chapter six thermochemistry

176

CHAPTER SIX

THERMOCHEMISTRY

For Review

1. Potential energy: energy due to position or composition

Kinetic energy: energy due to motion of an object

Path-dependent function: a property that depends on how the system gets from the initial

state to the final state; a property that is path-dependent

State function: a property that is independent of the pathway

System: that part of the universe on which attention is to be focused

Surroundings: everything in the universe surrounding a thermodynamic system

2. Plot a represents an exothermic reaction. In an exothermic process, the bonds in the product

molecules are stronger (on average) than those in the reactant molecules. The net result is that

the quantity of energy Δ(PE) is transferred to the surroundings as heat when reactants are

converted to products.

For an endothermic process, energy flows into the system from the surroundings as heat to

increase the potential energy of the system. In an endothermic process, the products have

higher potential energy (weaker bonds on average) than the reactants.

3. First law of thermodynamics: the energy of universe is constant. A system can change its

internal energy by flow of work, heat, or both (E = q + w). Whenever a property is added to

the system from the surroundings, the sign is positive; whenever a property is added to the

surroundings by the system, the sign is negative.

4. As a gas expands, the system does work on the surroundings so w is negative. When a gas

contracts, the surroundings do work on the system so w is positive. H2O(l) → H2O(g); To

boil water, heat must be added so q is positive. The molar volume of a gas is huge compared

to the molar volume of a liquid. As a liquid converts to a gas, the system will expand its

volume, performing work on the surroundings; w is negative.

5. qP = ΔH; qV = ΔE; A coffee-cup calorimeter is at constant (atmospheric) pressure. The heat

released or gained at constant pressure is ΔH. A bomb calorimeter is at constant volume. The

heat released or gained at constant volume is ΔE.

CHAPTER 6 THERMOCHEMISTRY 177

6. The specific heat capacities are: 0.89 J/°Cg (Al) and 0.45 J/°Cg (Fe)

Al would be the better choice. It has a higher heat capacity and a lower density than Fe.

Using Al, the same amount of heat could be dissipated by a smaller mass, keeping the mass

of the amplifier down.

7. In calorimetry, heat flow is determined into or out of the surroundings. Because ΔEuniv = 0 by

the first law of thermodynamics, ΔEsys = ΔEsurr; what happens to the surroundings is the

exact opposite of what happens to the system. To determine heat flow, we need to know the

heat capacity of the surroundings, the mass of the surroundings that accepts/donates the heat,

and the change in temperature. If we know these quantities, qsurr can be calculated and then

equated to qsys (qsurr = qsys). For an endothermic reaction, the surroundings (the calorimeter

contents) donates heat to the system. This is accompanied by a decrease in temperature of the

surroundings. For an exothermic reaction, the system donates heat to the surroundings (the

calorimeter) so temperature increases.

8. Hess’s law: in going from a particular set of products, the change in enthalpy is the same

whether the reaction takes place in one step or in a series of steps (ΔH is path independent).

When a reaction is reversed, the sign of ΔH is also reversed but the magnitude is the same. If

the coefficients in a balanced reaction are multiplied by a number, the value of ΔH is

multiplied by the same number while the sign is unaffected.

9. Standard enthalpy of formation: the change in enthalpy that accompanies the formation of

one mole of a compound from its elements with all substances in their standard states. The

standard state for a compound has the following conventions:

a. gaseous substances are at a pressure of exactly 1 atm.

b. for a pure substance in a condensed state (liquid or solid), the standard state is the pure

liquid or solid.

c. for a substance present in solution, the standard state is a concentration of exactly 1 M.

The standard state of an element is the form in which the element exists under conditions of

1 atm and 25C. ofHΔ values for elements in their standard state are, by definition, equal to

zero.

Step 1: reactants → elements in standard states )reactants(HΔnHΔ ofr1 −=

Step 2: elements in standard state → products )products(HΔnHΔ ofp2 =

_________________________________________________________________________________________________

reactants → products 21oreaction HΔHΔHΔ +=

oreactionHΔ )reactants(HΔn)products(HΔn o

frofp −=

10. Three problems are: there is only a finite amount of fossil fuels, fossil fuels can be expensive,

and the combustion and exploration of fossil fuels can add pollution to the biosphere whose

effects may not be reversible. Some alternative fuels are syngas from coal, hydrogen from the

breakdown of water, and ethanol from the fermentation of sugar.

