/*! 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 8 The following reaction is found ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 8

The following reaction is found to be at equilibrium at 25掳C: \(2 \mathrm{SO}_{3}(g) \leftrightarrow \mathrm{O}_{2}(g)+2 \mathrm{SO}_{2}(g) \quad \Delta H=-198 \mathrm{kJ} / \mathrm{mol}\) The value for \(K_{\mathrm{c}}\) at \(25^{\circ} \mathrm{C}\) is \(8.1 .\) What must happen in order for the reaction to reach equilibrium if the initial concentrations of all three species was 2.0 \(M\) ? (A) The rate of the forward reactions would increase, and \(\left[\mathrm{SO}_{3}\right]\) would decrease. (B) The rate of the reverse reaction would increase, and \(\left[\mathrm{SO}_{2}\right]\) would decrease. (C) Both the rate of the forward and reverse reactions would increase, and the value for the equilibrium constant would also increase. (D) No change would occur in either the rate of reaction or the concentrations of any of the species.

Short Answer

Expert verified
The correct answer is (A): The rate of the forward reaction would increase, and the concentration of \(SO_{3}\) would decrease.

Step by step solution

01

Write down the balanced equation and expression for Qc

The balanced reaction is: \(2 SO_{3}(g) \leftrightarrow O_{2}(g)+2 SO_{2}(g)\). The reaction quotient \(Q_{c}\) is given by: \[Q_{c} = \frac{[O_{2}][SO_{2}]^{2}}{[SO_{3}]^{2}}\]
02

Calculation of initial Qc

Based on the given initial concentrations, the reaction quotient Qc is calculated as follow: \[Q_{c} = \frac{(2.0)(2.0)^{2}}{(2.0)^{2}} = \frac{8}{4} = 2\]
03

Compare Qc with Kc

The value of Qc calculated in step 2 is less than the given Kc (2 < 8.1). This means the reaction will proceed to the right to achieve equilibrium.
04

Interpretation

Since the reaction moves to the right (forward direction), the concentration of the reactant \(SO_{3}\) will decrease, while the concentration of the products \(O_{2}\) and \(SO_{2}\) will increase. Hence, the forward reaction rate will increase.

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 91影视!

Key Concepts

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

Reaction Quotient
The Reaction Quotient, denoted as \(Q_c\), is a crucial concept in chemical equilibrium that helps predict the direction of a reaction. It is calculated using the initial concentrations of reactants and products in a reaction and is expressed through the equation: \[Q_c = \frac{[O_{2}][SO_{2}]^{2}}{[SO_{3}]^{2}}\]In this equation, the concentrations of the products \( [O_2] \) and \( [SO_2] \) are in the numerator, raised to the power of their respective coefficients in the balanced chemical equation, while the concentration of the reactant \( [SO_3] \) is in the denominator.
  • A \(Q_c\) value greater than the equilibrium constant \(K_c\) suggests the reaction will proceed in the reverse direction to reach equilibrium.
  • Conversely, if \(Q_c\) is less than \(K_c\), the reaction moves forward to form more products.
Understanding \(Q_c\) helps in predicting how a system not at equilibrium will shift to reach that state, making it an indispensable tool for solving chemical equilibrium problems.
Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept in understanding how a chemical system at equilibrium responds to external changes. The principle states that if a system at equilibrium experiences a change in concentration, temperature, or pressure, the system will adjust to minimize that change and re-establish equilibrium.
For instance, in our reaction, if the concentrations of reactants or products are altered, the system will react to counteract the change:
  • Increasing the concentration of \(SO_3\) would cause the system to shift towards the products \(O_2\) and \(SO_2\) to reduce the concentration of \(SO_3\).
  • If \(SO_2\) or \(O_2\) concentration increases, the reaction will shift towards the reactants to restore equilibrium.
Le Chatelier's Principle is not just restricted to concentration changes. It can also be applied to changes in temperature and pressure, offering a comprehensive framework for predicting and analyzing shifts in equilibrium.
Equilibrium Constant
The Equilibrium Constant, denoted \(K_c\), is a measure of the ratio of concentrations of products to reactants at equilibrium, each raised to the power of their coefficients in the balanced equation. In our example of \(2 SO_{3}(g) \leftrightarrow O_{2}(g)+2 SO_{2}(g)\), \(K_c\) is given as 8.1 at 25掳C.
It provides insight into the position of the equilibrium:
  • A large \(K_c\) (much greater than 1) indicates the reaction favors products at equilibrium, meaning there's a higher concentration of products compared to reactants.
  • A small \(K_c\) (much less than 1) suggests reactants are favored.
The value of \(K_c\) is temperature-dependent, meaning it will change if the reaction is subjected to a different temperature. Thus, understanding \(K_c\) helps in determining not only the direction a reaction will move towards to establish equilibrium when \(Q_c\) is not equal to \(K_c\), but also the extent to which products or reactants are formed at equilibrium.
Endothermic and Exothermic Reactions
Reactions can either absorb heat (endothermic) or release heat (exothermic). The given reaction \(2 \, SO_3(g) \leftrightarrow O_2(g) + 2 \, SO_2(g)\) has a \( \Delta H \) of -198 kJ/mol, indicating an exothermic reaction.
This means that the reaction releases heat as it proceeds, and understanding this aspect is crucial when considering external influences like changes in temperature.
  • For an exothermic reaction, increasing the temperature will shift the equilibrium towards the reactants as the system seeks to absorb the excess heat.
  • Conversely, decreasing the temperature favors the formation of products, as it maintains the release of heat preferred by the reaction dynamics.
Understanding whether a reaction is endothermic or exothermic helps in predicting how the equilibrium position will change with temperature fluctuations, directly linking it to Le Chatelier's Principle.

