Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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) \(\alpha=\left[\frac{K_{P}}{4+K_{P}}\right]^{1 / 2}\)

Short Answer

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

Step by step solution

01

- 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.
02

- 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, \( K_p = \frac{(P_{NO_2})^2}{P_{N2O4}} \).
03

- 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 \( P_{N2O4} = \frac{1-\alpha}{1+2\alpha}P \).
04

- 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.
05

- Solve for α

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

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

One mole of ethanol is treated with one mole of ethanoic acid at \(25^{\circ} \mathrm{C}\). Onefourth of the acid changes into ester at equilibrium. The equilibrium constant for the reaction will be (a) \(1 / 9\) (b) \(4 / 9\) (c) 9 (d) \(9 / 4\)

In equilibrium: \(\mathrm{SCN}(\mathrm{aq})+\mathrm{Fe}^{+3}(\mathrm{aq}) \rightleftharpoons\left[\mathrm{Fe}(\mathrm{SCN})^{2+}\right](\mathrm{aq})\) \(\begin{array}{lll}\text { colourless } & \text { yellow } & \text { deep red }\end{array}\) If thiocyanate ions are added in equilibrium mixture (a) The solution becomes colourless. (b) The yellow colour of solution deepens. (c) The red colour of the solution deepens. (d) Concentration of \([\mathrm{Fe}(\mathrm{SCN})]^{2+}\) ion will decrease.

An aqueous solution of volume \(500 \mathrm{ml}\), when the reaction: \(2 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s}) \rightleftharpoons\) \(\mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})\), reached equilibrium, the concentration of \(\mathrm{Cu}^{2+}\) ions was \(x \mathrm{M}\). To this solution, \(500 \mathrm{~m}\) of water is added. At the new equilibrium, the concentration of \(\mathrm{Cu}^{2+}\) ions would be (a) \(2 x \mathrm{M}\) (b) \(x \underline{M}\) (c) between \(x\) and \(0.5 x \mathrm{M}\) (d) less than \(0.5 x\) M

Forty per cent of a mixture of \(0.2 \mathrm{~mol}\) of \(\mathrm{N}_{2}\) and \(0.6 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) react to give \(\mathrm{NH}_{3}\) according to the equation: \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) at constant temperature and pressure. Then, the ratio of the final volume to the initial volumes of gases is (a) \(4: 5\) (b) \(5: 4\) (c) \(7: 10\) (d) \(8: 5\)

Amounts of \(0.8 \mathrm{~mol}\) of \(\mathrm{PCl}_{5}\) and \(0.2\) mole of \(\mathrm{PCl}_{3}\) are mixed in a \(1 \mathrm{i}\) flask. At equilibrium, \(0.4\) mole of \(\mathrm{PCl}_{3}\) is present. The equilibrium constant for the reaction, \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) will be (a) \(0.05 \mathrm{~mol}^{-1}\) (b) \(0.13 \mathrm{~mol} 1^{-1}\) (c) \(0.013 \mathrm{~mol} 1^{-1}\) (d) \(0.60 \mathrm{~mol} 1^{-1}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free