15
CHAPTER
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Energy and Chemical Change
Section 15.1 Energy
6. Challenge A 4.50-g nugget of pure gold
pages 516–522
absorbed 276 J of heat. The initial temperature
was 25.0C. What was the final temperature?
Practice Problems
q c m T
pages 519–521
1. A fruit and oatmeal bar contains 142 nutritional
Calories. Convert this energy to calories.
142 Calories 142 kcal
_
1000 cal
142,000 cal
142 kcal 1 kcal
2. An exothermic reaction releases 86.5 kJ. How
many kilocalories of energy are released?
1 kcal
86.5 kJ _ 20.7 kcal
4.184 kJ
3. Challenge Define a new energy unit, named
after yourself, with a magnitude of one-tenth
of a calorie. What conversion factors relate this
new unit to joules? To Calories?
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Unit X 0.1 cal
1 cal 4.184 J
X (0.1 cal)(4.184 J/cal) 0.4184 J
1 cal 0.001 Calorie
X (0.1 cal)(1 Cal/1000 cal) 0.0001 Calorie
4. If the temperature of 34.4 g of ethanol increases
from 25.0°C to 78.8°C, how much heat has been
absorbed by the ethanol? Refer to Table 15.2.
q c m T
q 2.44 J/(g °C) 34.4 g 53.8°C 4.52 103 J
5. A 155-g sample of an unknown substance was
heated from 25.0°C to 40.0°C. In the process,
the substance absorbed 5696 J of energy. What
is the specific heat of the substance? Identify
the substance among those listed in Table 15.2
on page 520.
q c m T
c
q
(5696 J)
_
___ 2.45 J/(g°C)
mT
T q
(276 J)
_
__ 475C
cm
(0.129 J/g·C)(4.50 g)
T Tf Ti Tf 25.0C 475C
Tf 5.00 102C
Section 15.1 Assessment
page 522
7. Explain how energy changes from one form to
another in an exothermic reaction. In an endothermic reaction.
Chemical potential energy changes to heat in
exothermic reactions and the heat is released.
In endothermic reactions, heat is absorbed and
changed to chemical potential energy.
8. Distinguish between kinetic and potential
energy in the following examples: two separated magnets; an avalanche of snow; books on
library shelves; a mountain stream; a stock-car
race; separation of charge in a battery.
Two separated magnets illustrate potential
energy. In a snow avalanche, positional potential
energy is changing to kinetic energy. Books on
a shelf illustrate positional potential energy. As
water races down a mountain stream, positional
potential energy is changing to kinetic energy.
In a stock-car race, chemical potential energy is
being changed to kinetic energy. The separation
of charge in a battery illustrates electrical
potential energy.
9. Explain how the light and heat of a burning
candle are related to chemical potential energy.
Chemical potential energy, contained in the
candle, is changed to energy in the form of light
and heat and released as the chemical combustion
reaction takes place.
(155 g)(40.0 25.0°C)
The specific heat is very close to the value for
ethanol.
Solutions Manual
Chemistry: Matter and Change • Chapter 15
297
15
10. Calculate the amount of heat absorbed when
5.50 g of aluminum is heated from 25.0C to
95.0C. The specific heat of aluminum is
0.897 J/(g-C).
q cmT
q (0.897 J/(gC))(5.50 g)(95.0C 25.0C)
q 345 J
11. Interpret Data Equal masses of aluminum,
gold, iron, and silver were left to sit in the Sun
at the same time and for the same length of
time. Use Table 15.2 on page 520 to arrange
the four metals according to the increase in their
temperatures from largest increase to smallest.
The temperature change is inversely proportional
to the specific heat: aluminum, iron, silver, gold.
Section 15.2 Heat
pages 525–528
Practice Problems
page 525
12. A 90.0-g sample of an unknown metal absorbed
25.6 J of heat as its temperature increased
1.18C. What is the specific heat of the metal?
q c m T
25.6 J c 90.0 g 1.18C
c 0.241 J/(gC)
13. The temperature of a sample of water increases
from 20.0C to 46.6C as it absorbs 5650 J of
heat. What is the mass of the sample?
q c m T
5650 J 4.184 J/(gC) m 26.6C
m 50.8 g
14. How much heat is absorbed by a 2.00 103g
granite boulder (cgranite 0.803 J/(gC)) as its
temperature changes from 10.0C to 29.0C?
q c m T
q 0.803 J/(gC) 2.00 103 g 19.0C
q 30,500 J
298
Chemistry: Matter and Change • Chapter 15
SOLUTIONS MANUAL
15. Challenge If 335 g of water at 65.5°C loses
9750 J of heat, what is the final temperature of
the water?
q c m T c m (Tf Ti )
Tf q
_
T
cm
Tf 9750 J
___
65.5C
i
(4.184 J/(gC))(335 g)
Tf 58.5C
Section 15.2 Assessment
page 528
16. Describe how you would calculate the amount
of heat absorbed or released by a substance
when its temperature changes.
The heat absorbed or released equals the specific
heat of the substance times its mass times its
change in temperature.
17. Explain why H for an exothermic reaction
has a negative value.
Hrxn Hproducts Hreactants
and Hproducts < Hreactants·
18. Explain why a measured volume of water is an
essential part of a calorimeter.
The water absorbs the energy released. The heat
released equals the mass of water multiplied by
the change in temperature and by the specific heat.
19. Explain why you need to know the specific
heat of a substance in order to calculate how
much heat is gained or lost by the substance as
a result of a temperature change.
The specific heat of a substance tells you the
number of joules that are lost or gained for every
degree change in temperature and for every gram
of the substance.
20. Describe what the system means in thermody-
namics, and explain how the system is related
to the surroundings and the universe.
The system is the particular part of the universe
that contains the reaction or process that is being
studied. The surroundings are everything in the
universe except the system. Thus the universe is
the system and its surroundings.
Solutions Manual
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CHAPTER
15
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SOLUTIONS MANUAL
21. Calculate the specific heat in J/(gC) of an
unknown substance if a 2.50g sample releases
12.0 cal as its temperature changes from
25.0C to 20.0C.
q cmT
c
q
(12 cal)(4.184 J/cal)
_
__ 4.02 J/(gC)
mT
(2.50 g)(5.0C)
22. Design an Experiment Describe a procedure
you could follow to determine the specific heat
of a 45-g piece of metal.
Put a known mass of water into a calorimeter
and measure its temperature. Heat a 45-g metal
sample to 100C in boiling water. Put the heated
metal sample into the water in the calorimeter
and wait until the temperature of the water is
constant. Measure the final temperature of the
water. Assume no heat is lost to the surroundings.
Calculate the specific heat of the metal by
equating the quantity of heat gained by the
water to the quantity of heat lost by the metal.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Section 15.3 Thermochemical
Equations
pages 529–533
1. Analyze each of the five regions of the graph,
which are distinguished by an abrupt change
in slope. Indicate how the absorption of heat
changes the energy (kinetic and potential) of the
water molecules.
From 20C to 0.0C, the water molecules in ice
gain kinetic energy as shown by the temperature
rise. While the temperature remains at 0.0C, the
water molecules gain potential energy as the ice
melts to liquid water in an endothermic process.
As the temperature rises from 0.0C to 100C, the
water molecules again gain kinetic energy. At
100C, the water molecules gain potential energy
in an endothermic process as they evaporate to
water vapor.
2. Calculate the amount of heat required to pass
through each region of the graph (180 g H2O 10 mol H2O, Hfus 6.01 kJ/mol, Hvap 40.7 kJ/mol, c 4.184 J/(g-C)). How does the
length of time needed to pass through each region
relate to the amount of heat absorbed?
