Problem 44 - A Complete Resource Book in Chemistry for JEE Main

Question

Standard enthalpy and standard entropy changes for the oxidation of ammonia at \(298 \mathrm{~K}\) are \(-382.64 \mathrm{~kJ}\) \(\mathrm{mol}^{-1}\) and \(-145.6 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\), respectively. Standard Gibbs energy change for the same reaction at \(268 \mathrm{~K}\) is (a) \(-221.1 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(-339.3 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(-439.3 \mathrm{kJmol}^{-1}\) (d) \(-523.2 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

Step-by-Step Solution

Answer

(b) -339.3 kJ mol⁻¹

Step 1: Understanding the Formula

To find the Gibbs energy change, we use the equation: \( \Delta G = \Delta H - T \Delta S \), where \( \Delta H \) is the enthalpy change, \( T \) is the temperature, and \( \Delta S \) is the entropy change.

Step 2: Inserting Values

We are given that \( \Delta H = -382.64 \mathrm{~kJ} \mathrm{~mol}^{-1} \), \( \Delta S = -145.6 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \) and the temperature \( T = 268 \mathrm{~K} \). Note that \( \Delta S \) needs to be converted to kJ by dividing by 1000, so \( \Delta S = -0.1456 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \).

Step 3: Calculating \( T \Delta S \)

Calculate \( T \Delta S \) by multiplying \( T = 268 \mathrm{~K} \) with \( \Delta S = -0.1456 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \). This gives \( T \Delta S = 268 \times -0.1456 = -39.0528 \mathrm{~kJ} \mathrm{~mol}^{-1} \).

Step 4: Substitute into the Gibbs Energy Equation

Using the values: \( \Delta G = \Delta H - T \Delta S = -382.64 - (-39.0528) = -382.64 + 39.0528 = -343.5872 \mathrm{~kJ} \mathrm{~mol}^{-1} \).

Step 5: Select the Closest Answer

Compare the calculated Gibbs energy with the given options: (a) -221.1, (b) -339.3, (c) -439.3, and (d) -523.2. The closest value to \(-343.5872 \mathrm{~kJ} \mathrm{~mol}^{-1} \) is (b) -339.3 \mathrm{~kJ} \mathrm{~mol}^{-1}.

Key Concepts

Standard Enthalpy Change

The concept of **Standard Enthalpy Change** is central to understanding energy transformations in chemical reactions. Enthalpy change, denoted as \( \Delta H \), represents the heat absorbed or released under constant pressure. When we discuss standard enthalpy change, we're specifically referring to conditions where substances are in their standard states, typically defined at a pressure of 1 bar and a temperature of 298 K (roughly room temperature).

Enthalpy changes can be:

• **Exothermic**, where heat is released and \( \Delta H \) is negative.
• **Endothermic**, where heat is absorbed and \( \Delta H \) is positive.

For the oxidation of ammonia in the given exercise, a standard enthalpy change of \(-382.64 \text{ kJ mol}^{-1}\) indicates that the reaction releases energy, making it exothermic.

Understanding enthalpy involves examining molecular interactions and bond energies. Breaking molecular bonds requires energy, while forming bonds releases energy. The net enthalpy change reflects these competing processes throughout the reaction.

Standard Entropy Change

**Standard Entropy Change** (9 \( \Delta S \)) is another crucial concept in thermodynamics. Entropy, a measure of disorder or randomness, quantifies the number of specific ways a thermodynamic system can be arranged. The standard entropy change considers conditions where all reactants and products are in their standard states. In chemical reactions, entropy will increase if the final state is more disordered than the initial state.

For the reaction provided, the given standard entropy change is \( -145.6 \text{ JK}^{-1} \text{ mol}^{-1} \). A negative sign suggests a decrease in disorder during the reaction. This can happen if the products of the reaction are more structured or ordered than the reactants.

In practical terms:

• Entropy changes help predict the feasibility of a reaction alongside enthalpy.
• Reactions generating more particles generally increase entropy.

Thus, understanding entropy change helps in assessing reaction spontaneity when paired with enthalpy and the Gibbs free energy equation.

Thermodynamics Equation

The **Thermodynamics Equation** relating Gibbs Free Energy, enthalpy, and entropy is a fundamental formula in chemistry:
\[ \Delta G = \Delta H - T \Delta S \]
Here, \( \Delta G \) is the Gibbs free energy change, \( \Delta H \) is the enthalpy change, \( T \) is temperature in Kelvin, and \( \Delta S \) is the entropy change. This equation is vital because it predicts reaction spontaneity:

• If \( \Delta G < 0 \), the reaction is spontaneous.
• If \( \Delta G = 0 \), the reaction is at equilibrium.
• If \( \Delta G > 0 \), the reaction is non-spontaneous.

Understanding this equation involves:

• Converting entropy from \( \text{J K}^{-1} \text{ mol}^{-1} \) to \( \text{kJ K}^{-1} \text{ mol}^{-1} \) by dividing by 1000.
• Multiplying temperature and entropy to find \( T \Delta S \).
• Subtracting \( T \Delta S \) from \( \Delta H \) to calculate \( \Delta G \).

This equation provides a comprehensive view of energy transformations in reactions, revealing how temperature influences spontaneity.