Problem 205 - A Complete Resource Book in Chemistry for JEE Main

Question

The following reaction is performed at \(298 \mathrm{~K}\). $$ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}) $$ 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 \(298 \mathrm{~K}\) ? \([2015]\) \(\left(\mathrm{K}_{\mathrm{p}}=1.6 \times 10^{12}\right)\) (a) \(\mathrm{R}(298) \ln \left(1.6 \times 10^{12}\right)-86600\) (b) \(86600+\mathrm{R}(298) \ln \left(1.6 \times 10^{12}\right)\) (c) \(86600-\frac{\operatorname{In}\left(1.6 \times 10^{12}\right)}{R(298)}\) (d) \(0.5\left[2 \times 86,600-\mathrm{R}(298) \operatorname{In} 1.6 \times 10^{12}\right]\)

Step-by-Step Solution

Answer

The correct answer is (a) \( \mathrm{R}(298) \ln \left(1.6 \times 10^{12}\right)-86600 \).

Step 1: Understand the Reaction

The reaction under consideration is \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})\). This is a chemical equilibrium where \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) are reactants, and \(\mathrm{NO}_{2}\) is the product.

Step 2: Calculate \( \Delta G^\circ \) for the Reaction

The equilibrium constant \( K_p \) is given as \(1.6 \times 10^{12}\). We use the relation between \( K_p \) and \( \Delta G^\circ \) (the standard change in Gibbs free energy for the reaction) given by \( \Delta G^\circ = -RT \ln K_p \). Substitute \(R = 8.314 \text{ J/mol K}\) and \(T = 298 \text{ K}\).

Step 3: Simplify \( \Delta G^\circ \) Expression

Substitute the values into the equation: \[ \Delta G^\circ = - (8.314 \, \text{J/mol K})(298 \, \text{K}) \ln (1.6 \times 10^{12}) \] Calculate \( \Delta G^\circ \) accordingly.

Step 4: Set Up Free Energy Formation Equation

The relationship between reaction free energies is: \[ \Delta G^\circ = \sum \Delta G^\circ_f(\text{products}) - \sum \Delta G^\circ_f(\text{reactants}) \] With \( \Delta G^\circ_f(\mathrm{NO}) = 86,600 \, \text{J/mol} \), we adjust the equation for this reaction to solve for \( \Delta G^\circ_f(\mathrm{NO}_2) \).

Step 5: Rearrange to Solve for \( \Delta G^\circ_f(\mathrm{NO}_2) \)

We rewrite the equation: \[ \Delta G^\circ = 2 \Delta G^\circ_f(\text{NO}_2) - 2 \times 86,600 \] Solve for \( \Delta G^\circ_f(\text{NO}_2) \).

Step 6: Conclusion and Answer Selection

After calculating \( \Delta G^\circ \) and solving for \( \Delta G^\circ_f(\text{NO}_2) \), compare with the given answer choices to find the correct one. Use the fact that the answer leads to a specific arrangement of terms.

Key Concepts

Gibbs Free Energy

Gibbs Free Energy is an important concept in thermodynamics and chemistry that indicates the amount of reversible work obtainable from a thermodynamic system at a constant temperature and pressure. It is denoted by \( \Delta G \) and helps predict the direction of chemical reactions.

The equation for Gibbs Free Energy is:

• \( \Delta G = \Delta H - T\Delta S \)

Here, \( \Delta H \) represents enthalpy change, \( T \) is the absolute temperature, and \( \Delta S \) stands for entropy change. When \( \Delta G \) is negative, a reaction is spontaneous, implying it can proceed without external energy input. Conversely, a positive \( \Delta G \) indicates a non-spontaneous reaction.

In equilibrium conditions, \( \Delta G \) is zero, indicating no net change as the forward and reverse reactions balance each other. This forms the basis for understanding chemical equilibrium and its relationship with Gibbs Free Energy.

Standard Free Energy of Formation

The Standard Free Energy of Formation (\( \Delta G_f^\circ \)) is a measure of the energy change when one mole of a compound is formed from its elements under standard conditions (298 K and 1 atm pressure). It reflects the compound's stability and its potential to release or absorb energy when formed.

To compute \( \Delta G_f^\circ \), we use the following relationship:

• \( \Delta G^\circ = \sum \Delta G_f^\circ(\text{products}) - \sum \Delta G_f^\circ(\text{reactants}) \)

In the given reaction of nitrogen monoxide \((\text{NO})\) forming \(\text{NO}_2\), determining \(\Delta G_f^\circ\) helps identify the energy profile of the reaction. Understanding this value also aids in comparing the energies of the reactants and products to predict the feasibility and equilibrium position of the reaction.

Equilibrium Constant

The Equilibrium Constant (\(K_p\) for gases) provides insight into the extent of a reaction at equilibrium. It quantifies the concentrations of reactants and products in a balanced chemical reaction.

The relationship between \(K_p\) and \( \Delta G^\circ \) is given by:

• \( \Delta G^\circ = -RT \ln K_p \)

Where \( R \) is the universal gas constant and \( T \) is temperature. A larger \( K_p \) value indicates a reaction that favors the formation of products, while a smaller \( K_p \) suggests reactants are more favored.

By calculating \( \Delta G^\circ \) using \( K_p \), we convert energy changes into relatable concentrations of the chemical species involved. This is crucial for understanding reaction dynamics and determining the thermodynamic favorability and progression of chemical reactions.