178 CHAPTER 6 THERMOCHEMISTRY

Questions

9. Path-dependent functions for a trip from Chicago to Denver are those quantities that depend

on the route taken. One can fly directly from Chicago to Denver or one could fly from

Chicago to Atlanta to Los Angeles and then to Denver. Some path-dependent quantities are

miles traveled, fuel consumption of the airplane, time traveling, airplane snacks eaten, etc.

State functions are path independent; they only depend on the initial and final states. Some

state functions for an airplane trip from Chicago to Denver would be longitude change,

latitude change, elevation change, and overall time zone change.

10. Products have a lower potential energy than reactants when the bonds in the products are

stronger (on average) than in the reactants. This occurs generally in exothermic processes.

Products have a higher potential energy than reactants when the reactants have the stronger

bonds (on average). This is typified by endothermic reactions.

11. 2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g); All combustion reactions are exothermic;

they all release heat to the surroundings so q is negative. To determine the sign of w, con-

centrate on the moles of gaseous reactants versus the moles of gaseous products. In this

combustion reaction, we go from 25 moles of reactant gas molecules to 16 + 18 = 34 moles

of product gas molecules. As reactants are converted to products, an expansion will occur.

When a gas expands, the system does work on the surroundings and w is negative.

12. H = E + PV at constant P; From the strict definition of enthalpy, the difference between

H and E is the quantity PV. Thus, when a system at constant P can do pressure-volume

work, then H ≠ E. When the system cannot do PV work, then H = E at constant

pressure. An important way to differentiate H from E is to concentrate on q, the heat flow;

the heat flow by a system at constant pressure equals H and the heat flow by a system at

constant volume equals E.

13. CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) H = 891 kJ

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) H = 803 kJ

H2O(l) + 1/2 CO2(g) → 1/2 CH4(g) + O2(g) H1 = 1/2(891 kJ)

1/2 CH4(g) + 2 O2(g) → 1/2 CO2(g) + H2O(g) H2 = 1/2(803 kJ) ______________________________________________________________________________________

H2O(l) → H2O(g) H = H1 + H2 = 44 kJ

The enthalpy of vaporization of water is 44 kJ/mol.

14. The zero point for ofHΔ values are elements in their standard state. All substances are meas-

ured in relationship to this zero point.

15. Fossil fuels contain carbon; the incomplete combustion of fossil fuels produces CO(g) instead

of CO2(g). This occurs when the amount of oxygen reacting is not sufficient to convert all the

carbon to CO2. Carbon monoxide is a poisonous gas to humans.

16. Advantages: H2 burns cleanly (less pollution) and gives a lot of energy per gram of fuel.

Disadvantages: Expensive and gas storage and safety issues


Exercises

Potential and Kinetic Energy

17. KE = 2

1mv2; Convert mass and velocity to SI units. 1 J =

2

2

s

mkg1

Mass = 5.25 oz × lb205.2

kg1

oz16

lb1 = 0.149 kg

Velocity = s

m45

yd094.1

m1

mi

yd1760

s60

min1

min60

hr1

hr

mi100.1 2

=

KE = 2

1 mv2 =

2

1× 0.149 kg ×

2

s

m45

= 150 J

18. KE = 2

1 mv2 =

2

1 ×

25

5

cm100

m1

sec

cm100.2

g1000

kg1g100.1

− = 2.0 × 210− J

19. KE = 2

1 mv2 =

2

1× 2.0 kg ×

2

s

m0.1

= 1.0 J; KE =

2

1 mv2 =

2

1× 1.0 kg ×

2

s

m0.2

= 2.0 J

The 1.0 kg object with a velocity of 2.0 m/s has the greater kinetic energy.

20. Ball A: PE = mgz = 2.00 kg × 2s

m81.9× 10.0 m =

2

2

s

mkg196 = 196 J

At Point I: All of this energy is transferred to Ball B. All of B's energy is kinetic energy at

this point. Etotal = KE = 196 J. At point II, the sum of the total energy will equal

196 J.

At Point II: PE = mgz = 4.00 kg × 2s

m81.9× 3.00 m = 118 J

KE = Etotal PE = 196 J 118 J = 78 J

Heat and Work

21 a. ΔE = q + w = 47 kJ + 88 kJ = 41 kJ

b. ΔE = 82 47 = 35 kJ c. ΔE = 47 + 0 = 47 kJ

d. When the surroundings deliver work to the system, w > 0. This is the case for a.