The more heat required, the longer the time in
the region.
For the region 20C to 0.0C, use the equation:
q c m T
Problem-Solving Lab
q 4.184 J/(gC) 180 g 20C 1.5 104 J or 15 kJ
page 531
For the region at 0.0C, Hfus 6.01 kJ/mol
Time and Temperature Data for Water
Time
(mm)
Temperature
(°C)
Time
(mm)
Temperature
(°C)
Heat absorbed 6.01 kJ/mol 10 mol 60 kJ
For the region 0.0C to 100C, use the equation:
q c m T
0.0
20
13.0
100
1.0
0
14.0
100
2.0
0
15.0
100
For the region at 100C, Hvap 40.7 kJ/mol
3.0
9
16.0
100
Heat absorbed 40.7 kJ/mol 10 mol 410 kJ
4.0
26
17.0
100
5.0
42
18.0
100
6.0
58
19.0
100
7.0
71
20.0
100
8.0
83
21.0
100
9.0
92
22.0
100
10.0
98
23.0
100
11.0
100
24.0
100
12.0
100
25.0
100
Solutions Manual
q 4.184 J/(gC) 180 g 100C 7.5 10 4J
or 75 kJ
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3. Infer What would the heating curve of ethanol
look like? Ethanol melts at 114C and boils
at 78C. Sketch ethanol’s curve from 120C
to 90C. What factors determine the lengths of
the flat regions of the graph and the slope of the
curve between the flat regions?
From 120C to 114C the curve rises linearly.
At 114C it becomes horizontal for a time and
then rises linearly again until it reach 78C where
it becomes horizontal again. After a time the
curve rises again to 90C. The lengths of the flat
regions depend on the amount of ethanol being
heated and the amount of heat being added with
time. Those factors and the specific heat of the
substance determine the slope of the upward
curve between the flat regions.
C2H5OH(l) 3O2(g)
Hcomb 1367
0 2CO2(g) + 3H2O(l)
27. Determine Which of the following processes
are exothermic? Endothermic?
a. C2H5OH(l) 0 C2H5OH(g)
b. Br2(l) 0 Br2(s)
c. C5H12(g) 8O2(g) 0 5CO2(g) 6H2O(l)
d. NH3(g) 0 NH3(l)
e. NaCl(s) 0 NaCl(l)
Reactions b, c, and d are exothermic. Reactions
a and e are endothermic.
28. Explain how you could calculate the heat
Practice Problems
released in freezing 0.250 mol water.
page 532
multiply 0.250 mol times the molar heat of fusion
of water, 6.01 kJ/mol.
solid methanol at its melting point. Refer to
Table 15.4.
25.7 g CH3OH 2.58 kJ
3.22 kJ
__ __
1 mol CH3OH
32.04 g CH3OH
1 mol CH3OH
ammonia gas condenses to a liquid at its boiling
point?
1 mol NH
23.3 kJ
__ _ 376 kJ
3
17.03 g NH3
combustion of 206 g of hydrogen gas? Hcomb
286 kJ/mol
The molar mass of hydrogen is 2.01 g/mol.
206 g 24. How much heat evolves when 275 g of
275 g NH3
29. Calculate How much heat is liberated by the
1 mol NH3
286 kJ
1 mol
_
_ 29,300 kJ
2.01 g
1 mol
30. Apply The molar heat of vaporization of
ammonia is 23.3 kJ/mol. What is the molar heat
of condensation of ammonia?
23.3 kJ/mol
25. Challenge What mass of methane (CH4) must
12,880 kJ m 1 mol CH
891 kJ
__
_
m 12,880 kJ 1 mol CH
16.04 g CH
__
×_
4
16.04 g CH4
1 mol CH4
4
1 mol CH4
4
A
Enthalpy
be burned in order to liberate 12,880 kJ of heat?
Refer to Table 15.3 on page 529.
ΔH
C
891 kJ
m 232 g CH4
Section 15.3 Assessment
page 533
26. Write a complete thermochemical equation for
the combustion of ethanol (C2H5OH) (Hcomb
kJ/mol).
31. Interpreting Scientific Illustrations The
reaction A 0 C is shown in the enthalpy
diagram at right. Is the reaction exothermic or
endothermic? Explain your answer.
The reaction is exothermic because the product
(C) has a lower enthalpy than the reactant (A).
1367
300
Chemistry: Matter and Change • Chapter 15
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23. Calculate the heat required to melt 25.7 g of
15
CHAPTER
SOLUTIONS MANUAL
Practice Problems
34. Show how the sum of enthalpy of formation
pages 537–541
32. Use Equations a and b to determine H for the
following reaction.
2CO(g) 2NO(g) 0 2CO2(g) N2(g) H ?
a. 2CO(g) O2(g) 0 2CO2(g) H 566.0 kJ
b. N2(g) O2(g) 0 2NO(g) H 180.6 kJ
Add the first equation to the second equation
reversed.
2CO(g) O2(g)
2NO(g)
0 2CO2(g)
0 N2(g) O2(g)
2NO O2
4Al(s) 3MnO2(s) 0 2Al2O3(s) 3Mn(s)
H 1789 kJ
a. 4Al(s) 3O2(g) 0 2Al2O3(s) H
3352 kJ
a. 4Al(s) 3O2(g) 0 2Al2O3(s) H 3352 kJ
b. Mn(s) O2(g) 0 MnO2(s) H ?
b. Mn(s) O2(g) 0 MnO2(s) H x kJ
Let x H for equation b.
Add the Equation a to Equation b reversed
and tripled.
4Al(s) 3O2(g) 0 2Al2O3(s) H 3352 kJ
3MnO2(s) 0 3Mn(s) 3O2(g) H 3x kJ
4Al(s) 3MnO2(s) 0 2Al2O3(s) 3Mn(s)
H 3352 3x kJ
3352 3x kJ 1789 kJ
Because the direction of Equation b was
changed, H for equation b x 3352 1789
521 kJ
3
0 2NO2
NO is a reactant in the problem, so add the
reversed NO formation equation to the NO2
formation equation:
H 180.6 kJ
1789 kJ. Use this and reaction a to determine
H for Reaction b.
0 2NO
Formation of NO2: N2 2O2
2NO N2 2O2
33. Challenge for H for the following reaction is
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Formation of NO: N2 O2
H 566.0 kJ
2CO(g) 2NO(g) 0 2CO2(g)
N2(g) H 385.4 kJ
__
equations produces each of the following reactions. You do not need to look up and include
H values.
a. 2NO(g) O2(g) 0 2NO2(g)
0 N2 O2 2NO2
0 2NO2
b. SO3(g) H2O(l) 0 H2SO4(l)
H2(g) S(s) 2O2(g)
0 H2SO4(l)
_
SO3(g)
0 S(s) 3 O2(g)
H2O(l)
0 H2(g) 1 O2(g)
2
SO3(g) H2O(l)
_
2
0 H2SO4(l)
35. Use standard enthalpies of formation from
Table R-11 on page 975, to calculate Hrxn
for the following reaction.
4NH3(g) 7O2 (g) 0 4NO2(g) 6H2O(l)
Hrxn [4Hf(NO2) 6Hf(H2O)] 4Hf(NO3)
Hrxn [4(33.18 kJ) 6(285.83 kJ)]
4(46.11) kJ Hrxn 1398 kJ
36. Determine Hcomb butanoic acid,
C3H7COOH(l) 5O2(g) 0 4CO2(g) 4H2O(l). Use data in Table R-11 on page 975
and the following equation.
