chapters 13–16 resources - a beauty whisperertaylorscience.us/hinton/chemistry/chemistry resources...

Chapters 13–16 Resources

Copyright © by The McGraw-Hill Companies, Inc. All rights reserved. Permission is granted to reproduce the material contained herein on the condition that such materials be reproduced only for classroom use; be provided to students, teachers, and families without charge; and be used solely in conjunction with the Glencoe Chemistry: Matter and Change program. Any other reproduction, for sale or other use, is expressly prohibited.

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ISBN: 978-0-07-878763-8MHID: 0-07-878763-7

Printed in the United States of America.

1 2 3 4 5 6 7 8 9 10 045 11 10 09 08 07

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To the Teacher . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv

Chapters 13-16 Resources

Reproducible Student Pages

Student Lab Safety Form . . . . . . . . . . . . . . . . . . . . . . . . . . vi

Chapter 13

Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Chapter 14

Mixtures and Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 31

Chapter 15

Energy and Chemical Change . . . . . . . . . . . . . . . . . . . . . . 57

Chapter 16

Reaction Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

Teacher Guide and Answers

Chapter 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

Chapter 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

Chapter 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

Chapter 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

Table ofContents

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To the Teacher

This booklet contains resource materials to help you teach more effectively. You will find the following in the chapters:

Reproducible Pages

Hands-on ActivitiesMiniLab and ChemLab Worksheets: Each activity in this book is an expanded version of each lab that appears in the Student Edition of Glencoe Chemistry: Matter and Change. All materials lists, procedures, and questions are repeated so that students can read and complete a lab in most cases without having a textbook on the lab table. All lab ques-tions are reprinted with lines on which students can write their answers. In addition, for student safety, all appropriate safety symbols and caution statements have been reproduced on these expanded pages. Answer pages for each MiniLab and ChemLab are included in the Teacher Guide and Answers section at the back of this book.

Transparency ActivitiesTeaching Transparency Masters and Worksheets: These transparencies relate to major concepts that will benefit from an extra visual learning aid. Most of the transparencies contain art or photos that extend the concepts put forth in the textbook. Others contain art or photos directly from the Student Edition. There are 73 Teaching Transparencies, provided here as black-and-white masters accompanied by worksheets that review the concepts presented in the transparencies. Answers to worksheet questions are provided in the Teacher Guide and Answers section at the back of this book.

Math Skills Transparency Masters and Worksheets: These transparencies relate to math-ematical concepts that will benefit from an extra visual learning aid. Most of the trans-parencies contain art or photos directly from the Student Edition, or extend concepts put forth in the textbook. There are 42 Math Skills Transparencies, provided here as black-and-white masters accompanied by worksheets that review the concepts presented in the transparencies. Answers to worksheet questions are provided in the Teacher Guide and Answers section at the back of this book.

Intervention and AssessmentStudy Guide: These pages help students understand, organize, and compare the main chemistry concepts in the textbook. The questions and activities also help build strong study and reading skills. There are six study guide pages for each chapter. Students will find these pages easy to follow because the section titles match those in the textbook. Italicized sentences in the study guide direct students to the related topics in the text.

The Study Guide exercises employ a variety of formats including multiple-choice, matching, true/false, labeling, completion, and short answer questions. The clear, easy-to-follow exercises and the self-pacing format are geared to build your students’ confi-dence in understanding chemistry. Answers or possible responses to all questions are provided in the Teacher Guide and Answers section at the back of this book.

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Chapter Assessment: Each chapter assessment includes several sections that assess stu-dents’ understandings at different levels.

• The Reviewing Vocabulary section tests students’ knowledge of the chapter’s vocabu-lary. A variety of formats is used, including matching, true/false, completion, and comparison of terms.

• The Understanding Main Ideas section consists of two parts: Part A tests recall and basic understanding of facts presented in the chapter, while Part B is designed to be more challenging and requires deeper comprehension of concepts than does Part A. Students may be asked to explain chemical processes and relationships or to make comparisons and generalizations.

• The Thinking Critically section requires students to use several different higher-order learning skills, such as interpreting data and discovering relationships in graphs and tables, as well as applying their understanding of concepts to solve problems, compare and contrast situations, and to make inferences or predictions.

• The Applying Scientific Methods section puts students into the role of researcher. They may be asked to read about an experiment, simulation, or model and then apply their understanding of chapter concepts and scientific methods to analyze and explain the procedure and results. Many of the questions in this section are open-ended, giving students the opportunity to demonstrate both reasoning and creative problem-solving skills.

Answers or possible responses to all questions are provided in the Teacher Guide and Answers section at the back of this book

STP Recording Sheet: STP Recording Sheets allow students to use the Standardized Test Practice questions in the Student Edition as a practice for standardized tests. STP Recording Sheets give them the opportunity to use bubble answer grids and numbers grids for recording answers. Answers for the STP Recording Sheets can be found in the Teacher Wraparound Edition on Standardized Test Practice pages.

Teacher Guide and Answers: Answers or possible answers for questions in this booklet can be found in the Teacher Guide and Answers section. Materials, teaching strate-gies, and content background, along with chapter references, are also provided where appropriate.

To the Teacher continued

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Lab Safety Form

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Name:

Date:

Lab type (circle one) : Launch Lab MiniLab ChemLab

Lab Title:

Read carefully the entire lab and then answer the following questions. Your teacher must initial this form before you begin the lab.

1. What is the purpose of the investigation?

2. Will you be working with a partner or on a team?

3. Is this a design-your-own procedure? Circle: Yes No

4. Describe the safety procedures and additional warnings that you must follow as you perform this investigation.

5. Are there any steps in the procedure or lab safety symbols that you do not understand? Explain.

Teacher Approval Initials

Date of Approval

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Chapter 13 GasesMinilab . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

Chem Lab . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3

Teaching TransparencyMaster and Worksheets . . . . . . . . . . . . . . . . . . . . . . . . . . . .6

Maths Skill TransparencyMaster and Worksheets . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

Study Guide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

Chapter Assessment. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

STP Recording Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

Table ofContents

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2 Chemistry: Matter and Change • Chapter 13 ChemLab and MiniLab Worksheets

mini LAB 13Model a Fire Extinguisher

Why is carbon dioxide used in fire extinguishers?

Materials masking tape, aluminum foil, metric ruler, beaker, candle, matches, ther-mometer, barometer or weather radio, baking soda (NaHCO3), vinegar (5% CH3COOH)

Procedure1. Read and complete the lab safety form.

2. Measure the temperature with a thermometer. Obtain the air pressure with a barometer or weather ratio. Record your observations.

3. Roll a 23-cm � 30-cm piece of aluminum foil into a cylinder that is 30 cm long androughly 6 cm in diameter. Tape the edges with masking tape.

4. Use matches to light a candle. WARNING: Run water over the extinguished matchbefore throwing it away. Keep hair and loose clothing away from the flame.

5. Place 30 g of baking soda (NaHCO3) in a large beaker. Add 40 mL of vinegar (5% CH3 COOH).

6. Quickly position the foil cylinder at about 45° up and away from the top of the can-dle flame. WARNING: Do not touch the end of the aluminum tube that is nearthe burning candle.

7. While the reaction in the beaker is actively producing carbon dioxide gas, carefullypour the gas, but not the liquid, out of the beaker and into the top of the foil tube.Record your observations.

Analysis

1. Apply Calculate the molar volume of carbon dioxide gas (CO2) at room temperatureand atmospheric pressure.

2. Calculate the room-temperature densities in grams per liter of carbon dioxide, oxygen,and nitrogen gases. Recall that you will need to calculate the molar mass of each gas inorder to calculate densities.

3. Interpret Do your observations and calculations support the use of carbon dioxide gas toextinguish fires? Explain.

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ChemLab and MiniLab Worksheets Chemistry: Matter and Change • Chapter 13 3

Safety Precautions • Read and observe all cautions listed on the aerosol can of

office equipment duster. • Do not have any open flames in the room.

CHEMLAB 13

ProblemHow much pressure isrequired to burst a kernelof popcorn?

Objectives• Collect and analyze data

hard or soft water• Calculate the pressure

needed to burst a kernelof popcorn

• Design steps to controlexperimental error

Materialspopcorn kernels

(18-20) vegetable oil (1.5

mL) wire gauzesquares (2)

Bunsen burner ring stand

small iron ring 10-mL graduated

cylinder250-mL beakerbeaker tongs balance distilled waterpaper towels

Determine Pressure inPopcorn KernelsWhen the water vapor pressure inside a popcorn kernel

is great enough, the kernel bursts and releases thewater vapor. The ideal gas law can be used to find the pres-sure in the kernel as it bursts.

Pre-Lab

1. Read the entire CHEMLAB.

2. Water forms a meniscus when poured into a grad-uated cylinder. To correctly record the volume,which part of the meniscus do you read? Howmany digits do you write down when you recordthe measured volume?

3. In step 5, why is it important to dry the kernelsbefore moving on with the procedure?

4. How do you convert from mass of a compound tomoles of a compound? If you have 5.00 g ofwater, how many moles of water do you have?

5. Define atmospheric pressure (1atm) in terms ofkilopascals, millimeters of mercury, torrs, andbars.

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Procedure

1. Read and complete the lab safety form.

2. Create a table to record your data.

3. Place approximately 5 mL of distilled water inthe graduated cylinder, and record the volume.

4. Place 18-20 popcorn cornels in the graduatedcylinder with the water. Tap the cylinder toforce any air bubbles off the kernels. Record thenew volume.

5. Remove the kernels from the graduated cylinder, and dry them.

6. Place the dry kernels and 1.0-1.5mL of vegetable oil into the beaker.

7. Measure the total mass of the beaker, oil, andkernels.

8. Set up a Bunsen burner with a ring stand, ring,and wire guaze.

9. Place the beaker on the wire guaze and ring.Place another piece of wire guaze on top of thebeaker.

10. Gently heat the beaker with the burner. Movethe burner back and forth to heat the kernelsevenly.

11. Observe the changes in the kernels and oilwhile heating, then turn off the burner when thepopcorn has popped and before any burningoccurs.

12. Using the beaker tongs, remove the beaker fromthe ring and allow it to cool completely.

13. Measure the final mass of the beaker, oil, andpopcorn once cooling is complete.

14. Post your data at glencoe.com

15. Cleanup and Disposal Dispose of the popcornand oil as directed by your teacher. Wash andreturn all lab equipment to its designated location.

Analyze and Conclude

1. Calculate the volume of the popcorn kernels, in liters, by the difference in the volumes of distilledwater before and after adding popcorn.

2. Calculate the total mass of water vapor released using the mass measurements of the beaker, oil, andpopcorn before and after popping.

3. Convert Use the molar mass of water and the volume of popcorn to find the number of moles ofwater released.

4. Use Formulas Use the temperature of the boiling oil (225°C) as your gas temperature, and calculatethe pressure of the gas using the ideal gas law.

CHEMLAB 13

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5. Compare and Contrast atmospheric pressure to the pressure of the water vapor in thekernels.

6. Infer why all the popcorn kernels did not pop.

7. Error Analysis Identify a potential source of error for this lab, and suggest a method tocorrect it.

Inquiry Extension

Design an experiment that tests the amount of pressure necessary to different types ofpopcorn kernels.

CHEMLAB 13

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6 Chemistry: Matter and Change • Chapter 13 Teaching Transparency Masters

Volume (L)

Pressure vs. Volume for a Gas at Constant Temperature

Pres

sure

(kP

a)

250

200

150

100

50

010 2 3 4 5

(V1, P1)

(V2, P2)

Pressure vs. Volume GraphPressure vs. Volume Graph

TEACHING TRANSPARENCY MASTER

Use with Chapter 13,Section 13.1

39

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Teaching Transparency Worksheets Chemistry: Matter and Change • Chapter 13 7

1. Based on this graph, how is the volume of a gas affected by increased pressure at constant temperature?

2. The relationship between the volume and pressure of a gas at constant temperature is an inversely proportional relationship. Based on the evidence in this graph, define theinversely proportional relationship.

3. What do you notice when you multiply the pressure by the volume for any point on theline on the graph?

4. Based on the mathematical relationship derived from this graph, what is the pressure ofthe gas at 3.00 L at constant temperature?

5. What gas law does this graph represent?

6. What mathematical expression is used to define this law? Define all symbols used.

7. A sample of gas is compressed from 3.25 L to 1.20 L at constant temperature. If the pres-sure of this gas in the 3.25-L volume is 100.00 kPa, what will the pressure be at 1.20 L?List all known and unknown variables. Show all your work.

Pressure vs. Volume GraphPressure vs. Volume Graph

TEACHING TRANSPARENCY WORKSHEET


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Temperature (K)

Volume vs. Kelvin Temperaturefor a Gas at Constant Pressure

Vo

lum

e (L

)

5

4

3

2

1

02000 400 600 800 1000

(T1, V1)

(T2, V2)

Volume vs. Temperature GraphVolume vs. Temperature Graph



40

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1. Notice that this graph shows kelvin temperatures. How are the kelvin scale and Celsiusscale related mathematically?

2. Based on this graph, how is the volume of a gas affected by increased temperature atconstant pressure?

3. The relationship between the volume and temperature of a gas at constant pressure is adirectly proportional relationship. Based on the evidence in this graph, define the directlyproportional relationship.

4. What do you notice when you divide the temperature by the volume for any point on theline on the graph?

5. What law does this graph represent?

6. What mathematical expression is used to define this law? Define all symbols used.

7. The kelvin temperature of a sample of gas is decreased from 460 K to 240 K at constantpressure. If the volume of this gas at 460 K is 2.50 L, what will the volume be at 240 K?List all known and unknown variables. Show all your work.

Volume vs. Temperature GraphVolume vs. Temperature Graph



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2 m

ol

2 vo

lum

es2

mo

l2

volu

mes

1 m

ol

1 vo

lum

e1

mo

l1

volu

me

� ��

0 0M

eth

ane

gas

CH

4 (g)

Oxy

gen

gas

2O2 (g

)C

arb

on

dio

xid

e g

asC

O2 (g

)W

ater

vap

or

2H

2O(g

)

TEACHING TRANSPARENCY MASTER 41

Burning of Methane GasBurning of Methane Gas Use with Chapter 13,Section 13.3

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1. What do the coefficients in chemical equations involving gases, solids, and liquids represent?

2. What does Avogadro’s principle state?

3. Based on Avogadro’s principle, the coefficients in chemical equations involving onlygases represent two types of quantities. Name the two quantities.

4. Based on the balanced equation for the complete combustion of methane, how manyliters of carbon dioxide, CO2(g), and water vapor, H2O(g), are produced by the completecombustion of 1 L of methane gas, CH4?

5. What volume of oxygen gas is needed for the complete combustion of 8.00 L of methanegas, CH4? Assume that the pressure and temperature of the reactants are the same. Showall your work.

6. Write a balanced equation for the complete combustion of propane gas, C3H8, with oxygen, O2, to form carbon dioxide, CO2, and water, H2O.

7. What volume of carbon dioxide gas, CO2, is produced when 7.00 L of propane gas,C3H8, undergoes complete combustion, as shown in your answer to question 6? Show allyour work.

Burning of Methane GasBurning of Methane Gas



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12 Chemistry: Matter and Change • Chapter 13 Math Skills Transparency Masters

Solving Gas Problems Using Boyle’s Law

MATH SKILLS TRANSPARENCY MASTER


19

Solving Gas Problems Using Boyle’s Law

How can you solve gas problems using Boyle’s law?ProblemA sample of helium gas in a balloon has a pressure of 240 kPa in a 1.4-L container. What will be the new pressure if the gas sample is transferred to a 3.0-L container?

Step 1. Analyze the problem.

Known VariablesV1 � 1.4 L V2 � 3.0 L P1 � 240 kPa

Unknown VariableP2 � ? kPa

Step 2. Solve for the unknown.Divide both sides of the equation for Boyle’s law by V2 to solve for P2.

P1V1 � P2V2 P2 � P1� �Substitute the known values into the rearranged equation.

P2 � 240 kPa � �Multiply and divide numbers and units to solve for P2.

P2 � 240 kPa � � � 110 kPa

Step 3. Evaluate the answer.

1.4 L�3.0 L

1.4 L�3.0 L

V1�V2

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Math Skills Transparency Worsheets Chemistry: Matter and Change • Chapter 12 13

1. What two variables are inversely proportional, as stated in Boyle’s law?

2. What variable must be kept constant if you are solving a problem using Boyle’s law?

3. What is the first step in solving pressure-volume gas problems?

4. What is the equation for Boyle’s law?

5. How is the equation for Boyle’s law rearranged before solving it?

6. What is done to similar units in the numerator and the denominator in the final solving step?

7. What unit remains at the end? Is this the desired unit? Of what quantity is it a unit?

8. If the volume is almost doubled in a container at a fixed temperature, what will happento the pressure?

9. In step 3, how would you evaluate the answer to see whether or not it is reasonable?

10. If the known variables in a new problem are V1, P1, and P2, rewrite the equation forBoyle’s law to solve for the unknown variable.

Solving Gas Problems Using Boyle’s LawSolving Gas Problems Using Boyle’s Law

MATH SKILLS TRANSPARENCY WORKSHEET


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Solving Gas Problems Using the Combined Gas LawSolving Gas Problems Using the Combined Gas Law



20

How can you solve gas problems using the combined gas law?ProblemA gas at 100.0 kPa and 25.0°C fills a flexible container withan initial volume of 2.00 L. If the temperature is raised to60.0° and the pressure increased to 320.0 kPa, what is thenew volume?


Known VariablesP1 � 100.0 k Pa P2 � 320.0 kPaT1 � 25.0°C T2 � 60.0°C V1 � 2.00 LUnknown VariableV2 � ? L

Step 2. Solve for the unknown.Add 273 to the Celsius temperature for T1 and T2 to obtain the kelvin temperature.

T1 � 273 � 25.0°C � 298 K; T2 � 273 � 60.0°C � 333 KMultiply both sides of the equation for the combined law by T2 and divide by P2 to solve for V2.

� V2 � V1 � �� Substitute the known values into the rearranged equation; multiply and divide numbers and units to solve for V2.

V2 � 2.00 L� �� 0.698 L


333 K�298 K

100.0 kPa��320.0 kPa

T2�T1

P1�P2

P2V2�T2

P1V1�T1

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Math Skills Transparency Worksheets Chemistry: Matter and Change • Chapter 13 15

1. Describe how pressure relates to volume and temperature in the combined gas law.

2. How does volume relate to temperature in the combined gas law?

3. What is the equation for the combined gas law?

4. Kelvin temperature is used in the combined gas law, not Celsius. How are Celsiusdegrees converted to Kelvin degrees?

5. What is the first step in solving a combined gas law problem?

6. How is the equation for the combined gas law rearranged before solving it?

7. What is done to similar units in the numerator and the denominator in the final solving step?



Solving Gas Problems Using the Combined Gas LawSolving Gas Problems Using the Combined Gas Law



20

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Solving Gas Problems Using the Ideal Gas LawSolving Gas Problems Using the Ideal Gas Law



21

ProblemCalculate the number of moles of gas contained in a 2.0-L container at 200.0 K with a pressure of 120 kPa.


Known VariablesV = 2.0 L T = 200.0 KP = 120 kPa R = 8.314 L•kPa/(mol•K)Unknown Variablen = ? mol

Step 2. Solve for the unknown.Divide both sides of the ideal gas law equation by RT to solve for n.

PV � nRT n �

Substitute the known values into the rearranged equation.

n �(120 kPa)(2.0 L)

( )(200.0 K)

Multiply and divide numbers and units to solve for n.

n �(120 kPa)(2.0 L)

� 0.14 mol( )(200.0 K)


8.314�L�kPa��mol�K

8.314�L�kPa��mol�K

PV�RT

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1. What four variables does the ideal gas law describe?

2. What is the equation for the ideal gas law?

3. What is the first step in solving a combined gas law problem?

4. What numerical value of the gas constant, R, is used to solve this problem? Explain your answer.



7. If you were asked to find the molar mass of a gas in an ideal gas law problem, what formof the ideal gas law equation would you use? Show how this equation is derived from theoriginal form of the ideal gas law equation.