22. Step 1: ΔE1 = q + w = 72 J + 35 J = 107 J; Step 2: ΔE2 = 35 J 72 J = 37 J

ΔEoverall = ΔE1 + ΔE2 = 107 J 37 J = 70. J

23. ΔE = q + w; Work is done by the system on the surroundings in a gas expansion; w is

negative.

300. J = q 75 J, q = 375 J of heat transferred to the system

24. a. ΔE = q + w = 23 J + 100. J = 77 J

b. w = P ΔV = 1.90 atm (2.80 L 8.30 L) = 10.5 L atm × atmL

J3.101 = 1060 J

ΔE = q + w = 350. J + 1060 = 1410 J

c. w = P ΔV = 1.00 atm (29.1 L 11.2 L) = 17.9 L atm × atmL

J3.101 = 1810 J

ΔE = q + w = 1037 J 1810 J = 770 J

25. w = PΔV; We need the final volume of the gas. Since T and n are constant, P1V1 = P2V2.

atm00.2

)atm0.15(L0.10

P

PVV

2

11

2 == = 75.0 L

w = PΔV = 2.00 atm (75.0 L 10.0 L) = 130. L atm × J1000

kJ1

atmL

J3.101

= 13.2 kJ = work

26. w = 210. J = PΔV, 210 J = P (25 L 10. L), P = 14 atm

27. In this problem q = w = 950. J

950. J × J3.101

atmL1= 9.38 L atm of work done by the gases.

w = PΔV, 9.38 L atm = 760

.650−atm × (Vf 0.040 L), Vf 0.040 = 11.0 L, Vf = 11.0 L

28. ΔE = q + w, 102.5 J = 52.5 J + w, w = 155.0 J × J3.101

atmL1 = 1.530 L atm

w = PΔV, 1.530 L atm = 0.500 atm × ΔV, ΔV = 3.06 L

ΔV = Vf – Vi, 3.06 L = 58.0 L Vi, Vi = 54.9 L = initial volume


29. q = molar heat capacity × mol × ΔT = molC

J8.20o

× 39.1 mol × (38.0 0.0) °C = 30,900 J

= 30.9 kJ

w = PΔV = 1.00 atm × (998 L 876 L) = 122 L atm × atmL

J3.101= 12,400 J = 12.4 kJ

ΔE = q + w = 30.9 kJ + (12.4 kJ) = 18.5 kJ

30. H2O(g) → H2O(l); ΔE = q + w; q = 40.66 kJ; w = PΔV

Volume of 1 mol H2O(l) = 1 mol H2O(l) × g996.0

cm1

mol

g02.18 3

= 18.1 cm3 = 18.1 mL

w = PΔV = 1.00 atm × (0.0181 L 30.6 L) = 30.6 L atm × atmL

J3.101= 3.10 × 103 J

= 3.10 kJ

ΔE = q + w = 40.66 kJ + 3.10 kJ = 37.56 kJ

Properties of Enthalpy

31. This is an endothermic reaction so heat must be absorbed in order to convert reactants into

products. The high temperature environment of internal combustion engines provides the

heat.

32. One should try to cool the reaction mixture or provide some means of removing heat since

the reaction is very exothermic (heat is released). The H2SO4(aq) will get very hot and

possibly boil unless cooling is provided.

33. a. Heat is absorbed from the water (it gets colder) as KBr dissolves, so this is an

endothermic process.

b. Heat is released as CH4 is burned, so this is an exothermic process.

c. Heat is released to the water (it gets hot) as H2SO4 is added, so this is an exothermic

process.

d. Heat must be added (absorbed) to boil water, so this is an endothermic process.

34. a. The combustion of gasoline releases heat, so this is an exothermic process.

b. H2O(g) → H2O(l); Heat is released when water vaper condenses, so this is an exothermic

process.

c. To convert a solid to a gas, heat must be absorbed, so this is an endothermic process.

d. Heat must be added (absorbed) in order to break a bond, so this is an endothermic

process.


