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Applied Practice in

Thermodynamics

AP* Chemistry Series RESOURCE GUIDE

Volume 11

*AP is a registered trademark of the College Entrance Examination Board, which was not involved in the production of, and does not endorse, this product. Pre-AP is a trademark owned by the College Entrance Examination Board.

© 2009 Applied Practice, Ltd., Dallas, TX. All rights reserved.

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Printed in the United States of America.

APPLIED PRACTICE Resource Guide Thermodynamics

Teacher Notes and Strategies

A Note for Teachers.............................................................. 5

Teaching Strategies ............................................................... 8

Glossary of Terms............................................................... 14

Student Practices

Multiple-Choice Questions ................................................. 17

Entropy ..................................................................... 18

Free Energy Calculations.......................................... 22

Spontaneity ............................................................... 27

Free Energy and Equilibrium ................................... 31

Descriptive Chemistry and Laboratory .................... 35

Free-Response Questions.................................................... 41

Answer Key and Explanations

Multiple-Choice Answer Key ............................................. 51

Multiple-Choice Answer Explanations............................... 55

Free-Response Answers and Scoring Guides ..................... 65

*AP is a registered trademark of the College Entrance Examination Board, which was not involved in the production of, and does not endorse, this product.

©2009 by Applied Practice, Ltd., Dallas, TX. All rights reserved.

Teacher Notes and Strategies

for

Thermodynamics

A NOTE FOR TEACHERS

The Applied Practice in AP Chemistry series was designed for use by teachers as an instructional supplement to major units in the AP Chemistry curriculum. This series was also conceived as a resource for teachers in preparing students for the AP Chemistry Exam. As you teach each unit, your students will have the opportunity to practice and to develop those skills required on the exams. Each book in the series includes:

• Teaching notes and strategies • Glossary of terms • 75 multiple-choice questions replicating Section I of the AP Chemistry Exam • Multiple-choice answer keys and answer explanations • 6 free-response questions replicating Section II of the AP Chemistry Exam • Free-response answer keys and scoring guide

We offer a few suggestions and explanations to help you receive the maximum benefit from our materials:

1. Applied Practice booklets do not purport to duplicate exactly an Advanced Placement Examination. However, questions are modeled on those typically encountered on these exams. Thus, students using these materials will become familiar and comfortable with the format, question types, and terminology of Advanced Placement Examinations.

2. Each Applied Practice booklet focuses on one topic within the AP Chemistry

curriculum. These booklets are excellent resources for teachers and their students. Their unique format includes questions designed for use during the initial teaching of the required topics. Other questions are exceptional for the review phase of the course, as students pull the entire year together leading up to the AP Chemistry Exam. The AP exam often will require knowledge in multiple content areas on the same question.

3. You have the option of using the Applied Practice booklets for your own lesson

and test preparation or, if you so choose, students may work through an Applied Practice test booklet on their own as they progress though the course. The students can check their own answers with the answer key and read the answer explanations provided in the teacher edition, conferring with the teacher as needed.

4. The order of topics in the Applied Practice booklets has been organized to follow a logical progression that is similar to the sequence in many of the most widely selected AP chemistry textbooks. You will find that they can easily be adapted to whatever sequence you find most productive at your school.

©2009 by Applied Practice, Ltd., Dallas, TX. All rights reserved. 5

5. The free response questions in each topic were created to provide practice questions similar to both those given in part A of the AP Chemistry Exam, which allows use of a calculator, and those given in part B, in which no calculator is allowed. In a few cases, the specific content is best assessed with a combination of both types.

6. Due to the emphasis on laboratory experience in the College Board’s AP Chemistry program, the Applied Practice booklets in AP Chemistry frequently include laboratory-based questions appropriate to the subtopic addressed. A required laboratory-based question does appear on the AP Chemistry Exam. While most Applied Practice booklets in the AP Chemistry series do contain laboratory-based free-response questions, some topics do not lend themselves to the College Board-recommended laboratory experiments. However, each Applied Practice booklet does contain multiple-choice questions related to both laboratory and descriptive chemistry. Only one of the six free-response questions included on the AP Chemistry Exam is laboratory based.

7. Each booklet includes a glossary of terms that applies to the vocabulary of that

particular topic. 8. If the teacher wishes to replicate the conditions under which students will take the

actual AP Chemistry Exam, he or she should understand the following about multiple-choice versus free-response questions when using Applied Practice booklets: When answering multiple-choice questions (AP Exam, Section I) students are not allowed the use of a calculator, and the only reference information available to them is a periodic table (with only symbol, mass number, atomic number) and a small table of abbreviations/symbols used in the questions. When answering free-response questions (AP Exam, Section II), much more information is available to the student. In addition to the periodic table, a table of standard reduction potentials in aqueous solutions and a relatively complete list of equations, constants, and abbreviations/symbols are provided.

6 ©2009 by Applied Practice, Ltd., Dallas, TX. All rights reserved.

COPYRIGHT NOTICE

The copyright law of the United States (Title 17, United States Code) governs the making of photocopies or other reproductions of copyrighted material. Reproduction of individual worksheets from this booklet, excluding content intended solely for teacher use, is permissible by an individual teacher for use by his or her students in his or her own classroom. Content intended solely for teacher use may not be reproduced, stored in a retrieval system, or transmitted in any way or by any means (electronic, mechanical, photocopying, or otherwise) without prior written permission from Applied Practice. Reproduction of any portion of this booklet for use by more than one teacher or for an entire grade level, school, or school system, is strictly prohibited. By using this booklet, you hereby agree to be bound by these copyright restrictions and acknowledge that by violating these restrictions, you may be liable for copyright infringement and/or subject to criminal prosecution.


TEACHING STRATEGIES

Following are some suggestions for using Applied Practice materials as you work to help your students develop mastery in answering AP-style questions. GENERAL The AP Chemistry Exam will be a challenge for all students. In this section, we have provided strategies for answering the questions themselves as well as some suggestions for teachers to keep in mind when presenting the material. 1. A certain amount of rote memorization is required. We offer the following

suggestions to help students prepare for this:

• Mnemonic devices are excellent tools for memorizing hard facts. For example, in Kings Play Chess On Funny Green Squares, the first letter of each word represents the divisions of the classification system in binomial nomenclature: kingdom, phylum, class, order, family, genus, and species.

• Students are also encouraged to create flashcards; a tried and true method that can help them be successful.

Items that students are required to know for the AP Chemistry Exam include: • names and formulas of polyatomic ions • names and formulas of strong acids and bases, as well as common acids and bases • the metric prefixes from pico to giga • the Greek prefixes in molecular nomenclature: mono, di, tri, tetra, penta, etc. • the solubility rules for ionic compounds • the identity of the seven diatomic elements • the phases of all elements at room temperature (25oC) • the SI quantities and their units • rules for significant figures • rules for assigning oxidation numbers to atoms in compounds

2. Students will see many graphs in both the multiple-choice and free-response

questions. In a few instances, they may be required to produce the graph from given information. Teachers should incorporate interpretation of scientific data written in graphs during instruction both in the classroom and the laboratory. The more graphs that students are required to produce during the year, the easier the questions containing this type of information will be for them on the exam.


MULTIPLE-CHOICE QUESTIONS (MC): 1. The questions in these booklets are valuable for both the initial teaching of the

concepts of chemistry and for preparing students for the style of questions and the level of difficulty encountered on the AP Chemistry Exam. In particular, some subtopics are more fundamental than others. While these subtopics are required in order for students to be able to comprehend more advanced material later on, they would not necessarily appear “as is” in a question on the AP exam. However, we have included multiple-choice questions over this material to support instruction, to provide an expanded test bank of questions, and to familiarize students with this type of assessment.

2. The question most commonly asked by students preparing for the AP exam is: “How

many multiple-choice questions should I answer in Section I?” To answer this question, we will consider what is required for a student to receive a score of “3” (which is labeled as “qualified” by the College Board and considered “passing”; thus, this score is a reasonable goal for most students.) Judging by released past exams for which full data is available, a student typically must score approximately 52–58 points out of the160 possible on Sections I and II combined in order to receive a “3.” Thus, it is reasonable to assume that to pass the AP Chemistry Exam, students need to earn 26–29 points or more from Section I. For each question answered incorrectly, a 0.25 point penalty will be subtracted from the score. Typically, if students are answering more questions correctly than incorrectly, they should continue to increase the number of questions they answer, for it takes four incorrect answers to cancel one correct answer. Most students reach a point at which answering additional questions merely lowers their score. Successful students over the years typically answer the following numbers of questions corresponding to these eventual scores:

• 35–50 questions: “3” • 45–60 questions: “4” • 55–70 questions: “5”

This formula by no means guarantees a particular score; each student is unique in his or her approach to answering questions. This is simply offered as a guideline; an average number of questions a student may choose to answer out of the 75 total questions in an effort to attain a particular achievement level.

