( \text{kJ/mol} \) by dividing by 1000, yielding \( 20 \text{ kJ/mol} \). This matches option (d).

Key Concepts

Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), is an essential concept in thermodynamics that measures the total heat content of a system. It provides insight into whether a reaction is exothermic or endothermic.
Exothermic reactions release heat, indicated by a negative \( \Delta H \), while endothermic reactions absorb heat, resulting in a positive \( \Delta H \).

For our exercise's reaction, \( \Delta H \) is \(-11.5 \times 10^3 \text{ J/mol} \), indicating that the reaction releases energy, thus, an exothermic process. Understanding this helps in predicting reaction behavior and energy changes.

• **Negative \( \Delta H \):** Exothermic, heat released.
• **Positive \( \Delta H \):** Endothermic, heat absorbed.

These principles guide how energy interacts with the environment, influencing reaction feasibility.

Entropy Change

Entropy change, represented by \( \Delta S \), quantifies a system's disorder or randomness. It predicts the spontaneity of a process. A positive \( \Delta S \) signifies increased disorder and is typically associated with spontaneous processes.
Conversely, a negative \( \Delta S \) indicates reduced randomness, usually making processes less spontaneous.

In the problem given, \( \Delta S = -105 \text{ J/K/mol} \), suggesting a decrease in entropy, which implies that the system becomes more ordered after the reaction.

• **Positive \( \Delta S \):** Increased disorder, potentially spontaneous.
• **Negative \( \Delta S \):** Decreased disorder, likely non-spontaneous under constant conditions.

Grasping entropy change helps evaluate the direction and nature of reactions.

Thermodynamic Equations

Thermodynamic equations are crucial tools in assessing the energetics of chemical reactions. The Gibbs Free Energy equation, \( \Delta G = \Delta H - T \Delta S \), is especially important as it combines both enthalpy and entropy changes to determine the feasibility of a reaction.
\( \Delta G \) enables predictions about whether a process will occur spontaneously at a given temperature.*
*When \( \Delta G \) is negative, the process is spontaneous and favorable. Positive \( \Delta G \) indicates a non-spontaneous process.

For the exercise, substituting the enthalpy, entropy, and temperature values into the equation yields a \( \Delta G \) of \( 20 \text{ kJ/mol} \), showing that the process, at this temperature, requires additional energy input to proceed.
This formula not only anticipates spontaneity but also guides in measuring reaction viability in various temperature and pressure conditions.