# Silberberg7e solution manual ch 06

CHAPTER 6 THERMOCHEMISTRY: ENERGY
FLOW AND CHEMICAL CHANGE
FOLLOW–UP PROBLEMS
6.1A

Plan: The system is the liquid. Since the system absorbs heat from the surroundings, the system gains heat and q is
positive. Because the system does work, w is negative. Use the equation E = q + w to calculate E. Convert E
to kJ.
Solution:
E (kJ) = q + w = +13.5 kJ + –1.8 kJ = 11.7 kJ
E (J) = 11.7 kJ

6.1B

1000 J
1 kJ

= 1.17 x 104 J

Plan: The system is the reactant and products of the reaction. Since heat is absorbed by the surroundings, the
system releases heat and q is negative. Because work is done on the system, w is positive. Use the equation

E = q + w to calculate E. Both kcal and Btu must be converted to kJ.
Solution:
 4.184 kJ 
q =  26.0 kcal  
 = –108.784 kJ
 1 kcal 

 1.055 kJ 
w =  +15.0 Btu  
 = 15.825 kJ
 1 Btu 
E = q + w = –108.784 kJ + 15.825 kJ = –92.959 = – 93 kJ
6.2A

Plan: Convert the pressure from torr to atm units. Subtract the initial V from the final V to find ΔV. Use w = -PΔV
to calculate w in atm•L. Convert the answer from atm•L to J.
Solution:
Vinitial = 5.68 L
Vfinal = 2.35 L
P = 732 torr
Converting P from torr to atm: (732 torr)

1 atm
760 torr

= 0.9632 = 0.963 atm

w (atm•L) = –PΔV = –(0.963 atm)(2.35 L – 5.68 L) = 3.2068 = 3.21 atm•L
w (J) = (3.21 atm•L)
6.2B

= 325 J

Plan: Subtract the initial V from the final V to find ΔV. Use w = –PΔV to calculate w in atm•L. Convert the answer
from atm•L to J.
Solution:
Vinitial = 10.5 L
Vfinal = 16.3 L
P = 5.5 atm
w (atm•L) = –PΔV = – (5.5 atm)(16.3 L – 10.5 L) = –31.90 = –32 atm•L

w (J) = (–32 atm•L)

6.3A

101.3 J
1 atm•L

101.3 J
1 atm•L

= –3.2 x 103 J

Plan: Since heat is released in this reaction, the reaction is exothermic (H < 0) and the reactants are above the
products in an enthalpy diagram.

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6-1

Solution:
Enthalpy, H

C3H5(NO3)3(l)
 = 5.72 x 103 kJ
3 CO2(g) + 5/2 H2O(g) + 1/4 O2(g) + 3/2 N2(g)

6.3B

Plan: Since heat is absorbed in this reaction, the reaction is endothermic (H > 0) and the reactants are below the
products in an enthalpy diagram.
Solution:

6.4A

Plan: Heat is added to the aluminum foil, so q will be positive. The heat is calculated using the equation
q = c x mass x T. Table 6.2 lists the specific heat of aluminum as 0.900 J/g•K.
Solution:
T = 375°C – 18°C = (357°C)

change in 1 K
change in 1 degreeC

= 357 K

q = c x mass x T = (0.900 J/gK) (7.65 g) (357 K) = 2.46 x 103 J
6.4B

Plan: Heat is transferred away from the ethylene glycol as it cools so q will be negative. The heat released is
calculated using the equation q = c x mass x T. Table 6.2 lists the specific heat of ethylene glycol as 2.42 J/gK.
The volume of ethylene glycol is converted to mass in grams by using the density.
Solution:
T = 25.0°C – 37.0°C = (–12.0°C)

change in 1 K
change in 1 degreeC

= –12.0 K

 1 mL   1.11 g 
Mass (g) of ethylene glycol =  5.50 L   3  
= 6105 = 6.10 x 103 g
 10 L   mL 

q = c x mass x T = (2.42 J/gK) (6.10 x 103 g) (–12.0 K)
6.5A

1 kJ
1000 J

= –177.1440 = –177 kJ

Plan: The heat absorbed by the water can be calculated with the equation c x mass x T; the heat absorbed by the
water equals the heat lost by the hot metal. Since the mass and temperature change of the metal is known, the
specific heat capacity can be calculated and used to identify the metal.
Solution:
TH 2 O = Tfinal – Tinitial = 27.25°C – 25.55°C = (1.70°C)

change in 1 K
change in 1 degreeC

Tmetal = Tfinal – Tinitial = 27.25°C – 65.00°C = (–37.75°C)
qH 2 O =  qmetal
cH 2 O x mass H 2 O x TH 2 O = cmetal x mass metal x Tmetal

–cmetal = 

cH 2O x mass H 2O x TH 2O
mass metal x Tmetal

= 

= 1.70 K

change in 1 K
change in 1 degreeC

 4.184 J/g  K  25.00 g 1.70 K 
12.18 g  37.75 K 

= –37.75 K

= 0.386738 = 0.387 J/gK

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6-2

From Table 6.2, the metal with this value of specific heat is copper.
6.5B

Plan: The heat absorbed by the titanium metal can be calculated with the equation c x mass x T; the heat
absorbed by the metal equals the heat lost by the water. Since the mass and temperature change of the water is
known, along with the specific heat and final temperature of the titanium, initial temperature of the metal can be
calculated.
Solution:
TH 2 O = Tfinal – Tinitial = 49.30°C – 50.00°C = (–0.70°C)

change in 1 K
change in 1 degreeC

= –0.70 K

Converting Tfinal from oC to K = 49.30oC + 273.15 = 322.45 K

Tmetal = Tfinal – Tinitial = 322.45 K – Tinitial

cH 2 O x mass H 2 O x TH 2 O = cmetal x mass metal x Tmetal

ΔTmetal =

–cH2 O x massH2 O x ∆TH2 O
cmetal x massmetal

=

–(4.184 J/gK) (75.0 g) (–0.70 K)
(0.228 J/gK) (33.2 g)

= 29.0187 = 29.0 K

ΔTmetal =29.0187 K = 322.45 K – Tinitial (using unrounded numbers to avoid rounding errors)
Tinitial = 293.4313 K – 273.15 = 20.2813 = 20.3oC

6.6A

Plan: First write the balanced molecular, total ionic and net ionic equations for the acid-base reaction. To find
qsoln, we use the equation q = c x mass x T, so we need the mass of solution, the change in temperature, and the
specific heat capacity. We know the solutions’ volumes (25.0 mL and 50.0 mL), so we find their masses with the
given density (1.00 g/mL). Then, to find qsoln, we multiply the total mass by the given c (4.184 J/g•K) and the
change in T, which we find from Tfinal – Tinitial. The heat of reaction (qrxn) is the negative of the heat of solution
(qsoln).
Solution:
a) The balanced molecular equation is: HNO3(aq) + KOH(aq)  KNO3(aq) + H2O(l)
The total ionic equation is: H+(aq) + NO3– (aq) + K+(aq) + OH– (aq)  K+(aq) + NO3– (aq) + H2O(l)
The net ionic equation is: H+(aq) + OH–(aq)  H2O(l)
b) Total mass (g) of solution = (25.0 mL + 50.0 mL) x 1.00 g/mL = 75.0 g
T = 27.05oC – 21.50oC = (5.15oC)

change in 1 K
change in 1 degreeC

= 5.15 K

qsoln (J) = csoln x masssoln x Tsoln = (4.184 J/gK)(75.0 g)(5.15 K) = 1620 J
qsoln (kJ) = 1620 J
qrxn = –qsoln
6.6B

1 kJ
1000 J

= 1.62 kJ
so

qrxn = –1.62 kJ

Plan: Write a balanced equation. Multiply the volume by the molarity of each reactant solution to find moles of each
reactant. Use the molar ratios in the balanced reaction to find the moles of water produced from each reactant; the
smaller amount gives the limiting reactant and the actual moles of water produced. Divide the heat evolved by the
moles of water produced to obtain the enthalpy in kJ/mol. Since qsoln is positive, the solution absorbed heat that was
released by the reaction; qrxn and ΔH are negative.
Solution:
Ba(OH)2(aq) + 2HCl(aq) → 2H2O(l) + BaCl2(aq)
 103 L   0.500 mol Ba(OH) 2 
Moles of Ba(OH)2 =  50.0 mL  
 = 0.0250 mol Ba(OH)2
 1 mL  
1L

 103 L   0.500 mol HCl 
Moles of HCl =  50.0 mL  
 = 0.0250 mol HCl
 1 mL  
1L

 2 mol H 2 O 
Moles of H2O from Ba(OH)2 =  0.0250 mol Ba(OH) 2  
 = 0.0500 mol H2O
 1 mol Ba(OH) 2 

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6-3

 2 mol H 2 O 
Moles of H2O from HCl =  0.0250 mol HCl  
 = 0.0250 mol H2O
 2 mol HCl 
HCl is the limiting reactant; 0.0250 mol of H2O is produced.
q
1.386 kJ
=
= –55.44 = – 55.4 kJ/mol
ΔH (kJ/mol) =
moles of H2O produced 0.0250 mol
6.7A