3. Another question related to multiple-choice questions on Section I of the AP Chemistry Exam is, “Should I guess?” Because of the penalty for wrong answers (-0.25) and the fact that there is no penalty for questions not answered (just no points earned), it is advisable not to answer questions if the student has little or no knowledge of the information addressed in the question. However, if students can eliminate one or more of the answer choices, it may be a good idea to select an answer, as statistically the odds are then in their favor overall for those questions.


4. A general strategy for multiple-choice questions involves time management. Since the exam limits students’ time to 90 minutes for Section I, they will need to learn to be efficient. While some students simply work more quickly than others, virtually all students find that time is a limiting factor on the AP Chemistry Exam. The ability to read and comprehend quickly is a tremendous advantage. Another skill that will help students overcome the time factor is reading and assessing the level of difficulty of each multiple-choice question quickly. Those questions that students believe they can answer but will take quite a bit of time to analyze should be skipped over initially. Students can mark those questions and return to them later, provided there is enough time. The vast majority of questions answered correctly by students are those for which they can determine the answer almost immediately. Those questions that consume the most time are also those that are most likely to be answered incorrectly.

5. As students approach AP-level question sets, teachers should prepare students to be

satisfied with a lower percentage of correct answers than is typical for them up to this point in their educational experience. Most schools around the country consider 90% an “A” grade, and many AP chemistry students are disappointed when earning test scores in the 70% or 80% range. However, the AP Chemistry Exam is scored in such a way that a successful “passing” score is between 34% and 50%, and a “5”—the highest score possible—is achieved with any point total above approximately 63%. Thus, it will take some time for students to adjust psychologically. Students should understand that a typical passing rate nationwide on the AP Chemistry Exam is approximately 56% of those taking the exam. It is definitely an achievement for a high school student to find success with a “3,” “4,” or “5” on one of the most challenging AP exams. One way to help students cope with a lower percentage of correctly-answered questions is to point out that identifying one’s weakest areas is actually very beneficial when taking the exam. Students who can identify these questions quickly and spend little time on them can accumulate precious minutes to use on other questions, thus raising their score.

6. The percentage scores on AP exams noted above refer to data published by the

College Board on their released questions and will vary from year to year, as individual students and classes of students vary in their abilities from year to year. Also keep in mind is that when students are being assessed on tests in class during the school year, they are often assessed on a concentrated, narrow amount of information – helpful for formative assessment and preparation, but not necessarily characteristic of a final AP exam administered after course has been completed. While students may achieve a higher degree of success on AP-level multiple-choice questions when they are tested immediately at the end of a unit; teachers are encouraged to factor in the level of difficulty and drastically reduced raw score that is regarded as highly successful by the College Board when assessing the students’ efforts.


7. AP chemistry students typically do struggle with AP-level questions early in the course. AP-level questions require a broad background in the overall chemistry curriculum, and are also more difficult than those most students have encountered up to this point. It is advisable to provide answers to practice questions for students, especially early in the school year. They will find it helpful to read the question, work out an answer, and then get immediate feedback. Providing answer keys allows students to work backward from the answer and develops their understanding of how to attack multiple-choice questions at this level of difficulty. The answer explanations provided in the Applied Practice booklets are particularly useful in this regard. In addition to allowing some access to answer keys (unless, of course, you will be using these questions for assessment), teachers should plan to include time in class for guided practice of these questions, working them out with the students. Collaboration among students in small learning groups is also a useful technique. Chemistry is much easier to learn when students are able to talk about the content and reason out answers together.

8. A simple but useful tip for students is to underline, or in some way highlight, key

terms in the questions. Many times, a student will choose an incorrect answer because he or she simply misread the question.

FREE-RESPONSE QUESTIONS (FRQs): 1. One of the most common reasons for producing an unsatisfactory response to a free-

response question on an AP Chemistry Exam is that the student did not answer the specific question that was asked. Often, students write information that is factually correct but does not apply to the question, or their response relates to information in the question without actually answering the question. Free-response questions often present a great deal of information when only small bits of that information are actually needed to answer each of many different parts of the question. Students should read the entire question before they begin to write. They should underline, or in some way highlight, exactly what is being asked in each part of the question. After writing their response, students should proofread their work to be sure they answered the question specifically. Often, students will look back over their answers when their papers are returned and realize their answers make little sense, often responding, “But I meant to say ____.” They will likely experience greater success when they take time to read over their answers to verify they actually wrote what they intended to say.

2. Students should be specific in their answers and should understand the difference

between a definition and information that justifies or explains their answer. In most cases, students should use complete sentences. A few questions have parts that merely require a list of items, and others ask for additions to diagrams, but for the most part,


students will need to write coherently. A one-word answer is not sufficient to earn credit.

3. Students should be sure they understand the difference between an answer that

justifies their point and one that merely cites a correlation. A common example on past tests is a question about what determines the relative strength of London dispersion forces between molecules. Larger-sized atoms and molecules have more electrons and are more easily “ionizable”; thus, they are more likely to form dispersion forces. As atoms and molecules get larger, their nuclei get heavier, and thus the atomic and molecular weights increase along with dispersion forces. However, while increasing mass correlates to increasing strength of attraction between molecules, it does not explain or justify the increase.

An analogy illustrating this concept would be a graph of children’s mass with their level of intelligence. As they age, their brains develop at the same time their bodies grow and become heavier. Thus, older children are both smarter and heavier than younger children. But it would be incorrect to say that the children are becoming smarter because they are becoming heavier.

4. When teaching concepts in chemistry, it is tempting to make atoms, ions, and

molecules anthropomorphic. Humans tend to more easily understand concepts that are explained in a way that is relevant to the way we think and behave. Teachers must be careful not to instruct the students about what the atoms “want” or “need.” Often, students will write answers about an atom that is “happy” because it got the electron it “needed.” Even though students may have come to an appropriate level of comprehension on this concept, they will not receive credit when citing the emotions of a piece of matter on the AP Chemistry Exam.

5. Students should write out words fully, unless the abbreviations are universally

accepted. While their teachers may have come to understand a particular student’s shorthand, students are likely to lose out when the AP exams are graded.

6. Students should develop the habit of avoiding pronouns in their written answers, such

as beginning their first sentence with “it.” In many questions, there are dozens of things to which “it” could refer, and students will receive no credit for their answer if they use vague pronouns.

7. Using the free-response questions in the Applied Practice books is a good way to

develop students’ writing skills. Answers, explanations, and scoring guides for each of the six questions in an Applied Practice booklet are provided. The more writing students do, the better they will become at answering the questions in Section II of the AP exam. Although it takes time to grade students’ writing, the time will be well worth it in terms of their improvement and eventual success. Free-response questions are of three general types: large calculation questions that require good problem solving skills, equation writing, and “essay” types that require good writing skills.


The first type of question will be seen on Section A of Part II of the exam, and the last two types will be found on Section B of Part II.

8. When answering free-response questions that require calculations (section II, part A,

questions 1, 2, and 3), students should develop these good habits from the beginning of the year:

Always list any mathematical formula to be used (i.e. PV=nRT) at the beginning of

the question. Never put numerical information in an equation without first writing that equation down first. The vast majority of all mathematical formulas to be used are listed in the reference sheets provided on the AP Chemistry Exam in section II. Some knowledge of mathematical formulas will be required on section I (multiple- choice), so it is good for students to know them and not always rely on the reference sheets.

While students’ work must be shown, typically much less work is required than that

required by most high school math teachers. If the student lists the mathematical formula, substitutes the correct numerical values from the question into the proper location, and clearly marks the answer, the student does not need to write out each of the various algebraic steps. However, there is no disadvantage in scoring if students are more confident showing their work. They simply need to be coached in the amount of time to use on each calculation question. Some students will need to write fewer steps if time becomes a factor overall.

Once the students calculate and write a final answer, they should do the following

each time:

• Double check the number of significant figures in the answer. They must be within one significant figure up or down from the correct number.

• Double check the calculations. • Make sure they have labeled their answer as to which quantity it is (do not leave

answer as “x=0.050”). A label helps ensure they are answering the question correctly.

• Always include units (except for a few that do not use units, such as specific gravity and the equilibrium constant, Keq, etc).

• Box in the final answer. This not only helps the grader, but it provides an easy way for the students to check their work to ensure they have finished each part of each question. Also, answers from previous sections are often used on subsequent parts of the question. It is easier to find this information if it is boxed.


