Chapter 7. Chemical Energetics: Enthalpy
Faculty Resource and Organizational Guide (FROG)
Table of Contents
Materials for Chapter 7 Activities ................................................................................................5
Reagents for Chapter 7 Activities: ...............................................................................................6
Section 7.1. Energy and Change ..................................................................................................7
Learning Objectives for Section 7.1 ............................................................................................7
Investigate This 7.1. What happens when foods are put in a flame? ..........................................7
Consider This 7.2. Is energy involved when foods are burned? .................................................8
Consider This 7.3. How is release of thermal energy related to chemical bond energy? ...........9
Section 7.2. Thermal Energy (Heat) and Mechanical Energy (Work)...................................10
Learning Objectives for Section 7.2 ..........................................................................................10
Investigate This 7.4. What happens when a pinwheel is held over a flame? ............................10
Consider This 7.5. What causes a pinwheel over a flame to turn? ...........................................12
Consider This 7.7. How do thermal and mechanical energies interact? ...................................13
Section 7.3. Thermal Energy (Heat) Transfer ..........................................................................14
Learning Objectives for Section 7.3 ..........................................................................................14
Investigate This 7.8. Can radiation change the temperature of water? .....................................14
Consider This 7.9. How does radiation change the thermal energy of water? .........................16
Consider This 7.10. How does radiation transfer energy to water? ..........................................17
Consider This 7.11. How are conduction and convection the same? different? .......................17
Section 7.4. State Functions and Path Functions .....................................................................18
Learning Objectives for Section 7.4 ..........................................................................................18
Investigate This 7.12. What happens when you rub your hands together?...............................18
Consider This 7.13. What change occurs when you rub your hands together? ........................19
Consider This 7.14. How do different pathways for descent of Pike's peak compare? ............20
Consider This 7.15. How does the energy of the skin molecules change in Figure 7.5? .........20
Section 7.5. System and Surroundings ......................................................................................21
Learning Objectives for Section 7.5 ..........................................................................................21
Consider This 7.18. How do systems and surroundings interact? ............................................21
Section 7.6. Calorimetry and Introduction to Enthalpy..........................................................23
Learning Objectives for Section 7.6 ..........................................................................................23
Investigate This 7.19. What happens when an acid and base are mixed? ................................23
Consider This 7.20. What causes the changes when an acid and base are mixed? ..................24
Investigate This 7.22. What happens when urea dissolves is water?........................................25
Consider This 7.23. What is T when urea dissolves in water? ...............................................26
Consider This 7.24. What is observed for an endothermic reaction in a calorimeter? .............26
Consider This 7.30. How can you get more accurate values for Hreaction?..............................27
Section 7.7. Bond Enthalpies ......................................................................................................28
Learning Objectives for Section 7.7 ..........................................................................................28
Investigate This 7.32. What happens when yeast is added to hydrogen peroxide? ..................29
February 2005
ACS Chemistry FROG
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Chemical Energetics: Enthalpy
Chapter 7
Consider This 7.33. What reaction occurs when yeast is added to hydrogen peroxide? ..........30
Investigate This 7.35. What is the gas formed in the hydrogen peroxide reaction? .................31
Consider This 7.36. Which bonds break and form in the hydrogen peroxide reaction?...........32
Consider This 7.38. How do bond enthalpies for single and multiple bonds compare?...........33
Section 7.8. Standard Enthalpies of Formation .......................................................................34
Learning Objectives for Section 7.8 ..........................................................................................34
Section 7.9. Harnessing Energy in Living Systems ..................................................................34
Learning Objectives for Section 7.9 ..........................................................................................34
Section 7.10. Pressure-Volume Work, Internal Energy and Enthalpy ..................................34
Learning Objectives for Section 7.10 ........................................................................................34
Consider This 7.55 How is the first law of thermodynamics used? .........................................34
Consider This 7.58. How can E (or H) be the same for a reaction at constant volume and
constant pressure? ......................................................................................................................35
Section 7.11. What Enthalpy Doesn't Tell Us ...........................................................................36
Learning Objectives for Section 7.11 ........................................................................................36
Section 7.13. Extension: Ideal Gases and Thermodynamics ...................................................36
Learning Objectives for Section 7.13 ........................................................................................36
Investigate This 7.63. What happens when a gas is compressed or heated? ............................36
Consider This 7.64. What causes changes when a gas is compressed or heated? ....................37
Consider This 7.65. How are manometer readings interpreted and quantified? .......................38
Consider This 7.66. How does kinetic-molecular theory apply to Investigate This 7.63? .......39
Investigate This 7.72. Do reactions in open containers and capped containers differ? ............40
Consider This 7.73. Why do reactions in open and capped containers give different results? 44
Consider This 7.74. What are qV and qP for the hydrogen carbonate-acid reaction? ...............45
Consider This 7.76. What is w for the hydrogen carbonate-acid reaction? ..............................46
Solutions for Chapter 7 Check This Activities ...........................................................................47
Check This 7.6. The difference between warm and cool gases ................................................47
Check This 7.16. The system in Figure 7.5 ..............................................................................47
Check This 7.17. Closed and open systems ..............................................................................47
Check This 7.21. Is the reaction of an acid with a base exothermic or endothermic? ..............47
Check This 7.25. Energy transfers in a calorimeter ..................................................................47
Check This 7.27. Determination of qP(reaction) for dissolution of urea in a calorimeter .............48
Check This 7.29. Determination of Hreaction for dissolution of urea in a calorimeter .............48
Check This 7.31. Effect of increasing the amount of another calorimetric reaction ................48
Check This 7.34. Properties of the possible H2O2 reaction products........................................48
Check This 7.37. Overall homolytic bond cleavage reaction of methane ................................48
Check This 7.39. Formation of polyethylene............................................................................49
Check This 7.40. Sulfur–sulfur bonds ......................................................................................49
Check This 7.42. Reaction enthalpy from average bond enthalpies .........................................50
Check This 7.43. The hydrogen peroxide decomposition reaction ..........................................52
Check This 7.45. Formation of compounds from their standard-state elements ......................52
Check This 7.47. Standard enthalpy change for an isomerization reaction ..............................52
Check This 7.48. Comparison of formation of 1-butene from its atoms and its elements .......53
Check This 7.50. Standard enthalpy change for a reaction.......................................................53
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ACS Chemistry FROG
Chapter 7
Check This 7.51.
Check This 7.52.
Check This 7.53.
Check This 7.54.
Check This 7.57.
Check This 7.60.
Check This 7.62.
Check This 7.67.
Check This 7.69.
Check This 7.71.
Check This 7.77.
Chemical Energetics: Enthalpy
Bond enthalpy and enthalpy of formation comparison ................................54
The phosphate bond to ribose in ATP4– and ADP3– .....................................54
Other biological fuels ...................................................................................55
Other pathways for ATP hydrolysis .............................................................55
Pressure–volume work at constant pressure .................................................55
Comparing E and H .................................................................................55
Comparing E and H quantitatively ..........................................................55
Relationships among the properties of a gas sample ....................................56
Using the ideal-gas equation.........................................................................56
Using the ideal-gas equation.........................................................................56
Ereaction and Hreaction for the hydrogen carbonate-acid reaction .................57
Solutions for Chapter 7 End-of-Chapter Problems....................................................................58
Problem 7.1. ...............................................................................................................................58
Problem 7.2. ...............................................................................................................................58
Problem 7.3. ...............................................................................................................................58
Problem 7.4. ...............................................................................................................................58
Problem 7.5. ...............................................................................................................................59
Problem 7.6. ...............................................................................................................................60
Problem 7.7. ...............................................................................................................................60
Problem 7.8. ...............................................................................................................................60
Problem 7.9. ...............................................................................................................................60
Problem 7.10. .............................................................................................................................61
Problem 7.11. .............................................................................................................................61
Problem 7.12. .............................................................................................................................62
Problem 7.13. .............................................................................................................................62
Problem 7.14. .............................................................................................................................62
Problem 7.15. .............................................................................................................................62
Problem 7.16. .............................................................................................................................63
Problem 7.17. .............................................................................................................................63
Problem 7.18. .............................................................................................................................63
Problem 7.19. .............................................................................................................................64
Problem 7.20. .............................................................................................................................64
Problem 7.21. .............................................................................................................................64
Problem 7.22. .............................................................................................................................65
Problem 7.23. .............................................................................................................................65
Problem 7.24. .............................................................................................................................66
Problem 7.25. .............................................................................................................................66
Problem 7.26. .............................................................................................................................67
Problem 7.27. .............................................................................................................................67
Problem 7.28. .............................................................................................................................68
Problem 7.29. .............................................................................................................................69
Problem 7.30. .............................................................................................................................69
Problem 7.31. .............................................................................................................................70
Problem 7.32. .............................................................................................................................70
Problem 7.33. .............................................................................................................................71
Problem 7.34. .............................................................................................................................72
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Chemical Energetics: Enthalpy
Chapter 7
Problem 7.35. .............................................................................................................................72
Problem 7.36. .............................................................................................................................73
Problem 7.37. .............................................................................................................................74
Problem 7.38. .............................................................................................................................75
Problem 7.39. .............................................................................................................................75
Problem 7.40. .............................................................................................................................75
Problem 7.41. .............................................................................................................................76
Problem 7.42. .............................................................................................................................77
Problem 7.43. .............................................................................................................................78
Problem 7.44. .............................................................................................................................78
Problem 7.45. .............................................................................................................................79
Problem 7.46. .............................................................................................................................79
Problem 7.47. .............................................................................................................................79
Problem 7.48. .............................................................................................................................80
Problem 7.49. .............................................................................................................................80
Problem 7.50. .............................................................................................................................81
Problem 7.51. .............................................................................................................................81
Problem 7.52. .............................................................................................................................81
Problem 7.53. .............................................................................................................................82
Problem 7.54. .............................................................................................................................82
Problem 7.55. .............................................................................................................................82
Problem 7.56. .............................................................................................................................82
Problem 7.57. .............................................................................................................................83
Problem 7.58. .............................................................................................................................83
Problem 7.59. .............................................................................................................................84
Problem 7.60. .............................................................................................................................84
Problem 7.61. .............................................................................................................................85
Problem 7.62. .............................................................................................................................85
Problem 7.63. .............................................................................................................................85
Problem 7.64. .............................................................................................................................86
Problem 7.65. .............................................................................................................................86
Problem 7.66. .............................................................................................................................87
Problem 7.67. .............................................................................................................................87
Problem 7.68. .............................................................................................................................87
Problem 7.69. .............................................................................................................................88
Problem 7.70. .............................................................................................................................88
Problem 7.71. .............................................................................................................................89
Problem 7.72. .............................................................................................................................89
Problem 7.73. .............................................................................................................................90
Problem 7.74. .............................................................................................................................90
Problem 7.75. .............................................................................................................................92
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ACS Chemistry FROG
Chapter 7
Chemical Energetics: Enthalpy
Materials for Chapter 7 Activities
Activity
Material
Total Quantity
7.1, 7.4
Bunsen burner or candle
1
7.1, 7.4
Matches or lighter
1
7.1
Tongs
1
7.1
Pan of water
1
7.4
Long metal cylinder
1
7.4
Plastic or paper pinwheel
1
7.8
Bright incandescent bulb
1
7.8
Lamp
1
7.8
Narrow flat-sided clear
colorless containers
2
7.8
Reflector
1
7.8
Thermometers or
temperature probes
2
7.19
Small test tubes
2
7.19
Plastic pipets
2
7.22
Calorimeter; Styrofoam®
cup, cover, and temperature
probe or thermometer
1
7.32
100 mL graduated cylinder
1
7.35
Long fireplace match or
wooden splint
1
7.35
Stirring rod
1
7.36
Molecular model kits
1/student
7.63
50-mL plastic syringe
1
7.63
Plastic tubing
7.72
2-L soft drink bottle
1
7.72
Watch or timer
1
7.72
50-mL plastic syringe
1
7.72
New 24/40 ribbed rubber
septum
1
7.72
Powder funnel
1
ACS Chemistry FROG
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Chemical Energetics: Enthalpy
Chapter 7
Reagents for Chapter 7 Activities:
Activity
7.1
Reagents
Spaghetti, minimarshmallow, potato chip
Total Quantity
1-3 of each.
Optional: Nuts (such as
walnuts, peanuts, and/or
pecans), corn chip, piece of
chocolate
6
7.19
1M NaOH
10 mL
7.19
1M HCl
10 mL
7.22
Urea, H2NC(O)NH2(s)
6.0 g
7.32
3% hydrogen peroxide
solution, H2O2
10 mL
7.32
Liquid dishwashing
detergent
1-2 mL
7.32
Yeast
2 to 3 g
7.63
Food coloring
several drops
7.72
Sodium hydrogen carbonate
(baking soda, NaHCO3(s))
2  8.4 g
7.72
6 M hydrochloric acid
(HCl) solution.
2  20.0 mL
ACS Chemistry FROG
Chapter 7
Chemical Energetics: Enthalpy
Section 7.1. Energy and Change
Learning Objectives for Section 7.1
• Identify some forms of energy observed in chemical changes and learn that energy is
conserved in the changes.
Investigate This 7.1. What happens when foods are put in a flame?
Goal:
Observe that energy is released (heat and light) when foods are burned and get a qualitative idea
of the energy content of various foods.
Set-up time:
• Approximately 15 minutes (assuming food products are already available).
Time for activity:
• Approximately 5-10 minutes (depending on how many food products you test).
SAFETY NOTE
Wear your safety goggles.
If someone in the class has a food allergy to one of the
food items you choose, do not burn that item.
Equipment:
• Bunsen burner or candle.
• Matches or lighter.
• Tongs (optional).
• Pan of water.
Reagents (foods):
• Spaghetti.
• Mini-marshmallow.
• Potato chip.
• Optional: Nuts (such as walnuts, peanuts, and/or pecans), corn chip, piece of chocolate.
Procedure:
• This activity is conducted as a class activity.
• Ask for a student volunteer and provide safety goggles.
• Have the student take a full-length spaghetti piece and hold it by one end over the pan of
water.
• Bring the flame under the free end of the spaghetti and hold it there until its appearance begins
to change. Then remove it from the flame. If necessary, have the student extinguish the
spaghetti by immersing it in water.
• The rest of the class should record their observations.
• Repeat with a marshmallow (or other food product) held on the end of another full-length
piece of spaghetti.
• Instead of using a full-length piece of spaghetti as a holder, tongs may be used.
Clean-up:
• All fully extinguished food products should be disposed in a trash container.
Anticipated results:
Here are photographs of a few foods burning:
ACS Chemistry FROG
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Chemical Energetics: Enthalpy
spaghetti
Chapter 7
mini-marshmallow
pecan
potato chip
corn chip
chocolate
Follow-up discussion:
• Use Consider This 7.2 and 7.3 to initiate and facilitate classroom discussion of this activity.
• Discuss this activity qualitatively — food as a source of energy, conversion of one form of
energy into another, etc.
• The discussion should introduce students to the first law of thermodynamics as well as the
vocabulary used in measuring energy.
Follow-up activities:
• End of chapter problems 7.1 through 7.3.
Consider This 7.2. Is energy involved when foods are burned?
Goal:
Students should conclude that energy is released differently when different foods are burned and
relate this observation qualitatively to the composition of various foods.
Classroom options:
• Allow 3-5 minutes for students, working in small groups, to answer these questions. Then, the
groups can share their conclusions with the class, summarizing answers on the chalkboard or
an overhead transparency.
• This activity can be conducted as an open class discussion.
Time for activity:
• Approximately 10 minutes.
Instructor's notes:
• This activity is a follow-up discussion to Investigate This 7.1 and serves to focus students on
the idea that energy is released when food is burned and that this is the energy that is provided
by our metabolic processes.
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ACS Chemistry FROG
Chapter 7
Chemical Energetics: Enthalpy
• When the foods burn in this activity, they combine with oxygen to produce mostly carbon
dioxide and water.
• Stress that the form of energy changes, but energy is always conserved as stated in the first
law of thermodynamics.
Students should reason and conclude:
(a) When the foods were placed in the flame some of them, such as marshmallows, chips,
and nuts, ignited and burned, even when removed from the flame. Others, such as spaghetti,
were less able to sustain a flame, but continued to glow and char when out of the flame. A
flame was required to ignite the foods and light and heat were released when the foods
burned on their own; these observations are evidence that energy was involved in the
changes. All of the products shown in the chapter opener can be ignited and will burn or
char, as we have observed for a few of them. The flames of each of the burning foods were
different. Yes, once our bodies metabolize these foods, they provide us with energy.
(b) The foods did not all burn the same way, as noted in part (a). Chips and marshmallows
burn quickly. Nuts burn steadily. Spaghetti burns slowly and not as vigorously. Chips contain
a good deal of fat as well as starch and the fat burns quickly. (Do baked chips burn the same
as fried chips?) Marshmallows are mainly sugar and gelatin (protein) and are puffed up,
giving them lots of surface area for reaction with oxygen from the air. Nuts are full of oil,
which burns well but has to ooze to the surface of the nut, in order to vaporize and burn, so
nuts don’t usually burn rapidly. Spaghetti is mostly starch (a polymer of glucose), very
similar to the cellulose in wood, and it burns much like wood. Perhaps these observations on
the foods are somewhat indicative of their energy release over time, but metabolic processes
are different than combustion, so we should be careful not to draw too many conclusions
about metabolism from these observations.
Follow-up activities:
• Consider This 7.3. How is release of thermal energy related to chemical bond energy?
• End of chapter problems 7.1 through 7.3.
Consider This 7.3. How is release of thermal energy related to chemical bond energy?
Goal:
Students should link the energy release observed in Investigate This 7.1 to the relative strengths
of the chemical bonds in the products and reactants.
Classroom options:
• Allow 3-5 minutes for students, working in small groups, to answer these questions. Then, the
groups can share their conclusions with the class, summarizing answers on the chalkboard or
an overhead transparency.
• This activity can be conducted as an open class discussion.
Time for activity:
• Approximately 10 minutes.
Instructor notes:
• Review the results from Investigate This 7.1 to be sure the class agrees on the observations.
• Discuss the fact that different compounds are being burned in each food, but that the resulting
products are pretty much the same in each sample (carbon dioxide + water). [Of course there
are other products produced by the atoms other than carbon and hydrogen in the foods, but
they do not affect the argument about relative bond strengths.]
ACS Chemistry FROG
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Chemical Energetics: Enthalpy
Chapter 7
Students should reason and conclude:
All the foods burned and released energy in the form of light and heat. Thus, the change from
the chemical bonds in the reactants, oxygen and the compounds in the food, to the chemical
bonds in the products, mostly carbon dioxide and water, releases energy. The bonds in the
products must be stronger, that is, lower in energy than those in the reactants, in order that
the change should release energy. Because we have no way to know how much fuel was
burned in each case, we cannot compare the relative energy output of the foods from our
qualitative observations in Investigate This 7.1.
Follow-up activities:
• End of chapter problems 7.1 through 7.3.
Section 7.2. Thermal Energy (Heat) and Mechanical Energy (Work)
Learning Objectives for Section 7.2
• Identify the forms of energy transferred in physical and chemical changes and show how we
can interpret energy conservation in the changes.
• Draw molecular level representations of thermal energy (undirected kinetic energy) and
mechanical energy (directed kinetic energy) transfers.
Investigate This 7.4. What happens when a pinwheel is held over a flame?
Goal:
Observe the motion of a pinwheel placed over a burner with and without a chimney.
Set-up time:
• At least 20 minutes initially. This activity is dependent on type of pinwheel, type of chimney,
and room ventilation. Practice this activity with the pinwheel, heat source, and chimney you
will use before doing it in your class.
Time for activity:
• Less than 10 minutes.
Equipment:
• Bunsen burner or microburner. A candle or an incandescent electric light bulb will work, if
the pinwheel is very lightweight and has little resistance to turning.
• Long metal cylinder. A 3-foot long aluminum conduit tube with a 3.5 cm inner diameter
seems to work with a microburner (see photograph below). A metal tube vacuum cleaner
attachment can be used. A glass cylinder could also be used. Soup cans (or similar shaped
cans) with both ends cut out can be taped together (duct tape) to make a cylinder, but not
everyone has success with this chimney.
• Metal, plastic or paper pinwheel. Commercial toy pinwheels generally are too heavy and exert
too much friction on their bearing surface to turn well when held "face down" over a flame.
Paper pinwheels work well.
NOTE: Here are two recipes for constructing paper pinwheels.
• Construct a simple pinwheel by cutting out a 3-inch circle from a 3” by 5” card, cutting slits
about an inch long along the radii at 60 intervals, bending one edge of each slit into a fin, and
suspending the pinwheel on a length of thread passing through a tiny hole in the center of the
pinwheel and knotted at one end to prevent the thread going through the hole. This is the set
up shown in the photograph accompanying Investigate This 7.4 in the textbook. When doing
the investigation, allow the pinwheel to hang from its support, a ring on a tall ring stand, for
example, until it stops rotating (due to twisting of the thread). When the source of heat is
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ACS Chemistry FROG
Chapter 7
Chemical Energetics: Enthalpy
moved below it, the pinwheel should turn. Without a chimney to direct the flow of hot gases,
the motion will probably be slow, but should speed up when the chimney is in place. The
thread will tend to wind up, so that the motion slows over time and, when the source of heat is
removed, the pinwheel reverses its rotation as the thread unwinds. The candle shown in the
photo can be replaced with a 75- or 100-watt incandescent light bulb as the source of heat for
gases in the air and students might be asked to draw a figure analogous to Figure 7.2 to
explain what is going on in this case.
• Construct a paper pinwheel using these materials: strong construction paper, long dressmaker
pin, penny, and a pencil with an eraser or a thin wood dowel. Cut the paper into a 6" by 4"
rectangle. Draw two diagonal lines, corner to corner. Trace a circle using the penny in the
center where the diagonal lines cross. Remove the penny and then cut along the diagonals
from each corner of the paper to the edge of the circle made with the penny. Fold (without
creasing) alternate corners into the center and fasten together with the pin. Stick the pin firmly
into the top of the dowel or the eraser of the pencil. Bend the pin so the pinwheel is parallel to
the pencil or dowel (acting as a handle). These directions are adapted from Gray, Judith, and
Ellison, Sheila, 365 Days of Creative Play: For Children 2 years and Up, Sourcebooks, Inc.
Procedure:
• Due to the experimental variability of this activity, you should conduct it yourself in class,
after practicing with your apparatus. The procedure described below uses a microburner, a
paper pinwheel, and an aluminum metal conduit tube.
SAFETY NOTE
Wear your safety goggles.
Be careful not to burn yourself or the pinwheel with the
burner flames or other heat sources.
• Set the burner to a low, flickering yellow flame.
• Hold your hand about 25 cm above the flame and note how warm the combustion gases
coming from the flame fee. Keep raising your hand until you cannot feel the heat from the
flame, noting the distance between your hand and the flame. Note: We could feel the heat up
to 2 ft above the microburner.
• Hold the pinwheel about 25 cm above the flame and observe its motion, if any. Try different
orientations and angles to find the one that causes the pinwheel to spin the fastest. Note: Our
paper pinwheel spun slowly at a slight angle about 3 ft above the bench top slightly off center
of the burner.
• Place the metal cylinder over the burner so that it serves as a chimney. Wait a few moments
for the metal to become warm.
• Hold your hand just above the top of the chimney and note how warm the combustion gases
from the flame feel. Move your hand upward until you do not feel any more heat. Compare
your results to holding your hand over the burner at the same distance without the chimney.
Note: We could feel heat 5 feet above the burner.
• Hold the pinwheel just above the top of the chimney and observe its motion, if any. Try
different orientations to find the one that causes the pinwheel to spin the fastest. Note: We
observed fast pinwheel spinning even at 2 feet above this type of chimney. Better results are
obtained with a longer, narrower tube.
• If your pinwheel does not spin, check to make sure there is no friction between the pinwheel
and its axle.
ACS Chemistry FROG
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Chemical Energetics: Enthalpy
Chapter 7
Anticipated results:
• This is a photograph of the set up described above.
Follow-up discussion:
• Use Consider This 7.5 to initiate discussion of this activity.
Follow-up activities:
• Check This 7.6. The difference between warm and cool gases.
• Consider This 7.7. How do thermal and mechanical energies interact?
• End of Chapter problems 7.4 through 7.8.
Consider This 7.5. What causes a pinwheel over a flame to turn?
Goal:
Discussion of the observations from Investigate This 7.4 should lead students to conclude that
the movement of hot gas can be directed (by the chimney) to perform work (cause a pinwheel to
spin).
Classroom options:
• Allow 3-5 minutes for students, working in small groups, to answer these questions. Then, the
