Chapter 15

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.SOLUTIONS MANUALCHAPTER 15

Energy and Chemical ChangeEnergy and Chemical Change

Solutions Manual Chemistry: Matter and Change • Chapter 15 297

Section 15.1 Energypages 516–522

Practice Problemspages 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

� 142,000 cal

2. An exothermic reaction releases 86.5 kJ. How many kilocalories of energy are released?

86.5 kJ � 1 kcal _ 4.184 kJ

� 20.7 kcal

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?

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 _

m�T �

(5696 J) ___

(155 g)(40.0 � 25.0°C) � 2.45 J/(g�°C)

The specific heat is very close to the value for ethanol.

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?

q � c � m � �T

�T � q _ cm �

(276 J) __

(0.129 J/g·�C)(4.50 g) � 475�C

�T � Tf � Ti � Tf � 25.0�C � 475�C

Tf � 5.00 � 102�C

Section 15.1 Assessmentpage 522

7. Explain how energy changes from one form to another in an exothermic reaction. In an endo-thermic 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 sepa-rated 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.

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298 Chemistry: Matter and Change • Chapter 15 Solutions Manual

SOLUTIONS MANUALCHAPTER 15

10. Calculate the amount of heat absorbed when 5.50 g of aluminum is heated from 25.0�C to 95.0�C. The specific heat of aluminum is 0.897 J/(g-�C).

q � cm�T

q � (0.897 J/(g��C))(5.50 g)(95.0�C � 25.0�C)

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 Heatpages 525–528

Practice Problemspage 525

12. A 90.0-g sample of an unknown metal absorbed 25.6 J of heat as its temperature increased 1.18�C. What is the specific heat of the metal?

q � c � m � �T

25.6 J � c � 90.0 g � 1.18�C

c � 0.241 J/(g��C)

13. The temperature of a sample of water increases from 20.0�C to 46.6�C as it absorbs 5650 J of heat. What is the mass of the sample?

q � c � m � �T

5650 J � 4.184 J/(g��C) � m � 26.6�C

m � 50.8 g

14. How much heat is absorbed by a 2.00 � 103�g granite boulder (cgranite � 0.803 J/(g��C)) as its temperature changes from 10.0�C to 29.0�C?

q � c � m � �T

q � 0.803 J/(g��C) � 2.00 � 103 g � 19.0�C

q � 30,500 J

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 _ cm � Ti

Tf � �9750 J ___ (4.184 J/(g��C))(335 g)

� 65.5�C

Tf � 58.5�C


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.

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21. Calculate the specific heat in J/(g��C) of an unknown substance if a 2.50�g sample releases 12.0 cal as its temperature changes from 25.0�C to 20.0�C.

q � cm�T

c � q _

m�T �

(�12 cal)(4.184 J/cal) __

(2.50 g)(�5.0�C) � 4.02 J/(g��C)

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 100�C 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.

Section 15.3 Thermochemical Equationspages 529–533

Problem-Solving Labpage 531

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 �20�C to 0.0�C, the water molecules in ice gain kinetic energy as shown by the temperature rise. While the temperature remains at 0.0�C, 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 100�C, the water molecules again gain kinetic energy. At 100�C, 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 �20�C to 0.0�C, use the equation: q � c � m � �T

q � 4.184 J/(g��C) � 180 g � 20�C � 1.5 � 104 J or 15 kJ

For the region at 0.0�C, �Hfus � 6.01 kJ/mol

Heat absorbed � 6.01 kJ/mol � 10 mol � 60 kJ

For the region 0.0�C to 100�C, use the equation: q � c � m � �T

q � 4.184 J/(g��C) � 180 g � 100�C � 7.5 � 1 0 4 J or 75 kJ

For the region at 100�C, �Hvap � 40.7 kJ/mol

Heat absorbed � 40.7 kJ/mol � 10 mol � 410 kJ

Time and Temperature Data for Water

Time (mm)

Temperature(°C)

Time (mm)

Temperature(°C)

0.0 �20 13.0 100

1.0 0 14.0 100

2.0 0 15.0 100

3.0 9 16.0 100

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

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3. Infer What would the heating curve of ethanol look like? Ethanol melts at �114�C and boils at 78�C. Sketch ethanol’s curve from �120�C to 90�C. What factors determine the lengths of the flat regions of the graph and the slope of the curve between the flat regions?