4C(s) 4H2(g) O2(g) 0 C3H7COOH(l) H
534 kJ
Hcomb [4Hf(H2O) 4Hf(CO2)]
Hf(C3H7COOH)
Hcomb [4(286 kJ 4(394 kJ)] (534 kJ)
Hcomb 2186 kJ
Solutions Manual
Chemistry: Matter and Change • Chapter 15
301
15
SOLUTIONS MANUAL
37. Challenge Two enthalpy of formation equa-
tions, a and b, combine to form the equation for
the reaction of nitrogen oxide and oxygen. The
product of the reaction is nitrogen dioxide:
1
NO(g) O2(g) 0 NO2(g) H°rxn
2
58.1 kJ
_
a.
_1 N (g) _1 O (g) 0 NO(g) H 91.3 kJ
b.
_1 N (g) O (g) 0 NO (g) H ?
2
2
2
2
2
2
f
2
2
f
What is H°f for equation b?
Reverse equation. a and change the sign of
Hf to obtain equation c:
_
_
c. NO(g) 0 1 N2(g) 1 O2(g) H°f 91.3 kJ
2
2
The enthalpy of the reaction under standard
conditions (1 atm and 298 K) equals the sum
of the standard enthalpies of formation of
the products minus the sum of the standard
enthalpies of formation of the reactants.
40. Describe how the elements in their standard
states are defined on the scale of standard
enthalpies of formations.
Elements in their standard states are assigned
enthalpies of formation of zero.
41. Examine the data in Table 15.5 on page 538.
What conclusion can you draw about the
stabilities of the compounds listed relative to
the elements in their standard states? Recall that
low energy is associated with stability.
All compounds listed in Table 15.5 are more stable
than their constituent elements.
42. Calculate Use Hess’s law to determine H for
Add equations b and c:
H°rxn 58.1 kJ H°f (c) H°f (b)
the reaction NO(g) O(g) 0 NO2(g) H ?
given the following reactions. Show your work.
a. O2(g) 0 2O(g) H 495 kJ
58.1 kJ 91.3 kJ H°f (b)
b. 2O3(g) 0 3O2(g) H 427 kJ
H°f (b) 58.1 kJ 91.3 kJ 33.2 kJ
c. NO(g) O3(g) 0 NO2(g) O2(g) H
NO(g) _1 O (g) 0 NO (g)
2
2
2
199 kJ
Section 15.4 Assessment
Multiply c by 2:
page 541
38. Explain what is meant by Hess’s law and how
it is used to determine Hrxn.
Hess’s law says that if two or more equations
add up to an overall equation, the Hrxn of the
overall equation is the sum of the Hrxn values
of the equations that were combined. The Hrxn
of a reaction can be determined by choosing
equations that contain the species in the overall
equation, reversing the equations if necessary,
and multiplying them and their Hrxn values by
whatever factors are necessary. Then add the
Hrxn values to obtain the value for the overall
equation.
2NO(g) 2O3(g) 0 2NO2(g) 2O2(g) H
2(199 kJ) 398 kJ
Reverse b and change the sign of H:
3O2(g) 0 2O3(g) H 427 kJ
Reverse a and change the sign of H:
2O(g) 0 O2(g) H 495 kJ
Add the three equations and their H values:
2NO(g) 2O(g) 0 2NO2(g) H 466 kJ
This is the equation and H for 2 moles of NO
reacting. Divide the equation and H by 2:
NO(g) O(g) 0 NO2(g) H 233 kJ
39. Explain in words the formula that can be used
to determine Hrxn when using Hess’s law.
Hrxn 302
Hf (products) Hf° (reactants)
Chemistry: Matter and Change • Chapter 15
Solutions Manual
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
CHAPTER
CHAPTER
15
SOLUTIONS MANUAL
43. Interpret Scientific Illustrations Use the
data below to draw a diagram of standard heats
of formation similar to Figure 15.15 on
page 538, and use your diagram to determine
the heat of vaporization of water at 298 K.
Liquid water: Hf 285.8 kJ/mol
Gaseous water: Hf 241.8 kJ/mol
Students diagrams will show a line representing
liquid water at 285.8 kJ/mol below 0.0 kJ and
a line representing gaseous water 241.8 kJ/mol
below 0.0 kJ. The heat of vaporization is the
energy difference between the two lines or
241.8 kJ (285.8 kJ) 44.0 kJ
H°f (kJ/mol)
d. C10H8(l) 0 C10H8(s)
Ssystem is negative because the system’s
entropy decreases. Solid particles have less
freedom to move around than liquid particles.
45. Challenge Comment on the sign of Ssystem
for the following reaction.
Fe(s) Zn2(aq) 0 Fe2(aq) Zn(s)
The states of the two reactants are the same on
both sides of the equation, so it is impossible from
the equation alone to predict the sign of Ssystem.
46. Determine whether each of the following reac-
0
241.8
H2O (g)
Hvap 44 kJ/mol
285.8
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Ssystem is positive because the system’s
entropy increases. Entropy increases when a
solid or liquid dissolves to form a solution.
H2O (l)
Hvap 44.0 kJ/mol
tions is spontaneous.
a. Hsystem 75.9 kJ, T 273 K,
Ssystem 138 J/K
Ssystem
138 J/K 0.138 kJ/K
Gsystem
Hsystem TSsystem
Gsystem
75.9 kJ (273 K)(0.138 kJ/K)
Gsystem
75.9 kJ 37.7 kJ 114 kJ
spontaneous reaction
Section 15.5 Reaction
Spontaneity
pages 542–548
Practice Problems
pages 545–548
44. Predict the sign of Ssystem for each of the
following changes.
a. ClF(g) F2(g) 0 ClF3(g)
Ssystem is negative because the system’s
entropy decreases. There are more gaseous
reactant particles than product particles.
b. NH3(g) 0 NH3(aq)
Ssystem is negative because the system’s
entropy decreases. Aqueous particles have less
freedom to move around.
c. CH3OH(l) 0 CH3OH(aq)
Solutions Manual
b. Hsystem 27.6 kJ, T 535 K,
Ssystem 55.2 J/K
Ssystem 55.2 J/K
0.0552 kJ/K
Gsystem
Hsystem TSsystem
Gsystem
27.6 kJ (535 K)(0.0552 kJ/K)
Gsystem
27.6 kJ 29.5 kJ 1.9 kJ
nonspontaneous reaction
c. Hsystem 365 kJ, T 388 K,
Ssystem 55.2 J/K
Ssystem
55.2 J/K 0.0552 kJ/K
Gsystem
Hsystem TSsystem
Gsystem
365 kJ (388 K)(0.0552 kJ/K)
Gsystem
365 kJ 21.4 kJ 386 kJ
nonspontaneous reaction
Chemistry: Matter and Change • Chapter 15
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CHAPTER
d. Hsystem 452 kJ, T 165 K,
Ssystem 55.7 J/K
Ssystem
55.7 J/K 0.0557 kJ
Gsystem
Hsystem TSsystem
Gsystem
452 kJ (165 K)(0.0557 kJ/K)
Gsystem
452 kJ 9.19 kJ 443 kJ
nonspontaneous reaction
47. Challenge Given Hsystem 144 kJ and
Ssystem 36.8 J/K for a reaction, determine
the lowest temperature in kelvins at which the
reaction would be spontaneous.