Plan: The bomb calorimeter gains heat from the combustion of graphite, so –qgraphite = qcalorimeter. Convert the mass
of graphite from grams to moles and use the given kJ/mol to find qgraphite. The heat lost by graphite equals the heat
gained by the calorimeter, or T multiplied by Ccalorimeter.
Solution:
 1 mol C 
Moles of graphite =  0.8650 g C  
 = 0.0720233 mol C
 12.01 g C 

 393.5 kJ 
qgraphite =  0.0720233 mol C  
 = –28.3412 kJ/mol
 1 mol C 
– qgraphite = Ccalorimeter Tcalorimeter
28.3412 kJ/mol = Ccalorimeter(2.613 K)
Ccalorimeter = 10.84623 = 10.85 kJ/K
6.7B

Plan: The bomb calorimeter gains heat from the combustion of acetylene, so –qrxn = qcalorimeter. Use the given heat
capacity from Follow-up Problem 6.7A to find qrxn: the amount of heat lost by the reaction of acetylene equals the
amount of heat gained by the calorimeter, or T multiplied by Ccalorimeter. Divide the heat produced by the
combustion of acetylene (in kJ) by the moles of acetylene to obtain the enthalpy in kJ/mol.
Solution:
ΔT = (11.50oC)

change in 1 K
change in 1 degree C

= 11.50 K

–qrxn = qcalorimeter = (ccalorimeter)(ΔT)
–qrxn = (10.85 kJ/K) (11.5 K) = 124.8 kJ
qrxn = –124.8 kJ
Moles of acetylene = (2.50 g C2H2)
ΔH (kJ/mol) =
6.8A

q
moles of acetylene

=

1 mol C2 H2
26.04 g C2 H2
–124.8 kJ

0.0960 moles

= 0.0960 mol C2H2

= –1300 = – 1.30 x 103 kJ/mol

Plan: To find the heat required, write a balanced thermochemical equation and use appropriate molar ratios to
solve for the required heat.
Solution:
C2H4(g) + H2(g)  C2H6(g)
H = –137 kJ
3
 10 g   1 mol C 2 H 6   137 kJ 
Heat (kJ) = 15.0 kg C 2 H 6  
 1 kg   30.07 g C H   1 mol C H 
2 6 
2 6 


= –6.83405 x 104 = –6.83 x 104 kJ

6.8B

Plan: Write a balanced thermochemical equation and use appropriate molar ratios to solve for the required heat.
Solution:
N2(g) + O2(g)  2NO(g) H = +180.58 kJ
Heat (kJ) = (3.50 t NO)

6.9A

103 kg

1000 g

1 mol NO

+180.58 kJ

1t

1 kg

30.01 g NO

2 mol NO

= 1.05 x 107 kJ

Plan: Manipulate the two equations so that their sum will result in the overall equation. Reverse the first equation
(and change the sign of H); reverse the second equation and multiply the coefficients (and H) by two.

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6-4

Solution:

6.9B

6.10A

2NO(g) + 3/2O2(g)  N2O5(s)
H = –(223.7 kJ)= –223.7 kJ
2NO2(g)  2NO(g) + O2(g)
H = –2(–57.1 kJ) = 114.2 kJ
Total: 2NO2(g) + 1/2O2(g)  N2O5(s)
H = –109.5 kJ
Plan: Manipulate the three equations so that their sum will result in the overall equation. Reverse the first equation
(and change the sign of H) and multiply the coefficients (and H) by 1/2. Multiply the coefficients of the second
equation (and H) by 1/2. Reverse the third equation (and change the sign of H).
Solution:
NH3(g)  1/2N2(g) + 3/2H2(g)
H = –1/2(–91.8 kJ)= 45.9 kJ
1/2N2(g) + 21/2H2(g) + 1/2Cl2(g)  NH4Cl(s)
H = 1/2(–628.8 kJ)= –314.4 kJ
NH4Cl(s)  NH3(g) + HCl(g)
H = –(–176.2 kJ) = 176.2 kJ
Total: 1/2H2(g) + 1/2Cl2(g)  HCl(g)
H = –92.3 kJ
Plan: Write the elements as reactants (each in its standard state), and place one mole of the substance formed on
the product side. Balance the equation with the following differences from “normal” balancing — only one mole
of the desired product can be on the right hand side of the arrow (and nothing else), and fractional coefficients are
allowed on the reactant side. The values for the standard heats of formation ( H f ) may be found in the appendix.
Solution:
a) C(graphite) + 2H2(g) + 1/2O2(g)  CH3OH(l)

H f = –238.6 kJ/mol

b) Ca(s) + 1/2O2(g)  CaO(s)

H f = –635.1 kJ/mol

c) C(graphite) + 1/4S8(rhombic)  CS2(l)

H f = 87.9 kJ/mol

6.10B Plan: Write the elements as reactants (each in its standard state), and place one mole of the substance formed on
the product side. Balance the equation with the following differences from “normal” balancing — only one mole
of the desired product can be on the right hand side of the arrow (and nothing else), and fractional coefficients are
allowed on the reactant side. The values for the standard heats of formation ( H f ) may be found in the appendix.
Solution:

6.11A

a) C(graphite) + 1/2H2(g) + 3/2Cl2(g)  CHCl3(l)

H f = –132 kJ/mol

b) 1/2N2(g) + 2H2(g) + 1/2Cl2(g)  NH4Cl(s)

H f = –314.4 kJ/mol

c) Pb(s) + 1/8S8(rhombic) + 2O2(g)  PbSO4(s)

H f = –918.39 kJ/mol

Plan: Look up H f values from the appendix and use the equation H rxn
= m H f(products)
– n H f(reactants)

to solve for H rxn
.
Solution:

H rxn
= m H f(products)
– n H f(reactants)

= {4 H f [H3PO4(l)]} – {1 H f [P4O10(s)] + 6 H f [H2O(l)]}
= (4 mol)( –1271.7 kJ/mol) – [(1 mol)( –2984 kJ/mol) + (6 mol)(–285.840 kJ/mol)]
= –5086.8 kJ – [–2984 kJ + –1714.8 kJ] = –388 kJ
6.11B

Plan: Apply the H rxn
to this reaction, substitute given values, and solve for the H f (CH3OH).
Solution:

H rxn
= m H f(products)
– n H f(reactants)

H rxn
= {1 H f [CO2(g)] + 2 H f [H2O(g)]} – {1 H f [CH3OH(l)] + 3/2 H f [O2(g)]}
–638.6 kJ = [(1 mol)(–393.5 kJ/mol) + (2 mol)(–241.826 kJ/mol)]

– [ H f [CH3OH(l)] + (3/2 mol)(0 kJ/mol)]
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6-5

–638.6 kJ = (–877.152 kJ) – H f [CH3OH(l)]
H f [CH3OH(l)] = –238.552 = –238.6 kJ

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6-6

B6.1

Plan: Convert the given mass in kg to g, divide by the molar mass to obtain moles, and convert moles to kJ of
energy. Sodium sulfate decahydrate will transfer 354 kJ/mol.
Solution:
 103 g   1 mol Na 2SO 4 •10H 2 O  

354 kJ
Heat (kJ) =  500.0 kg Na 2SO 4 •10H 2 O  
 1 kg   322.20 g Na SO •10H O   1 mol Na SO •10H O 
2
4
2 
2
4
2 


= –5.4935x105 = – 5.49x105 kJ

B6.2

Plan: Three reactions are given. Equation 1) must be multiplied by 2, and then the reactions

can be added, canceling substances that appear on both sides of the arrow. Add the H rxn

values for the three reactions to get the H rxn
for the overall gasification reaction of 2 moles of

coal. Use the relationship H rxn
= m H f (products) – n H f (reactants) to find the heat of combustion of 1 mole

of methane. Then find the H rxn
for the gasification of 1.00 kg of coal and H rxn
for the combustion of the
methane produced from 1.00 kg of coal and sum these values.
Solution:

a) 1) 2C(coal) + 2H2O(g)  2CO(g) + 2H2(g)
2) CO(g) + H2O(g)  CO2(g) + H2(g)

H rxn
= 2(129.7 kJ)

H rxn
= – 41 kJ

H rxn
3) CO(g) + 3H2(g)  CH4(g) + H2O(g)
= –206 kJ
2C(coal) + 2H2O(g)  CH4(g) + CO2(g)
b) The total may be determined by doubling the value for equation 1) and adding to the other two values.

H rxn
= 2(129.7 kJ) + (–41 kJ) + (–206 kJ) = 12.4 = 12 kJ
c) Calculating the heat of combustion of CH4:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

H rxn
= m H f (products) – n H f (reactants)

H rxn
= [(1 mol CO2)( H f of CO2) + (2 mol H2O)( H f of H2O)]

– [(1 mol CH4)( H f of CH4) + (2 mol O2)( H f of O2)]

H rxn
= [(1 mol)(–395.5 kJ/mol) + (2 mol)(–241.826 kJ/mol)]
– [(1 mol)(–74.87kJ/mol) + (2 mol )(0.0 kJ/mol)]

H rxn
= –804.282 kJ/mol CH4
Total heat for gasification of 1.00 kg coal:
 103 g   1 mol coal   12.4 kJ 
H° = 1.00 kg coal 
= 516.667 kJ
 1 kg   12.00 g coal   2 mol coal 



Total heat from burning the methane formed from 1.00 kg of coal:
 103 g   1 mol coal   1 mol CH 4   804.282 kJ 
H° = 1.00 kg coal 
= –33511.75 kJ
 1 kg   12.00 g coal   2 mol coal   1 mol CH 


4 



Total heat = 516.667 kJ + (–33511.75 kJ) = –32995.083 = –3.30x104 kJ

END–OF–CHAPTER PROBLEMS

6.1

The sign of the energy transfer is defined from the perspective of the system. Entering the system is positive, and
leaving the system is negative.