35. 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s) ΔH = -1652 kJ; Note that 1652 kJ of heat are released when

4 mol Fe react with 3 mol O2 to produce 2 mol Fe2O3.

a. 4.00 mol Fe × Femol4

kJ1652−= 1650 kJ; 1650 kJ of heat released

b. 1.00 ml Fe2O3 × 32OFemol2

kJ1652−= 826 kJ; 826 kJ of heat released

c. 1.00 g Fe × Femol4

kJ1652

g85.55

Femol1 − = 7.39 kJ; 7.39 kJ of heat released

d. 10.0 g Fe × g85.55

Femol1 = 0.179 mol Fe; 2.00 g O2 ×

g00.32

Omol1 2 = 0.0625 mol O2

0.179 mol Fe/0.0625 mol O2 = 2.86; The balanced equation requires a 4 mol Fe/3 mol O2

= 1.33 mol ratio. O2 is limiting since the actual mol Fe/mol O2 ratio is greater than the

required mol ratio.

0.0625 mol O2 × 2Omol3

kJ1652−= 34.4 kJ; 34.4 kJ of heat released

36. a. 1.00 mol H2O × OHmol2

kJ572

2

−= 286 kJ; 286 kJ of heat released

b. 4.03 g H2 × 22

2

Hmol2

kJ572

Hg016.2

Hmol1 − = 572 kJ; 572 kJ of heat released

c. 186 g O2 × 22

2

Omol

kJ572

Og00.32

Omol1 − = 3320 kJ; 3320 kJ of heat released

d. 2Hn =

RT

PV=

K298Kmol

atmL08206.0

L100.2atm0.1 8

= 8.2 × 106 mol H2

8.2 × 106 mol H2 × OHmol2

kJ572

2

− = 2.3 × 109 kJ; 2. 3 × 109 kJ of heat released

37. From Sample Exercise 6.3, q = 1.3 × 108 J. Since the heat transfer process is only 60.%

efficient, the total energy required is: 1.3 × 108 J × J.60

J.100= 2.2 × 108 J

mass C3H8 = 2.2 × 108 J × 83

83

3

83

HCmol

HCg09.44

J102221

HCmol1

= 4.4 × 103 g C3H8


38. a. 1.00 g CH4 × 44

4

CHmol

kJ891

CHg04.16

CHmol1 − = 55.5 kJ

b. n = ,RT

PV

K298Kmol

atmL08206.0

L1000.1atm760

.740 3

= 39.8 mol CH4

39.8 mol × mol

kJ891−= 3.55 × 104 kJ

39. When a liquid is converted into gas, there is an increase in volume. The 2.5 kJ/mol quantity is

the work done by the vaporization process in pushing back the atmosphere.

40. H = E + PV; From this equation, H > E when V > 0, H < E when V < 0, and

H = E when V = 0. Concentrate on the moles of gaseous products versus the moles of

gaseous reactants to predict V for a reaction.

a. There are 2 moles of gaseous reactants converting to 2 moles of gaseous products so

V = 0. For this reaction, H = E.

b. There are 4 moles of gaseous reactants converting to 2 moles of gaseous products so

V < 0 and H < E.

c. There are 9 moles of gaseous reactants converting to 10 moles of gaseous products so

V > 0 and H > E.

Calorimetry and Heat Capacity

41. Specific heat capacity is defined as the amount of heat necessary to raise the temperature of

one gram of substance by one degree Celsius. Therefore, H2O(l) with the largest heat

capacity value requires the largest amount of heat for this process. The amount of heat for

H2O(l) is:

energy = s × m × ΔT = Cg

J18.4o

× 25.0 g × (37.0°C 15.0°C) = 2.30 × 103 J

The largest temperature change when a certain amount of energy is added to a certain mass of

substance will occur for the substance with the smallest specific heat capacity. This is Hg(l),

and the temperature change for this process is:

ΔT = ms

energy

=

g.550Cg

J14.0kJ

J1000kJ7.10

o

= 140°C


42. a. s = specific heat capacity = Kg

J24.0

Cg

J24.0o

= since ΔT(K) = ΔT(°C).

energy = s × m × ΔT = Cg

J24.0o

× 150.0 g × (298 K - 273 K) = 9.0 × 102 J

b. molar heat capacity = Cmol

J26

Agmol

Agg9.107

Cg

J24.0oo

=

c. 1250 J = Cg

J24.0o

× m × (15.2°C 12.0°C), m = 2.324.0

1250

= 1.6 × 103 g Ag

43. s = specific heat capacity = C)2.251.55(g00.5

J133

TΔm

qo−

= = 0.890 J/Cg

From Table 6.1, the substance is aluminum.