groups can share their conclusions with the class, summarizing answers on the chalkboard or
an overhead transparency.
• This activity can be conducted as an open class discussion.
Time for activity:
• Approximately 10-15 minutes.
Instructor notes:
• Use this discussion to introduce students to thermal and mechanical energy as two different
forms of kinetic energy.
• The discussion should also elucidate the function of the chimney in Investigate This 7.4.
Without a chimney, the gases leaving the flame dissipate. Not enough of the hot gas goes
directly upward to spin the pinwheel. With the chimney, a greater upward-directed flow of hot
gas occurs and turns the pinwheel. The temperature difference noted with and without the
chimney is convincing evidence that more hot gas is directed upward with the chimney.
• Discuss the difference between directed and undirected kinetic energy and that both thermal
energy and mechanical energy are forms of kinetic energy.
• Figure 7.2 can be used to help explain the microscopic movement of gases in this activity.
Students should reason and conclude:
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ACS Chemistry FROG
Chapter 7
Chemical Energetics: Enthalpy
(a) A difference in temperature of the combustion gases without and with a chimney is
observed. The combustion gases with the chimney in place felt warmer than at the same
distance above the burner without the chimney. More of the hot gas from the combustion is
directed upward by the chimney. The result is that the gas feels warmer at a given distance
above the flame, because there is more hot gas to transfer energy to your hand.
(b) The pinwheel turned notably faster with the chimney placed over the burner. This must
mean that a larger amount of gas (more molecules) was hitting the pinwheel when the
chimney was in place.
(c) The gas at the outlet of the chimney felt warmer and the pinwheel in this position spun
faster. In part (a) we concluded that more hot gas was reaching our hand when the chimney
was in place to direct the flow. In part (b) we concluded that the pinwheel spun faster above
the chimney because more gas molecules were hitting it. Our correlation between warmer gas
and faster pinwheel spin is consistent with our conclusion (explanation) that more hot gas is
directed upward by the chimney.
Follow-up activities:
• Check This 7.6. The difference between warm and cool gases.
• Consider This 7.7. How do thermal and mechanical energies interact?
• End of chapter problems 7.4 through 7.8.
Consider This 7.7. How do thermal and mechanical energies interact?
Goal:
Conclude that thermal and mechanical energies can be interchanged.
Classroom options:
• Allow 3-4 minutes for students, working in small groups, to answer these questions. Then, the
groups can share their conclusions with the class, summarizing answers on the chalkboard or
an overhead transparency.
Time for activity:
• Approximately 10-15 minutes.
Instructor notes:
• Use the discussion to reinforce the idea that energy is conserved but the form of energy may
change.
• Try to be sure that students link the macroscopic level of a pinwheel spinning with the
microscopic motions of the hot gases that strike it to make it turn.
Students should reason and conclude:
To make the pinwheel spin, the rising gases have to collide with it and give up some of their
energy of motion (kinetic energy) to the pinwheel vanes. Thus, the gas molecules that have
collided with the pinwheel will, on the average, be moving more slowly after the collision.
Slower moving gas has less kinetic energy and, hence, a lower temperature. Collisions with
the pinwheel cool the rising gas more than if the pinwheel was not present.
Follow-up activities:
• End of chapter problems 7.4 through 7.8.
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Section 7.3. Thermal Energy (Heat) Transfer
Learning Objectives for Section 7.3
• Identify the forms of energy transferred in physical and chemical changes and show how we
can interpret energy conservation in the changes.
• Draw molecular level representations of thermal energy (undirected kinetic energy) and
mechanical energy (directed kinetic energy) transfers.
• Identify whether a thermal energy transfer occurs by radiation, conduction, and/or convection.
Investigate This 7.8. Can radiation change the temperature of water?
Goal:
Observe the temperature of water in containers with a bright light shining on them.
Set-up time:
• Approximately 10 minutes, if the light bulb and reflector set up are available.
Time for activity:
• From about 10 minutes to 20-30 minutes. (Some instructors have the class monitor the
temperatures for the entire class session.)
Equipment:
• Bright incandescent bulb (either 100, 150 or 200 watt). The light beam has to be unfiltered, so
that plenty of infrared radiation is included. Light from a slide projector is usually filtered to
prevent overheating the slides, so this is usually not a good source.
• Lamp holder (a lamp kit can be used as shown in the set-up below).
• 2 narrow flat-sided clear, colorless containers filled with room temperature water. Plastic, 75cm2 cell-culture flasks used in microbiology work very well.
• Reflector (one can be made by covering poster board with aluminum foil as shown in the setup below).
• Two standard laboratory thermometers. Digital thermometers or temperature probes interfaced
to a computer with the results projected for the whole class to observe are excellent.
Procedure:
• Conduct this investigation as a class activity, using students to read the temperatures (unless
computer interfaced probes with projected output are used).
SAFETY NOTE
Wear your safety goggles.
• The set-up is shown here:
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• In the set-up shown, the thermometer in the left bottle was placed 7 cm from the surface of the
light bulb and the thermometer in the right bottle 12 cm from the surface of the light bulb.
• Record the temperature of the water in the containers.
• Direct a bright, incandescent light beam through the containers.
• After 10 minutes in the light beam, read and record the temperature in the two containers.
• Remove the container nearer the light source and read and record the temperature in the
remaining container after another 10 minutes.
NOTE: The outcome of this simple activity is not obvious to students who have never
experienced the effect of infrared absorption by water (or never thought about the origin of the
heat absorption by water). The setup is designed to minimize conduction and convection effects
and students are asked to think about this as well.
Anticipated results:
• The results for three different conditions are given here.
100-watt bulb:
Initial reading
10 minutes
Temperature in Nearer Bottle
Temperature in Further Bottle
21.5 C
22 C
25 C
22.5 C
24 C
20 minutes
100-watt bulb;
Temperature in Nearer Bottle
Temperature in Further Bottle
Initial reading
22 C
21.7 C
15 minutes
27 C
23 C
25.5 C
30 minutes
150-watt bulb:
The temperature of the nearer container after 10 minutes is about 10 C warmer than at
the start, but the further container only 2-3 degrees warmer. When the nearer container is
removed, the water in the remaining container will be almost as warm (after 10 more
minutes) as the nearer container before it was removed.
Instructor notes:
• Use the “waiting time” in this activity to discuss temperature, thermal energy, and heat.
• Temperature is a measure of the average kinetic energy of the particles in a substance.
• Thermal energy is the sum of all the kinetic energy of the particles. For a given object or
amount of material, an increase in thermal energy will increase the temperature (because the
average particle is moving faster) and vice versa.
• Heat is a measure of the thermal energy transferred between two objects at different
temperatures (transfer from hot to cold).
Follow-up discussion:
• Use Consider This 7.9 to initiate discussion of the observations in this activity and try to get to
Consider This 7.10 and 7.11 while the experimental set-up is still available for testing.
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Follow-up activities:
• Consider This 7.10. How does radiation transfer energy to water?
• Consider This 7.11. How are conduction and convection the same? Different?
• End of chapter problems 7.9 through 7.11.
Consider This 7.9. How does radiation change the thermal energy of water?
Goal:
Students should conclude that radiation can be absorbed by and change the thermal energy of
water.
Classroom options:
• Allow 3-4 minutes for students, working in small groups, to answer these questions. Then, the
groups can share their conclusions with the class, summarizing answers on the chalkboard or
an overhead transparency.
• This activity can be conducted as an open class discussion.
Time for activity:
• Approximately 10 minutes.
Instructor notes:
• Try to be sure that the class understands the relationship of thermal energy and temperature of
a sample of water as this discussion is carried on.
• Try to get to Consider This 7.10 and 7.11 while the experimental set-up from Investigate This
7.8 is still available for testing.
Students should reason and conclude:
(a) The temperatures in both containers increased. The temperature of the water in the
container nearer the light bulb increased more than the temperature of the water in the one
behind it. Since temperature is a measure of thermal energy and the temperature of the water
in both containers increased, the thermal energy of the water increased in both containers.
The thermal energy in the container nearer the light source increased more, as evidenced by
its greater increase in temperature.
(b) After 10 minutes in the absence of the nearer container, the temperature of the water in
the remaining container increased to almost the same temperature as in the nearer container
when it was removed. Since temperature further increased, we conclude that the thermal
energy of the water has increased (to almost the same amount as in the container that was
removed). If radiation (light energy) is absorbed by water to increase its thermal energy and
hence its temperature, this would explain the increase in temperature of the water in the
containers. Since some of the radiant energy is absorbed by the sample of water in the nearer
container (in order to increase its thermal energy), this energy will be missing from the
radiation that goes into the second container. If there is not as much energy to be absorbed,
the water in the second container will not gain as much thermal energy and its temperature
will not increase as much as that in the first, as we observed. When the nearer container is
removed, all the radiant energy can reach the second container, the water will gain more
energy, and the temperature will increase, as observed. [Since the light beam is not really
focused through the samples, the container farther from the source will not receive quite as
much radiant energy and will not warm quite as much in the same time as the nearer one.]
Follow-up Activities:
• Consider This 7.10. How does radiation transfer energy to water?
• Consider This 7.11. How are conduction and convection the same? different?
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• End of chapter problems 7.9 through 7.11.
Consider This 7.10. How does radiation transfer energy to water?
Goal:
Conclude, based on their observations and data from Investigate This 7.8 and the discussion that
precedes this activity, that infrared radiation is absorbed by water and increases its thermal
energy.
Classroom options:
• This activity can be an extension of the discussion begun with Consider This 7.9.
Time for activity:
• Approximately 10 minutes.
Instructor notes:
• Remind students of the terms used to explain the concepts behind these activities. Chapter 4
may need to be reviewed, especially electromagnetic radiation wavelengths and energy
emission by glowing bodies.
• If possible, introduce conduction and convection as a follow-up to this activity while the setup for Investigate This 7.8 is still available to be analyzed. By all means have some students
try the experiment suggested at the end of part (b) of this activity. This is quite a good
demonstration of the infrared absorbing property of water.
Students should reason and conclude:
(a) Water molecules absorb infrared radiation from the light beam, which causes them to
move faster, that is, have greater thermal energy. We detect this increase as an increase in the
temperature of the water sample. We cannot see any difference in the light beam before and
after it enters the water, because infrared radiation is invisible to our eyes (see Figure 4.13,
page 227).
(b) You would feel a difference in your hand placed in the light beam before and after it
passes through a container of water. When you place your hand in the light beam before it
passed through the container of water, your hand will absorb the infrared radiation and you
will feel your hand getting warmer. When you place your hand in the beam after it passes
through the water, you will notice that your hand gets less warm because the water molecules
in the container will have absorbed much of the infrared radiation from the light beam.
Follow-up discussion:
• Use the discussion of radiation energy transfer to introduce conduction and convection as
modes of energy transfer that require the presence of matter.
Follow-up activities:
• Consider This 7.11. How are conduction and convection the same? Different?
• End of chapter problems 7.9 through 7.11.
Consider This 7.11. How are conduction and convection the same? different?
Goal:
Students should conclude that conduction requires particles/bodies in contact and convection
involves the movement of particles from one place to another, and should apply these ideas to
the Investigate This 7.8 set-up.
Classroom options:
• This activity can be an extension of the discussion begun with Consider This 7.9 and
continued with Consider This 7.10.
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Time for activity:
• Approximately 10-15 minutes.
Instructor notes:
• Show or refer students to Figure 7.3 as this activity is carried out.
Students should reason and conclude:
(a) The fundamental difference between conduction and convection is that the particles
originally energized by the thermal energy source are often the ones we sense in convection.
That is, the particles move from the source to the sensor. But, in conduction, there is a relay
or domino effect from the originally-energized particles through many intermediate particles
to the sensor. The particles do not move from the source to the sensor. The mass of particle
moving in a convection current can, of course, pass on their energy to other particles when
they collide with them, so some of particles that reach the sensor have been energized by
collisions along the way, not directly by the energy source.
(b) There could be some minor conduction and/or convection effects in Investigate This 7.8.
Convection would require motion of the air from the source to the containers of water. Since
the containers are not above the energy source, there is probably not very much convective
heating, as long as the light bulb apparatus is pretty well open so that heated air does not
contact the containers. Air (and other gases) is not a very good conductor of thermal energy,
because the molecules are far apart and transfer of energy between them is not nearly as rapid
as in a liquid or solid. A gentle flow of air between the light and the containers would make
sure that warmed air molecules were swept away before they could reach the containers of
water. If the results are still the same, with this intervention, then we can be more certain that
convection and conduction are not responsible for the observed temperature changes in the
investigation.
Follow-up activities:
• End of chapter problems 7.9 through 7.11.
Section 7.4. State Functions and Path Functions
Learning Objectives for Section 7.4
• Identify the forms of energy transferred in physical and chemical changes and show how
energy is conserved in the changes.
• Define and identify the variables that are functions of state and those that are functions of the
path for a given change.
Investigate This 7.12. What happens when you rub your hands together?
Goal:
Conclude that rubbing your hands together makes the palms feel warm.
Set-up time:
• None.
Time for activity:
• 1-2 minutes.
Procedure:
• Students place their palms of their hands together and rub them briskly for 3-4 seconds.
Follow-up discussion:
• Use Consider This 7.13 to initiate discussion of the results of this activity.
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Follow-up activities:
• Consider This 7.14--How do different pathways for descent of Pike's Peak compare?
• Consider This 7.15--How does the energy of the skin molecules change in Figure 7.5?
• End of Chapter problems -- 7.12-7.16.
Consider This 7.13. What change occurs when you rub your hands together?
Goal:
Students should conclude that mechanical motion can be converted to thermal energy and the
same change in thermal energy can be brought about in another way, thus introducing the
concepts of path function and state functions.
Classroom options:
• Allow 3-4 minutes for students, working in small groups, to answer these questions. Then, the
groups can share their conclusions with the class, summarizing answers on the chalkboard or
an overhead transparency.
• This activity could be conducted as an open class discussion.
Time for activity:
• About 5-10 minutes.
Instructor notes:
• Use this discussion as an introduction to the importance of defining thermodynamic variables
such as temperature, thermal energy, and mechanical energy.
• Students should begin to understand the variables involved with thermodynamics as well as
the difference between state functions and path functions.
Students should reason and conclude:
(a) Because temperature is a measure of the average motion of molecules, the molecules in
your skin must have been made to move faster by rubbing your hands together, since your
hands felt warmer after the rubbing.
(b) Mechanical energy supplied by your muscles to rub your hands together is transferred to
molecules at the surface of the skin (simply by pushing on them), thus increasing the kinetic
energy of the skin molecules, which we sense as an increase in temperature (larger amount of
thermal energy in the skin due to the rubbing).
(c) Yes, you could use a heater, a candle, or a campfire to warm your hands. Also, a pair of
mittens can be worn. Thermal energy that is continuously released by metabolic processes in
our bodies is trapped in the fibers and air pockets of the mittens. Since the mittens are then at
a higher temperature than the surrounding air, less thermal energy is transferred from the skin
(because the transfer depends on the difference in temperature between the object that is
releasing the energy and the body that is receiving it) and the thermal energy builds up in the
skin, making your hands inside the mittens feel warmer.
Follow-up discussion:
• Show or refer students to Figures 7.4 and 7.5 during this discussion, in order to reinforce that
thermal and mechanical energy transfers are path functions.
Follow-up activities:
• Consider This 7.14. How do different pathways for descent of Pike's Peak compare?
• Consider This 7.15. How does the energy of the skin molecules change in Figure 7.5?
• End of chapter problems 7.12 through 7.16.
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Consider This 7.14. How do different pathways for descent of Pike's peak compare?
Goal:
Conclude that the distance traveled on the ground for the same change in elevation is different
for different pathways.
Classroom options:
• Allow 2-3 minutes for students, working in small groups, to answer these questions. Then, the
groups can share their conclusions with the class, summarizing answers on the chalkboard or
an overhead transparency.
• This activity could be conducted as an open class discussion.
• This activity could also be assigned as a homework problem and then discussed at the next
class session.
Time for activity:
• About 5-10 minutes.
Instructor notes:
• Display Figure 7.4 (Two ways to get to the top of Pike's Peak) before conducting this activity.
Students should reason and conclude:
Even though your change in elevation (a state function) is the same by both pathways, your
car's odometer will indicate that you traveled more linear ground distance than your friend,
because you had to use a winding mountain road as compared to the more direct descent via
the cog railway.
Follow-up discussion:
• Use this discussion to help define and distinguish path functions from state functions, linking
the discussion to Investigate This 7.12, Consider This 7.13 and Figure 7.5.
Follow-up activities:
• Consider This 7.15. How does the energy of the skin molecules change in Figure 7.5?
• End of chapter problems 7.12 through 7.16.
Consider This 7.15. How does the energy of the skin molecules change in Figure 7.5?
Goal:
Conclude that different paths for warming hands produce the same energy change in the hands.
Classroom options:
• This activity can be conducted as an open class discussion.
• Allow 2-3 minutes for students, working in small groups, to answer these questions. Then, the
groups can share their conclusions with the class, summarizing answers on the chalkboard or
an overhead transparency.
• This activity could also be assigned as a homework problem and then discussed at the next
class session.
Time for activity:
• Approximately 5-10 minutes.
Instructor notes:
• Show or refer students to Figure 7.5 during the discussion in this activity.
Students should reason and conclude:
Although the paths are different, the energy change for the skin molecules is the same
because the temperature change, a measure of the thermal energy change, is the same.
Follow-up activities:
• End of chapter problems 7.12 through 7.16.
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Section 7.5. System and Surroundings
Learning Objectives for Section 7.5
• Identify the system and the relevant surroundings for a given change.
• Define and identify open, closed, and isolated systems.
Consider This 7.18. How do systems and surroundings interact?
Goal:
For a variety of examples, differentiate between the system and surroundings and determine
whether the systems are open, closed, or isolated.
Classroom options:
• Allow a couple of minutes for students, working in small groups, to answer each part. After
each part, the groups can share their conclusions with the class and at the end of the activity
summarize the answers on the chalkboard or an overhead transparency.
• This activity could be conducted as an open class discussion.
• This activity could also be assigned as a homework problem and then discussed at the next
class section.
Time for activity:
• Approximately 15-20 minutes, depending on the amount of discussion about interpretations of
systems and surroundings.
Instructor notes:
• Show or refer students to Figure 7.6 and Table 7.1 as you begin the discussion of these
examples, so they are reminded of the definitions of system and surroundings and the
characteristics of isolated, closed, and open systems.
• Help students see that there are different ways to look at these examples.
Students should reason and conclude:
(a) The relevant surroundings are the beaker, your hand, and the air surrounding the beaker.
Your hand feels cold because energy is being transferred from your hand to the beaker (as
well as from the warmer air to the cooler beaker) and the beaker is, in turn transferring
energy to the water as the NH4Cl(s) crystals dissolve. The system is closed because no matter
is exchanged with the surroundings.
(b) The relevant surroundings are the walls of the well-insulated container. The walls of the
container can transfer thermal energy to the system and become cooler. The system is closed.
If we assume that the walls of the container cannot transfer energy to or accept energy from
the system, then the system is isolated, because no exchange of matter, heat, or work occurs.
This is the assumption we make in using a simple calorimeter, like the one shown in Figure
7.7, where we assume that the expanded foam container neither gains nor gives up thermal
energy to its contents. This assumption cannot be strictly correct, but it is often good enough
to obtain results good to a few percent.
(c) For this case, we will assume that no energy transfers take place to or from the “wellinsulated container.” The water in the container is the surroundings for this example. The hot
piece of metal transfers thermal energy to the water. This transfer will continue until the
system (metal) and surroundings (water) reach the same temperature. When their
temperatures are the same, no net transfer of energy will occur between them. The system is
a closed system because only thermal energy is exchanged. No matter is exchanged.
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(d) If the water and piece of metal together are the system (and no thermal energy is
transferred to or from the container walls), this is an isolated system, because no matter, heat,
or work enters or leaves this system. The surroundings (rest of the world) do not interact with
the system. The hot piece of metal will still transfer thermal energy to the water until the
metal and water reach the same temperature. When their temperatures are the same, no net
transfer of energy will occur between them. Since no thermal energy has entered or left this
system, the sum of the thermal changes in the system (cooling of the metal and warming of
the water) has to be zero. [This is the way we will analyze our calorimetric measurements.]
(e) The picture here shows figure in the Web Companion, Chapter 7, Section 7.6, page 1,
after the labels have been moved to differentiate the system from the surroundings for this
calorimeter set-up. This figure is very similar to what is shown in Figure 7.7 in the textbook,
with the exception that no stirrer is shown here and a syringe is not used to add a reagent in
the textbook figure.
[Note that the differentiation between system and surroundings in this figure is like the
differentiation in part (c) above. Alternatively, the solvent water could be taken to be part of
the system, as in part (d) above. That is the way we will treat the calorimeter in Section 7.6.]
Follow-up discussion:
• Discuss important points in the "Reflection and Projection” and use the discussion of part (e)
to lead into Section 7.6 on calorimetry.
Follow-up activities:
• End of chapter problems 7.17 through 7.21.
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Section 7.6. Calorimetry and Introduction to Enthalpy
Learning Objectives for Section 7.6
• Use the data from calorimetric measurements to calculate the thermal energy transferred to or
from a reaction system.
• Use the data from constant pressure calorimetric measurements to calculate the enthalpy
change for the reacting system.
Investigate This 7.19. What happens when an acid and base are mixed?
Goal:
• Observe that thermal energy is transferred from the reaction system to the surroundings when
an acid and a base are mixed.
Set-up time:
• 5-10 minutes, if the solutions are already available.
Time for activity:
• Less than 5 minutes.
Equipment:
• Small test tubes.
• Long-stem plastic pipets.
Chemicals:
• 1 M NaOH (4.0 g per 100 mL of water). Dispense in properly labeled plastic pipets.
• 1 M HCl (8.5 mL concentrated HCl in 100 mL solution). Dispense in properly labeled plastic
pipets.
Procedure:
• If your classroom situation is not safe for distributing the reagents to groups, ask for a few
volunteers to do the activity and report the results to the rest of the class.
SAFETY NOTE
Wear safety goggles.
• Have each group/student add about 1 mL of the NaOH solution to the test tube and then about
1 mL of the HCl solution.
• Gently touch the test tube near the bottom and report/record observations.
Alternative procedures:
• Place 0.5 mL of NaOH on a liquid crystal thermometer at the position of the green stripe. Add
0.5 mL of HCl directly on the puddle of NaOH solution. Photographs of the results are:
base on thermometer
acid + base on thermometer
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NOTE: The color stripe on a liquid crystal thermometer is green when the temperature of the
surroundings matches the temperature printed on that stripe. If the temperature is more than
about 3-5 C hotter or colder than the temperature printed on a stripe, the stripe looks black
(because it is clear and the backing color shows through). If the stripe senses a temperature that
is just a little warmer than the printed temperature, it appears blue. If the stripe senses a
temperature that is just a little cooler than printed temperature, it appears red. In the example
shown in the photographs, the base appears to have a temperature of about toward the cooler
end of the 24-27 C range. When the acid is added the temperature of the liquid increases to
about the middle of the 27-29 C range. Thus, there is an increase in temperature of about 4 C.
• A thermometer, liquid in glass or digital, could also be used to semi-quantify the temperature
change from the heat of reaction.
• An extension or alternative activity might be to measure the temperature change when
ammonium nitrate or calcium chloride is added to water in a Styrofoam® cup, thus linking
back to Investigate This 2.22 in Chapter 2, Section 2.5. The measured temperature changes
could then be used for the calculations in this section, in place of the ones in the textbook. It is
not a good idea to replace Investigate This 7.22, if you intend to do the extension of the ureadissolution activity in Chapter 9, Section 9.7.
Anticipated results:
• The reaction between the NaOH and HCl solutions is exothermic, so the test tube should feel
warm (as we also see in the liquid crystal thermometer result above).
Follow-up discussion:
• Use Consider This 7.20 to initiate discussion of this activity and Check This 7.21 as an
assignment to check understanding.
Follow-up activities:
• Investigate This 7.22. What happens when urea dissolves in water?
• End of chapter problems 7.22 through 7.27.
Consider This 7.20. What causes the changes when an acid and base are mixed?
Goal:
Conclude that the familiar (form Chapter 6) reaction of an acid and a base releases thermal
energy to the surroundings.
Classroom options:
• This activity can be conducted as an open class discussion as Investigate This 7.19 is being
carried out.
Time for activity:
• Approximately 5-10 minutes.
Instructor notes:
Students should reason and conclude:
The test tube felt warm after the acid was mixed into the base. Thermal energy is transferred
from the contents of the test tube to the surroundings, including fingers touching the test
tube. We recall from Chapter 6 that hydrochloric acid solution contains hydronium and
chloride ions and sodium hydroxide solution contains sodium and hydroxide ions. We know
that hydronium ions and hydroxide ions combine with one another to form water, so it seems
reasonable to conclude that this combination is the reaction that occurs when the solutions
are mixed and our results suggest that energy is released by the reaction. The energy
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increases the temperature of the solution and we detect the increased temperature because
thermal energy is transferred to our skin.
Follow-up discussion:
• Discussion should help students reinforce or gain understanding of the difference between
systems and surroundings, thermal energy transfer, and the definition of enthalpy.
• This is a good opportunity to review endothermic and exothermic changes.
Follow-up activities:
• Check This 7.21. Is the reaction of an acid with a base exothermic or endothermic?
• Investigate This 7.22. What happens when urea dissolves in water?
• End of chapter problems 7.22 through 7.27.
Investigate This 7.22. What happens when urea dissolves is water?
Goal:
Determine the temperature change (decrease) when urea is dissolved in water.
Set-up time:
• 5-10 minutes.
Time for activity:
• 5-10 minutes.
Equipment:
• Constant pressure calorimeter, shown in Figure 7.7. Good results can be obtained without a lid
or stirrer using the thermometer or a computer-interfaced temperature probe to stir the
contents of the Styrofoam® cup.
• 100-mL graduated cylinder to measure the water.
Chemicals:
• 6.0 g of urea, H2NC(O)NH2(s).
Procedure:
• Do this as a class investigation with volunteers to carry out the activity and have students
work in small groups to discuss and analyze the results.
SAFETY NOTE
Wear your safety goggles
• Set up a simple constant-pressure calorimeter, like the one illustrated in Figure 7.7 (or the
simpler set-up suggested above), containing 100. mL of room temperature water.
• While gently stirring the water, record its temperature every 15 seconds for 1-2 minutes.
• Add 6.0 g of urea all at once to the water and stir vigorously to dissolve the urea as quickly as
possible, while continuing to record the temperature of the solution for another 3 minutes.
Anticipated results:
• In numerous trials of this activity, using a computer interfaced temperature probe, the
temperature drops 3.5-3.6 ºC.
Follow-up discussion:
• Use Consider This 7.23 to initiate discussion of the results of this activity as it is being carried
out.
Follow-up activities:
• Consider This 7.24. What is observed for an endothermic reaction in a calorimeter?
• Check This 7.25. Energy transfers in a calorimeter.
• Worked Example 7.26. Determination of qP(reaction) for a reaction in a calorimeter.
• Check This 7.27. Determination of qP(reaction) for dissolution of urea in a calorimeter.
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• Worked Example 7.28. Determination of Hreaction for a reaction in a calorimeter.
• Check This 7.29. Determination of Hreaction for dissolution of urea in a calorimeter.
• End of chapter problems 7.22 through 7.27.
Consider This 7.23. What is T when urea dissolves in water?
Goal:
Determine T for the solution when urea dissolves in water in a Styrofoam® cup constantpressure calorimeter.
Classroom options:
• Conduct this activity as an open class discussion as Investigate This 7.22 is done.
Time for activity:
• Approximately 2 minutes in addition to Investigate This 7.22 for the class to agree on the
value for the temperature change.
Instructor notes:
• Use the agreed upon results for Investigate This 7.22 to determine T.
Students should reason and conclude:
T = Tf – TI = –3.5 to –3.6 ºC.
Follow-up activities:
• Consider This 7.24. What is observed for an endothermic reaction in a calorimeter?
• Check This 7.25. Energy transfers in a calorimeter.
• Worked Example 7.26. Determination of qP(reaction) for a reaction in a calorimeter.
• Check This 7.27. Determination of qP(reaction) for dissolution of urea in a calorimeter.
• Worked Example 7.28. Determination of Hreaction for a reaction in a calorimeter.
• Check This 7.29. Determination of Hreaction for dissolution of urea in a calorimeter.
• End of chapter problems 7.22 through 7.27.
Consider This 7.24. What is observed for an endothermic reaction in a calorimeter?
Goal:
Treating the contents of an ideal calorimeter as an isolated system, give the signs for qP(reaction)
and qP(water) for an endothermic reaction in the calorimeter.
Classroom options:
• Allow 2-3 minutes for students, working in small groups, to answer these questions. Then, the
groups can share their conclusions with the class, summarizing answers on the chalkboard or
an overhead transparency.
• This activity could be conducted as an open class discussion.
Time for activity:
• Approximately 10 minutes.
Instructor notes:
• Introduce the constant-pressure calorimeter, using Figure 7.7 and/or the figure in the Web
Companion, Chapter 7, Section 7.6, page 1, as an example.
• Discuss the concept of an ideal constant-pressure calorimeter and introduce the relationship
that is at the heart of our approach, equation (7.3): qP(reaction) + qP(solution) = 0.
NOTE: You (and some of your students) may be more familiar with a different approach to
ideal adiabatic calorimetry in which the reaction is taken as the system and the solution in which
it occurs is the surroundings. The problem that often arises for students in this approach is what
signs to assign to the thermal energy transfers (heats) from or to the system and surroundings. In
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the approach we have taken in this text, the calorimeter contents are regarded as an isolated
system, which transfers no energy to nor receives energy from the surroundings (including in
this ideal case, the calorimeter components themselves). Thus, the sum of all energy transfers