8. What is Avogadro’s principle and how does it relate to the ideal gas law?

Solving Gas Problems Using the Ideal Gas LawSolving Gas Problems Using the Ideal Gas Law



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18 Chemistry: Matter and Change • Chapter 13 Study Guide

GasesGases

Section 13.1 The Gas LawsIn your textbook, read about the basic concepts of the three gas laws.

Use each of the terms below to complete the passage. Each term may be used more than once.

Boyle’s law relates (1) and (2) if

(3) and amount of gas are held constant. Charles’s law relates

(4) and (5) if (6)

and amount of gas are held constant. Gay-Lussac’s law relates (7)

and (8) if (9) and amount of gas are

held constant.

In your textbook, read about the effects of changing conditions on a sample of gas.

For each question below, write increases, decreases, or stays the same.

10. The room temperature increases from 20°C to 24°C. Whathappens to the pressure inside a cylinder of oxygen contained inthe room?

11. What happens to the pressure of the gas in an inflated expandableballoon if the temperature is increased?

12. An aerosol can of air freshener is sprayed into a room. Whathappens to the pressure of the gas if its temperature staysconstant?

13. The volume of air in human lungs increases before it is exhaled.What happens to the temperature of the air in the lungs to causethis change, assuming pressure stays constant?

14. A leftover hamburger patty is sealed in a plastic bag and placed inthe refrigerator. What happens to the volume of the air in the bag?

15. What happens to the pressure of a gas in a lightbulb a few minutesafter the light is turned on?

STUDY GUIDE CHAPTER 13

pressure temperature volume

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Study Guide Chemistry: Matter and Change • Chapter 13 19Study Guide Chemistry: Matter and Change • Chapter 13 19

Section 13.2 The Combined Gas Law and Avogadro’s PrincipleIn your textbook, read about the combined gas law.

Fill in the following table. State what gas law is derived from the combined gas law when the variable listed in the first column stays constant and the variables in the second column change.

In your textbook, read about the relationships among temperature, pressure, and vol-ume of a sample of gas.

Fill in the blanks between the variables in the following concept map to show whetherthe variables are directly or inversely proportional to each other. Write direct or inversebetween the variables.

In your textbook, read about the combined gas law and Avogadro’s principle.

Circle the letter of the choice that best completes the statement or answers the question.

7. The variable that stays constant when using the combined gas law is

a. amount of gas. b. pressure. c. temperature. d. volume.

8. The equation for the combined gas law can be used instead of which of the followingequations?

a. Boyle’s law b. Charles’s law c. Gay-Lussac’s law d. all of these

9. Which of the following expresses Avogadro’s principle?

a. Equal volumes of gases at the same temperature and pressure contain equal numbers of particles.

b. One mole of any gas will occupy a certain volume at STP.

c. STP stands for standard temperature and pressure.

d. The molar volume of a gas is the volume that one mole occupies at STP.

temperature

volume pressure

5.

6.

4.

STUDY GUIDECHAPTER 13

Stays constant Change Becomes this law

Volume Temperature, pressure 1.

Temperature Pressure, volume 2.

Pressure Temperature, volume 3.

Derivations from the Combined Gas Law

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20 Chemistry: Matter and Change • Chapter 13 Study Guide 20 Chemistry: Matter and Change • Chapter 13 Study Guide

Answer the following questions.

10. What is standard temperature and pressure (STP)?

11. What is the molar volume of a gas equal to at STP?

In your textbook, read about how to solve problems using the combined gas law andAvogadro’s principle.

Each problem below needs more information to determine the answer. List as many letters as are needed to solve the problem.

a. molar volume of the gas d. pressure of the gas

b. molar mass of the gas e. volume of the gas

c. temperature of the gas f. No further information is needed.

12. What volume will 1.0 g N2 gas occupy at STP?

13. What volume will 2.4 mol He occupy at STP?

14. A gas sample occupies 3.7 L at 4.0 atm and 25°C. What volume will thesample occupy at 27°C?

15. A sample of carbon dioxide is at 273 K and 244 kPa. What will its volumebe at 400 kPa?

16. A sample of oxygen occupies 10.0 L at 4.00 atm pressure. At whattemperature will the pressure equal 3.00 atm if the final volume is 8.00 L?

17. At what pressure will a sample of gas occupy 5.0 L at 25°C if it occupies3.2 L at 1.3 atm pressure and 20°C?

18. How many grams of helium are in a 2-L balloon at STP?

19. One mole of hydrogen gas occupies 22.4 L. What volume will the sampleoccupy if the temperature is 290 K and the pressure is 2.0 atm?

Section 13.2 continued


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Section 13.2 The Ideal Gas LawIn your textbook, read about the ideal gas law.


1. Why is the mathematical relationship among the amount, volume, temperature, and pres-sure of a gas sample called the ideal gas law?

2. Define the ideal gas constant, R.

3. In Table 14.1 in your textbook, why does R have different numerical values?

4. What variable is considered in the ideal gas law that is not considered in the combinedgas law?

In your textbook, read about real versus ideal gases.

For each statement below, write true or false.

5. An ideal gas is one whose particles take up space.

6. At low temperatures, ideal gases liquefy.

7. In the real world, gases consisting of small molecules are the only gasesthat are truly ideal.

8. Most gases behave like ideal gases at many temperatures and pressures.

9. No intermolecular attractive forces exist in an ideal gas.

10. Nonpolar gas molecules behave more like ideal gases than do gasmolecules that are polar.

11. Real gases deviate most from ideal gas behavior at high pressures and lowtemperatures.

12. The smaller the gas molecule, the more the gas behaves like an ideal gas.


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In your textbook, read about applying the ideal gas law.

Rearrange the ideal gas law, PV � nRT, to solve for each of the following variables.Write your answers in the table.

In your textbook, read about using the ideal gas law to solve for molar mass, mass, or density.

Use the following terms below to complete the statements. Each term may be used more than once.

The number of moles of a gas is equal to the (17) divided by the

(18) .

Density is defined as (19) per unit (20) .

To solve for M in the equation M � , the (21) and the

(22) of the gas must be known.

According to the equation D � , the (23) of the gas must be

known when calculating density.

MP�RT

mRT�PV



Variable to Find Rearranged Ideal Gas Law Equation

n 13.

P 14.

T 15.

V 16.

Rearranging the Ideal Gas Law Equation

mass molar mass volume

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Section 13.3 Gas StoichiometryIn your textbook, read about gas stoichiometry.

Balance the following chemical equation. Then use the balanced equation to answer thequestions.

1. H2(g) � O2(g) 0 H2O(g)

2. List at least two types of information provided by the coefficients in the equation.

3. If 4.0 L of water vapor is produced, what volume of hydrogen reacted? What volume of oxygen?

4. If it is known that 2 mol of hydrogen reacts, what additional information would you needto know to find the volume of oxygen that would react with it?

5. List the steps you would use to find the mass of oxygen that would react with a knownnumber of moles of hydrogen.

6. Find the mass of water produced from 4.00 L H2 at STP if all of it reacts. Show your work.


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24 Chemistry: Matter and Change • Chapter 13 Chapter Assessment

GasesGases

Reviewing VocabularyMatch the definition in Column A with the term in Column B.

Column A Column B

1. Equal volumes of gases at the same temperature andpressure contain equal numbers of particles.

2. One mole of any gas will occupy a volume of 22.4 L atSTP.

3. R represents the relationship among pressure, volume,temperature, and number of moles of gas present.

4. Temperature, pressure, and volume are related for a fixedamount of gas.

5. The physical behavior of an ideal gas can be expressed interms of the pressure, volume, temperature, and numberof moles of gas present.

6. The pressure of a given mass of gas varies directly withthe kelvin temperature when the volume remains constant.

7. The volume of a given amount of gas held at a constanttemperature varies inversely with the pressure.

8. The volume of a given mass of gas is directly proportionalto its kelvin temperature at constant pressure.

Answer the following question.

9. Explain why R can have different numerical values.

CHAPTER ASSESSMENTCHAPTER 13

a. Avogadro’s principle

b. Boyle’s law

c. Charles’s law

d. combined gas law

e. Gay-Lussac’s law

f. ideal gas constant

g. ideal gas law

h. molar volume

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Chapter Assessment Chemistry: Matter and Change • Chapter 13 25

Understanding Main Ideas (Part A)

Fill in the following table with one of these terms: increases, decreases, stays the same.Each term may be used more than once. Assume the amount of gas is constant.

Circle the letter of the choice that best answers the question.

13. What volume will one mole of a gas occupy under standard temperature and pressure?

a. 1 L b. 22.4 L c. 273 L d. 293 L

14. What information is NOT given by the coefficients in a balanced chemical equation?

a. the mass ratios of reactants and products

b. the mole ratios of reactants and products

c. the ratios of number of molecules of reactants and products

d. the volume ratios of gaseous reactants and products

15. What variable is mentioned in the ideal gas law that is assumed to be constant in theother gas laws?

a. number of moles b. pressure c. temperature d. volume


Volume Pressure Temperature

increases stays the same 1.

increases 2. stays the same

stays the same increases 3.

4. increases stays the same

5. stays the same increases

stays the same 6. increases

decreases stays the same 7.

decreases 8. stays the same

stays the same decreases 9.

increases decreases 10.

stays the same 11. decreases

12. stays the same decreases

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Understanding Main Ideas (Part B)

Each of the following examples gives a change in volume, temperature, amount, or pressure of a gas sample. Indicate whether the other variable mentioned would increaseor decrease as a result. If a variable is not mentioned, assume it is constant.

1. Additional gas is added to a soccer ball. The pressure .

2. An inflated balloon is placed in a refrigerator. The volume .

3. A piston in an engine compresses the gas into a smaller volume. The

pressure .

4. Dry ice (solid carbon dioxide) is sealed in a plastic bag. As the temperature increases, the

amount of gas present in the bag .

5. Compressed air in scuba tanks cools off as a diver swims at deeper levels. The pressure

in the tanks .

6. The volume of an inflated balloon increases when the amount of gas in the

balloon .

7. A person sits on an air mattress. The pressure .

Solve each of the following problems. Show your work.

8. The volume and amount of gas are constant in a tire. The initial pressure and temperatureare 1.82 atm and 293 K. At what temperature will the gas in the tire have a pressure of2.35 atm?

9. What is the volume of 2.3 mol Cl2 at 290 K and 0.89 atm? �R � �

10. An inflated balloon is left outside overnight. Initially, it has a volume of 1.74 L when thetemperature is 20.2°C and the pressure is 1.02 atm. At what temperature will the balloonhave a volume of 1.56 L if the pressure falls to 0.980 atm?

0.0821 L�atm��

mol�K


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Thinking CriticallyUse the following graph to answer the questions.

1. What gas law does this graph illustrate?

2. What is the pressure of Sample A if the temperature is 400 K?

3. What is the temperature of Sample B if the pressure is 2.5 atm?

4. Compare the pressures of Sample A and Sample B at 400 K.

5. If the samples contain the same amount of gas, which sample has a greater volume?Explain, using the ideal gas law.

6. If the samples have the same volume, which sample contains a greater amount of gas?Explain, using the ideal gas law.

Temperature (K)

Pres

sure

(at

m)

5.0

4.0

3.0

2.0

1.0

00 100 200 300 400 500 600 700

Sample A

Sample B


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Applying Scientific Methods Elemental nitrogen (N2) makes up approximately 79 percent of the air. Nitrogen is essential tolife, but plants cannot use it in its elemental form. In nature, lightning and bacteria in the soiland roots of certain plants convert atmospheric nitrogen into nitrogen-containing compoundsthat plants can use. However, due to the amount of agricultural crops planted, natural methodsof converting atmospheric nitrogen do not provide all the nitrogen compounds needed by thegrowing plants. For this reason, chemical fertilizers are sprayed over the soil, supplying nitro-gen compounds.

Ammonia (NH3) gas is a component in the manufacturing of fertilizers. One method usedto prepare the ammonia is called the Haber process. Fritz Haber first developed this process,which produces ammonia from elemental nitrogen and hydrogen gases. The following dia-gram summarizes the Haber process.

Nitrogen gas Hydrogen gas

Compressor

Catalyst Chamber Return Pump

Cooler

Ammonia gas


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Use what you know about gases and the Haber process to answer the following ques-tions. Show your work.

1. Write a balanced chemical equation for the chemical reaction that occurs in the Haberprocess. Show the state of each reactant and product.

2. Heated nitrogen and hydrogen gases are reduced in volume in the compressor. Whateffect do these changes in temperature and volume have on the pressure of the gases?

3. The return pump moves any unreacted hydrogen and nitrogen gases from the cooler backinto the compressor. What effect will removing gases from the cooler have on the pres-sure of the gases in the cooler? Explain.

4. Ammonia is a real gas. What will happen to it if the pressure continues to be increasedand the temperature continues to be decreased?

5. What volume of ammonia gas will be produced from 2400 L of hydrogen gas at the sametemperature and pressure?

6. The Haber process occurs at a pressure of approximately 800.0 atm and a temperature of400.0°C. Assume a gas sample occupies 25.0 L at these conditions. What volume will thesample occupy at STP?

7. If 126 L of ammonia is produced at 800.0 atm and 400.0°C in the Haber process, whatmass of hydrogen was used in the reaction?

Applying Scientific Methods, continued


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30 Chemistry: Matter and Change • Chapter 13

Student Recording Sheet

Name Date Class

Standardized Test PracticeMultiple Choice

Select the best answer from the choices given, and fill in the corresponding circle.

1. 3. 5. 7.

2. 4. 6.

Short Answer

Answer each question with complete sentences.

8.

9.

10.

Extended Response


11.

SAT Subject Test:Chemistry

12. 14.

13.

CHAPTER 13

Assessment

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Chapter 14 Mixtures and SolutionsMiniLab . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

ChemLab . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

Teaching Transparency Masters and Worksheets . . . . . . . . . . . . . . . . . . . . . . . . . . 36

Math Skills Transparency Masters and Worksheets . . . . . . . . . . . . . . . . . . . . . . . . . 40

Study Guide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44



Table ofContents

31

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mini LAB 14Examine Freezing Point Depression

How do you measure freezing point depression?

Materials 400-mL beakers (2), crushed ice, rock salt (NaCl), cold tap water, stirring rods (2), nonmercury thermometer.

Procedure1. Read and complete the lab safety form..

2. Fill two 400-mL beakers with crushed ice. Add 50 mL of cold tap water to each beaker.

3. Measure the temperature of each beaker using a nonmercury thermometer.

4. Stir the contents of each beaker with a stirring rod until both beakers are at a constant temperature approximately 1 min. Record the temperature.

5. Add 75 g of rock salt (NaCl) to one of the beakers. Continue stirring both beakers. Some of the salt will dissolve.

6. When the temperature in each beaker is constant, record the final readings.

7. To clean up, flush the contents of each beaker down the drain with excess water.

Analysis

1. Compare your readings taken for the ice water and the salt water. How do youexplain the observed temperature change?

2. Explain why salt was added to only one of the beakers.

3. Explain Salt is a strong electrolyte that produces two ions, Na� and Cl�, when it dissociates in water. Explain why this is important to consider when calculating thecolligative property of freezing point depression?

4. Predict if it would be better to use coarse rock salt or fine table salt when makinghomemade ice cream. Explain.

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CHEMLAB 14

Safety Precautions

ProblemHow do factors affect the rate of solution formation?

Objectives• Observe the formation of

solutions• Hypothesize how stirring,

surface area, and temperature affect solution formation

• Infer the reasons that stirring, surface area, andtemperature affect solution formation

• Collect and analyze data

Materialscopper (II) sulfate

pentahydratedistilled water test tubes (6)25-mL graduated

cylinderglass stirring rod forceps test-tube rackmortar and pestlespatula clock

Investigate FactorsAffecting SolubilityBackground: The process of making a solution

involves the solvent coming in contact with thesolute particles. When you add a soluble compoundto water, several factors affect the rate of solutionformation.

Pre-Lab


2. How is forming an aqueous solution differentfrom a chemical reaction that takes place in aque-ous solution? Do you think that there are anysimilarities between the two processes? Explain.

3. What is the definition of temperature? When thetemperature of a solution goes up, what happensto the particles that make up the solution?

4. What is the surface area of a cube that is 8 cm ona side? Predict how the surface area will changeif the 8-cm-a-side cube is divided into 8 cubesthat are each 4 cm on a side. Calculate the newsurface area to verify your prediction.

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Procedure


2. Create a table to record your data.

3. Write a hypothesis that uses what you knowabout reaction rates to explain what you mightobserve during the procedure.

4. Place the six test tubes in the test-tube rack.

5. Place one crystal of copper (II) sulfate pentahy-drate in each of the first two test tubes.

6. For the remaining test tubes, use the mortar andpestle to crush a crystal. Use the spatula toscrape it into the third test tube.

7. Measure 15-mL of room-temperature distilledwater. Pour the water into the first test tube andrecord the time.

8. Observe the solution in the test tube just afteradding the water and after fifteen minutes.

9. Leave the first test tube undisturbed in the rack.

10. Repeat steps 7 and 8 for the third and fourthtest tubes.

11. Use the glass stirring rod to agitate the secondtest tube for 1-2 min.

12. Leave the third test tube undisturbed.

13. Agitate the fourth test tube with the glass stirring rod for 1-2 min.

14. Repeat steps 7 and 8 for the fifth test tube usingcold water. Leave the fifth test tube undis-turbed.

15. Repeat steps 7 and 8 for the sixth test tubeusing hot water. Leave the test tube undis-turbed.

16. Cleanup and Disposal Dispose of the remaining solids and solutions as directed byyour teacher. Wash and return all lab equipmentto its designated location.

CHEMLAB 14


1. Compare and Contrast What effect did you observe due to the agitation of the second and fourth testtubes verses the solutions in the first and third test tubes?

2. Observe and Infer What factors caused the more rapid solution formation in the fourth test tube incomparison to the second test tube?

3. Recognize Cause and Effect Why do you think the results for the third, fifth, and sixth test tubeswere different?

4. Discuss whether or not your data supported your hypothesis.

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5. Error Analysis Identify a major potential source of error for this lab and suggest an easymethod to correct it.

Inquiry Extension

Think Critically The observations made in this lab were macroscopic in nature. Propose asubmicroscopic explanation to account for these factors that affected the rate of solutionformation. At the molecular level, what is occuring to speed solution formation in each case?

CHEMLAB 14

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0 10 20 30 40 50 60 70 80 90 100

Solu

bili

ty (

g s

olu

te/1

00 g

H2O

) 90

100

80

70

60

50

40

30

20

10

0

Temperature (oC)

Solubilities as a Function of Temperature

NaClKClO3

KCl

CaCl2

Ce2(SO4)3

Solubility–Temperature GraphsSolubility–Temperature Graphs



42

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1. What variables are plotted on the graph?

2. What is the unit of each variable?

3. Use the graph to complete the table below.

4. At what temperature are sodium chloride and potassium chloride equally soluble

in water?

5. How does the solubility of cerium(III) sulfate differ from the solubility of potassium chlorate over the temperature range 0°C–100°C?

6. How many grams of sodium chloride will dissolve in 1.0 kg of water at 20°C?

7. Explain whether increasing temperature has a greater effect on the solubility of KCl oron the solubility of NaCl.

8. Explain how you might make a solution containing 42 g KCl dissolved in 100 g H2O at atemperature of 40°C. What term describes this type of solution?

Solubility–Temperature GraphsSolubility–Temperature Graphs



42

Substance Solubility at 10°C

Calcium chloride (CaCl2)

Cerium(III) sulfate (Ce2(SO4)3

Potassium chloride (KCl)

Potassium chlorate (KClO3)

Sodium chloride (NaCl)

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Phase Diagram of Solvent and SolutionPhase Diagram of Solvent and Solution



43

Pure solventSolution

LIQUIDSOLID

GAS

P

Tf Tb

Increasingtemperature

Freezing pointof solution

1 atm

Boiling pointof solution

Normal freezing point of water

Normal boiling point of water

Incr

easi

ng

p

ress

ure

� �

�

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1. What variables are plotted on the phase diagram?

