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Energy and Chemical Change Section 15.1 Energy
6. Challenge A 4.50-g nugget of pure gold absorbed 276 J of heat. The initial temperature was 25.0 C. What was the final temperature?
pages 516–522
Practice Problems
q
c m
pages 519–521
1. A fruit and oatmeal bar contains 142 nutritional Calories. Convert this energy to calories.
_ _
142 Calories 142 kcal
142
kcal
1000 cal 1 kcal
1 kcal 4.184 kJ
142,000
20.7
4.184
kcal
0.001
J
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 c
__ (276 J) (0.129 J/g·C)(4.50 g)
T f T i T f 25.0C
475C
475C
102C
___
T
q (5696 J) mT (155 g)(40.0 25.0°C)
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. 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.
X (0.1 cal)(4.184 J/cal) 0.4184 J 1 cal
Section 15.1 Assessment
Unit X 0.1 cal 1 cal
q cm
cal
3. Challenge Define a new energy unit, named after yourself, with a magnitude of one-tenth of a calorie. What conversion conversion factors relate this new unit to joules? To Calories? . c n I , s e i n a p m o C l l i H w a r G c M e h T f o n o i s i v i d a , l l i H w a r G c M / e o c n e l G © t h g i r y p o C
T
_
T f 5.00
2. An exothermic reaction releases 86.5 kJ. How many kilocalories of energy are released? 86.5 kJ
T
T
2.45
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.
J/(g°C)
The specific heat is very close to the value for ethanol. Solutions Manual
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10. Calculate the amount of heat absorbed when 5.50 g of aluminum is heated from 25.0 C to 95.0C. The specific heat of aluminum is 0.897 J/(g-C). q
(0.897
q
345
J/(gC))(5.50 g)(95.0C
25.0C)
The temperature change is inversely proportional to the specific heat: aluminum, iron, silver, gold.
pages 525–528
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? 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? c m
T
5650 J 4.184 J/(gC) 50.8
m
26.6C
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?
298
q
c m
q
0.803
T
q
30,500
J/(gC)
_
c m
(T f
___ q cm
T i )
T i
9750 J (4.184 J/(gC))(335 g)
65.5C
T f 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.
H rxn H products H reactants
page 525
m
T f
T
17. Explain why H for for an exothermic reaction has a negative value.
Practice Problems
c m
The heat absorbed or released equals the specific heat of the substance times its mass times its change in temperature.
Section 15.2 Heat
q
T f
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.
q
cmT
q
q
15. Challenge If 335 g of water at 65.5°C loses 9750 J of heat, what is the final temperature of the water?
2.00 103 g 19.0C
J
Chemistry: Matter and Change • Chapter 15
and H products < H reactants·
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 thermodynamics, 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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21. Calculate the specific heat in J/(gC) of an unknown substance if a 2.50 g sample releases 12.0 cal as its temperature changes from 25.0C to 20.0C. q
c
_
cmT q
mT
__
(12 cal)(4.184 J/cal) (2.50 g)(5.0C)
4.02
J/(gC)
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.
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Section 15.3 Thermochemical Equations pages 529–533 529–533
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 0.0 C 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, H fus 6.01 kJ/mol, H vap 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 c m
20C
q
q
4.184 J/(gC) 104 J or 15 kJ
to 0.0C, use the equation:
T
180 g 20C
Problem-Solving Lab
1.5
page 531 531
For the region at 0.0C,
Time and Temperature Data for Water Time (mm)
Temperature (°C)
Time (mm)
Temperature (°C)
Heat absorbed
6.01
H fus 6.01
kJ/mol
kJ/mol 10 mol 60 kJ
For the region 0.0C to 100C, use the equation: q
c m
4.184
T
0.0
20
1 3 .0
100
q
1.0
0
1 4 .0
100
or 75 kJ
2.0
0
1 5 .0
100
For the region at 100C,
3.0
9
1 6 .0
100
4.0
26
1 7 .0
100
5.0
42
1 8 .0
100
6.0
58
1 9 .0
100
7.0
71
2 0 .0
100
8.0
83
2 1 .0
100
9.0
92
2 2 .0
100
10. 0
98
2 3 .0
100
11. 0
100
2 4 .0
100
12. 0
100
2 5 .0
100
Solutions Manual
J/(gC)
Heat absorbed
180 g 100C 7.5 1 10 0 4J
40.7
H vap 40.7
kJ/mol
kJ/mol 10 mol 410 kJ
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3. Infer What would the heating curve 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?