GLOSSARY OF TERMS cuvette—transparent tubular laboratory vessel or dish for holding a liquid endothermic—a chemical reaction in which heat is absorbed enthalpy—(H) the amount of energy contained in the substances in a reaction (system) available for transfer between itself and the environment (surroundings) entropy—(S) a measure of the randomness or disorder of a system (substances being studied); thermodynamic quantity that applies to the second law of thermodynamics equilibrium constant—(Ka, Kb, Kw, etc.) the constant value that expresses the relationship between the concentration of products and reactants in a reversible chemical reaction at equilibrium exothermic—a chemical reaction that produces heat Gibbs free energy—(G) a measure of the capacity of a system to do work, such as the likelihood of a chemical reaction to form products; a thermodynamic quantity defined by the equation: G = H – TS Henderson-Hasselbalch equation—used to determine the pH of a buffer system; pH =

pKa + log[conjugatebase][conjugate acid]

momentum—the product of mass and velocity, usually refers to linear momentum that includes direction and magnitude of motion precipitation—the formation of an insoluble solid compound during a chemical reaction in a solution spontaneous—a chemical or physical change that occurs by itself thermodynamics—the study of the relationship between matter and heat or other forms of energy, involved in a chemical or physical reaction


Student Practices

for

Thermodynamics

Multiple-Choice Questions


Entropy The following answer choices can be used in questions 1-3. Each answer may be used once, more than once, or not at all. (A) CO(g) + H2O(g) CO2(g) + H2(g) (B) Na2CO3(s) Na2O(s) + CO2(g) (C) N2(g) + 3H2(g) 2NH3(g) (D) KCl(s) K+

(aq) + Cl-(aq)

(E) H2O(s) H2O(l) 1. In which reaction would the entropy change be closest to zero? 2. In which reaction would ∆S have the most positive value? 3. In which reaction would ∆S have the most negative value? 4. Reactions where entropy decreases include I. (NH4)2CO3(s) 2NH3(g) + CO2(g) +H2O(g) II. NaCl(s) Na+

(aq) + Cl-(aq)

III. SO2(g) + CaO(s) CaSO3(s) (A) I only (B) II only (C) III only (D) I and II only (E) I, II, and III 5. Which of the following correctly lists the substances in terms of increasing standard

molar entropy? I. H2O(s) < H2O(l) < H2O(g) II. O2(g) < H2O(l) < CaCl2(s) III. Cl2(g) < PCl3(g) < PCl5(g) (A) I only (B) III only (C) I and II only (D) I and III only (E) I, II, and III


6. “The entropy of a pure crystalline substance at absolute zero, is zero,” is a statement of which of the following?

(A) The 1st law of thermodynamics (B) The 2nd law of thermodynamics (C) The 3rd law of thermodynamics (D) The law of conservation of energy (E) The law of conservation of matter 7. A typical unit for entropy is (A) J. (B) J-1 K-1. (C) J mol K. (D) J mol-1. (E) J mol-1 K-1. 8. Which of the following thermodynamic quantities defines the degree of disorder? (A) Enthalpy (B) Entropy (C) ∆H (D) Gibbs free energy (E) ∆G 9. Which of the following processes involves an increase in entropy? (A) The precipitation of lead(II) chloride when mixing solutions of lead(II) nitrate and sodium chloride (B) The freezing of ethanol (C) The condensation of water vapor (D) The decomposition of a solid, group 2 carbonate to yield carbon dioxide gas when heated (E) The formation of solid calcium sulfate when gaseous SO3 is passed over heated, solid CaO


10. Which of the following statements is NOT correct? (A) Solids have smaller entropy values than gases. (B) Gases have higher entropy values than liquids. (C) Liquids have higher entropy values than solids. (D) Particles in solution have lower entropy values than those in solids. (E) Gases have higher entropy values than solids. 11. Which of the following is a direct consequence of this statement: “The entropy of a

pure crystalline substance at absolute zero, is zero”? (A) All substances will have values for standard molar entropy that are greater than zero. (B) A substance at a temperature of 300 K will have a lower absolute entropy than

the same substance at a temperature of 273 K. (C) Entropy can be measured in units of kJ mol-1 K-1. (D) Absolute zero is equal to 0 K. (E) Entropy and enthalpy are directly proportional. 12. All of the following are true statements in relation to entropy EXCEPT (A) the change in entropy during a reaction is given the symbol ∆S (B) positive values of ∆S are associated with increases in disorder (C) entropy is a measure of the randomness or disorder of a system (D) ∆S can be calculated using the formula, ∆S = ∑S(products) – ∑S(reactants) (E) a positive value for ∆S is essential for a spontaneous reaction 13. Which of the following substances will have the largest value for standard molar

enthalpy? (A) H2O(l) at 298K (B) H2O(l) at 300K (C) H2O(g) at 380K (D) H2O(g) at 385 K (E) H2O(g) at 400 K


14. Which of the following is a statement of the 2nd law of thermodynamics? (A) Energy is conserved in a chemical reaction. (B) Mass is conserved in a chemical reaction. (C) The entropy of a pure crystalline substance at absolute zero, is zero. (D) The total entropy of the universe tends to increase. (E) Momentum is conserved in a collision. 15. Which of the following is the correct set-up for the calculation of the entropy change

in the reaction below? 2NO2(g) N2O4(g)

Standard molar entropies at 298 K: NO2(g) = 240 J mol-1 K-1, N2O4(g) = 304 J mol-1 K-1 (A) 240 - 304 (B) 304 - 240 (C) 304 – ((2)(240)) (D) 304 – ((298)(240)) (E) 304 – ((298)(2)(240))


Free Energy Calculations The following answer choices can be used in questions 16-18. Each answer may be used once, more than once, or not at all. (A) ∆S = ∑S(products) – ∑S(reactants) (B) ∆Go = ∑∆Gf

o(products) – ∑∆Gf

o(reactants)

(C) ∆Ho = ∑ΔH fo

(products) – ∑ΔH fo

reactants) (D) ∆Go = - RT lnK (E) ∆Go = ∆Ho - T∆So 16. Can be used directly to calculate the entropy change at a particular temperature, in a

reaction where the value for Gibbs free energy change and enthalpy change are known

17. Can be used directly to calculate the Gibbs free energy change at a particular

temperature, for a reaction where values for enthalpy and entropy changes are known 18. Can be used directly to calculate enthalpy change of a reaction where the entropy

change and Gibbs free energy change for the reaction at a particular temperature are known

19. For a reaction at a particular temperature, T, it is found that ∆Go = -10 kJ mol-1, ∆Ho

= -20 kJ mol-1, and ∆So = +0.1 kJ mol-1 K-1. Which of the following is true? I. The reaction will be spontaneous at all temperatures. II. T = 1000 K III. The reaction will be spontaneous at the temperature studied. (A) I only (B) II only (C) III only (D) I and III only (E) I, II, and III


20. In the expression ∆Go = - RT lnK, in order to calculate a value for ∆Go in units of J mol-1

I. T must be recorded in units of K II. R must have a value of 0.00831 III. R must have units of J mol-1 K-1 (A) I only (B) III only (C) I and II only (D) I and III only (E) I, II, and III 21. The standard Gibbs free energy of formation of ethene, hydrogen, and ethane are

68.1, 0.00, and -32.9 kJ mol-1, respectively. Which is the correct combination of calculated ∆Go value and statement regarding the spontaneity of the reaction?

C2H4 + H2 C2H6

(A) +101 kJ mol-1, spontaneous (B) -101 kJ mol-1, spontaneous (C) +35.2 kJ mol-1, nonspontaneous (D) -35.2 kJ mol-1, spontaneous (E) -32.9 kJ mol-1, spontaneous Questions 22-24 refer to the following information. The Haber process can be summarized by the reaction below, and data associated with each of the substances is shown in the table.

N2(g) + 3H2(g) 2NH3(g)

Substance ΔH fo in kJ mol-1 So in J mol-1 K-1

N2(g) 0 192 H2(g) 0 131

NH3(g) -46 193 22. What is the value of ∆Ho (in kJ) for the forward reaction as written? (A) +92 (B) +46 (C) 0 (D) -46 (E) -92


23. What is the value for ∆So for the forward reaction? (A) -199 J mol-1 K-1 (B) 0 J mol-1 K-1 (C) +199 J mol-1 K-1 (D) +386 J mol-1 K-1 (E) +484 J mol-1 K-1 24. For the reverse reaction ∆Go is found to be +33.2 kJ at 298 K. What is the value for

the equilibrium constant for the forward reaction? (A) 6.64 x 105 (B) 6.64 x 10-5 (C) 3.32 x 10-4 (D) 1.30 x 10-3 (E) 3.32 x 10-1 25. The single displacement reaction between zinc metal and copper(II) ions in solution is

summarized below. What can be said of the equilibrium constant and the spontaneity of this reaction at this temperature?

Zn(s) + Cu2+

(aq) Zn2+(aq) + Cu(s) ∆G = -212 kJ mol-1

(A) K > 1, and the reaction is nonspontaneous. (B) K = 1, and the reaction is nonspontaneous. (C) K > 1, and the reaction is spontaneous. (D) K < 1, and the reaction is spontaneous. (E) K = 0, and the reaction is spontaneous. 26. For a reaction that is known to be spontaneous at a particular temperature, it is also

found that the ∆So value is negative. What can be deduced about the reaction from this information?