2. What solvent is represented in the phase diagram?

3. What phases of the solvent are represented in the diagram?

4. What do the solid lines represent?

5. What is the term applied to a solution in which water is the solvent?

6. What do the dashed lines represent?

7. At each temperature, what does �P represent?

8. At any temperature, how does the vapor pressure of the aqueous solution compare withthe vapor pressure of the pure solvent?

9. Will a solution boil at the same temperature as the pure solvent under normal atmos-pheric pressure? Explain.

10. What must you do to the temperature of a solution to make it boil if it is at the boilingpoint of the pure solvent under normal atmospheric pressure?

11. How does the freezing point of a solution compare with the freezing point of the puresolvent at the same pressure?

Phase Diagram of Solvent and SolutionPhase Diagram of Solvent and Solution



43

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Percent by Mass

Percent by mass � � 100mass of solute

mass of solution

mass of solute � mass of solvent

Molality

Molality (m) �moles of solute

kilogram of solvent

Mole Fraction

XB �nB

nA � nB

mass of solvent (g)

molar mass of solvent (g/mol)

mass of solute (g)

molar mass of solute (g/mol)

1 kg

1000 gmass of solvent (g) �

mass of solute (g)

molar mass of solute (g/mol)

Calculating Percent by Mass,Mole Fraction, and Molalityfrom Mass Measurements




22

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1. Calculate the percent by mass, mole fraction, and molality of a solution that contains25.0 g of sodium chloride (NaCl) dissolved in 100.0 g of water.

Percent by Mass:

Mole Fraction:

Molality:





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Calculating the Boiling Point of a Solution

Tbsolution � Tbsolvent

� msoluteKbsolvent

Tbsolution � Tbsolvent

� msoluteKbsolvent

Tbsolution � 100°C � (2.00m)(0.515°C/m) � 100°C � 1.03°C

Tbsolution � 101.03°C

Find the boiling point of a 1.00m aqueous solution of KBr.

(Kbsolvent

� 0.515°C/m) Because a 1.00m aqueous solution of

potassium bromide contains 1 mol of K� ions and 1 mol of

Cl� ions per 1 kg of water, msolute � 2 � 1.00m, or 2.00m.

Boiling point of solution Boiling point elevation constantMolality of solute

�Tb: Boiling point elevationBoiling point of solvent

Calculating the Freezing Point of a Solution

Tfsolution � Tfsolvent

� msoluteKfsolvent

Tfsolution � Tfsolvent

� msoluteKfsolvent

Tfsolution � 0°C � (1.5m)(1.85°C/m) � 0°C � 2.78°C

Tfsolution � �2.78°C

Find the freezing point of a 1.50m aqueous solution of sucrose.

(Kfsolvent � 1.85°C/m) Because a 1.50m aqueous solution of sucrose

contains 1.50 mol of sucrose molecules per 1 kg of water, msolute � 1.50m.

Freezing point of solution Freezing point depression constant

Molality of solute particles

�Tf: Freezing point depressionFreezing point of solvent

Calculating Boiling Points andFreezing Points of SolutionsCalculating Boiling Points andFreezing Points of Solutions



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1. What is the boiling point of a 1.5m aqueous solution of fructose (C6H12O6)? The boilingpoint elevation constant of water is 0.515°C/m. Assume solute is a nonelectrolyte.

2. What is the freezing point of a solution of 1.5m aqueous solution of potassium bromide (KBr)?The freezing point depression constant of water is 1.86°C/m. Assume 100% dissociation.

3. Calculate the freezing point of a solution of a 0.975m aqueous solution of calcium nitrate(Ca(NO3)2). Assume 100% dissociation.

4. What is the boiling point of a solution of 20.0 g carbon tetrachloride (CCl4) dissolved in250.0 g of benzene (C6H6)? The boiling point of benzene is 80.10°C, and its boilingpoint elevation constant is 2.53°C/m. Assume solute is a nonelectrolyte.

5. Calculate the boiling point for a solution of 25.0 g calcium nitrate (Ca(NO3)2) dissolvedin 100.0 g of water. Assume 100% dissociation.

Calculating Boiling Points andFreezing Points of SolutionsCalculating Boiling Points andFreezing Points of Solutions



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Mixtures and SolutionsMixtures and Solutions


Section 14.1 Heterogeneous and Homogeneous MixturesIn your textbook, read about suspensions and colloids.


1. A solution is a mixture containing particles that settle out of the mixture ifleft undisturbed.

2. The most abundant substance in a colloid is the dispersion medium.

3. A colloid can be separated by filtration.

4. A solid emulsion consists of a liquid dispersed in a solid.

5. Whipped cream is an example of a foam.

6. In an aerosol, the dispersing medium is a liquid.

7. Brownian motion results from the collisions of particles of the dispersionmedium with the dispersed particles.

8. Dispersed particles in a colloid do not tend to settle out because they havepolar or charged atomic groups on their surfaces.

9. Stirring an electrolyte into a colloid stabilizes the colloid.

10. Colloids demonstrate the Tyndall effect.

The table below lists the characteristics of particles in colloids, solutions, and suspen-sions. Place a check in the column of each mixture whose particles have a particularcharacteristic.

Characteristics of Particles Colloid Solution Suspension

11. Less than 1 nm in diameter

12. Between 1 nm and 1000 nm in diameter

13. More than 1000 nm in diameter

14. Settle out if undisturbed

15. Pass through standard filter paper

16. Lower vapor pressure

17. Scatter light

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Study Guide Chemistry: Matter and Change • Chapter 14 45

Section 14.2 Solution ConcentrationIn your textbook, read about expressing concentration and using percent to describeconcentration.

Data related to aqueous solutions of sodium chloride (NaCl) and aqueous solutions ofethanol (C2H5OH) are provided in the table below. Use the table to answer the followingquestions. Circle the letter of the choice that best answers the question.

1. What is the percent by mass of NaCl in solution 1?

a. 0.030% b. 2.9% c. 3.0% d. 33%

2. Which of the following solutions is the most dilute?

a. Solution 1 b. Solution 2 c. Solution 3 d. Solution 4

3. What is the percent by volume of C2H5OH in Solution 5?

a. 0.2% b. 1.9% c. 2.0% d. 22%

4. Which of the following solutions is the most concentrated?

a. Solution 5 b. Solution 6 c. Solution 7 d. Solution 8

In your textbook, read about molarity and preparing molar solutions.

Read the following problem and then answer the questions.

An 85.0-mL aqueous solution contains 7.54 g iron(II) chloride (FeCl2). Calculate the molarityof the solution.

5. What is the mass of the solute?

6. What is the volume of the solution?

7. Write the equation that is used to calculate molarity.

8. In what unit must the amount of the solute be expressed to calculate molarity?

9. In what unit must the volume of the solution be expressed to calculate molarity?

10. Write the expression needed to convert the volume of the solution given in the problem

to the volume needed to calculate molarity.


Mass (g) Volume (mL)

Solution NaCl H2O Solution C2H5OH H2O

1 3.0 100.0 5 2.0 100.0

2 3.0 200.0 6 5.0 100.0

3 3.0 300.0 7 9.0 100.0

4 3.0 400.0 8 15.0 100.0

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11. What quantity must be used to convert the mass of the solute given in the problem to theamount of solute needed to calculate molarity?

12. Write the expression used to calculate the amount of solute.

13. Calculate the molarity of the solution. Show all your work.

In your textbook, read about molality and mole fractions.


14. How does molality differ from molarity?

15. Calculate the molality of a solution of 15.4 g sodium bromide (NaBr) dissolved in 125 gof water. Show all your work.

16. What is mole fraction?

17. Calculate the mole fraction of HCl in an aqueous solution that contains 33.6% HCl bymass. Show all your work.



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Section 14.3 Solvation and Solubility?In your textbook, read about the characteristics of solutions.

Use each of the terms below just once to complete the passage.

Air is a(n) (1) of oxygen gas dissolved in nitrogen

gas. The oxygen in air is the (2) , and nitrogen is the

(3) . Because oxygen gas dissolves in a solvent, oxygen gas

is a(n) (4) substance. A substance that does not dissolve is

(5) . (6) solutions are the most common

type of solutions. If one liquid is soluble in another liquid, such as acetic acid in water, the

two liquids are (7) . However, if one liquid is insoluble in another,

the liquids are (8) .

Read about solvation in aqueous solutions in your textbook.

The diagram shows the hydration of solid sodium chloride to form an aqueous solution.Use the diagram to answer the following questions.

9. Hydration is solvation in which the solvent is water. What is solvation?

Cl�

Cl�

Cl�

Cl� Cl�Na�

Na�

Na�

Na�

O HH

Na�

Cl�

Cl�

� �

�

immiscible liquid soluble solution

insoluble miscible solute solvent

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10. As sodium chloride dissolves in water, what happens to the sodium and chloride ions?

11. Explain the orientation of the water molecules around the sodium ions and chloride ions.

12. How does the strength of the attraction between water molecules and sodium and chloride ions compare with the strength of the attraction between the sodium ions and chloride ions? How do you know?

13. List three ways that the rate of solvation may be increased.

In your textbook, read about heat of solution, solubility, and factors that affect solubility.


14. The overall energy change that occurs when a solution forms is called theheat of solution.

15. Solubility is a measure of the minimum amount of solute that dissolves ina given amount of solvent at a specified temperature and pressure.

16. Solvation continues as long as the solvation rate is less than thecrystallization rate.

17. In a saturated solution, solvation and crystallization are in equilibrium.

18. Additional solute can be dissolved in an unsaturated solution.

19. The solubility of a gas dissolved in a liquid decreases as the temperatureof the solution increases.



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Section 14.4 Colligative Properties of SolutionsIn your textbook, read about electrolytes and colligative properties, vapor pressure low-ering, boiling point elevation, and freezing point depression.

Use the table to answer the following questions.

1. Which properties in the table are colligative properties?

2. What can you conclude about the relationship between colligative properties and thenumber of ions in solution from the 1.0m NaCl(aq) and 2.0m NaCl(aq) solutions?

3. What can you conclude about the relationship between colligative properties and the typeof ions in solution from the 1.0m HCl(aq) and 1.0m NaCl(aq) solutions?

Suppose that in a simple system, a semipermeable membrane is used to separate asucrose-water solution from its pure solvent, water. Match the descriptions of the systemin Column A with the terms in Column B.

Column A Column B

4. Cannot cross the semipermeable membrane

5. Can cross the semipermeable membrane

6. The side that exerts osmotic pressure

7. The diffusion of the solvent particles across thesemipermeable membrane from the area of higher solventconcentration to the area of lower solvent concentration

8. The barrier with tiny pores that allow some particles topass through but not others

9. The side from which more water molecules cross thesemipermeable membrane

10. A colligative property of solutions

Solution Density (g/L) Boiling Point (°C) Freezing Point (°C)

1.0m C2H5OH(aq) 1.05 100.5 �1.8

1.0m HCl(aq) 1.03 101.0 �3.7

1.0m NaCl(aq) 1.06 101.0 �3.7

2.0m NaCl(aq) 1.12 102.1 �7.4

a. osmotic pressure

b. water molecules

c. semipermeable membrane

d. sugar molecules

e. osmosis

f. solution side

g. pure solvent side

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Mixtures and SolutionsMixtures and Solutions


Column A Column B

1. The diffusion of solvent particles across a semipermeablemembrane from areas of lower solute concentration to areasof higher solute concentration

2. A mixture with particles that settle out if undisturbed

3. The erratic movement of colloid particles

4. The amount of additional pressure caused by watermolecules moving into a solution

5. A measure of how much solute is dissolved in a specificamount of solvent or solution

6. The overall energy change that occurs when a solution forms

7. A heterogeneous mixture of intermediate size particles

8. The process of surrounding solute particles with solventparticles to form a solution

9. The ratio of the number of moles of solute in solution to thetotal number of moles of solute and solvent

10. The scattering of light by dispersed colloid particles

11. The statement that the solubility of a gas in a liquid is directlyproportional to the pressure of the gas above the liquid

Describe each pair of related terms.

12. soluble, insoluble

13. miscible, immiscible

14. molarity, molality


a. Brownian motion

b. colloid

c. concentration

d. heat of solution

e. Henry’s law

f. mole fraction

g. osmosis

h. osmotic pressure

i. solvation

j. Tyndall effect

k. suspension

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In the space at the left, write true if the statement is true; if the statement is false,change the italicized word or phrase to make it true.

1. A solution may exist as a gas, a liquid, or a solid depending on thestate of the solvent.

2. Molar solutions are calculated and expressed in grams per liter.

3. The most common solvent among liquid solutions is ethanol.

4. Nonpolar solutes are more soluble in nonpolar solvents.

5. A supersaturated solution contains less dissolved solute than asaturated solution at the same temperature.

6. The lowering of the vapor pressure of a pure solvent when asolution is formed is a colligative property.

7. A 1M solution of a nonelectrolyte will have a lesser effect on thecolligative properties of its solution than a 1M solution of anelectrolyte will have on the colligative properties of its solution.

8. In an aerosol, the dispersing medium is a liquid.

9. A dilute solution contains a small amount of solute relative to thesolute’s solubility.

10. Attractions between the dispersed particles and the particles of thedispersing medium of a colloid produce magnetic layers that keepthe dispersed particles from settling out.

11. Boiling point depression is the temperature difference between asolution’s and a pure solvent’s boiling point.

Circle the letter of the response that best answers the question.

12. What term describes a solution in which the dissolved solute is in equilibrium with theundissolved solute?

a. dilute solution b. saturated solution c. supersaturated solution d. unsaturated solution

13. Which of the following statements explains the solubility of ionic substances in water?

a. The molar mass of water is 18.02 g/mol.

b. An oxygen atom has six electrons in its outermost energy level.

c. Water molecules are polar.

d. Water is a covalent substance.

14. Which of the following compounds provides the most solute particles when completelydissociated in water?

a. MgCl2 b. KBr c. NaCl d. Na3PO4


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1. Briefly describe the solvation of sodium chloride to form an aqueous solution.

2. How would you prepare each of the following solutions? Show your calculations.

a. 1.00 L of a 2.00M aqueous solution of sodium hydroxide (NaOH)

b. 90.0 mL of a 1.20M aqueous solution of sodium oxalate (Na2C2O4) from a 2.00M solution of Na2C2O4

3. What is the mole fraction of the solute in a 1.00m solution of barium chloride (BaCl2)?Show your calculations.


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Thinking CriticallyThe graph below shows the solubility versus temperature for two compounds, A and B.Use the graph to answer the questions below.

1. One of the curves represents carbon dioxide (CO2); the other represents tin(II) iodide(SnI2). Identify compounds A and B. Explain your reasoning.

2. A third substance, HgBr2, has a solubility of 0.50 g HgBr2/100.0 g H2O at 20°C. If thesolution is saturated at this temperature, calculate the molality of the solution.

3. Calculate the molarity of the HgBr2 solution. Assume the density of the solution is thesame as the density of the solvent.

Temperature (°C)

Solu

bili

ty (

g s

olu

te/1

00 g

H2O

)

10 200 30 40 50 60 70 80 90 100

1.0

2.0

0

3.0

4.0

5.0

A

B


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Applying Scientific MethodsA time-release capsule releases a drug at a constant rate so that the concentration of the drugin the body is not so high as to damage the body nor so low as to be ineffective. The diagrambelow shows such a capsule.

Notice that the capsule has two compartments sepa-rated by an impermeable, elastic membrane. One com-partment contains the drug, the other a saturated solutionof sodium chloride. The outer wall of the drug compart-ment consists of a rigid, selectively permeable material.The wall is designed to allow only molecules of the drugto pass outward through the wall. The outer wall of thecompartment containing the sodium chloride solutionconsists of a semipermeable membrane. A thin, protectivecoating that dissolves when the capsule enters the bodycovers the entire capsule.

The graph shows the concentration of the sodiumchloride solution in the capsule over time after the cap-sule has entered the body.

Use the diagram of the capsule and the graph to answer questions 1 and 2.

1. What happens to the concentration of the aqueous sodium chloride solution over time?

2. What process would account for your answer to question 1? Explain.


Saturated NaCl (aq)solution

Drug

Semipermeable membrane

Elastic, impermeable membrane

Rigid, selectivelypermeable membrane

Time

Co

nce

ntr

atio

n o

f N

aCl s

olu

tio

n

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3. The graph below shows the volume of the aqueous sodium chloride solution in the capsule after the capsule has entered the body. Explain whether the graph supports your answer to question 2.

4. Which of the following graphs represents the mass of the drug in the capsule after thecapsule has entered the body? Explain your choice.

5. What are two functions of the elastic, impermeable membrane separating the aqueoussodium chloride solution and the drug? How is each function related to a property of themembrane?

Time

Mas

s o

f d

rug

a

Time

Mas

s o

f d

rug

Time

Mas

s o

f d

rug

b c



Time

Vo

lum

e o

f N

aCl s

olu

tio

n

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CHAPTER 14

Assessment



1. 4. 7. 10.

2. 5. 8.

3. 6. 9.

Short Answer


11.

12.

13.

Extended Response


14.

15.


16. 18.

17. 19.

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Chapter 15 Energy and Chemical ChangeMiniLab . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

ChemLab . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59


Math Skills Transparency Masters and Worksheets . . . . . . . . . . . . . . . . . . . . . . . . . . 68

Study Guide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74



Table ofContents

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mini LAB 15Determine Specific Heat

How can you determine the specific heat of a metal? You can use a coffee-cupcalorimeter to determine the specific heat of a metal.

Materials distilled water, 250-mL beaker (2), hot plate, balance, metal cylinder, crucibletongs, graduated cylinder, polystyrene coffee cup, nonmercury thermometer


2. Make a table to record your data.

3. Pour approximately 150 mL of distilled water into a 250-mL beaker. Place the beaker ona hot plate set on high.

4. Use a balance to find the mass of a metal cylinder.

5. Using crucible tongs, carefully place the metal cylinder in the beaker on the hot plate.

6. Measure 90.0 mL of distilled water using a graduated cylinder.

7. Pour the water into a polystyrene coffee cup nested in a second 250-mL beaker.

8. Measure and record the temperature of the water using a nonmercury thermometer.

9. When the water on the hot plate begins to boil, measure and record the temperature asthe initial temperature of the metal.

10. Carefully add the hot metal to the cool water in the coffee cup with the crucible tongs.Do not touch the hot metal with your hands.

11. Stir, and measure the maximum temperature of the water after the metal was added.

Analysis

1. Calculate the heat gained by the water. The specific heat of H2O is 4.184 J/g·°C.Because the density of water is 1.0 g/mL, use the volume of water as the mass.

2. Calculate the specific heat of your metal. Assume that the heat absorbed by thewater equals the heat lost by the metal.

3. Compare this experimental value to the accepted value for your metal.

4. Describe major sources of error in this lab. What modifications could you make inthis experiment to reduce the error?

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CHEMLAB 15

Safety Precautions• Always wear safety goggles and a lab apron.• Tie back long hair.• Hot objects may not appear to be hot.• Do not heat broken, chipped, or cracked glassware.• Do not eat any items used in the lab.

ProblemHow many Calories are in a potato chip?

Objectives• Identify the reactants and

products in the reaction.• Measure mass and temper-

ature in order to calculatethe amount of heatreleased in the reaction.

• Propose changes in theprocedure and design ofthe equipment todecrease the percenterror.

Measure CaloriesThe burning of a potato chip releases heat stored in the

substances contained in the chip. Using calorimetry, youwill approximate the amount of energy contained in a potatochip.

Pre-Lab


2. Prepare all written materials that you will takeinto the laboratory. Be sure to include safety pre-cautions and procedure notes. Use the data tableon the next page.

3. Form a hypothesis about how the quantity of heatproduced by the combustion reaction will com-pare with the quantity of heat absorbed by thewater.

4. What formula will you use to calculate the quan-tity of heat absorbed by the water?

5. Assuming that the potato chip contains compoundsmade up of carbon and hydrogen, what gases willbe produced in the combustion reaction?