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

Questions 32-36 refer to the following. Two half-cells are set up as follows: Half-Cell A: Strip of \(\mathrm{Cu}(s)\) in \(\mathrm{CuNO}_{3}(a q)\) Half-Cell B: Strip of \(\mathrm{Zn}(s)\) in \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (aq) When the cells are connected according to the diagram below, the following reaction occurs: GRAPH CAN'T COPY $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ Correctly identify the anode and cathode in this reaction as well as where oxidation and reduction are taking place. (A) Cu is the anode where oxidation occurs, and Zn is the cathode where reduction occurs. (B) Cu is the anode where reduction occurs, and Zn is the cathode where oxidation occurs. (C) Zn is the anode where oxidation occurs, and Cu is the cathode where reduction occurs. (D) Zn is the anode where reduction occurs, and Cu is the cathode where oxidation occurs.

Which substance would have the highest vapor pressure? (A) Ethanol, because of the hybridization of its carbon atoms (B) Actone, because it exhibits the weakest intermolecular forces (C) Ethylene glycol, because it has the most lone pairs assigned to individual atoms (D) All three substances would have similar vapor pressure because they have a similar number of electrons.

Questions 32-36 refer to the following. Two half-cells are set up as follows: Half-Cell A: Strip of \(\mathrm{Cu}(s)\) in \(\mathrm{CuNO}_{3}(a q)\) Half-Cell B: Strip of \(\mathrm{Zn}(s)\) in \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (aq) When the cells are connected according to the diagram below, the following reaction occurs: GRAPH CAN'T COPY $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ If the \(\mathrm{Cu}^{+}+e^{-} \rightarrow \mathrm{Cu}(s)\) half-reaction has a standard reduction potential of \(+0.52 \mathrm{V},\) what is the standard reduction potential for the \(\mathrm{Zn}^{2+}+2 e^{-} \rightarrow \mathrm{Zn}(s)\) half-reaction? (A) \(+0.76 \mathrm{V}\) (B) \(-0.76 \mathrm{V}\) (C) \(+0.24 \mathrm{V}\) (D) \(-0.24 \mathrm{V}\)

The following mechanism is proposed for a reaction: \(\begin{array}{ll}{2 \mathrm{A} \rightarrow \mathrm{B}} & {\text { (fast equilibrium) }} \\ {\mathrm{C}+\mathrm{B} \rightarrow \mathrm{D}} & {\text { (slow) }} \\ {\mathrm{D}+\mathrm{A} \rightarrow \mathrm{E}} & {\text { (fast) }}\end{array}\) Which of the following is the correct rate law for compete reaction? (A) Rate \(=k[\mathrm{C}]^{2}[\mathrm{B}]\) (B) Rate \(=k\left[\mathrm{Cl}[\mathrm{A}]^{2}\right.\) (C) Rate \(=k[\mathrm{C}][\mathrm{A}]^{3}\) (D) Rate \(=k[\mathrm{D}][\mathrm{A}]\)

Which compound, \(\mathrm{CaCl}_{2}\) or \(\mathrm{CaO}\) , would you expect to have a high melting point? Why? (A) \(\mathrm{CaCl}_{2}\) because there are more ions per lattice unit (B) \(\mathrm{CaCl}_{2}\) because a chlorine ion is smaller than an oxygen ion (C) Cao, because the charge of oxygen ion exceeds that of chlorine ion (D) CaO, because the common charges of calcium and oxygen ions are identical in magnitude
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation
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.