C2H5OH(l) 3O2(g) 0 2CO2(g) + 3H2O(l) H comb 1367
27. Determine Which of the following processes are exothermic? Endothermic? a. C2H5OH(l) 0 C2H5OH(g)
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.
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 released in freezing 0.250 mol water.
Practice Problems
multiply 0.250 mol times the molar heat of fusion of water, 6.01 kJ/mol.
page 532
23. Calculate the heat required to melt 25.7 g of solid methanol at its melting point. Refer to Table 15.4.
25.7 g CH3OH 2.58
__ __ 1 mol CH3OH
32.04 g CH3OH
kJ
3.22 kJ 1 mol CH3OH
24. How much heat evolves when 275 g of ammonia gas condenses to a liquid at its boiling point? 275 g NH3
_ __ _ __ _
__ 1 mol NH3
17.03 g NH3
23.3 kJ 1 mol NH3
376
kJ
25. Challenge What mass of methane (CH 4) must be burned in order to liberate 12,880 kJ of heat? Refer to Table 15.3 on page 529. 12,880 kJ
m
m 12,880
kJ
m 232
1 mol CH4
16.04 g CH4
16.04 g CH4 1 mol CH4
×
891 kJ 1 mol CH4
1 mol CH4
Calculate
How much heat is liberated liberated by the combustion of 206 g of hydrogen gas? H comb 286 kJ/mol
__
The molar mass of hydrogen is 2.01 g/mol. 206 g
1 mol 2.01 g
286 kJ 1 mol
29,300
kJ
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 A y p l a h t n E
∆H
C
891 kJ
g CH4
Section 15.3 Assessment page 533
26. Write a complete thermochemical equation for the combustion of ethanol (C 2H5OH) ( H comb 1367 kJ/mol). 300
29.
Chemistry: Matter and Change • Chapter 15
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 answer.. The reaction is exothermic because the product (C) has a lower enthalpy than the reactant (A).
Solutions Manual
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Practice Problems pages 537–541
32. Use Equations a and b to determine H for 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
a. 2NO(g) O2(g) 0 2NO2(g) Formation of NO: N2
O2 0 2NO
Formation of NO2: N2
2O 2 0 2NO2
Add the first equation to the second equation reversed.
NO is a reactant in the problem, so add the reversed NO formation equation to the NO 2 formation equation:
2CO(g) O2(g) 0 2CO2(g)
H 566.0
kJ
2NO N2
2NO(g) 0 N2(g)
H 180.6
kJ
2NO O2 0 2NO2
O 2(g)
2CO(g) 2NO(g) 0 2CO2(g) N 2(g) H 385.4 kJ
4Al(s) 3MnO2(s) 0 2Al2O3(s) 3Mn(s) H 1789 kJ
a. 4Al(s) 3O2(g) 0 2Al2O3(s) H 3352 kJ b. Mn(s) O2(g) 0 MnO2(s) H ?
Add the Equation a to Equation b reversed and tripled.
3MnO2(s) 0 3Mn(s)
H 3352
3352 3 x kJ kJ 1789
__
kJ
3O2(g) H 3 x kJ kJ
4Al(s) 3MnO2(s) 0 2Al2O3(s) H 3352 3 x kJ kJ
3Mn(s)
kJ
Because the direction of Equation b was changed, H for for equation b x 3352 1789 521 kJ 3
SO3(g) 0 S(s)
H2O(l) 0 H2(g)
_3 O (g) 2
2
_1 O (g) 2
2
H2O(l) 0 H2SO4(l)
35. Use standard enthalpies of formation from Table R-11 on page 975, to calculate H rxn for the following reaction.
H rxn [4(33.18
kJ) 6(285.83 kJ)] kJ H rxn 1398 kJ
4(46.11)
36. Determine H comb 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 H comb [4Hf (H2O) 4Hf (CO2)] Hf (C3H7COOH) H comb [4(286 H comb 2186
Solutions Manual
H rxn [4Hf (NO2) 6Hf (H2O)] 4Hf (NO3)
equation b.