Materialslarge potato chipor other snack food250-mL beaker 100-mL graduated

cylinder evaporating dishnonmercury

thermometerring stand with ringwire gauzematches

stirring rodbalance

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Procedure


2. Measure the mass of a potato chip and record itin a data table.

3. Place the potato chip in an evaporating dish onthe metal base of the ring stand. Position the ringand wire gauze so that they will be 10 cm abovethe top of the potato chip.

4. Measure the mass of an empty 250-mL beakerand record it in the data table.

5. Using a graduated cylinder, measure 50 mL ofwater and pour it into the beaker. Measure themass of the beaker and water and record it inyour data table.

6. Measure and record the initial temperature of thewater.

7. Place the beaker on the wire gauze on the ring stand. Use a match to ignite the bottom ofthe potato chip.

8. Gently stir the water in the beaker while the chipburns. Measure and record the highest tempera-ture attained by the water.

9. Cleanup and Disposal Wash all lab equipmentand return it to its designated place.

CHEMLAB 15


1. Classify Is the reaction exothermic or endothermic? Explain how you know.

2. Observe and Infer Describe the reactant and products of the chemical reaction. Was the reactant(potato chip) completely consumed? What evidence supports your answer?

3. Calculate Determine the mass of the water and its temperature change. Use the equation q � c � m � �T to calculate how much heat, in joules, was transferred to the water in thebeaker by the burning of one chip.

Mass of beaker and 50 mL of water

Mass of empty beaker

Mass of water in beaker

Mass of potato chip

Highest temperature of water

Initial temperature of water

Change in temperature

Observations of the Burning of a Potato Chip

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4. Calculate Convert the quantity of heat in joules/chip to Calories/chip.

5. Calculate From the information on the chip’s container, determine the mass in grams ofone serving. Determine how many Calories are contained in one serving. Use your data tocalculate the number of Calories released by the combustion of one serving.

6. Error Analysis Compare your calculated Calories per serving with the value on thechip’s container. Calculate the percent error.

7. Compare your class results with other sutdents by posting your data at glencoe.com

Inquiry Extension

Predict Do all potato chips have the same number of calories? Make a plan to test severaldifferent brands of potato chips.

CHEMLAB 15

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Thermometer

Stirrer

Ignitionterminals

Insulation

Sealedreactionchambercontainingsubstanceand oxygen

Water

Using a CalorimeterUsing a Calorimeter



44

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1. The calorimeter shown on the transparency is used to measure the caloric content offoods. To do this, a sample of food is burned inside the reaction chamber of the calorime-ter. What is the system? What are the surroundings?

2. What besides food must be added to the chamber? Explain why.

3. What are the products of the reaction that takes place in the reaction chamber?

4. Why is the calorimeter insulated?

5. What does the thermometer measure?

6. Describe the movement of heat as the reaction takes place inside the chamber.

7. Assuming that no heat escapes from the calorimeter, what equation would you use todetermine the amount of heat released by the burning food in the reaction chamber?Define all variables in the equation.

8. Does the answer obtained from the equation in question 7 have a positive or negativevalue? Explain why. What is the sign of �H for the reaction?

Using a CalorimeterUsing a Calorimeter



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120

100 80 60 40 20

2010

020

0H

eat

Ad

ded

(kc

al)

Liq

uid

wat

er a

nd

wat

er v

apo

r

Liq

uid

wat

erIc

e an

dliq

uid

wat

er

Ice

Water vapor

740

Temperature (�C)

0 0

–20

–40

Temperature Changes of WaterTemperature Changes of Water



45

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1. The graph shows what happens when ice, at �40°C, is gradually heated to more than120°C. What is happening in the region between �40°C and 0°C?

2. At 0°C, the temperature does not change even though 80 kcal of heat is added. Why doesthe temperature not change?

3. What is happening in the region between 0°C and 100°C?

4. When the water reaches 100°C, the temperature does not change even though 540 kcal ofheat is added. Why does the temperature not change?

5. Compare the amount of heat needed to convert liquid water to water vapor with theamount needed to convert ice to liquid water. Explain the difference.

6. What is happening in the region above 100°C?

7. If you continue to add heat, what will happen to the water vapor?

Temperature Changes of WaterTemperature Changes of Water



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66 Chemistry: Matter and Change • Chapter 15 Teaching Transparency Master

H2(

g) �

1 O

2(g

)2

�H

� �

242

kJ

�H

� �

286

kJ

�H

� �

44 k

J

H2O

(g)

H2O

(l)

Changes in Enthalpy and EntropyChanges in Enthalpy and Entropy


Use with Chapter 15,Sections 15.4 and 15.5

46

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1. What do the arrows on the transparency represent?

2. Why do the arrows vary in direction?

3. What can you conclude about the transitions and the magnitudes of the enthalpies shownon the transparency?

4. How does Hess’s law apply to your answer to Question 3?

5. What do the ball models for liquid water and gaseous water on the transparency show?

6. What do the ball models indicate about the overall order of the molecules?

7. When molecules become more ordered or disordered, what happens to the entropy?

Changes in Enthalpy and EntropyChanges in Enthalpy and Entropy


Use with Chapter 15,Sections 15.4 and 15.5

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Determining the Heat Capacityof a CalorimeterDetermining the Heat Capacityof a Calorimeter



24

• Place 100.0 g of room temperature water in the calorimeter. Let thewater and calorimeter come to a constant temperature. Record thistemperature as: Ti cold water.

• Then add 100.0 g of hot water to the calorimeter. Record thetemperature of the hot water as: Ti hot water.

• Measure the temperature in the calorimeter every 15 seconds for 10 minutes.

• Plot temperature versus time. From the graph, determine the finaltemperature, Tf.

To determine the heat capacity of the calorimeter, use therelationship:

�Heathot water � �Heatcold water � Heatcalorimeter

�(masshot water � (specific heat of water) � (Tf � Ti hot water)) �

�masscold water � (specific heat of water) � (Tf � Ti cold water )

� Heat Capacity � (Tf � Ti cold water)

Example:

�(100.0 g � 4.184 J/g�°C � (29.8°C � 40.0°C)) �

(100.0 g � 4.184 J/g�°C � (29.8°C � 20.0°C)) � (Heat Capacity � (29.8°C � 20.0°C))

4267.68 J � 4100.32 J � (Heat Capacity � (9.8°C))

167.36 J / 9.8°C � Heat Capacity

17.1 J/°C � Heat Capacity

A calorimeter relies on the fact that the amount of heat lost by ahot body equals the amount of heat gained by a cold body. Theamount of heat absorbed or lost by the calorimeter apparatus isrelated to its heat capacity and the temperature change.

Mass Initial Temperature Final Temperature

Hot water 100.0 g 40.0°C 29.8°C

Cold water 100.0 g 20.0°C 29.8°C

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1. Why is it unnecessary to know the mass of the calorimeter apparatus to determine itsheat capacity?

2. You are using a calorimetry experiment to determine the amount of heat absorbed when asolid melts. How would the results be affected if the heat capacity of the calorimeter wasnot included in your calculations? Why?

3. What is the mathematical relationship for heat from the hot water?

4. What is the mathematical relationship for heat from the cold water?

5. Why does the calorimeter apparatus use a heat capacity rather than a specific heat?

6. Determine the heat capacity of a calorimeter using the following data.

Determining the Heat Capacityof a CalorimeterDetermining the Heat Capacityof a Calorimeter



24

Mass Initial Temperature Final Temperature

Hot water 150.0 g 45.0°C 36.8°C

Cold water 100.0 g 25.0°C 36.8°C

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Hess’s LawHess’s Law



25

How can you find the enthalpy of a specificreaction from a series of other reactions?

Here is a series of chemical reactions and their enthalpies.

H2(g) � Cl2(g) 0 2HCl(g) �H° � �185 kJ

2H2(g) � O2(g) 0 2H2O(g) �H° � �483.7 kJ

Calculate �H° for the reaction: 4HCl(g) � O2(g) 0 2Cl2(g) � 2H2O(g)

Reverse the reaction that contains HCl(g) to make HCl(g) a reactant.Change the sign of its enthalpy value because the reaction is reversed.

2HCl(g) 0 H2(g) � Cl2(g) �H° � 185 kJ

Multiply the reversed equation and enthalpy by 2 to obtain 4 mol HCl reacting.

2(2HCl(g) 0 H2(g) � Cl2(g) �H° � �185 kJ)

Reversing the equation and the sign of the enthalpy gives equation 3.

4HCl(g) 0 2H2(g) � 2Cl2(g) �H° � �370 kJ

Reaction 2 already has O2(g) as a reactant and the coefficients have thecorrect values. The following two equations and their enthalpies can beadded to obtain the desired equation.

4HCl(g) 0 2H2(g) � 2Cl2(g) �H° � �370 kJ

2H2(g) � O2(g) 0 2H2O(g) �H° � �483.7 kJ

4HCl(g) � 2H2(g) � O2(g) 0 2H2(g) � 2Cl2(g) � 2H2O(g) �H° � �370 kJ �(�483.7 kJ)

The term 2H2(g) appears on both sides of the equation so it can be cancelled. Follow the rules for significant digits for the final enthalpyvalue.

4HCl(g) � O2(g) 0 2Cl2(g) � 2H2O(g) �H° � �114 kJ

2

3

3

2

1

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1. In the example, what type of reactions are described in the starting equations?

2. What must be done to the formation reaction for hydrogen chloride to obtain the targetequation?

3. What must be done to the formation reaction for water to obtain the target equation?

4. How can the enthalpy for the target reaction be obtained?

5. Calculate �H° for the target reaction: 2Al(s) � Fe2O3(s) 0 2Fe(s) � Al2O3(s). Showyour work. Use the following reactions and enthalpies.

4Al(s) � 3O2(g) 0 2Al2O3(s) �H° � �3351.4 kJ

4Fe(s) � 3O2(g) 0 2Fe2O3(s) �H° � �1648.4 kJ

Hess’s LawHess’s Law



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Determining SpontaneityDetermining Spontaneity



26

Determine whether a chemical reaction isspontaneous or nonspontaneous.

From the relationship, �G � �H � T�S, it can be seen that when �H isnegative and �S is positive the reaction will always be spontaneous.When �H is positive and �S is negative, the reaction will always benonspontaneous.

What happens when both �H and �S are positive or when they are bothnegative? In these cases, spontaneity depends on the temperature.

The condensation of a gas identified as A has an enthalpy of �2.00 kJ/mol (�2000 J/mol) and an entropy of �12.0 J/mol�K. When will this condensation be spontaneous and when will it be nonspontaneous?

Use the relationship �G � �H � T�S to determine spontaneity. Find thetemperature at which equilibrium exists, that is, when �G � 0. At lower temperatures, the condensation is spontaneous. At highertemperatures, the condensation is nonspontaneous.

At equilibrium, �G � 0. Therefore, �H � T�S.

Solving for T gives: T �

Substituting the values for the condensation of A gives: T �

T � 167 K

To be spontaneous, �G must be negative, and to be nonspontaneous, �Gmust be positive.

To check this, perform test calculations.

�G � �2.00 kJ/mol � � (166 K � (�12.0 J/mol�K)) � �8 J/mol

(spontaneous below 167 K)

�G � �2.00 kJ/mol � � (168 K � (�12.0 J/mol�K)) � �16 J/mol

(nonspontaneous above 167 K)

1000 J1 kJ

1000 J1 kJ

(�2.00 kJ/mol)(1000 J)(�12.0 J/mol�K)(1 kJ)

�H��S

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1. The term T�S always has the same sign as �S. Explain why this is so.

2. A gas has a very large �H and a very small �S. Both values have negative signs. Whatcan you say about the temperature below which condensation will be spontaneous?

3. A chemical reaction has a �H of �135.5 kJ/mol and a �S of �148.9 J/ mol�K. Will thisreaction be spontaneous at a temperature of 750°C? Show your work.

4. Use the data given in question 3 to determine the temperature at which the reaction willbe in equilibrium. Give your answer in kelvins. Show your work.

Determining SpontaneityDetermining Spontaneity



26

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74 Chemistry: Matter and Change • Chapter 15 Study Guide74 Chemistry: Matter and Change • Chapter 15 Study Guide

Energy and Chemical ChangeEnergy and Chemical Change

Section 15.1 EnergyIn your textbook, read about the nature of energy.


1. Energy is the ability to do work or produce heat.

2. The law of conservation of energy states that energy can becreated and destroyed.

3. Chemical potential energy is energy stored in a substance becauseof its composition.

4. Heat is a form of energy that flows from a warmer object to acooler object.

5. A calorie is the amount of energy required to raise the temperatureof one gram of pure water by one degree Celsius.

6. A calorie is the SI unit of heat and energy.

7. The specific heat of a substance is the amount of heat required toraise the temperature of one gram of that substance by one degreeCelsius.

8. Kinetic energy is energy of motion.

9. Chemicals participating in a chemical reaction contain only potential energy.

10. One nutritional Calorie is equal to 100 calories.

11. One calorie equals 4.184 joules.

12. When a fuel is burned, some of its chemical potential energy islost as heat.

13. To convert kilojoules to joules, divide the number of kilojoules by1000 joules/1 kilojoule.

Answer the following question. Show all your work.

14. If the temperature of a 500.0-g sample of liquid water is raised 2.00°C, how much heat isabsorbed by the water? The specific heat of liquid water is 4.184 J/(g�°C).


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a. system

b. calorimeter

c. thermochemistry

d. universe

e. enthalpy

f. enthalpy (heat) ofreaction

g. surroundings


Section 15.2 Heat in Chemical Reactions and ProcessesIn your textbook, read about measuring heat and about chemical energy and the universe.

For each item in Column A, write the letter of the matching item in Column B.

Column A Column B

1. An insulated device used to measure the amount of heatabsorbed or released during a chemical or physicalprocess

2. The study of heat changes that accompany chemicalreactions and phase changes

3. The specific part of the universe that contains the reactionor process you wish to study

4. The change in enthalpy in a chemical reaction

5. A system plus its surroundings

6. The heat content of a system at constant pressure

7. Everything in the universe except the system beingstudied

Use the illustration to answer the following questions.

8. A scientist is studying the solution in the flask. What is the system?

9. What are the surroundings?

10. What is the universe?

Solution ofBa(OH)2

andNH4NO3

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Section 15.3 Thermochemical EquationsIn your textbook, read about writing thermochemical equations and about changes of state.

Use the following terms to complete the statements. Some terms will be used more than once.

1. A(n) is a balanced chemical equation thatincludes the physical states of all reactants and products and the energy changethat accompanies the reaction.

2. The enthalpy change for the complete burning of one mole of a substance is the

.

3. The is the heat required to vaporize one moleof a liquid.

4. The is the heat required to melt one mole of asolid substance.

5. Converting two moles of a liquid to a solid requires an amount of energy that is twice

the .

6. 2H2(g) � O2(g) 0 2H2O(g) �H � �572 kJ is a(n) .

7. The conversion of a gas to a liquid involves the

8. When a gas condenses to a liquid, heat is to thesurroundings.

9. Sweating makes you feel cooler because, as it evaporates, the water on your skin

heat from your body.

10. If you put an ice cube in a glass of soda pop, the heat absorbed by the ice will cause the

ice to melt, and the soda pop will become .

11. If it takes 100 joules to melt a piece of ice, must be absorbed by the ice.

12. In the equation H2O(s) 0 H2O(l) �H � 600 kJ, the positive value for �H means that

is absorbed in the reaction.


thermochemical equation enthalpy of combustion released

molar enthalpy of vaporization molar enthalpy of fusion absorbs

cool heat

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Section 15.4 Calculating Enthalpy ChangeIn your textbook, read about Hess’s law and standard enthalpy (heat) of formation.


1. Hess’s law states that if two or more thermochemical equationscan be added to produce a final equation for a reaction, then thesum of all the enthalpy changes for the individual reactions is theenthalpy change for the final reaction.

2. The standard enthalpy of formation is the change in enthalpy thataccompanies the formation of one gram of a compound in itsstandard state from its constituent elements in their standardstates.

3. The standard state of iron is solid.

4. For a pure gas, the standard state is the gas at a pressure of oneatmosphere.

5. The symbol used to represent standard enthalpy of formation is�Hf°.

6. The standard state of a substance is the normal state of thesubstance at 0 K and one atmosphere pressure.

7. The standard enthalpy of formation of a free element in itsstandard state is 0.0 kJ.

8. A standard enthalpy of formation that has a negative value meansthat energy is absorbed during the reaction.

9. The standard state of oxygen is gas.

10. Standard enthalpies of formation provide data for calculating theenthalpies of reactions under standard conditions using Hess’slaw.

11. The standard state of mercury is solid.


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Use the table to answer the following questions.

12. What does a formation equation show?

13. What does the negative sign on the value of an enthalpy of formation indicate?

14. Using the formation equations for CH4(g), CH3OH(g), and H2O(g), calculate �Hrxn forthe following equation. Show and explain all your work.

CH4(g) � H2O(g) 0 CH3OH(g) � H2(g)


Compound Formation Equation �Hf° (kJ/mol)

CH4(g) C(graphite) � 2H2(g) 0 CH4 (g) 75

CH3OH(g) C(graphite) � 2H2(g) � O2(g) 0 CH3OH(g) 239

H2O(g) O2(g) � H2(g) 0 H2O(g) 2421�2

1�2


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Section 15.5 Reaction SpontaneityIn your textbook, read about spontaneous processes and about entropy, the universe,and free energy.

Use each of the terms below to complete the statements.

1. A(n) is a physical or chemical change that occurs with no

outside intervention.

2. A measure of disorder or randomness of the particles that make up a system is called

.

3. The states that spontaneous processes always proceed in such

a way that the entropy of the universe increases.

4. is the energy that is available to do work.


5. A process cannot be spontaneous if it is exothermic and there is anincrease in disorder.

6. A process cannot be spontaneous if it is endothermic and there is adecrease in disorder.

7. A process cannot be spontaneous if it is exothermic and there is a decreasein disorder as long as the temperature remains low.

8. A process cannot be spontaneous if it is endothermic and there is anincrease in disorder as long as the temperature remains high.

9. A process can never be spontaneous if the entropy of the universeincreases.

10. When �G for a reaction is negative, the reaction is spontaneous.

11. When �G for a reaction is positive, the reaction is not spontaneous.

12. When �H for a reaction is negative, the reaction is never spontaneous.

13. When �H for a reaction is large and positive, the reaction is not expectedto be spontaneous.


spontaneous process entropy second law of thermodynamics free energy

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Energy and Chemical ChangeEnergy and Chemical Change


Column A Column B

1. The ability to do work or produce heat

2. States that energy cannot be created or destroyed

3. Energy flowing from a warmer to a cooler object

4. The study of heat changes from chemical reactions andphase changes

5. Energy stored in a substance because of its composition

6. Heat required to raise the temperature of one gram of asubstance by one degree Celsius

7. An insulated device measuring the heat absorbed orreleased during a chemical or physical process

8. The system plus the surroundings

9. States that spontaneous processes always proceed in sucha way that the entropy of the universe increases

10. Everything in the universe other than the system

11. The change in enthalpy in a chemical reaction

12. A balanced chemical equation that includes the physicalstates of all reactants and products and the energy changethat accompanies the reaction

13. A system’s heat content at constant pressure

14. Energy required to vaporize one mole of a liquid

15. Enthalpy change occurring when one mole of a compoundin its standard state forms from its constituent elements intheir standard states

16. Energy required to melt one mole of a solid

17. A physical or chemical change without outside intervention

18. The enthalpy change for the complete burning of onemole of a substance

19. Energy that is available to do work

20. The SI unit of heat and energy


a. calorimeter

b. standard enthalpy(heat) of formation

c. chemical potentialenergy

d. heat

e. law of conservation ofenergy

f. specific heat

g. free energy

h. energy

i. thermochemicalequation

j. enthalpy (heat) ofreaction

k. molar enthalpy (heat)of fusion

l. molar enthalpy (heat)of vaporization

m. second law ofthermodynamics

n. spontaneous process

o. joule

p. thermochemistry

q. enthalpy (heat) ofcombustion

r. universe

s. enthalpy

t. surroundings

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1. A negative sign for �G indicates that, at constant temperature andpressure, the reaction is spontaneous.

