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)

103 L 0.500 mol Ba(OH) 2

Moles of Ba(OH)2 = 50.0 mL

= 0.0250 mol Ba(OH)2

1 mL

1L

103 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

CHEMICAL CONNECTIONS BOXED READING PROBLEMS

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 + PV (constant P)

a) H < E, PV is negative.

b) H = E, a fixed volume means PV = 0.

c) H > E, PV 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 + 55C

=

= 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) + 1/2Hf(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 25.0C

g•

C

1 mL

3.50 J

9.352x104 J =

x 1060. g x Tfinal 25.0C

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 = –PV. 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 = –PV = 1.00 atm (24.8763 23.6448) L

= –124.75 = –1.2x10 J

1

L•atm

d) qP = E + PV = (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 + PV =

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/gC 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

1.50x102 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)

103 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 = 1.47145x103 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

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)

103 L 0.500 mol Ba(OH) 2

Moles of Ba(OH)2 = 50.0 mL

= 0.0250 mol Ba(OH)2

1 mL

1L

103 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

CHEMICAL CONNECTIONS BOXED READING PROBLEMS

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 + PV (constant P)

a) H < E, PV is negative.

b) H = E, a fixed volume means PV = 0.

c) H > E, PV 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 + 55C

=

= 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) + 1/2Hf(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 25.0C

g•

C

1 mL

3.50 J

9.352x104 J =

x 1060. g x Tfinal 25.0C

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 = –PV. 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 = –PV = 1.00 atm (24.8763 23.6448) L

= –124.75 = –1.2x10 J

1

L•atm

d) qP = E + PV = (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 + PV =

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/gC 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

1.50x102 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)

103 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 = 1.47145x103 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

Copyright © McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution

in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

6-25

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