From �120�C to �114�C the curve rises linearly. At �114�C it becomes horizontal for a time and then rises linearly again until it reach 78�C where it becomes horizontal again. After a time the curve rises again to 90�C. 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.

Practice Problemspage 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 � 1 mol CH3OH

__ 32.04 g CH3OH

� 3.22 kJ __ 1 mol CH3OH

� 2.58 kJ

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 (CH4) must be burned in order to liberate 12,880 kJ of heat? Refer to Table 15.3 on page 529.

12,880 kJ � m � 1 mol CH4 __

16.04 g CH4 � 891 kJ _

1 mol CH4

m � 12,880 kJ � 16.04 g CH4 __ 1 mol CH4

× 1 mol CH4 _

891 kJ

m � 232 g CH4


26. Write a complete thermochemical equation for the combustion of ethanol (C2H5OH) (�Hcomb � �1367 kJ/mol).

C2H5OH(l) � 3O2(g) 0 2CO2(g) + 3H2O(l) �Hcomb � �1367

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 released in freezing 0.250 mol water.

multiply 0.250 mol times the molar heat of fusion of water, 6.01 kJ/mol.

29. Calculate How much heat is liberated by the combustion of 206 g of hydrogen gas? �Hcomb � �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

C

ΔH

Enth

alpy

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).

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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) 0 2CO2(g) �H � �566.0 kJ

2NO(g) 0 N2(g) � O2(g) �H � �180.6 kJ

2CO(g) � 2NO(g) 0 2CO2(g) � N2(g) �H � �385.4 kJ

33. Challenge for �H for the following reaction is �1789 kJ. Use this and reaction a to determine �H for Reaction b.

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 __ 3 � �521 kJ

34. Show how the sum of enthalpy of formation equations produces each of the following reac-tions. You do not need to look up and include �H values.

a. 2NO(g) � O2(g) 0 2NO2(g)

Formation of NO: N2 � O2 0 2NO

Formation of NO2: N2 � 2O2 0 2NO2

NO is a reactant in the problem, so add the reversed NO formation equation to the NO2 formation equation:

2NO � N2 � 2O2 0 N2 � O2 � 2NO2

2NO � O2 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 _ 2 O2(g)

H2O(l) 0 H2(g) � 1 _ 2 O2(g)

SO3(g) � 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.

4NH3(g) � 7O2 (g) 0 4NO2(g) � 6H2O(l)

�H�rxn � [4Hf�(NO2) � 6Hf�(H2O)] � 4Hf�(NO3)

�H�rxn � [4(33.18 kJ) � 6(�285.83 kJ)] �4(�46.11) kJ �H�rxn � �1398 kJ

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 kJ � 4(�394 kJ)] � (�534 kJ)

�H�comb � �2186 kJ

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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:

NO(g) � 1 _ 2 O2(g) 0 NO2(g) �H°rxn

� �58.1 kJ

a. 1 _ 2 N2(g) � 1 _

2 O2(g) 0 NO(g) �H�f � 91.3 kJ

b. 1 _ 2 N2(g) � 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:

c. NO(g) 0 1 _ 2 N2(g) � 1 _

2 O2(g) �H°f � �91.3 kJ

Add equations b and c:

NO(g) � 1 _ 2 O2(g) 0 NO2(g)

�H°rxn � �58.1 kJ � �H°f (c) � �H°f (b)

�58.1 kJ � �91.3 kJ � �H°f (b)

�H°f (b) � �58.1 kJ � 91.3 kJ � 33.2 kJ


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, 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.