4Al(s) 3O2(g) 0 2Al2O3(s)
S(s) 2O 2(g) 0 H2SO4(l)
4NH3(g) 7O2 (g) 0 4NO2(g) 6H2O(l)
b. Mn(s) O2(g) 0 MnO2(s) H x kJ kJ H for for
H2(g)
SO3(g)
a. 4Al(s) 3O2(g) 0 2Al2O3(s) H 3352 kJ
Let x
2O 2 0 N2 O 2 2NO 2
b. SO3(g) H2O(l) 0 H2SO4(l)
33. Challenge for H for for the following reaction is 1789 kJ. Use this and reaction a to determine for Reaction b. H for
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34. Show how the sum of enthalpy of formation equations produces each of the following reactions. You do not need to look up and include values. H values.
kJ 4( 394 kJ)] ( 534 kJ)
kJ
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37. Challenge Two enthalpy of formation equations, a and b, combine to form the equation for the reaction of nitrogen oxide and oxygen. The product of the reaction is nitrogen dioxide:
NO(g)
_1 O (g) 2
2 58.1 kJ
NO2(g) H °rxn
0
a.
_1 N (g) _1 O (g)
b.
_1 N (g)
2
2
2
2
2
2
NO(g) H f 91.3 kJ
0
O2(g) 0 NO2(g) H f ?
What is H °f for equation
b?
Reverse equation. a and change the sign of H f to obtain equation c:
_
_
2
2
c. NO(g) 0 1 N2(g) 1 O2(g) H °f 91.3 kJ Add equations b and c: NO(g)
_1 O (g) 2
2
NO2(g)
0
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 for the reaction NO(g) O(g) 0 NO2(g) H ? given the following reactions. Show your work.
H °rxn 58.1 kJ H °f ( c ) H °f ( b)
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 199 kJ
Section 15.4 Assessment page 541
38. Explain what is meant by Hess’s law and how it is used to determine H rxn. Hess’s law says that if two or more equations add up to an overall equation, the H rxn of the overall equation is the sum of the H rxn values of the equations that were combined. The H rxn of a reaction can be determined by choosing equations that contain the species in the overall equation, reversing the equations if necessary, necessary, and multiplying them and their H rxn values by whatever factors are necessary. Then add the H rxn values to obtain the value for the overall equation.
Multiply c by 2: 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: values: 2NO(g) 2O(g) 0 2NO2(g) H 466 kJ This is the equation and H for for 2 moles of NO reacting. Divide the equation and H by by 2: NO(g) O(g) 0 NO2(g) H 233 kJ
39. Explain in words the formula that can be used to determine H rxn when using Hess’s law. H rxn H f (products) H f ° (reactants)
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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: H f
285.8
kJ/mol
Gaseous water: H f 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
) l o m / J 241.8 k (
H2O (g)
f
°
H vap 44
H
285.8
Hvap 44.0
kJ/mol
H2O (l)
kJ/mol
Section 15.5 Reaction Spontaneity pages 542–548
Practice Problems pages 545–548
44. Predict the sign of S system 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
positive because the system’s entropy increases. Entropy increases when a solid or liquid dissolves to form a solution.
d. C10H8(l) 0 C 10H8(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 S system 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 reactions is spontaneous.
0
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Ssystem is
a. H system 75.9 kJ, T 273 K, S system 138 J/K S system
138 J/K 0.138 kJ/K
Gsystem
Hsystem T S system
Gsystem
75.9 kJ (273 K)(0.138 kJ/K)
Gsystem
75.9 kJ 37.7 kJ 114 kJ
spontaneous reaction
b. H system 27.6 kJ, T 535 K, S system 55.2 J/K S system 55.2
J/K 0.0552 kJ/K
Gsystem
Hsystem T S system
Gsystem
27.6 kJ (535 K)(0.0552 kJ/K)
Gsystem
27.6 kJ 29.5 kJ 1.9 kJ
nonspontaneouss reaction nonspontaneou
c. H system 365 kJ, T 388 K, S system 55.2 J/K S system
55.2 J/K 0.0552 kJ/K
Gsystem
Hsystem T S system
Gsystem
365 kJ (388 K)(0.0552 kJ/K)
Gsystem
365 kJ 21.4 kJ 386 kJ
nonspontaneouss reaction nonspontaneou
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d. H system 452 kJ, T 165 K, S system 55.7 J/K S system
55.7 J/K 0.0557 kJ
Gsystem
Hsystem T S system
Gsystem
452 kJ (165 K)(0.0557 kJ/K)
Gsystem
452 kJ 9.19 kJ 443 kJ
nonspontaneous reaction
47. Challenge Given H system 144 kJ and S system 36.8 J/K for a reaction, determine the lowest temperature in kelvins at which the reaction would be spontaneous. Gsystem
Hsystem T S system
_
For the reaction to be spontaneous: Gsystem < 0: Hsystem T S system < 0 T > > T > >
___ Hsystem
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 H system 20.5 kJ, T 298 K , and S system 35.0 J/K is spontaneous or nonspontaneous. S system 35.0 Gsystem 20.5 10.1
J/K 0.0350 kJ/K kJ (298 K)(0.0350 kJ/K)
kJ
The system is spontaneous.
