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 approximately \(5.39 \mathrm{~kJ}\) (option a).

Step 1: Understand the Given Information

We're given that the standard enthalpy change for the decomposition of \(\mathrm{N}_2\mathrm{O}_4\) to \(\mathrm{NO}_2\) is \(58.04 \mathrm{~kJ}\). The standard entropy change for the reaction is \(176.7 \mathrm{~J} \mathrm{~K}^{-1}\). We need to find the standard free energy change (\(\Delta G^\circ\)) for the reaction at \(25^{\circ} \mathrm{C}\).

Step 2: Convert Temperature to Kelvin

The reaction temperature is provided as \(25^{\circ} \mathrm{C}\). Convert this to Kelvin using the formula \(T(K) = T(^{\circ}C) + 273.15\): \[T = 25 + 273.15 = 298.15 \mathrm{~K}\]

Step 3: Use Gibbs Free Energy Equation

The formula to calculate the standard free energy change is: \[\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ\] Where: \(\Delta G^\circ\) is the standard free energy change, \(\Delta H^\circ\) is the standard enthalpy change, \(T\) is the temperature in Kelvin, and \(\Delta S^\circ\) is the standard entropy change.

Step 4: Substitute Given Values into the Formula

Place the values into the free energy equation: \[\Delta G^\circ = 58.04 \mathrm{~kJ} - 298.15 \mathrm{~K} \times \left(\frac{176.7 \mathrm{~J} \mathrm{~K}^{-1}}{1000 \mathrm{~J/kJ}}\right)\] Converting entropy from J to kJ (since 1 kJ = 1000 J): \[\Delta S^\circ = \frac{176.7}{1000} = 0.1767 \mathrm{~kJ} \mathrm{~K}^{-1}\]

Step 5: Calculate

Calculate the value of \(T \Delta S^\circ\): \[T \Delta S^\circ = 298.15 \cdot 0.1767 = 52.711155 \mathrm{~kJ}\] Then calculate \(\Delta G^\circ\): \[\Delta G^\circ = 58.04 - 52.711155 = 5.328845 \mathrm{~kJ}\]

Step 6: Round Appropriately and Compare

Round the calculated \(\Delta G^\circ\) to two decimal places for clarity: \[\Delta G^\circ \approx 5.33 \mathrm{~kJ}\] Comparing with the given options, the closest value is \(5.39 \mathrm{~kJ}\) (option a).

Key Concepts

Standard Enthalpy Change

The Standard Enthalpy Change, denoted by \( \Delta H^\circ \), is a crucial concept in thermodynamics. It represents the heat absorbed or released during a chemical reaction at constant pressure, under standard conditions (1 atm pressure and 298.15 K temperature).
For decomposition reactions, like the decomposition of \( \mathrm{N}_2\mathrm{O}_4\) into \( \mathrm{NO}_2 \), it measures the energy required to break the bonds.

• For endothermic reactions, \( \Delta H^\circ \) is positive, indicating that energy is absorbed.
• For exothermic reactions, \( \Delta H^\circ \) is negative, meaning energy is released.

In the example problem, the standard enthalpy change is \( 58.04 \mathrm{~kJ} \), indicating that the decomposition reaction requires energy input.

Standard Entropy Change

Standard Entropy Change, \( \Delta S^\circ \), is a measure of the disorder or randomness in a system during a chemical reaction. It's an important factor in understanding how spontaneous a reaction can be.
- Entropy tends to increase with temperature because the particles move more chaotically.
- An increase in entropy (positive \( \Delta S^\circ \)) often means the system becomes more disordered.
- A decrease (negative \( \Delta S^\circ \)) means more order.In our scenario, \( \Delta S^\circ = 176.7 \mathrm{~J} \, \mathrm{~K}^{-1} \), suggesting an increase in disorder as \( \mathrm{N}_2\mathrm{O}_4 \) molecules break apart into \( \mathrm{NO}_2 \) molecules.

Thermodynamics

Thermodynamics is the branch of physical science that deals with the relations between heat and other forms of energy such as mechanical, electrical, or chemical energy. The primary goal is to understand and predict how energy transfers can affect material properties and reactions.
- One main focus is determining how energy is conserved and transferred in various systems, which is crucial for calculating changes in thermodynamic quantities such as enthalpy, entropy, and free energy.
- Gibbs Free Energy (\( \Delta G^\circ \)), which combines enthalpy and entropy changes, predicts the spontaneity of chemical reactions.
- A negative \( \Delta G^\circ \) indicates a spontaneous reaction, while a positive \( \Delta G^\circ \) suggests non-spontaneity.
In essence, thermodynamics helps chemists and engineers design processes that effectively utilize energy changes during reactions.

Decomposition Reaction

A decomposition reaction involves breaking down a compound into simpler components or elements. This type of reaction is important for understanding how substances interact and change.

Common characteristics include:

• The original compound is typically a single substance that splits into two or more products.
• Energy, often in the form of heat, electricity, or light, is necessary to break the bonds holding the elements together.

In the given exercise, the decomposition of \( \mathrm{N}_2\mathrm{O}_4 \) into \( \mathrm{NO}_2 \) is studied. Such reactions are integral to many chemical processes, including combustion and digestion.

Temperature Conversion

Temperature Conversion is a basic yet crucial step when dealing with scientific calculations. Most thermodynamic equations require temperature in Kelvin. Kelvin is the absolute temperature scale used in science because it starts at absolute zero, simplifying certain calculations.

To convert from Celsius to Kelvin:
\( \text{Temperature in Kelvin (K)} = \text{Temperature in Celsius (\(^\circ\)C)} + 273.15 \)
This conversion ensures consistency in thermodynamic equations, such as when calculating Gibbs Free Energy or other temperature-dependent properties.
In our exercise, converting 25\(^\circ\)C to Kelvin gave us 298.15 K, which was necessary for the Gibbs Free Energy calculation.