(A) ∆H is negative. (B) ∆H is positive. (C) ∆H = 0. (D) ∆H is positive at high temperatures and negative at low temperatures. (E) Nothing can be deduced about ∆H from this information.


Questions 27-28 refer to the following information. A reaction is known to have ∆Ho = +190 kJ, ∆So = +50 J K-1, and ∆Go = +90 kJ. Assuming that ∆Ho and ∆So are fixed, answer the questions that follow. 27. Calculate the value of T in Kelvin for the reaction at this temperature. (A) 100 K (B) 200 K (C) 500 K (D) 1000 K (E) 2000 K 28. Which expression shows the correct set up for finding the Kelvin temperature at

which the reaction becomes spontaneous? (A) +290 = +190 – T(+50) (B) +290 = +190 – T(+50/1000) (C) 0 = +190 – T(+50) (D) 0 = +190 – T(+50/1000) (E) 0 = +50 – T(+190/1000) 29. A reaction that occurs in the gaseous phase can be summarized by the equation

below. X(g) Y(g) + Z(g)

Kp is found to be 3.0 x 102 at 273 K. If a mixture of gases with the partial pressures

shown in the table below is introduced into a flask at 273 K, what will be true at this instantaneous point in time?

Gas Partial Pressures X 10 Y 10 Z 50

(A) The reaction is spontaneous in the reverse direction. (B) The reaction is spontaneous in the forward direction. (C) The reaction is at equilibrium. (D) ∆Go for the forward reaction is positive. (E) ∆Go = 0


30. The equilibrium constant for a single displacement reaction between a metal and a solution of another metal’s ions is found to be 4.0 x 1020. When equilibrium has been established, which of the following correctly describes the free energy change and the relative amounts of reactants and products found in the reaction mixture?

(A) The free energy change is positive, and large amounts of products are found. (B) The free energy change is negative, and large amounts of products are found. (C) The free energy change is positive, and large amounts of reactants are found. (D) The free energy change is zero, and large amounts of reactants are found. (E) The free energy change is zero, and large amounts of products are found.


Spontaneity The following answer choices can be used in questions 31-33. Each answer may be used once, more than once, or not at all. (A) ∆Go = - , ∆Ho = - , ∆So = + (B) ∆Go = - , ∆Ho = + , ∆So = + (C) ∆Go = - , ∆Ho = - , ∆So = - (D) ∆Go = + , ∆Ho = + , ∆So = - (E) ∆Go = + , ∆Ho = - , ∆So = + 31. Reactions with these characteristics are spontaneous at all temperatures. 32. Reactions with these characteristics are never spontaneous at any temperature. 33. Reactions with these characteristics are spontaneous because of their favorable

entropies rather than their non-favorable enthalpies 34. An exothermic reaction where the number of moles of gas increases is I. One that has a negative ∆Ho. II. One that has a positive ∆So. III. One that is spontaneous at all temperatures. (A) I only (B) II only (C) III only (D) I and II only (E) I, II, and III 35. Which of the following is true for the ΔH f

o , ΔS fo , and ΔGf

o for any element? I. ΔH f

o = 0 II. ΔS f

o = 0 III. ΔGf

o≠ 0 (A) I only (B) III only (C) I and II only (D) II and III only (E) I, II, and III


36. When water vapor condenses, the process involves which change combination below?

(A) ∆Ho = + , ∆So = + (B) ∆Ho = + , ∆So = - (C) ∆Ho = - , ∆So = + (D) ∆Ho = - , ∆So = - (E) ∆Ho = + , ∆So = 0 37. Which combination of ∆Ho, ∆So, and T∆So leads to a nonspontaneous reaction at all

temperatures? (A) ∆Ho = + , ∆So = - , T∆So = + (B) ∆Ho = - , ∆So = + , T∆So = - (C) ∆Ho = + , ∆So = + , T∆So = + (D) ∆Ho = - , ∆So = - , T∆So = + (E) ∆Ho = + , ∆So = - , T∆So = - 38. If ∆Go for the forward reaction in an equilibrium process is greater than 0, all of the

following statements are true EXCEPT (A) the reaction will have a negative ∆Go for the reverse process (B) the forward reaction is nonspontaneous (C) ∆Go = -RT lnK (D) K > 1 (E) the reverse reaction is spontaneous 39. For a spontaneous reaction, (A) ∆Go = - , K > 1, lnK = + (B) ∆Go = - , K > 1, lnK = - (C) ∆Go = - , K < 1, lnK = + (D) ∆Go = + , K < 1, lnK = + (E) ∆Go = + , K > 1, lnK = +


40. For a reaction with a positive ∆So and a positive ∆Ho, what can be said of the process?

(A) ∆So favors spontaneity, but it can only be spontaneous if temperatures are

relatively high. (B) ∆So favors spontaneity, but it can only be spontaneous if temperatures are

relatively low. (C) ∆So does not favor spontaneity, but it can be spontaneous if temperatures are

relatively high. (D) ∆So does not favor spontaneity, but it can be spontaneous if temperatures are

relatively low. (E) ∆So favors spontaneity, and it will be spontaneous at all temperatures. 41. For the reaction shown below, which statement correctly predicts the change in the

value of ∆Go and associated change in spontaneity as temperature increases? (Assume that ∆Ho and ∆So are temperature independent.)

C(s) + O2(g) CO2(g) ∆H = -393 kJ, ∆S = 3.3 J mol-1 K-1

(A) ∆Go is increasingly negative, and fewer products will be present at equilibrium. (B) ∆Go is increasingly positive, and fewer products will be present at equilibrium. (C) ∆Go is increasingly negative, and more products will be present at equilibrium. (D) ∆Go is increasingly positive, and more products will be present at equilibrium. (E) ∆Go does not change, and the amount of products at equilibrium will not change. 42. A spontaneous reaction that produces products from a position where only reactants

are present initially definitely has (A) ∆Ho = + (B) ∆Ho = - (C) ∆Go = + (D) ∆Go = - (E) ∆So = - 43. For a reaction that is nonspontaneous at any temperature, what is true? (A) The reaction must be exothermic, and disorder increases. (B) The reaction must be endothermic, and disorder increases. (C) The reaction must be exothermic, and disorder decreases. (D) The reaction must be endothermic, and disorder decreases. (E) The reaction could be endothermic or exothermic, but disorder must increase.


44. All of the following statements about a reaction that is spontaneous at all temperatures are true EXCEPT

(A) it is exothermic (B) it has a positive ∆So (C) it has a negative ∆Go at all temperatures (D) it will have a T∆So term that is positive (E) it will have products that are less disordered than reactants 45. Which of the following combinations is impossible? (A) ∆Ho = + , ∆So = - , T∆So = - (B) ∆Ho = - , ∆So = + , T∆So = + (C) ∆Ho = + , ∆So = + , T∆So = + (D) ∆Ho = - , ∆So = - , T∆So = - (E) ∆Ho = + , ∆So = - , T∆So = +


Free Energy and Equilibrium The following answer choices can be used in questions 46-48. Each answer may be used once, more than once, or not at all. (A) K > 1 (B) K = 1 (C) K < 1 (D) K = 0 (E) K < 0 46. All spontaneous reactions have this value for K. 47. The value of K when ∆Go is a negative number. 48. When K has this value, the value for ∆Go is 0. 49. Which of the following is a mathematical consequence of ∆Go = - RT lnK? I. K > 1 means ∆Go is negative. II. lnK for a spontaneous reaction must be positive. III. R, the gas constant, in this expression must have the value 8.31 if ∆Go is to be

measured in J mol-1. (A) I only (B) II only (C) III only (D) I and II only (E) I, II, and III 50. For a system where ∆G = ∆Go, which of the following statements is true? I. Equilibrium has been established. II. Q = K III. The term RT lnQ = 1 (A) I only (B) III only (C) I and II only (D) II and III only (E) I, II, and III


51. When equilibrium is established, what can be said of the free energy present in the system?

(A) It is at a maximum. (B) It is at a minimum. (C) ∆Go is positive. (D) ∆Go is negative. (E) It is between the maximum and minimum values. 52. The rusting of iron at 298 K has a ∆Go value approximately equal to -1.5 x 106 J.

What does this value alone tell us about the chemical reaction involved in the rusting of iron?