within the calorimeter must be zero. The observable variable in this system is the temperature of
the solution. The direction of the temperature change immediately provides the sign of
qP(solution), and, therefore, from equation (7.3), the sign of qP(reaction). This approach seems
effective in reducing amount of sign confusion in calorimetric calculations.
Students should reason and conclude:
(a) An endothermic reaction is one in which thermal energy is transferred to the reaction, so
qP(reaction) is positive.
(b) Since qP(reaction) is positive and qP(reaction) + qP(solution) = 0, qP(solution) must be negative, that is
thermal energy will leave the solution. This means that the solution will become cooler.
(c) The dissolution of urea in water is endothermic. The decrease in temperature observed in
Investigate This 7.22 is the evidence that the dissolution of urea is endothermic.
Follow-up discussion:
• Review specific heat (Chapter 1, Worked Example 1.57, page 55) with respect to linking
thermal energy change to temperature change.
Follow-up activities:
• Check This 7.25. Energy transfers in a calorimeter.
• Worked Example 7.26. Determination of qP(reaction) for a reaction in a calorimeter.
• Check This 7.27. Determination of qP(reaction) for dissolution of urea in a calorimeter.
• Worked Example 7.28. Determination of Hreaction for a reaction in a calorimeter.
• Check This 7.29. Determination of Hreaction for dissolution of urea in a calorimeter.
• Consider This 7.30. How can you get more accurate values for Hreaction?
• Check This 7.31. Effect of increasing the amount of another calorimetric reaction.
• End of chapter problems 7.22 through 7.27.
Consider This 7.30. How can you get more accurate values for Hreaction?
Goal:
Students conclude that doubling the amount of reaction coupled with doubling the heat capacity
of the calorimeter contents gives the same temperature change and same precision for H(reaction)
as in the original trial.
Time for activity:
• 10-15 minutes.
Classroom options:
• Allow 2-3 minutes for students, working in small groups, to answer these questions. Then, the
groups can share their conclusions with the class, summarizing answers on the chalkboard or
an overhead transparency.
• This activity could be conducted as an open class discussion.
• This activity could also be assigned as a homework problem and then discussed at the next
class section.
Instructor notes:
• All the reasoning in this activity is based upon experimental values, not on any table values.
• It is likely for students to reason that doubling the amount of reaction will double the energy
released by the reaction and hence should double the temperature change of the solution. The
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“experimental” evidence contradicting this conclusion for this reaction system is designed to
produce the sort of cognitive dissonance that will put students in a receptive mode to learn
why their reasoning was wrong.
Students should reason and conclude:
(a) As indicated above, the result (same temperature change with double the amount of
reactants) is likely not to be what is expected.
(b) Doubling the amount of each reactant volume doubles both the energy released and the
heat capacity of the system. This leads to:
qP(reaction) = (m)(c)(T) = (400. g)(4.18 J·g–1·ºC–1)(0.66 ºC–1) = 1.10  103 J
qP(reaction) = –qP(solution) = –1.10  103 J
The acetic acid, (0.200 L)(0.105 M) = 0.0210 mol, is still the limiting reactant
3
–1.10  10 J
Hreaction =
= –52 kJ·mol–1
0.0210 mol
This result is given to two significant figures because the energy value has an uncertainty of
about 2% (from the uncertainty in T), so Hreaction will have an uncertainty of about 2%.
This result (within the uncertainty) is the same as in Worked Example 7.28. There has been
no improvement in the precision of the result, because the temperature change is the same
and it is the limiting factor on the uncertainty in Hreactio.
(c) The example and questions in the Web Companion Chapter 7, Section 7.6, page 3, are
directly related to parts (a) and (b). Scaling the calorimeter up by a factor of three, as
suggested in the Companion, would result in the same temperature change although the
thermal energy transferred would increase by three. Thus, Hreaction would remain the same,
as it does in our example of a doubling of the scale of the experiment.
Follow-up discussion:
• Discuss how to improve calorimetric measurements, such as making more precise temperature
measurements or changing the amount of reaction without changing the heat capacity, as
exemplified in Check This 7.31.
Follow-up activities:
• Check This 7.31. Effect of increasing the amount of another calorimetric reaction.
• End of chapter problems 7.22 through 7.27.
Section 7.7. Bond Enthalpies
Learning Objectives for Section 7.7
• Define and give examples of homolytic bond cleavage reactions.
• Write equations for homolytic bond cleavage and homolytic bond formation for compounds,
use bond enthalpies to calculate the enthalpy changes associated with these processes, and
obtain the enthalpy change for gas phase reactions.
• Draw enthalpy level diagrams that show how bond enthalpies combine to give the enthalpy
change for a reaction.
• Use bond enthalpies to predict whether, for given atoms, reactions will favor singly-bonded or
multiply-bonded products.
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Investigate This 7.32. What happens when yeast is added to hydrogen peroxide?
Goal:
Conclude that a gas (or gases) is produced in the exothermic (graduated cylinder gets warm)
reaction when yeast is added to hydrogen peroxide.
Set-up time:
• 10 minutes if all equipment and chemicals are already prepared.
Time for activity:
• Approximately 10 minutes.
Equipment:
• 100 mL graduated cylinder.
Chemicals:
• 10 mL 3% hydrogen peroxide solution, H2O2.
• 1-2 mL liquid dishwashing detergent.
• Approximately 2.0 g dried yeast. Instead of yeast, a small piece of liver or other fresh red
meat can be used.
Procedure:
• This activity should be done by the instructor.
SAFETY NOTE
Wear your safety goggles.
• Add hydrogen peroxide to the graduated cylinder followed by dishwashing detergent. Gently
swirl contents without creating suds.
• Carefully add yeast to the solution. Try not to spill yeast granules on the sides of the cylinder.
Gently swirl the contents.
• Feel the cylinder to determine if the cylinder feels warm, cool, or has not changed in
temperature.
Anticipated results:
• Gas is produced which gradually fills the cylinder with soap-bubble foam. The detergent is
added so the bubbles will remain while Consider This 7.33 and Check This 7.34 are
completed before conducting Investigate This 7.35 to identify the gas.
• The cylinder gets warm at the bottom where the reaction occurs, indicating that the reaction is
exothermic.
• This photograph shows that the cylinder essentially fills with gas (foam) from the reaction.
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Follow-up discussion:
• Use Consider This 7.33 to initiate discussion about what reaction occurred and how to test for
the products of the reaction.
• Students need to understand that chemical reactions involve breaking and making bonds.
Thus, energy is involved. In this example, the decomposition of hydrogen peroxide is slightly
exothermic. This means that the stronger bonds of the products replace the weaker bonds of
the reactants.
• Figure 7.8 (Bonding rearrangements for possible reactions of hydrogen peroxide) can be
displayed to discuss the possible reaction scenarios for the decomposition of hydrogen
peroxide.
• Discuss how we can test for the possible products, leading students to observe Investigate
This 7.35.
Follow-up activities:
• Check This 7.34. Properties of the possible H2O2 reaction products.
• Investigate This 7.35. What is the gas formed in the hydrogen peroxide reaction?
• Consider This 7.36. Which bonds break and form in the hydrogen peroxide reaction?
Consider This 7.33. What reaction occurs when yeast is added to hydrogen peroxide?
Goal:
Provide reasonable suggestions, based on the observations from Investigate This 7.32, about the
possible products of the reaction and ways to test these hypotheses.
Classroom options:
• Allow 2-3 minutes for students, working in small groups, to answer these questions. Then, the
groups can share their conclusions with the class, summarizing answers on the chalkboard or
an overhead transparency.
• This activity could be conducted as an open class discussion.
Time for activity:
• Approximately 10-20 minutes, depending on how much of Check This 7.34 you wish to
include in the discussion.
Instructor notes:
• You may need to introduce this discussion by pointing out that hydrogen peroxide is produced
in most aerobic organisms and can harm the organism, if it is not decomposed quickly.
Therefore, organisms, including yeast, contain enzymes (proteins) specifically designed to
decompose hydrogen peroxide. If you wish to demonstrate the presence of these enzymes in
other organisms, try small pieces of horseradish root, a higher organism than yeast. [Pieces of
red meat contain hemoglobin from the blood and the iron in this protein (see Figure 6.10) also
catalyzes the peroxide decomposition, as we will further explore in Investigate This 11.6,
Chapter 11, Section 11.1.] The bottom line here is that the product(s) are formed solely from
hydrogen peroxide as it decomposes in the presence of the yeast enzymes.
Students should reason and conclude:
• We know a reaction occurred when yeast was added to hydrogen peroxide because bubbles of
gas formed and the graduated cylinder felt slightly warm to the touch.
• Since only H and O atoms are available from H2O2, reasonable suppositions about the gas or
gases produced are H2 and/or O2. Molecular level models of formation of these gases are
shown in Figure 7.8 and symbolic equations for the reactions are given in the figure and in
equations (7.6) and (7.7).
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• Students may know from previous study or can be prompted to figure out how to test for these
products, knowing that hydrogen burns in air (or explodes if it is mixed with air or oxygen)
and oxygen supports combustion. Figuring this out is the gist of Check This 7.34 and
Investigate This 7.35 involves carrying out the test.
Follow-up discussion:
• Show or refer to Figure 7.8 to illustrate how the bonds might be broken and formed in the
decomposition reaction of hydrogen peroxide.
Follow-up activities:
• Check This 7.34. Properties of the possible H2O2 reaction products.
• Investigate This 7.35. What is the gas formed in the hydrogen peroxide reaction?
Investigate This 7.35. What is the gas formed in the hydrogen peroxide reaction?
Goal:
Students conclude, based on the results of a glowing splint test, that oxygen gas is the product of
the hydrogen peroxide decomposition reaction.
Set-up time:
• A couple of minutes to arrange the safety shields for the test.
Time for activity:
• 3-5 minutes, including discussion of the outcome of the test.
Materials:
• Foam in graduated cylinder from Investigate This 7.32.
• Long fireplace match or wooden splint.
• Stirring rod.
Procedure:
• After discussing the background for testing for hydrogen gas or oxygen gas (covered in Check
This 7.34), test the flammability of the gas trapped in the foam from Investigate This 7.32.
This test should be carried out by the instructor following all safety precautions.
SAFETY NOTES
Wear your safety goggles.
Place the graduated cylinder behind a safety shield that
protects others as well as yourself when the gas is tested.
• Very gently stir the foam to break up some of the bubbles, while keeping the gas in the
cylinder.
• Light a long fireplace match or wooden splint.
• Blow it out and quickly insert the glowing tip a few centimeters into the mouth of the
cylinder.
• Observe and record the results.
Anticipated results:
• The product is oxygen gas alone. Depending on how the glowing match is inserted in the
graduated cylinder, it might glow more brightly or reignite in the oxygen atmosphere, but it
could be wet from the soap bubbles and simply die out, which is why you try to get rid of as
much soap film as possible before the test. In any case, there is no explosion, so no
hydrogen/oxygen mixture has been produced. Here is a photograph of the result (with no
safety shields, because it was done with knowledge of the outcome and not in a classroom).
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Clean-up:
• After confirming that it is completely extinguished, the match can be thrown away. Rinse the
graduated cylinder with copious amounts of water.
Follow-up discussion:
• Based upon the results of this activity, the correct reaction for the decomposition of hydrogen
peroxide is shown in Figure 7.8(a) and equation (7.6).
• The glowing match glows more brightly or bursts into flame, indicating the presence of
oxygen gas. There is no "pop" (small explosion), so hydrogen gas is not present in the gas.
• Use Consider This 7.36 to initiate discussion of which bonds are broken and which bonds are
formed in the reaction.
Follow-up activities:
• Check This 7.37. Overall homolytic bond cleavage reaction of methane.
• End of chapter problems 7.28 through 7.37.
Consider This 7.36. Which bonds break and form in the hydrogen peroxide reaction?
Goal:
Find, by modeling reactions (7.6) and (7.7), that the same number of two-electron bonds is
formed as is broken in each reaction.
Classroom options:
• The activity could be conducted as an open class discussion with students building and
rearranging their models individually.
• This activity could also be assigned as a homework problem and then discussed at the next
class section.
Time for activity:
• 10-15 minutes.
Instructor notes:
• Show or refer students to reactions (7.6) and (7.7) as the activity is conducted.
• Make classroom-size models of the reactants and products before the class session, as a
possible aid in the discussion.
Students should reason and conclude:
(a) For the reactants in reaction (7.6), three bonds must be broken: the two H–O bonds in one
of the H2O2 molecules and the O–O bond in the other. For the products, three bonds are
formed: two H–O bonds (one each between a free H and the O in an OH to form H2O) and
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one O–O bond to make the second bond between the O–O left after the O–H bonds were
broken. There are six two-electron bonds (represented by the plastic connectors) in both the
reactants and the products.
(b) For the reactants in reaction (7.7), two bonds are broken: both H–O bonds. For the
products, two bonds are formed: one bond between the free Hs to form H2 and one O–O bond
to make the second bond between the O–O left after the O–H bonds were broken.. There are
three two-electron bonds (represented by the plastic connectors) in both the reactants and the
products.
Follow-up discussion:
• Use the discussion to introduce bond enthalpy and homolytic bond cleavage.
Follow-up activities:
• Check This 7.37. Overall homolytic bond cleavage reaction of methane.
• Consider This 7.38. How do bond enthalpies for single and multiple bonds compare?
• Check This 7.39. Formation of polyethylene.
• Check This 7.40. Sulfur-sulfur bonds.
• Worked Example 7.41. Reaction enthalpy from average bond enthalpies.
• Check This 7.42. Reaction enthalpy from average bond enthalpies.
• Check This 7.43. The hydrogen peroxide decomposition reaction.
• End of chapter problems 7.28 through 7.37.
Consider This 7.38. How do bond enthalpies for single and multiple bonds compare?
Goal:
Conclude that there are two patterns of relative strengths of multiple compared to single bonds.
Classroom options:
• Allow 2-3 minutes for students, working in small groups, to answer these questions. Then, the
groups can share their conclusions with the class, summarizing answers on the chalkboard or
an overhead transparency.
• This activity could be conducted as an open class discussion.
Time for activity:
• Approximately 10 minutes.
Instructor notes:
• Remind students that only elements for which multiple bond data are given in Table 7.3 (that
is, CC, NN, OO, and SS) can be analyzed for this activity.
• Table 7.4 and Figure 7.9 display the patterns to be found in this activity. Show or refer
students to Figure 7.9 after they have discussed their results and see how the pattern(s) they
found compare.
Students should reason and conclude:
(a) There are two patterns of relative strengths of multiple bonds compare to single bonds in
the data in Table 7.3. For carbon and sulfur, the average bond enthalpies of multiple bonds
are less than the sum of the average bond enthalpies of single bonds with an equivalent
number of bonding electrons. For example, a S=S bond with four bonding electrons has an
average bond enthalpy of only 352 kJ·mol–1, but two S–S bonds with a total of four bonding
electrons have a total bond enthalpy of 2  268 kJ·mol–1 = 536 kJ·mol–1. The double bond is
much weaker than two single bonds. For nitrogen and oxygen, the opposite pattern is
observed, the average bond enthalpies of multiple bonds are greater than the sum of the
average bond enthalpies of single bonds with an equivalent number of bonding electrons.
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(b) Two patterns emerge, as indicated in part (a). These patterns are given quantitatively for
carbon, nitrogen, and oxygen in Table 7.4 and illustrated graphically in Figure 7.9.
Follow-up activities:
• Check This 7.39. Formation of polyethylene.
• Check This 7.40. Sulfur-sulfur bonds.
• Worked Example 7.41. Reaction enthalpy from average bond enthalpies.
• Check This 7.42. Reaction enthalpy from average bond enthalpies.
• End of chapter problems 7.28 through 7.37.
Section 7.8. Standard Enthalpies of Formation
Learning Objectives for Section 7.8
• Define standard states for elements and compounds and write the equations whose enthalpy
changes are the standard enthalpies of formation of the compounds.
• Use standard enthalpies of formation to calculate the standard enthalpy change for a reaction.
• Draw enthalpy level diagrams that show how standard enthalpies of formation combine to
give the standard enthalpy change for a reaction.
Section 7.9. Harnessing Energy in Living Systems
Learning Objectives for Section 7.9
• Define coupled reactions and identify examples based on the definition.
• Show whether coupling between two reactions would be an energetically favorable
combination.
Section 7.10. Pressure-Volume Work, Internal Energy and Enthalpy
Learning Objectives for Section 7.10
• Identify the forms of energy transferred in physical and chemical changes and show how
energy is conserved in the changes.
• Draw molecular level representations of thermal energy (undirected kinetic energy) and
mechanical energy (directed kinetic energy) transfers.
• Define and identify the variables that are functions of state and those that are functions of the
path for a given change.
• State the first law of thermodynamics in terms of internal energy, heat, and work and use it to
analyze a change that occurs by different pathways.
• Use pressure and volume data or the ideal gas equation to calculate the pressure-volume work
done in a reaction that involves gases.
Consider This 7.55 How is the first law of thermodynamics used?
Goal:
Use the first law of thermodynamics to analyze the operation of a small engine.
Classroom options:
• Allow 2-3 minutes for students, working in small groups, to work on this activity. Then, the
groups can share their conclusions with the class, summarizing answers on the chalkboard or
an overhead transparency.
• This activity could be conducted as an open class discussion.
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• This activity could also be assigned as a homework problem and then discussed at the next
class section.
Time for activity:
• From 5 to 15 minutes, depending on the depth of the discussion of the first law and the
relationship of the state function, E, and the path dependent variables, q and w.
Instructor notes:
•
E
is the same for the original and the redesigned engine because both engines are working
between the same initial and final states.
Students should reason and conclude:
For the original engine, heat and work are produced in the surroundings and we have q = –
5 kJ (per power stroke) and w = –2 kJ (per power stroke). Thus E = q + w = –7 kJ (per
power stroke).
For the redesigned engine, more work is performed, w = –3 kJ (per power stroke) but E is
unchanged, because the redesigned engine operates between the same initial and final states
as the original engine and E is a function of state (dependent only on the initial and final
state of the system -- the engine). Therefore,
E = –7 kJ (per power stroke) = q + w
q = E – w
q = [–7 kJ (per power stroke)] – [ –3 kJ (per power stroke)] = –4 kJ (per power stroke).
The same amount of energy is released in the redesigned engine. More work is done, but less
heat is released. Overall, the redesigned engine is more efficient (more of the energy is
released as work to do useful things in the surroundings).
Follow-up discussion:
• Students should be able to explain why the redesigned engine is more efficient and why E is
the same for both engines.
Follow-up activities:
• Worked Example 7.56. Pressure-volume work at constant pressure.
• Check This 7.57. Pressure-volume at constant pressure.
• Consider This 7.58. How can E (or H) be the same for a reaction at constant volume and
constant pressure
• End of Chapter problems 7.50 through 7.58.
Consider This 7.58. How can E (or H) be the same for a reaction at constant volume
and constant pressure?
Goal:
Students show how q and w combine to give the same E (or H) for constant volume and
constant pressure reactions.
Classroom options:
• Allow 3-5 minutes for students, working in small groups, to work on this activity. Then, the
groups can share their conclusions with the class, summarizing answers on the chalkboard or
an overhead transparency.
• This activity could be conducted as an open class discussion.
Instructor notes:
• As you introduce this activity, remind students of the sign conventions for q and w.
Students should reason and conclude:
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(a) The reaction is exothermic, that is, thermal energy leaves the reacting system, so thermal
energy transfer to the system is negative (< 0).
(b) More thermal energy enters the solution when the reaction is run at constant volume (the
temperature change in the solution is larger) than when run at constant pressure. Thus, more
thermal energy leaves the reacting system at constant volume than at constant pressure. We
can write this relationship as |qV| > |qP|. Or we can say that, if qP < 0, then qV << 0, that is, qV
is more negative than qP or, conversely, qP is more positive than qV.
(c) Since a gas is produced by the reaction, the change in volume at constant pressure is
positive, V > 0. Thus, the work done on the system, w = –PV, is negative, w < 0.
(d) Under constant volume conditions, E = qV. Under constant pressure conditions,
E = qP + w. If E is the same for both reaction conditions, then we must have qV = qP + w.
If we substitute the “numerical values” in this equation, we get: (<< 0) = (< 0) + (< 0). This
equation can be satisfied because two negative quantities, qP and w, combine additively to
give a larger negative quantity. Thus, E can be (and is) the same for both reaction
conditions. The same sort of analysis is possible for H, which can be written H = qP or
H = E – w = qV – w. You can show how H can be a constant.
Follow-up activities:
• End of Chapter problems 7.57 through 7.63.
Section 7.11. What Enthalpy Doesn't Tell Us
Learning Objectives for Section 7.11
• Determine whether a process is consistent with (allowed) by the first law of thermodynamics.
Section 7.13. Extension: Ideal Gases and Thermodynamics
Learning Objectives for Section 7.13
• Use the kinetic-molecular model of gases to explain the observed effects of changes in P, V, n,
or T for gases.
• Calculate the final value of P, V, n, or T for an ideal gas, given their initial values and the
changes in three of the variables.
• Use the ideal gas equation to calculate the P-V work done on or by a chemical reaction
system.
Investigate This 7.63. What happens when a gas is compressed or heated?
Goal:
Observe what happens when a sample of gas is compressed or heated.
Set-up time:
• 15 minutes.
Time for activity:
• 10-15 minutes.
Materials:
• 50-mL plastic syringe.
• U-tube made of plastic tubing (see set-up in text). A completely adequate U-tube manometer
is readily made from about 1.5 m of clear plastic tubing such as Tygon®. The tubing can be
clamped into a U shape or taped in a U shape to an appropriate length of 1”-by-4” lumber.
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About 20-30 cm of one end of the tubing should not be part of the U, but left, as in the sketch,
for attaching the sample syringe.
NOTE: It is, of course, possible to use pressure transducers interfaced to a computer to do this
investigation and project the pressures. However, the action of the gas in pushing down the
liquid in the manometer is then lost and the impact of the investigation is probably diminished.
Chemicals:
• Food coloring.
• Water.
Procedure:
• This is a class activity that can be done with student volunteers.
SAFET NOTE
Wear your safety goggles.
• Draw enough colored water (red is usually best for visibility) into the tube to fill the U to
about the half-way point in each arm.
• When the syringe is attached, the equal pressure in the arms will probably be upset, so you
have to manipulate the plunger to bring the pressure on the syringe side to atmospheric.
• To avoid making a fountain, take care not to push the plunger in too rapidly or too far when
carrying out the investigation.
• Push the syringe plunger in to decrease the volume of air in the syringe by several milliliters.
Record your observations on the liquid levels in the U-tube.
• Readjust the syringe plunger until the U-tube liquid levels are the same again. While holding
the volume constant, warm the syringe by holding it in your hand or directing a warm stream
of air onto it. Record your observations on the liquid levels in the U-tube.
Clean-up:
• Rinse the tube with water, so as not to permanently discolor it and make it harder to use again
for this activity.
Follow-up discussion:
• Use Consider This 7.64 to initiate discussion, which can be begun while the activity is carried
out.
Follow-up activities:
• Consider This 7.65. How are manometer readings interpreted and quantified?
• Consider This 7.66. How does kinetic-molecular theory apply to Investigate this 7.63?
• Check This 7.67. Relationships among the properties of a gas sample.
• Worked Example 7.68. Using the ideal-gas equation.
• Check This 7.69. Using the ideal-gas equation.
• Worked Example 7.70. Using the ideal-gas equation.
• Worked Example 7.71. Using the ideal-gas equation.
• End of chapter problems 7.63 through 7.70.
Consider This 7.64. What causes changes when a gas is compressed or heated?
Goal:
Connect changes in fluid levels in the manometer in Investigate This 7.63 to changes in the
pressure of the gas sample trapped on one side of the manometer and suggest a molecular level
explanation for why compression or warming of the gas changes the pressure in the observed
direction.
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Classroom options:
• Allow 3-5 minutes for students, working in small groups, to work on this activity. Then, the
groups can share their conclusions with the class, summarizing answers on the chalkboard or
an overhead transparency.
• This activity could be conducted as an open class discussion.
Instructor notes:
• Be sure the class agrees on the observations from Investigate This 7.63 as they are discussed.
Students should reason and conclude:
(a) The liquid level in the sample (syringe) arm of the manometer moves down when plunger
is pushed in decreasing the volume of gas in the syringe. This must mean that more pressure
is being exerted on the surface of the liquid in this arm, compared to the pressure before the
gas volume was decreased. When the plunger is pushed in, the volume available to the gas
decreases, so there are more molecules per unit volume. The number of molecules colliding
with the walls, including the surface of the manometer liquid, is increased and, hence, the
pressure is increased.
(b) The liquid level in the sample (syringe) arm of the manometer moves down when the gas
is warmed while keeping its volume the same. This must mean that more pressure is being
exerted on the surface of the liquid in this arm, compared to the pressure before the gas was
warmed. When the gas in the syringe is warmed, the average kinetic energy of the molecules
increases and the average speed with which they collide with the walls, including the surface
of the manometer liquid is increased, and, hence, the pressure is increased. The liquid moves
down in the manometer arm.
Follow-up discussion:
• Show or refer students to Figure 7.21 to discuss how a manometer works as this activity is
being carried out. This discussion should lead into Consider This 7.65 to see how manometer
readings are interpreted and quantified.
Follow-up activities:
• Consider This 7.65. How are manometer readings interpreted and quantified?
• Consider This 7.66. How does kinetic-molecular theory apply to Investigate this 7.63?
• Check This 7.67. Relationships among the properties of a gas sample.
• Worked Example 7.68. Using the ideal-gas equation.
• Check This 7.69. Using the ideal-gas equation.
• Worked Example 7.70. Using the ideal-gas equation.
• Worked Example 7.71. Using the ideal-gas equation.
• End of chapter problems 7.63 through 7.70.
Consider This 7.65. How are manometer readings interpreted and quantified?
Goal:
Calculate the pressure represented by the reading of a water manometer.
Classroom options:
• Allow 3-5 minutes for students, working in small groups, to work on this activity. Then, the
groups can share their conclusions with the class, summarizing answers on the chalkboard or
an overhead transparency.
• This activity could be conducted as an open class discussion.
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Chapter 7
Chemical Energetics: Enthalpy
Instructor notes:
• Show or refer students to the picture of the set up from Investigate This 7.63 to begin this
activity and then Figure 7.21 as they work on the activity.
Students should reason and conclude:
(a) The liquid level in the manometer is higher in the sample arm than in the arm open to the
atmosphere. This is the opposite situation to that shown in Figure 7.21 and must mean that
the pressure of the gas sample in the syringe is lower than atmospheric pressure. You must
have moved the syringe plunger out, increasing the volume of air in the syringe and thus
lowering its pressure. There are fewer molecules per unit volume and, therefore fewer
molecules colliding with the walls of the container and the surface of the fluid in the
manometer. Hence, the pressure is lower than before the plunger was moved.
(b) There is more than one strategy for finding the pressure of gas in the syringe. Here is one
approach. The quantity h [see Figure 7.21(b)] measures the level of liquid in one arm of the
manometer compared to the level in the other. Let us define h = Lopen – Lsample. where Lopen is
the level in the arm open to the atmosphere and Lsample is the level in the arm attached to the
sample. When Lopen > Lsample, then h > 0, and the pressure of the sample given by the equation
in Figure 7.21(a) is greater than atmospheric, which we know is true under these conditions.
For the problem here, we are told that Lopen < Lsample, so h < 0, and the data in the problem
give h = –15.7 cm = –0.157 m. Thus, we have:
Psample = Patm + gh
 1 kg 
Psample = 100.4 kPa + 
(9.80 m·s–2)(–0.157 m) = 98.9 kPa
 1 L 
The calculated sample pressure is lower than atmospheric pressure, which we know is true
from our reasoning in part (a).
Follow-up discussion:
• Use the discussion and reasoning for this problem as an opening to introduce the kineticmolecular model of gases.
Follow-up activities:
• Consider This 7.66. How does kinetic-molecular theory apply to Investigate this 7.63?
• Check This 7.67. Relationships among the properties of a gas sample.
• Worked Example 7.68. Using the ideal-gas equation.
• Check This 7.69. Using the ideal-gas equation.
• Worked Example 7.70. Using the ideal-gas equation.
• Worked Example 7.71. Using the ideal-gas equation.
• End of chapter problems 7.63 through 7.70.
Consider This 7.66. How does kinetic-molecular theory apply to Investigate This 7.63?
Goal:
Use the kinetic-molecular theory to explain the results of Investigate This 7.63.
Classroom options:
• Allow 3-5 minutes for students, working in small groups, to work on this activity. Then, the
groups can share their conclusions with the class, summarizing answers on the chalkboard or
an overhead transparency.
• This activity could be conducted as an open class discussion.
ACS Chemistry FROG
39
Chemical Energetics: Enthalpy
Chapter 7
Instructor notes:
• Introduce and discuss the kinetic-molecular theory before conducting this activity.
• The previous explanations in Consider This 7.64 and 7.65 have been given in terms of the
kinetic-molecular theory because some students will probably know something about the
behavior of gases from previous courses and you can capitalize on this by interweaving an
introduction to kinetic-molecular theory as these activities are discussed.
Students should reason and conclude:
(a) The number of molecules per unit volume increases as the volume of air in the syringe
decreases. The number of molecules hitting the walls per unit time increases, which we
observe as an increase in pressure of the gas.
(b) Warming the gas sample causes the average speed of molecules to increase, which means
that their collisions with the wall will be more energetic. We observe this increase in the
energy of the wall collisions as an increase in pressure of the gas.
(c) The pressure of the gas will decrease. Removing some of the molecules while keeping the
volume constant means that there will be fewer molecules per unit volume [as in part (a)
where we decreased the density of the gas by increasing its volume]. Fewer molecules will
collide with the walls of the container, thus decreasing the push on the walls and hence
decreasing the pressure.
Follow-up discussion:
• Use the discussion of this activity as a lead-in to the ideal-gas equation.
Follow-up activities:
• Check This 7.67. Relationships among the properties of a gas sample.
• Worked Example 7.68. Using the ideal-gas equation.
• Check This 7.69. Using the ideal-gas equation.
• Worked Example 7.70. Using the ideal-gas equation.
• Worked Example 7.71. Using the ideal-gas equation.
• End of chapter problems 7.63 through 7.70.
Investigate This 7.72. Do reactions in open containers and capped containers differ?
Goal:
Observe that the temperature change of the solution in the reaction of hydrochloric acid with
sodium hydrogen carbonate (baking soda) is different when the reaction takes place in a closed
compared to an open container.
NOTE: The analysis of this system is relatively straightforward, but does demand keeping close
track of signs and directionality of effects. Since the bottom line message of the body of the
chapter is that E  H for reactions in solution, you and your class might find it confusing that
effort is expended on analysis of a gas-producing system where finding the difference between
E and H, providing an application of the ideal gas equation, is the objective. If you think that
the pedagogical value of the analysis is worthwhile, the deviation of this system from the
mainline message should be discussed. Here, a gas is produced, so the volume change is
considerable, and the energy change is rather small. As a consequence the PV work term is
relatively important compared to the heat term and a difference between E and H, about
10%, is rather easy to observe. You can do the activity and analyze it only qualitatively
(directionality) as an assessment of student understanding of the directionality of the changes in
thermodynamic variables.
40
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Chapter 7
Chemical Energetics: Enthalpy
Set-up time:
• 15-20 minutes. You should add the weighed sodium hydrogen carbonate (baking soda) to dry
plastic soda bottles before class. There is no point taking up class time for this part of the set
up. Two versions of the set up are given here. Dr. Jonathan Mitschele, Saint Joseph’s College,
Standish, ME, found that the set up described and pictured in the text is more elaborate than it
needs to be. Both the original and his alternative version are outlined.