6.2

No, an increase in temperature means that heat has been transferred to the surroundings, which makes q negative.

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6-7

6.3

E = q + w = w, since q = 0.
Thus, the change in work equals the change in internal energy.

6.4

Plan: Remember that an increase in internal energy is a result of the system (body) gaining heat or having work
done on it and a decrease in internal energy is a result of the system (body) losing heat or doing work.
Solution:
The internal energy of the body is the sum of the cellular and molecular activities occurring from skin level
inward. The body’s internal energy can be increased by adding food, which adds energy to the body through the
breaking of bonds in the food. The body’s internal energy can also be increased through addition of work and
heat, like the rubbing of another person’s warm hands on the body’s cold hands. The body can lose energy if it
performs work, like pushing a lawnmower, and can lose energy by losing heat to a cold room.

6.5

a) electric heater
e) battery (voltaic cell)

6.6

Plan: Use the law of conservation of energy.
Solution:
The amount of the change in internal energy in the two cases is the same. By the law of energy conservation, the
change in energy of the universe is zero. This requires that the change in energy of the system (heater or air
conditioner) equals an opposite change in energy of the surroundings (room air). Since both systems consume the
same amount of electrical energy, the change in energy of the heater equals that of the air conditioner.

6.7

Heat energy; sound energy

Kinetic energy

Potential energy

Mechanical energy

Chemical energy

b) sound amplifier

c) light bulb

d) automobile alternator

(impact)
(falling text)
(raised text)
(raising of text)
(biological process to move muscles)

6.8

Plan: The change in a system’s energy is E = q + w. If the system receives heat, then its qfinal is greater than
qinitial so q is positive. Since the system performs work, its wfinal < winitial so w is negative.
Solution:
E = q + w
E = (+425 J) + (–425 J) = 0 J

6.9

q + w = –255 cal + (–428 cal) = –683 cal

6.10

Plan: The change in a system’s energy is E = q + w. A system that releases thermal energy has a negative
value for q and a system that has work done on it has a positive value for work. Convert work in calories to
work in joules.
Solution:
 4.184 J 
Work (J) = 530 cal 
 = 2217.52 J
 1 cal 
E = q + w = –675 J + 2217.52 J = 1542.52 = 1.54x103 J

 103 cal   4.184 J  
3
 0.247 kcal 

 1 kcal   1 cal   = 1648.4 = 1.65x10 J

6.11

 103 J 
E = q + w = 0.615 kJ 
+
 1 kJ 

6.12

Plan: Convert 6.6x1010 J to the other units using conversion factors.

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6-8

Solution:
C(s) + O2(g)  CO2(g) + 6.6x1010 J
(2.0 tons)
 1 kJ 
a) E (kJ) = (6.6 x 1010 J)  3  = 6.6x107 kJ
 10 J 

 1 cal   1 kcal 
7
7
b) E (kcal) = (6.6 x 1010 J) 
 = 1.577x10 = 1.6x10 kcal
  3
 4.184 J   10 cal 
 1 Btu 
7
7
c) E (Btu) = (6.6 x 1010 J) 
 = 6.256x10 = 6.3x10 Btu
1055
J

6.13

CaCO3(s) + 9.0x106 kJ  CaO(s) + CO2(g)
(5.0 tons)
 103 J 
= 9.0x109 J
a) E (J) = (9.0x106 kJ) 
 1 kJ 

 103 J   1 cal 
b) E (cal) = (9.0x106 kJ) 
= 2.15105x109 = 2.2x109 cal
 1 kJ   4.184 J 


 103 J   1 Btu 
c) E (Btu) = (9.0x106 kJ) 
= 8.5308x106 = 8.5x106 Btu
 1 kJ   1055 J 



6.14

 103 cal   4.184 J 
= 1.7154x107 = 1.7x107 J
E (J) = (4.1x103 Calorie) 
 1 Calorie   1 cal 


 103 cal   4.184 J   1 kJ 
E (kJ) = (4.1x103 Calorie) 
= 1.7154x104 = 1.7x104 kJ
 1 Calorie   1 cal   103 J 

6.15

Plan: 1.0 lb of body fat is equivalent to about 4.1x103 Calories. Convert Calories to kJ with the appropriate
conversion factors.
Solution:
 4.1x103 Cal   103 cal   4.184 J   1 kJ  

h
Time = 1.0 lb  
= 8.79713 = 8.8 h
 1.0 lb   1 Cal   1 cal   103 J   1950 kJ 







6.16

The system does work and thus its internal energy is decreased. This means the sign will be negative.

6.17

Since many reactions are performed in an open flask, the reaction proceeds at constant pressure. The
determination of H (constant pressure conditions) requires a measurement of heat only, whereas E requires
measurement of heat and PV work.

6.18

The hot pack is releasing (producing) heat, thus H is negative, and the process is exothermic.

6.19

Plan: An exothermic process releases heat and an endothermic process absorbs heat.
Solution:
a) Exothermic, the system (water) is releasing heat in changing from liquid to solid.
b) Endothermic, the system (water) is absorbing heat in changing from liquid to gas.
c) Exothermic, the process of digestion breaks down food and releases energy.
d) Exothermic, heat is released as a person runs and muscles perform work.
e) Endothermic, heat is absorbed as food calories are converted to body tissue.
f) Endothermic, the wood being chopped absorbs heat (and work).

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6-9

g) Exothermic, the furnace releases heat from fuel combustion. Alternatively, if the system is defined as the air
in the house, the change is endothermic since the air’s temperature is increasing by the input of heat energy from
the furnace.
6.20

The internal energy of a substance is the sum of kinetic (EK) and potential (EP) terms.
EK (total) = EK (translational) + EK (rotational) + EK (vibrational)
EP = EP (atom) + EP (bonds)
EP (atom) has nuclear, electronic, positional, magnetic, electrical, etc., components.

6.21

H = E + PV (constant P)
a) H < E, PV is negative.
b) H = E, a fixed volume means PV = 0.
c) H > E, PV is positive for the transformation of solid to gas.

6.22

Plan: Convert the initial volume from mL to L. Subtract the initial V from the final V to find ΔV. Calculate w in
atm•L. Convert the answer from atm•L to J.
Solution:
Vinitial = 922 mL
Vfinal = 1.14 L
P = 2.33 atm
Converting Vinitial from mL to L: (922 mL)

1L
103 mL

= 0.922 L

w (atm•L) = -PΔV = -(2.33 atm)(1.14 L – 0.922 L) = -0.51 atm•L
w (J) = (-0.51 atm•L)
6.23

1J

= -52 J

9.87 x 10-3 atm•L

Plan: Convert the pressure from mmHg to atm. Subtract the initial V from the final V to find ΔV. Calculate w in
atm•L. Convert the answer from atm•L to kJ.
Solution:
Vinitial = 0.88 L
Vfinal = 0.63 L
P = 2660 mmHg
Converting P from mmHg to atm: (2660 mmHg)

1 atm
760 mmHg

= 3.50 atm

w (atm•L) = -PΔV = -(3.50 atm)(0.63 L – 0.88 L) = 0.88 atm•L
w (J) = (0.88 atm•L)
6.24

101.3 J

1 kJ

1atm•L

103 J

= 0.089 kJ

Plan: An exothermic reaction releases heat, so the reactants have greater H (Hinitial) than the products (Hfinal).
H = Hfinal – Hinitial < 0.
Solution:
Reactants
H = ( ), (exothermic)
Products

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6-10

6.25

Plan: Combustion of hydrocarbons and related compounds require oxygen (and a heat catalyst) to yield
carbon dioxide gas, water vapor, and heat. Combustion reactions are exothermic. The freezing of liquid water is
an exothermic process as heat is removed from the water in the conversion from liquid to solid. An exothermic
reaction or process releases heat, so the reactants have greater H (Hinitial) than the products (Hfinal).
Solution:
a) Combustion of ethane: 2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(g) + heat
2C2H6 + 7O2 (initial)
Increasing, H

6.26

H = (), (exothermic)
4CO2 + 6H2O (final)

b) Freezing of water: H2O(l)  H2O(s) + heat

a) Na(s) + 1/2Cl2(g)  NaCl(s) + heat
Na(s) + 1/2Cl2(g)
Increasing, H

6.27

exothermic
NaCl(s)

b) C6H6(l) + heat  C6H6(g)

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6-11

Plan: Combustion of hydrocarbons and related compounds require oxygen (and a heat catalyst) to yield carbon
dioxide gas, water vapor, and heat. Combustion reactions are exothermic. An exothermic reaction releases heat, so
the reactants have greater H (Hinitial) than the products (Hfinal). If heat is absorbed, the reaction is endothermic and
the products have greater H (Hfinal) than the reactants (Hinitial).
Solution:
a) 2CH3OH(l) + 3O2(g)  2CO2(g) + 4H2O(g) + heat
2CH3OH + 3O2 (initial)
Increasing, H

6.28

H = (), (exothermic)
2CO2 + 4H2O (final)

Increasing, H

b) Nitrogen dioxide, NO2, forms from N2 and O2.
1/2N2(g) + O2(g) + heat  NO2(g)
NO2 (final)

1/2N2 + O2 (initial)

a) CO2(s) + heat  CO2(g)
CO2(g)
Increasing, H

6.29

H = (+), (endothermic)

H = (+), (endothermic)
CO2(s)

Increasing, H

b) SO2(g) + 1/2O2(g)  SO3(g) + heat
SO2(g) + 1/2O2(g)

6.30

H = (), (exothermic)
SO3(g)

Plan: Recall that qsys is positive if heat is absorbed by the system (endothermic) and negative if heat is released
by the system (exothermic). Since E = q + w, the work must be considered in addition to qsys to find ΔEsys.
Solution:
a) This is a phase change from the solid phase to the gas phase. Heat is absorbed by the system so qsys is positive
(+).
b) The system is expanding in volume as more moles of gas exist after the phase change than were present before
the phase change. So the system has done work of expansion and w is negative. ΔEsys = q + w. Since q is
positive and w is negative, the sign of ΔEsys cannot be predicted. It will be positive if q > w and negative if
q < w.

