/*! 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 17 Part 1: You run the chemical rea... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Part 1: You run the chemical reaction \(\mathrm{C}(a q)+\mathrm{D}(a q) \rightleftharpoons\) \(2 \mathrm{E}(a q)\) at \(25^{\circ} \mathrm{C}\). The equilibrium constant \(K_{c}\) for the reaction at this temperature is \(2.0 .\) a. Write the equilibrium-constant expression for the reaction. b. Can you come up with some possible concentrations of \(\mathrm{C}\), \(\mathrm{D}\), and \(\mathrm{E}\) that you might observe when the reaction has reached equilibrium at \(25^{\circ} \mathrm{C} ?\) What are these values? c. A student says that only a very limited number of concentrations for \(\mathrm{C}, \mathrm{D}\), and \(\mathrm{E}\) are possible at equilibrium. Is this true? State why you think this is true or is not true. d. If you start with \(1.0 M\) concentrations of both \(\mathrm{C}\) and \(\mathrm{D}\) and allow the reaction to come to equilibrium, would you expect the concentration of \(\mathrm{C}\) to have decreased to zero? If not, what would you expect for the concentration of \(\mathrm{C} ?\) (An approximate value is fine.)

Short Answer

Expert verified
No, only limited concentrations are possible at equilibrium. Concentrations can vary and \([C]\) does not decrease to zero.

Step by step solution

01

Writing the Equilibrium Expression

For the reaction \( \mathrm{C}(aq) + \mathrm{D}(aq) \rightleftharpoons 2 \mathrm{E}(aq) \), the equilibrium constant \( K_c \) is expressed in terms of the concentrations of the products and reactants. The formula is: \[ K_c = \frac{[\mathrm{E}]^2}{[\mathrm{C}][\mathrm{D}]} \] where \([\mathrm{E}]\), \([\mathrm{C}]\), and \([\mathrm{D}]\) are the molar concentrations of \(\mathrm{E}\), \(\mathrm{C}\), and \(\mathrm{D}\) at equilibrium.
02

Determining Possible Equilibrium Concentrations

Since \( K_c = 2.0 \), we want values such that \( \frac{[\mathrm{E}]^2}{[\mathrm{C}][\mathrm{D}]} = 2.0 \). If we assume \([\mathrm{C}] = 0.5\, M\) and \([\mathrm{D}] = 0.5\, M\), then \([\mathrm{E}]\) can be found using: \[ 2.0 = \frac{[\mathrm{E}]^2}{0.5 \times 0.5} \] \[ 2.0 = \frac{[\mathrm{E}]^2}{0.25} \] \[ [\mathrm{E}]^2 = 0.5 \] \[ [\mathrm{E}] = \sqrt{0.5} \approx 0.71\, M \].
03

Assessing the Variety of Equilibrium Concentrations

Numerous combinations of concentrations for \(\mathrm{C}\), \(\mathrm{D}\), and \(\mathrm{E}\) can satisfy the equation \( \frac{[\mathrm{E}]^2}{[\mathrm{C}][\mathrm{D}]} = 2.0 \). Therefore, it is not true that only a very limited number of concentrations can exist. Variability in initial conditions or reaction pathways can lead to different concentrations at equilibrium.
04

Evaluating Concentration Changes from Initial Conditions

Starting with \([\mathrm{C}] = 1.0\, M\) and \([\mathrm{D}] = 1.0\, M\), we posit some \(x\) moles of \(\mathrm{C}\) react to reach equilibrium, making \([\mathrm{E}] = 2x\). At equilibrium, \([\mathrm{C}] = 1-x\) and \([\mathrm{D}] = 1-x\). Using the equilibrium constant expression: \[ 2.0 = \frac{(2x)^2}{(1-x)(1-x)} \] Solve for \(x\) to find that \([\mathrm{C}]\) will not decrease to zero, and calculating this evenly gives an approximate change. After solving, \([\mathrm{C}] \approx 0.5\, M \) at equilibrium.

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.

