/*! 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 84 An equilibrium mixture of \(\mat... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 84

An equilibrium mixture of \(\mathrm{SO}_{3}, \mathrm{SO}_{2}\), and \(\mathrm{O}_{2}\) at \(727^{\circ} \mathrm{C}\) is \(0.0160 \mathrm{MSO}_{3}, 0.0056 \mathrm{MSO}_{2}\), and \(0.0021 \mathrm{M} \mathrm{O}_{2}\). What is the value of \(K_{c}\) for the following reaction? $$ \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) $$

Short Answer

Expert verified
The value of \( K_c \) is approximately 62.29.

Step by step solution

01

Understanding the Reaction

The chemical reaction provided is \( \text{SO}_2 (g) + \frac{1}{2} \text{O}_2 (g) \rightleftharpoons \text{SO}_3(g) \). We need to find the equilibrium constant \( K_c \) for this reaction given the equilibrium concentrations of \( \text{SO}_3 \), \( \text{SO}_2 \), and \( \text{O}_2 \).
02

Writing the Expression for Kc

The expression for the equilibrium constant \( K_c \) is given by:\[ K_c = \frac{[\text{SO}_3]}{[\text{SO}_2][\text{O}_2]^{1/2}} \]Here, the concentration terms \([\text{SO}_3]\), \([\text{SO}_2]\), and \([\text{O}_2]\) are in molarity (M).
03

Substitute the Given Concentrations

Plug the given concentrations into the \( K_c \) expression:- \([\text{SO}_3] = 0.0160\, \text{M}\)- \([\text{SO}_2] = 0.0056\, \text{M}\)- \([\text{O}_2] = 0.0021\, \text{M}\)Thus, the formula becomes:\[ K_c = \frac{0.0160}{0.0056 \times (0.0021)^{1/2}} \]
04

Calculating the Value of Kc

Calculate the value by solving the expression:- First, calculate \((0.0021)^{1/2} \approx 0.0458\)- Then calculate \(0.0056 \times 0.0458 \approx 0.00025688\)- Finally, \( K_c = \frac{0.0160}{0.00025688} \approx 62.29\)

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Key Concepts

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

Equilibrium Constant (Kc)
The equilibrium constant, denoted as \( K_c \), is a fundamental concept in chemical equilibrium. It describes the ratio of the concentrations of products to reactants at equilibrium, each raised to the power of their coefficients in the balanced equation. For the reaction \( \text{SO}_2 (g) + \frac{1}{2} \text{O}_2 (g) \rightleftharpoons \text{SO}_3 (g) \), the expression for \( K_c \) is formulated as:\[ K_c = \frac{[\text{SO}_3]}{[\text{SO}_2][\text{O}_2]^{1/2}} \]
  • The square root for \([\text{O}_2]\) appears because its coefficient is \( \frac{1}{2} \) in the reaction.

Concentration values are substituted to calculate \( K_c \), reflecting the balance of forward and reverse reactions at this temperature. A higher \( K_c \) suggests products are favored, while a lower one indicates reactants predominate.
Equilibrium Concentrations
Understanding equilibrium concentrations is critical in evaluating the state of a reversible reaction. In this scenario, the equilibrium concentrations of \( \text{SO}_3 \), \( \text{SO}_2 \), and \( \text{O}_2 \) are key to determining the \( K_c \). Given values are:
  • \([\text{SO}_3] = 0.0160\, \text{M}\)
  • \([\text{SO}_2] = 0.0056\, \text{M}\)
  • \([\text{O}_2] = 0.0021\, \text{M}\)

These concentrations represent the specific point at which the forward and reverse reactions occur at the same rate. Calculating \( K_c \) from these values provides insight into how the system is balanced between the reactants and the products. This can show how conditions like temperature might affect where equilibrium lies, though other factors outside concentration are constant.
Reaction Quotients
A reaction quotient, denoted as \( Q \), is similar to \( K_c \) but is calculated using initial concentrations before a system has reached equilibrium. The expression for \( Q \) has the same format as \( K_c \):\[ Q = \frac{[\text{SO}_3]}{[\text{SO}_2][\text{O}_2]^{1/2}} \]
  • If \( Q = K_c \), the system is at equilibrium.
  • If \( Q < K_c \), the reaction will proceed forward toward products to reach equilibrium.
  • If \( Q > K_c \), the reaction will shift backward to form more reactants.

For the given exercise, calculating \( Q \) with initial concentrations can predict the direction the reaction will shift as it moves toward equilibrium. Often, calculating the \( K_c \) values and understanding \( Q \) serves as the cornerstone for determining how different conditions influence a chemical system's equilibrium.

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Most popular questions from this chapter

Iodine, \(\mathrm{I}_{2}\), is a blue-black solid, but it easily vaporizes to give a violet vapor. At high temperatures, this molecular substance dissociates to atoms: $$ \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) $$ An absent-minded professor measured equilibrium constants for this dissociation at two temperatures, \(700^{\circ} \mathrm{C}\) and \(800^{\circ} \mathrm{C}\). He obtained the \(K_{p}\) values \(0.01106\) and \(0.001745\), but he couldn't remember which value went with what temperature. Can you please help him assign temperatures to the equilibrium constants? State your argument carefully.

The equilibrium constant \(K_{c}\) for the synthesis of methanol, \(\mathrm{CH}_{3} \mathrm{OH}\) $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ is \(4.3\) at \(250^{\circ} \mathrm{C}\) and \(1.8\) at \(275^{\circ} \mathrm{C}\). Is this reaction endothermic or exothermic?

Gaseous acetic acid molecules have a certain tendency to form dimers. (A dimer is a molecule formed by the association of two identical, simpler molecules.) The equilibrium constant \(K_{c}\) at \(25^{\circ} \mathrm{C}\) for this reaction is \(3.2 \times 10^{4}\). a. If the initial concentration of \(\mathrm{CH}_{3} \mathrm{COOH}\) monomer (the simpler molecule) is \(4.0 \times 10^{-4} M\), what are the concentrations of monomer and dimer when the system comes to equilibrium? (The simpler quadratic equation is obtained by assuming that all of the acid molecules have dimerized and then some of it dissociates to monomer.) b. Why do acetic acid molecules dimerize? What type of structure would you draw for the dimer? c. As the temperature increases, would you expect the percentage of dimer to increase or decrease? Why?

For the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ show that $$ K_{c}=K_{p}(R T)^{2} $$ Do not use the formula \(K_{p}=K_{c}(R T)^{\Delta n}\) given in the text. Start from the fact that \(P_{i}=[i] R T\), where \(P_{i}\) is the partial pressure of substance \(i\) and \([i]\) is its molar concentration. Substitute into \(K_{c}\).

The equilibrium-constant expression for a gas reaction is $$ K_{c}=\frac{\left[\mathrm{CO}_{2}\right]^{3}\left[\mathrm{H}_{2} \mathrm{O}\right]^{4}}{\left[\mathrm{C}_{3} \mathrm{H}_{8}\right]\left[\mathrm{O}_{2}\right]^{5}} $$ Write the balanced chemical equation corresponding to this expression.
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