39. Explain in words the formula that can be used to determine �H�rxn when using Hess’s law.

�H�rxn � �Hf� (products) � �Hf° (reactants)

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 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

b. 2O3(g) 0 3O2(g) �H ��427 kJ

c. NO(g) � O3(g) 0 NO2(g) � O2(g) �H � �199 kJ

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: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

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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: �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

/mo

l)

�241.8 H2O (g)

�285.8 H2O (l)

�Hvap � �44 kJ/mol

0

�Hvap � �44.0 kJ/mol

Section 15.5 Reaction Spontaneitypages 542–548


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)

�Ssystem is positive because the system’s entropy increases. Entropy increases when a solid or liquid dissolves to form a solution.

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-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 � T�Ssystem

�Gsystem � �75.9 kJ � (273 K)(0.138 kJ/K)

�Gsystem � �75.9 kJ � 37.7 kJ � � 114 kJ

spontaneous reaction

b. �Hsystem � �27.6 kJ, T � 535 K,�Ssystem � �55.2 J/K

�Ssystem � �55.2 J/K � �0.0552 kJ/K


�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 � 365 kJ � (388 K)(�0.0552 kJ/K)

�Gsystem � 365 kJ � 21.4 kJ � 386 kJ


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d. �Hsystem � 452 kJ, T � 165 K,�Ssystem � 55.7 J/K

�Ssystem � 55.7 J/K � 0.0557 kJ


�Gsystem � 452 kJ � (165 K)(0.0557 kJ/K)

�Gsystem � 452 kJ � 9.19 kJ � 443 kJ


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.


For the reaction to be spontaneous:�Gsystem < 0: �Hsystem � T�Ssystem < 0

T > �Hsystem

_ �Ssystem

T > �144 kJ ___ (�36.8 J/K)(1 kJ/1000 J)

T > 3910 K

At any temperature above 3910 K, the reaction is spontaneous.


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 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.

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.

Writing in Chemistrypage 549

Write thermochemical equations for the 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.

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.

1 mol of ethanol � 46.07 g/mol

1 mol of octane � 114.23 g/mol

�1367 kJ _ 1 mol

� 1 mol _ 46.07 g

� 1000 g _ 1 kg

� �29,670 kJ/kg ethanol

� 5471 kJ _ 1 mol

� 1 mol _ 114.23 g

� 1000 g

_ 1 kg

� �47,890 kJ/kg octane

Octane is the better fuel based on the mass burned.

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Chapter 15 Assessmentpages 552–555

Mastering Concepts 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 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 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 relation-ship between the Calorie and a kilocalorie?

One nutritional Calorie equals 1000 calories.

One nutritional Calorie equals 1 kilocalorie.

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.

60. Ethanol has a specific heat of 2.44 J/(g��C). What does this mean?

It means that 2.44 J is required to raise the temperature of one gram of ethanol by one degree Celsius.

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.

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?

1367 kJ � 1000 J _ 1 kJ

� 1 cal _ 4.184 J

� 1 Calorie _ 1000 cal

� 327 Calories

65. To vaporize 2.00 g of ammonia 656 calories are 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

� 1367 kJ

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67. Metallurgy A 25.0�g bolt made of an alloy absorbed 250 joules of heat as its temperature changed from 25.0�C to 78.0�C. What is the specific heat of the alloy?

�T � 78.0�C − 25.0�C � 53.0�C

c � q _

m � �T � 250 J __

25.0 g � 53.0�C

c � 0.189 J/g��C

Section 15.2

Mastering Concepts 68. Why is a foam cup used in a student calorim-

eter 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?

Products

Reactants

Enth

alpy

ΔH = 233 kJ

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

Student answers will vary. One example: universe � my body (the system) � everything else (the surroundings);

another example: a beaker in which a reaction is going on (the system) � everything else (the surroundings)

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?

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 41.2�C to �12.9�C? The specific heat of granite is 0.803 J/(g��C).