52. Outline Use the blue and 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. Summary.
S system
144
kJ (36.8 J/K)(1 kJ/1000 J)
T > > 3910 K
At any temperature above 3910 K, the reaction is spontaneous.
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 the system becomes more disordered during a process. The system’s entropy increases.
50. Decide Does the entropy 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.
Writing in Chemistry page 549
Write thermochemical equations for the complete combustion of 1 mol octane (C 8H18), a component of gasoline, and 1 mol ethanol ( H comb of C 8H18 5471 kJ/mol; H comb 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. C2H5OH(l) 3O2(g) 0 2CO2(g) 1367 kJ/mol
3H2O(l) Hcomb
C8H18(l) 25/2O2(g) 0 8CO2(g) 5471 kJ/mol
9H2O(l) Hcomb
Octane releases the greater amount of energy per mol.
___ ___
1 mol of ethanol 1 mol of octane
46.07
g/mol
114.23
g/mol
1367
kJ 1000 g 1 mol 1 mol 46.07 g 1 kg 29,670 kJ/kg ethanol
1000 g 5471 kJ 1 mol 114.23 g 1 mol 1 kg 47,890 kJ/kg octane
Octane is the better fuel based on the mass burned.
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SOLUTIONS MANUAL
60. Ethanol has a specific heat of 2.44 J/(gC).
pages 552–555
What does this mean?
Mastering Concepts
It means that 2.44 J is required to raise the temperature of one gram of ethanol by one degree Celsius.
53. Compare and contrast temperature and heat. 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 converted
and released when it burns in an automobile engine? . c n I , s e i n a p m o C l l i H w a r G c M e h T f o n o i s i v i d a , l l i H w a r G c M / e o c n e l G © t h g i r y p o C
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 nutri-
tional Calories. How many calories does the food item contain? 124 Calories
__ 1000 calories 1 Calorie
124,000 calories
__
63. How many joules are absorbed in a process that
absorbs 0.5720 kcal? 0.5720 kcal
1000 cal 1 kcal
4.184 J cal
2,393 J
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?
__
1000 J 1 kJ 327 Calories 1367 kJ
1 cal 4.184 J
_ 1 Calorie 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
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
__
required. How many kilojoules are required to vaporize the same mass of ammonia? 656 cal
4.184 J 1 cal
1 kJ 1000 J
2.74 kJ
66. The combustion of one mole of ethanol releases
___
326.7 Calories of energy. How many kilojoules are released? 326.7 Cal
1000 cal 1 Cal
4.184 J 1 cal
1 kJ 1000 J
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67. Metallurgy A 25.0g bolt made of an alloy absorbed 250 joules of heat as its temperature changed from 25.0C to 78.0C. What is the specific heat of the alloy?
_ __
T 78.0C
c c
q
− 25.0C
53.0C
250 J m T 25.0 g 53.0C 0.189 J/gC
71. Under what condition is the heat ( q) evolved or absorbed in a chemical reaction equal to a change in enthalpy ( H )? )? when the reaction is carried out at constant pressure
72. The enthalpy change for a reaction, H , is negative. What does this indicate about the chemical potential energy of the system before and after the reaction?
Section 15.2
The system’s chemical potential energy is less after the reaction than before the reaction.
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 endothermic or exothermic? How do you know?