(A) The reaction is fast. (B) The reaction is nonspontaneous. (C) The reaction will definitely have a positive ∆So. (D) The reaction will definitely have a positive ∆Ho. (E) The reaction will have a value for K that is greater than 1. 53. When equilibrium in a system has been established, all of the following are true

EXCEPT (A) the formation of the reactants and the products are equally favored (B) the term RT lnK = 0 (C) Q = K (D) ∆Go = 0 (E) free energy has reached a maximum 54. For a spontaneous reaction, the expression ∆Go = - RT lnK tells us that as conditions

are changed to make K larger, the reaction will also (A) create a more likely forward reaction (B) become less spontaneous (C) yield a “lnK term” that is smaller (D) produce an increasingly positive ∆Go (E) create more reactants


55. Mathematically, the correct expression for the calculation of K, the equilibrium constant, is

(A) K = -∆Go RT (B) K = ln ∆Go RT

(C) K = e(−

ΔGRT

)

(D) K = ln ∆Go

(E) K = e−

RTΔG

⎛ ⎝ ⎜

⎞ ⎠ ⎟

56. The ∆Go value for the self-ionization of water at 298 K is +80.0 kJ mol-1. What is the

approximate value of K at this temperature? (A) 1.0 x 1025 (B) 1.0 x 1020 (C) 1.0 x 107 (D) 1.0 (E) 1 x 10-14 57. The reaction below has a K value that is approximately equal to 4.0 x 1038. What does

this value suggest about the reaction?

Fe3+(aq) + 3OH-

(aq) Fe(OH)3(s) (A) That the K value for the reverse reaction will also be very large (B) That the reaction is unlikely to occur (C) That the ∆Go for the reaction will be negative (D) That iron(III) hydroxide will not precipitate (E) That the reaction is nonspontaneous


58. Consider the equilibrium that is set up between water and steam as in the equation below. It is found that the equilibrium is established at a temperature of 373 K. Which of the following statements is false?

H2O(l) H2O(g)

(A) Water boils at 373 K. (B) ∆Go for the forward reaction at 298 K will be positive. (C) ∆Go for the forward reaction at 400 K will be negative. (D) ∆Go for the forward reaction at 373 K will be positive. (E) At equilibrium, ∆Go is zero. 59. The reaction below has a value for lnK that is equal to 110 at 298 K. What does this

suggest about the reaction?

2CO(g) + O2(g) 2CO2(g) (A) That K is less than 1 (B) That the reaction is nonspontaneous (C) That ∆Go for the reaction is a negative number (D) That many more reactants than products are present in the equilibrium mixture (E) That ∑ΔGf

o(products) – ∑ΔGf

o(reactants) = a positive number

60. An extremely large negative value for ∆Go suggests all of the following EXCEPT (A) K is a very large, positive number (B) large quantities of products are formed (C) the reaction is essentially irreversible (D) the reverse reaction is likely (E) the reaction is spontaneous


Descriptive Chemistry and Laboratory The following answer choices can be used in questions 61-63. Each answer may be used once, more than once, or not at all. (A) Ka (B) Kb (C) Kc (D) Kp (E) Ksp 61. The specific equilibrium constant that could be determined directly by performing an

analysis of data collected in a titration between aqueous sodium hydroxide and an organic acid.

62. The equilibrium constant that could be determined directly by the analysis of the

partial pressures of NH3, H2, and N2 in an equilibrium mixture. 63. The equilibrium constant that could be determined by the analysis of an equilibrium

mixture of barium ions, sulfate ions, and barium sulfate. 64. In order to avoid death, living systems need I. a source of energy II. a molecular system to allow energy to be transferred into useful work III. to avoid a state of thermodynamic equilibrium (A) I only (B) II only (C) III only (D) I and II only (E) I, II, and III


65. In a procedure to determine the equilibrium constant for the reaction below that uses a colorimeter, what steps are necessary during the experiment?

Fe3+

(aq) + SCN-(aq) FeSCN2+

(aq) I. Calibration of the colorimeter with solutions of known concentration II. Ensuring that the cuvette is clean and free from fingerprints III. Plotting a graph in order to determine concentrations (A) I only (B) III only (C) I and II only (D) I and III only (E) I, II, and III Questions 66-68 refer to the following information. The Ksp value for a relatively insoluble salt such as PbI2 can be determined by performing an experiment that measures the concentration of ions present in a saturated solution of the salt. 66. Which expression allows the correct determination of Ksp? (A) Ksp = [Pb2+][I-] (B) Ksp = [Pb2+][I-]2 (C) Ksp = [Pb2+]2[I-]2

(D) Ksp = [Pb2+][I−]2

[PbI2]

(E) [ ][ ][ ]2

2

PbIIPbKsp

−+

=


67. One method of measuring the concentration of iodide ions present in the saturated solution is to convert them to I2 which is colored, and then to use a colorimeter to determine the concentration of I2 present. Which of the following would be suitable to convert iodide ions to I2? (A) A reagent that causes precipitation (B) An oxidizing agent (C) A reducing agent (D) A reagent that causes a gas to be produced (E) A reagent that increases the solubility of PbI2 68. Which of the following salts could be analyzed in a way similar to that in question

#67 in order to determine a meaningful Ksp value? (A) MgCl2 (B) NaI (C) NaBr (D) PbBr2 (E) Pb(NO3)2 Questions 69-71 refer to the following information. A weak acid, CH3COOH, is titrated with a strong base, KOH, and it is found that 20.00 mL of 0.001 M acid requires 15.00 mL of base for complete neutralization. 69. At what point in the titration will the concentration of salt produced in the

neutralization reaction be equal to the concentration of the acid remaining in the mixture?

(A) When 0.00 mL of KOH has been added (B) When 7.50 mL of KOH has been added (C) When 15.00 mL of KOH has been added (D) When 20.00 mL of KOH has been added (E) Never


70. When the concentration of the salt produced in the neutralization reaction is equal to the concentration of the acid remaining in the mixture, which of the following relationships is true?

(A) pKa = 1 x 10-7 (B) pH = -log pKa (C) pH = Ka (D) pH = pKa (E) pH = 7 71. A similar experiment is conducted, this time using a different acid, HCl, but the same

base. It was not possible to determine a Ka value. Which of the following is a feasible reason for the failure of this experiment?

(A) No suitable indicator can be found for this titration. (B) It is impossible to plot a titration curve for this titration. (C) HCl is a strong acid and as such no end point can be determined with KOH. (D) HCl is a strong acid and as such cannot be neutralized by KOH. (E) HCl is a strong acid and as such has no meaningful value for Ka. Questions 72-75 refer to the following information. In an experiment to determine the Kc for an esterification reaction, a known number of moles of ethanoic acid are brought together with an equal number of moles of ethanol and a known, small amount of sulfuric acid at a known temperature. There is no ester or water present initially. The reaction mixture is allowed to reach equilibrium.

CH3COOH + C2H5OH CH3CO2C2H5 + H2O The equilibrium mixture is then titrated with NaOH solution. 72. What is the role of the sulfuric acid? (A) A reactant (B) An intermediate (C) A product (D) A catalyst (E) An oxidizing agent


73. What is the purpose of titrating the mixture with NaOH? (A) To determine only the moles of CH3COOH present (B) To determine only the moles of C2H5OH present (C) To determine only the moles of sulfuric acid present (D) To determine only the moles of CH3CO2C2H5 present (E) To determine the total moles of CH3COOH and sulfuric acid present 74. Once the number of moles of CH3COOH at equilibrium has been established, what

can also be established? (A) Only the moles of ethanol at equilibrium (B) Only the moles of ester at equilibrium (C) The moles of ethanol and water but not the moles of ester at equilibrium (D) The moles of ester and water but not the moles of ethanol at equilibrium (E) The moles of ethanol, ester, and water at equilibrium 75. The experiment yields a Kc value = 3.9. The experiment is then repeated using larger

initial quantities of reactants at the same temperature. What is the consequence for the value of Kc in this new trial?

(A) It will be larger than 3.9. (B) It will be smaller than 3.9. (C) It will be equal to 3.9 because the temperature has not changed. (D) It will be equal to 3.9 since it is a constant and has the same magnitude at all temperatures. (E) It is not possible to tell without more data.


Free-Response

Questions


1. The thermodynamics of the combustion reaction between oxygen and butane has an important application in the igniters used for outdoor grills. Use the data below to answer the questions that follow.

Substance ΔH fo Standard Molar Entropy

in J mol-1 K-1 CO2(g) -393 kJ mol-1 214 H2O(l) -286 kJ mol-1 70.0 O2(g) 0.00 kJ mol-1 205

C4H10(g) -125 kJ mol-1 310 (a) Write a chemical equation to summarize the combustion reaction between butane

and oxygen. (b) Explain why the ΔH f

o for oxygen is zero. (c) Calculate the value of enthalpy of combustion of 1 mole of butane. (d) Without doing any calculations, predict the sign of ∆S for the combustion of

butane. Explain your answer. (e) Calculate ∆G for the combustion of butane. (f) In light of your answer to (e), comment on the fact that butane will not

instantaneously combust in oxygen gas.


2. Calcium metal will react with fluorine gas to form a stable compound with the formula CaF2. The theoretical compound CaF is not formed when calcium and fluorine react, but a value for its enthalpy of formation can still be calculated.

(a) Using the data below, calculate the enthalpies of formation of CaF2(s) and CaF(s).

