/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 Consider the equilibrium system ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider the equilibrium system \(3 \mathrm{~A} \rightleftharpoons \mathrm{B}\). Sketch the change in concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) with time for these situations: (a) Initially only A is present; (b) initially only B is present; (c) initially both A and \(\mathrm{B}\) are present (with \(\mathrm{A}\) in higher concentration). In each case, assume that the concentration of \(\mathrm{B}\) is higher than that of \(\mathrm{A}\) at equilibrium.

Short Answer

Expert verified
A decreases, B increases to equilibrium with B > A in all cases.

Step by step solution

01

Understanding the Equilibrium System

The equilibrium system given is \(3 \mathrm{~A} \rightleftharpoons \mathrm{B}\). This indicates that three molecules of \(\mathrm{A}\) convert into one molecule of \(\mathrm{B}\) and vice versa. At equilibrium, the rates of the forward and reverse reactions are equal, which fixes the concentrations of \(\mathrm{A}\) and \(\mathrm{B}\).
02

Case (a) - Initially Only A is Present

In this scenario, at time zero, the concentration of \(\mathrm{A}\) is high and \(\mathrm{B}\) is zero. As the system moves towards equilibrium, \(\mathrm{A}\) decreases and \(\mathrm{B}\) increases. Eventually, \(\mathrm{B}\) reaches a higher concentration than \(\mathrm{A}\) as given by the equilibrium conditions.
03

Case (b) - Initially Only B is Present

Here, initially, the concentration of \(\mathrm{B}\) is high while \(\mathrm{A}\) is zero. \(\mathrm{B}\) will convert to \(\mathrm{A}\) until equilibrium is reached, resulting in a decrease of \(\mathrm{B}\) and an increase of \(\mathrm{A}\). Again, \(\mathrm{B}\) will end up with higher concentration than \(\mathrm{A}\) at equilibrium.
04

Case (c) - Initially, High Concentration A and Some B

In this setup, both substances are present initially with \(\mathrm{A}\) at a higher concentration. \(\mathrm{A}\) decreases and \(\mathrm{B}\) increases as the system moves to equilibrium. Nevertheless, the system will adjust such that \(\mathrm{B}\)'s concentration surpasses that of \(\mathrm{A}\) at equilibrium.
05

Graphical Representation of Concentrations

Plot graphs with time on the x-axis and concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) on the y-axis. For all cases, \(\mathrm{A}\) decreases and \(\mathrm{B}\) increases until equilibrium is reached. Case (a) starts with \(\mathrm{A}\) only, case (b) starts with \(\mathrm{B}\) only, and case (c) starts with more \(\mathrm{A}\) than \(\mathrm{B}\). Equilibrium will always reflect \(\mathrm{B}\) at higher levels than \(\mathrm{A}\).

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 91Ó°ÊÓ!

Key Concepts

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

Reaction Rates
Understanding reaction rates is essential when analyzing chemical equilibrium. Reaction rates refer to the speed at which reactants convert into products over time. In a chemical reaction like \(3 \mathrm{~A} \rightleftharpoons \mathrm{B}\), the rate at which \(A\) becomes \(B\) and vice versa is crucial for achieving equilibrium.

- **Forward Reaction Rate:** Initially, when only \(A\) is present, \(A\) begins to transform into \(B\). The rate of this conversion is known as the forward reaction rate.

- **Reverse Reaction Rate:** Conversely, the reverse reaction rate represents how \(B\) can convert back into \(A\).

These rates are initially unequal but change as the reaction progresses. They adjust until both rates become equal, indicating the system has reached equilibrium.
Equilibrium Concentrations
Once a chemical reaction reaches equilibrium, the concentrations of the reactants and products become stable. In the exercise of \(3 \mathrm{~A} \rightleftharpoons \mathrm{B}\), establishing equilibrium means the concentrations stop changing with time because the reaction rates are balanced.

- If only \(A\) starts in the system, \(A\) will reduce, and \(B\) will increase until the concentration of \(B\) surpasses that of \(A\), achieving equilibrium conditions specified by the problem.

- When the system begins solely with \(B\), it must transform some of \(B\) back into \(A\), stabilizing at an equilibrium where \(B\) is again higher than \(A\).

- In cases where both \(A\) and \(B\) are present initially, the concentrations shift until \(B\) once again is in greater concentration than \(A\), creating the equilibrium stated for the exercise. Understanding these concentration dynamics is key to analyzing equilibrium states.
Dynamic Equilibrium
Dynamic equilibrium plays a significant role in how equilibrium is maintained in a chemical reaction. When a system reaches this state, reactions continue to occur, but the overall concentrations stay constant.

- **Continuous Reaction Activity:** In dynamic equilibrium, the reactions have not stopped. Instead, the forward and reverse reactions occur at the same rate, meaning that \(3 \mathrm{~A} \rightarrow \mathrm{B}\) and \(\mathrm{B} \rightarrow 3 \mathrm{~A}\) happen simultaneously, keeping concentrations unchanged.

- **Constant Concentrations:** This particular dynamic means that although molecules are continuously reacting, the amounts of \(A\) and \(B\) in the system do not fluctuate.

The concept of dynamic equilibrium helps understand why systems can stay stable externally while remaining active at the molecular level.

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

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.]

Consider the following reaction at equilibrium: $$\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)$$ From the following data, calculate the equilibrium constant (both \(K_{P}\) and \(K_{\mathrm{c}}\) ) at each temperature. Is the reaction endothermic or exothermic? $$\begin{array}{ccc}\text { Temperature }\left({ }^{\circ} \mathbf{C}\right) & {[\mathbf{A}]} & {[\mathbf{B}]} \\\\\hline 200 & 0.0125 & 0.843 \\\300 & 0.171 & 0.764 \\\400 & 0.250 & 0.724\end{array}$$

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 HCl solution are added to the mixture (ignore the reaction between \(\mathrm{CO}_{2}\) and water), \((\mathrm{g})\) the temperature is increased?

Consider this reaction: $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$The equilibrium constant \(K_{P}\) for the reaction is \(1.0 \times\) \(10^{-15}\) at \(25^{\circ} \mathrm{C}\) and 0.050 at \(2200^{\circ} \mathrm{C}\). Is the formation of nitric oxide endothermic or exothermic? Explain your answer.

The equilibrium constant \(K_{P}\) for the following reaction is found to be \(4.31 \times 10^{-4}\) at \(375^{\circ} \mathrm{C}\) : $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ In a certain experiment a student starts with 0.862 atm of \(\mathrm{N}_{2}\) and \(0.373 \mathrm{~atm}\) of \(\mathrm{H}_{2}\) in a constant-volume vessel at \(375^{\circ} \mathrm{C}\). Calculate the partial pressures of all species when equilibrium is reached.

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.