Problem 16 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The following reaction is performed at \(298 \mathrm{~K}\). [Main 2015] $$2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})$$. The standard free energy of formation of \(\mathrm{NO}(\mathrm{g})\) is \(86.6 \mathrm{KJ} / \mathrm{mol}\) at \(298 \mathrm{~K}\). What is the standard free energy of formation of \(\mathrm{NO}_{2}(\mathrm{~g})\) at \(298 \mathrm{~K} ?\left(K_{\mathrm{p}}\right.\) \(\left.=1.6 \times 10^{12}\right)\). (a) \(86600-\frac{\ln \left(1.6 \times 10^{12}\right)}{\mathrm{R}(298)}\) (b) \(\quad 0.5\left[2 \times 86,600-R(298) \ln \left(1.6 \times 10^{12}\right)\right]\) (c) \(R(298) \ln \left(1.6 \times 10^{12}\right)-86600\) (d) \(86600+R(298) \ln \left(1.6 \times 10^{12}\right)\)

Step-by-Step Solution

Answer

The correct answer is (b).

Step 1: Write the Expression for ΔG° of the Reaction

The change in standard Gibbs free energy for the reaction, \( \Delta G^\circ \), can be calculated using the relation \( \Delta G^\circ = -RT \ln K_{p} \), where \( R \) is the universal gas constant with a value of \( 8.314 \ \text{J/mol K} \), \( T \) is the temperature in Kelvin, and \( K_{p} \) is the equilibrium constant.

Step 2: Convert the Gas Constant

Before substitution, ensure that \( R \) is correctly used in compatible units with \( \Delta G^\circ \), converting it from \( \text{J} \) to \( \text{kJ} \) by dividing it by 1000. Thus, \( R = 0.008314 \ \text{kJ/mol K} \).

Step 3: Substitute Values into the ΔG° Formula

Substitute the provided values into the formula: \[ \Delta G^\circ = -0.008314 \times 298 \times \ln(1.6 \times 10^{12}) \].

Step 4: Calculate ΔG°

Perform the calculation to find \( \Delta G^\circ \). First, calculate the natural logarithm \( \ln(1.6 \times 10^{12}) \), and then multiply by the numbers to find that \( \Delta G^\circ = -70.45 \ \text{kJ/mol} \).

Step 5: Use the Reaction's ΔG° Expression to Find ΔG°f of NO₂

The standard Gibbs free energy of formation of \( \text{NO}_2 \) is derived from \( \Delta G^\circ = 2 \cdot \Delta G^\circ_{f, \text{NO}_2} - (2 \cdot 86.6) \). Rearrange the equation to solve for \( \Delta G^\circ_{f, \text{NO}_2} \): \[ \Delta G^\circ_{f, \text{NO}_2} = \frac{1}{2} ((70.45 + 2 \cdot 86.6)) \].

Step 6: Select the Correct Option

Now, verify which of the given options raw matches our derived formula. The correct option is (b): \( 0.5[2 \times 86,600 - R(298) \ln (1.6 \times 10^{12})] \), given that it reflects the form used to derive \( \Delta G^\circ_{f, \text{NO}_2} \).

Key Concepts

Chemical Equilibrium

Chemical equilibrium refers to the state in a chemical reaction where the concentrations of reactants and products remain constant over time. This happens because the forward and reverse reactions occur at the same rate. Imagine a busy highway, where cars enter and exit at equal rates, maintaining a consistent number of cars on the road. Similarly, in a chemical equilibrium, no net change occurs despite ongoing processes.

Key characteristics of chemical equilibrium include:

• Both forward and reverse reactions are occurring.
• The concentrations of reactants and products do not change.
• It is a dynamic state, not a static one.

Equilibrium is crucial for understanding reactions in nature and industry. In the given reaction of nitrogen monoxide (NO) with oxygen (O₂) to form nitrogen dioxide (NO₂), it achieves equilibrium. This ensures the predictable behavior of the reaction under constant conditions.

Standard Free Energy Change

The standard free energy change, denoted as \( \Delta G^\circ \), is the difference in Gibbs free energy between the reactants and products under standard conditions (1 bar, 298 K, and 1 M concentration for solutions). It provides insights into a reaction's spontaneity. When \( \Delta G^\circ \) is negative, the reaction is spontaneous, meaning it can occur on its own.

Important aspects of \( \Delta G^\circ \) include:

• Indicates the amount of energy available to do work or energy required.
• Calculated using the relation: \( \Delta G^\circ = -RT \ln K_p \), where \( R \) is the gas constant and \( T \) is the temperature in Kelvin.
• A negative \( \Delta G^\circ \) suggests products are favored at equilibrium.

In our context, we calculated \( \Delta G^\circ \) for the proposed reaction, arriving at \(-70.45 \ \text{kJ/mol}\), which implies that forming NO₂ is energetically favorable under standard conditions.

Equilibrium Constant

The equilibrium constant, denoted \( K_p \) for gaseous reactions, is a dimensionless number that expresses the ratio of product pressures to reactant pressures at equilibrium. It offers a quantitative measure of where the equilibrium position lies in a reaction. A very large \( K_p \) signals that, at equilibrium, the reaction greatly favors product formation.

Vital points about the equilibrium constant include:

• Values greater than 1 indicate the products are favored.
• For the given reaction, \( K_p = 1.6 \times 10^{12} \), showing a strong tendency toward NO₂.
• Changes in \( K_p \) are influenced by temperature variations.

Understanding \( K_p \) helps predict how changes in conditions like pressure or temperature can affect the equilibrium state in practical scenarios, such as chemical manufacturing or environmental processes.