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6-12

c) ΔEuniv = 0. If the system loses energy, the surroundings gain an equal amount of energy. The sum of the
energy of the system and the energy of the surroundings remains constant.
6.31

a) There is a volume decrease; Vfinal < Vinitial so ΔV is negative. Since wsys = –PΔV, w is positive, +.
b) ∆Hsys is – as heat has been removed from the system to liquefy the gas.
c) ∆Esys = q + w. Since q is negative and w is positive, the sign of ΔEsys and ΔEsurr cannot be predicted. ΔEsys
will be positive and ΔEsurr will be negative if w > q and ΔEsys will be negative and ΔEsurr will be positive if
w < q.

6.32

The molar heat capacity of a substance is larger than its specific heat capacity. The specific heat capacity of a
substance is the quantity of heat required to change the temperature of 1 g of a substance by 1 K while the molar
heat capacity is the quantity of heat required to change the temperature of 1 mole of a substance by 1 K.
The specific heat capacity of a substance is multiplied by its molar mass to obtain the molar heat capacity.

6.33

To determine the specific heat capacity of a substance, you need its mass, the heat added (or lost), and the change
in temperature.

6.34

Specific heat capacity is an intensive property; it is defined on a per gram basis. The specific heat capacity of
a particular substance has the same value, regardless of the amount of substance present.

6.35

Specific heat capacity is the quantity of heat required to raise 1g of a substance by 1 K. Molar heat
capacity is the quantity of heat required to raise 1 mole of substance by 1 K. Heat capacity is also the quantity of
heat required for a 1 K temperature change, but it applies to an object instead of a specified amount of a
substance. Thus, specific heat capacity and molar heat capacity are used when talking about an element or
compound while heat capacity is used for a calorimeter or other object.

6.36

In a coffee-cup calorimeter, reactions occur at constant pressure. qp = H.
In a bomb calorimeter, reactions occur at constant volume. qv = E.

6.37

Plan: The heat required to raise the temperature of water is found by using the equation
q = c x mass x T. The specific heat capacity, cwater, is found in Table 6.2. Because the Celsius degree is the same
size as the Kelvin degree, T = 100°C – 25°C = 75°C = 75 K.
Solution:

J 
3
q (J) = c x mass x T =  4.184
 22.0 g 75 K  = 6903.6 = 6.9x10 J
g K

6.38
6.39

J 
q (J) = c x mass x T =  2.087
 0.10 g 75  10.K  = –17.7395 = –18 J
g K

Plan: Use the relationship q = c x mass x T. We know the heat (change kJ to J), the specific heat capacity, and
the mass, so T can be calculated. Once T is known, that value is added to the initial temperature to find the
final temperature.
Solution:
q (J) = c x mass x T
Tinitial = 13.00°C
Tfinal = ? mass = 295 g c = 0.900 J/g•K
3
 10 J 
q = 75.0 kJ 
= 7.50x104 J
 1 kJ 

7.50x104 J = (0.900 J/g•K)(295 g)(T)

7.50x10 J 
4

T =

 0.900 J 

 gK 

 295 g  

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6-13

T = 282.4859 K = 282.4859°C

(Because the Celsius degree is the same size as the Kelvin degree, T is the
same in either temperature unit.)

T = Tfinal – Tinitial
Tfinal T + Tinitial
Tfinal = 282.4859°C + 13.00°C = 295.49 = 295°C
6.40

q (J) = c x mass x T
–688 J = (2.42 J/g•K)(27.7 g)(T)
688 J 
(T) =
= –10.26345 K = –10.26345°C
 2.42 J 
27.7 g 

 gK 
T = Tfinal – Tinitial
Tinitial Tfinal – T
Tinitial = 32.5°C – (–10.26345°C) = 42.76345 = 42.8°C

6.41

Plan: Since the bolts have the same mass and same specific heat capacity, and one must cool as the other heats
(the heat lost by the “hot” bolt equals the heat gained by the “cold” bolt), the final temperature is an average of the
two initial temperatures.
Solution:
 T1 + T2   100.C + 55C 

 = 
 = 77.5°C
2
2

 


6.42

–qlost = qgained
– 2(mass)(cCu)(Tfinal – 105)°C = (mass)(cCu)(Tfinal – 45)°C
– 2(Tfinal – 105)°C = (Tfinal – 45)°C
2(105°C) – 2Tfinal = Tfinal – 45°C
210°C + 45°C = Tfinal + 2Tfinal = 3Tfinal
(255°C)/3 = Tfinal = 85.0°C

6.43

Plan: The heat lost by the water originally at 85°C is gained by the water that is originally at 26°C. Therefore
–qlost = qgained. Both volumes are converted to mass using the density.
Solution:
 1.00 g 
 1.00 g 
Mass (g) of 75 mL = 75 mL 
Mass (g) of 155 mL = 155 mL 
 = 75 g
 = 155 g
1
mL

 1 mL 
–qlost = qgained
c x mass x T (85°C water)= c x mass x T (26°C water)
– (4.184 J/g°C)(75 g)(Tfinal – 85)°C =(4.184 J/g°C)(155 g)(Tfinal – 26)°C
– (75 g)(Tfinal – 85)°C = (155 g) (Tfinal – 26)°C
6375 – 75Tfinal = 155Tfinal – 4030
6375 + 4030 = 155Tfinal + 75Tfinal
10405 = 230.Tfinal
Tfinal = (10405/230.) = 45.24 = 45°C

6.44

–qlost = qgained
– [24.4 mL(1.00 g/mL)](4.184 J/g°C)(23.5 – 35.0)°C = (mass)(4.184 J/g°C)(23.5 – 18.2)°C
– (24.4)(23.5 – 35.0) = (mass)(23.5 – 18.2)
– (24.4)(–11.5) = (mass)(5.3)
280.6 = (mass)(5.3)
52.943 g = mass
 1 mL 
Volume (mL) = 52.943 g 
 = 52.943 = 53 mL
 1.00 g 

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6-14

6.45

Plan: Heat gained by water and the container equals the heat lost by the copper tubing so
qwater + qcalorimeter = –qcopper.
Solution:
T = Tfinal – Tinitial
Specific heat capacity in units of J/g•K has the same value in units of J/g°C since the Celsius and Kelvin unit are
the same size.
–qlost = qgained = qwater + qcalorimeter
– (455 g Cu)(0.387 J/g°C)(Tfinal – 89.5)°C
= (159 g H2O)(4.184 J/g°C)(Tfinal – 22.8)°C + (10.0 J/°C)(Tfinal – 22.8)°C
– (176.085)(Tfinal – 89.5) = (665.256)(Tfinal – 22.8) + (10.0)(Tfinal – 22.8)
15759.6075 – 176.085Tfinal = 665.256Tfinal – 15167.8368 + 10.0Tfinal – 228
15759.6075 + 15167.8368 + 228 = 176.085Tfinal + 665.256Tfinal + 10.0Tfinal
31155.4443 = 851.341Tfinal
Tfinal = 31155.4443/(851.341) = 36.59573 = 36.6°C

6.46

–qlost = qgained = qwater + qcalorimeter
– (30.5 g alloy)(calloy)(31.1 – 93.0)°C = (50.0 g H2O)(4.184 J/g°C)(31.1 – 22.0)°C + (9.2 J/°C)(31.1 – 22.0)°C
– (30.5 g)(calloy)(–61.9°C) = (50.0 g)(4.184 J/g°C)(9.1°C) + (9.2 J/°C)(9.1°C)
1887.95(calloy) = 1903.72 + 83.72 = 1987.44
calloy = 1987.44/1887.95 = 1.052697 = 1.1 J/g°C

6.47

Benzoic acid is C6H5COOH, and will be symbolized as HBz.
–qreaction = qwater + qcalorimeter
 1 mol HBz   3227 kJ   103 J 
4
–qreaction = – 1.221 g HBz 

  1 kJ  = 3.226472x10 J
122.12
g
HBz
1
mol
HBz




qwater = c x mass x T = 4.184 J/g°C x 1200 g x T
qcalorimeter = C x T = 1365 J/°C x T
–qreaction = qwater + qcalorimeter
3.226472x104 J = 4.184 J/g°C x 1200 g x T + 1365 J/°C x T
3.226472x104 J = 5020.8(T) + 1365(T)
3.226472x104 J = 6385.8(T)
T = 3.226472x104/6385.8 = 5.052573 = 5.053°C