2. For a given substance, the entropy always increases in the followingorder: gas 0 liquid 0 solid.

3. For the reaction NH4Cl(s) 0 NH3(g) � HCl(g), the entropy change isnegative.

4. Hess’s law states that if two or more thermochemical equations can beadded to produce a final equation for a reaction, then the sum of all theenthalpy changes for the individual reactions is the enthalpy change forthe final reaction.

Use the illustration of three systems to answer the following questions.

5. How do the bottles in the three systems differ?

6. What do the arrows in the illustration indicate?

7. Which of the systems will show the greatest change in enthalpy and entropy as time pro-gresses? Which will show the least change? Explain your answers.

8. What are the surroundings in the illustration?


Water

Heat

A

Water vapor

Water

C

Seal

Water

B

Heat

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1. Consider the following reaction under standard conditions.

AgCl(s) 0 Ag�(aq) � Cl�(aq) �G � �54.1 kJ

Will this reaction occur spontaneously? Explain.

2. Which of the following species has the highest entropy at 25°C? Explain your answer.

a. CH3OH(l) b. CO(g) c. MgCO3(s) d. H2O(l) e. Ni(s)

3. Predict the change in entropy (�S) for the following reactions. Explain your answer.

a. H2O(g) 0 H2O(l) b. 2NO(g) 0 N2(g) � O2(g) c. MgCO3(s) 0 MgO(s) � CO2(g)

Use the illustration to answer the following questions.

4. Determine if each process shown in the illustration isendothermic or exothermic. Explain.

5. What happens to the entropy of each system? Explainyour answer.


Heat

500kJ

Heat

Iron, 1.0 kg, 0°C

Ice, 2.0 kg, 0°C Ice, 0.5 kg, 0°C

1100°C

500kJ

A

B

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Entropy ofvaporization

Entropy offusion

Temperature (T )

Entr

op

y (S

)

Meltingpoint

Boilingpoint

Gas

Liquid

Solid

Thinking CriticallyUse the graph to answer the following questions.

1. What does the graph show?

2. Why does the entropy increase dramatically at the melting and boiling points, but the temperaturedoes not increase at all?

Use the nutrition label to answer the following questions.

3. How much energy is contained in the six-cookie serving size recommended on the label?

4. How was the number of Calories contained in the cookies determined?

5. How much energy in joules is provided by eating six cookies? (1 cal � 4.184 J)


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Applying Scientific MethodsRachel and Ravi are lab partners in Ms. Santo’s chemistry class. After completing a calorime-try experiment to determine the specific heat of lead, they discovered that the results had apercent error of 17.3%. They thought that the large percent error wasn’t a good indication oftheir lab technique. They knew that for each of the three trials they had carefully measuredthe masses and the temperatures of the dry, lead sinker and the water in the calorimeter beforeand after they submerged the hot sinker. Perhaps the error wasn’t a result of their techniquebut of other factors, including the design of the experiment, so they decided to investigate further.

They recalled that the equation they used to determine c(lead) stemmed from the equationshowing that the quantity of heat lost by the lead sinker equals the quantity of heat gained bythe water.

qlead � qwater � 0

qlead � �qwater

Ravi grabbed a pencil and drew a quick sketch of the process represented by the equation.

heat transfer from hot sinker to cool water

System (before) System (after)

After a few seconds of thought, he said, “We’re forgetting that the calorimeter absorbssome heat,” and he wrote,

qlead � �(qwater � qcalorimeter)

“Sure!” agreed Rachel. “Here’s what the process should look like.”

1. Sketch the process represented by the equation that Ravi wrote.

System (before) System (after)

The lab partners decided to design an experiment to determine the effect of the calorime-ter on calorimetry experiments. They determined the amount of heat lost by the sinker frommass and temperature measurements and used the accepted value of the specific heat of lead(c(lead) � 0.129 J/(g�°C)). They also determine the heat gained by the water from mass andtemperature measurements and used the accepted value of the specific heat of water (c(lwater) � 4.18 J/(g�°C)).


hot sinker � cool water warm sinker and warm water

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The table shows data taken from a typical trial of the experiment.

2. From the data, determine the value of qcalorimeter. Interpret its meaning.

Ravi assumes that qcalorimeter for their calorimeter doesn’t change, that is, qcalorimeter � constant.

However, Rachel thinks that qcalorimeter might vary with the temperature change of the water,�Twater. She thinks that for their calorimeter, qcalorimeter � constant�Twater.

3. From the results of question 2, what are the value and units of Ravi’s constant and ofRachel’s constant?

4. Explain if you think Ravi or Rachel is correct.



Initial temperature of sinker (°C) 160.0

Initial temperature of water (°C) 21.3

Final temperature of sinker and water (°C) 31.4

Mass of sinker (g) 142.81

Mass of water (g) 50.0

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1. 4. 7.

2. 5. 8.

3. 6. 9.

Short Answer


10.

11.

12.

Extended Response


13.

14.


15. 17.

16. 18.

CHAPTER 15

Assessment

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Chapter 16 Reaction RatesMiniLab . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

ChemLab . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89


Math Skills Transparency Masters and Worksheets . . . . . . . . . . . . . . . . . . . . . . . . . . 98

Study Guide . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

Chapter Assessment. . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

STP Recording Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

Table ofContents

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mini LAB 16Examine Reaction Rate and Temperature

What is the effect of temperature on a common chemical reaction?

Materials: nonmercury thermometer, hot plate, 250-mL beaker, balance, water, effervescent tablet, stopwatch or clock with second hand


2. Break a single effervescent tablet into four equal pieces.

3. Use a balance to measure the mass of one piece of the tablet. Measure 50 mL of room-temperature water (approximately 20°C) into a 250-mL beaker. Use a nonmercurythermometer to measure the temperature of the water.

4. With a stopwatch or a clock with seond hand ready, add the piece of tablet to the water.Record the amount of time elapsed between when the tablet hits the water and when allof the solid has dissolved.

5. Repeat Steps 3 and 4, this time gradually warming the 50 mL of water to about 50°C on ahot plate. Maintain the temperature (equilibrate) throughout the run.

Analysis

1. Identify the initial mass the final mass, and t1 and t2 for each run.

2. Calculate the reaction rate by finding the mass of reactant consumed per second for each run.

3. Describe the relationship between reaction rate and temperature for this reaction.

4. Predict what the reaction rate would be if the reaction rate were carried out at 40°Cand explain the basis for your prediction. To test your prediction, repeat the reactionat 40°C using another piece of tablet.

5. Evaluate how well your prediction for the reaction rate at 40°C compares to themeasured reaction rate.

Mass of Reaction Reaction Temperature Tablet (g) time (s) rate (g/s) (°C)

Data Table

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CHEMLAB 16

Safety Precautions• Always wear safety goggles and a lab apron.• Never pipette any chemical by mouth.• Hydrochloric acid is corrosive. Avoid contact with skin and eyes.

ProblemHow does the concentra-tion of a reactant affectthe reaction rate?

Objectives• Sequence the acid con-

centrations from the mostto the least concentrated.

• Observe which concentra-tion results in the fastestreaction rate.

Materials10-mL graduated

pipette safety pipette filler6M hydrochloric

aciddistilled water25 mm � 150 mm

test tubes (4)test-tube rack

magnesium ribbonemery cloth or fine

sandpaperscissorsplastic rulertongswatch with second

hand or stop-watch

stirring rod

Observe How ConcentrationAffects Reaction RateThe collision theory describes how the change in concentration of

one reactant affects the rate of chemical reactions.

Pre-Lab

1. Read the entire CHEMLAB. Prepare all writtenmaterials that you will take into the laboratory.Be sure to include safety precautions and proce-dure notes. Use the data table on the next page.

2. Use emery paper or sandpaper to polish the mag-nesium ribbon until it is shiny. Use scissors to cutthe magnesium into four 1-cm pieces.

3. Place the four test tubes in the test-tube rack.Label the test tubes #1 (6.0M HCl), #2 (3M HCl), #3 (1.5M HCl), and #4 (0.75M HCl).

4. Form a hypothesis about how the chemical reac-tion rate is related to reactant concentration.

5. What reactant quantity is held constant? What arethe independent and dependent variables?

6. What gas is produced in the reaction betweenmagnesium and hydrochloric acid? Write the balanced formula equation for the reaction.

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7. Why is it important to clean the magnesium rib-bon? If one of the four pieces is not thoroughlypolished, how will the rate of the reaction involv-ing that piece be affected?

Procedure


2. Use a safety pipette filler to draw 10 mL of 6.0Mhydrochloric acid (HCl) into a 10-mL graduatedpipette.

3. Dispense the 10 mL of 6.0M HCl into Test Tube #1.

4. Draw 5.0 mL of the 6.0M HCl from Test Tube #1with the empty pipette. Dispense this acid intoTest Tube #2 and use the pipette to add an addi-tional 5.0 mL of distilled water to the acid. Usethe stirring rod to mix thoroughly. This solution is 3.0M HCl.

5. Draw 5.0 mL of the 3.0M HCl from Test Tube #2with the empty pipette. Dispense this acid intoTest Tube #3 and use the pipette to add an addi-tional 5.0 mL of distilled water to the acid. Usethe stirring rod to mix thoroughly. This solution is 1.5M HCl.

6. Draw 5.0 mL of the 1.5M HCl from Test Tube #3with the empty pipette. Dispense this acid intoTest Tube #4 and use the pipette to add an addi-tional 5.0 mL of distilled water to the acid. Usethe stirring rod to mix thoroughly. This solutionis 0.75M HCl.

7. Draw 5.0 mL of the 0.75M HCl from Test Tube#4 with the empty pipette. Neutralize and discardit in the sink.

8. Using the tongs, place a 1-cm length of magne-sium ribbon into Test Tube #1. Record the time inseconds that it takes for the bubbling to stop.

9. Repeat step 8 using the remaining three TestTubes of HCl and the three remaining pieces ofmagnesium ribbon. Record in the data table thetime (in seconds) it takes for the bubbling to stop.

9. Cleanup and Disposal Place acid solutions inan acid discard container. Thoroughly wash alltest tubes and lab equipment. Discard other mate-rials as directed by your teacher. Return all labequipment to its proper place.

CHEMLAB 16

Test tube [HCl] (M) Time (s)

1

2

3

4

Reaction Time Data

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1. Make a Graph Plot the concentration of the acid on the x-axis and the reaction time on the y-axis. Draw a smooth curve through the data points.

2. Conclude Based on your graph, what is the relationship between the acid concentration and the reaction rate?

3. Hypothesis Write a hypothesis using collision theory, reaction rate, and reactantconcentration to explain your results.

4. Error Analysis Compare your experimental results with those of several other students inthe laboratory. Explain the differences.

Inquiry Extension

Design an Experiment Based on your observations and results, would temperature variationsaffect reaction rates. Plan an experiment to test your hypothesis.

CHEMLAB 16

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Factors That Affect Reaction RateFactors That Affect Reaction Rate



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1. Look at the sequence of pictures. What happened to the apple over time?

2. No chemical was added to the apple. Explain why the apple changed.

3. How could you test your answer to Question 3?

4. What effect would increasing the amount of air surrounding the apple have on the apple?Explain your answer.

5. How would slicing the apple into more pieces affect the apple? Explain your answer.

6. The apple in the pictures is raw. A cooked apple would not change the same way. Give apossible reason why.

Factors That Affect Reaction RateFactors That Affect Reaction Rate



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Reaction OrderReaction Order



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1. What is formed when the two solutions mix?

2. Explain how the substance is formed in terms of the particles in the two solutions.

3. On the basis of the image in the transparency, what can you conclude about the rate ofthe reaction? Explain your answer.

4. On the basis of your answer to Question 3, what can you conclude about the activationenergy for the reaction?

5. What can you conclude about the reaction order for the reaction?

6. If the rate equation is determined to be k[Pb2�][I�], what is the reaction order?

7. Would you classify the reaction as exothermic, endothermic, nonspontaneous, or sponta-neous. Explain your answer.

Reaction OrderReaction Order



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1.00

0.80

0.60

[H2O

2] (

mo

l/L)

�[H

2O2]

0.40

0.20

00 1 2 3 4 5 6 7 8 9 10

Relative time

Change in [H2O2] with Time

�t

Reaction RateReaction Rate



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1. What does the graph show?

2. Is the hydrogen peroxide a reactant or product in the reaction? How do you know?

3. The hydrogen peroxide is part of a decomposition reaction. Write a balanced equation forthe reaction.

4. From the graph, what can you conclude about the rate of this reaction?

5. Define the term instantaneous rate.

6. How can you use the graph to find the instantaneous rate of the reaction that you identi-fied in your answer to Question 3? Give your answer in mathematical terms.

7. To show the complete relationship of reactants to products in this reaction, what elsewould you need to plot on the graph?

8. What would you expect your answer to Question 7 to show?

Reaction RateReaction Rate



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Determining Reaction Orders Determining Reaction Orders



27

Trial Initial [A] (M) Initial [B] (M) Initial Rate (mol/L�s)

1 0.30 0.10 1.20 � 10�2

2 0.60 0.10 4.80 � 10�2

3 0.60 0.20 9.60 � 10�2

Step 1 Write the rate equations for trials 1

and 2.

Step 2 Compare Rate 2 with Rate 1. (Hint:

Use the initial rate data.)

Step 3 Substitute the rate equations in

step 1 into the equation in step 2.

Step 4 Compare the concentrations of each

reactant. (Hint: Use the concentration

data.)

Step 5 Substitute the equations from step 4

into the equation of step 3 and

simplify.

Rate 1 � k[A1]m[B1]n

Rate 2 � k[A2]m[B2]n

Rate 2 � 4 Rate 1

k[A2]m[B2]n � 4k[A1]m[B1]n

[A2] � 2[A1] [B2] � [B1]

k(2[A1])m[B1]n � 4k[A1]m[B1]n

k(2)m[A1]m[B1]n � 4k[A1]m[B1]n

(2)m � 4

m � 2

What is the order of the reaction for reactant A?

Assume that the general rate law for this type of reaction is

Rate � k[A]m[B]n

Determine m by comparing trials 1 and 2.

Experimental Initial Rates for aA � bB 0 products

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Use the following data to determine the order of the reaction for each reactant.

Assume that the general rate law for this type of reaction is Rate � k[A]m[B]n

1. Determine n by comparing trials 1 and 2. Follow Steps 1–5 shown on the transparency.

Step 1.

Step 2.

Step 3.

Step 4.

Step 5.

2. Determine m by comparing trials 2 and 3. Follow Steps 1–5 shown on the transparency.

Step 1.

Step 2.

Step 3.

Step 4.

Step 5.

Determining Reaction Orders Determining Reaction Orders



27

Experimental Initial Rates for aA � bB 0 products

Trial Initial [A] (M) Initial [B] (M) Initial Rate (mol/L�s)

1 0.10 0.30 7.20 � 10�3

2 0.10 0.60 1.44 � 10�2

3 0.20 0.90 8.64 � 10�2

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Reaction RatesReaction Rates

Section 16.1 A Model for Reaction RatesIn your textbook, read about expressing reaction rates and explaining reactions andtheir rates.

Use each of the terms below just once to complete the passage.

According to the (1) , atoms, ions, and molecules must collide in

order to react. Once formed, the (2) is a temporary, unstable

arrangement of atoms that may then form products or may break apart to reform the reactants.

Every chemical reaction requires energy, and the minimum amount of energy that reacting

particles must have to form the activated complex is the (3) . In a

chemical reaction, the (4) is the change in concentration of a reactant

or product per unit time. It may be expressed using the units of (5) .

Use the energy diagram for the rearrangement reaction of methyl isonitrile to acetoni-trile to answer the following questions.

6. What kind of reaction is represented by this diagram,endothermic or exothermic?

7. What is the chemical structure identified at the top ofthe curve on the diagram?

8. What does the symbol Ea represent?

9. What does the symbol �E represent?


collision theory activated complex mol/(L�s)

activation energy reaction rate

Reaction Progress

H3C�N�C

H3C�C�N

a

Ener

gy

�E

Ea

H3C...C

N

�

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For each item in Column A, write the letter of the matching item in Column B.

Column A Column B

10. Expresses the average rate of loss of a reactant a. average reaction rate

11. Expressed as �quantity/�time b. positive number

12. Expresses the average rate of formation of a product c. negative number

Use the figure below to answer the following questions.

13. What molecules collided in collisions A, B, and C?

14. What do the arrows represent?

15. Which collision(s) formed products? What were the products?

16. Explain why the other collision(s) did not form products.

17. Which collision(s) formed an activated complex? Identify the activated complex.

Incorrect orientationRebound

Correct orientationCollisionCollision Activated complex Products

Correct orientationInsufficient energy

ReboundCollision

�A B

C

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Section 16.2 Factors Affecting Reaction RatesIn your textbook, read about the factors that affect reaction rates (reactivity, concentra-tion, surface, area, temperature, and catalysts).

In the space at the left, write true if the statement is true; if the statement is false,change the italicized word to make it true.

1. Decreasing the concentration of reactants increases the collisionfrequency between reacting particles.

2. A heterogeneous catalyst exists in a different physical state thanthe reaction it catalyzes.

3. Increasing the concentration of a substance increases the kineticenergy of the particles that make up the substance.

4. Catalysts increase the rates of chemical reactions by raising theactivation energy of the reactions.

5. Increasing the surface area of a reactant increases the rate of thereaction.

6. Raising the temperature of a reaction increases the rate of thereaction by increasing the energy of the collisions betweenreacting particles.


7. A chemist heated a sample of steel wool in a burner flame exposed to oxygen in the air.He also heated a sample of steel wool in a container of nearly 100% oxygen. The steel-wool sample in the container reacted faster than the other sample. Explain why.

8. Would the chemist have observed the same results if he used a block of steel instead ofsteel wool? Explain your answer.

9. How would the reaction have differed if the steel wool was not heated?


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Section 16.3 Reaction Rate LawsIn your textbook, read about reaction rate laws and determining reaction order.

Use each of the terms below to complete the statements.

Equation 1 aA + bB 0 cC + dD

Equation 2 � � k[A]m[B]n

1. Equation 1 describes a .

2. Equation 2 expresses the mathematical relationship between the rate of a chemical

reaction and the concentrations of the reactants. This is known as the

.

3. The variable k in equation 2 is the , a numerical value that relates

the reaction rate and the concentration at a given temperature.

4. The variables m and n are the . These define how the rate is affected

by the concentrations of the reactants.

5. The square brackets [ ] represent .

6. The variable t represents .

Answer the questions about the following rate law.

Rate � k [A]1[B]2

7. What is the reaction order with respect to A?

8. What is the reaction order with respect to B?

9. What is the overall reaction order for the rate law?

10. Doubling the concentration of A will cause the rate to double. What would happen if you dou-

bled the concentration of B?

11. A reaction rate can be expressed as Rate 5 k[A]2. What is the reaction order for this reaction?

�[A]�

�t

chemical reaction rate law specific rate constant

reaction orders concentration time


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Section 16.4 Instantaneous Reaction Rates and ReactionMechanismsIn your textbook, read about instantaneous reaction rates.

Circle the letter of the choice that best completes the statement.

1. is determined by finding the slope of the straight line tangent to the curve of aplot of the change in concentration of a reactant versus time.

a. Instantaneous rate c. Reaction mechanism

b. Change in temperature d. Reaction order

2. A(n) consists of two or more elementary steps.

a. complex reaction c. reaction mechanism

b. elementary step d. reaction order

3. A(n) is a substance produced in an elementary step and consumed in anotherelementary step.

a. instantaneous rate c. reaction mechanism

b. intermediate d. rate-determining step

4. A(n) is the complete sequence of elementary reactions that make up a complexreaction.

a. instantaneous rate c. reaction mechanism

b. elementary step d. reaction order

5. The is the slowest of the elementary steps in a complex reaction.

a. instantaneous rate c. rate-determining step

b. intermediate d. reaction order

6. The can be used to determine the instantaneous rate for a chemical reaction.

a. rate-determining step c. products

b. intermediates d. rate law

7. An element or compound that reacts in one step of a complex reaction and reforms in aanother step of the complex reaction is

a. an intermediate.

b. a catalyst.

c. not part of the reaction mechanism.

d. shown in the net chemical equation for the reaction.


