At \(25^{\circ} \mathrm{C}\), a mixture of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) gases are in equilibrium in a cylinder fitted with a movable piston. The concentrations are: \(\left[\mathrm{NO}_{2}\right]=0.0475 \mathrm{M}\) and \(\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]=0.491 \mathrm{M}\). The volume of the gas mixture is halved by pushing down on the piston at constant temperature. Calculate the concentrations of the gases when equilibrium is reestablished. Will the color become darker or lighter after the change? [Hint: \(K_{\mathrm{c}}\) for the dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is \(4.63 \times\) \(10^{-3} . \mathrm{N}_{2} \mathrm{O}_{4}(g)\) is colorless and \(\mathrm{NO}_{2}(g)\) has a brown color.]

Short Answer

Expert verified
Given that the quadratic equation has real and positive roots, the new concentrations will be positive real numbers. And by observing the shift in equilibrium and the properties of the gases involved, it can be concluded that the color will become darker.

Step by step solution

01

Understand the equilibrium

The equilibrium under consideration here is the dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}.\) As per the ideal gas law, the molar concentrations can be used in place of partial pressures to represent the equilibrium in the following manner: \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \leftrightarrow 2 \mathrm{NO}_{2}(g).\) Now understand that if the volume of the container is decreased, the system will respond so as to increase the volume. Since this reaction produces more moles of gas on the right side, the system will shift in the direction of the products, increasing the concentration of \(\mathrm{NO}_{2}.\)
02

Write expressions for initial, change, and equilibrium concentrations

Let the change in concentration of \(\mathrm{NO}_{2}\) be x. Then the change in concentration of \(\mathrm{N}_{2} \mathrm{O}_{4}\) will be half as much \(\frac{-x}{2})\) because one mole of \(\mathrm{N}_{2} \mathrm{O}_{4}\) produces 2 moles of \(\mathrm{NO}_{2}.\) Then the equilibrium concentrations of these compounds will be \(\left[\mathrm{NO}_{2}\right]=0.0475+2x,\) and \(\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]=0.491-x.\) Note that volume is halved so the concentrations are doubled.
03

Substitute values into the equation for Kc

Substitute the expressions for the concentrations into the equation for Kc, which is given by \(K_{c}=\frac{\left[\mathrm{NO}_{2}\right]^{2}}{\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]}.\) The resultant equation is: \(4.63 x 10^{-3} = {(0.0475 M + 2x)^2}} \over {0.491 - x).\)
04

Solve for x

The equation from step 3 is quadratic and thus can be solved using the quadratic formula or by using numerical solver. After simplifying, x will give the change in concentration.
05

Calculate the new equilibrium concentrations and describe the color change

Substitute x back into the expressions for equilibrium concentrations from step 2 to get the new equilibrium concentrations. Since here NO2 has a brown color, when the equilibrium shifts to the right, i.e., more NO2 is produced. Hence, it can be inferred that the color of the gas mixture would become darker on decreasing the volume.

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.

Equilibrium Concentrations
Understanding the concept of equilibrium concentrations is key to unlocking the behavior of chemical reactions that have reached a state of balance. Equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction, and thus the concentrations of the reactants and products remain constant over time.

For example, consider the dissociation of dinitrogen tetroxide (2O4) into nitrogen dioxide (NO2). At equilibrium, even though both the forward and reverse reactions are still occurring, the concentration of N2O4 and NO2 does not change. When an external pressure is applied, such as by halving the volume of the gas mixture, it's crucial to understand that the concentrations of the gases double as a response. By systematically calculating the change in these concentrations, one can predict the behavior of the system when it reaches a new equilibrium state.
Le Chatelier's Principle
Le Chatelier's Principle provides a qualitative prediction of how a system at equilibrium responds to changes in concentration, temperature, and pressure. It states that if an external change is applied to a system at equilibrium, the system will adjust itself in order to counteract that change and a new equilibrium will be established.