6.48

a) Energy will flow from Cu (at 100.0°C) to Fe (at 0.0°C).
b) To determine the final temperature, the heat capacity of the calorimeter must be known.
c) – qCu = qFe + qcalorimeter assume qcalorimeter = 0.
– qCu = qFe + 0
– (20.0 g Cu)(0.387 J/g°C)(Tfinal – 100.0)°C = (30.0 g Fe)(0.450 J/g°C)(Tfinal – 0.0)°C + 0.0 J
– (20.0 g)(0.387 J/g°C)(Tfinal – 100.0°C) = (30.0 g)(0.450 J/g°C)(Tfinal – 0.0°C)
– (7.74)(Tfinal – 100.0) = (13.5)(Tfinal – 0.0)
774 – 7.74 Tfinal = 13.5Tfinal
774 = (13.5 + 7.74) Tfinal = 21.24Tfinal
Tfinal = 774/21.24 = 36.44068 = 36.4°C

6.49

–qhydrocarbon = qwater + qcalorimeter
–qhydrocarbon = (2.550 L H2O)(1mL /10–3L)(1.00g/mL)(4.184 J/g°C)(23.55 – 20.00)°C
+ (403 J/°C)(23.55 – 20.00)°C
–qhydrocarbon = (2550. g)(4.184 J/g°C)(3.55°C) + (403 J/°C)(3.55°C)
–qhydrocarbon = (37875.66 J) + (1430.65 J) = 39306.31 J
qhydrocarbon = –3.930631x104 J
qhydrocarbon/g = (–3.930631x104 J)/1.520 g = –2.5859x104 = –2.59x104 J/g

6.50

The reaction is: 2KOH(aq) + H2SO4(aq)  K2SO4(aq) + 2H2O(l)

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6-15

q (kJ) = (25.0 + 25.0) mL(1.00 g/mL)(4.184 J/g°C)(30.17 – 23.50)°C(1 kJ/103 J) = 1.395364 kJ
(The temperature increased so the heat of reaction is exothermic.)
Amount (moles) of H2SO4 = (25.0 mL)(0.500 mol H2SO4/L)(10–3 L/1 mL) = 0.0125 mol H2SO4
Amount (moles) of KOH = (25.0 mL)(1.00 mol KOH/L)(10–3 L/1 mL) = 0.0250 mol KOH
The moles show that both H2SO4 and KOH are limiting.
The enthalpy change could be calculated in any of the following ways:
H = –1.395364 kJ/0.0125 mol H2SO4 = – 111.62912 = – 112 kJ/mol H2SO4
H = –1.395364 kJ/0.0250 mol KOH = –55.81456 = –55.8 kJ/mol KOH
(Per mole of K2SO4 gives the same value as per mole of H2SO4, and per mole of H2O gives the same
value as per mole of KOH.)
6.51

Reactants  Products + Energy
Hrxn = (–)
Thus, energy is a product.

6.52

Plan: Recall that H is positive for an endothermic reaction in which heat is absorbed, while H is negative for an
exothermic reaction in which heat is released.
Solution:
The reaction has a positive Hrxn, because this reaction requires the input of energy to break the oxygen-oxygen
bond in O2:
O2(g) + energy  2O(g)

6.53

Plan: Recall that H is positive for an endothermic reaction in which heat is absorbed, while H is negative for an
exothermic reaction in which heat is released.
Solution:
As a substance changes from the gaseous state to the liquid state, energy is released so H would be negative for
the condensation of 1 mol of water. The value of H for the vaporization of 2 mol of water would be twice the
value of H for the condensation of 1 mol of water vapor but would have an opposite sign (+H).
H2O(g)  H2O(l) + Energy
2H2O(l) + Energy  2H2O(g)
Hcondensation = (–)
Hvaporization = (+)2[Hcondensation]
The enthalpy for 1 mole of water condensing would be opposite in sign to and one-half the value for the
conversion of 2 moles of liquid H2O to H2O vapor.

6.54

Plan: Recall that H is positive for an endothermic reaction in which heat is absorbed, while H is negative for an
exothermic reaction in which heat is released. The Hrxn is specific for the reaction as written, meaning that
20.2 kJ is released when one-eighth of a mole of sulfur reacts. Use the ratio between moles of sulfur and H to
convert between amount of sulfur and heat released.
Solution:
a) This reaction is exothermic because H is negative.
b) Because H is a state function, the total energy required for the reverse reaction, regardless of how the change
occurs, is the same magnitude but different sign of the forward reaction. Therefore, H = +20.2 kJ.
 20.2 kJ 
c) Hrxn = 2.6 mol S8 
= –420.16 = –4.2x102 kJ
 1/ 8  mol S8 

d) The mass of S8 requires conversion to moles and then a calculation identical to part c) can be performed.
 1 mol S8   20.2 kJ 
Hrxn =  25.0 g S8  
 = –15.7517 = –15.8 kJ
 
 256.48 g S8   1 / 8  mol S8 

6.55

MgCO3(s)  MgO(s) + CO2(g)
a) Absorbed
b) Hrxn (reverse) = –117.3 kJ

Hrxn = 117.3 kJ

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6-16

 117.3 kJ 
c) Hrxn = 5.35 mol CO 2 
 = –627.555 = –628 kJ
 1 mol CO 2 
 1 mol CO 2   117.3 kJ 
d) Hrxn = 35.5 g CO 2 

 = –94.618 = –94.6 kJ
 44.01 g CO 2   1 mol CO 2 

6.56

Plan: A thermochemical equation is a balanced equation that includes the heat of reaction. Since heat is absorbed
in this reaction, H will be positive. Convert the mass of NO to moles and use the ratio between NO and H to
find the heat involved for this amount of NO.
Solution:
a) 1/2N2(g) + 1/2O2(g)  (g)
H = 90.29 kJ
 1 mol NO   90.29 kJ 
b) Hrxn = 3.50 g NO 

 = –10.5303 = –10.5 kJ
 30.01 g NO   1 mol NO 

6.57

Hrxn = 394 kJ
a) KBr(s)  K(s) + 1/2Br2(l)
 103 g   1 mol KBr   394 kJ 
b) Hrxn = 10.0 kg KBr 
= –3.3109x104 = –3.31x104 kJ
 1 kg   119.00 g KBr   1 mol KBr 




6.58

Plan: For the reaction written, 2 moles of H2O2 release 196.1 kJ of energy upon decomposition. Use this
ratio to convert between the given amount of reactant and the amount of heat released. The amount of H2O2 must
be converted from kg to g to moles.
Solution:
2H2O2(l)  2H2O(l) + O2(g)
Hrxn = –196.1 kJ
 103 g   1 mol H 2 O 2   196.1 kJ 
Heat (kJ) = q = 652 kg H 2 O 2 
= –1.87915x106 = –1.88x106 kJ
 1 kg   34.02 g H O   2 mol H O 
2 2 
2 2 



6.59

For the reaction written, 1 mole of B2H6 releases 755.4 kJ of energy upon reaction.
Hrxn = –755.4 kJ
B2H6(g) + 6Cl2(g)  2BCl3(g) + 6HCl(g)
3
 10 g   1 mol B2 H 6   755.4 kJ 
Heat (kJ) = q = 1 kg 
= –2.73003x104 = –2.730x104 kJ/kg
 1 kg   27.67 g B H   1 mol B H 
2 6 
2 6 



6.60

4Fe(s) + 3O2(g)  2Fe2O3(s)

Hrxn = –1.65x103 kJ
 103 g   1 mol Fe   1.65x103 kJ 
a) Heat (kJ) = q =  0.250 kg Fe  
= –1846.46 = –1850 kJ
 1 kg   55.85 g Fe   4 mol Fe 




 2 mol Fe 2 O 3   159.70 g Fe 2 O 3 
b) Mass (g) of Fe2O3 = 4.85x103 kJ 
 = 938.84 = 939 g Fe2O3

3
 1.65x10 kJ   1 mol Fe 2 O 3 

6.61

2HgO(s)  2Hg(l) + O2(g)

Hrxn = 181.6 kJ
 1 mol HgO   181.6 kJ 
a) Heat (kJ) = q =  555 g HgO  

 = 232.659 = 233 kJ
 216.6 g HgO   2 mol Hg 
 2 mol Hg   200.6 g Hg 
b) Mass (g) of Hg =  275 kJ  

 = 607.544 = 608 g Hg
 181.6 kJ   1 mol Hg 

6.62

Plan: A thermochemical equation is a balanced equation that includes the heat of reaction. Heat is released in this
reaction so H is negative. Use the ratio between H and moles of C2H4 to find the amount of C2H4 that must
react to produce the given quantity of heat.
Solution:

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6-17

a) C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l)
Hrxn = –1411 kJ
 1 mol C 2 H 4   28.05 g C 2 H 4 
b) Mass (g) of C2H4 =  70.0 kJ  
 = 1.39157 = 1.39 g C2H4

 1411 kJ   1 mol C 2 H 4 
Hrxn = –5.64x103 kJ
 1 mol C12 H 22 O11   5.64x103 kJ 
b) Heat (kJ) = q = 1 g C12 H 22 O11  
 = –16.47677 = –16.5 kJ/g
 
 342.30 g C12 H 22 O11   1 mol C12 H 22 O11 

6.63

a) C12H22O11(s) + 12O2(g)  12CO2(g) + 11H2O(l)

6.64

Hess’s law: Hrxn is independent of the number of steps or the path of the reaction.