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8. To determine the instantaneous rate, you must know the specificrate constant, the concentrations of the reactants, and the reactionorders for the reaction.

9. A reaction rate that is defined as k[A][B] and that has a specificrate constant of 1.0 � 101 L/(mol�s), [A] � 0.1M, and [B] � 0.1M would have an instantaneous rate of 0.01 mol/(L�s).

In your textbook, read about reaction mechanisms.

Answer the following questions about the proposed reaction mechanism for the complexreaction below.

2NO(g) + 2H2(g) 0 N2(g) + 2H2O(g)

Proposed Mechanism 2NO 0 N2O2 (fast)

N2O2 + H2 0 N2O + H2O (slow)

N2O + H2 0 N2 + H2O (fast)

10. How many elementary steps make up the complex reaction?

11. What is the rate-determining step for this reaction?

12. What are N2O2 and N2O in the reaction?

13. Is there a catalyst involved in the reaction? Explain your answer.

14. What can you conclude about the activation energy for the rate-determining step?

15. If you wanted to increase the rate of the overall reaction, what would you do?



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Reaction RatesReaction Rates


Column A Column B

1. States that atoms, ions, or molecules must collide in order to react

2. Minimum amount of energy required to form the activated complex

3. A substance that slows down reaction rates

4. Substance that increases the rate of a chemical reaction withoutbeing used up

5. Mathematical relationship between the rate of a chemical reactionand the concentrations of the reactants

6. Numerical value that relates reaction rate and concentration at agiven temperature

Use the following terms to complete the statements.

7. The is a specific reaction rate determined by finding theslope of the straight line tangent to the curve of a plot of the change in concentration of areactant versus time.

8. A(n) consists of two or more elementary steps.

9. A substance produced in one elementary step and consumed in a subsequent elementary

step is a(n) .

10. The for a reactant defines how the reaction rate is affectedby the concentration of that reactant.

11. A(n) is the complete sequence of elementary steps thatmake up a complex reaction.

12. The is the slowest of the elementary steps in a complexreaction.


instantaneous rate intermediate rate-determining step

complex reaction reaction mechanism reaction order

a. activationenergy

b. inhibitor

c. collision theory

d. rate law

e. specific rateconstant

f. catalyst

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1. The rate of a chemical reaction is unrelated to the spontaneity ofthe reaction.

2. In order to react, the particles participating in a chemical reactionmust collide in the correct orientation.

3. Reaction rates are generally calculated and expressed in grams perliter per second.

4. Factors that influence the rate of chemical reactions includecatalysts, temperature, and the reactivity, concentration, andsurface area of the reactants.

5. Catalysts decrease the rates of chemical reactions by lowering theactivation energies of the reactions.

6. Raising the temperature of a reaction increases the rate of thereaction by increasing the collision frequency and the energy ofthe collisions forming the activated complex.

7. The reaction order for a chemical reaction is determinedexperimentally using the method of initial rates.

8. The instantaneous rate for a chemical reaction is calculated fromthe rate law, the specific rate constant, and the concentrations ofall reactants.

9. A complex reaction’s rate-determining step determines theinstantaneous rate of the overall reaction.

10. The formula to determine the rate of consumption of a reactantwill include a negative sign.

11. Increasing the temperature by 10 K can decrease the reaction rateby a factor of 2.

12. The most effective catalyst for automobile catalytic converters is ahomogeneous catalyst.

13. In the expression Rate � k[A], the reaction rate quadruples whenthe concentration of A doubles.

14. The spontaneity of a chemical reaction is related to the activationenergy for the reaction.


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Identify the substances at positions 1 through 4 of the energy level diagram. Choosefrom the substances listed below.

1.

2.

3.

4.


5. The reaction represented by the following balanced equation is catalyzed by a certainchemical species.

S2O82� � 2I� 0 2SO4

2� � I2

What is the catalyst in its reaction mechanism?

Step 1 Fe3� � 2I� 0 Fe2� � I2

Step 2 S2O82� � Fe2� 0 2SO4

2� � Fe3�

6. In the reaction 2N2O 0 2N2 � O2, describe how much oxygen gas is formed in relationto the amount of nitrogen gas.

7. A reaction is carried out with water as a solvent. How would additional water affect thereaction rate?

8. In the reaction below, why does the equation for the rate of reaction for CO expresseduse a negative sign, but the equation for the rate of reaction for NO does not?

CO � NO2 0 CO2 � NO


activated complex products reactants intermediates

Reaction progress

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1

2

3

4

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Thinking CriticallyAnswer the following questions.

1. For each situation below, identify the reaction that will happen faster and explain why.

a. You have 50 pounds each of salt in block form and salt in granular form. You want to dissolve the salt in water.

b. You have two 50-pound bags of salt. You also have a tub of water at 20˚C and a tub of water at 70˚C. You want to dissolve a bag of salt in each tub of water.

c. You have two 50-pound bags of salt. You also have a tub of water at 20˚C and a pail of water at 20˚C. You want to dissolve a bag of salt in each container of water.

2. For the overall chemical equation 2H2S(g) + O2(g) 0 2S(s) + 2H2O(l), what can youconclude about the rate law?

3. For the chemical reaction system described by the diagram, what can you conclude aboutthe magnitude of the activation energy with respect to reaction direction?

Reaction progress

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Name Date Class


4. Using the graph of a hypothetical reaction A 0 B, answer the following questions aboutthe reaction.

a. How long did it take for the reaction to produce equal amounts of A and B?

b. What is the average rate of loss of A over the time of the test?

c. What is the average rate of gain of B over the same period?

d. How does the rate of loss of A compare to the rate of gain of B?

5. For an exothermic reaction, the rate law at 298 K is Rate = k[H2][I2]. Predict the effect ofeach of the following changes on the initial rate of the reaction and give the reason foryour prediction.

a. increasing the volume of the reaction vessel at constant temperature

b. adding a catalyst

c. increasing the temperature

Time (min)

Nu

mb

er o

f m

ole

s1

0.8

0.6

0.4

0.2

00 10 20 30 40 50 60

1.00

0.74

0.540.46

0.26

0.40

0.30 0.22

0.84

0.780.70

0.60

0.16

A

B


Thinking Critically, continued

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Applying Scientific MethodsConsider the following laboratory situations and answer the questions.

1. Copper metal dissolves in hydrochloric acid according to the following reaction.

Cu(s) � 2HCl(aq) 0 CuCl2(aq) � H2(g)

Explain how you would determine the rate law for the production of hydrogen gas.

2. If you have two reactants that will combine to form products, what is the least number oftest experiments you must perform to determine the reaction rate order for this experiment?Explain your answer.


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CHAPTER 16

Assessment



1. 4. 7. 10.

2. 5. 8.

3. 6. 9.

Short Answer


11.

12.

Extended Response


13.

14.

15.


16. 18.

17. 19.


Chemistry: Matter and Change Teacher Guide and Answers 113Fast Files, Chapters 5-8 Resources

TEACHER GUIDE AND ANSWERS

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CHAPTER 13

MiniLab 13 – Model a FireExtinguisher

Analysis

1. Using V2= P1V1T2/(T1P2) at 298 K and 98.6kPa, the molar volume is 25.1 L.

2. CO2 44.0 g/25.1 L = 1.75 g/L; O2 32.0 g/25.1 L = 1.27 g/L; N2 28.0 g/25.1 L = 1.12 g/L

3. Yes, the heavier CO2 gas moves down thecylinder, displacing the air and extinguishingthe candle flame.

ChemLab 13 – Determine Pressure inPopcorn Kernels

Pre-Lab

2. To correctly read the volume, take a levelsighting of the bottom of the meniscus. Whenyou record the volume, write down all the digitsyou are sure of, plus one estimated digit. Eachmarking on the graduated cylinder represents aknown digit, while the estimated digit comesfrom estimating the position of the bottom ofthe meniscus between two markings.

3. To get the mass of the water inside the kernels,you subtract the mass of the beaker, oil, andpopcorn after popping from the mass beforepopping. Any water on the outside of thekernels will cause the before popping mass to betoo high and will introduce error.

4. Multiply mass by the inverse of molar mass toconvert to moles. In 5.00 g of water:

5.00 g H2O � 1 mol/18.02 g = 0.277 mol H2O

5. 1 atm equals 101.3 kPa, 760 mm Hg, 760 torr,and 1.01 bar.


1. Sample calculation: 7.8 mL - 5.0 mL = 2.8 mL;2.8 mL = 0.0028 L

2. Sample calculation: 110.943 g - 109.952 g = 0.991 g

3. Sample calculation: 0.991 g H2O � (1 mol H2O)/(18.02 g H2O) = 0.0550 mol H2O

4. Sample calculation: P = nRT/V = ((0.0550 mol)(0.0821 (L.atm)/(mol.K))(498 K))/0.0028 L = 803 atm

5. The calculated value is much higher than theatmospheric pressure.

6. The unpopped kernels could be due to lowerwater content, which would result in lesspressure as the water vaporizes.

7. Answers will vary. Sources of error include thepresence of water vapor on the beaker, loss ofsome of the oil, and the fact that some kernelsdo not pop.

Inquiry Extension

Students might discover that different types ofpopcorn require different pressures to pop. The amount of water content will have the biggest effect on the rate of popping.

Teaching Transparency 39 – Pressurevs. Volume Graph

1. At constant temperature, the volume of a gasdecreases with increased pressure.

2. As one variable (either volume or pressure)increases, the other variable decreases.

3. At constant temperature, the volume of a gaswhen multiplied by its pressure always equalsthe same number. For example, at one liter ofvolume, the pressure of this gas is 200 kPa. Oneliter multiplied by 200 kPa equals 200 L.kPa.Similarly, 2L.100 kPa and 4L.50 kPa both equal200 L.kPa.

4. The pressure is P.3.00 L = 200 L.kPa; P = 66.7kPa.

5. This graph represents Boyle’s law.

6. P1V1 = P2V2, where P1 represents the initialpressure of the gas, P2 represents the finalpressure of the gas, V1 represents the initialvolume of the gas, and V2 represents the finalvolume of the gas.

7. Known variables: V1 = 3.25 L; V2 = 1.20 L;P1 = 100.00 kPa

Unknown variable: P2 = ? kPa

Solve for the unknown: P1V1 = P2V2;P2 = P1(V1/V2);

P2 = 100.00 kPa (3.25 L/1.20 L); P2 = 271 kPa

Teaching Transparency 40 – Volumevs. Temperature Graph

1. TK = 273 + TC, where TK is a temperature

113_131_CMC_FF_878763.qxd 04/14/2007 12:17 PM Page 113 Printer PDF

114 Teacher Guide and Answers Chemistry: Matter and ChangeFast Files, Chapters 13-16 Resources


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measurement using the kelvin scale, and TC is atemperature measurement using the Celsiusscale.

2. At constant pressure, the volume of a gasincreases with increased temperature.

3. As one variable (either volume or temperature)increases, the other variable increases.

4. At constant pressure, the kelvin temperature ofa gas when divided by its volume always equalsthe same number. For example, at 200 K, thevolume of this gas is 1 L. 200 K divided by 1 Lequals 200 K/L. Similarly, 600 K/3 L and 1000K/5 L both equal 200 K/L.

5. This graph represents Charles’s law.

6. V1/T1 = V2/T2, where V1 represents the initialvolume of the gas, T1 represents the initialkelvin temperature of the gas, V2 represents thefinal volume of the gas, and T2 represents thefinal kelvin temperature of the gas.

7. Known variables: T1 = 460 K; T2 = 240 K; V1 =2.50 L

Unknown variable: V2 = ? L

Solve for the unknown: V1/T1 = V2/T2; V2 = V1(T2/T1);

V2 = 2.50 L (240 K/460 K); V2 = 1.3 L

Teaching Transparency 41 – Burningof Methane Gas

1. The coefficients in chemical equations representall relative numbers of particles.

2. Avogadro’s principle states that each mole ofany gas occupies 22.4 L at STP.

3. The coefficients in chemical equations involvingonly gases represent all relative numbers ofparticles and relative volumes.

4. When 1 L of methane gas reacts with oxygen, 1L of carbon dioxide and 2 L of water vapor areproduced.

5. Known: VCH4 = 8.00 L Unknown: VO2 = ? L

Using the balanced equation given, 2 L O2/1 LCH4

Multiply the known volume of CH4 by thevolume ratio to find the volume of O2.

VO2 = (8.00 L CH4) (2 L O2/1 L CH4) = 16.0 LO2

6. C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

7. Known: VC3H8 = 7.00 L Unknown: VCO2= ? L

Using the balanced equation from question 6, 3L CO2/1 L C3H8

Multiply the known volume of C3H8 by thevolume ratio to find the volume of CO2

VCO2= (7.00 L C3H8)(3 L CO2/1 L C3H8)

= 21.0 L O2

Math Skills Transparency 19 –Solving Gas Problems Using Boyle’sLaw

1. Volume and pressure are inversely proportional.

2. Temperature must be kept constant.

3. Analyze the problem, and list the knownvariables and the unknown variable.

4. P1V1 = P2V2

5. The unknown variable is isolated on one side ofthe equation; that is P2 = P1(V1/V2)

6. Similar units in the numerator and thedenominator are cancelled out.

7. The remaining unit is kPa, which is the desiredunit, describing pressure.

8. If the volume is almost doubled, the pressurewill be decreased by almost half if thetemperature does not change.

9. When the volume is approximately doubled,the pressure is expected to decrease byapproximately half. The calculated value of 110kPa is reasonable. The unit in the answer is kPa,a pressure unit.

10. To solve for the unknown variable V2, use V2 =V1(P1/P2)

Math Skills Transparency 20 –Solving Gas Problems Using theCombined Gas Law

1. Pressure is inversely proportional to volumeand directly proportional to temperature.

2. Volume is directly proportional to temperature.

3. P1V1/T1 = P2V2/T2

4. Use the equation, TK = 273 + TC°

5. Analyze the problem, and list the knownvariables and the unknown variables.




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6. The unknown variable is isolated on one side ofthe equation; that is, V2 = V1(P1/P2)(T2/T1)

7. Similar units in the numerator and thedenominator are cancelled out.

8. The remaining unit is L, which is the desiredunit, describing volume.

9. Increasing the temperature causes the volumeto increase, but increasing pressure causes thevolume to decrease. Because the greaterpressure change is much greater than thetemperature change, the volume undergoes anet decrease. The calculated answer agrees withthis. The unit is liters, a volume unit.

Math Skills Transparency 21 –Solving Gas Problems Using theIdeal Gas Law

1. The ideal gas law describes pressure, volume,temperature, and the number of moles.

2. PV = nRT

3. Analyze the problem and list the knownvariables and unknown variables.

4. The R value used in this problem is 8.314L•kPa/(mol•K) because the pressure of the gasin this problem is expressed in kPa, not in atmor mm Hg.

5. The remaining unit is mol, which is the desiredunit, describing the number of moles, a quantity.

6. The unit in the answer is moles.

7. PV = nRT; because the number of moles of agas (n) is equal to the mass (m) divided by themolar mass (M), then the ideal gas law equationcan be rewritten as PV = mRT/M or M = mRT/PV.

8. Avogadro’s principle states that equal volumesof gases at the same temperature and pressurecontain equal numbers of particles, Thisprinciple translates into the conversion factor of22.4 L/1 mol gas that may be used in the idealgas law.

Study Guide - Chapter 13 – Gases

Section 13.1 The Gas Laws

1. pressure

2. volume

3. temperature

4. temperature

5. volume

6. pressure

7. pressure

8. temperature

9. volume

10. increases

11. stays the same

12. decreases

13. increases

14. decreases

15. increases

Section 13.2 The Combined Gas Law andAvogadro’s Principle

1. Gay-Lussac’s

2. Boyle’s

3. Charles’s

4. direct

5. direct

6. inverse

7. a

8. d

9. a

10. 1.00 atm pressure and 0.00°C

11. 22.4 L

12. a, b

13. a

14. d

15. e, c

16. c

17. f

18. a, b

19. f

Section 13.2 - The Ideal Gas Law

1. It works best when applied to problemsinvolving ideal gases, those that have minimalattraction between particles and occupy anegligible volume.



2. R is a constant that relates pressure, volume,amount, and temperature for a gas sample.

3. The numerical value of R depends on what unitis used for the pressure variable in the ideal gaslaw equation.

4. n, the number of moles of gas present

5. false

6. false

7. false

8. true

9. true

10. true

11. true

12. true

13. n =

14. p =

15. T =

16. V =

17. mass

18. molar mass

19. mass

20. volume

21. mass

22. volume

23. molar mass

Section 13.3 Gas Stoichiometry

1. 2, 2

2. relative numbers of moles, molecules, volumesof gases

3. 4.0 L, 2.0 L

4. mole ratio for hydrogen and oxygen from thebalanced equation, conditions of temperatureand pressure, molar volume if using thecombined gas law (If using the ideal gas law,molar volume is not used.)

5. Use the mole ratio of hydrogen and oxygen tofind the corresponding number of moles ofoxygen that react. Find the molar mass of

oxygen. Multiply the molar mass of oxygen bythe number of moles of oxygen.

6. 4.00 L H2/22.4 L/mol = 0.179 mol H2; H2 andH2O are in a 1:1 ratio, so 0.179 mol H2O isproduced; 0.179 mol H2O 18.02 g/mol = 3.23 gH2O

Chapter Assessment - Chapter 13 –Gases

Reviewing Vocabulary

1. a

2. h

3. f

4. d

5. g

6. e

7. b

8. c

9. The numerical value depends on the unit usedto measure the pressure of a sample of gas.


1. increases

2. decreases

3. increases

4. decreases

5. increases

6. increases

7. decreases

8. increases

9. decreases

10. stays the same

11. decreases

12. decreases

13. b

14. a

15. a


1. increases

2. decreases

3. increases


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PV

RT

nRT

V

PV

nR

nRT

P




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4. increases

5. decreases

6. increases

7. increases

8. T2 = P2T1/P1 = (2.35 atm)(293 K)/1.82 atm 5378 K

9. V = nRT/P = (2.3 mol)(0.0821 L?atm/mol?K)(290 K)/0.89atm = 62 L

10. T1 = 273 + 20.2°C = 293.2 K; T2 = P2V2T1/P1V1= (0.980 atm)(1.56 L)(293.2 K)/(1.02 atm)(1.74 L) = 253 K

Thinking Critically

1. Gay-Lussac’s law

2. 4.0 atm

3. 500 K

4. The pressure of Sample A is twice that ofSample B.

5. Sample B; compare two pressures at the sametemperature. With n, R, and T being constant,the ideal gas law equation shows that P isinversely proportional to V for each sample. Thesample with the lower pressure will have thegreater volume.

6. Sample A; compare two pressures at the sametemperature. With V, R, and T being constant,the ideal gas law equation shows that P isdirectly proportional to n for each sample. Thesample with the greater pressure will have thegreater amount of gas.

Applying Scientific Methods

1. N2(g) + 3H2(g) 0 2NH3(g)

2. Pressure increases.

3. According to the ideal gas law equation, iftemperature and volume are constant, pressuredecreases when amount of gas decreases.