�T � 41.2�C � (�12.9�C) � 54.1�C

qgranite � [0.803 J/(g��C)](3.58 � 106 g)(54.1�C)

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 18.4�C, 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

volume of water � (2.00 � 103 cm)(1.25 � 103 cm) (3.75 � 102 cm) � 9.38 � 108 cm3 � 9.38 � 108 mL

mass of water � (9.38 � 108 mL)(1.000 g/mL)� 9.38 � 108 g

q � c � m � �T

�T � (29.0�C � 18.4�C) � 10.6�C

q � [4.184 J/(g��C)](9.38 � 108 g)(10.6�C) � 4.16 � 1010 J

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76. How much heat is absorbed by a 44.7-g piece of lead when its temperature increases by 65.4�C?

q � c � m � �T

q � 0.129 J/(g·�C) � 44.7 g � 65.4�C � 377 J

77. Food Preparation When 10.2 g of canola oil at 25.0�C is placed in a wok, 3.34 kJ of heat is required to heat it to a temperature of 196.4�C. What is the specific heat of canola oil?

3.34 kJ � 1000 J _ 1 kJ

� 3340 J

�T � Tf � Ti � 196.4�C � 25.0�C � 171.4�C

q � c � m � �T

c � q _

m � �T � 3340 J __

10.2 g � 171.4�C � 1.91 J/(g��C)

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 10.5�C. What is the specific heat of the alloy?

q � c � m � �T; qwater � qalloy

4.184 J/(g��C) � 125 g � 10.5�C � calloy � 58.8 g � 106.1�C

calloy � (4.184 J/g·�C)(125 g)(10.5�C)

___ (58.8 g)(106.1�C)

calloy � 0.880 J/(g��C)

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.

80. The molar enthalpy of fusion of methanol is 3.22 kJ/mol. What does this mean?

It means that 3.22 kJ of energy is required to melt one mole of methanol.

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

Mastering Problems 83. Use information from Figure 15.24 to calculate

how much heat is required to vaporize 4.33 mol of water at 100�C?

H2O(g)

H2O(l)

H2O(s)

ΔHvap = +40.7 kJ

ΔHfus = +6.01 kJ

ΔHcond = -40.7 kJ

ΔHsolid = -6.01 kJ

Enth

alpy

Phase Changes for Water

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 H2O

__ 18.0 g

� 0.656 mol H2O

q � mol � �Hsolid

q � 0.656 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 �Hcomb of propane is �2219 kJ/mol.

q � mol � �Hcomb

moles of propane � 4560 kJ __ 2219 kJ/mol

� 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.

mcarbon � mcoal � 0.962 � 1000 g

_ 1 kg

� (5.00 kg)(0.962)( 1000 g

_ 1 kg

) � 4810 g

mol C � 4810 g C � 1 mol _ 12.0 g C

� 401 mol 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 � 1 mol _ 18.02 g

� 40.7 kJ _ 1 mol

� 2830 kJ

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.

90. How does �H for a thermochemical equation change when the amounts of all substances are tripled and the equation is reversed?

�H is tripled, and its sign is changed.

Al(s), Cl2(g)

AlCl3 (s)

ΔH

f° (k

J/m

ol)

-704

0.0

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 _ 2 Cl2(g) 0 AlCl3(s) �H�f � �704 kJ

Mastering Problems 92. Use standard enthalpies of formation from

Table R-11 on page 975 to calculate �H�rxn for the following reaction. P4O6(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 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

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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 _ 2 C2 0 A2C3 �H � �1874 kJ

2B � 3 _ 2

C2 0 B2C3 �H � �285 kJ

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 chem-

ical 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. 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’s entropy increases.

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 �G�system � �H�system � T�S�system. 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

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 �G�system � �H�system � T�S�system.

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

�G�system � �H�system � T�S�system

�G�system � 145 kJ � (293K)(0.195 kJ/K)

� 87.9 kJ

nonspontaneous

b. �Hsystem � �232 kJ, T � 273 K, �Ssystem

� 138 J/K

�Ssystem � 0.138 kJ/K

�G�system � �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

�G�system � �15.9 kJ � (373K)(�0.268 kJ/K)

� 84.1 kJ

nonspontaneous

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100. Calculate the temperature at which �Gsystem � 0if �Hsystem � 4.88 kJ and �Ssystem � 55.2 J/K.