NOTE: If neither of these options appeals to you, the animation in the Web Companion, Chapter
7, Section 7.11, is an idealized representation of this activity with data for further analysis.
Time for activity:
• 5-10 minutes. It’s best to have two people available to carry out this activity.
Materials:
• Two dry 2-L soft drink bottles. Smaller 1-L bottles would work just as well, if a thermometer
short enough to fit completely in them is available.
• Cap for a soft drink bottle (for the alternative set up).
• Two –10 to 110 C thermometers. One of these thermometers must fit completely into the soft
drink bottles. Shorter range thermometers are OK as well, if the range includes 0 to 25 C.
• Watch or timer.
• 50-mL plastic syringe.
• Needle for the syringe (for the original set up).
• Short length of small diameter Tygon® tubing to attach to the syringe outlet (for the
alternative set up).
• New 24/40 ribbed rubber septum (for the original set up).
• Length of 18-gauge copper wire to hold the septum on the bottle (for the original set up).
• Powder funnel.
• A 0.1-g (or better) balance.
Chemicals:
• Two 8.4-g samples of sodium hydrogen carbonate [baking soda, NaHCO3(s)].
• Two 20-mL samples of 6 M hydrochloric acid (HCl) solution.
• This picture shows the materials and reagents for the original version of the constant volume
activity:
Original procedure:
SAFETY NOTE
Wear your safety goggles.
ACS Chemistry FROG
41
Chemical Energetics: Enthalpy
Chapter 7
• Before class, add 8.4 g of NaHCO3(s) [= NaHOCO2(s)] to each of the two dry 2-L bottles.
Use a powder funnel and take care not to get the solid on the walls of the bottle. It should all
be at the bottom of the bottles.
• Place a thermometer in one of the bottles (with the thermometer bulb in one of the depressions
in the bottom of the bottle, if this is the style of bottle you have).
• Use the other thermometer to measure the temperature of the 6 M HCl(aq) to the nearest
0.1 C. Record the temperature. (It is wise have two people carrying out this activity. While
one manipulates the 2-L bottle, the other can measure and record the temperature of the HCl
solution, draw 20 mL into the syringe, and add it to the reaction vessel.)
• Use the syringe to dispense 20 mL of the HCl solution into the 2-L bottle.
• Record the temperature and time as you swirl the mixture to be sure all of the solid contacts
the acid solution.
• Tilt the bottle to bring all the solution together around the thermometer bulb and make sure it
is submerged in the solution.
• Record the temperature of the mixture approximately every ten seconds for about two
minutes.
• Repeat the experiment with the second sample with following changes:
• After the NaHCO3(s) and thermometer are in the bottle, seal the bottle with the septum. Wire
the septum to the bottle. Although these septa fit rather tightly and the collar is stretched over
the bottle threads, a modest pressure builds up in the bottle and wiring the septum on helps
insure that there will be no leaks as well as that it will stay put. For best results, a new septum
should be used each time the closed reaction is conducted.
• Put the syringe needle through the septum to inject the acid into the bottle. The picture in the
textbook shows the set up at this point (without the septum wired on, as it should have been).
• IMPORTANT: When the reaction is complete, carefully remove the needle from the syringe.
Insert the needle through the septum to release the gas pressure from inside the bottle before
removing the septum.
Alternative procedure:
SAFETY NOTE
Wear your safety goggles.
• Before class, add 8.4 g of NaHCO3(s) to each of the two dry 2-L bottles. Use a powder funnel
and take care not to get the solid on the walls of the bottle. It should all be at the bottom of the
bottles.
• Place a thermometer in one of the bottles.
• Use the other thermometer to measure the temperature of the 6 M HCl(aq) to the nearest
0.1 C. Record the temperature. (It is wise have two people carrying out this activity. While
one manipulates the 2-L bottle, the other can measure and record the temperature of the HCl
solution, draw 20 mL into the syringe, and add it to the reaction vessel.)
• Carefully tilt the 2-L bottle on its side, keeping all the solid at the bottom, with the bottom
inclined a little above the top.
• Use the syringe with attached short length of Tygon® tubing to add 20 mL of HCl solution
near the top of the bottle, so it does not contact the solid. Note that it would not be necessary
to take such care to add the acid to this uncapped reaction, but it is done to be sure that the
conditions are essentially all the same for the uncapped and capped reactions.
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Chapter 7
Chemical Energetics: Enthalpy
• Turn the bottle upright, swirl to get all the solid and acid mixed, and read and record the
temperature of the solution as soon as possible.
• Tilt the bottle to bring all the solution together around the thermometer bulb and make sure it
is submerged in the solution.
• Record the temperature of the mixture approximately every ten seconds for about two
minutes.
• Repeat the experiment with the second sample with following changes:
• After the HCl solution has been added to the bottle, seal the bottle with the bottle cap while
the bottle is still inclined and the acid and solid have not come in contact. When the cap is
securely in place, continue the procedure as above.
• When the readings have all been taken, carefully loosen the bottle cap a bit to release the gas
pressure in the bottle.
Clean-up:
• Rinse the bottles and syringe with copious amounts of water.
Sample results:
• In a sample set of reactions, the initial temperature of the 6 M HCl solution was 19.5 C
G
are
Qraphics
uickTim
needed
decom
e™
toand
see
pressor
athis
picture.
(temperatures read to the nearest
0.5
C).
Temperature
vs. time data, beginning 10 seconds
after addition of the HCl solution, are given in this graph.
10
Temperature, deg C
9
8
7
6
capped
uncapped
0
20
40 60 80 100 120
5 Time, sec
• We see that there is a difference of 2.0 C between the capped and uncapped reactions. The
maximum changes in temperature, T for the capped and uncapped reactions are –11.5 C and
–13.5 C, respectively.
NOTE: These results were obtained using an earlier version of the activity in which only 6.8 g
of NaHCO3(s), 0.081 mol, was reacted instead of the 0.10 mol in the present version.
Extrapolating to 0.10 mol of reaction, we should observe T for the capped and uncapped
reactions as –14.2 C and –16.7 C, respectively. These are the values we will use for further
calculations based on Investigate This 7.72. The values from the Web Companion are –15.5 C
and –18.8 C, respectively, for the capped and uncapped reactions and we will also give the
results for these data, but not show the calculations.
ACS Chemistry FROG
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Chemical Energetics: Enthalpy
Chapter 7
Follow-up discussion:
• Use Consider This 7.73 to initiate discussion of the results from this investigation. The
analysis is carried on through all the rest of the activities in this section, which are scaffolded
to lead in steps to the final result.
Follow-up activities:
• Consider This 7.74. What are qV and qP for the hydrogen carbonate-acid reaction?
• Worked Example 7.75. Converting qP (or H) to qV (or E).
• Consider This 7.76. What is w for the hydrogen carbonate-acid reaction?
• Check This 7.77. Ereaction and Hreaction for the hydrogen carbonate-acid reaction.
• End of chapter problems 7.63 through 7.70.
Consider This 7.73. Why do reactions in open and capped containers give different
results?
Goal:
Identify the constant volume and constant pressure systems in Investigate This 7.72 and reason
that this difference causes a difference in the observed temperature changes.
Classroom options:
• Allow 3-5 minutes for students, working in small groups, to work on this activity. Then, the
groups can share their conclusions with the class, summarizing answers on the chalkboard or
an overhead transparency.
Instructor notes:
• Be sure that the class agrees on the results from Investigate This 7.72 as you begin this
discussion. You should use your own results, if you have them, or those from the reactions
reported above for Investigate This 7.72, or the data from the Web Companion, Chapter 7,
Section 7.11.
Students should reason and conclude:
(a) The maximum change in temperature in the open bottle was –16.7 C (or whatever the
class observed in their activity). The maximum change in temperature in the open bottle was
–14.2 C (or whatever the class observed in their activity). The endothermic nature of this
reaction is a bit surprising, because other acid-base reactions observed or discussed earlier in
the chapter were exothermic. A difference in the hydrogen carbonate-acid reaction might be
that a gas is produced, as well as water, from the acid-base reaction.
(b) The reaction in the capped bottle was carried out at constant volume, the volume of the
bottle in which all products were trapped. The reaction in the open bottle was carried out at
constant pressure. In this latter case, the gas produced by the reaction could push back the air
in the bottle, which could leave the bottle, thus keeping the pressure in the bottle at
atmospheric pressure.
(c) In a constant volume reaction system, no work is done by or on the system, but in a
constant pressure system that produces a gas, work can be done by the system on its
surroundings by pushing them back. Perhaps it is this difference in the work produced that
causes the difference in the temperature changes in the two systems.
Follow-up discussion:
• Try to be sure the discussion clearly makes these points: (1) the reaction in the capped bottle
is carried out at constant volume (no work is done) and (2) the reaction in the uncapped bottle
is carried out at constant pressure and the carbon dioxide does work by pushing out the air,
that is, pushing back the atmosphere.
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Chapter 7
Chemical Energetics: Enthalpy
Follow-up activities:
• Consider This 7.74. What are qV and qP for the hydrogen carbonate-acid reaction?
• Worked Example 7.75. Converting qP (or H) to qV (or E).
• Consider This 7.76. What is w for the hydrogen carbonate-acid reaction?
• Check This 7.77. Ereaction and Hreaction for the hydrogen carbonate-acid reaction.
• End of Chapter problems 7.63 through 7.70.
Consider This 7.74. What are qV and qP for the hydrogen carbonate-acid reaction?
Goal:
Calculate qV and qP for the hydrogen carbonate-acid reaction observed in Investigate This 7.72.
Classroom options:
• Allow 3-5 minutes for students, working in small groups, to work on this activity. Then, the
groups can share their conclusions with the class, summarizing answers on the chalkboard or
an overhead transparency.
• This activity could also be assigned as a homework problem and then discussed at the next
class session.
Instructor notes:
• If necessary, briefly review calorimetric calculations.
Students should reason and conclude:
In addition to the assumptions that the solution exchanges no thermal energy with the
surroundings during the reaction and that the solution has a specific heat = 4.18 J·g-1·C-1, we
will assume that the density of the hydrochloric acid solution is 1.00 g mL-1. Thus, the mass
of solution is 20 g plus the mass of the added sodium hydrogen carbonate (8.3 g) minus the
mass of gas, carbon dioxide, lost (0.1 mol = 4.4 g), which is a total of 24 g. For our analysis
we know that qreaction + qsolution = 0, or rearranging, we have qreaction = –qsolution. For the
constant volume system, qV = qreaction and for the constant pressure system, qP = qreaction,.
For the constant volume system, T = –14.2 C and we get
qV = –qsolution = –(m)(c)(T) = –(24.0 g) (4.18 J·C-1·g-1) (–14.2 C) = 1.42 kJ
For the constant pressure system, T = –16.7 C and we get
qP = –qsolution = –(24.0 g) (4.18 J·C-1·g-1) (–16.7 C) = 1.68 kJ
[The difference between these two values, 0.26 kJ, represents the pressure-volume work done
by the system on the surroundings (as the gas is produced and pushes the atmosphere back).
The work done on the system is –0.26 kJ. This is energy lost by the system, which is why
more energy has to be supplied to the system from the thermal surroundings (the solution) for
the constant pressure reaction. The results from the Web Companion give qV = 1.55 kJ and qP
= 1.86 kJ.]
Follow-up activities:
• Worked Example 7.75. Converting qP (or H) to qV (or E).
• Consider This 7.76. What is w for the hydrogen carbonate-acid reaction?
• Check This 7.77. Ereaction and Hreaction for the hydrogen carbonate-acid reaction.
• End of Chapter problems 7.63 through 7.70.
ACS Chemistry FROG
45
Chemical Energetics: Enthalpy
Chapter 7
Consider This 7.76. What is w for the hydrogen carbonate-acid reaction?
Goal:
Use ideal gas behavior to calculate w for the hydrogen carbonate reaction and compare the
result with the difference between qV and qP from Consider This 7.74.
Classroom options:
• Allow 3-5 minutes for students, working in small groups, to work on this activity. Then, the
groups can share their conclusions with the class, summarizing answers on the chalkboard or
an overhead transparency.
• This activity could also be assigned as a homework problem and then discussed at the next
class session.
Instructor notes:
• If necessary, review limiting reactant calculations as an introduction to this activity.
Students should reason and conclude:
(a) The molar mass of sodium hydrogen carbonate, NaHCO3, is 84 g·mol–1, so for our
activity we have
8.4 g
8.4 g NaHCO3 =
–1 = 0.10 mol NaHCO3
84 g  mol
(b) We used 20 mL of 6 M HCl (= 6 M H3O+), which gives
0.020 L of 6 M H3O+ = (0.20 L)(6 mol·L–1) = 0.12 mol of H3O+
(c) Equation (7.57) shows that the hydronium ion and sodium hydrogen carbonate react in a
one-to-one molar ratio to produce one mole of carbon dioxide gas for each mole of each
reactant that reacts. The limiting reactant in this activity is the sodium hydrogen carbonate,
since there are fewer moles of this reactant. Thus, 0.10 mol of NaHCO3(s) produces 0.10 mol
of CO2(g).
(d) Using the ideal gas equation to determine PV for the production of 0.10 mol of gas, we
get
PV = nRT = (0.10 mol) (8.314 J mol-1 K-1) (293 K) = 0.24 kJ
(e) Substituting our result from part (d) into equation 7.58 gives
qV = qP – PV = qP – nRT = qP – 0.24 kJ
Rearranging, we get
qP – qV = 0.24 kJ
The value we got for this difference in Consider This 7.74 was 0.26 kJ. These results are
essentially the same, especially considering the crudeness of the experimental set-up in
Investigate This 7.72, and show that equation 7.58 is satisfied in this reaction system. [The
results from the Web Companion give qP – qV = 0.31 kJ, which is within about 25% of the
value calculated using the ideal gas equation.]
Follow-up activities:
• End of Chapter problems 7.63 through 7.70.
46
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Chapter 7
Chemical Energetics: Enthalpy
Solutions for Chapter 7 Check This Activities
Check This 7.6. The difference between warm and cool gases
The “motion tails” on the molecules represented in Figure 7.2 are supposed to suggest their
relative speeds. The tails on the gases inside and at the top of the chimney have longer tails than
those outside at the bottom. Thus, the molecules that have not reacted or entered the chimney
are moving more slowly and we have said that molecular motion and temperature are
proportional, so the slower moving molecules are cooler than those in the chimney or leaving it.
Check This 7.16. The system in Figure 7.5
The molecules in the skin are system. They are made to move faster by being pushed on as the
hands are rubbed briskly together or by absorbing infrared energy emitted by the flame.
Check This 7.17. Closed and open systems
(a) The fuel and flame are an open system, because mass, oxygen for the combustion, must be
supplied from the surroundings.
(b) Explosions like this are a closed system. In an explosion, thermal energy is released and
work (mechanical energy) is done on the surrounding (by pushing them back). However, no
mass has to be transferred to the system, because all of the reactant atoms are included in the
ammonium nitrate molecules. The products can, of course, mix with the surroundings, but this
is not relevant to the reaction and its consequences.
Check This 7.21. Is the reaction of an acid with a base exothermic or endothermic?
(a) The solutions used in Investigate This 7.19 contain ions formed by the reagents dissolved in
water: NaOH(aq) is a solution of Na+(aq) and OH–(aq) and HCl(aq) is a solution of H3O+(aq)
and Cl–(aq). The reactive species are H3O+(aq) and OH–(aq) and the reaction is:
H3O+(aq) + OH-(aq)  2H2O(aq)
(b) The reaction is exothermic. When the reactant are mixed, the test tube becomes warm to the
touch, so thermal energy must be leaving the reacting system and warming the surroundings,
including fingers touching the test tube. The enthalpy of the system decreases, Hfinal < Hinitial and
H is negative.
Check This 7.25. Energy transfers in a calorimeter
The thermometer indicates that the temperature of the solution increases. Solvent water
molecules (and other molecules and ions dissolved in the water) next to reacting ions, have
absorbed energy by contact with the products of the chemical reaction, which have been
energized by the reaction. These energized water molecules come in contact with other water
molecules and transfer some of their energy to them and so on. Warmer water is less dense than
cooler water (above 4 ˚C), so convection currents are set up in the liquid that mix the warmer
with the cooler water and enhance further contacts between molecules for energy exchange.
Eventually, water molecules that have gained energy come in contact with the thermometer.
Transfer of thermal energy via contact with the thermometer occurs, causing the mercury atoms
to vibrate faster, increasing its volume. The expanding volume of the mercury results in the
mercury filling more of the volume of its container (the narrow tube of the thermometer) and
indicating that an increase in temperature has occurred. Both conduction (via contact) and
convection occur to transfer energy in this system.
ACS Chemistry FROG
47
Chemical Energetics: Enthalpy
Chapter 7
Check This 7.27. Determination of qP(reaction) for dissolution of urea in a calorimeter
The information from Consider This 7.23 indicates that T for the dissolution of urea in
Investigate This 7.22 is –3.5 to –3.6 C. Let us take T = –3.55 C, but remember that we only
really measure the change to two significant figures, so the best we can do is a precision of
about 3% in our result. The simple experimental setup probably makes it even more uncertain,
so we report two significant figures in the result.
qP(solution) = (m)(c)(T) = (106. g)(4.18 J·g–1·C–1)(–3.55 C) = –1.6  103 J
Rearrange equation (7.3) to get qP(reaction):
qP(reaction) = –qP(solution) = 1.6  103 J
Check This 7.29. Determination of Hreaction for dissolution of urea in a calorimeter
The molar mass of urea, NH2C(O)NH2, is 60 g·mol–1 and 6.0 g were dissolved in 100. mL of
water in Investigate This 7.22. Thus, 0.10 mol of urea was dissolved and gave a 1.0 M solution.
The molar enthalpy change for the dissolution is:
3
1.6  10 J
Hreaction = qP(reaction)(per mole) =
= 16 kJ·mol–1
0.1 mol
Check This 7.31. Effect of increasing the amount of another calorimetric reaction
(a) In the previous example, Consider This 7.30, we found that doubling the amount of reactants
did not change the temperature change for their reaction, so perhaps it is surprising to find that,
in this case, doubling the amount of urea dissolved in the same volume of water doubles the
temperature change.
(b) The difference between the two examples is that here the system (urea dissolving) was
doubled, but the surroundings (the solution) remained essentially the same. The surroundings
are a little different when double the amount of urea dissolves, because the mass of the solution
and perhaps its specific heat will be a bit different, but these are relatively small changes and we
expect the change in temperature to about double, as it does. In the example in Consider This
7.30, both the reacting system and the surroundings (total amount of solution) were doubled, so
the doubled energy released had to warm twice as much solution.
Check This 7.34. Properties of the possible H2O2 reaction products
(a) Reaction (7.6) produces a gas that consists of oxygen. The gas will support combustion. If a
glowing splint is placed in the gas, the splint will glow more brightly and may burst into flame.
(b) Reaction (7.7) produces a mixture of hydrogen and oxygen gases. This is an explosive
mixture. Hydrogen is flammable and when mixed with oxygen will react so rapidly that an
explosion occurs. If the amount of gas is small the explosion will not be very dangerous and is
likely to produce only a “pop” sound instead of a deafening blast.
(c) Insert a glowing splint in the cylinder containing the product gas(es).
If reaction (7.6) produces the gas, the glowing splint will glow brightly or even re-ignite.
If reaction (7.7) produces the gas, the glowing splint will ignite the mixture of hydrogen and
oxygen and a small explosion will be heard.
Check This 7.37. Overall homolytic bond cleavage reaction of methane
Each reaction in Table 7.2 produces a carbon-containing species that is the reactant in the next
reaction. If we add the reactions together, these products and reactants cancel, as shown here:
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There are eight bonding electrons in methane, represented as four pairs of bonding electrons by
the strokes in the CH4 structure. The products also contain eight valence electrons, four on the
carbon atom and one each on the four hydrogen atoms, so all the bonding electrons in methane
are accounted for in the products.
Check This 7.39. Formation of polyethylene
The polyethylene "product" should be similar to the structure shown in Figure 7.10, except that
it will contain more monomer, -(-CH–CH-)- units.
Check This 7.40. Sulfur–sulfur bonds
(a) Lewis structures for S8 and S2 are:
S
S
(b) A molecular model of the S8 ring (using oxygen atoms) is shown here:
Four S2 models will have two connectors (probably bent) between each pair of sulfurs,
representing a total of eight two-electron (electron pair) bonds. The model here shows that there
are eight two-electron bonds in S8. The eight connectors used for the S2s could have been used
to build the S8. The reactants and products in the model reaction have the same number of twoelectron bonds. We can compare the reactants and products, using bond enthalpies (as in Figure
7.8), because we are comparing species with the same number of bonding electrons.
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(c) A bond formation enthalpy diagram for the reactants, 4S2, and product, S8, of the reaction
represented by the models in part (b) is:
The H for 8S(g)  4S2(g) is –1408 kJ·mol-1 [= –4·(352 kJ·mol-1)]. The H for 8S(g)  S8(g)
is –2144 kJ·mol-1 [= –8·(268 kJ·mol-1
H = –736 kJ·mol-1.
(d) The results in part (c), based on bond enthalpies, show that the singly-bonded S8 molecule is
a great deal more stable than four doubly-bonded S2 molecules. This is because the double bond
is weaker than two single bonds. The opposite is true for oxygen where a double bond is
stronger than two single bonds in oxygen, so the doubly-bonded species is the more stable.
Thus, a good deal of the difference between the two elements can be attributed to the relative
strengths of their single and multiple bonds.
Check This 7.42. Reaction enthalpy from average bond enthalpies
(a) We can use equation (7.14) to get Hreaction from bond enthalpies (all bond enthalpies in
kJ·mol–1):
Hreaction = (BHreactants) + (-BHproducts)
= BH(O–O) + 2BH(O–H) – BH(O=O) – BH(H–H)
= 142 + 2(460) – 498.7 – 436.4
= (1062 kJ·mol–1) + (-935 kJ·mol–1) = 127 kJ·mol–1
(b) The enthalpy diagram for this reaction is:
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(c) The interactive exercise in the Web Companion, Chapter 7, Section 7.7, page 3, shows the
cumulative result of breaking all the bonds in the reactants and then making all the bonds in the
products, in order to find the overall Hreaction for the reaction of water with ethene to form
ethanol. This is exactly the procedure we have used in part (a) here and in Worked Example
7.41. Figure 7.11 and our figure in part (b) are just like the enthalpy level diagram in the
Companion, except that our figures do not show the components of diagram, as the Companion
does. On page 2 of this section of the Companion, a shorter method for doing these kinds of
problems is hinted at, but not fully carried out. In this approach, you focus only on the bonds in
the reactants that are broken going from reactants to products and only those bonds in the
product that must be made to create the product from the pieces on the reactant side. The bonds
that remain as they are when going from reactants to products are ignored in bond enthalpy
calculations. For the water-ethene reaction, only the carbon-carbon double bond and one of the
hydrogen-oxygen bonds needs to be broken in the reactants; if you try to break other bonds, you
get this error message:
To get the product, you need to make carbon-carbon, carbon-hydrogen, and carbon-oxygen
single bonds. The bonds that have to be broken in the reactants require a total of 1080 kJ·mol–1.
Making the bonds in the product releases a total of 1112 kJ·mol–1. The reaction is exothermic
and Hreaction = –32 kJ·mol–1. This is the same numerical value you get from the diagram on
page 3 of this section of the Companion, demonstrating that this shorter method gives the same
result as you get by going through the complete atomization. This screen from page 3 illustrates
why the two methods give the same result:
Here you can see that all of the bonds in the reactants have been broken; the first two broken are
the two that must be broken to produce fragments that can be recombined to yield the product.
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In the product, only those bonds that are common to the reactants and products have been
reformed. Note that the enthalpy regained (released) in this reformation is exactly the same as
the enthalpy required for the bond breaking, as it must be. The only bonds remaining to be
formed are those that combine the fragments (H, CH2, CH2, and OH) to give the ethanol
product. These bond enthalpies are the ones to compare to the first two broken to find the net
change for the process. Try this yourself with the Companion to see what you get when the final
three bonds are formed.
Check This 7.43. The hydrogen peroxide decomposition reaction
(a) The enthalpy change as calculated from Check This 7.42 is endothermic. However, the
enthalpy change for this reaction is exothermic, as observed in Investigate This 7.32. The
graduated cylinder felt warm after the reaction.
(b) Reaction (7.6) is responsible for the observations in Investigate This 7.32 and 7.35. The
reaction was exothermic. The glowing splint confirmed the presence of oxygen gas. No small
explosion was detected, so the presence of a mixture of hydrogen and oxygen gases is unlikely.
(The absence of a positive test cannot always be taken as proving the absence of the system
being tested for, because other circumstances might prevent the test working correctly. In this
case, this is one more piece of evidence that is consistent with all the others and strengthens the
final conclusion.) The calculated enthalpy change for reaction (7.6) is exothermic.
Check This 7.45. Formation of compounds from their standard-state elements
The reaction equation for forming dimethyl amine from its elements is:
2C(graphite) + 7/2H2(g) + 1/2N2(g)  (CH3)2NH(g)
The reaction equation for forming ethylamine from its elements is:
2C(graphite) + 7/2H2(g) + 1/2N2(g)  CH3CH2NH2(g)
The reactants in both examples are the same. We conclude that the reactants in equations for
forming isomers from the elements in their standard states are the same.
Check This 7.47. Standard enthalpy change for an isomerization reaction
(a) The molecular model for the reaction of cyclopentane to form one of its isomers, 1-pentene,
is:
(b) The enthalpy change for this reaction is the difference between the initial and final states,
that is, the enthalpy of formation of the product minus the enthalpy of formation of the reactant:
Hreaction = Hf(1-pentene) – Hf(cyclopentane)
= (–20.92 kJ mol-1) – (–77.24 kJ mol-1) = 56.32 kJ mol-1
(c) The enthalpy level diagram (modeled after Figure 7.13) for this reaction is:
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This diagram shows graphically what we found in part (b), the reaction is endothermic.
Check This 7.48. Comparison of formation of 1-butene from its atoms and its elements
(a) The enthalpy level diagram for formation of 1-butene from its atoms and its elements is:
(b) Although cyclo-C4H8 and 1-butene have the same chemical formula, their Hºfs from
graphite and hydrogen gas (and Hºreactions from gaseous carbon and hydrogen atoms) are
different because the structures are different and the bonding is different. In both cases,
however, their enthalpies of formation are positive, that is, they are less stable than the standard
state elements from which they are formed.
Check This 7.50. Standard enthalpy change for a reaction
(a) The enthalpy of formation of the products is:
Hºf(products) = 6Hºf[CO2(g)] + 6Hºf[H2O(l)]
= 6(–393.51 kJ·mol-1) + 6(–285.83 kJ·mol-1) = –4076.0 kJ·mol-1
The enthalpy of formation of the reactants is:
Hºf(reactants) = Hºf[C6H12O6(s)] + 6Hºf[O2(g)]
= (–1274.4 kJ·mol-1) + 6(–0 kJ·mol-1) = –1274.4 kJ·mol-1
Thus, Hºreaction = Hºf(products) – Hºf(reactants) = –2801.6 kJ·mol-1.
(b) The enthalpy level diagram representing this reaction and the values in part (a) is:
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Check This 7.51. Bond enthalpy and enthalpy of formation comparison
(a) To go from reactants to products, you need to break a C–C double bond and the Br–Br bond
in the reactants and form two C–Br bonds a one C–C single bond in the products. (The C–H
bonds are not considered because there are the same number in the reactants and products.) The
enthalpy of reaction (all enthalpies in kJ·mol-1)
Hreaction = (BHreactants) + (-BHproducts)
= BH(C=C) + BH(Br–Br) – 2BH(C–Br) – BH(C–C)
= 620 + 193 – 2(276) – 347 = –86 kJ·mol-1
(b) We can use equation 7.30 to find the standard enthalpy of reaction from standard enthalpies
of formation (all enthalpies in kJ·mol-1):
Hreaction = [nj(H f)j]products – [(nj(Hf)j]reactants
= Hf[CH2BrCH2Br(g)] – Hf[CH2CH2(g)] – Hf[Br2(g)]
= –43.1 – (52.3) – (30.9) = –126.3 kJ·mol-1
(c) Although, as expected, the results are not the same, they are in the same direction: both say
that the reaction is exothermic. As we have said, reaction enthalpies based on bond enthalpies
are simply estimates, but are often useful in providing the direction of enthalpy change. Bond
enthalpy calculations assume that when two atoms make a covalent bond, this bond is
independent of other bonds in the molecule. Average bond enthalpy data are given as reasonable
averages, assumed independent of the molecule. For example, ethene and 1,2-dibromoethane
have different electronic geometries about the carbon atoms, but we assumed in part (a) that the
same C–H bond enthalpy applied to both molecules. This assumption is likely to be a large part
of the source of the difference between the values in parts (a) and (b). The reaction enthalpies
based on standard enthalpies of formation are more accurate because chemical bonding
approximations are not involved; the results are based on experimental values for the actual
molecules involved.
Check This 7.52. The phosphate bond to ribose in ATP4– and ADP3–
The bond that links the first phosphate group to ribose in ATP4– and ADP3– is a phosphate ester
bond. The bond is formed by loss of water between an acidic phosphate group and the alcohol
group on the ribose.
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Check This 7.53. Other biological fuels
The potato chip in Investigate This 7.1 contains fat molecules from the way it is prepared and
fats are a good source of metabolic energy, providing more energy per gram than the
metabolism of sugars. Fatty foods like the chips or nuts burn better and more smoothly in
investigations like these and seem to be better fuels, as their caloric value suggests. The
marshmallow contains protein, gelatin, and protein is also a source of metabolic energy,
providing about the same energy per gram as sugars. The starches in the pasta and the potato
chip are broken down metabolically to their monomers, which are sugars, mainly glucose, so
these are also a source of metabolic energy. Starches provide us a delayed energy input, because
it takes time to break them down to their sugar monomers.
Check This 7.54. Other pathways for ATP hydrolysis
(a) Since each of the individual reactions releases about the same amount of enthalpy as the
hydrolysis of ATP to ADP, the combined reaction (the sum) releases twice as much enthalpy,
 ≈ –42 kJ·mol–1. The advantage to the organism is twice the enthalpy of hydrolysis of ATP
alone, so the dual hydrolysis can be coupled with reactions that require more driving force than
furnished by the single hydrolysis reaction.
(b) Phosphoric anhydride bonds are hydrolyzed in each case, so the enthalpy released is
approximately the same.
Check This 7.57. Pressure–volume work at constant pressure
In this change, the volume of the system decreases, that is, V < 0, so the work done on the
system is:
w = –PV = –(103 kPa)(–0.521 L) = 53.7 kJ
Check This 7.60. Comparing E and H
E and H are almost the same for reactions that either do not involve gases or in which the
V is
approximately zero. In reaction (b) there are six moles of gaseous products and reactants and
reaction (d) involves no gases, so E and H are almost the same for reactions (b) and (d). In
reaction (a), V is negative (three moles of gaseous reactants produce zero moles of gaseous
product) and –PV is positive. Work is done on the system in reaction (a). In reaction (c), V is
positive (zero moles of gas reactant produces one mole of gaseous product) and –PV is
negative. Work is done by the system on the surroundings in reaction (c).
Check This 7.62. Comparing E and H quantitatively
(a) In Check This 7.51(b) you calculated the standard enthalpy of reaction, Horeaction = –126.3
kJ (per mole of reaction), for the bromination of ethene with gaseous bromine. The reaction here
is the reverse, so Hreaction = 126.3 kJ. The decomposition is endothermic. The reaction
produces two moles of gaseous product for every mole of gaseous reactant decomposed, so the
system does work on the surroundings. Work leaves the system, so wP (work entering the
system) is negative: wP = –2.5 kJ. We can write:
E = qp + wp = Hreaction + wp = 126.3 kJ + (–2.5 kJ) = 123.8 kJ
(b) For this case, we see that Hreaction and Ereaction differ by a little less than 2%. Thus, the
approximation that Ereaction  Hreaction is quite good, but not as good as when the thermal
change in the reaction was much larger. You can see from Worked Example 7.61 and this
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Chapter 7
Check This that, for a one mole change in the amount of gaseous products vs. reactants, the
work done by or on the system under standard conditions at 1 bar (= 100 kPa) pressure is 2.5 kJ.
This is a rather small amount of energy that will be swamped by large thermal changes for most
reactions. The approximation that Ereaction  Hreaction breaks down when the thermal change
in the reaction is relatively small, say no more than 50 kJ.
Check This 7.67. Relationships among the properties of a gas sample
nT
The equation, P 
, shows that pressure is inversely proportional to volume (that is, it is
V
1
proportional to ) and directly proportional to temperature and the number of molecules.
V
These are the relationships stated in the text.
Check This 7.69. Using the ideal-gas equation
For the change that is made, the number of moles of gas in the syringe and the temperature of
the gas remain the same, so the right-hand side of the ideal-gas equation is constant and we can
write:
PiVi = PfVf
Solving for Vf and substituting known values gives:
P V 1.00  10 kPa 44.7 mL 
Vf = i i =
 29.8 mL
2
Pf
1.50  10 kPa
2
Check This 7.71. Using the ideal-gas equation
(a) In this problem, the volume remains the same while n2 moles of a second gas are added to n1
moles of the first gas. In the process, the pressure and temperature of the gas sample changes.
We know the initial conditions, P1, T1, n = n1, and the final conditions, P2, T2, n = (n1 + n2), so
we can write:
R
P
P2
 1 
V n1T1 (n1  n2 )T2
57.8 kPa
95.8 kPa