4. It will liquefy.

5. volume ratio for NH3:H2 is 2:3; 2/3 = X/2400 L;X = 1600 L

6. T1 = 273 + 400.0°C = 673 K; T2 = 273 + 0.00°C= 273 K; V2 = V1(P1/P2)(T2/T1) = 25.0 L (800.0 atm/1.00 atm)(273 K/673 K) = 8110 L

7. T1 = 273 + 400.0°C = 673 K; T2 = 273 + 0.00°C= 273 K; V2 5 V1(P1/P2)(T2/T1) = 126 L (800.0 atm/1.00 atm)(273 K/673 K) = 40 900 L at STP; 40 900 L NH3/22.4 L/mol= 1830 mol NH3;mol ratio of H2 and NH3 = 3:2; 3 mol H2/2 molNH3 = X/1830 mol NH3;X = 2750 mol H2; 2750 mol H2 � 2.02 g/mol = 5560 g H2

CHAPTER 14

MiniLab 14 – Examine Freezing PointDepression

Analysis

1. The freezing point of the water was lowered4°–6°C by the addition of the salt. The ionsinterfere with the attractive forces betweenwater molecules, thus preventing the water from freezing at its normal freezing temperature of 0°C.

2. The beaker containing only ice serves as thecontrol.

3. The number of particles in the solution affectsthe colligative properties of a solution. Because1 mol of sodium chloride produces 2 mol ofions in solution, it has a greater effect on thefreezing point and boiling point than a solutethat produces only 1 mol of particles insolution.

4. Fine table salt is a better choice because itdissolves more quickly in the cold waterthan coarse rock salt thus producing thegreatest freezing point depression as quicklyas possible.

Expected Results:

The salt will lower the freezing point of the water4° to 6°C.

ChemLab 14 – Investigate FactorsAffecting Solubility

Pre-Lab

2. As with any chemical reaction, reactions inaqueous solution involve the rearrangement ofone or more substances to form different




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product(s). In solutions, the identity of bothsolute and solvent remain the same. Forexample, if a solution of sugar water in a beakerevaporates, the sugar will be left behind just asit was before the solution was made. Answersabout similarities will vary. Students mightpoint out that collisions are involved in bothreactions and solution formation. Substancescollide in chemical reactions; similarly, watercolliding with the surface of a NaCl crystal canpull sodium ions and chloride ions intosolution (though no new substances areformed).

3. Temperature is defined as a measure of theaverage kinetic energy of the particles in a sampleof matter. Since KE = mv2, the velocities of theparticles that make up a solution must increase onaverage as temperature increases. This will resultin more collisions between particles.

4. The surface area of a cube 8 cm on a

side: (6 sides) = 384 cm2

The surface area of 8 cubes that are each 4 cmon a side:

(8 cubes) = 768 cm2


1. The test tubes that were agitated showed darkersolutions than the first and third tubes thatwere left undisturbed. The fourth test tubecontained the darkest solution.

2. The fourth test tube had a much faster rate ofdissolving due to the crushing of the crystal. Bycrushing the solid, the surface area of the solidis increased and this causes the solid to dissolvemore readily.

3. The solid in the sixth test tube dissolved thefastest due to the increased temperature. Thecold water in the fifth test tube was very slow todissolve the solid.

4. Answers will vary.

5. The major potential source of error is the sizeof the crystal. If the crystals are very different insize, the results can be altered.

Inquiry Extension

The process of solution formation is based on theinteractions of the solvent and solute. Any processthat increases these interactions will increase therate of solution formation.

Teaching Transparency 42 –Solubility-Temperature Graphs

1. solubility and temperature

2. solubility: g solute/100 g H2O; temperature: °C

3.

4. 30°C

5. The solubility of cerium(III) sulfate decreasesover the temperature range 0°C–100°C, whereasthe solubility of potassium chlorate increasesover the same temperature range.

6. (1.0 kg H2O)(1000 g/1 kg)(36 g NaCl/100 gH2O) = 360 g NaCl

7. The line of the solubility graph of KCl shows asharper rise than the line of the solubility graphof NaCl as temperature is increased. Thesharper rise indicates that the solubility of KClincreases more than the solubility of NaCl.

8. Heat 100 g of water to a temperature of 50°C.Add 42 g KCl and stir until dissolved. Slowlycool the solution, without disturbing it, to 40°C.The resulting KCl solution is a supersaturatedsolution.

Teaching Transparency 43 – PhaseDiagram of Solvent and Solution

1. pressure and temperature

2. water

3. gas, liquid, and solid

4. The solid lines represent the pressures andtemperatures at which two phases of the puresolvent (water) coexist.

5. aqueous solution

6. The dashed lines represent the pressures andtemperatures at which two phases of theaqueous solution coexist.

(4 cm � 4 cm)

1 side

6 sides

1 cubes( )

(8 cm � 8 cm)

1 side

Substance Solubility 100C

Calcium chloride (CaCl2) 64 g CaCl2/100 g H2O

Cerium(III) sulfate (Ce2(SO4)3 10 g Ce2(SO4)3/100 g H2O

Potassium chloride (KCl) 30 g KCl/100 g H2O

Potassium chlorate (KClO3) 5 g KClO3/100 g H2O

Sodium chloride (NaCl) 36 g NaCl/100 g H2O




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7. �P represents the lowering of the vaporpressure due to the addition of the solute.

8. The vapor pressure of the aqueous solution isalways lower than the vapor pressure of thepure solvent.

9. No; a liquid boils when its vapor pressureequals atmospheric pressure. Because the vaporpressure of a solution is always less than thevapor pressure of the pure solvent at the sametemperature, the solution can never boil undernormal atmospheric pressure at the sametemperature as the pure solvent.

10. raise its temperature

11. The freezing point of the solution is lower.

Math Skills Transparency 22 –Calculating Percent by Mass, MollFraction, and Molality, from MassMeasurements

1. Percent by mass= mass of solute/mass of solution x 100

= mass of solute/(mass of solute + mass ofsolvent) x 100

= 25.0 g/(25.0 g + 100.0 g)=(25.0 g/125.0 g) x 100 = 20.0%

Mole Fraction:

XNaCl = nNaCl/nNaCl + nH2O = (mass ofsolute/molar mass of solute)/[(mass ofsolute/molar mass of solute) + (mass ofsolvent/molar mass of solvent)]

= (25.0 g NaCl/58.44 g NaCl/mol NaCl)/[(25.0g NaCl/58.44 g NaCl/mol NaCl) + (100.00 gH2O/18.02 g H2O/mol H2O)]

+ 0.428 mol/0.428 mol + 5.55 mol + 7.16 3 10-2

XNaCl + XH2O = 1

XH2O = 1 XNaCl

XH2O = 1 0.0716

XH2O = 0.9284

Molality:

Molality = moles of solute/kilogram of solvent= 0.428 mol NaCl/(100.0 g H2O � 1 kg/1000 g)

= 4.28 mol NaCl/kg H2O

= 4.28m

Math Skills Transparency 23 –Calculating Boiling points andFreezing points of Solutions

1. Tbsolution= Tbsolvent

+ msoluteKbsolvent= 100°C + (1.5m)(0.515°C/m) = 100°C + 0.77°C= 100.77°C

2. Tfsolution= Tfsolvent

msoluteKfsolvent= 0°C (2)(1.5m)(1.86°C/m) = 0°C 5.6°C= 5.6°C

3. Tfsolution= Tfsolvent

msoluteKfsolvent= 0°C (3)(0.975m)(1.86°C/m) = 0°C 5.44°C= 25.44°C

4. 20.0 g CCl4 1 mol CCl4/153.81 g CCl4 = 0.130mol CCl4Molality = moles of solute/kilograms of solvent

= 0.130 mol CCl4/0.250 kg C6H6 = 0.520m

Tbsolution= Tbsolvent

+ msoluteKbsolvent= 80.10°C + (0.520m)(2.53°C/m) = 80.10°C + 1.32°C = 81.42°C

5. 25.0 g Ca(NO3)2 � (1 mol Ca(NO3)2/164.10 gCa(NO3)2) = 0.152 mol Ca(NO3)2

Molality = moles of solute/kilograms of solvent

= 0.152 mol Ca(NO3)2/0.100 kg H2O = 0.152m

Tbsolution= Tbsolvent

+ msoluteKbsolvent= 100°C + (3)(0.152m)(0.515°C/m) = 100°C 1 0.235°C = 100.235°C

Study Guide - Chapter 14 – Mixturesand Solutions

Section 14.1 Heterogeneous andHomogeneous Mixtures

1. false

2. true

3. false

4. true

5. true

6. false

7. true

8. true




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9. false

10. true

Characteristics of Colloid Solution SuspensionParticles

11. Less than 1 nm in diameter ✔

12. Between 1 nm and 1000 nm in diameter ✔

13. More than 1000 nm in diameter ✔

14. Settle out if undisturbed ✔

15. Pass through standard filter paper ✔ ✔

16. Lower vapor pressure ✔

17. Scatter light ✔ ✔

Section 14.2 Solution Concentration

1. a

2. d

3. c

4. d

5. 7.54 g

6. 85.0 mL

7. Molarity = moles of solute/liters of solution

8. mole

9. liter

10. 85.0 mL � 1L/ 1000 mL

11. the molar mass of FeCl2

12. 7.54 g FeCl2 � 1 mol FeCl2/126.75 g FeCl213. 85 mL H2O � 1 L/1000 ml = 0.085 L H2O

7.54 g FeCl2 � 1 mol FeCl2/126.75 g FeCl2= 0.0595 mol FeCl2Molarity = moles of solute/liters of solution= 0.0595 mol FeCl2/0.0850 L H2O

Molarity = 0.700 mol FeCl2/L H2O = 0.700MFeCl2 (aq)

14. Molality is a measure of the number of molesof solute dissolved in 1 kilogram of solvent.Molarity is a measure of the number of molesof solute dissolved in 1 liter of solution.

15. 15.4 g NaBr � 1 mol NaBr/102.89 g NaBr =0.150 mol NaBr

125 g H2O � 1Kg/1000 g = 0.125 kg H2O

Moality = moles of solute/kilograms of solvent

= 0.150 mol NaBr/0.125 kg H2O = 1.20 molNaBr/kg H2O

Molality = 1.20m NaBr(aq)

16. Mole fraction is the ratio of the number ofmoles of solute in a solution to the totalnumber of moles of solute and solvent.

17. mHCl = 100 g 0.336 = 33.6 g HCl mH2O =100 g 33.6 = 66.4 g H2O

33.6 g HCl � 1 mol HCL/36.46 g HCL = 0.922mol HCl

66.4 g H2O � 1 mol H2O/18.02 g H2O = 3.68mol H2O

XHCl = nHCL/nHCL + nH2O = 0.922 mol/0.922mol + 3.68 mol = 0.922 mol/4.60 mol

XHCl = 0.200

Section 14.3 Solvation and Solubility

1. solution

2. solute

3. solvent

4. soluble

5. insoluble

6. Liquid

7. miscible

8. immiscible

9. Solvation is the process of surrounding soluteparticles with solvent particles to form asolution.

10. The sodium and chloride ions are separatedand surrounded by the water molecules.

11. Because the sodium ion is positively charged, itattracts the negatively charged portion of thewater molecule (the oxygen atom) and repels thepositively charged portion of the water molecule(the hydrogen atoms). Because the chloride ionis negatively charged, it attracts thepositively charged portion of the water moleculeand repels the negatively charged portion.

12. The attraction between the water molecules andthe sodium and chloride ions is greater than theattraction between the sodium and chlorideions. The greater strength of attraction betweenthe water molecules and the ions is what causesthe solvation process to occur.




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13. stirring or shaking the solution, breaking thesolute into smaller pieces, and heating thesolvent

14. true

15. false

16. false

17. true

18. true

19. true

Section 14.4 Colligative Properties ofSolutions

1. boiling point and freezing point

2. Colligative properties depend on the number ofelectrolytes in solution.

3. Colligative properties are independent of thetype of electrolytes in solution.

4. d

5. b

6. f

7. e

8. c

9. g

10. a

Chapter Assessment - Chapter 14 –Mixtures and Solutions


1. g

2. k

3. a

4. h

5. c

6. d

7. b

8. i

9. f

10. j

11. e

12. If a substance dissolves in another substance,the first substance is soluble. If a substance doesnot dissolve in another substance, the firstsubstance is insoluble.

13. Miscible liquids are soluble in each other, andimmiscible liquids are not.

14. Molarity is the number of moles of solutedissolved per liter of solution. Molality is theratio of the number of moles of solute dissolvedin one kilogram of solvent.


1. true

2. moles per liter

3. water

4. true

5. more

6. true

7. true

8. gas

9. true

10. electrostatic

11. boiling point elevation

12. b

13. c

14. d


1. When a sodium chloride crystal is placed inwater, the charged ends of the polar watermolecules attract the positive sodium ions andthe negative chloride ions. Because theattraction between the water molecules and theions is greater than the attraction between theions in the crystal, the ions break away from thecrystal. The water molecules surround the ionsand keep them separated, forming a solution.

2. a. Molarity = moles of solute/liters of solution

2.00M NaOH = 2.00 mol NaOH/1.00 L ofsolution

(2.00 mol NaOH)(40.00 g NaOH/1 mol NaOH)= 80.00 g NaOH

Add 80.00 g of NaOH to a 1-L volumetric flask.Add distilled water to the flask to completelydissolve the NaOH. Carefully add additionaldistilled water to bring the solution up to the 1-L calibration line.

b. M1V1 = M2V2

V1 = V2(M2/M1) = (90.0 mL)(1.20M/2.00 M) =54.0 mL

Add 54.0 mL 2.00M Na2C2O4 to a graduatedcylinder. Carefully add distilled water to bringthe solution up to the 90.0-mL calibration line.




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Observations of the Burning of a Potato Chip

Mass of beaker and 50 mL of water 148.26 gMass of empty beaker 98.00 gMass of water in beaker 50.26 gMass of potato chip 1.63 gHighest temperature of water 37.4°CInitial temperature of water 22.1°CChange in temperature 15.3°C

3. Molality = moles of solute/kilograms of solvent

1.00m BaCl2 = 1.00 mol BaCl2/1.00 kg H2O

(1.00 � 103 g H2O)(1 mol H2O/18.02 g H2O) =55.5 mol H2O

X BaCl2= nBaCl2

/( nBaCl2+ nH2O) = 1.00 mol

BaCl2/(1.00 mol BaCl2 + 55.5 mol H2O)

X BaCl2= 0.0177

Thinking Critically

1. Compound A is tin(II) iodide, and compoundB is carbon dioxide. The curve for compound Aindicates that the solubility of the substanceincreases with increasing temperature. Thistrend is characteristic of most solid solutesdissolved in liquid solvents, such as tin(II)iodide dissolved in water. The curve forcompound B indicates that the solubility of thesubstance decreases with increasingtemperature. This trend is characteristic ofgases dissolved in liquid solvents, such ascarbon dioxide dissolved in water.

2. (0.50 g HgBr2)(1 mol HgBr2/360.30 g HgBr2) =1.4 103 mol HgBr2

(100.0 g H2O)(1 kg/1000 g) = 0.1000 kg H2O

Molality = moles of solute/kilograms of solvent= 1.4 103 mol HgBr2/0.1000 kg H2O

= 1.4 102 mol HgBr2/kg H2O = 1.4 102m HgBr2

3. Mass of solution = 1000 g + 0.50 g = 1000.50 g

Volume of solution = (1000.50 g)(1 kg/1000 g)(1L/1kg) = 0.1005 L

Molarity = moles of solute/liters ofsolution = 1.4 10-3 mol HgBr2/0.1005 L

= 1.4 10-2 mol HgBr2/L = 1.4 10-2M HgBr2


1. The solution becomes dilute.

2. Osmosis; water diffuses through thesemipermeable membrane from the body,which has a lower concentration of sodiumchloride, to the capsule, which has a higherconcentration of sodium chloride.

3. Yes, the graph supports the explanation becausethe graph shows that the volume of the sodiumchloride solution increases. The increasedvolume means that water is diffusing into thecompartment through the semipermeablemembrane.

4. Graph c; if the drug is entering the body at aconstant rate, it must be leaving the capsule at aconstant rate. So, as time passes, the mass of thedrug inside the capsule must decrease at aconstant rate.

5. Because the membrane is impermeable, it keepswater from diffusing through the membraneinto the drug and diluting it. Because themembrane is elastic, it allows the volume of thesodium chloride to expand into the drugcompartment and push the drug through thecapsule wall into the body.

CHAPTER 15

MiniLab 15 – Determine Specific Heat

Analysis

1. – 3. Student results will vary depending on themetal used.

4. Students might cite loss of heat to thesurroundings, hot water clinging to the metal orloss of heat from the metal in the process oftransfer, measurement error, and spillage.Students should suggest that the greatest sourceof error, the loss of heat from the calorimeter,can be reduced by improving the insulation.

ChemLab 15 – Measure Calories

Pre-Lab

3. The quantity of heat produced by the reactionwill exceed the quantity of heat absorbed by thewater.

4. q = c � m � �T

5. carbon dioxide and water vapor




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1. The reaction is exothermic because heat andlight were visible and the temperature of thewater increased.

2. The potato chip burns in oxygen from the air toproduce carbon dioxide gas, water vapor, andunburned carbon. The chip was not completelyconsumed as indicated by the presence of sootand ash.

3. 50.26 g water, 15.3 °C; q = c � m � �T;q = 4.184 J/(g·°C) � 50.26 g � 15.3°C = 3220J/chip

4. 0.770 Calories

5. 28 g per serving; 0.770 Calories/1.63 g � 28 g/1serving = 13.2 Calories

6. 75 Calories per serving, but answers will varyaccording to the snack used (75 Calories 13.2Calories)/75 Calories 100 = 82% error.

Inquiry Extension

Answers will vary. All experiments shouldinclude the elements of the ChemLab includingSafety and testing precautions.

Teaching Transparency 44 – Using aCalorimeter

1. The system is the reaction inside the chamber.The surroundings are everything except thesystem.

2. Oxygen gas must be added because combustionrequires oxygen.

3. The products are principally carbon dioxide gas,water vapor, and heat.

4. The insulation prevents heat from escaping.

5. The thermometer measures the temperature ofthe water surrounding the reaction chamber.

6. The heat released from the combustion reactionmoves from the chamber to the watersurrounding the chamber.

7. q = c � m � ΔT, where q is the amount of heatabsorbed by the water, c is the specific heat ofwater, m is the mass of the water in thecalorimeter, and ΔT is the change in thetemperature of the water.

8. The answer has a positive value because ΔT ispositive. Combustion is an exothermic reaction.

Teaching Transparency 45 –Temperature Changes of Water

1. 20 kcal of heat is being added to the ice. The iceremains a solid even though its temperaturerises 40 degrees.

2. During this period, the ice is melting, andliquid and solid water coexist. All the heat thatis absorbed is converting the solid to the liquid,so the temperature does not change as long asboth states coexist.

3. Because all the solid water has been convertedto liquid water, the 100 kcal of heat that is beingadded increases the temperature of the water.

4. When the temperature reaches 100°C, the liquidwater boils, and liquid and gaseous watercoexist. All the heat that is being absorbed isconverting the liquid to the gas, so thetemperature does not change as long as bothstates coexist.

5. More heat is needed to convert liquid water towater vapor than is needed to convert ice toliquid water. That is because it takes moreenergy to separate the particles in liquid farenough to form a gas than it takes for watermolecules in a solid to slide past each other toform a liquid.

6. All the water is in the gaseous state, and thetemperature of the gas rises as more heat isadded.

7. Its temperature will continue to increase.

Teaching Transparency 46 – Changesin Enthalpy and Entropy

1. The arrows represent the enthalpy changes inthe formation of liquid water from theelemental forms of hydrogen and oxygen, in the vaporization of liquid water, and in thedecomposition of gaseous water to elementalhydrogen and oxygen.

2. Arrows that point upward indicate that energyis absorbed, or the process is endothermic. Thedownward arrow indicates that energy isreleased, or the process is exothermic.

3. The enthalpy of transition from liquid water tothe elemental forms of hydrogen and oxygen(286 kJ) is equal to the sum of the enthalpy fortransition from liquid water to gaseous water




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(44 kJ) and the enthalpy for transition ofgaseous water to the elemental forms ofhydrogen and oxygen (242 kJ).

4. Hess’s law states that the enthalpy for an overallreaction (the transition from liquid water to theelemental forms of hydrogen and oxygen) is thesum of the enthalpies for the individualcomponent reactions (the transition fromliquid water to gaseous water plus the transitionof gaseous water to the elemental forms ofhydrogen and oxygen).