6.65

Hess’s law provides a useful way of calculating energy changes for reactions which are difficult or impossible to
measure directly.

6.66

Plan: Two chemical equations can be written based on the description given:
C(s) + O2(g)  CO2(g)
H1
(1)
CO(g) + 1/2O2(g)  CO2(g)
H2
(2)
The second reaction can be reversed and its H sign changed. In this case, no change in the coefficients is
necessary since the CO2 cancels. Add the two H values together to obtain the H of the desired reaction.
Solution:
C(s) + O2(g)  CO2(g)
H1
CO2(g)  CO(g) + 1/2O2(g)
–H2 (reaction is reversed)
Hrxn = H1 + – (H2)
Total C(s) + 1/2O2(g)  CO(g)
How are the H values for each reaction determined? The H1 can be found by using the heats of formation in
Appendix B:
H1 = [Hf(CO2)] – [Hf(C) + Hf(O2)] = [–393.5 kJ/mol] – [0 + 0] = –393.5 kJ/mol.
The H2 can be found by using the heats of formation in Appendix B:
H2 = [Hf(CO2)] – [Hf(CO) + 1/2Hf(O2)] = [–393.5] – [–110.5 kJ/mol + 0)] = –283 kJ/mol.
Hrxn = H1 + – (H2) = –393.5 kJ + – (–283.0 kJ) = –110.5 kJ

6.67

Plan: To obtain the overall reaction, add the first reaction to the reverse of the second. When the second reaction
is reversed, the sign of its enthalpy change is reversed from positive to negative.
Solution:
Ca(s) + 1/2O2(g)  CaO(s)
H = –635.1 kJ
CaO(s) + CO2(g)  CaCO3(s)
H = –178.3 kJ (reaction is reversed)
Ca(s) + 1/2O2(g) + CO2(g) CaCO3(s)
H = –813.4 kJ

6.68

2NOCl(g)  2NO(g) + Cl2(g)
2NO(g)  N2(g) + O2(g)
2NOCl(g)  N2(g) + O2(g) + Cl2(g)

6.69

Plan: Add the two equations, canceling substances that appear on both sides of the arrow. When matching the
equations with the arrows in the Figure, remember that a positive H corresponds to an arrow pointing up while a
negative H corresponds to an arrow pointing down.
Solution:
1)
N2(g) + O2(g)  2NO(g)
H = 180.6 kJ
2)
2NO(g) + O2(g)  2NO2(g)
H = –114.2 kJ

H = –2(–38.6 kJ)
H = –2(90.3 kJ)
H = 77.2 kJ + (– 180.6 kJ) = –103.4 kJ

3)
N2(g) + 2O2(g)  2NO2(g)
Hrxn = +66.4 kJ
In Figure P6.67, A represents reaction 1 with a larger amount of energy absorbed, B represents reaction 2 with
a smaller amount of energy released, and C represents reaction 3 as the sum of A and B.
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6-18

6.70

1)
2)

P4(s) + 6Cl2(g)  4PCl3(g)
4PCl3(g) + 4Cl2(g)  4PCl5(g)

H1 = –1148 kJ
H2 = – 460 kJ

P4(s) + 10Cl2(g)  4PCl5(g)
3)
Equation 1) = B, equation 2) = C, equation 3) = A

6.71

Plan: Vaporization is the change in state from a liquid to a gas: H2O(l)  H2O(g) . The two equations describing
the chemical reactions for the formation of gaseous and liquid water can be combined to yield the equation for
vaporization.
Solution:
1) Formation of H2O(g):
H2(g) + 1/2O2(g)  H2O(g)
H = –241.8 kJ
2) Formation of H2O(l):
H2(g) + 1/2O2(g)  H2O(l)
H = –285.8 kJ
Reverse reaction 2 (change the sign of H) and add the two reactions:
H2(g) + 1/2O2(g)  H2O(g)
H = –241.8 kJ
H2O(l)  H2(g) + 1/2O2(g)
H = +285.8 kJ
H2O(l)  H2O(g)

6.72

Hoverall = –1608 kJ

Hvap = 44.0 kJ

C(s) + 1/4S8(s)  CS2(l)
CS2(l)  CS2(g)

H = +89.7 kJ
H = +27.7 kJ

C(s) + 1/4S8(s)  CS2(g)

H = +117.4 kJ

6.73

C (diamond) + O2(g)  CO2(g)
CO2(g)  C(graphite) + O2(g)
C(diamond)  C(graphite)

H = –395.4 kJ
H = –(–393.5 kJ)
H = – 1.9 kJ

6.74

, is the enthalpy change for any reaction where all substances are in their
The standard heat of reaction, H rxn

standard states. The standard heat of formation, H f , is the enthalpy change that accompanies the formation of
one mole of a compound in its standard state from elements in their standard states. Standard state is 1 atm for
gases, 1 M for solutes, and the most stable form for liquids and solids. Standard state does not include a specific
temperature, but a temperature must be specified in a table of standard values.
6.75

The standard heat of reaction is the sum of the standard heats of formation of the products minus the sum of the
standard heats of formation of the reactants multiplied by their respective stoichiometric coefficients.

H rxn
= m H f (products) – n H f (reactants)

6.76

Plan: H f is for the reaction that shows the formation of one mole of compound from its elements in their
standard states.
Solution:
a) 1/2Cl2(g) + Na(s)  NaCl(s) The element chlorine occurs as Cl2, not Cl.
b) H2(g) + 1/2O2(g)  H2O(g) The element hydrogen exists as H2, not H, and the formation of water is written
with water as the product.
c) No changes

6.77

Plan: Formation equations show the formation of one mole of compound from its elements. The elements must be
in their most stable states ( H f = 0).
Solution:
a) Ca(s) + Cl2(g)  CaCl2(s)
b) Na(s) + 1/2H2(g) + C(graphite) + 3/2O2(g)  NaHCO3(s)

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6-19

c) C(graphite) + 2Cl2(g)  CCl4(l)
d) 1/2H2(g) + 1/2N2(g) + 3/2O2(g)  HNO3(l)
6.78

a) 1/2H2(g) + 1/2I2(s)  HI(g)
b) Si(s) + 2F2(g)  SiF4(g)
c) 3/2O2(g)  O3(g)
d) 3Ca(s) + 1/2P4(s) + 4O2(g)  Ca3(PO4)2(s)

6.79

Plan: The enthalpy change of a reaction is the sum of the heats of formation of the products minus the sum of the
heats of formation of the reactants. Since the H f values (Appendix B) are reported as energy per one mole, use
the appropriate stoichiometric coefficient to reflect the higher number of moles.
Solution:

H rxn
= m H f (products) – n H f (reactants)

a) H rxn
= {2 H f [SO2(g)] + 2 H f [H2O(g)]} – {2 H f [H2S(g)] + 3 H f [O2(g)]}
= [(2 mol)(–296.8 kJ/mol) + (2 mol)(–241.826 kJ/mol)] – [(2 mol)(–20.2 kJ/mol) + (3 mol)(0.0 kJ/mol)]
= –1036.9 kJ
b) The balanced equation is CH4(g) + 4Cl2(g)  CCl4(l) + 4HCl(g)

H rxn
= {1 H f [CCl4(l)] + 4 H f [HCl(g)]} – {1 H f [CH4(g)] + 4 H f [Cl2(g)]}

H rxn
= [(1 mol)(–139 kJ/mol) + (4 mol)(–92.31 kJ/mol)] – [(1 mol)(–74.87 kJ/mol) + (4 mol)(0 kJ/mol)]
= –433 kJ

6.80

H rxn
= m H f (products) – n H f (reactants)

a) H rxn
= {1 H f [SiF4(g)] + 2 H f [H2O(l)]} – {1 H f [SiO2(s)] + 4 H f [HF(g)]}
= [(1 mol)(–1614.9 kJ/mol) + (2 mol)(–285.840 kJ/mol)]
– [(1 mol)(–910.9 kJ/mol) + (4 mol)(–273 kJ/mol)]
= –184 kJ
b) 2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(g)

H rxn
= {4 H f [CO2(g)] + 6 H f [H2O(g)]} – {2 H f [C2H6(g)] + 7 H f [ O2(g)]}
= [(4 mol)(–393.5 kJ/mol) + (6 mol)(–241.826 kJ/mol)] – [(2 mol)(–84.667 kJ/mol) + (7 mol)(0 kJ/mol)]
= –2855.6 kJ (or –1427.8 kJ for reaction of 1 mol of C2H6)

6.81

Plan: The enthalpy change of a reaction is the sum of the heats of formation of the products minus the sum of the
heats of formation of the reactants. Since the H f values (Appendix B) are reported as energy per one mole, use

the appropriate stoichiometric coefficient to reflect the higher number of moles. In this case, H rxn
is known and

H f of CuO must be calculated.
Solution:

H rxn
= m H f (products) – n H f (reactants)

Cu2O(s) + 1/2O2(g)  2CuO(s)