44. s = C)0.205.53(g6.125

J585o−

= 0.139 J/g °C

Molar heat capacity = Cmol

J9.27

Hgmol

g6.200

Cg

J139.0oo

=

45. | Heat loss by hot water | = | Heat gain by cooler water |

The magnitude of heat loss and heat gain are equal in calorimetry problems. The only

difference is the sign (positive or negative). To avoid sign errors, keep all quantities positive

and, if necessary, deduce the correct signs at the end of the problem. Water has a specific

heat capacity = s = 4.18 J/°Cg = 4.18 J/Kg (ΔT in °C = ΔT in K).

Heat loss by hot water = s × m × ΔT = Kg

J18.4× 50.0 g × (330. K Tf)

Heat gain by cooler water = Kg

J18.4× 30.0 g × (Tf 280. K); Heat loss = Heat gain, so:

K

J209× (330. KTf) =

K

J125× (Tf 280. K), 6.90 × 104 209 Tf = 125 Tf 3.50 × 104

334 Tf = 1.040 × 105, Tf = 311 K

Note that the final temperature is closer to the temperature of the more massive hot water,

which is as it should be.

46. Heat loss by hot water = heat gain by cold water; Keeping all quantities positive to avoid sign

errors:


Cg

J18.4o

× mhot × (55.0 °C 37.0°C) = Cg

J18.4o

× 90.0 g × (37.0 °C 22.0°C)

mhot = C0.18

C0.15g0.90o

o= 75.0 g hot water needed

47. Heat loss by Al + heat loss by Fe = heat gain by water; Keeping all quantities positive to

avoid sign error:

Cg

J89.0o

× 5.00 g Al × (100.0°C Tf) + Cg

J45.0o

× 10.00 g Fe × (100.0 Tf)

= Cg

J18.4o

× 97.3 g H2O × (Tf 22.0°C)

4.5(100.0 - Tf) + 4.5(100.0 Tf) = 407(Tf 22.0), 450 4.5 Tf + 450 4.5 Tf

= 407 Tf 8950

416 Tf = 9850, Tf = 23.7°C

48. heat released to water = 5.0 g H2 × 2Hg

J.120 + 10. g methane ×

methaneg

J.50 = 1.10 × 103 J

heat gain by water = 1.10 × 103 J = Cg

J18.4o

× 50.0 g × T

T = 5.26°C, 5.26°C = Tf 25.0°C, Tf = 30.3°C

49. Heat gain by water = heat loss by metal = s × m × ΔT where s = specific heat capacity.

Heat gain = Cg

J18.4o

× 150.0 g × (18.3°C - 15.0°C) = 2100 J

A common error in calorimetry problems is sign errors. Keeping all quantities positive helps

eliminate sign errors.

heat loss = 2100 J = s × 150.0 g × (75.0°C - 18.3°C), s = C7.56g0.150

J2100o

= 0.25 J/g°C

50. Heat gain by water = heat loss by Cu; Keeping all quantities positive to avoid sign errors:

Cg

J18.4o

× mass × (24.9°C 22.3°C) = Cg

J20.0o

× 110. g Cu × (82.4°C 24.9°C)

11 × mass = 1300, mass = 120 g H2O


51. 50.0 × 310− L × 0.100 mol/L = 5.00 × 310− mol of both AgNO3 and HCl are reacted. Thus,

5.00 × 310− mol of AgCl will be produced since there is a 1:1 mole ratio between reactants.

Heat lost by chemicals = Heat gained by solution

Heat gain = Cg

J18.4o

× 100.0 g × (23.40 22.60)°C = 330 J

Heat loss = 330 J; This is the heat evolved (exothermic reaction) when 5.00 × 310− mol of

AgCl is produced. So q = 330 J and ΔH (heat per mol AgCl formed) is negative with a value

of:

ΔH = J1000

kJ1

mol1000.5

J3303

−−

= 66 kJ/mol

Note: Sign errors are common with calorimetry problems. However, the correct sign for ΔH

can easily be determined from the ΔT data, i.e., if ΔT of the solution increases, then the

reaction is exothermic since heat was released, and if ΔT of the solution decreases, then the

reaction is endothermic since the reaction absorbed heat from the water. For calorimetry

problems, keep all quantities positive until the end of the calculation, then decide the sign for

ΔH. This will help eliminate sign errors.