Ca(s) Ca(g) ∆Ho1 = +193 kJ

Ca(g) Ca+(g) + e- ∆Ho

2 = +590 kJ Ca+

(g) Ca2+(g) + e- ∆Ho

3 = +1150 kJ F2(g) 2F(g) ∆Ho

4 = +158 kJ F(g) + e- F-

(g) ∆Ho5 = -348 kJ

Ca2+(g) + 2F-

(g) CaF2(s) ∆Ho6 = -2580 kJ

Ca+(g) + F-

(g) CaF(s) ∆Ho7 = -798 kJ

(b) Use your answers in (a) above to suggest why the compound that is formed when

calcium reacts with fluorine has the formula CaF2 and not CaF. (c) What is the name given to the enthalpy change ∆Ho

5 above? (d) Given that the standard molar entropies for calcium metal, fluorine gas, and

calcium fluoride (CaF2) are 42, 203, and 69 J mol-1 K-1 respectively, calculate the value for ∆Go for the formation of CaF2.


3. The Born-Haber process is a very important industrial and commercial chemical reaction where ammonia is produced from the reaction between nitrogen gas and hydrogen gas. The equation for the reaction is given below.

N2(g) + 3H2(g) 2NH3(g); ∆H = -92.0 kJ mol-1

Data associated with the process are given in the table below.

Substance Standard Molar Entropy in J mol-1 K-1 N2 192 H2 131

NH3 193 (a) Without doing a calculation, predict the sign of the ∆So for the forward reaction.

Explain your answer. (b) Calculate ∆Go for the reaction at 298 K. (c) Calculate K for the reaction at 298 K. (d) Briefly comment on the relationship between the values you have calculated for ∆G and K in parts (b) and (c) above. (e) Calculate the temperature at which the sign of ∆Go changes. (f) Is the change in enthalpy or the change in entropy the driving force behind this

reaction? Explain your answer.


4. The standard molar entropies of three halogen elements and two of their compounds at 298 K are given below.

Substance Standard molar entropy in J mol-1 K-1

Cl2(g) 223 Br2(l) 152 I2(s) 117

ICl3(s) 172 BrF5(l) 225

(a) Explain the trend amongst the standard molar entropies of the elements. (b) Many inter-halogen compounds exist (compounds where elemental halogens

react with one another to produce new compounds). Two such compounds are ICl3(s) and BrCl(g).

(i) Write a chemical equation to shown the formation of ICl3(s) from its elements. (ii) Calculate the ΔS f

o for the reaction you have written in (b)(i). (iii) Given that the ∆Ho for the formation of ICl3(s) is -88.0 kJ mol-1, calculate the

value of ∆Go for the formation of ICl3(s) at 298K. (c) Using data in the table above, and the fact that ΔS f

o for the formation of BrF5(l) = -155 J mol-1 K-1, calculate the standard molar enthalpy of fluorine gas. Comment on the value that you calculate in relation to the tabulated value for chlorine gas.


5. Consider the decomposition of phosphorous pentabromide into phosphorous tribromide and bromine. An equation for the reaction is shown below.

PBr5(g) PBr3(g) + Br2(g)

It is found that at higher temperatures a greater proportion of the phosphorous

pentabromide decomposes.

(a) At which temperature, 900K or 1000K, would you expect the entropy in the equilibrium system to be the greatest? Explain your answer.

(b) Does the enthalpy change associated with the forward reaction favor the

formation of reactants? Explain your answer. (c) Does the entropy change associated with the forward reaction favor the formation

of reactants? Explain your answer. (d) Carefully explain the mathematical relationship between the value of K, the

equilibrium constant, and the value of ∆Go for the reaction.


6. Use thermodynamic principles to explain each of the following observations. (a) A significant increase in entropy occurs when a sample of a group 2 carbonate is

subjected to strong heating. (b) Some reactions are found to be spontaneous at one temperature but

nonspontaneous at another, different temperature. (c) The value of the absolute entropy for ClF3(g) is significantly higher than the

absolute entropy for Cl2(g). (d) The values for ΔGf

o , ΔH fo , and ΔS f

o for the formation of hydrogen gas are all found to be zero.


Answer Key and Explanations

for

Thermodynamics

Multiple-Choice

Answer Key

©2009 by Applied Practice, Ltd., Dallas, TX. All rights reserved. 51Spiral-bound teacher pages may not be shared or reproduced.

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MULTIPLE-CHOICE ANSWER KEY

1. A

2. B

3. C

4. C

5. D

6. C

7. E

8. B

9. D

10. D

11. A

12. E

13. E

14. D

15. C

16. E

17. E

18. E

19. D

20. D

21. B

22. E

23. A

24. A

25. C

26. A

27. E

28. D

29. B

30. E

31. A

32. D

33. B

34. E

35. C

36. D

37. E

38. D

39. A

40. A

41. C

42. D

43. D

44. E

45. E

46. A

47. A

48. B

49. E

50. C

51. B

52. E

53. E

54. A

55. C

56. E

57. C

58. D

59. C

60. D

61. A

62. D

63. E

64. E

65. E

66. B

67. B

68. D

69. B

70. D

71. E

72. D

73. E

74. E

75. C


Multiple-Choice

Answer Explanations


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ANSWER EXPLANATIONS ENTROPY

1. (A) CO(g) + H2O(g) CO2(g) + H2(g). Two moles of gas on each side of the equation means that there is no significant change in entropy. 2. (B) Na2CO3(s) Na2O(s) + CO2(g). Zero moles of gas become one mole of gas. Since gases have standard molar entropies that are huge compared to solids and liquids, any such change will result in a large increase in entropy. 3. (C) N2(g) + 3H2(g) 2NH3(g). Four moles of gas become two moles of gas. Since gases have standard molar entropies that are huge compared to solids and liquids, any such change will result in a large decrease in entropy. 4. (C) III only. In statement I, the number of moles of gas increases, thus increasing entropy. In statement II, a solid becomes an aqueous solution with greater disorder and therefore increases entropy. In statement III, one mole of gas becomes zero moles of gas. Since gases have standard molar entropies that are huge compared to solids and liquids, any such change will result in a large decrease in entropy. 5. (D) I and III only. Gases have greater standard molar entropies than liquids, which in turn have greater entropies than solids; the more complex the molecule (greater number of atoms and bonds), the greater the entropy. 6. (C) The 3rd law of thermodynamics. Statement of fact. 7. (E) J mol-1 K-1. Statement of fact. 8. (B) Entropy. Statement of fact. 9. (D) The decomposition of a solid, group 2 carbonate to yield carbon dioxide gas when heated. An equation to illustrate the generic process (where M is the group 2 element) would be: MCO3(s) MO(s) + CO2(g). Zero moles of gas become one mole of gas. Since gases have standard molar entropies that are huge compared to solids and liquids, any such change will result in a large increase in entropy. 10. (D) Particles in solution have lower entropy values than those in solids. Solids have particles held in fixed positions with very little freedom of movement. In solutions, the particles are free to move, and with that random movement comes relative disorder and therefore greater entropy. 11. (A) All substances will have values for standard molar entropy that are greater than zero. Since absolute zero is a theoretical temperature, and therefore all substances will have temperatures greater than it, all substances will have energies and entropies greater than zero.

56 ©2009 by Applied Practice, Ltd., Dallas, TX. All rights reserved.Spiral-bound teacher pages may not be shared or reproduced.

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12. (E) a positive value for ∆So is essential for a spontaneous reaction. Since spontaneity is dependent on a negative value for ∆Go, and ∆Go = ∆Ho – T∆So, it is possible for ∆So to be negative and still result in a negative ∆G value as long as ∆Ho is sufficiently negative. 13. (E) H2O(g) at 400 K. Gases have greater standard molar entropies than liquids, which in turn have greater entropies than solids. An increase in temperature (and with it energy) leads to greater disorder (entropy). 14. (D) The total entropy of the universe tends to increase. Statement of fact. 15. (C) 304 – ((2)(240)). Application of the formula, ∆So = ∑So

(products) – ∑So(reactants),

remembering that standard molar entropies are quoted “per mole.”


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ANSWER EXPLANATIONS FREE ENERGY CALCULATIONS

16. (E) ∆Go = ∆Ho - T∆So. ∆Go, ∆Ho, and T are known and ∆So is required. 17. (E) ∆Go = ∆Ho - T∆So. ∆So, ∆Ho, and T are known and ∆Go is required. 18. (E) ∆Go = ∆Ho - T∆So. ∆Go, ∆So, and T are known and ∆Ho is required. 19. (D) I and III only. ∆Go = ∆Ho - T∆So = -10 kJ = -20 kJ + T(0.1 kJ mol-1 K-1). T = 100 K. If ∆Go is negative then the reaction is spontaneous. ∆Go will always be negative with a positive ∆So and a negative ∆Ho. 20. (D) I and III only. Dimensional analysis tells us that units of R must be J mol-1 K-1, and temperature must be measured in K for ∆Go to be in J mol-1. The value for R with these units is 8.31, making statement II incorrect. 21. (B) -101 kJ mol-1, spontaneous. ∆Go = ∑ΔGf ( prod )

o – ∑ΔGf (rxts)o = -32.9 – (68.1 + 0).

