/*! 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 60 Suppose sulfur dioxide reacts wi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose sulfur dioxide reacts with oxygen at \(25^{\circ} \mathrm{C}\). $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ The equilibrium constant \(K_{c}\) equals \(8.0 \times 10^{35}\) at this temperature. From the magnitude of \(K_{c}\), do you think this reaction occurs to any great extent when equilibrium is reached at room temperature? If an equilibrium mixture is \(1.0 \mathrm{M}\) \(\mathrm{SO}_{3}\) and has equal concentrations of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\), what is the concentration of \(\mathrm{SO}_{2}\) in the mixture? Does this result agree with what you expect from the magnitude of \(K_{c} ?\)

Short Answer

Expert verified
The reaction proceeds extensively to form \(\mathrm{SO}_{3}\), as expected. \([\mathrm{SO}_{2}] = 1.0 \times 10^{-12} \mathrm{M}\), consistent with \(K_{c} = 8.0 \times 10^{35}\).

Step by step solution

01

Understanding the Reaction and Equilibrium Constant

We have the reaction \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\) with an equilibrium constant \(K_{c} = 8.0 \times 10^{35}\) at \(25^{\circ} \mathrm{C}\). The exceedingly large value of \(K_{c}\) suggests that at equilibrium, the concentration of products (\(\mathrm{SO}_{3}\)) is much higher relative to the concentrations of reactants (\(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\)). This implies that the reaction proceeds almost completely to the right.
02

Setting Up the Expression for Kc

The equilibrium expression for the reaction is \[K_{c} = \frac{[\mathrm{SO}_{3}]^2}{[\mathrm{SO}_{2}]^2 [\mathrm{O}_{2}]}\]. Given \([\mathrm{SO}_{3}] = 1.0 \mathrm{M}\), and \([\mathrm{SO}_{2}] = [\mathrm{O}_{2}]\), let \([\mathrm{SO}_{2}] = [\mathrm{O}_{2}] = x\). Substituting these in, we have \[8.0 \times 10^{35} = \frac{(1.0)^2}{x^2 \cdot x} = \frac{1}{x^3}\].
03

Solving for x, the Concentration of SO2

Re-arrange the equation to solve for \(x\): \(x^3 = \frac{1}{8.0 \times 10^{35}}\). Taking the cube root of both sides, \(x = \left(\frac{1}{8.0 \times 10^{35}}\right)^{1/3}\). This simplifies to \(x \approx 1.0 \times 10^{-12} \mathrm{M}\), which means \([\mathrm{SO}_{2}] = [\mathrm{O}_{2}] \approx 1.0 \times 10^{-12} \mathrm{M}\).
04

Analyzing the Results

The calculated equilibrium concentration of \(\mathrm{SO}_{2}\) is extremely low, \(1.0 \times 10^{-12} \mathrm{M}\). This result is consistent with the large \(K_{c}\) value and indicates that almost all reactants convert to the product \(\mathrm{SO}_{3}\), as expected given the high favorability of the forward reaction under these conditions.

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.