52. NH4NO3(s) → NH4+(aq) + NO3

− (aq) ΔH = ?; mass of solution = 75.0 g + 1.60 g = 76.6 g

Heat lost by solution = Heat gained as NH4NO3 dissolves. To help eliminate sign errors, we

will keep all quantities positive (q and ΔT), then deduce the correct sign for ΔH at the end of

the problem. Here, since temperature decreases as NH4NO3 dissolves, heat is absorbed as

NH4NO3 dissolves, so it is an endothermic process (ΔH is positive).

Heat loss by solution = Cg

J18.4o

× 76.6 g × (25.00 23.34)°C = 532 J = heat gain as

NH4NO3 dissolves

ΔH = J1000

kJ1

NONHmol

NONHg05.80

NONHg60.1

J532

34

34

34

= 26.6 kJ/mol NH4NO3 dissolving

53. Since ΔH is exothermic, the temperature of the solution will increase as CaCl2(s) dissolves.

Keeping all quantities positive:

Heat loss as CaCl2 dissolves = 11.0 g CaCl2 × 22

2

CaClmol

kJ5.81

CaClg98.110

CaClmol1 = 8.08 kJ

Heat gain by solution = 8.08 × 103 J = Cg

J18.4o

× (125 + 11.0) g × (Tf 25.0°C)

Tf 25.0°C = 13618.4

1008.8 3

= 14.2°C, Tf = 14.2°C + 25.0°C = 39.2°C


54. 0.100 L × L

HClmol500.0= 5.00 × 210− mol HCl

0.300 L × L

)OH(Bamol100.0 2 = 3.00 × 210− mol Ba(OH)2

To react with all the HCl present, 5.00 × 210− /2 = 2.50 × 210− mol Ba(OH)2 are required.

Since 3.00 × 210− mol Ba(OH)2 are present, HCl is the limiting reactant.

5.00 × 210− mol HCl × HClmol2

kJ118= 2.95 kJ of heat is evolved by reaction.

Heat gain by solution = 2.95 × 103 J = Cg

J18.4o

× 400.0 g × ΔT

ΔT = 1.76°C = Tf Ti = Tf 25.0°C, Tf = 26.8°C

55. a. heat gain by calorimeter = heat loss by CH4 = 6.79 g CH4mol

kJ802

g04.16

CHmol1 4

= 340. kJ

heat capacity of calorimeter = C8.10

kJ.340o

= 31.5 kJ/°C

b. heat loss by C2H2 = heat gain by calorimeter = 16.9°C × C

kJ5.31o

= 532 kJ

ΔEcomb = 2222 HCmol

g04.26

HCg6.12

kJ532

− = 1.10 × 103 kJ/mol

56. Heat gain by calorimeter = C

kJ56.1o

× 3.2°C = 5.0 kJ = heat loss by quinine

Heat loss = 5.0 kJ, which is the heat evolved (exothermic reaction) by the combustion of

0.1964 g of quinone.

ΔEcomb = g1964.0

kJ0.5−= 25 kJ/g; ΔEcomb =

mol

g09.108

g

kJ25

−= 2700 kJ/mol

Hess's Law

57. Information given:

C(s) + O2(g) → CO2(g) ΔH = 393.7 kJ

CO(g) + 1/2 O2(g) → CO2(g) ΔH = 283.3 kJ


Using Hess’s Law:

2 C(s) + 2 O2(g) → 2 CO2(g) ΔH1 = 2(393.7 kJ)

2 CO2(g) → 2 CO(g) + O2(g) ΔH2 = 2(283.3 kJ)

__________________________________________________________

2 C(s) + O2(g) → 2 CO(g) ΔH = ΔH1 + ΔH2 = 220.8 kJ

Note: The enthalpy change for a reaction that is reversed is the negative quantity of the

enthalpy change for the original reaction. If the coefficients in a balanced reaction are

multiplied by an integer, the value of ΔH is multiplied by the same integer while the sign

stays the same.

58. C4H4(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l) ΔHcomb = 2341 kJ

C4H8(g) + 6 O2(g) → 4 CO2(g) + 4 H2O(l) ΔHcomb = 2755 kJ

H2(g) + 1/2 O2(g) → H2O(l) ΔHcomb = 286 kJ

By convention, H2O(l) is produced when enthalpies of combustion are given and, since per

mole quantities are given, the combustion reaction refers to 1 mole of that quantity reacting

with O2(g).