Negative ∆G values are associated with spontaneous reactions. 22. (E) -92. ∆Ho = ∑ΔH f ( prod )

o – ∑ΔH f (rxts)o = (2(-46)) – ((0) + 2(0)) = -92.

23. (A) -199 J mol-1 K-1. ∆So = ∑So

(products) – ∑ So(reactants) = (2(193)) – (192 + 3(131)) = -

199. 24. (A) 6.64 x 105. The number cannot be generated without a calculator, but it is the only choice greater than 1. This is consistent with a negative ∆Go for the forward reaction, which is turn is consistent with a positive ∆Go for the reverse reaction. 25. (C) K >1 and the reaction is spontaneous. This choice is consistent with a negative ∆G. 26. (A) ∆H is negative. Since ∆Go = ∆Ho - T∆So, the only way that ∆G can be negative (spontaneous reaction) with a negative ∆So is for ∆Ho to be negative. 27. (E) 2000 K. ∆Go = ∆Ho - T∆So = +90 kJ mol-1 = +190 kJ mol-1 - T(0.05 kJ mol-1 K-1). T = 2000 K. 28. (D) 0 = +190 – T(+50/1000). ∆Go = ∆Ho - T∆So and when ∆Go = 0 the reaction “flips” from nonspontaneous to spontaneous and vice versa. ∆So must be in units of kJ. 29. (B) The reaction is spontaneous in the forward direction. Calculation of Q gives a value of 50. In order to reach equilibrium, a value of K of 300 is required. The reaction must go forward to produce more products.


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30. (E) The free energy change is zero, and large amounts of products are found. ∆Go = 0 at equilibrium and since K is such a large number it suggests that products will predominate.


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ANSWER EXPLANATIONS SPONTANEITY

31. (A) ∆Go = - , ∆Ho = - , ∆So = + . Negative ∆Go is associated with spontaneous reactions, and since ∆G = ∆H - T∆S, the only way that ∆Go can always be negative (at any temperature) is with a positive ∆So and a negative ∆Ho. 32. (D) ∆Go = + , ∆Ho = + , ∆So = - . Positive ∆Go is associated with nonspontaneous reactions, and since ∆Go = ∆Ho - T∆So, the only way that ∆Go can always be positive (at any temperature) is with a negative ∆So and a positive ∆Ho. 33. (B) ∆Go = - , ∆Ho = + , ∆So = + . Negative ∆G is associated with spontaneous reactions. Since ∆Go = ∆Ho - T∆So, a negative ∆Go is favored by a positive ∆So and a negative ∆Ho. 34. (E) I, II, and III. Exothermic means a negative enthalpy change, and increasing moles of gas means positive entropy change. Reactions with that combination are always spontaneous (yield negative ∆G values) according to ∆Go = ∆Ho - T∆So. 35. (C) I and II only. All “∆ formation” data are equal to zero for elements, since when an element is formed from its elements (the definition of formation) there is no change. 36. (D) ∆Ho = - , ∆So = - . Energy is lost on cooling, and entropy decreases as the system becomes more ordered. 37. (E) ∆Ho = + , ∆So = - , T So = - . Since Go = Ho – T So, the only way that G can always be positive (at any temperature) and therefore nonspontaneous, is with a negative So and a positive Ho. Since temperature is measured in Kelvin (always a positive number), then T So will take the sign of So. 38. (D) K > 1. Positive ∆Go is associated with a nonspontaneous reaction. K > 1 is associated with a spontaneous reaction. 39. (A) G = - , K > 1, lnK = + . Statement of fact associated with Go = -RT lnK. 40. (A) So favors spontaneity, but it can only be spontaneous if temperatures are relatively high. Positive ∆So means greater disorder that favors spontaneity, but positive ∆Ho does the opposite. Since ∆Go = ∆Ho – T ∆So, it is still possible for ∆Go to be negative (spontaneous reaction) as long as T is sufficiently large. 41. (C) ∆Go is increasingly negative, and more products will be present at equilibrium. Since ∆Go = ∆Ho - T∆So = -393 - T∆So and ∆So is positive, then as T increases the term T∆So will also become increasingly positive and ∆Go increasingly negative. More negative values for ∆Go are associated with increased spontaneity.


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42. (D) ∆Go = - . Definition. 43. (D) The reaction must be endothermic, and disorder decreases. Positive ∆Go is associated with nonspontaneous reactions, and since ∆Go = ∆Ho - T∆So, the only way that ∆Go can always be positive (at any temperature) is with a negative ∆So and a positive ∆Ho. 44. (E) it will have products that are less disordered than reactants. Negative ∆Go is associated with spontaneity, and since ∆Go = ∆Ho - T∆So, the only way that ∆Go can always be negative (at any temperature) is with a positive ∆So and a negative ∆Ho. 45. (E) ∆Ho = + , ∆So = - , T∆So = + . Since T is measured in Kelvin (always positive), the term T∆So must take on the same sign as ∆So.


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ANSWER EXPLANATIONS FREE ENERGY AND EQUILIBRIUM

46. (A) K > 1. Statement of fact associated with ∆Go = -RT lnK. 47. (A) K > 1. Statement of fact associated with ∆Go = -RT lnK. 48. (B) K = 1. Statement of fact associated with ∆Go = -RT lnK. 49. (E) I, II, and III. Statement I is a mathematical fact. Statement II is a mathematical fact and associated with negative ∆Go values representing spontaneous reactions. Statement III is true via dimensional analysis. 50. (C) I and II only. At this point equilibrium has been established and RT lnQ = 0. 51. (B) It is at a minimum. Statement of fact. 52. (E) The reaction will have a value for K that is greater than 1. Negative ∆Go values mean spontaneous reactions but tell us nothing about the speed of a reaction. Negative ∆G can be achieved with either a negative ∆So or a negative ∆Ho. 53. (E) free energy has reached a maximum. By definition, at equilibrium free energy has reached a minimum. 54. (A) create a more likely forward reaction. By math, as K gets larger, ∆Go gets more negative, i.e., the reaction becomes increasingly likely.

55. (C) K = ΔG-RTe

⎛ ⎞⎜ ⎟⎝ ⎠ . Algebraic consequence of re-arranging ∆Go = -RT lnK.

56. (E) 1 x 10-14. ∆G value here is largely irrelevant since the self-ionization of water at 298 is represented by Kw which = 1 x 10-14. 57. (C) The ∆G for the reaction will be negative. Large, positive values of K are associated with negative ∆G values. 58. (D) ∆G for the forward reaction at 373 K will be positive. ∆G at equilibrium is 0. 59. (C) That ∆G for the reaction is a negative number. If the term lnK is positive, then ∆G must be negative (according to ∆G = -RT lnK). These values are associated with spontaneous reactions. 60. (D) the reverse reaction is likely. Since ∆G = -RT lnK a positive value for lnK means a negative value for ∆G and a reaction that is spontaneous.


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ANSWER EXPLANATIONS DESCRIPTIVE CHEMISTRY AND LABORATORY

61. (A) Ka. Titration of a strong base with a weak acid will allow the calculation of the weak acid dissociation constant via the Henderson-Hasselbalch equation and knowing that pH = pKa half-way to the equivalence point. 62. (D) Kp. Kp is a measure of the ratio of partial pressure of gaseous products and reactants. 63. (E) Ksp. Ksp is a measure of the product of ion concentrations for sparingly soluble salts such as barium sulfate. 64. (E) I, II, and III. The combination of all three statements defines the thermodynamics of living systems. 65. (E) I, II, and III. Calibration of a colorimeter with known concentrations and the associated plots are the means to determine unknown concentrations when using Beer’s law. A dirty cuvette will cause erroneous absorption readings. 66. (B) Ksp = [Pb2+][I-]2. Definition of Ksp with pure solids not included in the expression. 67. (B) An oxidizing agent. Iodide ions must lose electrons to become I2, i.e., they must be oxidized. 68. (D) PbBr2. A Ksp calculation/value is only meaningful for a sparingly soluble salt. All other choices are completely soluble. 69. (B) When 7.50 mL of KOH has been added. [salt] = [acid] occurs half-way to the equivalence point. 70. (D) pH = pKa. Application of the Henderson-Hasselbalch equation tells us that when [salt] = [acid] then pH = pKa + log (1) = pKa. This occurs half-way to the equivalence point. 71. (E) HCl is a strong acid and as such has no, meaningful value for Ka. Statement of fact. 72. (D) A catalyst. Statement of fact, and the substance does not appear in the balanced, chemical equation. 73. (E) To determine the total moles of CH3COOH and sulfuric acid present. NaOH will react with any acid present and allow the total moles of acid present at equilibrium to be calculated.