Equilibrium Constant Expression
In chemistry, the Equilibrium Constant Expression (Kc) is crucial for describing the state of a reversible reaction at equilibrium. For the given reaction, \( \mathrm{C}(aq) + \mathrm{D}(aq) \rightleftharpoons 2 \mathrm{E}(aq) \), the equilibrium constant expression is essential to determine the relationship between the concentrations of reactants and products. The formula for this specific reaction is:
  • \( K_c = \frac{[\mathrm{E}]^2}{[\mathrm{C}][\mathrm{D}]} \)
Here, [E], [C], and [D] represent the molar concentrations of substances E, C, and D at equilibrium. The numerical value of \( K_c \) gives insight into the position of equilibrium. If \( K_c \) is greater than 1, the reaction favors product formation. On the other hand, if \( K_c \) is less than 1, the reactants are favored when equilibrium is established.
This expression allows us to predict the concentration of one participant if the others are known, helping chemists understand how equilibria shift under different conditions.
Reaction Quotient
The Reaction Quotient, represented as \( Q_c \), is calculated similarly to the equilibrium constant but applies to reactions not yet at equilibrium. It helps to predict which direction a reaction will shift to reach equilibrium. For the given reaction, the reaction quotient would be:
  • \( Q_c = \frac{[\mathrm{E}]^2}{[\mathrm{C}][\mathrm{D}]} \)
By comparing \( Q_c \) with \( K_c \):
  • If \( Q_c < K_c \), the forward reaction is favored, producing more products to achieve equilibrium.
  • If \( Q_c > K_c \), the reverse reaction is favored, converting products back into reactants.
  • If \( Q_c = K_c \), the system is already at equilibrium.
Understanding \( Q_c \) is fundamental for predicting how a reaction will behave immediately after it has started or if changes are made to the concentration of reactants or products.
Le Chatelier's Principle
Le Chatelier's Principle is a key concept in predicting how a change in conditions affects the position of equilibrium in a chemical reaction. According to this principle, if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium will shift to counteract the effect of the change.
  • Concentration: Adding more of a reactant or product causes the system to shift towards the side that will consume the added substance. If a substance is removed, the system shifts to produce more of what was removed.
  • Temperature: Increasing the temperature of an endothermic reaction shifts equilibrium to the right, favoring the formation of products, whereas for an exothermic reaction, it favors reactants. Decreasing the temperature has the opposite effect.
  • Pressure: Alterations in pressure (considering gaseous reactions) affect equilibrium based on the number of moles of gas on each side. Increasing pressure shifts equilibrium to the side with fewer gas molecules, while decreasing pressure does the opposite.
Le Chatelier's Principle is a powerful tool in chemical engineering and laboratory settings for controlling reactions to optimize the yield and ensure safety by understanding the effects of manipulating external 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

The equilibrium constant \(K_{c}\) for the reaction $$ \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) $$ equals \(4.1\) at \(300^{\circ} \mathrm{C}\). a. A sample of \(35.8 \mathrm{~g}\) of \(\mathrm{PCl}_{5}\) is placed in a 5.0-L reaction vessel and heated to \(300^{\circ} \mathrm{C}\). What are the equilibrium concentrations of all of the species? b. What fraction of \(\mathrm{PCl}_{5}\) has decomposed? c. If \(35.8 \mathrm{~g}\) of \(\mathrm{PCl}_{5}\) were placed in a \(1.0\) - \(\mathrm{L}\) vessel, what qualitative effect would this have on the fraction of \(\mathrm{PCl}_{5}\) that has decomposed (give a qualitative answer only; do not do the calculation)? Why?

A gaseous mixture containing \(1.00 \mathrm{~mol}\) each of \(\mathrm{CO}\), \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CO}_{2}\), and \(\mathrm{H}_{2}\) is exposed to a zinc oxide-copper oxide catalyst at \(1000^{\circ} \mathrm{C}\). The reaction is $$ \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) $$ and the equilibrium constant \(K_{c}\) is \(0.58\) at \(1000^{\circ} \mathrm{C}\). What is the direction of reaction (forward or reverse) as the mixture attains equilibrium?

The following reaction has an equilibrium constant \(K_{c}\) equal to \(3.59\) at \(900^{\circ} \mathrm{C}\). $$ \mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g) $$ For each of the following compositions, decide whether the reaction mixture is at equilibrium. If it is not, decide which direction the reaction should go. a. \(\left[\mathrm{CH}_{4}\right]=1.26 M,\left[\mathrm{H}_{2} \mathrm{~S}\right]=1.32 M,\left[\mathrm{CS}_{2}\right]=1.43 M\) \(\left[\mathrm{H}_{2}\right]=1.12 \mathrm{M}\) b. \(\left[\mathrm{CH}_{4}\right]=1.25 M,\left[\mathrm{H}_{2} \mathrm{~S}\right]=1.52 M,\left[\mathrm{CS}_{2}\right]=1.15 M\) \(\left[\mathrm{H}_{2}\right]=1.73 \mathrm{M}\) c. \(\left[\mathrm{CH}_{4}\right]=1.30 M,\left[\mathrm{H}_{2} \mathrm{~S}\right]=1.41 M,\left[\mathrm{CS}_{2}\right]=1.10 M\) \(\left[\mathrm{H}_{2}\right]=1.20 \mathrm{M}\) d. \(\left[\mathrm{CH}_{4}\right]=1.56 M,\left[\mathrm{H}_{2} \mathrm{~S}\right]=1.43 M,\left[\mathrm{CS}_{2}\right]=1.23 M\) \(\left[\mathrm{H}_{2}\right]=1.91 M\)

The amount of nitrogen dioxide formed by dissociation of dinitrogen tetroxide, $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ increases as the temperature rises. Is the dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) endothermic or exothermic?

The following reaction is important in the manufacture of sulfuric acid. $$ \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) $$ At \(900 \mathrm{~K}, 0.0216 \mathrm{~mol}\) of \(\mathrm{SO}_{2}\) and \(0.0148 \mathrm{~mol}\) of \(\mathrm{O}_{2}\) are sealed in a 1.00-L reaction vessel. When equilibrium is reached, the concentration of \(\mathrm{SO}_{3}\) is determined to be \(0.0175 \mathrm{M}\). Calculate \(K_{c}\) for this reaction.

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.