Gsystem
Hsystem TSsystem
For the reaction to be spontaneous:
Gsystem < 0: Hsystem TSsystem < 0
H
T>_
system
SOLUTIONS MANUAL
The system’s entropy increases. The system
consists of the sugar and tea. Randomness or
disorder increases as sugar molecules, which were
originally locked into position in the solid structure
of the sugar cube, disperse throughout the tea.
51. Determine whether the system Hsystem 20.5 kJ, T 298 K, and Ssystem 35.0
J/K is spontaneous or nonspontaneous.
Ssystem 35.0 J/K 0.0350 kJ/K
Gsystem 20.5 kJ (298 K)(0.0350 kJ/K) 10.1 kJ
The system is spontaneous.
52. Outline Use the blue and red headings to
outline the section. Under each heading,
summarize the important ideas discussed.
Students outlines chould include all important
ideas expressed in the Section Summary.
Ssystem
144 kJ
___
(36.8 J/K)(1 kJ/1000 J)
Writing in Chemistry
page 549
T > 3910 K
Write thermochemical equations for the
At any temperature above 3910 K, the reaction is
spontaneous.
complete combustion of 1 mol octane (C8H18),
a component of gasoline, and 1 mol ethanol
(Hcomb of C8H18 5471 kJ/mol; Hcomb
of C2H5OH 1367 kJ/mol). Which releases
the greater amount of energy per mole of fuel?
Which releases more energy per kilogram of
fuel? Discuss the significance of your findings.
Section 15.5 Assessment
page 548
48. Compare and contrast spontaneous and
nonspontaneous reactions.
A reaction occurs spontaneously only when the
temperature, entropy change within the system,
and energy exchanged between the system and
surroundings cause the entropy of the universe to
increase.
49. Describe how a system’s entropy changes if
C2H5OH(l) 3O2(g) 0 2CO2(g) 3H2O(l) Hcomb
1367 kJ/mol
C8H18(l) 25/2O2(g) 0 8CO2(g) 9H2O(l) Hcomb
5471 kJ/mol
Octane releases the greater amount of energy
per mol.
the system becomes more disordered during a
process.
1 mol of ethanol 46.07 g/mol
The system’s entropy increases.
1367 kJ
1000 g
_
1 mol _
_
50. Decide Does the entropy of a system increase
or decrease when you dissolve a cube of sugar
in a cup of tea? Define the system, and explain
your answer.
1 mol of octane 114.23 g/mol
1 mol
46.07 g
1 kg
29,670 kJ/kg ethanol
1000 g
5471 kJ
1 mol
_
__
114.23 g
1 mol
47,890 kJ/kg octane
1 kg
Octane is the better fuel based on the mass burned.
304
Chemistry: Matter and Change • Chapter 15
Solutions Manual
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T>
CHAPTER
15
Chapter 15 Assessment
SOLUTIONS MANUAL
60. Ethanol has a specific heat of 2.44 J/(gC).
pages 552–555
What does this mean?
Mastering Concepts
53. Compare and contrast temperature and heat.
It means that 2.44 J is required to raise the
temperature of one gram of ethanol by one
degree Celsius.
Heat is a form of energy that flows from a
warmer object to a cooler object. Temperature is
a measure of the average kinetic energy of the
particles in a sample of matter.
54. How does the chemical potential energy of a
61. Explain how the amount of energy required to
raise the temperature of an object is determined.
The amount of energy required equals the
product of the object’s specific heat, its mass, and
its change in temperature.
system change during an endothermic reaction?
It increases.
55. Describe a situation that illustrates potential
energy changing to kinetic energy.
Student answers will vary. A typical answer is:
During an avalanche, the potential energy of
snow at a higher altitude is converted to kinetic
energy as the snow cascades down a mountain.
56. Cars How is the energy in gasoline converted
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
and released when it burns in an automobile
engine?
Some is converted to work to move pistons in the
engine; much is released as heat.
57. Nutrition How does the nutritional Calorie
compare with the calorie? What is the relationship between the Calorie and a kilocalorie?
One nutritional Calorie equals 1000 calories.
Mastering Problems
62. Nutrition A food item contains 124 nutritional Calories. How many calories does the
food item contain?
124 Calories 1000 calories
__
124,000 calories
1 Calorie
63. How many joules are absorbed in a process that
absorbs 0.5720 kcal?
0.5720 kcal 4.184 J
1000 cal
_
_ 2,393 J
1 kcal
cal
64. Transportation Ethanol is being used as an
additive to gasoline. The combustion of 1 mol
of ethanol releases 1367 kJ of energy. How
many Calories are released?
1367 kJ 1000 J
1 cal
1 Calorie
_
__
1 kJ
327 Calories
4.184 J
1000 cal
One nutritional Calorie equals 1 kilocalorie.
65. To vaporize 2.00 g of ammonia 656 calories are
58. What quantity has the units J/(g°C)?
specific heat
59. Describe what might happen when the air above
required. How many kilojoules are required to
vaporize the same mass of ammonia?
656 cal 4.184 J
1 kJ
_
_ 2.74 kJ
1 cal
1000 J
the surface of a lake is colder than the water.
If the air is cool enough, water vapor from the
lake might condense and form fog. Heat will be
transferred from the warmer water to the cooler
air. The air immediately above the water will be
slightly warmer than the surrounding air, and the
fog might appear to rise off the lake somewhat
like steam.
Solutions Manual
66. The combustion of one mole of ethanol releases
326.7 Calories of energy. How many kilojoules
are released?
326.7 Cal 1000 cal
4.184 J
1 kJ
_
_ _ 1367 kJ
1 Cal
1 cal
1000 J
Chemistry: Matter and Change • Chapter 15
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SOLUTIONS MANUAL
67. Metallurgy A 25.0g bolt made of an alloy
71. Under what condition is the heat (q) evolved
absorbed 250 joules of heat as its temperature
changed from 25.0C to 78.0C. What is the
specific heat of the alloy?
or absorbed in a chemical reaction equal to a
change in enthalpy (H)?
T 78.0C − 25.0C 53.0C
c
q
250 J
_
__
when the reaction is carried out at constant
pressure
72. The enthalpy change for a reaction, H, is
m T
25.0 g 53.0C
c 0.189 J/gC
negative. What does this indicate about the
chemical potential energy of the system before
and after the reaction?
Section 15.2
Mastering Concepts
68. Why is a foam cup used in a student calorimeter rather than a typical glass beaker?
The foam cup is better insulated than a glass
beaker, so that a minimal amount of heat is
transferred into or out of the calorimeter.
69. Is the reaction shown in Figure 15.23 endo-
thermic or exothermic? How do you know?
The system’s chemical potential energy is less
after the reaction than before the reaction.
73. What is the sign of H for an exothermic
reaction? An endothermic reaction?
H is negative for an exothermic reaction and
positive for an endothermic reaction.
Mastering Problems
74. How many joules of heat are lost by 3580 kg
granite as it cools from 41.2C to 12.9C?
The specific heat of granite is 0.803 J/(gC).
T 41.2C (12.9C) 54.1C
Products
Enthalpy
qgranite [0.803 J/(gC)](3.58 106 g)(54.1C)
ΔH = 233 kJ
qgranite 1.56 108 J
75. Swimming Pool A swimming pool
Reactants
The reaction is endothermic because the enthalpy
of the products is 233 kJ higher than the enthalpy
of the reactants.
70. Give two examples of chemical systems and
define the universe in terms of those examples.
measuring 20.0 m 12.5 m is filled with water
to a depth of 3.75 m. If the initial temperature
is 18.4C, how much heat must be added to
the water to raise its temperature to 29.0C?