Chemical Equilibrium
Chemical equilibrium occurs in a reversible chemical reaction when the rate of the forward reaction equals the rate of the reverse reaction. This results in the concentrations of the reactants and products remaining constant over time. When a system reaches this state, it does not mean that reactions have stopped; rather, they are occurring at equal and opposite rates.
Understanding this concept is crucial in predicting the behavior of a chemical reaction. The position of equilibrium gives us insight into the concentrations of reactants and products when equilibrium is achieved. The equilibrium constant, denoted as \( K_c \), is vital in quantifying this state. It helps us determine the ratio of product to reactant concentrations at equilibrium.
The magnitude of the equilibrium constant can tell us a lot about the reaction's characteristics at equilibrium. If \( K_c \) is large, it suggests that the products are favored, and the reaction proceeds significantly to the right. Conversely, a small \( K_c \) indicates that reactants are favored and little reaction occurs.
Reaction Extent
The extent of a reaction refers to how far a reaction progresses toward forming products before reaching equilibrium. It is closely related to the concept of the equilibrium constant \( K_c \).
In reactions with a large \( K_c \), like the reaction of sulfur dioxide (\( \ ext{SO}_2 \)) with oxygen (\( \ ext{O}_2 \)) to form sulfur trioxide (\( \ ext{SO}_3 \)), the extent of the reaction is significant. This occurs because a large \( K_c \) indicates that the equilibrium position is far to the right, favoring a nearly complete conversion of reactants to products.
This information is crucial when predicting the concentrations of substances in a system at equilibrium. In practical scenarios, knowing the extent of a reaction helps chemists optimize conditions to maximize the desired products.
Sulfur Dioxide
Sulfur dioxide (\( \ ext{SO}_2 \)) is a colorless gas known for its sharp smell. It is a significant air pollutant and is produced by volcanic eruptions and industrial processes, such as the burning of fossil fuels.In the reaction \( 2\ \ ext{SO}_2(g) + \ ext{O}_2(g) \rightleftharpoons 2\ \ ext{SO}_3(g) \), sulfur dioxide reacts with oxygen to form sulfur trioxide. This reaction is notable for its large equilibrium constant at room temperature, indicating that when equilibrium is reached, the concentration of \( \ ext{SO}_2 \) is extremely low.Understanding the behavior of \( \ ext{SO}_2 \) in chemical reactions allows us to better comprehend its impact on the environment and its role in industrial chemical processes. This knowledge is helpful in controlling emissions and minimizing its impact on air quality.
Reaction Quotient
The reaction quotient, denoted as \( Q_c \), is similar to the equilibrium constant but applies to non-equilibrium conditions. It is calculated using the concentrations of reactants and products at any stage prior to reaching equilibrium.
For a reaction \( a\ A + b\ B \rightleftharpoons c\ C + d\ D \), the reaction quotient \( Q_c \) is represented as: \[Q_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}\]
Comparing \( Q_c \) to \( K_c \) helps us predict the direction in which a reaction must proceed to reach equilibrium:
  • If \( Q_c < K_c \), the reaction will proceed to the right, favoring products.
  • If \( Q_c > K_c \), the reaction will shift to the left, favoring reactants.
  • If \( Q_c = K_c \), the system is at equilibrium.
In our exercise, we find that almost all sulfur dioxide and oxygen are converted to sulfur trioxide, consistent with the very large \( K_c \). This indicates that the initial \( Q_c \) was much lower than \( K_c \), driving the reaction strongly toward product formation.

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

Consider the production of ammonia from its elements: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ In a pilot experiment, a mixture of 1.00 part nitrogen gas and 3.00 parts hydrogen gas at a total pressure of \(10.00 \mathrm{~atm}\) was passed through a hot tube over a catalyst at \(450.0^{\circ} \mathrm{C}\). What are the total moles of gas in a 1.000 - \(\mathrm{L}\) volume of this equilibrium mixture? If the equilibrium mixture contains a partial pressure of 0.204 atm of ammonia, how many moles of ammonia are in this 1.000 -L volume? How many moles of \(\mathrm{N}_{2}\) are in this same volume? What is the partial pressure of \(\mathrm{N}_{2}\) in this equilibrium mixture? What is the equilibrium constant \(K_{p}\) for this reaction?

Ammonium hydrogen sulfide, \(\mathrm{NH}_{4} \mathrm{HS}\), is unstable at room temperature and decomposes: $$ \mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g) $$ You have placed some solid ammonium hydrogen sulfide in a closed flask. Which of the following would produce less hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S},\) which is a poisonous gas? a. Removing some \(\mathrm{NH}_{3}\) from the flask b. Adding some \(\mathrm{NH}_{3}\) to the flask c. Removing some of the \(\mathrm{NH}_{4} \mathrm{HS}\) d. Increasing the pressure in the flask by adding helium gas Explain each of your answers.

Suppose \(1.000 \mathrm{~mol} \mathrm{CO}\) and \(3.000 \mathrm{~mol} \mathrm{H}_{2}\) are put in a 10.00 - \(\mathrm{L}\) vessel at \(1200 \mathrm{~K}\). The equilibrium constant \(K_{c}\) for $$ \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ equals 3.92. Find the equilibrium composition of the reaction mixture.

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 equilibrium was studied by analyzing the equilibrium mixture for the amount of \(\mathrm{HCl}\) produced. $$ \mathrm{LaCl}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{LaOCl}(s)+2 \mathrm{HCl}(g) $$ A vessel whose volume was \(1.25 \mathrm{~L}\) was filled with \(0.0125 \mathrm{~mol}\) of lanthanum(III) chloride, \(\mathrm{LaCl}_{3}\), and \(0.0250 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\). After the mixture came to equilibrium in the closed vessel at \(619^{\circ} \mathrm{C}\), the gaseous mixture was removed and dissolved in more water. Sufficient silver ion was added to precipitate the chloride ion completely as silver chloride. If \(3.59 \mathrm{~g} \mathrm{AgCl}\) was obtained, what is the value of \(K_{c}\) at \(619^{\circ} \mathrm{C} ?\)

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.