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74. (E) The moles of ethanol, ester, and water at equilibrium. An ICE table will reveal relationships between all of the substances. 75. (C) It will be equal to 3.9 because the temperature has not changed. Equilibrium constants are constant at constant temperature.


Free-Response Answers and

Scoring Guides


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FREE-RESPONSE ANSWERS THERMODYNAMICS

1. (a) 2C4H10(g) + 13O2(g) 8CO2(g) + 10H2O(l) (1 point) (b) Oxygen is an element. When an element is formed from its elements (the

definition of formation), no change takes place so the ∆Ho (change in enthalpy) is zero.

i.e. O2(g) O2(g) represents no change. (1 point) (c) ∆Ho = ∑ΔH f

o(products) – ∑ΔH f

o(reactants)

= (8(-393) + 10(-286)) – (2(-122) + 13(0)) = -5754 kJ This value is for the combustion of two moles of butane, so divide by 2. ∆H�

c = -2877 kJ mo1-1 (2 points) (d) Since 15 moles of gas are becoming 8 moles of gas and 10 moles of liquid, the

degree of disorder will decrease and ∆So will be negative. (1 point) (e) ∆Go = ∆Ho – T∆So ∆So = ∑So

(products) – ∑So(reactants)

= (8(214) + 10(70)) – (2(310) + 13(205)) = -873 J mol-1 K-1 -873 J mol-1 K-1 is for two moles, so divide by 2 = 436.5 J mol-1 K-1 ∆G = -2877 kJ mol-1 – (298 K (-0.4365 kJ mol-1 K-1)) ∆G = -2747 kJ mol-1 (2 points)

(f) Despite the fact that there is a negative ∆Go value for this reaction, and the reaction is described as being feasible or spontaneous, the reaction has an activation energy that must be overcome before it can proceed. (1 point)

66 © 2009 Applied Practice, Ltd., Dallas, TX. All rights reserved.Spiral-bound teacher pages may not be shared or reproduced.

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2. (a) Chemical reaction is Ca(s) + F2(g) CaF2(s)

Ca(s) Ca(g) +193 kJ Ca(g) Ca+

(g) + e- +590 kJ Ca+

(g) Ca2+(g) + e- +1150 kJ

F2(g) 2F(g) +158 kJ 2F(g) + 2e- 2F-

(g) 2(-348 kJ) Ca2+

(g) + 2F-(g) CaF2(s) -2580 kJ

============================= Ca(s) + F2(g) CaF2(s) -1185 kJ (2 points)

============================= Chemical reaction is Ca(s) + ½F2(g) CaF(s)

Ca(s) Ca(g) +193 kJ Ca(g) Ca+

(g) + e- +590 kJ ½F2(g) F(g) (+158 kJ)/2 F(g) + e- F-

(g) -348 kJ Ca+

(g) + F-(g) CaF(s) -798 kJ

============================ Ca(s) + ½F2(g) CaF(s) -284 kJ (2 points) ============================ (b) The enthalpy decrease for the formation CaF2 is greater than the enthalpy

decrease for the formation CaF. The larger the magnitude for an exothermic process, the more negative the value for ∆G and the more spontaneous the reaction. (2 points)

(c) First electron affinity of fluorine. (1 point) (d) Chemical reaction is Ca(s) + F2(g) CaF2(g) ∆Go = ∆Ho – T∆So ∆So = ∑So


= (69) – (42 + 203) = -176 J mol-1 K-1 ∆Go = -1185 kJ mol-1 – (298 K (-0.176 kJ mol-1 K-1) = - 1133 kJ mol-1(2 points)

© 2009 Applied Practice, Ltd., Dallas, TX. All rights reserved. 67Spiral-bound teacher pages may not be shared or reproduced.

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3. (a) Negative. 4 moles of gas become 2 moles of gas, the disorder decreases, and

∆So will be negative. (1 point) (b) ∆Go = ∆Ho – T∆So ∆So = ∑So


= (2(193)) – (192 + 3(131)) = -199 J mol-1 K-1 ∆Go = -92.0 kJ mol-1 – (298 K (-0.199 kJ mol-1 K-1)) = - 32.7 kJ mol-1 (2 points) (c) ∆Go = -RT lnK -32.7 kJ mol-1 = - (.008314 kJ mol-1 K-1)(298 K) lnK K = 5.39 x 105 (1 point) (d) ∆Go for the forward reaction is a negative number suggesting that the

reaction is feasible (spontaneous) and goes to the product side. Since a large number of products are formed, one would expect the K value to be greater than 1. The numbers are consistent with a feasible reaction that tends toward the product side of the equation. (1 point)

(e) ∆Go = ∆Ho – T∆So 0 = -92.0 kJ mol-1 – (T (-0.199 kJ mol-1 K-1)) T = 462 K (1 point) (f) Enthalpy change, ∆Ho. Feasible reactions (those with negative ∆Go values)

are favored by having negative ∆Ho values and positive ∆So values. (1 point)


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4. (a) On passing from chlorine to bromine to iodine there is a change of state

from gas to liquid to solid. The states become increasingly ordered, and as a result the stand molar entropies will decrease. (1 point)

(b) (i) ½I2(s) + 1½Cl2(g) ICl3(s) (1 point)

(ii) ∆So = So(products) –So

(reactants) = ((172)) – (½(117) + 1½(223)) = -221 J mol-1 K-1 (1 point) (iii) ∆Go = -88.0 kJ mol-1 – (298 K (-0.221 kJ mol-1 K-1)) = - 22.1 kJ mol-1

(1 point) (c) ½Br2(l) + 2½F2(g) BrF5(s)

∆So = ∑So(products) – ∑So

(reactants) = ((225)) – (½(152) + 2½(X)) = -155 J mol-1 K-1

X = 122 J mol-1 K-1

The value for standard molar entropy for fluorine gas is smaller than that of chlorine gas since chlorine is a more complicated, larger molecule with a greater number of electrons and as such will be more disordered. (3 points: 2 for calculation, 1 for explanation)


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5. (a) Since the forward reaction results in an increase in disorder (2 moles of gas

as opposed to 1 mole of gas), 1000 K causes a greater number of gaseous moles and therefore an increase in disorder. (Even if the equilibrium were not shifted by a change in temperature, disorder still increases with increased temperature). (2 points)

(b) Yes. The ∆Ho is positive (endothermic) for the forward reaction that does

not favor a negative ∆Go for the forward reaction, since ∆Go = ∆Ho – T∆So. If the forward reaction is not favored then reactants ARE favored. (2 points)

(c) No. The ∆S is positive for the forward reaction that favors a negative ∆Go

for the forward reaction, since ∆Go = ∆Ho – T∆So. If the forward reaction IS favored then reactants are not favored. (2 points)

(d) Since ∆Go = -RT lnK, then larger values for K (proportionally larger

numbers of products) will result in more negative values of ∆Go. This is consistent with a reaction that is feasible (spontaneous). (2 points)


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6. (a) MCO3(s) MO(s) + CO2(g) (where M is a group 2 metal) 1 mole of solid becomes 1 mole of solid and 1 mole of gas, so the disorder

increases and ∆So will be positive. Gases are significantly more disordered than solids. (2 points)

(b) Since ∆Go = ∆Ho – T∆So, a combination of positive ∆Ho and positive ∆So or a

combination of negative ∆Ho and negative ∆So can lead to negative OR positive ∆Go depending on the value of T. (2 points)

(c) A more complicated molecule (one with more atoms and more bonds) will be

more disordered and will have a greater molar entropy than a relatively simple molecule. The bending and stretching of bonds lead to disorder. (2 points)

(d) The standard enthalpy of formation is defined as the formation of a

substance from its elements in their standard state. For hydrogen that means:

H2(g) H2(g)

In this reaction there is no change, so ΔGf

o , ΔH fo , and ΔS f

o values are all equal to zero. (2 points)


Contributors Series Editor David Emmerson is a veteran teacher with more than twenty years’ experience teaching AP Chemistry. He has previously worked on developing an online AP Chemistry course. He holds a B.S. in Biology from Cornell University and an M.A. in Science Education from the State University of New York Writer Adrian Dingle is a chemistry educator with 18 years’ experience and is creator of the award-winning chemistry Web site www.adriandingleschemistrypages.com. He holds a B.Sc. (Hons.) Chemistry and a Postgraduate Certificate in Education from the University of Exeter, England.


Applied Practice Chemistry Series

Volume 1: Chemistry Fundamentals Volume 2: Reactions & Stoichiometry Volume 3: Elements and Periodicity Volume 4: Chemical Bonding Volume 5: Gases Volume 6: IMF, Liquids, and Solids Volume 7: Reaction Energy Volume 8: Solutions and Equilibrium Volume 9: Kinetics and Nuclear Chemistry Volume 10: Acids, Bases, and Other Equilibria Volume 11: Thermodynamics Volume 12: REDOX and Electrochemistry Volume 13: Coordination and Organic Chemistry

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