H rxn
= –146.0 kJ

H rxn
= {2 H f [CuO(s)]} – {1 H f [Cu2O(s)] + 1/2 H f [O2(g)]}

–146.0 kJ = {(2 mol) H f [CuO(s)]} – {(1 mol)(–168.6 kJ/mol) + (1/2 mol)(0 kJ/mol)}
–146.0 kJ = 2 mol H f [CuO(s)] + 168.6 kJ
H f [CuO(s)] = 

314.6 kJ
= –157.3 kJ/mol
2 mol

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6-20

6.82

H rxn
= m H f (products) – n H f (reactants)

H rxn
= –1255.8 kJ

C2H2(g) + 5/2O2(g)  2CO2(g) + H2O(g)

H rxn
= {2 H f [CO2(g)] + 1 H f [H2O(g)]} – {1 H f [C2H2(g)] + 5/2 H f [O2(g)]}
–1255.8 kJ = {(2 mol)(–393.5 kJ/mol) + (1 mol)(–241.826 kJ/mol)}

– {(1 mol) H f [C2H2(g)] + (5/2 mol)(0.0 kJ/mol)}
–1255.8 kJ = –787.0 kJ – 241.8 kJ – (1 mol) H f [C2H2(g)]
H f [C2H2(g)] =

6.83

227.0 kJ
= 227.0 kJ/mol
 1 mol

Plan: The enthalpy change of a reaction is the sum of the heats of formation of the products minus the sum of
the heats of formation of the reactants. Since the H f values (Appendix B) are reported as energy per one mole,
use the appropriate stoichiometric coefficient to reflect the higher number of moles. Hess’s law can also be used
to calculate the enthalpy of reaction. In part b), rearrange equations 1) and 2) to give the equation wanted.

) and multiply the coefficients (and H rxn
) of the second
Reverse the first equation (changing the sign of H rxn
reaction by 2.
Solution:
2PbSO4(s) + 2H2O(l)  Pb(s) + PbO2(s) + 2H2SO4(l)

H rxn
= m H f (products) – n H f (reactants)

a) H rxn
= {1 H f [Pb(s)] + 1 H f [PbO2(s)] + 2 H f [H2SO4(l)]}

– {2 H f [PbSO4(s)] + 2 H f [H2O(l)]}
= [(1 mol)(0 kJ/mol) + (1 mol)(–276.6 kJmol) + (2 mol)(–813.989 kJ/mol)]
– [(2 mol)(–918.39 kJ/mol) + (2 mol)(–285.840 kJ/mol)]
= 503.9 kJ
b) Use Hess’s law:
PbSO4(s)  Pb(s) + PbO2(s) + 2SO3(g)

6.84

H rxn
= –(–768 kJ) Equation has been reversed.

2SO3(g) + 2H2O (l)  2H2SO4(l)

H rxn
= 2(–132 kJ)

2PbSO4(s) + 2H2O(l)  Pb(s) + PbO2(s) + 2H2SO4(l)

H rxn
= 504 kJ

Plan: The enthalpy change of a reaction is the sum of the heats of formation of the products minus the sum of the
heats of formation of the reactants. Since the H f values (Appendix B) are reported as energy per one mole, use
the appropriate stoichiometric coefficient to reflect the higher number of moles. Convert the mass of stearic acid

to moles and use the ratio between stearic acid and H rxn
to find the heat involved for this amount of acid. For
part d), use the kcal/g of fat relationship calculated in part c) to convert 11.0 g of fat to total kcal and compare to
the 100. Cal amount.
Solution:
a) C18H36O2(s) + 26O2(g)  18CO2(g) + 18H2O(g)

= m H f (products) – n H f (reactants)
b) H rxn

H rxn
= {18 H f [CO2(g)] + 18 H f [H2O(g)]} – {1 H f [C18H36O2(s)] + 26 H f [O2(g)]}
= [(18 mol)(–393.5 kJ/mol) + (18 mol)(–241.826 kJ/mol)] – [(1 mol)(–948 kJ/mol) + (26 mol)(0 kJ/mol)]
= –10,487.868 = –10,488 kJ
 1 mol C18 H36 O2  10, 487.868 kJ 
c) q (kJ) = 1.00 g C18 H36 O2  

 = –36.8681 = –36.9 kJ
 284.47 C18 H36 O2  1 mol C18 H36 O2 

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6-21

 1 kcal 
q (kcal) =  36.8681 kJ  
 = –8.811688 = –8.81 kcal
 4.184 kJ 

 8.811688 kcal 
d) q (kcal) = 11.0 g fat  
 = 96.9286 = 96.9 kcal
1.0 g fat

Since 1 kcal = 1 Cal, 96.9 kcal = 96.9 Cal. The calculated calorie content is consistent with the package
information.

6.85

a) H2SO4(l)  H2SO4(aq)

H rxn
= {1 H f [H2SO4(aq)]} – {1 H f [H2SO4(l)]}
= [(1 mol)(–907.51 kJ/mol)] – [(1 mol)(–813.989 kJ/mol)]
= –93.52 kJ
b) q (J) = c x mass x T
 3.50 J 
 1.060 g 
93.52 kJ x 103 J/kJ = 
 x 1000. mL  
 x Tfinal  25.0C 
g•

C
 1 mL 

 3.50 J 
9.352x104 J = 
 x 1060. g  x Tfinal  25.0C 
 g•C 
9.352 x 104 J = (Tfinal)3710 J/ºC – 9.2750x104 J
Tfinal = 50.1995 = 50.2°C
c) Adding the acid to a large amount of water releases the heat to a large mass of solution and thus, the potential
temperature rise is minimized due to the large heat capacity of the larger volume.
6.86

Plan: Use the ideal gas law, PV = nRT, to calculate the volume of one mole of helium at each temperature.
Then use the given equation for ΔE to find the change in internal energy. The equation for work, w = –PΔV, is
needed for part c), and qP = ΔE + PΔV is used for part d). For part e), recall that ΔH = qP.
Solution:
nRT
a) PV = nRT or V =
P
T = 273 + 15 = 288 K
and
T = 273 + 30 = 303 K
L•atm 

 0.0821 mol•K   288 K 
nRT

= 
= 23.6448 = 23.6 L/mol
Initial volume (L) = V =
P
1.00 atm 
L•atm 

 0.0821 mol • K   303 K 
nRT

Final volume (L) = V =
= 
= 24.8763 = 24.9 L/mol
P
1.00 atm 
b) Internal energy is the sum of the potential and kinetic energies of each He atom in the system (the balloon). The
energy of one mole of helium atoms can be described as a function of temperature, E = 3/2nRT, where n = 1 mole.
Therefore, the internal energy at 15°C and 30°C can be calculated. The inside back cover lists values of R with
different units.
E = 3/2nRT = (3/2)(1.00 mol) (8.314 J/mol•K)(303 – 288)K = 187.065 = 187 J
c) When the balloon expands as temperature rises, the balloon performs PV work. However, the problem specifies
that pressure remains constant, so work done on the surroundings by the balloon is defined by the equation:
w = –PV. When pressure and volume are multiplied together, the unit is L•atm, so a conversion factor is needed
to convert work in units of L•atm to joules.
 101.3 J 
2
w = –PV =  1.00 atm  (24.8763  23.6448) L  
 = –124.75 = –1.2x10 J
1
L•atm

d) qP = E + PV = (187.065 J) + (124.75 J) = 311.815 = 3.1x102 J
e) H = qP = 310 J.

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6-22

f) When a process occurs at constant pressure, the change in heat energy of the system can be described by a state
function called enthalpy. The change in enthalpy equals the heat (q) lost at constant pressure: H = E + PV =
E – w = (q + w) – w = qP
6.87

a) Respiration:
C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(g)

H rxn
= m H f (products) – n H f (reactants)

= {6 H f [CO2(g)] + 6 H f [H2O(g)]} – {1 H f [C6H12O6(s)] + 6 H f [O2(g)]}
= [(6 mol)(–393.5 kJ/mol) + (6 mol)(–241.826 kJ/mol)] – [(1 mol)(–1273.3 kJ/mol) + (6 mol)(0.0 kJ/mol)]
= –2538.656 = – 2538.7 kJ
Fermentation:
C6H12O6(s)  2CO2(g) + 2CH3CH2OH(l)

H rxn
= {2 H f [CO2(g)] + 2 H f [CH3CH2OH(l)]} – [1 H f [C6H12O6(s)]}
= [(2 mol)(–393.5 kJ/mol) + (2 mol)(–277.63 kJ/mol)] – [(1 mol)(–1273.3 kJ/mol)] = –68.96 = – 69.0 kJ
b) Combustion of ethanol:
CH3CH2OH(l) + 3O2(g)  2CO2(g) + 3H2O(g)

H rxn
= {2 H f [CO2(g)] + 3 H f [H2O(g)]} – {1 H f [CH3CH2OH(l)] + 3 H f [O2(g)]}

H rxn
= [(2 mol)(–393.5 kJ/mol) + (3 mol)(–241.826 kJ/mol)] – [(1 mol)(–277.63 kJ/mol) + (3 mol)(0.0 kJ/mol)]
= –1234.848 = – 1234.8 kJ
Heats of combustion/mol C:
 2538.656 kJ   1 mol C6 H12 O 6 
Sugar: 

 = –423.1093 = –423.11 kJ/mol C
6 mol C

 1 mol C6 H12 O 6  

 1234.848 kJ   1 mol CH 3 CH 2 OH 
Ethanol: 

 = –617.424 = –617.42 kJ/mol C
2 mol C

 1 mol CH 3 CH 2 OH  
Ethanol has a higher value.