Using Hess’s Law to solve:

C4H4(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l) ΔH1 = 2341 kJ

4 CO2(g) + 4 H2O(l) → C4H8(g) + 6 O2(g) ΔH2 = (2755 kJ)

2 H2(g) + O2(g) → 2 H2O(l) ΔH3 = 2(286 kJ) _______________________________________________________________________

C4H4(g) + 2 H2(g) → C4H8(g) ΔH = ΔH1 + ΔH2 + ΔH3 = 158 kJ

59. 2 N2(g) + 6 H2(g) → 4 NH3(g) ΔH = 4(46 kJ)

6 H2O(g) → 6 H2(g) + 3 O2(g) ΔH = 3(-484 kJ)

___________________________________________________

2 N2(g) + 6 H2O(g) → 3 O2(g) + 4 NH3(g) ΔH = 1268 kJ

No, since the reaction is very endothermic (requires a lot of heat), it would not be a practical

way of making ammonia due to the high energy costs.

60. ClF + 1/2 O2 → 1/2 Cl2O + 1/2 F2O ΔH = 1/2 (167.4 kJ)

1/2 Cl2O + 3/2 F2O → ClF3 + O2 ΔH = 1/2 (341.4 kJ)

F2 + 1/2 O2 → F2O ΔH = 1/2 (43.4 kJ)

__________________________________________________________

ClF(g) + F2(g) → ClF3 ΔH = 108.7 kJ

61. NO + O3 → NO2 + O2 ΔH = 199 kJ

3/2 O2 → O3 ΔH = 1/2(-427 kJ) O → 1/2 O2 ΔH = 1/2(495 kJ) ________________________________________________

NO(g) + O(g) → NO2(g) ΔH = 233 kJ


62. C6H4(OH)2 → C6H4O2 + H2 ΔH = 177.4 kJ

H2O2 → H2 + O2 ΔH = (191.2 kJ)

2 H2 + O2 → 2 H2O(g) ΔH = 2(241.8 kJ)

2 H2O(g) → 2 H2O(l) ΔH = 2(43.8 kJ) ________________________________________________________________

C6H4(OH)2(aq) + H2O2(aq) → C6H4O2(aq) + 2 H2O(l) ΔH = 202.6 kJ

63. CaC2 → Ca + 2 C ΔH = (62.8 kJ)

CaO + H2O → Ca(OH)2 ΔH = 653.1 kJ

2 CO2 + H2O → C2H2 + 5/2 O2 ΔH = (1300. kJ)

Ca + 1/2 O2 → CaO ΔH = 635.5 kJ

2 C + 2 O2 → 2 CO2 ΔH = 2(393.5 kJ)

______________________________________________________________________________________________________

CaC2(s) + 2 H2O(l) → Ca(OH)2(aq) + C2H2(g) ΔH = 713 kJ

64. P4O10 → P4 + 5 O2 ΔH = (2967.3 kJ)

10 PCl3 + 5 O2 → 10 Cl3PO ΔH = 10(285.7 kJ)

6 PCl5 → 6 PCl3 + 6 Cl2 ΔH = 6(84.2 kJ)

P4 + 6 Cl2 → 4 PCl3 ΔH = 1225.6 __________________________________________________________________________________________

P4O10(s) + 6 PCl5(g) → 10 Cl3PO(g) ΔH = 610.1 kJ

Standard Enthalpies of Formation

65. The change in enthalpy that accompanies the formation of one mole of a compound from its

elements, with all substances in their standard states, is the standard enthalpy of formation for

a compound. The reactions that refer to Hof are:

Na(s) + 1/2 Cl2(g) → NaCl(s); H2(g) + 1/2 O2(g) → H2O(l)

6 C(graphite, s) + 6 H2(g) + 3 O2(g) → C6H12O6(s)

Pb(s) + S(rhombic, s) + 2 O2(g) → PbSO4(s)

66. a. aluminum oxide = Al2O3; 2 Al(s) + 3/2 O2(g) → Al2O3(s)

b. C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)

c. NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq)

d. 2 C(graphite, s) + 3/2 H2(g) + 1/2 Cl2(g) → C2H3Cl(g)

e. C6H6(l) + 15/2 O2(g) → 6 CO2(g) + 3 H2O(l)

Note: ΔHcomb values assume one mole of compound combusted.

f. NH4Br(s) → NH4+(aq) + Br-(aq)


67. In general: ΔH° = np products,fHΔ nr reactants,fΔH and all elements in their standard state

have fHΔ = 0 by definition.

a. The balanced equation is: 2 NH3(g) + 3 O2(g) + 2 CH4(g) → 2 HCN(g) + 6 H2O(g)