n1 289.2 K (n1  n2 ) 302.7 K
n1  n2
n
(95.8 kPa)(289.2 K)
=1+ 2 =
= 1.584
n1
n1
(57.8 kPa) (302.7 K)
n2
= 0.584
n1
(b) The easy way to do find the number of moles is to use the ideal gas equation for the initial
conditions to find the initial number of moles, n1, and then use the result in part (a) to find n2.
We can check the result by using the ideal gas equation to find n1 + n2 from the final conditions.
PV
57.8 kPa 0.547L) 
n1  1 
= 0.0131 mol
RT1 8.314 J  mol -1  K-1 289.2 K
n2 = 0.584n1 = 0.584(0.013 mol) = 0.0077 mol
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The sum, n1 + n2 = 0.0208 mol, and checking, we find:
PV
95.8 kPa 0.547L )
n1 + n2 = 2 
= 0.0208 mol
RT2 8.314 J  mol -1  K-1 302.7 K 
Check This 7.77. Ereaction and Hreaction for the hydrogen carbonate-acid reaction
From Consider This 7.76, we know that 0.10 mol of reactants reacted in Investigate This 7.72,
so we can convert qP and qP from Consider This 7.74 to Hreaction and Ereaction:
qP
1.68kJ
Hreaction =
=
= 16.8 kJ mol-1
0.10 mol
0.10 mol
qV
1.68kJ
Ereaction =
=
= 14.2 kJ mol-1
0.10 mol
0.10 mol
[The data from the Web Companion give Hreaction = 18.6 kJ mol-1 and Ereaction = 15.5 kJ mol–
1
.]
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Chapter 7
Solutions for Chapter 7 End-of-Chapter Problems
Problem 7.1.
(a) Assuming all the food energy, 180 Cal, from eating a slice of pizza goes to keep your heart
beating at 1 J of energy for each beat, the number of beats that can be sustained is:
 1 kcal   4184 kJ   1000 J   1 beat 
180 Cal = (180 Cal) 
= 7.5  108 beats
 1 Cal   1 kcal   1 kJ   1 J 
At a pulse rate (heartbeat rate) of about 80 per minute, this would keep you going for about:
 1 min 
7.5  108 beats = (7.5  108 beats) 
= 9.4  106 min [ 5 yr]
 80 beats 
(b) If the efficiency of use of this food energy is about 27%, the value calculated for capturing
the energy from glucose oxidation in Section 7.9, the answers in part (a) would be only 27% as
large: 2.0  108 beats; 2.5  106 min [ 18 yr]
Problem 7.2.
(a) Once they have been ignited both the pasta and potato chip, in Investigate This 7.1, continue
to burn on their own. They release energy in the form of light and enough thermal energy to
keep their temperatures high enough to continue burning. You might find that a potato chip
burns somewhat better than pasta, because the fat is a better fuel than starch. This is reflected in
their nutritional Calorie values as well. A gram of starch provides about 4 Calories (17 kJ) and a
gram of fat about 9 Calories (38 kJ).
(b) The oxidation of glucose is used as an example throughout this chapter. You know that the
oxidation is quite exothermic, which is shown by the burning pasta, potato chip, and
marshmallow (mainly sugars including glucose). It doesn’t matter that the glucose units are
strung together as a polymer in starch or separated in a marshmallow -- each still oxidizes to
produce a large output of enthalpy.
Problem 7.3.
If a baked, rather than a fried, potato chip were used in Investigate This 7.1, the baked chip will
still burn, but probably not to burn as vigorously as the fried chip because there is less residual
fat available as fuel on the baked chip. The raw potato slice may not burn at all because it has a
high water content. The Olestra® chip will also burn quite well, for the Olestra is a long-chain
polymeric fatty acid. It does not produce heat value in the body because it cannot be
metabolized.
Problem 7.4.
Potential energy and kinetic energy are the two types of energy. Mechanical energy and thermal
energy are both examples of kinetic energy. [Chemical energy stored in chemical bonds is a
form of potential energy not included in this listing.] This is a visual representation of the
relationship.
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energy
potential energy
kinetic energy
mechanical energy
thermal energy
Problem 7.5.
[NOTE: For an excellent discussion of the forces acting on a helium-filled balloon see, J. E.
Harriman, “On the Buoyancy of a Helium-Filled Balloon,” J. Chem. Educ. 2005, 82, 246-7. The
discussion requires use of the barometric formula for the variation of atmospheric pressure with
height above some reference level, but should be accessible to students who know the formula
or who will accept it.]
(a) This diagram illustrates a tethered helium-filled balloon represented similarly to the model
for the movement of combustion gases in Figure 7.2. The smaller dots inside the balloon (circle)
represent He atoms and the larger dots outside the balloon represent the molecules in the air,
mainly N2 and O2.
(b) The helium atoms in the balloon are in constant, random motion. They do not exhibit any
directed motion. The same is true for the gases in the air outside the balloon (assuming constant
temperature). What is represented crudely in the sketch is the variation in pressure (hence,
number of particles) in the gravitational field of the Earth. Pressure decreases with height, so
more air molecules are shown below the balloon than above and similarly, more He atoms are
shown nearer the bottom of the balloon. The net result is a force (from collisions of the
molecules and atoms with the outside and inside walls of the balloon) pushing the balloon up.
The balloon is motionless, because this upward force pulls the string up, which results in an
exactly balanced downward force in the string that is held motionless by its connection to some
heavy object.
If the string is cut or untied, there is no longer a downward force counteracting the net upward
buoyant force of the gases, and the balloon will rise. The helium atoms in the balloon are still in
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Chapter 7
constant, random motion, keeping the balloon fully inflated. If we were inside the balloon and
could observe the He atoms in motion, we would see no difference between the tethered and
untethered motion. However, if we make our observations from a fixed location outside the
balloon, the He atoms would have a nonrandom upward component to their motion because
they are rising. You might think this directionality would decrease the number of collisions
inside the balloon at the bottom and increase them at the top. But this is not the case, because
the balloon itself is moving and the top is moving away from the atoms inside as the bottom is
moving toward them. The directionality of the atomic movements is just balanced by the
directionality of the balloon moving with the same net velocity.
Problem 7.6.
(a) Since the temperature of the water in a waterfall increases when it falls and hits the pool at
the bottom, its internal energy must have increased.
(b) Work has been done on the water to increase its internal energy (and, hence, its
temperature). The work is done on the water when its potential energy at the top of the fall is
converted to kinetic (directed) motion as it is falling and then this directed motion is changed to
random motion when it hits the pool at the bottom and rapidly decelerates due to the force of the
collision.
Problem 7.7.
Water molecules in ice are held in place by hydrogen bonds and move only slightly within the
crystal lattice. When heat, thermal energy, is added to the solid, the molecules vibrate in place
more rapidly and take up a little more space (on the average), but the change is quite small and
the solid does only a tiny bit of work pushing back on its surroundings. Molecules in liquid
water molecules are more free to move about, but are also hydrogen bonded and held rather
close to one another. As with the solid, addition of thermal energy causes more rapid motion
and the molecules take up a little more space (water expands a bit when heated), but, again, the
change is rather small (although larger than in ice), so the liquid does only a small amount of
work pushing back on its surroundings. In the vapor phase, where water molecules are on
average, quite distant from their neighbors, hydrogen bonding is negligible. Almost all the heat
energy added to the vapor phase water is transformed into more rapid translational motion,
kinetic energy, and the average volume per molecule increases substantially. The increase in
volume of the gas can do a substantial amount of work as it pushes back on its surroundings.
Problem 7.8.
Economic and environmental factors are important to every consumer, so it is important for
automotive engineers to design engines that convert the highest possible percentage of energy to
work. Whatever the price of fuel, consumers want to obtain the maximum number of miles
possible per unit of fuel. This can only be done if the chemical energy from the fuel is converted
efficiently as possible to work, and not wasted as heat. Efficient conversion of fuel energy to
work also means release of lower concentrations of possible pollutants (carbon dioxide, carbon
monoxide, nitrogen oxides, and hydrocarbons from fossil fuels) into the atmosphere.
Problem 7.9.
(a) A silver spoon at room temperature, when placed in a cup of hot water, becomes too warm
to touch, because thermal energy is transferred by conduction. The water molecules are moving
more rapidly, on the average, than the silver atoms in the spoon. Thermal energy is transferred
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from the warmer, more energetic water molecules to the cooler, less energetic silver atoms in
contact with the hot liquid. The energized silver atoms then transfer some of their energy to
their neighbors and eventually the whole spoon becomes uncomfortably warm.
(b) When an apple pie is baked in an electric oven, thermal energy transfer is taking place
through both conduction and convection. The electric heating unit will transfer thermal energy
to molecules of gases in the air directly in contact with the unit; this is conduction. The warmer
air rises, allowing the denser, cooler air to circulate near the heating coil; this sets up a
convection current. (In ovens advertised as “convection ovens”, there is at least one fan to help
distribute the air evenly.) The molecules of hot gas that are in contact with the pie will conduct
thermal energy into the pie, eventually resulting in baking the pie. The rack on which the pie
pan sits will also get thermal energy by conduction from the air molecules and transfer some of
this thermal energy to the pie by conduction.
(c) When the water in an outdoor swimming pool cools from 25 C to 20 C as the summer
season changes into autumn, the water temperature is going down because thermal energy is
being lost by contact between warmer water molecules and the cooler air molecules.
Conduction is involved directly at the water/air interface, but convection currents in both water
and air are likely the major factors in transferring the thermal energy to the surface and away
from the surface.
(d) When you sunbathe with your back exposed to the sun, until your back feels very warm,
thermal energy is being transferred from the sun to your back by means of long wavelength
infrared electromagnetic radiation. As the molecules in the skin on your back absorb radiation,
the molecules move faster. This means that their temperature increases, which is what you
sense. [Infrared radiation is accompanied by the more energetic ultraviolet rays, so remember to
use sunscreen to avoid skin damage.]
Problem 7.10.
Since the cooler air near the ice rink floor has a higher density, it will stay near the floor for a
relatively long time. The ice rink floor will remain cool until it is warmed up by heat radiating
from the sunlit roof and walls and conduction from the roof and the walls to the floor materials.
Although convection currents will not be set up with the higher density gas below the lower
density gas, diffusion of gas molecules will occur and warmer molecules will transfer thermal
energy to cooler molecules over time. None of these processes is going to be fast so bringing the
floor and roof to the same temperature will take a long time.
Problem 7.11.
To warm an individual whose core temperature is too low, a convection method might be to
place the person on a hammock or cot in a small room with a heated floor (much like an oven,
but only warm, not hot). Warm air displaced from the floor by cooler air from the ceiling of the
room would contact the person and transfer thermal energy from the air molecules to her/his
body. A radiation method may consist of placing the person under a bank of “heat lamps,”
electric lamps that emit a large fraction of infrared electromagnetic radiation. Most of the
thermal transfer to the patient will be radiative. A disadvantage of this set up is that the radiation
travels in straight lines and will transfer energy only to the side of the body facing the lamps.
Using banks of lamps on both sides or putting reflective material on the side opposite the lamps
will help with this problem. Wrapping the victim in an electric blanket will be the most efficient
process of warming. In this method, heat is transferred by conduction, direct contact with the
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warm surface, and it will be relatively easy to control the amount of thermal energy transferred
can be so the individual will be heated at an appropriate rate and uniformly.
Problem 7.12.
In your bank statement, the difference between the ending and initial balance corresponds to a
state function:
Balance = Balancefinal – Balanceinitial
All of the transactions correspond to path-dependent functions. They describe all the different
expenditures and deposits that are responsible for the overall change in balance. The order in
which they occur does not matter, that is, there are many possible paths to the final balance.
Problem 7.13.
This football play has moved the team from its own 35-yard line, past midfield, to the
opponent’s 40-yard line. The team has moved the ball 25 yards closer to the end zone where
they can score. The net gain in yardage is an example of a state function, a function that
depends only on the initial and final states. The path taken does not matter. Only the gain in
yardage counts as far as the game is concerned. (Of course the record books will note a
completed pass by the quarterback and receiver, because fans of the game—as well as
coaches—are interested in the paths taken as well as the outcome.)
Problem 7.14.
A significant fraction of the heat energy in a fossil-fuel burning power plants producing
electricity is transferred to the water used to cool the turbines, and there are many sources of
friction in the turbines and generator that lower their efficiency. Even when the electricity has
been generated, there will be losses over long-distance power lines usually required to reach the
power grid. Most students will intuitively recognize that 100% conversion of thermal energy to
work seems unlikely and the introduction to this chapter tells them it is not possible. Heat losses
must present serious limitations, given that the best engineers have only found a 42% efficiency
rate possible. A more complete answer to this question will be possible after studying Chapter 8,
where the second law of thermodynamics and entropy are studied in more detail.
Problem 7.15.
(a) The change from a battery being fully charged battery to being totally discharged battery is a
state function no matter how the process of discharge is carried out. The paths may be quite
different, but the end result is exactly the same, a dead battery. State functions do not depend on
the path taken to achieve the present state.
(b) When a battery is used to power a flashlight, much of the energy lost by the battery will be
used to heat the filament in the light bulb and some of this energy is released as radiant energy,
light (both visible and infrared). Batteries do get warm as they are being used, so there is some
thermal energy transfer by contact with the surroundings. When a battery is used to power a toy
car, there will be a small amount of thermal energy transfer by friction between the toy car and
the roadbed on which the car runs. Much of the energy lost will be changed to mechanical work
done in moving the toy car and some of it will be lost in the friction within the motor and other
moving parts of the car.
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Problem 7.16.
The state functions will be identical for a plane, a helicopter, a parachutist, and a hawk
descending from 1892 feet to 892 feet, as each object will make an altitude change of 1000 feet
to land safely on the ground. The path functions will be quite different. The airplane and
helicopter are obliged to follow certain traffic patterns, with designated altitudes and descent
rates into the airport. The helicopter can make a more direct descent, but still has to stay out of
the way of other traffic. The parachute jumper will likely be following a spiral pattern but
dropping pretty much vertically to land safely. The bird will take whatever path it wants to, with
little regard to the airplane, helicopter, or parachute jumper. Interestingly enough, the bird also
lands into the wind, just as the other airborne objects plan to do!
Problem 7.17.
(a) The enthalpy change for a system open to the atmosphere is dependent on the identity of the
reactants and products. The change in enthalpy for any process or reaction depends on the
initial and final states of the system, that is, on the identity of the reactants and products. This is
a correct statement.
(b) The enthalpy change for a system open to the atmosphere is zero. An enthalpy change may
be zero for a reaction, but this is not, in general the case, so this is not a correct statement.
(c) The enthalpy change for a system open to the atmosphere is negative. An enthalpy change
may be negative for a reaction, but this is not necessarily true in general, so this is not a correct
statement.
(d) The enthalpy change for a system open to the atmosphere is qP. A measure of the enthalpy
change for a process is the heat exchanged with the surroundings at constant pressure. This is a
correct statement.
(e) The enthalpy change for a system open to the atmosphere is positive. An enthalpy change
may be positive for a reaction, but this is not necessarily true in general, so this is not a correct
statement.
Problem 7.18.
(a) An open system refers to a part of the universe being studied that can exchange matter and
energy with its surroundings. Some open system examples include an open aquarium, a cup of
coffee, and hot springs. Your body is an open system since you exchange both energy (in the
form of heat) and matter (food, O2, CO2, H2O) with the surroundings.
(b) In a closed system, only energy is exchanged with its surroundings. An example of such a
system is a cup with hot coffee that is tightly closed so no vapors may escape, but the cup is not
insulated so heat energy may be exchanged with surroundings.
(c) An isolated system does not exchange matter or energy with its surroundings. An example of
a nearly isolated system might be a properly stoppered Thermos® bottle of coffee. A stopper
prevents water vapor from escaping (or other matter from entering), while the vacuum
construction keeps heat from being lost to (or gained from) the surroundings.
(d) An exothermic process is a process that releases heat to its surroundings. Reactions of
sodium metal with water, burning of a match, or condensation of steam are examples of
exothermic processes.
(e) An endothermic process is a process that absorbs heat from its surroundings. Melting ice or
evaporation of rubbing alcohol are examples of endothermic processes.
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Problem 7.19.
If we pour coffee (presumably hot) from a properly stoppered Thermos bottle into a mug, the
coffee, which was nearly an isolated system in the stoppered and insulated Thermos bottle,
becomes an open system that can exchange matter and energy with its surroundings. A closed
system can exchange energy, but not matter, with its surroundings. If, in our coffee example, we
transferred the coffee to a mug with a tight fitting lid, energy could be exchanged through the
walls of the mug, but no matter, including water vapor and the odor producing molecules from
the coffee, could escape (and no matter, such as sugar or cream, could be added).
Problem 7.20.
From the standpoint of thermodynamics, Earth is an open system, because matter and energy
are continually exchanged with the surroundings. The most obvious example is electromagnetic
radiation from the sun, which maintains the temperature of the planet. Sunlight is also
responsible for the production of essentially all the food on the planet via photosynthesis in
green plants. The oxygen that maintains aerobic life is a by-product of photosynthesis. Matter in
the form of meteorites and atoms and ions from the sun and other sources fall to the earth from
outer space. The Earth, of course, radiates some energy back to space and atoms and molecules
in the upper atmosphere are continuously lost to space.
Problem 7.21.
The diagram shows an experiment in which a hot silver bar is added to a beaker of water at
room temperature.
(a) If the object of the experiment is to find the specific heat of the silver, we have a couple of
ways to define the system and the surroundings. In one ideal case, the silver bar and the water
form the collection of atoms and molecules under consideration, the system, and we assume that
the system is isolated, so no energy or matter can enter from the surroundings, including the
beaker, the thermometer, the air above the water (and everything else in the universe). This
definition neglects the obvious necessity for energy transfer between the water and the
thermometer required to make the thermometer respond to the changing water temperature, but
we assume this is a tiny effect. For this case, no thermal energy is gained or lost by the system,
so we set the sum of any energy changes within the system equal to zero, that is,
Esilver + Ewater = 0
To use this equation to find the specific heat of the silver, you need to know the mass of the
silver bar, the mass of the water, the specific heat of water, and the temperature changes for the
silver and water when they have come to the same final temperature.
A second ideal division of the world into system and surroundings is quite similar, except that
you treat the silver bar as the system and the water as the surroundings and assume nothing else
in the universe changes. In this case, the energy change in the silver bar and the energy change
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in the water are the same, except the silver loses energy and the water gains energy, that is, they
have opposite signs:
Esilver = –Ewater
To find the specific heat of the silver, you need the same information as in the previous case.
Note that the equations you have to work with are identical, simply rearranged. It is probably
conceptually simpler to choose the silver and water together as an isolated system (which is
what we have done in Section 7.6 of the text).
(b) If you already know the specific heat of the silver, you can use this experimental set up to
determine how “ideal” the experiment actually is. That is, you can test if there is significant