6.88

a) Reactions:
1) C21H44(s) + 32O2(g)  21CO2(g) + 22H2O(g)
2) C21H44(s) + 43/2O2(g)  21CO(g) + 22H2O(g)
3) C21H44(s) + 11O2(g)  21C(s) + 22H2O(g)
Heats of combustion:

= {21 H f [CO2(g)] + 22 H f [H2O(g)]} – {[1 H f [C21H44(s)] + 32 H f [O2(g)]}
1) H rxn
= [(21 mol)(–393.5 kJ/mol) + (22 mol)(–241.826 kJ/mol)]
– [(1 mol)(–476 kJ/mol) + (32 mol)(0.0 kJ/mol)]
= –13,107.672 = –13,108 kJ

2) H rxn
= {21 H f [CO(g)] + 22 H f [H2O(g)]} – {1 H f [C21H44(s)] + 43/2 H f [O2(g)]}
= [(21 mol)(–110.5 kJ/mol) + (22 mol)(–241.826 kJ/mol)]
– [(1 mol)(–476 kJ/mol) + (43/2 mol)(0.0 kJ/mol)] = –7164.672 = –7165 kJ

3) H rxn
= {21 H f [C(s)] + 22 H f [H2O(g)]} – {1 H f [C21H44(s)] + 11 H f [O2(g)]}
= [(21 mol)(0.0 kJ/mol) +(22 mol)(–241.826 kJ/mol)] – [(1 mol)(–476 kJ/mol) + (11 mol)(0.0 kJ/mol)]
= –4844.172 = –4844 kJ
 1 mol C 21H 44   13107.672 kJ 
4
4
b) q (kJ) =  254 g C 21H 44  

 = –1.12266x10 = –1.12x10 kJ
296.56
g
C
H
1
mol
C
H
21 44  
21 44 

c) The moles of C21H44 need to be calculated one time for multiple usage. It must be assumed that the remaining
87.00% of the candle undergoes complete combustion.
Moles C21H44 = (254 g C21H44)(1 mol C21H44/296.56 g C21H44) = 0.856488 mol
q = (0.87)(0.856488 mol)(–13107.672 kJ/mol) + (0.0800)(0.856488 mol)(–7164.672 kJ/mol)
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6-23

+ (0.0500)(0.856488 mol)(–4844.172 kJ/mol)] = –1.04655x104 = –1.05x104 kJ
6.89

H vap
= 569.4 J/g(44.05 g/mol)(1 kJ/1000 J) = 25.08 kJ/mol

a) EO(l) → EO(g)

H vap
= {1 H f [EO(g)]} – {1 H f [EO(l)]}

25.08 kJ/mol = { H f [EO(g)]} – [(1 mol)(–77.4 kJ/mol)]
H f [EO(g)] = –52.32 kJ/mol
EO(g) → CH4(g) + CO(g)

H rxn
= {1 H f [CH4(g)] + 1 H f [CO(g)]} – {1 H f [EO(g)]}

H rxn
= [(1 mol)(–74.87 kJ/mol) + (1 mol)(–110.5 kJ/mol)] – [(1 mol)(–52.32 kJ/mol)]

H rxn
= –133.0 kJ/mol
b) Assume that you have 1.00 mole of EO(g). 1.00 mole of EO(g) produces 1.00 mole or 16.04 g of CH4(g) and
1.00 mole or 28.01 g of CO(g). There is a total product mass of 16.04 g + 28.01 g = 44.05 g.
q = c x mass x T
 1000 J 
133.0 kJ  

q
 1 kJ 
ΔT =
=
c x mass
 2.5 J/gC  44.05 g 

ΔT = 1207.7ºC
ΔT = Tfinal – Tinitial
1207.7ºC = Tfinal – 93ºC
Tfinal = 1300.72 = 1301°C
6.90

a) 3N2O5(g) + 3NO(g) → 9NO2(g)

H rxn
= {9 H f [NO2(g)]} – {3 H f [N2O5(g)] + 3 H f [NO(g)]}
= [(9 mol)(33.2 kJ/mol)] – [(3 mol)(11 kJ/mol) + (3 mol)(90.29 kJ/mol)]
 = –5.07 = –5 kJ
 1.50x102 mol  
  103 J 
5.07 kJ
b)  9 molecules product  
= –76.05 = –76.0 J

 1 molecule product  9 moles product   1 kJ 




6.91

 4 qt   1 L   1 mL   0.692 g   1 mol C8 H18   5.44x103 kJ 
a) Heat (kJ) =  20.4 gal  


  3  
 

 1 gal   1.057 qt   10 L   mL   114.22 g   1 mol C8 H18 
= –2.54435678x106 = –2.54x106 kJ
1h

  65 mi   1 km 
3
3
b) Distance (km) = 2.54435678x106 kJ 


 = 4.84995x10 = 4.8x10 km
4
 5.5x10 kJ   1 h   0.62 mi 
c) Only a small percentage of the chemical energy in the fuel is converted to work to move the car; most of the
chemical energy is lost as waste heat flowing into the surroundings.

6.92

q = c x mass x T
In this situation, all of the samples have the same mass, 50. g, so mass is not a variable.
All also have the same q value, 450. J. So, 450. J α (c x ΔT). c, specific heat capacity, and ΔT are inversely
proportional. The higher the ΔT, the lower the value of specific heat capacity:
ΔT: B > D > C > A
Specific heat capacity: B < D < C < A

6.93

ClF(g) + 1/2O2(g)  1/2Cl2O(g) + 1/2OF2(g)

H rxn
= 1/2(167.5 kJ) =

F2(g) + 1/2O2(g)  OF2(g)

H rxn

83.75 kJ

= 1/2(–43.5 kJ) = –21.75 kJ

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6-24

6.94

1/2Cl2O(g) + 3/2OF2(g)  ClF3(l) + O2(g)

H rxn
= –1/2(394.1 kJ) = –197.05 kJ

ClF(g) + F2(g)  ClF3(l)

H rxn
=

–135.1 kJ

a) AgNO3(aq) + NaI(aq)  AgI(s) + NaNO3(aq)
 103 L   5.0 g AgNO3   1 mol AgNO3 
Moles of AgNO3 =  50.0 mL  


 1 mL  
1L
  169.9 g AgNO3 


= 1.47145x10–3 mol AgNO3
 10 3 L   5.0 g NaI   1 mol NaI 
Moles of NaI =  50.0 mL  


 1 mL   1 L
  149.9 g NaI 


= 1.6677785x10–3 mol NaI
The AgNO3 is limiting, and will be used to finish the problem:
 1 mol AgI   234.8 g AgI 
Mass (g) of AgI = 1.47145x103 mol AgNO 3 


 1 mol AgNO 3   1 mol AgI 

= 0.345496 = 0.35 g AgI
b) Ag+(aq) + I–(aq)  AgI(s)

H rxn
= {1 H f [AgI(s)]} – {1 H f [Ag+(aq)] + 1 H f [I–(aq)]}
= [(1 mol)(–62.38 kJ/mol)] – [(1 mol)(105.9 kJ/mol) + (1 mol)(–55.94 kJ/mol)]
= –112.3 kJ

c) H rxn
= q = c x mass x T

 112.3 kJ   1 mol AgI 
3

 1.47145 x10 mol AgNO3

mol
AgI
1
mol
AgNO

3

/c x mass = 
T = H rxn
 4.184 J  
 1.00 g  
 g•K   50.0  50.0  mL  mL  




= 0.39494 = 0.39 K

6.95

  10 J 

3



 1 kJ 

Plan: Use conversion factors to solve parts a) and b). For part c), first find the heat of reaction for the combustion
of methane by using the heats of formation of the reactants and products. The enthalpy change of a reaction is the
sum of the heats of formation of the products minus the sum of the heats of formation of the reactants. Since the
H fo values (Appendix B) are reported as energy per one mole, use the appropriate stoichiometric coefficient to
reflect the higher number of moles. For part e), convert the amount of water in gal to mass in g and use the
relationship q = c x mass x T to find the heat needed; then use the conversion factors between joules and therms
and the cost per therm to determine the total cost of heating the water.
Solution:
 1 cal   4.184 J   453.6 g   1.0 C   1.00 lb F 
3
a) 




 = 1054.368 = 1.1x10 J/Btu
 g C   1 cal   1 lb   1.8 F   1 Btu 
 100, 000 Btu   1054.368 J 
8
8
b) E = 1.00 therm  

 = 1.054368x10 = 1.1x10 J
Btu

 1 therm  
c) CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

H rxn
= {1 H f [CO2(g)] + 2 H f [H2O(g)]} – {1 H f [CH4(g)] + 2 H f [O2(g)]}
= [(1 mol)(–393.5 kJ/mol) + (2 mol)(–241.826 kJ/mol)] – [(1 mol)(–74.87 kJ/mol) + (2 mol)(0.0 kJ/mol)]
= –802.282 = –802.3 kJ/mol CH4
 1.054368x108 J   1 kJ   1 mol CH 4 
Moles of CH4 = 1.00 therm  

  103 J   802.282 kJ 
1 therm





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6-25

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