�Ssystem � 55.2 J/K � 0.0552 kJ/K

�G � �H � T�S

0 � 4.88 kJ � T(0.055.2 kJ/K)

T � 4.88 kJ __ 0.0552 kJ/K

� 88.4 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?

�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). �H0

rxn � � 718.3 kJ, and �S0rxn

� �368 J/K. Explain.

�G� � �H� � T�S�

�G� � �718.3 kJ � (298 K)(�0.368 kJ/K); �G� � −609 kJ

Yes. The reaction is spontaneous under standard conditions because �G0

rxn � �609 kJ, and a negative �G0

rxn indicates spontaneity.

103. Calculate the temperature at which �Gsystem � �34.7 kJ if �Hsystem � �28.8 kJ and �Ssystem � 22.2 J/K.

�G�system � �H�system � T�S�system

�34.7 kJ � �28.8 kJ � T (0.0222 kJ/K)

T � 266 K

Mixed Review

Tem

pera

ture

(ºC)

100

0

Heating Curve for Water

12

3

4

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.

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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 = ?

�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.

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.

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 phos-phorous 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) �H � �1774 kJ

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

Add equations c and d and their �H values.

PCl3(l) � Cl2(g) 0 PCl5(s) �H � �124 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 90.0�C and 50.0�C, 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

�c m 1 ( T 1 � T f ) = c m 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 ) � 2 m 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 _ 3 ( T 1 � 2 T 2 ) � 1 _

3 (50�C � 2(90�C)) � 76.7�C

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 combus-tion. Explain your answer. (Hint: Write and compare the balanced chemical equations for the two combustion reactions.)

C H 4 (g) � 2 O 2 (g) 0 C O 2 (g) � 2 H 2 O(l)

C H 2 O(g) � O 2 (g) 0 C O 2 (g) � H 2 O(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

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�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) �H0

comb � �1599.7 kJ/mol

1.000 kg of natural gas contains 884 g CH4 and 116 g C2H6.

884 g � 1 mol _ 16.0 g

� 55.2 mol CH4

116 g C2H6 � 1 mol _ 30.1 g

� 3.86 mol C2H6.

(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 phos-phorus atom? (Chapter 4)

number of protons � 15; number of electrons � 15; number of neutrons � mass number � number of protons � 16

115. What element has the electron configuration [Ar]4s13d5? (Chapter 5)

chromium

116. Name the following molecular compounds. (Chapter 8)

a. S2Cl2disulfur 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)

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 110�C ? (Chapter 13)

T 1

_ V 1

� T 2

_ V 2

; V 2 � T 2 � V 1

_ T 1

� (383 K)(20.0 c m 3 )

__ 263 K

� 29.1 c m 3

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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.

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.

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

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 unsatu-rated. 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 satura-tion 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

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126. 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 � 40.81 kJ/g � 498 kJ

q � c � m � �T

498,000 J � 4.184 J/(g��C) � 1,600 g � �T; �T � 74.4�C

�T � Tf � Ti; 74.4�C � Tf � 20.0�C; Tf � 94.4�C

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

25.0 g H2O � 1 mol H2O

_ 18.02 g

� 1.39 mol H2O

1.39 mol � 40.7 kJ/mol � 56.6 kJ

56.6 kJ � 1 g _

41.45 kJ � 1.37 g canola oil

Standardized Test Practicepages 556–557

Multiple ChoiceUse the graph below to answer Questions 1 to 3.