thermal energy transfer to and/or through the walls of the glass beaker or to the thermometer. In
this case it is useful to define the system as the silver bar and the water and everything else, the
beaker, thermometer, air around the beaker, and the table the apparatus is on, as the
surroundings. Energy must still be conserved, so the sum of the energy changes in each part of
the system and the surroundings must be zero:
Esilver + Ewater + Esurroundings = 0
To use this equation, we need the same information as in part (a), with the addition of the
specific heat of the silver. With all this information, we can calculate the first two terms in the
equation and then solve for Esurroundings. If the set up is reasonably close to ideal, Esurroundings
will be close to zero. It’s possible that Esurroundings could be a significant fraction, perhaps five
to ten percent, of the size of the other energy changes, thus making this not a particular good set
up for the determination discussed in part (a). It is virtually certain that Esurroundings > 0, since
the surroundings are at a lower temperature than the system (except for the water at the very
beginning), thermal energy will leave the system and go into the surroundings, a positive
change in its thermal energy.
(c) The experimental diagram illustrates an open system, because matter and energy can be
exchanged with the surroundings. For example, water molecules can evaporate from or
condense onto the surface of the water. Heat may be exchanged between the system and the
surroundings. Therefore, this cannot be a closed system because matter can be exchanged. It is
not an isolated system as there is no attempt to control thermal exchanges between the system
and the surroundings.
Problem 7.22.
To accurately measure the thermal energy change for reaction, we need to account for heat that
goes to heating (in the case of exothermic reaction) or cooling (in the case of endothermic
reaction) the calorimeter itself (that is the container, thermometer, and so on, exclusive of the
water, which we explicitly account for in our calculations). The heat lost or gained to the
calorimeter must be experimentally determined for each calorimeter to assure the accuracy of
calorimetric measurements.
Problem 7.23.
[NOTE: There is a typographical error in the problem statement in the text (1st and 2nd
printings). The sign of the enthalpy change should be positive (an endothermic reaction) not
negative. Energy is required to decompose ammonia to its elements.]
The equation for decomposition of gaseous ammonia, NH3(g), is:
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Chemical Energetics: Enthalpy
Chapter 7
NH3(g)  1/2N2(g) + 3/2 H2(g)
∆H = 45.9 kJ
This reaction equation and enthalpy change indicate that the formation of gaseous ammonia
(a) evolves 45.9 kJ for each mole of ammonia formed. This completion for the sentence is
correct. The reaction equation and enthalpy change are for the decomposition of ammonia. The
sign of the enthalpy change will be reversed when the reaction goes in the opposite direction to
form ammonia from its elements. For formation of a mole of ammonia, ∆H = –45.9 kJ, that is,
energy is evolved—the reaction is exothermic.
(b) evolves 23 kJ for each mole of nitrogen used. This completion for the sentence is incorrect.
Energy will be evolved [see part (a)], but for each mole of nitrogen that reacts, twice as much
energy is evolved, that is ∆H = –91.8 kJ, because the enthalpy change shown with the equation
is for only one-half mole of nitrogen.
(c) absorbs 45.9 kJ for each mole of ammonia formed. This completion for the sentence is
incorrect, because the formation evolves energy [see part (a)].
(d) absorbs 23 kJ for each mole of nitrogen used. This completion for the sentence is incorrect,
because the formation evolves energy [see part (a)].
(e) is an exothermic process. This completion for the sentence is correct [see part (a)].
Problem 7.24.
The setup for this experiment is like that shown in Problem 7.21, except that the container here
is an insulated calorimeter, so no energy is transferred in or out of the calorimeter after the hot
metal is added. The net change in energy in the calorimeter is zero and we can write:
Emetsl + Ewater = 0 = mmetalcmetalTmetal + mwatercwaterTwater
All of the quantities in this equation are known or can be calculated from the data given in the
problem. We have Tmetal = 25.0 C – 100.0 C = –75 C and Twater = 25.0 C – 23.0 C =
2.0 C. Rearrange the equation to solve for cmetal:
cmetal =
mwater cwater Twater
(20.0 g)(4.184 J  g –1  o C1 )(2.0 o C)
=
= 0.22 J·g–1·C–1
o
mmetal Tmetal
(10.0 g)(–75.0 C)
In the table, the specific heat of silver metal is 0.22 J·g–1·C–1 (and none of the other
possibilities is close to this value), so it is likely that the metal in this experiment is silver.
Problem 7.25.
The student in this problem twice added a mass of hot metal to a calorimeter containing water.
The temperature of the hot water bath was essentially the same in both trials and we assume that
there was the same amount of water in the calorimeter for each trial. The temperature change of
the water in the first trial was 1.2 C and, in the second trial was 2.2 C. His instructor had noted
a problem with his first trial that the student corrected in the second trial. If the second trial
gives a correct (or at least more correct result), then we can conclude that the temperature
change in the first trial was in error and must have been too small. The only way the
temperature change could be smaller than it should have been is if the metal was not as hot as in
should have been, that is, not as hot as the water bath used to heat it. The problem statement
says that “he transferred it [the hot metal] slowly to a calorimeter.” Here’s the student’s error.
He has to transfer the hot metal as fast as possible from the hot bath to the calorimeter, so it
does not lose thermal energy to the surroundings (the air and the tongs used to pick it up). Our
analysis of this sort of problem (see the solutions to Problems 7.21 and 7.24) assumes that all
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Chemical Energetics: Enthalpy
the thermal energy change in the metal, after it leaves the hot water bath, occurs in the
calorimeter. If some of it actually is lost to the surroundings, then the thermal energy change of
the water in the calorimeter will be lower than it should be, as observed in the first trial here.
[The data from the second trial give cmetal = 0.23 J·g–1·C–1.]
Problem 7.26.
Assume that the total volume of the solution formed when 100. mL of 0.5M HCl and 100. mL
of 0.5M of NaOH are mixed is 200. mL. The mass of the solution, assuming that its density is
1.0 g·mL–1, is 2.0  102 g [= (200. mL)(1.0 g·mL–1)]. The temperature change caused by the
acid-base reaction is Tsolution = 22.2 C – 19.0 C = 3.2 C. The temperature of the solution
increased, which means that thermal energy is released from the reaction system, so the reaction
is exothermic: ∆H < 0. The enthalpy change for the reaction and the thermal energy (or
enthalpy) change for the solution in the calorimeter have to sum to zero, since no energy is
transferred in or out of the calorimeter. (We assume that the acid and base solutions began at the
same temperature, so that the only change in the mixed solution temperature is caused by the
reaction.) We can write the energy conservation as:
∆H + msolutioncsolutionTsolution = 0 = ∆H + (2.0  102 g)(4.18 J·g–1·C–1)(3.2 C)
∆H = –2.7  103 J = –2.7 kJ
To find the enthalpy change per mole of reactants, we have to calculate the number of moles of
reactants that reacted and convert our result to kJ·mol–1. The reaction that occurs is:
HCl(aq) + NaOH(aq)  H2O(aq) + NaCl(aq)
A mole of NaOH(aq) reacts for each mole of HCl(aq) that reacts and we mixed the same
number of moles of each, so the number of moles that react is:
mol reactant = (0.100 L)(0.5 mol·L) = 0.05 mol
–2.7 kJ
∆H(per mole) =
= 54 kJ·mol–1 = 5  101 kJ·mol–1
0.05 mol
The data are only good enough to justify one significant figure in the final result.
Problem 7.27.
(a) An endothermic dissolution process takes energy from its surroundings, the water in which
the dissolution occurs. Since energy leaves the water, it gets cooler, so when the cold pack in
which the dissolution occurred is placed on your skin, energy is transferred from your warm
skin to the cool/cold solution. When energy leaves your skin, you feel cold, as we discussed in
Chapter 1, Section 1.10 when analyzing evaporation of sweat from your skin.
(b) If the cold pack contains ample water to dissolve 50.0 g of ammonium nitrate, then the
number of moles that dissolve is:
 1 mol NH4 NO3 
mol NH4NO3 = (50.0 g NH4NO3) 
= 0.625 mol
 80.0 g NH4 NO3 
The enthalpy change for this dissolution is:
H = [H(per mole)](mol NH4NO3) = (25.7 kJ·mol–1)(0.625 mol) = 16.1 kJ = 1.61  104 J
ACS Chemistry FROG
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Chapter 7
For the ideal case, we assume that no thermal energy leaves or enters the cold pack as the
dissolution occurs, so the sum of the enthalpy/energy changes that occur is zero in the 175 g
[= (125 g) + (50.0 g)] of solution:
∆H + msolutioncsolutionTsolution = 0 = ∆H + (175 g)(4.184 J·g–1·C–1)(Tfinal – 25 °C)
Tfinal =
–(1.61  10 4 J) – (175 g)(4.184 J  g –1  o C)(–25 o C)
= 3 °C
(175 g)(4.184 J  g –1  o C)
(c) To get the result in part (b), we assumed that there is no heat loss or gain between the heat
pack and the surroundings, that is, that there are no thermal energy transfers. This assumption
cannot be correct, since the cold pack is not insulated (and would be useless to us, if it were).
(d) If the final temperature of the cold pack is about 7 °C, we know that the enthalpy change
calculated in part (b) is either incorrect or not all the change resulted in change in the solution
temperature. The most likely explanation is that there was probably energy transfer from the
warmer surroundings to the solution as dissolution was occurring, so the maximum cooling
calculated in part (b) for the ideal case is not observed
Problem 7.28.
(a) The balanced equation for the complete combustion of ethanol using full Lewis structures
for each reactant and product molecule is:
H H
H C C O H
3O O
2O C O
H H
3 O H
H
(b) Check the models you have made of each reactant and product molecule in this reaction to
be sure they have the appropriate number and kind of single and double bonds.
(c) The total enthalpy input that will be required to break all the bonds in the reactants is the
sum of all the bond enthalpies (each multiplied by the appropriate stoichiometric factor to
account for the number of this kind of bond in the reactant molecules):
∑BHreactants = (5 BHC–H + BHC-C + BHC-O + BHO-H) + (3 BHO=O)
= [(5 
(351 kJ) + (460 kJ)] + [(3 
(The stoichiometric coefficients represent numbers of moles of bonds broken for each mole of
reaction. To make the expression easier to read, we have cancelled the stoichiometric moles
with the per mole units of the bond enthalpies.) The sign of ∑BHreactants is positive, which shows
that bond breaking is an endothermic process.
(d) The total enthalpy that will be released in forming all bonds in the products is the sum of all
the bond enthalpies (each multiplied by the appropriate stoichiometric factor to account for the
number of this kind of bond in the reactant molecules and each with a negative sign to denote
the release of energy as a bond is formed):
∑(–BHproducts) = 2·2·(–BHC=O) + 3·2·(–BHO-H)
=4
–799 kJ·mol-1) + (6  –460 kJ·mol-1) = –5956 kJ
The sign of ∑(–BHproducts) is negative, which shows that bond formation is an exothermic
process.
(e) The net enthalpy change, ∆H, in this combustion reaction is:
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∆H = ∑BHreactants + ∑(–BHproducts) = +4725 kJ + (–5956 kJ) = –1231 kJ
The combustion of alcohol is an exothermic process overall, just as is expected, if the substance
is being used as a fuel. The experimental reaction enthalpy will involve a reactant (ethanol) and
product (water) in their liquid states, but the calculations based on bond enthalpies assume all
species are gases. Since this is an exothermic reaction, the ethanol is at a higher energy than the
products. Converting gaseous ethanol to liquid will lower its energy, so the effect would be to
reduce the experimental relative to calculated energy. Converting gaseous water to liquid will
lower its energy and the effect would be to widen the energy gap between the reactants and
products and thus increase the experimental relative to calculated energy. Which effect would
predominate? Three moles of water are liquefied, but only one mole of ethanol is liquefied. The
molar enthalpy of vaporization (or condensation) of ethanol and water are about the same
(Table 1.2 in Chapter 1), so the water effect would be larger and the calculated energy would
likely be an underestimate. (The value calculated from standard enthalpies of formation –
Section 7.8 – is about -1368 kJ·mol–1, which, as predicted, is somewhat larger than the value,
-1231 kJ·mol–1, from bond enthalpies.)
(f) Values for food energy are determined with combustion reactions carried out in calorimetric
experiments, so the value calculated here is an estimate for ethanol’s fuel value in our bodies.
We need to be mindful of the caveat from part (e) that the calculation of the enthalpy of
combustion using average bond enthalpies for gaseous compounds at 25 °C is only an estimate.
Average bond energies are certainly likely to be somewhat different than the actual values in the
molecules under consideration and there error is introduced when the compounds are not gases.
Problem 7.29.
The more stable isomer, ethanol or dimethyl ether, will require greater energy to dissociate it
into atoms. We can use bond enthalpies to calculate the enthalpy change for atomization of each
isomer and compare the results.
ethanol
bond
# bonds
C-H
C-C
O-H
C-O
5
1
1
1
Total
dimethyl ether
BH, kJ
# bonds
BH, kJ
2070
347
460
351
3228
6
2484
2
Total
702
3186
Ethanol requires more enthalpy to atomize to the same collection of atoms, so it is the more
stable of these two isomers by about 42 kJ (per mole of isomer).
Problem 7.30.
We can use average bond enthalpies to estimate the enthalpy change for the reaction in which
two amino acids join to form a peptide bond, releasing a water molecule:
H
H
O
H
N C C
H
R
H
O
N C C
O H
H
R'
O H
H
H
O
N C C
H
R
N C C
H
ACS Chemistry FROG
H
R'
O
H
O
O H
H
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Chemical Energetics: Enthalpy
Chapter 7
One way to approach this problem is to sum the enthalpy change for breaking all the bonds in
the reactants and the enthalpy change for making all the bonds in the products. This is more
work than we need to do, because only a few bonds are broken and made going from reactants
to products. If all the others are included, they just cancel out in the breaking and making. In
this case, a C–O and N–H bond need to be broken in the reactants and a C–N and O–H bond
made in the reactants:
∆H = ∑BHreactants + ∑(–BHproducts) = (BHC–O + BHN–H) + [(–BHC–N) + (–BHO–H)]
–
–
736 kJ = +4 kJ
(Stoichiometric coefficients, all 1 mol in this case, represent numbers of moles of bonds broken
for each mole of reaction. To make the expression easier to read, we have cancelled the
stoichiometric moles with the per mole units of the bond enthalpies.)
Problem 7.31.
All the bonds in the open-chain and cyclic forms of glucose are the same, except those involved
in making the cyclic form from the open-chain form. This change involves breaking a C=O on
carbon-1 and making two C-O bonds in its place. An O-H bond also has to be broken, but
another is made, so they cancel out. Thus, the difference is 745 kJ required to break the C=O
compared to –702 kJ released when 2 C–O bonds are made. Thus, the change from the open
chain to cyclic form is endothermic by about 43 kJ (per mole of glucose) and the open-chain
form is the more stable. In nature, almost all glucose molecules exist in cyclic forms. There
must be another stabilizing influence (or bond enthalpy calculations for glucose, which is a
solid, are not accurate enough to give us the actual enthalpy difference).
Problem 7.32.
(a) Note that, for the reactions of both diazomethane and diazirine reacting to give ethene and
nitrogen gas, we can ignore the C–H bonds, since there are four in reactants and four in the
products. We have to focus on the carbon-nitrogen, and nitrogen-nitrogen bonds. For the
reaction of two moles of diazomethane, we have:
Hrxn = {2BH(C=N) + 2BH(N=N)} + {[–BH(C=C)] + 2[–BH(NN)]}
Hrxn = 2·(615 kJ) + 2·(418 kJ) + [(– 620 kJ) + 2·(–941.4 kJ)] = –437 kJ
For the reaction of two moles of diazirine, we have:
Hrxn = {4BH(C–N) + 2BH(N=N)} + {[–BH(C=C)] + 2[–BH(NN)]}
Hrxn = 4·(276 kJ) + 2·(418 kJ) + [(– 620 kJ) + 2·(–941.4 kJ)] = –563 kJ
These reactions are highly exothermic with the predominant contribution in both cases coming
from the great stability of the triple-bonded nitrogen product. Note that that only difference
between these two calculations is the breaking of different kinds of carbon-nitrogen bonds. For
diazomethane, two C=N bonds are broken and, for diazirine, four C–N bonds are broken. Recall
that the enthalpy change required to break one C=N bond is greater than twice the enthalpy
change to break two C–N bonds. The difference is 63 kJ. For two moles of diazomethane and
diazirine, the difference in reaction enthalpies is 126 kJ [= 2·(63 kJ)], reflecting the difference in
carbon-nitrogen bonds broken.
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(b) An enthalpy-level diagram (like Figure 7.11) that combines the two reactions is:
Diazomethane is the more stable molecule, relative to diazirine. Complicating factors that might
make the conclusions based on average bond enthalpies suspect include the strain of making the
three-membered ring (which distorts the usual bonding angles for carbon and nitrogen and will
change the bond enthalpies) in diazirine and/or the formal separation of charge in the
diazomethane molecule that could make it less stable than a comparable molecule without
formal charge.
(c) You might argue that the strain energy in diazirine is responsible for its higher energy
(63 kJ·mol–1 higher, based on these bond enthalpy calculations), but the argument is not wholly
satisfactory. To complicate matters, we find experimentally that diazirine can be kept as a gas in
the dark at room temperature indefinitely, whereas samples of gaseous diazomethane react as
shown here and on the walls of their containers within a few hours. Based on exothermicity, the
reactions shown here seem very favorable, although we know that exothermicity is not a good
criterion for deciding the direction of a reaction. Diazomethane does react, but some factor
slows the diazirine reaction so much that it is undetectable.
Problem 7.33.
As you see, the molecules in this table are isomers,
and their complete combustion will give identical
products in each case: 6 molecules of CO2 and 6
molecules of H2O for each molecule of sugar
oxidized. The identical values of their enthalpies of
combustion (to within 0.3%) indicates that the
bonding must be essentially identical in these
isomers. That is, there must be the same number of
ACS Chemistry FROG
Sugar
Formula
fructose C6H12O6
galactose C6H12O6
glucose C6H12O6
Hcombustion
–1
kJ mol
–2812
–2803
–2803
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Chemical Energetics: Enthalpy
Chapter 7
C–C, C–H, C–O, C=O, and O–H bonds in all three molecules. This is true if you compare their
open-chain forms with one another or their cyclic forms with one another (in nature, the cyclic
forms vastly predominate).
Problem 7.34.
(a) Disaccharides are sugars formed by the combination of two simpler sugars, for example:
H
H HO H
C
O
H OC
C H
H C
O
HO C
H
H
C OH
H
H
H HOH
C
O
H OC
C H
H C
O
HO C
H
H
H
O
H CH
O
H O
C H
H H
H C
O C
O
C
C
H
O
H
H
H
C
C OH
H
H
O
H
C
O
C H
H C
O C
O
H H
H
H
H2 O
COH
H
In this reaction a C–O bond is broken in one sugar and an O–H bond in the other. Then to make
the products, a C–O bond is made between the two sugars and an O–H bond is made to produce
the water (from the OH that had left a sugar). Exactly the same number and kind of bonds are
broken and made. Bond enthalpy calculations (which we do not have to do, if there is no change
in the kind and number of bonds) give an estimated Hreaction  0 kJ for this reaction.
(b) One way to estimate the enthalpy of combustion of maltose is to break the reaction into two
reactions: hydrolysis of maltose to give two glucose and then combustion of the glucose:
maltose + H2O  2 glucose
Hrxn = 0 kJ
2glucose + 12O2  12CO2 + 12H2O
Hrxn = – 5606 kJ (from 7.17)
The sum of these two reactions is:
maltose + 12 O2  12 CO2 + 11 H2O Hrxn = – 5606 kJ
This is the combustion reaction for maltose; we predict Hrxn = – 5606 kJ to be compared with
the experimental value of –5644 kJ. The difference is small (only about 0.6%), but the
combustion enthalpies are experimental values that are good to at least four significant figures.
This means that the estimated value of 0 kJ for the formation of maltose (or the reverse, its
hydrolysis) is incorrect. The actual value for the hydrolysis above must be about –36 kJ, in
order to make the sum of the two reaction enthalpies above equal to the measured enthalpy of
combustion of maltose. We know that bond enthalpies are averages that can be off by a few
percent for any particular bond in a molecule, so we shouldn’t be surprised that this small
discrepancy occurs. It is a warning, however, not to trust bond enthalpies when the quantity of
interest is the difference between two large values — small errors in the large values can lead to
large percentage discrepancies in their difference.
Problem 7.35.
(a) We can use average bond enthalpies to estimate the standard enthalpy change (in kJ per
mole of hydrazine) for the reaction between hydrazine and hydrogen peroxide:
H2NNH2(g) + 2HOOH(g)  N2(g) + 4H2O(g)
Let’s cancel the H–O bonds before going through all the arithmetic. Four are broken on the left
and eight are made on the right, so we will include four in the products, but ignore them in the
reactants. The Hrxn for this reaction is given by:
Hrxn = –BH(NN) – 4BH(H–O) + 4BH(H–N) + BH(N–N) + 2BH(O–O)
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Chapter 7
Chemical Energetics: Enthalpy
Hrxn = – (941 kJ) – (1840 kJ) + (1572 kJ) + (193 kJ) + (284 kJ) = – 732 kJ
(b) We can use average bond enthalpies to estimate the standard enthalpy change for the
reaction between two moles of ammonia and hydrogen peroxide (in kJ per two moles of
ammonia):
2NH3(g) + 3HOOH(g)  N2(g) + 6H2O(g)
In this case, canceling the H–O bonds leaves a net of six in the products:
Hrxn = –BH(NN) –6BH(H–O) + 6BH(H–N) + 3BH(O–O)
Hrxn = – (941 kJ) – (2760 kJ) + (2358 kJ) + (426 kJ) = –917 kJ
Two moles of ammonia reacting release about 25% more enthalpy than a mole of hydrazine.
(c) A mole of hydrazine is 32.0 g of hydrazine, so the enthalpy change from hydrazine is –
22.9 kJ·g–1   (–732 kJ) (32.0 g) . The comparable value for ammonia is –26.97 kJ·g–1