73. What is the sign of H for for an exothermic reaction? An endothermic reaction? H is 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 41.2 C to 12.9C? The specific heat of granite is 0.803 J/(g C). T 41.2C (12.9C) 54.1C
Products
qgranite [0.803 J/(gC)](3.58 106 g)(54.1C)
y p l a h t n E
∆H = 233 kJ
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. universe system
surroundings
qgranite 1.56
108 J
75. Swimming Pool A swimming pool 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 29.0 C? Assume that the density of water is 1.000 g/mL. Change the dimensions of the pool’s water from meters to centimeters. 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 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
J/(gC)](9.38 108 g)(10.6C) 4.16 1010 J q
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76. How much heat is absorbed by a 44.7-g piece of lead when its temperature increases by 65.4C? q c m
T
q 0.129 J/(g·C)
44.7
g 65.4C
377
J
77. Food Preparation When 10.2 g of canola 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 1 kJ
T f T i 196.4C
q c m
25.0C 171.4C
T
3340 J 10.2 g 171.4C
1.91
J/(gC)
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 106.1 C while the temperature of the water increases by 10.5C. What is the specific heat of the alloy? q c m
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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.
CH4(g)
2O2(g) 0 CO2(g) 2H2O(l) H 891
T ;
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)
qwater qalloy
___
4.184 J/(gC) 125 g 10.5C c alloy 58.8 g 106.1C c alloy
kJ
T
q
m
It means that 3.22 kJ of energy is required to melt one mole of methanol.
82. Write the thermochemical equation for the combustion of methane. Refer to Table 15.3.
J
_ __
T
c
3340
80. The molar enthalpy of fusion of methanol is 3.22 kJ/mol. What does this mean?
∆H vap = +40.7 kJ y p l a h t n E
(4.184 J/g·C)(125 g)(10.5C) (58.8 g)(106.1C)
c alloy 0.880 J/(gC)
∆H cond = -40.7 kJ
H2O(l)
Section 15.3
∆H fus = +6.01 kJ
Mastering Concepts
∆H solid = -6.01 kJ
79. Write the sign of H system for each of the following changes in physical state. a. C2H5OH(s) 0 C 2H5OH(l) H system is
positive.
b. H2O(g) 0 H2O(l) H system is
negative.
c. CH3OH(l) 0 CH3OH(g) H system is
positive.
d. NH3(l) 0 NH3(s) H system is
Solutions Manual
negative.
H2O(s) q
mol H vap
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 H2O 18.0 g
q
mol H solid
q
0.656
0.656
mol H2O
mol ( 6.01 kJ/mol) 3.94 kJ
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85. Grilling What mass of propane (C3H8) must be burned in a barbecue grill to release 4560 kJ of heat? The H comb of propane is 2219 kJ/mol. q mol H comb moles of propane
__
H is is tripled, and its sign is changed.
4560 kJ 2.055 mol 2219 kJ/mol
2.055 mol C3H8 44.09 g C3H8/mol C3H8 90.60 g
86. Heating with Coal How much heat is liberated 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? H comb of carbon 394 kJ/mol.
_
_
1000 g mcarbon mcoal 0.962 1 kg 1000 g (5.00 kg)(0.962)( ) 4810 g 1 kg
_
1 mol mol C 4810 g C 401 mol C 12.0 g C q mol Hcomb 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 100 C? 1255 g
90. How does H for for a thermochemical equation change when the amounts of all substances are tripled and the equation is reversed?
1 mol 40.7 kJ 2830 kJ 18.02 g 1 mol
88. A sample of ammonia ( H solid 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.
0.0
Al(s), Cl2(g)
) l o m / J k (
° f
H
∆
-704
AlCl3 (s)
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)
_3 Cl (g) 2
2
AlCl3(s) H f 704 kJ
0
Mastering Problems 92. Use standard enthalpies of formation from Table R-11 on page 975 to calculate H rxn for the following reaction. P 4O6(s) 2O2(g) 0 P4O10(s) H rxn H f (products) H f (reactants) H rxn [1(2984.0 kJ)] [1(1640.1 kJ)] 1343.9 kJ
93. Use Hess’s law and the following thermochemical equations to produce the thermochemical equation for the reaction C(s, diamond) 0 C(s, graphite). What is H for 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
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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 H system is negative for this reaction, it will tend to be spontaneous at lower temperatures.