In the context of the dissociation equilibrium of N2O4 and NO2, applying pressure by reducing the volume causes the equilibrium to shift in the direction that would reduce pressure. Since 2 moles of NO2 are produced for each mole of N2O4 that dissociates, the system can decrease pressure by consuming NO2 and producing N2O4. However, the external pressure change due to the volume reduction encourages the system to shift towards the side with more gas particles—toward NO2 in this case—thus, the concentration of NO2 will increase.
Equilibrium Constant Kc
The equilibrium constant Kc is a numerical value that characterizes the ratio of the concentrations of products to reactants at equilibrium for a reversible chemical reaction at constant temperature. For the given reaction, the equilibrium constant is expressed as \(K_c=\frac{[NO_{2}]^{2}}{[N_{2}O_{4}]}\). It reflects the extent to which a reaction will proceed before reaching equilibrium.

When the volume of the reaction vessel is halved, we face a change in concentration. The mathematical expression for Kc allows us to set up an equation to solve for the new equilibrium concentrations. Despite the change in volume, Kc remains constant for a given temperature—a crucial point allowing us to predict the concentrations of reactants and products after any disturbance to the system.
Dissociation of N2O4
The dissociation of N2O4 into NO2 is an equilibrium reaction that can be represented as \(N_{2}O_{4} (g) \leftrightarrow 2 NO_{2} (g)\). This reaction is endothermic, meaning that it absorbs heat. As N2O4 gas dissociates, it produces NO2 gas, which is brown in color, thereby increasing the color intensity of the gas mixture.

When the dissociation is subjected to a change in conditions, such as an increase in volume or a decrease in pressure, the dissociation reaction's direction will shift to re-establish equilibrium. Since N2O4 is colorless, its increase leads to a lighter color, whereas an increase in NO2 concentration results in a darker color. Calculating the changes in the equilibrium concentrations can also help in predicting the visual color change, as observed in our exercise.

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

The equilibrium constant \(K_{P}\) for the reaction $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ is \(5.60 \times 10^{4}\) at \(350^{\circ} \mathrm{C} . \mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) are mixed initially at 0.350 atm and 0.762 atm, respectively, at \(350^{\circ} \mathrm{C}\). When the mixture equilibrates, is the total pressure less than or greater than the sum of the initial pressures, 1.112 atm?

The equilibrium constant \(K_{P}\) for the reaction $$ \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$ is 1.05 at \(250^{\circ} \mathrm{C}\). The reaction starts with a mixture of \(\mathrm{PCl}_{5}, \mathrm{PCl}_{3},\) and \(\mathrm{Cl}_{2}\) at pressures of \(0.177 \mathrm{~atm}\) 0.223 atm, and 0.111 atm, respectively, at \(250^{\circ} \mathrm{C}\). When the mixture comes to equilibrium at that temperature, which pressures will have decreased and which will have increased? Explain why.

List four factors that can shift the position of an equilibrium. Which one can alter the value of the equilibrium constant?

Consider this reaction at equilibrium in a closed container: $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ What would happen if (a) the volume is increased, (b) some \(\mathrm{CaO}\) is added to the mixture, (c) some \(\mathrm{CaCO}_{3}\) is removed, (d) some \(\mathrm{CO}_{2}\) is added to the mixture, (e) a few drops of an \(\mathrm{NaOH}\) solution are added to the mixture, (f) a few drops of an \(\mathrm{HCl}\) solution are added to the mixture (ignore the reaction between \(\mathrm{CO}_{2}\) and water \(),(\mathrm{g})\) the temperature is increased?

In this chapter we learned that a catalyst has no effect on the position of an equilibrium because it speeds up both the forward and reverse rates to the same extent. To test this statement, consider a situation in which an equilibrium of the type $$ 2 \mathrm{~A}(g) \rightleftharpoons \mathrm{B}(g) $$ is established inside a cylinder fitted with a weightless piston. The piston is attached by a string to the cover of a box containing a catalyst. When the piston moves upward (expanding against atmospheric pressure), the cover is lifted and the catalyst is exposed to the gases. When the piston moves downward, the box is closed. Assume that the catalyst speeds up the forward reaction \((2 \mathrm{~A} \longrightarrow \mathrm{B})\) but does not affect the reverse process \((\mathrm{B} \longrightarrow 2 \mathrm{~A})\). Suppose the catalyst is suddenly exposed to the equilibrium system as shown below. Describe what would happen subsequently. How does this "thought" experiment convince you that no such catalyst can exist?

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