Assume that the density of water is 1.000 g/mL.
Change the dimensions of the pool’s water from
meters to centimeters.
universe system surroundings
20.0 m 2.00 103 cm; 12.5 m 1.25 103 cm;
3.75 m 3.75 102 cm
Student answers will vary. One example:
universe my body (the system) everything
else (the surroundings);
volume of water (2.00 103 cm)(1.25 103 cm)
(3.75 102 cm) 9.38 108 cm3
9.38 108 mL
another example: a beaker in which a reaction
is going on (the system) everything else (the
surroundings)
mass of water (9.38 108 mL)(1.000 g/mL)
9.38 108 g
q c m T
T (29.0C 18.4C) 10.6C
q [4.184 J/(gC)](9.38 108 g)(10.6C)
4.16 1010 J
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CHAPTER
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CHAPTER
76. How much heat is absorbed by a 44.7-g piece of
SOLUTIONS MANUAL
80. The molar enthalpy of fusion of methanol is
lead when its temperature increases by 65.4C?
3.22 kJ/mol. What does this mean?
q c m T
It means that 3.22 kJ of energy is required to melt
one mole of methanol.
q 0.129 J/(g·C) 44.7 g 65.4C 377 J
77. Food Preparation When 10.2 g of canola oil
at 25.0C is placed in a wok, 3.34 kJ of heat is
required to heat it to a temperature of 196.4C.
What is the specific heat of canola oil?
3.34 kJ 1000 J
_
3340 J
1 kJ
T Tf Ti 196.4C 25.0C 171.4C
81. Explain how perspiration can help cool your
body.
Your body is cooled as it supplies the heat
required to vaporize water from your skin.
82. Write the thermochemical equation for the
combustion of methane. Refer to Table 15.3.
CH4(g) 2O2(g) 0 CO2(g) 2H2O(l) H 891 kJ
q c m T
c
q
3340 J
_
__ 1.91 J/(gC)
m T
10.2 g 171.4C
78. Alloys When a 58.8–g piece of hot alloy is
placed in 125 g of cold water in a calorimeter,
the temperature of the alloy decreases by 106.1C
while the temperature of the water increases by
10.5C. What is the specific heat of the alloy?
Mastering Problems
83. Use information from Figure 15.24 to calculate
how much heat is required to vaporize 4.33 mol
of water at 100C?
Phase Changes for Water
H2O(g)
4.184 J/(gC) 125 g 10.5C
calloy 58.8 g 106.1C
calloy (4.184 J/g·C)(125 g)(10.5C)
___
(58.8 g)(106.1C)
calloy 0.880 J/(gC)
ΔHvap = +40.7 kJ
Enthalpy
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
q c m T; qwater qalloy
ΔHcond = -40.7 kJ
H2O(l)
Section 15.3
Mastering Concepts
79. Write the sign of Hsystem for each of the
following changes in physical state.
a. C2H5OH(s) 0 C2H5OH(l)
Hsystem is positive.
b. H2O(g) 0 H2O(l)
Hsystem is negative.
c. CH3OH(l) 0 CH3OH(g)
Hsystem is positive.
d. NH3(l) 0 NH3(s)
Hsystem is negative.
Solutions Manual
ΔHfus = +6.01 kJ
ΔHsolid = -6.01 kJ
H2O(s)
q mol Hvap
q 4.33 mol 40.7 kJ/mol 176 kJ
84. Agriculture Water is sprayed on oranges
during a frosty night. If an average of 11.8 g of
water freezes on each orange, how much heat is
released?
11.8 g H2O 1 mole H O
__
0.656 mol H O
2
18.0 g
2
q mol Hsolid
q 0.656 mol (6.01 kJ/mol) 3.94 kJ
Chemistry: Matter and Change • Chapter 15
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SOLUTIONS MANUAL
85. Grilling What mass of propane (C3H8) must be
90. How does H for a thermochemical equation
burned in a barbecue grill to release 4560 kJ of
heat? The Hcomb of propane is 2219 kJ/mol.
change when the amounts of all substances are
tripled and the equation is reversed?
q mol Hcomb
H is tripled, and its sign is changed.
4560 kJ
__
2.055 mol
2.055 mol C3H8 44.09 g C3H8/mol C3H8 90.60 g
86. Heating with Coal How much heat is liber-
ated when 5.00 kg of coal is burned if the
coal is 96.2% carbon by mass and the other
materials in the coal do not react in any way?
Hcomb of carbon 394 kJ/mol.
1000 g
m
m
0.962 _
1 kg
1000 g
_
) 4810 g
(5.00 kg)(0.962)(
AlCl3 (s)
coal
1 kg
mol C 4810 g C 1 mol
_
401 mol C
12.0 g C
q 401 mol C (394 kJ/mol C) 158,000 kJ
87. How much heat is evolved when 1255 g of
water condenses to a liquid at 100C?
1 mol
40.7 kJ
_
_ 2830 kJ
18.02 g
91. Use Figure 15.25 to write the thermochemical
equation for the formation of 1 mol of aluminum
chloride (a solid in its standard state) from its
constituent elements in their standard states.
Al(s) q mol Hcomb
1255 g -704
1 mol
88. A sample of ammonia (Hsolid 5.66 kJ/mol)
liberates 5.66 kJ of heat as it solidifies at its
melting point. What is the mass of the sample?
Mass mass of 1 mol ammonia 17.03 g
Section 15.4
Mastering Concepts
89. For a given compound, what does the standard
enthalpy of formation describe?
Standard enthalpy of formation describes the
change in enthalpy when one mole of the
compound in its standard state is formed from its
constituent elements in their standard states.
_3 Cl (g) 0 AlCl (s) H 704 kJ
2
2
3
f
Mastering Problems
92. Use standard enthalpies of formation from
Table R-11 on page 975 to calculate Hrxn
for the following reaction. P4O6(s) 2O2(g) 0
P4O10(s)
Hrxn Hf (products) Hf (reactants)
Hrxn [1(2984.0 kJ)] [1(1640.1 kJ)]
1343.9 kJ
93. Use Hess’s law and the following thermochem-
ical equations to produce the thermochemical
equation for the reaction C(s, diamond) 0 C(s,
graphite). What is H for the reaction?
a. C(s, graphite) O2(g) 0 CO2(g) H
394 kJ
b. C(s, diamond) O2(g) 0 CO2(g) H
396 kJ
Reverse Equation a, and add to Equation b.
CO2(g) 0 C(s, graphite) O2(g) H 394 kJ
C(s, diamond) O2(g) 0 CO2(g) H 396 kJ
C(s, diamond) 0 C(s, graphite). H 2 kJ
308
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carbon
Al(s), Cl2(g)
0.0
2219 kJ/mol
ΔHf° (kJ/mol)
moles of propane 15
CHAPTER
SOLUTIONS MANUAL
94. Use Hess’s law and the changes in enthalpy for
the following two generic reactions to calculate
H for the reaction 2A B2C3 0 2B A2C3.
What is H for the reaction?
2A _3 C
2
0 A2C3
H 1874 kJ
2B _3 C
2
0 B2C3
H 285 kJ
2
2
Reverse the second equation and change the sign
of its H value. Add the resulting equation to the
first equation. Add the H values. The resulting
thermochemical equation is 2A B2C3 0 2B A2C3 H 1589 kJ
Section 15.5
Mastering Concepts
95. Under what conditions is an endothermic chemical reaction in which the entropy of the system
increases likely to be spontaneous?