Use Hess’s law and the changes in enthalpy for the following two generic reactions to calculate for the reaction 2A B2C3 0 2B A2C3. H for What is H for for the reaction?
_3 C
2B
_3 C
2A
2
2
2 0 A2C3
2
H 1874
kJ 98.
B2C3
0
H 285
kJ
Reverse the second equation and change the sign of its H value. value. Add the resulting equation to the first equation. Add the H values. values. The resulting thermochemical equation is 2A B2C3 0 2B A2C3 H 1589 kJ
The heat released by an exothermic reaction increases the entropy of the surroundings surroundings.. Such a reaction decreases Gsystem because H system is negative in the equation Gsystem H system T Ssystem.
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.
96. . c n I , s e i n a p m o C l l i H w a r G c M e h T f o n o i s i v i d a , l l i H w a r G c M / e o c n e l G © t h g i r y p o C
Predict how the entropy of the system changes for the reaction CaCO3(s) 0 CaO(s) CO2(g). Explain. Because a gaseous product is formed, it’s likely that the system’ system’ss entropy increases.
97.
Which of these reactions would one expect to be spontaneous at relatively high temperatures? At relatively low temperatures? a.
b.
c.
2NH3(g) 0 N2(g) 3H2(g) H system 92 kJ 2NO2(g) 0 N2O4(g) H system 58 kJ CaCO3(s) 0 CaO(s) CO2(g) H system 178 kJ
For a spontaneous reaction, Gsystem must be negative as calculated in the expression Gsystem H system T Ssystem. Reactions a and c both have a positive H system. 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
Explain how an exothermic reaction changes the entropy of the surroundings. Does the enthalpy change for such a reaction increase or decrease Gsystem? Explain.
Mastering Problems 99.
Calculate Gsystem for each process, and state if the process is spontaneous or nonspontaneous. a. H system 145
195
kJ,
T 293
S system
J/K
Ssystem 195
J/K 0.195 kJ/K
Gsystem H system Gsystem 145 87.9
K,
T Ssystem
kJ (293K)(0.195 kJ/K)
kJ
nonspontaneous b. H system 232
138
kJ,
T 273
K,
S system
J/K
Ssystem 0.138
kJ/K
Gsystem 232 270
kJ
(273K)(0.138
kJ/K)
kJ
spontaneous c. H system 15.9
268
S system
J/K
Ssystem 268
J/K
Gsystem 15.9 84.1
kJ, T = = 373 K, 0.268
kJ/K
kJ (373K)(0.268 kJ/K)
kJ
nonspontaneous
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100. Calculate the temperature at which Gsystem 0 if H system 4.88 kJ and S system 55.2 J/K. Ssystem 55.2
__
4.88
T
Heating Curve for Water
J/K 0.0552 kJ/K
kJ T ((0.055.2 0.055.2 kJ/K)
4.88 kJ 0.0552 kJ/K
4
100
G H T S
0
Mixed Review
88.4
K
101. For the change H2O(l) 0 H2O(g), G0system is 8.557 kJ and H 0system is 44.01 kJ, What is S 0system for the change?
__
) C º ( e r u t a r e p m e T
3
0
2 1
G H T S
8.557 kJ 44.01 kJ (298 K) S S 8.557
kJ 44.01 kJ 298 K
0.119
kJ/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). H 0rxn 718.3 kJ, and S 0rxn 368 J/K. Explain. G H T S G 718.3 −609
kJ (298 K)(0.368 kJ/K); G
kJ
Yes. The reaction is spontaneous under standard conditions because G 0rxn 609 kJ, and a negative G 0rxn i indicates ndicates spontaneity. spontaneity.
103. Calculate the temperature at which Gsystem 34.7 kJ if H system 28.8 kJ and Ssystem 22.2 J/K. Gsystem Hsystem T Ssystem 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 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 planning 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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107. Differentiate between the enthalpy of formation of H2O(l) and H2O(g). Why is it necessary to specify the physical state of water in the following thermochemical equation: CH 4(g) 2O2(g) CO2(g) 2H2O(l or g) H = = ? H f 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.