290 300 310 320 330 340 350

ΔG

(kJ/

mol

)

7.00

6.00

5.00

4.00

3.00

2.00

1.00

0

Temperature (K)

ΔG for the Vaporization of Cyclohexaneas a Function of Temperature

1. In the range of temperatures shown, the vapor-ization of cyclohexane

a. does not occur at all.b. will occur spontaneously.c. is not spontaneous.d. occurs only at high temperatures.

c

2. What is the standard free energy of vaporization, �G�vap, of cyclohexane at 300 K?

a. 5.00 kJ/molb. 3.00 kJ/molc. 3.00 kJ/mold. 2.00 kJ/mol

a

3. When �G�vap is plotted as a function of temper-ature, 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 vapor-ization of cyclohexane?

a. �50.0 J/mol-Kb. �10.0 J/mol-Kc. �5.0 J/mol-Kd. �100 J/mol-K

d

slope � rise _ run

� (6.00 � 5.00)kJ/mol __ (290 � 300)K

� 0.1 kJ/mol�K

0.1 kJ/mol�K � 1000 J _ 1 kJ

� 100 J/mol�K

4. The metal yttrium, atomic number 39, forms

a. positive ions.b. negative ions.c. both positive and negative ions.d. no ions at all.

a

5. Given the reaction 2Al � 3FeO 0 Al2O3 � 3Fe, what is the mole-to-mole ratio between iorn (II) oxide and aluminum oxide?

a. 2:3b. 1:1c. 3:2d. 3:1

d

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Use the table below to answer Question 6.

Electronegativity of Selected Elements

H

2.20

Li 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

6. Which bond is the most electronegative?

a. H-Hb. H-Cc. H-Nd. H-O

d

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. Q2M3b. M2Q3c. Q3M2d. M3Q2

c

8. Wavelengths of light shorter than about 4.00 � 10�7 m are not visible to the human eye. What is the energy of a photon of ultra-violet light having a frequency of 5.45 � 1016 s�1? (Planck’s constant is 6.626 � 10�34 J�s.)

a. 3.61 � 1 0 � 17 Jb. 1.22 � 1 0 � 50 Jc. 8.23 � 1049 Jd. 3.81 � 1 0 � 24 J

a

(5.45 � 1016 s�1)(6.626 � 10�34 J�s) � 3.61 � 10�17

Use the graph below to answer Question 9.

250 260 270 280 290 300

Pres

ure

(kPa

)

1200

1000

800

600

400

200

0

Gas C

Gas A

Gas B

Temperature (K)

Pressures of Three Gasesat Different Temperatures

9. What is the predicted pressure of Gas B at 310 K?a. 500 kPab. 600 kPac. 700 kPad. 900 kPa

b

Short AnswerUse the figure below to answer Questions 11 to 13.

S CI K CaAr

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�.

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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 ResponseUse the information below to answer Questions 13 and 14.

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.

1 atm

1 atm

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.

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.

T1 _ V1

� T2 _ V2

300.0 K _ 2.1 L

� T2 _

5.4 L

T2 � (300.0 K)(5.4 L)

__ 2.1 L

T2 � 770 K

SAT Subject Test: Chemistry 15. 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 �20.0�C to 68.0�C?

a. 10.7 kJb. 8.30 kJc. 2.44 kJd. 1.22 kJe. 5.86 kJ

a

q � cm�T � (2.44 J/(g��C)) � (50.0 g) � (88.0�C) � 1kj _

1000 J � 10.7 kJ

16. 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?

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)

c

q � cm�T

c � q _

m�T � (1728 J)

__ (3.00 g) (642.0 °C)

� 0.897 J/(g�°C)

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Use the table below to answer Questions 17 and 18.

Density and Electronegativity Data for Elements

Elements Density (g/ml) Electronegativity

Aluminum 2.698 1.6

Fluorine 1.696 � 10�3 4.0

Sulfur 2.070 2.6

Copper 8.960 1.9

Magnesium 1.738 1.3

Carbon 3.513 2.6

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. aluminumb. magnesiumc. carbond. coppere. sulfur

� m _ V

� 9.250 g

_ 5.250 mL

� 1.762 g/mL

b

18. Which pair is most likely to form an ionic bond?

a. carbon and sulfurb. aluminum and magnesiumc. copper and sulfurd. magnesium and fluorinee. aluminum and carbon

d

Chapter 15

Documents

Transcript of Chapter 15