Such a reaction is likely to be spontaneous only at
higher temperatures.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
96. Predict how the entropy of the system changes
spontaneous (a negative value for Gsystem). On
the other hand, reaction b has fewer moles of
gas on the products side, which means entropy
decreases as products form. But because Hsystem
is negative for this reaction, it will tend to be
spontaneous at lower temperatures.
98. Explain how an exothermic reaction changes
the entropy of the surroundings. Does the
enthalpy change for such a reaction increase or
decrease Gsystem? Explain.
The heat released by an exothermic reaction
increases the entropy of the surroundings. Such
a reaction decreases Gsystem because Hsystem is
negative in the equation Gsystem
Hsystem TSsystem.
Mastering Problems
99. Calculate Gsystem for each process, and state if
the process is spontaneous or nonspontaneous.
a. Hsystem 145 kJ, T 293 K, Ssystem
195 J/K
Ssystem 195 J/K 0.195 kJ/K
Gsystem Hsystem TSsystem
for the reaction CaCO3(s) 0 CaO(s) CO2(g).
Explain.
Gsystem 145 kJ (293K)(0.195 kJ/K)
Because a gaseous product is formed, it’s likely
that the system’s entropy increases.
nonspontaneous
97. Which of these reactions would one expect to
be spontaneous at relatively high temperatures?
At relatively low temperatures?
a. 2NH3(g) 0 N2(g) 3H2(g)
Hsystem 92 kJ
b. 2NO2(g) 0 N2O4(g)
Hsystem 58 kJ
c. CaCO3(s) 0 CaO(s) CO2(g)
Hsystem 178 kJ
For a spontaneous reaction, Gsystem must be
negative as calculated in the expression Gsystem
Hsystem TSsystem. Reactions a and c both
have a positive Hsystem. However, both reactions
also have more moles of gaseous products
than gaseous reactants, which suggests that
entropy increases as products form. So, higher
temperatures will tend to make these reactions
Solutions Manual
87.9 kJ
b. Hsystem 232 kJ, T 273 K, Ssystem
138 J/K
Ssystem 0.138 kJ/K
Gsystem 232 kJ (273K)(0.138 kJ/K)
270 kJ
spontaneous
c. Hsystem 15.9 kJ, T = 373 K, Ssystem
268 J/K
Ssystem 268 J/K 0.268 kJ/K
Gsystem 15.9 kJ (373K)(0.268 kJ/K)
84.1 kJ
nonspontaneous
Chemistry: Matter and Change • Chapter 15
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15
100. Calculate the temperature at which Gsystem 0
if Hsystem 4.88 kJ and Ssystem 55.2 J/K.
SOLUTIONS MANUAL
Mixed Review
Heating Curve for Water
Ssystem 55.2 J/K 0.0552 kJ/K
T
4.88 kJ
__
88.4 K
0.0552 kJ/K
101. For the change H2O(l) 0 H2O(g), G0system
is 8.557 kJ and H0system is 44.01 kJ, What is
S0system for the change?
Temperature (ºC)
0 4.88 kJ T(0.055.2 kJ/K)
4
100
G H TS
3
0
1
2
G H TS
__
S 8.557 kJ 44.01 kJ 0.119 kJ/K
298 K
102. Is the following reaction to convert copper(II)
sulfide to copper(II) sulfate spontaneous under
standard conditions? CuS(s) 2O2(g) 0
CuSO4(s). H0rxn 718.3 kJ, and S0rxn
368 J/K. Explain.
G H TS
G 718.3 kJ (298 K)(0.368 kJ/K); G
−609 kJ
Yes. The reaction is spontaneous under standard
conditions because G0rxn 609 kJ, and a
negative G0rxn indicates spontaneity.
103. Calculate the temperature at which
Gsystem 34.7 kJ if Hsystem 28.8 kJ
and Ssystem 22.2 J/K.
Gsystem Hsystem TSsystem
34.7 kJ 28.8 kJ T (0.0222 kJ/K)
T 266 K
104. Heat was added consistently to a sample of
water to produce the heating curve in
Figure 15.26. Identify what is happening in
Sections 1, 2, 3, and 4 on the curve.
Section 1: The kinetic energy of the water (ice) is
increasing as the temperature rises.
Section 2: Potential energy is increasing as the
system absorbs energy in the process of melting.
Section 3: The kinetic energy of the water is
increasing as the temperature rises.
Section 4: Potential energy is increasing as
the system absorbs energy in the process of
evaporating.
105. Bicycling Describe the energy conversions
that occur when a bicyclist coasts down a long
grade, then struggles to ascend a steep grade.
As the bicyclist coats down a long grade,
potential energy of position is converted to
kinetic energy of motion. As the bicycle and
rider ascend a steep grade, chemical potential
energy and kinetic energy are converted to
potential energy of position.
106. Hiking Imagine that on a cold day you’re plan-
ning to take a thermos of hot soup with you on
a hike. Explain why you might fill the thermos
with hot water before filling it with hot soup.
The hot water will transfer energy to the
thermos in the form of heat, raising the
temperature of the thermos to nearly that of
the hot soup. Because the temperatures of the
thermos and soup are similar, the soup will lose
little heat to the thermos when placed inside.
310
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8.557 kJ 44.01 kJ (298 K) S
CHAPTER
15
SOLUTIONS MANUAL
107. Differentiate between the enthalpy of forma-
tion of H2O(l) and H2O(g). Why is it necessary
to specify the physical state of water in the
following thermochemical equation: CH4(g) 2O2(g) CO2(g) 2H2O(l or g) H = ?
Hf for H2O(l) and H2O(g) differ by
approximately the enthalpy of vaporization of
water. Because water in the liquid state has an
enthalpy of formation that differs from that of
water in the gaseous state, the enthalpy change
for the reaction depends upon the physical
states of all reactants and products.
Think Critically
108. Analyze each image in Figure 15.27 in terms
of potential energy of position, chemical
potential energy, kinetic energy, and heat.
By virtue of its position high on the mountain,
the snow has positional potential energy.
When the snow slides down the mountain, its
positional potential energy changes to kinetic
energy of motion.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Wood has chemical potential energy stored in its
bonds. This energy is being converted to heat,
light, and kinetic energy.
109. Apply Phosphorous trichloride is a starting
material for the preparation of organic phosphorous compounds. Demonstrate how
thermochemical equations a and b may be
used to determine the enthalpy change for the
reaction described by the equation PCl3(l) Cl2(g) 0 PCl5(s).
a. P4(s) 6Cl2(g) 0 4PCl3(l) H 1280 kJ
b. P4(s) 10Cl2(g) 0 4PCl5(s)
1774 kJ
H Reverse equation a and divide it by 4, yielding
equation c.
Equation c: PCl3(l) 0 1/4P4(s) 6/4Cl2(g) H 320 kJ
Divide equation b by 4, yielding equation d.
Equation d: 1/4P4(s) 10/4Cl2(g) 0 PCl5(s) H
444 kJ
110. Calculate Suppose that two pieces of iron,
one with a mass exactly twice the mass of the
other, are placed in an insulated calorimeter.
If the original temperatures of the larger piece
and the smaller piece are 90.0C and 50.0C,
respectively, what is the temperature of the
two pieces when thermal equilibrium has been
established? Refer to Table R-9 on page 975
for the specific heat of iron.
Let subscript 1 refer to the smaller, cooler piece.
Let subscript 2 refer to the larger, hotter piece.