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. 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 PCl 3(l) Cl2(g) 0 PCl5(s). a. P4(s) 6Cl2(g) 0 4PCl3(l) 1280 kJ
H
b. P4(s) 10Cl2(g) 0 4PCl5(s) 1774 kJ
H
Equation c: PCl3(l) 0 1/4P4(s) 320 kJ
6/4Cl2(g) H
Divide equation b by 4, yielding equation d. 10/4Cl2(g) 0 PCl5(s)
Add equations c and d and their PCl3(l)
Cl2(g) 0 PCl5(s)
Solutions Manual
Let subscript 2 refer to the larger, hotter piece.
q 1
H
heat
gained by
q 2
cm 1(T 1
T f ) = cm 2(T 2 temperature
T f ); T f = final
Eliminate the specific heat c from from this equation: From the problem statement: m2 = 2m 2 m1: m 1(T 1
T f )
2 2m m 1(T 2
T f )
Eliminate mass m1 from this equation: (T 1
T f )
2 2((T 2
T f )
Solve for the unknown T f :
_
T f 1 (T 1 3
2 2T T 2)
_1 (50 C 3
2(90C)) 76.7C
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)
Reverse equation a and divide it by 4, yielding equation c.
Equation d: 1/4P4(s) 444 kJ
Let subscript 1 refer to the smaller, cooler piece.
Heat lost by the hotter piece cooler piece:
Think Critically
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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.
2 2O O 2(g) 0 C CO O 2(g) 2 2H H 2O(l)
CH 2O(g)
O 2(g) 0 C CO O 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
H values. values.
H 124
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116.
the greater value, it’s likely that methane has the greater molar enthalpy of combustion.
a.
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.
b.
c.
16.0 g 1 mol 3.86 mol C2H6. 116 g C2H6 30.1 g
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.
1.000 kg of natural gas contains 884 g CH4 and 116 g C2H6.
_ _
CS2
carbon disulfide
C2H6(g) 7/2O2(g) 0 2CO2(g) 3H2O(l) H0comb 1599.7 kJ/mol
884 g 1 mol 55.2 mol CH4
S2Cl2
disulfur dichloride
Challenge Problem 112.
Name the following molecular compounds. (Chapter 8)
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
115.
Hydrogen bonds 119.
A sample of oxygen gas has a volume of 20.0 cm3 at 10.0C. What volume will this sample occupy if the temperature rises to 110C ? (Chapter 13)
_ _; V T1
V1
T 2
V 2
2
29.1 c cm m3
_ __ T 2 V 1 T1
(383 K)(20.0 cm c m 3) 263 K
What element has the electron configuration [Ar]4s13d5? (Chapter 5) chromium
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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 81.1 g
0.308
mol;
_ 0.308 mol 0.500 L
0.616M
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 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.
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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 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 research group burned four cooking oils in a bomb calorimeter to determine if a relatio relationship nship 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 Pennsylvan Pennsylvania. ia.
Combustion Results for Oils Types of Oil
H comb (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 however numbers of spinning blades that create sound and disturb the natural beauty of the landscape. Another concern is that flocks of migrating
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15
CHAPTER
126.
SOLUTIONS MANUAL
1.
Assuming that 12.2 g of soy oil is burned and that all the energy released is used to heat 1.600 kg of water, initially at 20.0 C, what is the final temperature of the water? Energy released 12.2 g q
c m
40.81
kJ/g
a. b. c.
kJ
d.
T
498,000 J 4.184 J/(gC) 1,600 g 74.4C T
498
T f T i ; 74.4C
2.
T f 20.0C; T f 94.4C
1.39 mol 56.6 kJ
1 mol H2O 18.02 g
_
40.7
kJ/mol
What is the standard free energy of vaporization, Gvap, of cyclohexane at 300 K? a.
5.00 b. 3.00 c. 3.00 d. 2.00
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? H vap 40.7 kJ/mol
_
does not occur at all. will occur spontaneously. is not spontaneous. occurs only at high temperatures. temperatures.
c
T ; T
127. Oils
25.0 g H2O
In the range of temperatures shown, the vaporization of cyclohexane
kJ/mol kJ/mol kJ/mol kJ/mol
a 1.39 56.6
mol H2O
3.
kJ
1g 1.37 g canola oil 41.45 kJ
When Gvap is plotted as a function of temperature, the slope of the line equals S vap and the y-intercept of the line equals H vap. What is the approximate standard entropy of the vaporization of cyclohexane? a. 50.0 b.
Standardized Test Practice
c.
pages 556–557
d.