  (–917 kJ)

(34.0 g) . Since more enthalpy is released by the ammonia reaction, why isn’t

ammonia used, instead of hydrazine as the fuel? There are quite practical reasons for using
hydrazine. Probably the most important, from an engineering point of view, is that hydrazine is
a liquid at room temperature, Table 7.4, and can be easily stored and pumped around in the
rocket. Ammonia, on the other hand, is a gas at room temperature and has to be refrigerated
and/or kept under high pressure to store it as a liquid. This just increases the complexity of the
rocket. Consider also the oxidant, hydrogen peroxide. About one-and-one-half times as much
peroxide is required for the oxidation of ammonia as for an equal mass of hydrazine. Therefore,
the amount of fuel and oxidant together is less for hydrazine. The same size rocket can carry a
somewhat greater payload with the hydrazine fuel because less mass of fuel and oxidant are
needed. Work out for yourself the enthalpy change for reaction of one gram of the
stoichiometric mixture of fuel and oxidant for each fuel.
Problem 7.36.
(a) The two bonds that have to be broken in the reactants and the three that have to be made in
the product, in order to react ethene and water to produce ethanol, are indicated here:
H H
H C
C
H H
H
H O H
H C
C
H
H O H
Using the data in the Web Companion, Chapter 7, Section 7.7, page 3, breaking a mole of the
H–O and C=C bonds requires:
Hbreak = (460 kJ) + (620 kJ) = 1080 kJ
Making the C–H, C–O, and C–C bonds releases energy, so the enthalpy change is negative:
Hmake = – [(414 kJ) + (351 kJ) + (347 kJ)] = –1112 kJ
The net change in enthalpy is the sum of these two values
Hnet = 1080 + (–1112) = –32 kJ
ACS Chemistry FROG
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Chemical Energetics: Enthalpy
Chapter 7
(b) In the Web Companion, you find that breaking all the bonds in the reactants requires 3196 kJ
(per mole of reaction) and making all bonds in the product releases –3228 (per mole of
reaction). The net change for the reaction is –32 kJ. The results for the two methods are
identical.
(c) Finding that the results of the two methods are identical is expected. When all the bonds in
the reactants are broken (atomizing the molecules), we are breaking several kinds of bonds,
several C–H bonds, for example, that will simply be remade in the product. Our assumption in
using average bond enthalpies is that the bond enthalpy for a given kind of bond is the same in
all molecules. Thus when the bonds are remade in the product, exactly the same amount of
enthalpy is released as was required to break the bonds in the first place -- the breaking and
making simply cancel out. The only bond breaking and making that contribute to a net
difference for the reaction are those that are different in the reactants and product, which is what
we focused on in part (a). You can see this argument graphically in the Web Companion,
Chapter 7, Section 7.7, page 3, if you first break the two bonds selected in the Companion,
Chapter 7, Section 7.7, page 2, and then break all the others to atomize the reactants. Now when
you make the product bonds, first make all those that are the same in the product as in the
reactants and then, finally, make the three new bonds. Compare the lengths of the bars on the
graph for the breaking and making the same set of bonds.
Problem 7.37.
Carbon–hydrogen bond-dissociation enthalpies are shown here for two different hydrogen
atoms in the reactions of propene and 2-butene:
(1)
H
H
H C C C
H H H (2)
(3)
H
H H
H C C C C H
H H
H
(4)
dissociation enthalpy
kJ mol –1
H
H C C C
H H H
H
361
H
H
H C C C
H
H
H
430
H H
H C C C C H
H H
H
H
358
H
H
H C C C C H
H H
H
H
430
(a) This enthalpy level diagram represents the enthalpies of the reactant and the products from
reactions (1) and (2). It takes less enthalpy to produce the products from reaction (1). Thus, the
free radicals formed in reaction (1) are more stable by 69 kJ·mol–1 than those formed in reaction
(2). A similar diagram can be drawn for the reactant and products from reactions (3) and (4) to
show that the products from reaction (3) are 72 kJ·mol–1 more stable than those from reaction
(4).
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ACS Chemistry FROG
Chapter 7
Chemical Energetics: Enthalpy
(b) The “equivalent” Lewis structures for the three- and four-carbon free radicals formed in
reactions (1) and (3) are:
H
H C C C
H H H
H
C C C H
H H H
H H
H C C C C H
H H
H
H H
H C C C C H
H H
H
The structures for the three-carbon radical are identical in energy, so we know from our
discussion in Chapter 5, Section 5.7, that the electrons involved are delocalized and of lower
energy than they would be if the two pi electrons and single free-radical electron did not
interact. The two four-carbon structures are not quite identical, but the sigma bonding to the
carbon atoms with the free radical electron is very similar in the two structures (the bonding is
to two hydrogen atoms in the first and to a hydrogen atom and carbon atom in the second
structure), so we can treat the structures as almost energy-equivalent and again see that there is
electron delocalization to lower the energy of this free radical, compared to the one with the
free-radical electron on the doubly-bonded carbon.
Problem 7.38.
The rhombic allotrope of sulfur, the familiar light yellow powder, is the more stable allotrope at
room temperature. At higher temperatures, near the melting point, the triclinic allotrope must be
more stable, since it is the form that crystallizes from the molten sulfur. Upon standing at room
temperature, the rhombic allotrope forms slowly on the surface of the triclinic needles, which
indicates a slow transformation to the more stable form. If we wait long enough, all the needles
change back to the light yellow powder.
Problem 7.39.
We can use standard enthalpies of formation to find the enthalpy for the reaction producing urea
from ammonia and carbon dioxide:
2NH3(g) + CO2(g)  H2NC(O)NH2(s) + H2O(l)
The standard enthalpies of formation for ammonia, carbon dioxide, urea, and water are – 46.1, –
393.5, –333.0, and –285.8 kJ·mol–1, respectively. The standard enthalpy change for the reaction
is (where the stoichiometric coefficients – units of mol – are not shown, if they are unity):
Horxn = Hof(H2O(l)) + Hof(H2NC(O)NH2(s)) – 2Hof(NH3(g)) – Hof(CO2(g))
Horxn = (–285.8 kJ) + (–333.0 kJ) – (– 92.2 kJ) – (– 393.5 kJ) = –133.1 kJ
Problem 7.40.
[NOTE: The problem is misstated when it asks for the enthalpy of vaporization (change from
liquid to gas) of urea. We can get the enthalpy of sublimation (change from solid to gas) but
ACS Chemistry FROG
75
Chemical Energetics: Enthalpy
Chapter 7
have no way to estimate the enthalpy change from liquid to gas, because we have no way to
estimate the enthalpy change from solid to liquid.]
To estimate the enthalpy of sublimation of urea, we recall that bond enthalpy values are
applicable to gas phase molecules. We could use bond enthalpies to calculate the enthalpy
change for the reaction in Problem 7.39 with the urea and water as gases and compare the result
with the enthalpy change calculated in the solution for Problem 7.39 with the urea as a solid and
water as a liquid. The difference is an estimate of the enthalpy of sublimation of urea plus the
enthalpy of vaporization of water. For the gas phase reaction we have:
Horxn = BHreactants + (–BHproducts)
Horxn = {6BH(H–N) + 2BH(C=O in CO2)}
+{2[–BH(H–O)] + 4[–BH(H–N)] + 2[–BH(C–N)] + [–BH(C=O)]
o
H rxn = {(2358 kJ) + (1598 kJ)} + {(–920 kJ) + (–1572 kJ) + (–552 kJ) + (–745 kJ)}
Horxn = {3956 kJ} + {–3789 kJ} = 167 kJ
These changes are represented on this enthalpy diagram sketch (not to scale):
Enthalpy is required to make gaseous urea and water from ammonia and carbon dioxide. The
enthalpy change going from solid urea and liquid water to the gases is 300 kJ [= (133 kg) +
(167 kJ)]. The enthalpy of vaporization of water is 44 kJ·mol–1, so the condensation of water
releases enthalpy, –44 kJ. This leaves 256 kJ as the enthalpy required to change solid urea to
gaseous urea. Our estimate for the enthalpy of sublimation of urea is about 256 kJ·mol–1.
Problem 7.41.
(a) The fermentation of glucose to ethanol and carbon dioxide is represented by this reaction:
C6H12O6(s)  2C2H5OH(l) + 2CO2(g)
From Appendix B, the standard enthalpies of formation of glucose, ethanol, and carbon
dioxides are –1274.4 kJ·mol–1, –277.69 kJ·mol–1, and –393.51 kJ·mol–1, respectively. The
standard enthalpy change for the fermentation reaction is:
76
ACS Chemistry FROG
Chapter 7
Chemical Energetics: Enthalpy
Hreaction = npH0f(products) – nrH0f(reactants)
Hreaction = 2Hf[C2H5OH(l)] + 2Hf[C2H5OH(l)] – Hf[C6H12O6(s]
Hreaction = 2(–277.69 kJ) + 2(–393.51 kJ) – (–1274.4 kJ) = –68.0 kJ
(Stoichiometric coefficients represent numbers of moles of bonds broken for each mole of
reaction. To make the expression easier to read, we have cancelled the stoichiometric moles
with the per mole units of the bond enthalpies.)
(b) Since Hreaction is negative (enthalpy is released), the fermentation reaction is exothermic.
(c) If a reaction is exothermic, the reactant (glucose) has higher enthalpy than the products
(ethanol and carbon dioxide).
(d) The standard enthalpy change we calculated for the reaction is the change for the reaction as
represented by the equation in part (a), where all quantities are molar quantities. That is, 68.0 kJ
are released when one mole of glucose gives two moles of ethanol (and two moles of carbon
dioxide). If fewer than two moles of ethanol are produced, this means that less than one mole of
glucose has reacted. We calculate the standard enthalpy change for the reaction producing 5.0 g
of ethanol (EtOH = C2H5OH), H, like this:
 1 mol EtOH   –68.0 kJ 
5.0 g EtOH = (5.0 g EtOH) 

 = –3.7 kJ
 46.0 g   2 mol EtOH 
(e) The quantity of heat released when 95.0 g of C2H5OH(l) is formed at constant pressure is qP,
which is equal to H for the reaction at one atmosphere pressure:
 1 mol EtOH   –68.0 kJ 
95.0 g EtOH = (95.0 g EtOH) 

 = –70.2 kJ
 46.0 g   2 mol EtOH 
Problem 7.42.
(a) The gas phase reaction of methanoic (formic) acid with ammonia produces formamide
(which contains a peptide bond) and water:
O
O
H
H C
N H
O H
H
H C
N H
H
H
O
H
In this reaction, a C–O and N–H bond are broken and a C–N and O–H bond are made. We can
use average bond enthalpies to estimate the enthalpy change for the reaction:
H = ∑BHreactants + ∑(–BHproducts) = (BHC–O + BHN–H) + [(–BHC–N) + (–BHO–H)]
–
–
736 kJ = +4 kJ
(Stoichiometric coefficients, all 1 mol in this case, represent numbers of moles of bonds broken
for each mole of reaction. To make the expression easier to read, we have cancelled the
stoichiometric moles with the per mole units of the bond enthalpies.) This value and one we
calculated for peptide bond formation in Problem 7.30 for a different reaction are identical,
+4 kJ. This is not surprising, because the equation used to get the enthalpies of reaction is
identical in the two cases. The equation is the same, because the reaction is the same in terms of
bonds broken and made in forming the peptide bond from a carboxylic acid and an amine.
(b) We can also use standard enthalpies of formation to calculate the change in enthalpy for this
reaction, assuming reactants and products are gases in their standard states. The enthalpies of
formation of gaseous methanoic acid and formamide are –379 and –186 kJ·mol–1, respectively.
ACS Chemistry FROG
77
Chemical Energetics: Enthalpy
Chapter 7

Hreaction = npH0f(products) – nr H0f(reactants)

Hreaction = H0f[HC(O)NH2] + H0f[H2O] – H0f[HC(O)OH] – H0f[NH3]

Hreaction = (–186 kJ) + –242 kJ – –379 kJ – –46 kJ–
(Stoichiometric coefficients, all 1 mol in this case, represent numbers of moles of reactants and
products for each mole of reaction. To make the expression easier to read, we have cancelled
the stoichiometric moles with the per mole units of the enthalpies of formation.)
(c) We used average bond enthalpies for the calculation in part (a) and standard enthalpies of
formation in part (b) and both give values close to zero kilojoules for the formation of one mole
of peptide bonds. This excellent agreement is probably somewhat fortuitous, since average bond
enthalpies are not likely to be exactly correct for the bonds in any particular molecule. Although
the bond enthalpy calculation predicts a slightly endothermic reaction and the enthalpy of
formation gives a slightly exothermic reaction, this is not a major discrepancy when the values
are so close to zero and we know that bond enthalpy calculations are almost certain to be off a
bit from the experimental value obtained from enthalpy of formation data.
Problem 7.43.
(a) The equation for the combustion of ammonia and its standard enthalpy change are:
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) Hreaction = – 905 kJ
We can use the standard enthalpies of formation for NO(g) and H2O(g) and the standard
enthalpy change for the reaction to calculate the standard enthalpy of formation for NH3(g).
Hreaction = npHf(products) – nr Hf(reactants)
–905 kJ = [4(90.25 kJ) + 6(–241.82 kJ)] – [4(Hf(NH3) + 5(0 kJ)]
–905 kJ = –1090 kJ – 4(Hf(NH3)
Hf(NH3) = –46 kJ·mol-1
(Stoichiometric coefficients represent numbers of moles of reactants and products for each mole
of reaction. To make the expression easier to read, we have cancelled the stoichiometric moles
with the per mole units of the enthalpies of formation.)
(b) The value for Hf(NH3) from Appendix B in the textbook is –46.11 kJ·mol-1. This is
excellent agreement with the result from part (a) and points up the fact that the table values are
often determined by means of values for combustion reactions, which are relatively
straightforward to run.
Problem 7.44.
Three reactions that can take place in the process of coal gasification are:
Reaction 1: C(s) + H2O(g)  H2(g) + CO(g)
Reaction 2: C(s) + 1/2O2(g)  CO(g)
Reaction 3: C(s) + 2H2O(g)  CH4(g) + O2(g)
The enthalpy of reaction for each of these three reactions is:
Reaction 1:

Horxn = Hof(H2) + Hof(CO) – Hof(C) – Ho(H2O)
Horxn = (0 kJ) + (–110.53 kJ) – (0 kJ) – (–241.82 kJ)
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Chapter 7

Chemical Energetics: Enthalpy
Horxn = +131.29 kJ
Reaction 2:
Horxn = Hof(CO) – Hof(C) – 1/2Hof(O2)
Horxn = (–110.53 kJ) – (0 kJ) – (0 kJ) = –110.53 kJ
Reaction 3:

Horxn = Hof(CH4) + Hof(O2) – Hof(C) – 2Hof(H2O)
Horxn = (–74.81 kJ) + (0 kJ) – (0 kJ) – 2(–241.82 kJ)

Horxn = +408.83 kJ
The most (only) exothermic reaction is Reaction 2 and the most endothermic is Reaction 3.
Thus, Reaction 2 is the most energetically favorable and Reaction 3 is the least. This does not
imply that chemists should work to control the reaction conditions to maximize Reaction 2. The
main goal of coal gasification is to produce fuels and starting materials for synthesis. CO and H2
are the most desirable products, because they can be used as starting materials for several
syntheses as well as being a good fuel mixture (called water gas).
Problem 7.45.
In a coupled reaction, the reaction providing the energy does not proceed unless the reaction
requiring the energy occurs. For example, the exothermic oxidation of glucose is coupled with
the endothermic formation of adenosine triphosphate, ATP4-, from adenosine diphosphate,
ADP3-. The reverse reaction, the hydrolysis of ATP4- to ADP3- is exothermic and is coupled to
other energy-requiring biological functions such as locomotion, information processing, and
synthesizing new biological molecules.
Problem 7.46.
The principal role of ATP is to act as an energy carrier from reactions that produce energy, such
as the oxidation of glucose, to reactions that require energy, such as the synthesis of proteins,
movement of muscles, and transport of molecules and ions in and out of cells. ATP is not a
storage molecule for energy because there is only a small amount of it in a cell that rapidly
cycles between energy producing and energy requiring processes. One analogy that is often
used is that ATP is like the money in your wallet or purse. The storage place for your money is
the savings account in your bank (this money is like fuel molecules in your body), but you can’t
purchase anything with that money, because it resides in the bank. If you take some of the
money out of the bank and have it in your wallet or purse, you can spend it for things you need
or want (just as you synthesize new molecules or move about using the energy from ATP
hydrolysis). When the money you carry is used up, you can return to the bank for more. (All of
this assumes that you have a savings account with a reasonable amount of money, just as you
must have fuel—from your food—in order to continue to produce ATP to sustain yourself.)
Problem 7.47.
(a) Waste heat from power plants is generally carried in the form of steam that has passed
through the turbines that run the generators to produce electricity. To capture energy from this
steam to evaporate seawater (to desalt it), requires that the power plant be sited on the seacoast.
To be useful in producing a lot of water, this has to be a large power plant, which could be a
limitation, if such a large power facility is not needed. Although the steam could be used
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directly to heat seawater in a distillation apparatus, such a single stage process is not very
efficient and is difficult to make into a continuous process. At present, most of the coupled
power/desalting facilities use a continuous multistage flash evaporation technology in which the
brine is evaporated at reduced pressure in a series of stages. The waste heat from power plants is
often used for heating buildings in cold and temperate climates and is not available for
desalting. The largest power/desalting facilities are in the warm countries of the Middle East
where the steam would be wasted, if not used for desalting. Systems that couple power
generation with desalting are 10-20% more energy efficient than those that carry out the
processes separately (that is, use some of the power generated to run a desalting process). A
schematic diagram of a multistage flash evaporation system is shown here.
[NOTE: This diagram is credited to desalinization.com by the author of the report found at:
http://www.twdb.state.tx.us/Desalination/The%20Future%20of%20Desalination%20in%20Tex
as%20-%20Volume%202/documents/C4.pdf.]
(b) Examples of other energetically coupled processes will vary depending on the background
of the students. They may present examples from engineering, other biological applications, or
from ecology courses. [Note that this question could be assigned as a web-based research
question.]
Problem 7.48.
The function of Mg2+ in many ATP4--coupled reactions is to complex with the negatively
charged ATP ions. (Complexation of cations by Lewis bases was the subject of Chapter 6,
Section 6.6.) The complex has a three dimensional structure that fits appropriately in the folds
of the enzymes that catalyze these reactions, whereas, the free ATP4- does not.
Problem 7.49.
(a) The equation for the uncoupled oxidation of pyruvic acid is:
2C3H4O3(l) + 5O2(g)  6CO2(g) + 4H2O(l)
Hpyruv oxid = –2332 kJ
Hpyruv oxid has been calculated from the standard heats of formation:
Hpyruv oxid = [6(–393.5 kJ) + 4(–285.8 kJ)] – [2(–585.8 kJ)] = –2332 kJ
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(Stoichiometric coefficients represent numbers of moles of reactants and products for each mole
of reaction. To make the expression easier to read, we have cancelled the stoichiometric moles
with the per mole units of the enthalpies of formation.)
(b) Recall that Hreaction= 21 kJ·mol-1 for the formation of one ATP4- from ADP3- and HOPO32. Therefore, the energy required to form 30 ATP4- will be 30 times as great.
30ADP3-(aq) + 30HOPO32-(aq) + 30H3O+(aq)  60H2O(l) + 30ATP4-(aq)
Hreaction = (30 mol)(21 kJ·mol-1) = 630 kJ
For the reaction of pyruvic acid oxidation coupled to ATP4- formation, we have:
Hcoupled reaction = (–2332 kJ) + 630kJ = –1703kJ.
The efficiency (percent) energy recovered in the form of ATP4- is:
 630 kJ 
% efficiency = 
100% = 27%
 2332 kJ 
If the metabolic reactions actually occurred with all species in their standard states, only 630 kJ
of the 2332 kJ released in the oxidation would be captured in the production of ATP4-. The
remaining 1703 kJ would be lost as heat.
Problem 7.50.
You should, by all means, invest in an inventor who claims to have built a device that is able to
do about 0.8 kJ of work for every 1 kJ of thermal energy put into it (after you have seen proof
that his device can actually do what he says). This device is about 80% efficient in converting
random thermal motion to the directed motion of work. This is a good deal better than the best
engines now available. There is no violation of the first law here, since he is not getting out
more than he puts in. The “lost” energy presumably remains in the device in some form or is
expelled as thermal energy to some other part of the surroundings. The inventors you have to
look out for are those who claim to get more work out of a thermal engine than the thermal
energy they put in; that is a violation of the first law.
Problem 7.51.
The work on a gas, w, associated with change in volume of a gas at constant pressure is –Pext∆V.
For a change from 10. L to 1.0 L at a constant pressure of 5.0 atmospheres the work is:
w = –Pext∆V = –(5 atm) (1.0 L – 10.0 L) = 45 L·atm
 101.3 J 
45 L·atm = (45 L·atm) 
= 4.6  103 J = 4.6 kJ
 1 L  atm 
Work is being done on the gas, as the volume is decreasing from 10.0 L to 1.0 L. A positive
sign for w is associated with work done on the system.
Problem 7.52.
The value of qP, the thermal (heat) energy required to cause the volume of a hot-air balloon to
change from 5.0  106 L to 5.5  106 L at a constant pressure of 1.0 atm, is given in the problem
statement: 1.5  108 J. This is the thermal energy added to the system at constant pressure. The
pressure-volume work done in this process is:
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 101.3 J 
w = –Pext∆V = –(1.0 atm)[(5.5  106 L) – (5.0  106 L)] 
= –5  107 J
 1 L  atm 
The negative sign for the work makes sense, because the gas is expanding and doing work on
the surroundings (pushing back the atmospheric gases), so work is leaving the system. The
change in internal energy, ∆E, for this process is:
∆E = qP + w = qP + (–Pext∆V)
∆E = (1.5  108 J) – (5  107 J) = 1.0  108 J
Problem 7.53.
The chemical energy of reaction of the metal with oxygen in a photographic flash bulb is
transformed to light energy and heat energy in the course of the reaction. As long as the bulb
does not explode, this is a constant volume reaction. This means no pressure-volume work is
done on or by the gas, so w = 0. Thus, E = qV + w = qV, where the subscript reminds us that the
reaction is carried out at constant volume.
Problem 7.54.
Assume that the gaseous chemical reaction in which the system loses heat and contracts takes
place at constant pressure (because we have been given no way to analyze a reaction in which
the pressure changes). Since H = qP and qP is negative (heat leaves the system), we know that
H is negative for this process. We also know that H = E + PV. The system contracts so
Vfinal < Vinitial and V = Vfinal – Vinitial is negative. This means that PV will be negative, because
P is always a positive value. E may be either positive or negative depending on the magnitude
of H and PV. In summary: PV is “–”; H is “–”; and, E may be “+” or “–”, depending on
the relative magnitudes of H and PV.
Problem 7.55.
(a) The pressure-volume work performed on the gas in the syringe in Experiment A is less than
that performed in Experiment B. Given that the cross-sectional area is the same for both
syringes and the pressure is the same in both experiments, the pressure-volume work is directly
proportional to the value of ∆x, which is smaller for Experiment A. The mathematical
relationship is:
w = –Pext  A  ∆x
The sign of the work is positive, because ∆x is negative, which is correct for work done on the
system.
(b) Although ∆x is the same for both Experiment A and the third experiment with a wider bore
syringe, ∆V is not the same because the cross-sectional area of the syringe is greater in the third
experiment. Because pressure-volume work is a product of the pressure, which is the same in
both experiments, and the change in volume, the work done in the third experiment is greater
than that done in Experiment A.
Problem 7.56.
Assume that this reaction occurs at constant temperature and pressure:
3H2(g) + N2(g)  2NH3(g)
To analyze the changes taking place, we note that 3 volumes of H2(g) react with 1 volume of
N2(g) to produce 2 volumes of NH3(g). Thus, V = –2 volumes [= (2 volumes) – (4 volumes)]
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and work, w = –Pext∆V, is greater than zero, indicating that work must be done on (enter) the
system during this change.
Problem 7.57.
(a) A driver’s side airbag might typically contain about 95 g of sodium azide that reacts to
produce the nitrogen used to inflate the airbag:
2[Na+][N3–](s)  2Na(g) + 3N2(g)
This stoichiometric equation shows that 3 moles of nitrogen gas are produced for every two
moles of the azide that react. The azide has a molar mass of 65 g. Therefore:
 1 mol azide   3 mol N 2 
95 g azide = (95 g azide) 
  2 mol azide  = 2.2 mol N2 produced

