Problem 39 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

For the reaction: \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})\), the relation between the degree of dissociation, \(\alpha\), of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) at pressure, \(P\), with its equilibrium constant \(K_{\mathrm{P}}\) is (a) \(\alpha=\frac{K_{P} / P}{4+K_{P} / P}\) (b) \(\alpha=\frac{K_{P}}{4+K_{P}}\) (c) \(\alpha=\left[\frac{K_{P} / P}{4+K_{P} / P}\right]^{1 / 2}\) (d) \(\alpha=\left[\frac{K_{P}}{4+K_{P}}\right]^{1 / 2}\)

Step-by-Step Solution

Answer

The correct relationship is (a) \(\alpha=\frac{K_{P} / P}{4+K_{P} / P}\).

Step 1 - Write the Initial and Equilibrium Moles

Assume we have one mole of N2O4 at the start. Initially, there are no moles of NO2. At equilibrium, we have (1-α) moles of N2O4, since α fraction of N2O4 dissociates, and 2α moles of NO2 because two moles of NO2 form per mole of N2O4 dissociated.

Step 2 - Write the Expression for Kp

Kp is defined by the ratio of the products of the partial pressures of the products to the reactants, each raised to the power of their stoichiometric coefficients in the balanced equation. Thus, \( K_p = \frac{(P_{NO_2})^2}{P_{N2O4}} \).

Step 3 - Express Partial Pressures in Terms of α and P

Partial pressure can be expressed in terms of moles and total pressure. The mole fraction of NO2 is \(\frac{2\alpha}{1+2\alpha}\) and that of N2O4 is \(\frac{1-\alpha}{1+2\alpha}\). So, \( P_{NO_2} = \frac{2\alpha}{1+2\alpha}P \) and \( P_{N2O4} = \frac{1-\alpha}{1+2\alpha}P \).

Step 4 - Substitute the Partial Pressures in the Expression for Kp

Substitute the expressions for \( P_{NO_2} \) and \( P_{N2O4} \) into the equation for Kp from Step 2 to get an equation that relates Kp, α, and P.

Step 5 - Solve for α

Rearrange the equation obtained in Step 4 to solve for α in terms of Kp and P.

Key Concepts

Degree of Dissociation

The degree of dissociation, represented by \( \alpha \), is a quantitative measure of the extent to which a compound separates into its constituent particles. In the context of the given exercise, where \( \mathrm{N}_{2}\mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}) \), \( \alpha \) represents the fraction of \( \mathrm{N}_{2}\mathrm{O}_{4} \) molecules that dissociate into \( \mathrm{NO}_{2} \) molecules at equilibrium.

To visualize this, imagine starting with 100 molecules of \( \mathrm{N}_{2}\mathrm{O}_{4} \); if \( \alpha \) is 0.25, it means 25 of the original \( \mathrm{N}_{2}\mathrm{O}_{4} \) molecules have dissociated to form \( \mathrm{NO}_{2} \). The calculation of \( \alpha \) is tied directly to the stoichiometry of the reaction, equilibrium constant, and the partial pressures of the reactants and products. Understanding \( \alpha \) is critical as it gives insight into the position of equilibrium and the concentrations (or pressures) of the substances present in a mixture at equilibrium.

Equilibrium Constant

The equilibrium constant, denoted by \( K_{p} \) for gases when given in terms of partial pressures, is a value that represents the ratio of the multiplication of the partial pressures of the products to that of the reactants, both raised to their respective stoichiometric coefficients. The value of \( K_{p} \) is crucial because it indicates whether the reaction favors the formation of products or reactants at a given temperature.

In our equation \( K_{p} = \frac{(P_{\mathrm{NO}_{2}})^{2}}{P_{\mathrm{N}_{2}\mathrm{O}_{4}}} \), \( K_{p} \) remains constant at a given temperature regardless of the initial amounts of reactants or products. Therefore, knowing \( K_{p} \) allows us to predict the direction of the reaction and calculate the degree of dissociation, as well as the concentrations or partial pressures of the involved gases.

Partial Pressure

In the realm of gas-phase reactions, the partial pressure is the pressure an individual gas in a mixture contributes to the total pressure. It is analogous to concentration but is used for gases when they are in a mixture.

In our step-by-step solution, partial pressures of \( \mathrm{NO}_{2} \) and \( \mathrm{N}_{2}\mathrm{O}_{4} \) are expressed as functions of \( \alpha \) and the total pressure \( P \) using the mole fraction. This relationship is essential for linking the observable, measurable quantities (pressure) to the degree of dissociation and eventually determining the equilibrium constant. By mastering the concept of partial pressure, students can accurately predict the behavior of gases in chemical reactions under various conditions.

At the heart of chemical reactions is stoichiometry, the numerical relationship between quantities of reactants and products. It springs from the balanced chemical equation, providing the proportions in which chemicals combine and form.

In our example, \( \mathrm{N}_{2}\mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2} \) demonstrates a stoichiometry of 1:2—the dissociation of one molecule of \( \mathrm{N}_{2}\mathrm{O}_{4} \) yields two molecules of \( \mathrm{NO}_{2} \). Stoichiometry is not only about counting molecules but also involves calculating masses, volumes, and pressures, thereby bridging the gap between the molecular scale and the real-world measurements. A deep understanding of stoichiometry is necessary to solve for \( \alpha \), and thereby fully comprehend the other aspects like equilibrium constant and partial pressures in chemical equilibrium.