Multiple Choice
slope rise run
_
0.1 kJ/molK
d
Use the graph below to answer Questions 1 to 3. for ∆G for
the Vaporization Vaporization of Cyclohexane as a Function of Temperature
7.00
J/mol-K 10.0 J/mol-K 5.0 J/mol-K 100 J/mol-K
6.00
4.
5.00
__
_
(6.00 5.00)kJ/mol (290 300)K 1000 J 1 kJ
100
0.1 kJ/molK
J/molK
The metal yttrium, atomic number 39, forms a.
positive ions. b. negative ions. c. both positive and negative ions. d. no ions at all.
) l o 4.00 m / J k ( 3.00
G
∆ 2.00
a
1.00
5.
0 290
300
310
320
330
340
Temperature (K)
350
Given the reaction 2Al 3FeO 0 Al2O3 3Fe, what is the mole-to-mole ratio between iorn (II) oxide and aluminum oxide? a. b. c. d.
2:3 1:1 3:2 3:1
d
314
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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
Gas C
2.20
1000
Li
Be
B
C
N
O
F
0 .9 8
1.57
2 .0 4
2.55
3 .0 4
3 .4 4
3.98
Na
Mg
Al
Si
P
S
Cl
0 .9 3
1.31
1 .6 1
1.90
2 .1 9
2 .5 8
3.16
) a P k ( e r u s e r P
Gas A
800 600
Gas B
400 200
6.
Which bond is the most electronegative? a. b. c. d.
0 250
H-H H-C H-N H-O
9.
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. b. c.
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d.
Q2M3 M2Q3 Q3M2 M3Q2
300
What is the predicted pressure of Gas B at 310 K? a. 500 kPa b. 600 kPa c. 700 kPa d. 900 kPa
Use the figure below to answer Questions 11 to 13. S
3.61 1 10 0 17 J 10 0 50 J b. 1.22 1 c. 8.23 1049 J 10 0 24 J d. 3.81 1
10.
11.
a 3.61 1017
CI
Ar
K
Ca
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.
a.
Solutions Manual
290
Short Answer
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.)
(5.45 1016 s1)(6.626 1034 Js)
280
b
c 8.
270
Temperature (K)
d 7.
260
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 .
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CHAPTER
12.
15
SOLUTIONS MANUAL
Use electron-dot models to explain what charge sulfur will most likely have when it forms an ion.
SAT Subject Test: Chemistry 15.
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 S 2.
The specific heat of ethanol is 2.44 J/(g C). How many kilojoules of energy are required to heat 50.0 g of ethanol from 20.0C to 68.0C? 10.7 b. 8.30 c. 2.44 d. 1.22 e. 5.86 a.
Extended Response Use the information below to answer Questions 13 and 14.
kJ kJ kJ kJ kJ
a (88.0C) 16.
1 atm
1 atm
State the gas law that describes why the gas in the second canister occupies a greater volume than the gas in the first canister.
1kj 10.7 kJ 1000 J
If 3.00 g of aluminum foil, placed in an oven and heated from 20.0°C to 662.0°C, absorbs 1728 J of heat, what is the specific heat of aluminum? 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) a.
c
13.
_
q cmT (2.44 J/(gC)) (50.0 g)
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
c
_
cmT
__
q (1728 J) mT (3.00 g) (642.0 °C)
0.897 J/(g°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. T T _ _
_ 1
V 1
2
V 2
316
_ T 2
__
300.0 K 2.1 L
5.4 L
(300.0 K)(5.4 L) 2.1 L
T 2
T 2
770
K
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Solutions Manual
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CHAPTER
SOLUTIONS MANUAL
Use the table below to answer Questions 17 and 18. Density and Electronegativity Data for Elements Elements
Density (g ( g/ m l )
Aluminum
Electronegativity
2.698 3
18.
Which pair is most likely to form an ionic bond? a. b.
1.6
c.
4.0
d.
Fluorine
1.696 10
Sulfur
2.070
2.6
e.
Copper
8.960
1.9
d
Magnesium
1.738
1.3
Carbon
3.513
2.6
17.
carbon and sulfur aluminum and magnesium magnesium copper and sulfur magnesium and fluorine fluorine aluminum and carbon
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
_ m
V
_ 9.250 g 5.250 mL
1.762
g/mL
b
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