Problem 62 - A Complete Resource Book in Chemistry for JEE Main

Question

The standard enthalpy of decomposition of \(\mathrm{N}_{2} \mathrm{O}_{4}\) to \(\mathrm{NO}_{2}\) is \(58.04 \mathrm{~kJ}\) and standard entropy of this reaction is \(176.7 \mathrm{~J} \mathrm{~K}^{-1}\). The standard free energy change for this reaction at \(25^{\circ} \mathrm{C}\), is (a) \(5.39 \mathrm{~kJ}\) (b) \(-5.39 \mathrm{~kJ}\) (c) \(539 \mathrm{~kJ}\) (d) \(53.9 \mathrm{~kJ}\)

Step-by-Step Solution

Answer

The standard free energy change is 5.39 kJ, option (a).

Step 1: Convert Temperature to Kelvin

Standard free energy change calculations require temperature to be in Kelvin. Convert the temperature from Celsius to Kelvin using the formula: \[ T (\text{Kelvin}) = T (\text{Celsius}) + 273.15 \] Thus, at \(25^{\circ} \text{C}\), \(T = 25 + 273.15 = 298.15\, \text{K}\).

Step 2: Convert Entropy to kJ/K

The standard entropy given is in \(\text{J K}^{-1}\). Convert this to \(\text{kJ K}^{-1}\) by dividing by 1000:\[ 176.7 \, \text{J K}^{-1} = \frac{176.7}{1000} = 0.1767 \, \text{kJ K}^{-1} \]

Step 3: Use Gibbs Free Energy Equation

The Gibbs free energy change (\(\Delta G^{\circ}\)) can be calculated using the equation:\[ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \]Where \(\Delta H^{\circ} = 58.04 \, \text{kJ}\), \(T = 298.15 \, \text{K}\), and \(\Delta S^{\circ} = 0.1767 \, \text{kJ K}^{-1}\).

Step 4: Calculate the Free Energy Change

Substitute the values into the Gibbs free energy equation:\[ \Delta G^{\circ} = 58.04 - 298.15 \times 0.1767 \]Calculate \(298.15 \times 0.1767 = 52.70\) kJ.

Step 5: Solve for the Final Value of ΔG°

Subtract the calculated entropy term from the enthalpy:\[ \Delta G^{\circ} = 58.04 - 52.70 = 5.34 \, \text{kJ} \]The closest answer choice, considering significant figures, is option \((a) \; 5.39 \, \text{kJ}\).

Key Concepts

Enthalpy

Enthalpy is a key concept in thermodynamics, representing the total energy contained in a chemical system. This energy includes both internal energy and the energy required to make room for it by displacing its environment. Enthalpy is symbolized by the letter \( H \) and is expressed in units of Joules or kilojoules.

• Enthalpy change (\( \Delta H \)) is often measured during chemical reactions, indicating whether heat is absorbed or released.
• For reactions occurring under constant pressure, the change in enthalpy tells us how much energy is transferred as heat.

In the context of the given reaction, the enthalpy of decomposition of \( \mathrm{N}_{2} \mathrm{O}_{4} \) into \( \mathrm{NO}_{2} \) is a positive \( 58.04 \, \text{kJ} \), meaning the reaction absorbs heat from its surroundings and is endothermic.
By understanding enthalpy, we can predict how different reactions will behave and how they will affect their environment.

Entropy

Entropy is a measure of disorder or randomness in a system. It provides insight into the number of possible arrangements that are available in a system's particles. In thermodynamics, entropy is denoted by the letter \( S \) and is typically measured in units of Joules per Kelvin (\( \text{J K}^{-1} \)).

• Increase in entropy reflects an increase in disorder; molecules spread out or disperse, occupying more space or creating more random arrangements.
• Entropy is often associated with the second law of thermodynamics, which states that the total entropy of an isolated system always increases over time.

In the given decomposition reaction, the standard entropy change is \( 176.7 \, \text{J K}^{-1} \), which suggests that the disorder in the system increases as \( \mathrm{N}_{2} \mathrm{O}_{4} \) undergoes decomposition. The molecules become more randomly distributed, reflecting a natural tendency towards higher entropy.

Thermodynamics

Thermodynamics is the branch of physical science concerned with heat and its relation to energy and work. It describes the principles governing the energy transformations that take place in chemical reactions and other physical processes.

• The four laws of thermodynamics (zeroth, first, second, and third) establish fundamental concepts such as temperature, energy conservation, entropy, and absolute zero.
• Thermodynamics helps us understand how energy is transferred within chemical reactions or phases of matter, and guides us in calculating changes such as Gibbs Free Energy (\( \Delta G \)).

In the given exercise, thermodynamics introduces us to the Gibbs Free Energy equation, \( \Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} \). This equation serves as a crucial tool, allowing us to determine whether reactions occur spontaneously. Here, after substituting the values into this equation, we calculated that the free energy change for the given reaction is \( 5.34 \, \text{kJ} \). Such calculations help evaluate and predict the feasibility of chemical transformations.