Heat lost by the hotter piece heat gained by
cooler piece:
q 1 q 2
cm 1(T 1 T f) = cm 2(T 2 T f); T f = final
temperature
Eliminate the specific heat c from this equation:
From the problem statement: m2 = 2m1:
m 1(T 1 T f) 2m 1(T 2 T f)
Eliminate mass m1 from this equation:
(T 1 T f) 2(T 2 T f)
Solve for the unknown Tf:
_
_
T f 1 (T 1 2T 2) 1 (50C 2(90C)) 76.7C
3
3
The result is a mass-weighted average of the
two temperatures.
111. Predict which of the two compounds,
methane gas (CH4) or methanal vapor (CH2O),
has the greater molar enthalpy of combustion. Explain your answer. (Hint: Write and
compare the balanced chemical equations for
the two combustion reactions.)
CH 4(g) 2O 2(g) 0 CO 2(g) 2H 2O(l)
CH 2O(g) O 2(g) 0 CO 2(g) H 2O(l)
Methane likely has the greater molar enthalpy of
combustion The chemical equations for the two
reactions reveal that the combustion of one mole
of methane yields one mole of carbon dioxide
and two moles of water, whereas the combustion
of one mole of methanal yields one mole of
carbon dioxide and one mole of water. Because
Add equations c and d and their H values.
PCl3(l) Cl2(g) 0 PCl5(s)
Solutions Manual
H 124 kJ
Chemistry: Matter and Change • Chapter 15
311
15
Hf(products) for the combustion of methane has
the greater value, it’s likely that methane has the
greater molar enthalpy of combustion.
Challenge Problem
112. A sample of natural gas is analyzed and found
to be 88.4% methane (CH4) and 11.6% ethane
(C2H6) by mass. The standard enthalpy of
combustion of methane to gaseous carbon
dioxide and liquid water is −891 kJ/mol. Write
the equation for the combustion of gaseous
ethane to carbon dioxide and water. Calculate
the standard enthalpy of combustion of ethane
using standard enthalpies of formation from
Table R-11 on page 975. Using that result and
the standard enthalpy of combustion of methane
in Table 15.3, calculate the energy released by
the combustion of 1 kg of natural gas.
C2H6(g) 7/2O2(g) 0 2CO2(g) 3H2O(l)
H0comb 1599.7 kJ/mol
1.000 kg of natural gas contains 884 g CH4 and
116 g C2H6.
_
16.0 g
1 mol
116 g C2H6 _
3.86 mol C2H6.
884 g 1 mol 55.2 mol CH4
30.1 g
SOLUTIONS MANUAL
116. Name the following molecular compounds.
(Chapter 8)
a. S2Cl2
disulfur dichloride
b. CS2
carbon disulfide
c. SO3
sulfur trioxide
d. P4O10
tetraphosphorus decoxide
117. Determine the molar mass for the following
compounds. (Chapter 10)
a. Co(NO3)2·6H2O
molar mass (58.93 g/mol) 2(14.01 g/mol) 12(16.00 g/mol) 12(1.01 g/mol) 291.07 g/mol
b. Fe(OH)3
molar mass 55.85 g/mol 3(16.00 g/mol) 3(1.01 g/mol) 106.88 g/mol
118. What kind of chemical bond is represented by
the dotted lines in Figure 15.28? (Chapter 12)
(55.2 mol CH4) ( 891 kJ/mol)) (3.86 mol
C2H6) (1599.7 kJ/mol) 55,400 kJ
Cumulative Review
113. Why is it necessary to perform repeated
experiments in order to support a hypothesis?
(Chapter 1)
Experiments must be repeated to be sure that
they yield similar results each time.
114. Phosphorus has the atomic number 15 and an
atomic mass of 31 amu. How many protons,
neutrons, and electrons are in a neutral phosphorus atom? (Chapter 4)
number of protons 15; number of electrons
15; number of neutrons mass number number of protons 16
Hydrogen bonds
119. A sample of oxygen gas has a volume of
20.0 cm3 at 10.0 C. What volume will
this sample occupy if the temperature rises to
110C ? (Chapter 13)
T
T V
T
(383 K)(20.0 cm )
_
_; V 2 _ __
1
2
V1
V2
29.1 cm 3
3
1
2
T1
263 K
115. What element has the electron configuration
[Ar]4s13d5? (Chapter 5)
chromium
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CHAPTER
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SOLUTIONS MANUAL
120. What is the molarity of a solution made by
dissolving 25.0 g of sodium thiocyanate
(NaSCN) )in enough water to make 500 mL of
solution? (Chapter 14)
_
_
25.0 g 1 mol 0.308 mol; 0.308 mol 0.616M
81.1 g
0.500 L
121. List three colligative properties of solutions.
(Chapter 14)
vapor pressure lowering, boiling point
elevations, freezing point elevation
Writing in Chemistry
122. Alternate Fuels Use library and internet
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
sources to explain how hydrogen might be
produced, transported, and used as a fuel for
automobiles. Summarize the benefits and
drawbacks of using hydrogen as an alternative
fuel for internal combustion engines.
Students may write that hydrogen could best
be used as an automobile fuel in fuel cells.
A large supply of the gas would need to be
produced, transported and handled. Much of
the technology now used for handling methane
and propane gases could be adapted for use
with hydrogen. Much of the hydrogen now
available is a byproduct of the petrochemical
industry. For full-scale use of hydrogen as a fuel
for automobiles and for other energy needs,
hydrogen would probably be produced by the
electrolysis of water using renewable sources of
energy such as wind power or solar energy. The
only product of the combustion of hydrogen is
water, so it is a nonpolluting source of power.
However, issues of safe use and handling must
be carefully considered.
123. Wind Power Research the use of wind as a
source of electrical power. Explain the possible
benefits, disadvantages, and limitations of its
use.
birds may inadvertently fly into the blades and
be destroyed. When windmills are located off
shore, fish could be adversely affected by the
structures.
Document-Based Questions
Cooking Oil A university research group
burned four cooking oils in a bomb calorimeter
to determine if a relationship exists between
the enthalpy of combustion and the number of
double bonds in an oil molecule. Cooking oils
typically contain long chains of carbon atoms
linked by either single or double bonds. A chain
with no double bonds is said to be saturated.
Oils with one or more double bonds are unsaturated. The enthalpies of combustion of the four
oils are shown in Table 15.7. The researchers
calculated that the results deviated by only
0.6% and concluded that a link between saturation and enthalpy of combustion could not be
detected by the experimental procedure used.
Data obtained from: http: Heat of Combustion
Oils. April 1998. University of Pennsylvania.
Combustion Results for Oils
Types of Oil
Hcomb (kJ/g
Soy oil
40.81
Canola oil
41.45
Olive oil
39.31
Extra-virgin olive oil
40.98
124. Which of the oils tested provided the greatest
amount of energy per unit mass when burned?
canola oil: 41.45 kJ/g
125. According to the data, how much energy
would be liberated burning 0.554 kg of
olive oil?
0.554 kg 1000 g/kg 39.31 kJ/g 21,800 kJ
Students will note that the wind is not a steady
source of energy and there will always be a
need for a backup. The advantage of wind
power is that it is nonpolluting. Many people,
however, object to the presence of large
numbers of spinning blades that create sound
and disturb the natural beauty of the landscape.
Another concern is that flocks of migrating
Solutions Manual
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313
15
SOLUTIONS MANUAL
126. Assuming that 12.2 g of soy oil is burned and
1. In the range of temperatures shown, the vapor-
that all the energy released is used to heat
1.600 kg of water, initially at 20.0C, what is
the final temperature of the water?
ization of cyclohexane
a. does not occur at all.
b. will occur spontaneously.
c. is not spontaneous.
d. occurs only at high temperatures.