65 g
If a mole of nitrogen gas occupies about 25 L at 25 C, the volume occupied by 2.2 mol N2 is
55 L [= (2.2 mol)(25 L·mol–1)] so the airbag is about 55 L in volume, which is approximately 2
cubic feet (or you can imagine about 25 empty, two-liter soft drink bottles suddenly appearing
in front of you when the airbag deploys).
(b) The external pressure against which the airbag pushes to inflate is atmospheric pressure,
Patm, 1.01  105 Pa (N·m–2). The volume change, V, of the gas is 55  10–3 m3 (since we go
from no gas to a final volume of 55 L and a liter is 10–3 m3 = a cubic decimeter). Thus, w, the
work done on the gas is:
w = –Patm·V = –(1.01  105 N·m–2)·(55  10–3 m3) = –5.6 kJ
The work done on the gas is negative, because the gas actually does work on its surroundings
(that is, work leaves the system) pushing back the atmospheric gases.
(c) We know that H = E +PextV and that the pressure-volume product is a positive 5.6 kJ.
Therefore, the measured thermal energy release at constant volume, qV = E, will be smaller by
5.6 kJ than that measured at constant pressure, qP = H.
Problem 7.58.
We are asked to calculate the amount of work that was done and the standard change in internal
energy, E, when 0.5 mol of H2(g) was reacted with 0.3 mol of O2(g), at a constant pressure of
1.0 atm, with a change in volume of –6.1 liters. The stoichiometry and standard enthalpy change
for the reaction are:
2H2(g) + O2(g)  2H2O(g) H = –484 kJ
Since E = H – PextV, we will first need to calculate both H and PextV (for the amount
of reaction that actually occurs) to obtain the value for E. The initial amounts of H2(g) and
O2(g) are not in the 2:1 molar ratio of the balanced equation. H2(g) is the limiting reagent and is
consumed completely while 0.05 mol of O2(g) [= (0.3 mol initial) – (0.25 mol reacted)] remains
unreacted. The enthalpy value given with the reaction equation is for the reaction of two moles
of H2(g) with one mole of O2(g). H for the reaction of 0.5 mol of H2(g) is:
 –484 kJ 
H[for 0.5 mol H2(g)] = [0.5 mol H2(g)] 
= –121 kJ
 2 mol H 2 (g) 
The work, w, done on the gas is
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 101 J 
w = –PextV = –(1.0 atm)(–6.1 L) 
= 6.2  102 J = 0.62 kJ
 1 L  atm 
The work is positive, so work is done on the system, which is consistent with the observation
that the volume of the system decreased (because the number of moles of gas decreased from
0.8 mol to 0.55 mol). Finally, we have the change in internal energy:
E = H – PextV= –121 kJ + 0.62 kJ = –120 kJ
Problem 7.59.
(a) Consider a 10 mL sample of pure water at 25 C and 1 atm pressure (state A). The sample is
cooled to 1 C and then the pressure is reduced to 0.5 atm (state B). It takes 5 hours to carry out
the change from state A to state B. The sample is then heated and the pressure is raised to 1 atm.
In one minute the water is back at 25 C ( the sample is back at state A). The internal energy
change in going from state A to B is equal to, but opposite in sign to the internal energy change
going from state B to A. This statement is true. Internal energy is a state function. The
difference in the internal energies of two states is independent of the path taken between the
states.
(b) The work done in changing from state A to B and the work done in changing from state B to
A in part (a) are numerically the same but opposite in sign. This statement is false. Work
depends on the path, that is, the method of going from one state to another. In this case the
volume changes are the same (but opposite in sign) going from A to B and from B to A. If each
change is carried out at constant external pressure the path from A to B is at 0.5 atm and the
path from B to A is at 1 atm, The pressures are not the same, so the numerical values of the
work cannot be the same.
(c) The enthalpy change for a change of state of a system is independent of the exact state of the
reactants or products. This statement is false. The enthalpy change for a change of state depends
on the exact state of the reactants or products.
(d) At constant pressure the amount of heat absorbed or evolved by a system is called the
enthalpy change, H. This statement is true. The heat absorbed or evolved by a system in a
constant pressure change is equal to the enthalpy change for the process: H = qP.
(e) If volume does not change, the amount of heat released during a change of state of a system
is equal to the decrease in internal energy of that system. This statement is true. The heat
released by a system in a constant volume change is equal to the internal energy change for the
process: E = qV.
(f) If a reaction is spontaneous, it is always exothermic. This statement is false. Endothermic
reactions which are spontaneous include freezing water at temperatures below 0 C and
dissolving many salts (for example NH4NO3).
(g) If the enthalpy change for this reaction, N2(g) + O2(g)  2NO(g), is 180.5 kJ, then the
enthalpy change for this reaction, 1/2N2(g) + 1/2O2(g)  NO(g), is 90.2 kJ. This statement is
true. The enthalpy change for a reaction is directly proportional to the number of moles of
reaction that occur.
Problem 7.60.
(a) The NH4Cl solution is cool to touch, because NH4Cl(s) dissolution is endothermic, H =
14.8 kJ·mol-1. Heat is transferred from the surroundings (including the solution, the beaker, and
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your hand) to the reaction. The CaCl2 solution is warm to touch, because CaCl2(s) dissolution is
exothermic, H = –82.9 kJ·mol-1. Heat is transferred from the reaction to the surroundings
(including the solution, the beaker, and your hand).
(b) In order to maintain the temperature of the beaker of solution as dissolution occurs, we have
to add 15.1 kJ of heat per mole of NH4Cl(s) that dissolves. For the CaCl2 solution,, we have to
remove 82.9 kJ of heat per mole of CaCl2(s) that dissolves.
(c) To determine H for dissolution of salts in water, the concentration of the dissolved salt
must be considered. The above values for the enthalpy changes are obtained by measuring
enthalpy changes for different final concentrations and extrapolating to the limit of infinite
dilution.
Problem 7.61.
[NOTE: Data for Hf of NH4Cl(s), –314.43 kJ·mol-1, and of CaCl2(s), –795.8 kJ·mol-1, from
Appendix B are required for the is problem, as well as the data from Problem 7.60. The Hf for
Cl–(aq) given in this problem statement is slightly different than the value in Appendix B. We
will use the value in the problem statement for the calculations here.]
We can represent the dissolution reaction for NH4Cl(s) as:
NH4Cl(s)  NH4+(aq) + Cl–(aq)
Using Hf[NH4Cl(s)] = –314.43 kJ·mol-1, from Appendix B, the enthalpy of dissolution from
Problem 7.60, and Hf for Cl–(aq) = –167.4 kJ·mol-1, from this problem statement, we can find
Hf[NH4+(aq)]:
Hdissolution = {Hf[NH4+(aq)] + Hf[Cl–(aq)]} – {Hf[NH4Cl(s)]}
Hf[NH4+(aq)] = Hdissolution – Hf[Cl–(aq)] + Hf[NH4Cl(s)]
Hf[NH4+(aq)] = (15.1 kJ·mol-1) – (–167.4 kJ·mol-1) + (–314.43 kJ·mol-1)
Hf[NH4+(aq)] = –131.9 kJ·mol-1
Similarly, we represent the dissolution reaction for CaCl2(s) as:
CaCl2(s) Ca2+(aq) + 2Cl–(aq)
The analysis to find Hf[Ca2+(aq)] is just like that for Hf[NH4+(aq)], except that we have to
account for the two moles of Cl–(aq) produced in this dissolution:
Hdissolution = {Hf[Ca2+(aq)] + 2Hf[Cl–(aq)]} – {Hf[CaCl2(s)]}
Hf[Ca2+(aq)] = Hdissolution – 2Hf[Cl–(aq)] + Hf[CaCl2(s)]
Hf[Ca2+(aq)] = (–82.9 kJ·mol-1) – 2(–167.4 kJ·mol-1) + (–795.8 kJ·mol-1)
Hf[Ca2+(aq)] = –543.9 kJ·mol-1
Problem 7.62.
Berthelot’s suggestion that all chemical processes that proceed spontaneously are exothermic
was incorrect. Endothermic processes that are spontaneous include freezing water at
temperatures below 0 C and dissolving many salts (for example NH4NO3 ).Problem 7.63.
Consider two samples of gas in identical size containers.
(a) If the temperature of the two samples is the same, then the pressure of gas in each container
is the same. This statement is false. Although the temperature and volume of the gas samples
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are the same, we have no information about the quantity of gas, that is, the number of moles of
gas, in each container. The one with more moles will have the higher pressure, since P =
n(RT/V) and all the quantities inside the parentheses on the right are the same for both samples.
We can amend the statement in two ways: (1) If the temperature and number of moles of gas is
the same, then the pressure of gas in each container is the same or (2) If the temperature of the
two samples is the same, then the pressure will be higher in the container with the greater
number of moles of gas.
(b) If the temperature and pressure of the two samples are the same, then the number of moles
of gas in each container is the same. This statement is true, since n = PV/RT and all the
quantities on the right are the same for both samples.
(c)
If the gas in each container is the same and the temperature and number of moles of gas
in each container are the same, then the number of collisions with the wall per unit time is the
same in both containers. This statement is true. To see why this is so, we note that the pressure
of the gas in each container is the same, since P = nRT/V and all the quantities on the right are
the same for both samples. Pressure is a result of collisions of the gas molecules with the walls.
Since the gas molecules and the pressure are the same in each container, the number of
collisions per unit time must be the same in each container.
Problem 7.64.
(a) The rubber balloon filled with helium decreased in size and buoyancy overnight because the
helium atoms inside were able to escape (slowly) through tiny holes in the stretched rubber.
After enough helium had escaped, the mass of the rubber balloon plus the remaining helium was
greater than the mass of air displaced by the now smaller balloon and the balloon no longer
floated. This is a density phenomenon. When there is a lot of helium in a balloon, its volume is
large and displaces air with a larger mass than the combined mass of the helium plus the rubber.
Thus the balloon is buoyed up (and will keep going up, as you know, if you have ever let go of a
helium balloon outdoors).
(b) Two identical rubber balloons, one filled with helium gas and the other with air decreased in
size overnight, but not to the same size. The larger balloon contains air. Molecules of air and
atoms of helium leak from the balloons through the tiny holes, but the helium atoms have lower
mass, so are moving faster than the air molecules, hit the holes more often, and leak out faster.
Air molecules from the outside can also leak into the balloons, but the rate is slower because the
pressure inside the balloons is slightly above atmospheric pressure and the net flow of gas is
generally outward.
Problem 7.65.
From the ideal gas equation we know that V = nRT/P. If the pressure remains the same, but the
volume of gas in the cylinder increases (the sort of change illustrated in Figure 7.20), the change
that occurs must involve an increase in number of moles of gas, nf > ni, at constant temperature,
an increase in the temperature of the gas, Tf > Ti, without a change in number of moles, or some
combination of changes in the number of moles and temperature, such that the product of their
final values, nf·Tf is greater than the product of their initial values, ni·Ti. A reaction that
produces more moles of gaseous products than were present initially, such as the reaction in
Investigate This 7.72 is a possibility for increasing the number of moles. Some gas phase
combustion reactions begin and end with same number moles of gas, but release a great deal of
energy which increases the temperature of the products. One example is:
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C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(g)
Problem 7.66.
We can justify the statement that “a mole of nitrogen gas occupies about 25 L at 25 C and
1 atm pressure” by calculating the volume of an ideal gas under these conditions. The value of R
in appropriate units is 8.21  10–2 L·atm·mol–1·K–1. With n = 1 mol, P = 1 atm, and T = 298 K,
we get:
–2
–1
–1
nRT
(1 mol)(8.21 10 L  atm  mol  K )(298 K)
V=
=
= 24.5 L  25 L
P
1 atm
This calculation does not depend on the identity of the gas, so all gases that we can approximate
as behaving ideally occupy about 25 L at 25 C and 1 atm pressure. Gases well above the
boiling point of their liquid at 25 C and 1 atm pressure can be treated as ideal, so oxygen also
fits this rule of thumb: molar volume is 25 L at 25 C and 1 atm.
[NOTE: Although not as accurate as the molar volume of 22.4 L at STP, this 25-at-25 rule of
thumb is quite adequate for many calculations that do not involve precise stoichiometry.]
Problem 7.67.
(a) We can use the ideal gas equation to calculate the expected pressure for a gas in fixed
volume container that is heated from –3 C (270 K) to, where its pressure is 38.3 kPa to 267 C
(540 K). At constant V and n, we know that P/T = nR/V = constant. Thus, we can write:
38.3 kPa
P2
P1
P
= 2 =
=
270 K
540 K
T1 T 2
P2 = 76.6 kPa
We could also have observed that the absolute temperature doubled, so the pressure also should
have doubled, as the calculations indicate.
(b) The measured pressure in the container at 540 K is 137.2 kPa, which is almost twice as high
(actual factor is 1.79 times as high) as we would have predicted from the ideal gas equation.
Evidently the number of moles of gas in the container has increased, since the volume has
remained constant. The simplest explanation is that some of the gas molecules we added to the
container might have dissociated to give two or more molecules for each molecule initially
present. An example of such a case is the dissociation of an N2O4 molecule to give two NO2
molecules (see Chapter 9, Section 9.11).
Problem 7.68.
(a) We use the ideal gas equation to find, n, the number of moles of gas in the 327-mL flask at
150 C (423 K) and 33.2 kPa pressure:
PV
(33.2 kPa)(0.327 L)
n=
=
= 3.09  10–3 mol
–1
–1
RT
(8.315 L·kPa·mol ·K )(423 K)
(b) Now we know that a 0.237 g sample of the hydrocarbon is 3.09  10–3 mol, so we have:
molar mass of the compound = (0.237 g)
= 76.8 g·mol–1
–3
(3.09  10 mol)
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We know that the ratio of carbon to hydrogen atoms in the hydrocarbon is one-to-one, so its
molecular formula is CzHz and:
molar mass of the compound = z(12.0 g·mol–1) + z(1.01 g·mol–1) = z(13.0 g·mol–1)
We equate our two values for the molar mass and solve for z to get the molecular formula:
z(13.0 g·mol–1) = 76.8 g·mol–1
z = 5.9 = 6
The molecular formula of the hydrocarbon is C6H6. The most common hydrocarbon with this
formula is benzene, which is a liquid at room temperature and a gas at 150 C, so its properties
are consistent with the compound in this problem.
Problem 7.69.
(a) We use the ideal gas equation to find, n, the number of moles of CO2 gas in the 127.5-mL
container at a pressure of 83.35 kPa and temperature of 295.3 K:
PV
(83.35 kPa)(0.1275 L)
n=
=
= 4.328  10–3 mol
–1
–1
RT
(8.315 L·kPa·mol ·K )(295.3 K)
(b) The reaction equation for the decomposition of CaCO3(s) is:
CaCO3(s)  CaO(s) + CO2(g)
Thus, one mole of CaCO3(s) yields one mole of CO2(g), so there are 4.328  10–3 mol of
CaCO3(s) in the original sample.
(c) To get the percentage of CaCO3(s) in the original sample, we need the mass of CaCO3(s) in
the original sample, which we get by converting the known number of moles to mass:
mass CaCO3(s) = (4.328  10–3 mol)(100.08 g·mol–1) = 0.4331 g
The percentage of the original sample that is CaCO3(s) is:
 0.4331 g 
% CaCO3(s) = 
100 % = 27.29 %
 1.587 g 
Problem 7.70.
(a) E and H are different for the fermentation process because the number of moles of gas
changes during the reaction from zero to two:
C6H12O6(s)  2C2H5OH(l) + 2CO2(g)
(b) The internal energy change, E, for this reaction is equal to, qV, the thermal energy
transferred into or out of the system if the reaction is carried out at constant volume. If the
reaction is carried out at constant pressure, the change in internal energy is still the same (it is a
state function), but now work is done by the system in pushing back the atmosphere. Under this
condition, the system loses energy as pressure-volume work, so E = qP – PV = H – PV,
where H is substituted for qP, the amount of thermal energy transferred at constant pressure.
Since PV is positive for the fermentation reaction (gas is formed, so V > 0), the equation
relating E and H shows that E will be more negative than H. Using the ideal gas
equation, we showed that PV = (n)RT , equation (7.59). Thus:
E = H – (n)RT
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Since n = 2 mol, we have (n)RT = (2 mol)(8.315 J·mol–1·K–1)(298 K) = 4.96  10–3 J,
assuming reaction at 298 K. Thus, E will be more negative than Ho by 4.96 kJ.
Problem 7.71.
(a) The balanced stoichiometric equation for the reaction of magnesium with hydronium ion is:
Mg(s) + 2H3O+(aq)  Mg2+(aq) + H2(g) + 2H2O(l)
If the magnesium reacts completely in this experiment, the number of moles that react is:
 1 mol Mg 
0.1372 g Mg = (0.1372 g) 
= 5.644  10–3 mol Mg

 24.31 g 
The original acidic solution contained 5  10–2 mol H3O+ [= (0.5 M)(0.1 L)], which is about five
times as much as needed to react with all the Mg. Assume that the specific heat of the solution
is 4.184 J·K–1·g–1, that 100.0 mL is 100.0 g, and that the heat capacity of the calorimeter is zero.
Since no thermal energy leaves or enters the calorimeter, the sum of the thermal energy changes
in the calorimeter is zero:
0 = qP(reaction) + (100.0 g)[25.69 – 19.32) K](4.184 J·K–1·g–1)
qP(reaction) = –(100.0 g)[25.69 – 19.32) K](4.184 J·K–1·g–1) = –2.67 kJ


–2.67 kJ
Hreaction = 
= –473 kJ·mol–1
–3
 5.644  10 mol Mg 
(b) Since a gas is formed, we need to correct for the pressure-volume work done. One mole of
gas is formed for every mole of Mg reacted, so n = 1 and we have:
E = H – nRT = (–473 kJ·mol–1) – (1 mol)(8.314 J·K–1·mol–1)(298 K)
E = (–473 kJ·mol–1) – (2.47 kJ·mol–1)  –475 kJ·mol–1
(c) You could measure E directly by carrying out the reaction in a sealed container (like a
bomb calorimeter) with some arrangement to drop the Mg in the acid when the container had
been sealed. In theory, the precision of the data above is good enough to measure the half
percent difference between E and H. However, in order actually to get data that are accurate
enough to observe this difference, you would have to take account of the actual mass of the
calorimetric liquid, its true specific heat, and (probably most important) the heat capacity of the
calorimeter.
Problem 7.72.
[NOTE: The illustration that accompanies this problem is modeled after one from Science 1998,
282, 1844. A similar illustration plus 4 movies of the motion of the filament (made to be
fluorescent) is available at: http://www.res.titech.ac.jp/seibutu/nature/f1rotate.html. There are
also other sources of such an illustration.]
(a) Recall that work is equal to force times the distance through which the force acts. In one
revolution the shaft of the molecular motor illustrated in this problem moves about 1 nm =
1  10–9 m while exerting a force of about 100  10–12 N, so:
work in a revolution = (100  10–12 N)(1  10–9 m) = 1  10–19 N·m (or J)
(b) Our estimate in this chapter was that the hydrolysis of ATP produces about 21 kJ·mol–1.
Hydrolysis of one molecule of ATP would produce:
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1 mol
 1000 J  

21 kJ·mol–1 = (21 kJ·mol–1) 
= 3.5  10–20 J·molec–1
 1 kJ   6.02  10 23 molec 
Since hydrolysis of three molecules of ATP is required for each revolution of the motor, the
energy provided by the ATP hydrolysis is about (3 molec)(3.5  10–20 J·molec–1) ≈ 1  10–19 J.
(c) The enthalpy supplied by the ATP is essentially the same as the work done by the motor.
The motor is 100% efficient in converting its energy source into work. Even if our estimates are
off somewhat, this molecular motor appears to be an extremely efficient device.
Problem 7.73.
The balanced equation for the combustion of glycylglycine is:
C4H8O3N2 + 13/2O2  2NO2 + 4CO2 + 4H2O
Hcombustion = –1966 kJ
The balanced equation for the combustion of glycine is:
C2H5O2N + 13/4O2  NO2 + 2CO2 + 5/2H2O
Hcombustion = –981 kJ
(Don’t worry at this stage about the somewhat strange stoichiometric coefficients required to
write the equations for one mole of glycylglycine and glycine.) In order to use the combustion
data to find the enthalpy change for the hydrolysis of glycylglycine to glycine, we need to
combine the combustion reactions in a way that will give us the hydrolysis equation as the result
of the combination. If we double the glycine combustion reaction, reverse it, and add it to the
glycylglycine combustion reaction, we will get the desired hydrolysis reaction:
2NO2 + 4CO2 + 5H2O  2C2H5O2N + 13/2O2
–2Hcombustion = –2(–981 kJ)
C4H8O3N2 + 13/2O2  2NO2 + 4CO2 + 4H2O
Hcombustion = –1966 kJ
C4H8O3N2 + H2O  2C2H5O2N
Hhydrolysis= –4 kJ
[The bond between the two glycine amino acids in glycylglycine is a peptide bond, which we
have encountered previously in Problems 7.30 and 7.42. In Problem 7.42, standard enthalpy of
formation data were used to find that formation of a peptide bond (plus a molecule of water) is
endothermic, Hreaction = 4 kJ·mol–1 (of peptide bonds formed). Here, we have calculated that
hydrolysis of a peptide bond, the reverse of the formation, is exothermic by –4 kJ·mol–1 (of
peptide bonds hydrolyzed). The two values are consistent with one another.]
Problem 7.74.
[NOTE: The illustration in this problem is modeled after one found at
http://www.physics.umn.edu/groups/demo/demo_gifs/4B70_10.GIF where its use as a physics
demonstration is described. The history of the fire piston can be found at
http://www.geocities.com/CollegePark/4201/piston.html. Other sites (all lists of physics demos
done at various colleges—some have photos of the actual syringe) are:
http://www.wfu.edu/Academic-departments/Physics/demolabs/demos/4/4B7010.jpg,
http://www.physics.umd.edu/deptinfo/facilities/lecdem/i5-22.htm,
http://www.ph.unimelb.edu.au/lecdem/hd2.htm.]
(a) To get E for the gas compressed in a fire syringe that is 15 cm long and has a crosssectional area of 0.20 cm2, we need the temperature change upon compression, the number of
moles of gas in the syringe, and the heat capacity of air, 21 kJ·mol–1·K–1. The temperature
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change is from room temperature, about 300 K, to about 1000 K (value given in the problem
statement), so T = 700 K. The initial volume of gas in the cylinder is:
V = (15 cm)(0.20 cm2) = 3.0 cm3 (mL)
The number of moles of gas is:
 1 mol 
3.0 mL = (3.0 mL) 
= 1.2  10–4 mol
 25000 mL 
The internal energy change is:
E = n  CV  T = (1.2  10–4 mol)(21 J·mol–1·K–1)(700 K) ≈ 1.8 J
(b) If no thermal energy is exchanged with the surroundings during the compression, q is zero
for the compression and the work done on the gas is equal to its change in internal energy:
w = E = 1.8 J. This work is the result of a decrease in volume of the gas to 1/20th of its initial
volume (part of the description in the problem statement). The change in volume, V, is:
V = (1/20)(3.0 mL) – (3.0 mL) = –(19/20)(3.0 mL) = –2.9 mL = –2.9  10–3 L
The work on the gas is:
 1 L  atm 
w = 1.8 J = (1.8 J) 
= 1.8  10–2 L·atm = – PV = – P(–2.9  10–3 L)
 101.3 J 
P = (1.8  10–2 L·atm)/(2.9  10–3 L) = 6.2 atm
This is the pressure that has to be applied to the gas in the piston to obtain the amount of work
required to compress the gas in the fire syringe.
(c) To get pressure in pound·in–2, multiply pressure in atmospheres by 15 pound·in–2·atm–1. The
result for the pressure in part (b) is about 93 pound·in–2. Since 1 in = 2.54 cm, a square inch, in2,
is (2.54 cm)2 = 6.5 cm2. The area of the piston in square inches is:
 1 in 2 
0.20 cm2 = (0.20 cm2) 
= 0.030 in2
 6.5 cm 2 
About 3 pounds of force [= (93 pound·in–2)(3.0  10–2 in2)] is required to produce a pressure of
93 pound·in–2 on the tiny piston. Three pounds of force is easy for most teenagers or adults to
exert, so the fire piston should be easy to use.
(d) Before the compression, the molecules are moving randomly and relatively slowly (short
arrows). During the compression, the molecules have a large component of motion in the
direction of the compression superimposed on their random motion. When the compression
stops, the large component of motion gets randomized and the molecules are again moving
randomly but at much greater speed (longer arrows) corresponding to a higher temperature.
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(e) The maximum temperature of the compressed gas is not the relevant variable here. The
amount of energy available in the compressed gas is what you have to consider. The calculation
in part (a) shows that the internal energy change of the gas is a little less than 2 J. Handbooks
show that the specific heat of paper or cotton (both made of cellulose, a glucose polymer) is in
the ballpark of 1 J·g–1·K–1. The kindling temperature of paper is 451 F (which is the source of
the title, “Fahrenheit 451,” of a Ray Bradbury story about book burning) or about 233 C. Thus,
you have to raise the temperature of paper about 200 degrees from room temperature to get it to
burn. One gram of paper would require about 200 J of energy to increase its temperature this
much. Since you only have about 2 J of energy in the compressed air of the fire piston, you have
enough energy to raise the temperature of about 10 mg of paper or cotton to its kindling point.
That is why you use a wisp of cotton in the demonstration. To start a larger fire, you use the
glowing tinder to ignite some further small pieces of material and finally build up to a real fire.
The match was a great invention and quickly made fire pistons obsolete.
Problem 7.75.
Begin this problem by writing the balanced equation for the reaction whose standard enthalpy
change you are to calculate:
2PH3(g) + 4O2(g)  P2O5(g) + 3H2O(g)
The atomization reactions need to be combined in such a way that their sum is equal to the
balanced equation. You can think of the solution as atomizing the reactants and then
recombining the atoms to give the products. Some of the atomization reactions given in the
problem statement will need to be used more than once. For example, the atomization of PH3(g)
must be used twice, so 2[PH3(g)  P(g) + 3H(g)] becomes 2PH3(g)  2P(g) + 6H(g) and H
for the reaction is also doubled as shown below. The reactions for recombination to give the
products are needed in the reverse direction, which changes the sign of H from (+) to (–).
The reactions, their standard enthalpy changes, and their sum are (species that cancel out in the
summation are lined out):
2PH3(g)  2P(g) + 6H(g)
H = 2(965 kJ) = 1930 kJ
4O2(g)  8O(g)
H = 4(490 kJ) = 1960 kJ
2P(g) + 5O(g)  P2O5(g)
H = –3382 kJ
6H(g) + 3O(g)  3H2O(g)
H = 3(–930 kJ) = –2790 kJ
2PH3(g) + 4O2(g)  P2O5(g) + 3H2O(g)
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H = –2282 kJ
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This H value is for 2 moles of PH3 reacting. Therefore, for the combustion of 1 mole of PH3,
H = –1141 kJ.
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