5. The molecules of liquid water are closertogether than the molecules of gaseous water.

6. As the molecules move farther apart, theybecome more disordered.

7. Increased disorder means that entropy hasincreased. Decreased disorder means thatentropy has decreased.

Math Skills Transparency 24 –Determining the Heat Capacity of aCalorimeter

1. The heat capacity of the calorimeter is the totalamount of heat absorbed or released by theapparatus and does not depend on how muchmass is involved.

2. The amount of heat absorbed would bedetermined to be less than it should be. Theheat released by the calorimeter was notaccounted for in the determination.

3. Heat = mass of hot water � specific heat ofwater � temperature change

4. Heat = mass of cold water � specific heat ofwater � temperature change.

5. Specific heat is used to describe a puresubstance. A calorimeter apparatus is composedof many substances. Rather than using manyspecific heats to describe the apparatus, theseare combined into a single relationship knownas the heat capacity.

6. 17.7 J/°C = Heat Capacity

Solution

–(150.0 g � 4.184 J/g?°C � (36.8°C 2 45.0°C)) =

(100.0 g � 4.184 J/g?°C � (36.8°C 2 25.0°C)) +(Heat Capacity � (36.8°C 2 25.0°C))

5146.32 J = 4937.12 J+(Heat Capacity �(11.8°C))

209.2 J/11.8°C = Heat Capacity

17.7 J/°C = Heat Capacity

Math Skills Transparency 25 – Hess’sLaw

1. They are examples of formation reactions.

2. This equation must be reversed, the equationand enthalpy multiplied by 2, and the sign ofthe enthalpy changed.

3. This equation does not change.

4. The resulting values for the enthalpy for eachreaction are added after all mathematicalrearrangements have been completed.

5. �H° = 851.5 kJ

Solution:

2Fe2O3(s)04Fe(s) + 3O2(g)�H° = 11648.4 kJ

4Al(s) + 3O2(g)02Al2O3(s)�H° = 23351.4 kJ

4Al(s) + 3O2(g) + 2Fe2O3(s)02Al2O3(s) +4Fe(s) + 3O2(g)

�H° = (23351.4 kJ) 1 (11648.4 kJ)

Cancelling terms gives:

4Al(s) + 2Fe2O3(s)02Al2O3(s) + 4Fe(s)�H° = 21703.0 kJ

This is twice the target reaction. Dividing by 2gives:

2Al(s) + Fe2O3(s)0Al2O3(s) + 2Fe(s)�H° = 2851.5 kJ

Math Skills Transparency 26 –Determining Spontaneity

1. Temperature, T, is in kelvins and kelvintemperatures have only positive values. Apositive value multiplied by another valueassumes the sign of that other value.

2. The temperature below which condensation isspontaneous will be rather high. A very largenumber divided by a very small one will give arelatively large result. The result in this case isthe temperature.

3. The reaction will be spontaneous.

Solution:

T = 750°C + 273 = 1023 K

�G = �H T�S = (135.5 kJ/mol) (1023 K �(148.9 J/molK)(1 kJ/1000 J))

�G = (135.5 kJ/mol) (152.3 kJ/mol) = 216.8 kJ




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The sign of �G indicates heat is released, so thereaction is spontaneous at this temperature.

4. The equilibrium temperature is 910.0 K.

Solution:

�G = 0 at equilibrium, so �H = T�S

T = �H ÷ �S = (135.5 kJ/mol) ÷ (148.9J/molK)(1 kJ/1000 J)

T = 910.0 K

Study Guide - Chapter 15 – Energyand Chemical Change

Section 15.1 Energy

1. true

2. cannot be

3. true

4. true

5. true

6. joule

7. true

8. true

9. both potential and kinetic energy

10. 1000 calories

11. true

12. true

13. multiply

14. q = c � m � �T

q = 4.184 J/(g·°C) � (5.000 � 102 g) � 2.00°C

q = 4.184 � 103 J = 4.18 kJ

Section 15.2 Heat in Chemical Reactionsand Processes

1. b

2. c

3. a

4. f

5. d

6. e

7. g

8. The system is the solution of Ba(OH)2 andNH4NO3.

9. The surroundings include everything except the solution.

10. The universe is the solution plus thesurroundings.

Section 15.3 Thermochemical Equations

1. thermochemical equation

2. enthalpy of combustion

3. molar enthalpy of vaporization

4. molar enthalpy of fusion

5. molar enthalpy of fusion

6. thermochemical equation

7. molar enthalpy of vaporization

8. released

9. absorbs

10. cool

11. heat

12. heat

Section 15.4 Calculating Enthalpy Change

1. true

2. one mole

3. true

4. true

5. true

6. 298 K

7. true

8. positive

9. true

10. true

11. liquid

12. A formation equation shows how a compoundis formed from the component elements. Theenthalpy of formation for the elements is 0.0kJ/mol. Therefore, the enthalpy change for theformation reaction is the enthalpy of formationfor the compound.

13. The negative sign indicates that the formationof the compound from the elements is anexothermic process.

14. CH4(g) and H2O(g) are the reactants andCH3OH(g) and H2(g) are the products in thefinal equation. Rearrange the threethermochemical equations in the table to obtainthis relationship.




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�Hf° (kJ/mol)

CH4(g) 0 C(graphite) + 2H2(g)75

C(graphite) + 2H2(g) + O2(g) 0 CH3OH(g)– 239

H2O(g) 0 O2(g) + H2(g) 242

Add all the equations.

CH4(g) + C(graphite) + 2H2(g) + O2(g) +H2O (g) 0 C(graphite) + 2H2(g) + CH3OH(g) + O2(g) + H2(g)

Remove all terms that are on both sides of theequation.

CH4(g) + H2O(g) 0 CH3OH(g) + H2(g)

Determine �Hrxn by adding the enthalpies ofthe reactions.

�Hrxn = 75 kJ 239 kJ + 242 kJ = 78 kJ

Section 15.5 Reaction Spontaneity

1. spontaneous

2. entropy

3. second law of thermodynamics

4. Free energy

5. false

6. true

7. false

8. false

9. false

10. true

11. true

12. false

13. true

Chapter Assessment - Chapter 15 –Energy and Chemical Change


1. h

2. e

3. d

4. p

5. c

6. f

7. a

8. r

9. m

10. t

11. j

12. i

13. s

14. l

15. u

16. k

17. n

18. q

19. g

20. o


1. true

2. false

3. false

4. true

5. The bottle in system A is open. The bottle insystem B is stoppered. The bottle in system C issealed.

6. The arrows indicate that both heat and watervapor can be added to or removed from systemA. In system B, the stopper prevents watervapor from entering or leaving the bottle, soonly heat can be added to or removed from thesystem. No arrows point to or from system C.Because of the seal around system C, neitherheat nor water vapor can be added to orremoved from system C.

7. System A will show the greatest change becauseboth heat and water vapor can enter or leavethe bottle. System C will show the least changebecause neither heat nor water vapor can enteror leave the bottle.

8. The surroundings are the bottles and all thingsoutside the bottles.


1. The reaction has a free energy value that isnegative. This indicates that the reaction isspontaneous.

2. CO has the highest entropy because the gaseousstate has the highest entropy.




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3. The gaseous state has higher entropy than thesolid state. In a, gas is converted to a liquid soentropy decreases. The reaction in b starts withtwo moles of gas and ends with two moles ofgas, so entropy does not change. In c, thereactant in the solid state forms a gas, soentropy increases.

4. Each process is endothermic; heat is beingadded to the system.

5. The entropy of the iron increases slightly. Thespacing between the iron atoms is slightlylarger, but the iron remains a solid. The entropyof the ice increases greatly because themolecules have moved farther apart and canmove more independently. Most of the solid hasbeen converted to liquid.

Thinking Critically

1. The graph shows how entropy increases as heatis added. A solid is transformed into a liquid,and then the liquid is transformed into a gas.

2. At the melting point, the solid and liquidcoexist. At the boiling point, the liquid and gascoexist. The temperature does not change atthese points as long as both states of matter arepresent. At the melting point and boiling point,the added heat does not change the temperatureof the substance but increases the disorder ofthe solid’s particles as they become liquid andincreases the disorder of the liquid’s particles asthey become a gas.

3. 120 Calories

4. Samples of the cookies were burned in acalorimeter and the amount of heat releasedwas measured.

5. 120 Calories � (1000 calories)/(1 Calorie) �(4.184 J)/(1 calorie) = 502,000 J


1.

2. �Tlead = 160.0°C – 31.4°C = 128.6°C

�Twater = 31.4°C – 21.3°C = 10.1°C

heat gained by calorimeter = heat lost by leadheat gained by water

qcalorimeter = qlead – qwater

qlead = c(lead)m�Tlead = (0.129 J/(g·°C))(142.81g)(128.6°C) = 2370 J

qwater = c(water)m�Twater = (4.18 J/(g·°C))(50.0g)(10.1°C) = 2110 J

qcalorimeter = qlead – qwater

= 2370 J – 2110 J = 260 J

The calorimeter absorbs 260 J of heat.

3. Ravi’s constant = 260 J

Rachel’s constant = 260 J/10.1°C = 26 J/°C

4. Student answers should support Rachel citingthat the heat absorbed by an object made of apure material is proportional to its change intemperature. So, the heat gained by an object(made of many parts) should also beproportional to its temperature change if everypart has the same temperature change.

CHAPTER 16

MiniLab 16 – Examine Reaction Rateand Temperature

Analysis

1. Results will vary depending on the exacttemperature and mass of the tablet.

2. The reaction rate should be higher at 50˚C thanit is at 20˚C.

3. The reaction proceeds faster at the highertemperature.

4. A ten degree increase in temperature shouldresult in doubling the rate of the reaction. Thus,a reasonable prediction is that the reaction ratewould quadruple.

5. The prediction was not validated by theexperimental data possibly because ofdifferences in the masses of the tablets used. Butthe rate at 40˚C was higher than the rate at20˚C.

hot sinkerand cool water and cool calorimeter

warmsinker and warm water and warm calorimeter

heat transferfrom hot sinker to

cool waterand cool

calorimeter




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Expected Results:

Data Table

ChemLab 16 – Observe HowConcentration Affects Reaction Rate

Pre-Lab

4. Students’ hypotheses should reflect that thereaction rate is faster with higher concentration;slower with lower concentration.

5. The amount of Mg is the constant. [HCl] is theindependent variable. Time is the dependentvariable.

6. hydrogen gas; Mg(s) + 2HCl(aq) 0 MgCl2(s) +H2(g)

7. Cleaning removes any deposits on the metaland exposes the entire magnesium surface forreaction. If a piece of magnesium is not clean,the reaction rate will be slower.


1.

2. Reaction time decreases with acidconcentration.

3. As the concentration of the HCl reactantincreases in the test tube, the probability of aneffective collision with the Mg metal increases.These collisions displace H2 gas from the acid,increasing the rate of the reaction as isevidenced by increased bubbling.

4. Student results should be close. Discrepanciescould arise from inaccurate dilutions of the acidsamples and variability in the length of themagnesium ribbon.

Inquiry Extension

Student experiments might involve choosing onedilution and repeating the experiment at two orthree different temperatures.

Teaching Transparency 47 – FactorsThat Affect Reaction Rate

1. The color of the apple changed from white todark brown.

2. A chemical in the apple reacted with the air,producing a brown substance.

3. Answers will vary. A possible answer is to put asliced apple in a vacuum. If the apple does notdarken, the test indicates that air is needed forthe reaction to occur.

4. Increasing the amount of air would cause theapple to change color faster because more airparticles would be colliding with the chemicalparticles in the apple and therefore the reactionwould occur faster.

5. Slicing the apple into more pieces wouldincrease the apple’s surface area. Therefore,more air molecules would react with thechemicals in the apple, and more of the applewould turn brown.

0.00

0.01

0.02

0.03

0.04

Rxn

Rat

e (g

/s)

0.05

0.06

0.07

0 10 20 30Temp (Celsius)

Rxn Rate vs. Temp

40 50 60 70

Mass of Reaction Reaction Temperature Tablet (g) time (s) rate (g/s) (°C)

0.92 27.56 0.033 23.6

0.79 13.22 0.060 53.4

0.89 13.77 0.065 64.0

0.54 11.38 0.047 40.0

Tim

e(s)

Concentration (M)

Reaction Time vs Concentration

Test tube [HCl] (M) Time (s)

1 6.0 5

2 3.0 25

3 1.5 40

4 0.75 110

Reaction Time Data




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6. Answers may vary. The cooking process involvesheat, which changes the chemical makeup of theapple. Because the chemicals in the apple havechanged, their reaction with air will also change.

Teaching Transparency 48 – ReactionOrder

1. A yellow precipitate, lead iodide, is formed.

2. The particles in the two solutions collide whenthe solutions mix. Some of these particles sticktogether and fall out of solution.

3. The reaction rate is very fast. The precipitateforms as soon as the two solutions mix.

4. The activation energy must be low.

5. Reaction orders are determined by experimentalmeasurements. Because the informationprovided does not contain any experimentalinformation, no conclusions can be made aboutthe reaction order.

6. The reaction order will be first order in Pb2+,first order in I, and second order overall.

7. The reaction is spontaneous because it producesa precipitate immediately when the two startingsolutions mix. The formation of a precipitatemeans that energy is lost from the system, sothe reaction is exothermic.

Teaching Transparency 49 – ReactionRate

1. The graph shows a decrease in hydrogenperoxide concentration over time.

2. The hydrogen peroxide is a reactant because itsconcentration decreases over time.

3. 2H2O2(aq) 0 2H2O(l) + O2(g)

4. The rate of the reaction decreases over time.

5. Instantaneous rate is the rate of a reaction at aspecific time.

6. The instantaneous rate can be found by findingthe slope of the straight line that is tangent tothe curve at a specific point. The slope isdefined mathematically as Δ[H2O2]/Δt.

7. The concentrations of the products, water andoxygen, as a function of time also are needed.

8. The concentrations of the products shouldshow an increase over time.

Math Skills Transparency 27 –Determining Reaction Orders

1. Step 1. Rate 1 = k[A1]m[B1]n; Rate 2 = k[A2]m[B2]n

Step 2. Rate 2 = 2 Rate 1

Step 3. k[A2]m[B2]n = 2k[A1]m[B1]n

Step 4. [A2] = [A1] [B2] = 2[B1]

Step 5. k[A1]m(2[B1])n = 2k[A1]m[B1]n

k[A1]m(2)n[B1]n = 2k[A1]m[B1]n

(2)n = 2; n = 1

2. Step 1. Rate 2 = k[A2]m[B2]n; Rate 3 = k[A3]m[B3]n

Step 2. Rate 3 = 6 Rate 2

Step 3. k[A3]m[B3]n = 6k[A2]m[B2]n

Step 4. [A3] = 2[A2] [B3] = 1.5[B2]

Step 5. k(2[A2])m(1.5[B2])n = 6k[A2]m[B2]n

k(2)m[A2]m(1.5)n[B2]n = 6k[A2]m[B2]n;From Question 1, you know n = 1.(2)m(1.5)1 = 6; (2)m = 6/1.5 = 4; m = 2

Study Guide - Chapter 16 – ReactionRates

Section 16.1 A Model for Reaction Rates

1. collision theory

2. activated complex

3. activation energy

4. reaction rate

5. mol/L·s.

6. exothermic

7. the activated complex

8. the activation energy

9. the net energy released from the exothermicreaction

10. c

11. a

12. b

13. CO and NO2

14. The arrows represent the direction and theamount of energy of the moving molecules.

15. Collision A; CO2 and NO



16. Collision A did not form products because thecarbon atom in the CO molecule did notcontact an oxygen atom in the NO2 molecule.Collision C did not form products because theCO molecule and the NO2 molecule did notcollide with sufficient energy.

17. Collision B; the activated complex is anOCONO molecule.

Section 16.2 Factors Affecting ReactionRates

1. Increasing

2. true

3. temperature

4. lowering

5. true

6. true

7. There was a greater concentration of oxygen inthe container. Increasing the concentration of areactant increases the rate of a reaction.

8. No; a block of steel would react more slowlybecause it has less surface area.

9. Not heating the steel wool would decrease therate of the reaction.

Section 16.3 Reaction Rate Laws

1. chemical reaction

2. rate law

3. specific rate content

4. reaction orders

5. concentration

6. time

7. the exponent to A, first order

8. the exponent to B, second order

9. the sum of 1 and 2, or third order

10. the rate would quadruple. The square of 2 is 4.

11. This is a second-order reaction

Section 16.4 Instantaneous Reaction Ratesand Reaction Mechanisms

1. a

2. a

3. b

4. c

5. c

6. d

7. b

8. true

9. 0.1 mol/ (L·s)

10. three

11. The slow step is the rate-determining step.

12. intermediates

13. There is no catalyst because no moleculereacted in one step and then was reformed in asubsequent step.

14. Of all the steps, the rate-determining step hasthe highest activation energy.

15. Speed up the rate-determining step.

Chapter Assessment - Chapter 16 –Reaction Rates


1. c

2. a

3. b

4. f

5. d

6. e

7. instantaneous rate

8. complex reaction

9. intermediate

10. reaction order

11. reaction mechanism

12. rate-determining step


1. true

2. true

3. moles

4. true

5. increase

6. true

7. true

8. true

9. true


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10. true

11. increase

12. heterogeneous

13. doubles

14. true


1. reactants

2. intermediates

3. activated complex

4. products

5. Fe3+

6. The rate of oxygen production is one-half thatof nitrogen production.

Rate = (1/2)(�[N2]/�t) = (�[O2]/�t)

7. The added water would reduce theconcentrations of the reactants and thereforeslow the reaction rate.

8. The CO is consumed during the reaction. TheNO is produced during the reaction.

Thinking Critically

1. a. The salt granules will dissolve faster becausethere is more surface area in the granuleform than in the block form.

b. The salt will dissolve faster in the 70°C waterbecause there is more energy in the hotterwater.

c. The salt will dissolve faster in the tubbecause it has a greater amount of water,which participates in the process.

2. No conclusions about the rate law can be madefrom the information given because rate lawsare determined only by experimentation.

3. The activation energy for the reverse reaction isgreater than the activation energy for theforward reaction.

4. a. about 24 minutes

b. Average rate = (0.16 – 1.00) mol/(60 – 0)min = 20.84 mol/60 min

= – 0.014 mol/min Average rate of loss =0.014 mol/min

c. Average rate = (0.84 – 0.00) mol/(60 – 0)

min = 0.84 mol/60 min

= 0.014 mol/min Average rate of gain =0.014 mol/min

d. The rates are the same.

5. a. The reaction rate will decrease because theconcentrations of the reactants aredecreased.

b. The reaction rate will increase because thecatalyst lowers the activation energy for thereaction.

c. The rate will increase because the kineticenergy of the reactants is increased.


1. You are measuring the amount of hydrogen gasproduced to get the reaction rate. But the ratelaw expression is related only to the startingconcentrations of the reactants. The rate lawexpression is related to the rate, so to determinethe rate law expression, you would proceed inthree steps. Step 1: Start with a known amountof copper and a known amount of hydrochloricacid and measure the rate under theseconditions. Step 2: Keep the amount of copperfrom Step 1 constant and vary the amount ofhydrochloric acid to determine the reactionorder for the acid under these conditions. Step3: Keep the amount of hydrochloric acid fromStep 1 constant and vary the amount of copperto determine the reaction order for copperunder these conditions. By combining thisinformation and using simultaneous equations,you will form the rate law.

2. The general rate equation for the chemicalreaction will be Rate = k[reactant 1]m[reactant2]n. You know the concentrations of the tworeactants because those are what you mustmeasure to start the reaction. You will alsomeasure the rate of change as these reactantsare used up in the reaction. Thus you know therate, the [reactant 1], and the [reactant 2]. Youdo not know the rate constant, k, or the orders,m and n. You have three variables and must usethree simultaneous equations to solve for threevariables. To get the three equations, you mustperform three experiments.


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