Energy released 12.2 g 40.81 kJ/g 498 kJ
q c m T
c
498,000 J 4.184 J/(gC) 1,600 g T; T
74.4C
2. What is the standard free energy of
T Tf Ti; 74.4C Tf 20.0C; Tf 94.4C
127. Oils can be used as fuels. How many grams of
canola oil would have to be burned to provide
the energy to vaporize 25.0 g of water? Hvap
40.7 kJ/mol
a
1 mol H O
25.0 g H2O _ 1.39 mol H2O
2
3. When Gvap is plotted as a function of temper-
18.02 g
ature, the slope of the line equals Svap and the
y-intercept of the line equals Hvap. What is
the approximate standard entropy of the vaporization of cyclohexane?
a. 50.0 J/mol-K
b. 10.0 J/mol-K
c. 5.0 J/mol-K
d. 100 J/mol-K
1.39 mol 40.7 kJ/mol 56.6 kJ
56.6 kJ 1g
_
1.37 g canola oil
41.45 kJ
Standardized Test Practice
pages 556–557
Multiple Choice
Use the graph below to answer Questions 1 to 3.
d
_ __
1000 J
0.1 kJ/molK _ 100 J/molK
slope rise (6.00 5.00)kJ/mol 0.1 kJ/molK
run
(290 300)K
ΔG for the Vaporization of Cyclohexane
as a Function of Temperature
7.00
1 kJ
6.00
4. The metal yttrium, atomic number 39, forms
5.00
ΔG (kJ/mol)
vaporization, Gvap, of cyclohexane at 300 K?
a. 5.00 kJ/mol
b. 3.00 kJ/mol
c. 3.00 kJ/mol
d. 2.00 kJ/mol
a.
b.
c.
d.
4.00
3.00
2.00
positive ions.
negative ions.
both positive and negative ions.
no ions at all.
a
1.00
0
290
300
310
320
330
340
Temperature (K)
350
5. Given the reaction 2Al 3FeO 0 Al2O3 3Fe, what is the mole-to-mole ratio between
iorn (II) oxide and aluminum oxide?
a. 2:3
b. 1:1
c. 3:2
d. 3:1
d
314
Chemistry: Matter and Change • Chapter 15
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Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
CHAPTER
CHAPTER
15
SOLUTIONS MANUAL
Use the table below to answer Question 6.
Use the graph below to answer Question 9.
Pressures of Three Gases
at Different Temperatures
Electronegativity of Selected Elements
H
1200
2.20
Be
B
C
N
O
F
0.98
1.57
2.04
2.55
3.04
3.44
3.98
Na
Mg
Al
Si
P
S
Cl
0.93
1.31
1.61
1.90
2.19
2.58
3.16
Presure (kPa)
Li
Gas C
1000
Gas A
800
Gas B
600
400
200
6. Which bond is the most electronegative?
a.
b.
c.
d.
H-H
H-C
H-N
H-O
d
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
7. Element Q has an oxidation number of 2,
while element M has an oxidation number
of 3. Which is the correct formula for a
compound made of elements Q and M?
a. Q2M3
b. M2Q3
c. Q3M2
d. M3Q2
c
8. Wavelengths of light shorter than about
4.00 107 m are not visible to the human
eye. What is the energy of a photon of ultraviolet light having a frequency of 5.45 1016
s1? (Planck’s constant is 6.626 1034 Js.)
a. 3.61 10 17 J
b. 1.22 10 50 J
c. 8.23 1049 J
d. 3.81 10 24 J
a
(5.45 1016 s1)(6.626 1034 Js) 3.61 1017
Solutions Manual
0
250
260
270
280
290
300
Temperature (K)
9. What is the predicted pressure of Gas B at
310 K?
a. 500 kPa
b. 600 kPa
c. 700 kPa
d. 900 kPa
b
Short Answer
Use the figure below to answer Questions 11 to 13.
S
CI
Ar
K
Ca
10. Explain why argon is not likely to form a
compound.
Argon already has a full outer energy level (eight
valence electrons) and is not likely to form an ion.
It does not need to gain or lose any electrons in
order to become chemically stable.
11. What is the chemical formula for calcium
chloride? Explain the formation of this ionic
compound using the election-dot structures
above.
CaCl2; a calcium atom becomes Ca2, losing its
two valence electrons to two chlorine atoms,
which each become Cl.
Chemistry: Matter and Change • Chapter 15
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CHAPTER
SOLUTIONS MANUAL
12. Use electron-dot models to explain what
charge sulfur will most likely have when it
forms an ion.
Sulfur has six valence electrons. Because atoms are
more stable when they have 8 valence electrons
completing their outer energy levels, sulfur tends
to gain two electrons to become the ion S2.
Extended Response
Use the information below to answer
Questions 13 and 14.
SAT Subject Test: Chemistry
15. The specific heat of ethanol is 2.44 J/(gC).
How many kilojoules of energy are required
to heat 50.0 g of ethanol from 20.0C to
68.0C?
a. 10.7 kJ
b. 8.30 kJ
c. 2.44 kJ
d. 1.22 kJ
e. 5.86 kJ
a
A sample of gas occupies a certain volume at a
pressure of 1 atm. If the pressure remains constant,
heating causes the gas to expand, as shown below.
q cmT (2.44 J/(gC))
1kj
(88.0C) _ 10.7 kJ
(50.0 g) 1000 J
16. If 3.00 g of aluminum foil, placed in an oven
1 atm
c
q cmT
13. State the gas law that describes why the gas in
the second canister occupies a greater volume
than the gas in the first canister.
c
q
(1728 J)
_
__ 0.897 J/(g°C)
mT
(3.00 g) (642.0 °C)
This is Charles’s law: at a constant pressure,
the volume of a given mass of gas is directly
proportional to its kelvin temperature.
14. If the volume in the first container is 2.1 L at
a temperature of 300.0 K, to what temperature
must the second canister be heated to reach
a volume of 5.4 L? Show your setup and the
final answer.
T2
T1
_
_
V1
V2
T2
300.0 K
_
_
2.1 L
T2 5.4 L
(300.0 K)(5.4 L)
__
2.1 L
T2 770 K
316
Chemistry: Matter and Change • Chapter 15
Solutions Manual
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
1 atm
and heated from 20.0°C to 662.0°C, absorbs
1728 J of heat, what is the specific heat of
aluminum?
a. 0.131 J/(g°C)
b. 0.870 J/(g°C)
c. 0.897 J/(g°C)
d. 2.61 J/(g°C)
e. 0.261 J/(g°C)
15
CHAPTER
SOLUTIONS MANUAL
Use the table below to answer Questions 17 and 18.
18. Which pair is most likely to form an ionic
Aluminum
2.698
1.6
Fluorine
1.696 103
4.0
Sulfur
2.070
2.6
bond?
a. carbon and sulfur
b. aluminum and magnesium
c. copper and sulfur
d. magnesium and fluorine
e. aluminum and carbon
Copper
8.960
1.9
d
Magnesium
1.738
1.3
Carbon
3.513
2.6
Density and Electronegativity Data for Elements
Elements
Density (g/ml)
Electronegativity
17. A sample of metal has a mass of 9.250 g and
occupies a volume of 5.250 mL. Which metal
is it?
a. aluminum
b. magnesium
c. carbon
d. copper
e. sulfur
9.250 g
_m _
1.762 g/mL
V
5.250 mL
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
b
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Chemistry: Matter and Change • Chapter 15
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