ΔH° = [ 2 mol HCN × HCN,fHΔ

+ 6 mol H2O(g) × OH,f

2HΔ

]

[2 mol NH3 × 3NH,fHΔ + 2 mol CH4 ×

4CH,fHΔ ]

ΔH° = [2(135.1) + 6(242)] [2(46) + 2(75)] = 940. kJ

b. Ca3(PO4)2(s) + 3 H2SO4(l) → 3 CaSO4(s) + 2 H3PO4(l)

ΔH° =

−+

−

mol

kJ1267)l(POHmol2

mol

kJ1433CaSOmol3 434

−+

−

mol

kJ814)l(SOHmol3

mol

kJ4126)PO(Camol1 42243

ΔH° = 6833 kJ (6568 kJ) = 265 kJ

c. NH3(g) + HCl(g) → NH4Cl(s)

ΔH° = [1 mol NH4Cl × ClNH,f4

HΔ ] [1 mol NH3 ×

3NH,fHΔ + 1 mol HCl × HCl,fHΔ

]

ΔH° =

−+

−−

−

mol

kJ92mol1

mol

kJ46mol1

mol

kJ314mol1

ΔH° = 314 kJ + 138 kJ = 176 kJ

68. a. The balanced equation is: C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g)

ΔH° =

−−

−+

−

mol

kJ278mol1

mol

kJ242mol3

mol

kJ5.393mol2

ΔH° = = 1513 kJ (278 kJ) = 1235 kJ

b. SiCl4(l) + 2 H2O(l) → SiO2(s) + 4 HCl(aq)

Since HCl(aq) is H+(aq) + Cl−(aq), then fHΔ = 0 167 = 167 kJ/mol.


ΔH° =

−+

−−

−+

−

mol

kJ286mol2

mol

kJ687mol1

mol

kJ911mol1

mol

kJ167mol4

ΔH° = 1579 kJ (1259 kJ) = 320. kJ

c. MgO(s) + H2O(l) → Mg(OH)2(s)

ΔH° =

−+

−−

−

mol

kJ286mol1

mol

kJ602mol1

mol

kJ925mol1

ΔH° = 925 kJ (888 kJ) = 37 kJ

69. a. 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g); ΔH° = np products,fHΔ nr reactants,fΔH

ΔH° =

−−

−+

mol

kJ46mol4

mol

kJ242mol6

mol

kJ.90mol4 = 908 kJ

2 NO(g) + O2(g) → 2 NO2(g)

ΔH° =

−

mol

kJ.90mol2

mol

kJ34mol2 = 112 kJ

3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g)

ΔH° =

−+

−

+

−

mol

kJ286mol1

mol

kJ34mol3

mol

kJ.90mol1

mol

kJ207mol2

= 140. kJ

Note: All fHΔ values are assumed ± 1 kJ.

b. 12 NH3(g) + 15 O2(g) → 12 NO(g) + 18 H2O(g)

12 NO(g) + 6 O2(g) → 12 NO2(g)

12 NO2(g) + 4 H2O(l) → 8 HNO3(aq) + 4 NO(g)

4 H2O(g) → 4 H2O(l) __________________________________________________ 12 NH3(g) + 21 O2(g) → 8 HNO3(aq) + 4 NO(g) + 14 H2O(g)

The overall reaction is exothermic since each step is exothermic.

70. 4 Na(s) + O2(g) → 2 Na2O(s), ΔH° = 2 mol × mol

kJ416−= -832 kJ


2 Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g)

ΔH° =

−−

−

mol

kJ286mol2

mol

kJ470mol2 = 368 kJ

2 Na(s) + CO2(g) → Na2O(s) + CO(g)

ΔH° =

−−

−+

−

mol

kJ5.393mol1

mol

kJ.5.110mol1

mol

kJ416mol1 = 133 kJ

In both cases, sodium metal reacts with the "extinguishing agent." Both reactions are

exothermic and each reaction produces a flammable gas, H2 and CO, respectively.

71. 3 Al(s) + 3 NH4ClO4(s) → Al2O3(s) + AlCl3(s) + 3 NO(g) + 6 H2O(g)

ΔH° =

−+

−+

+

−

mol

kJ1676mol1

mol

kJ704mol1

mol

kJ.90mol3

mol

kJ242mol6

−−

mol

kJ295mol3 = 2677 kJ

72. 5 N2O4(l) + 4 N2H3CH3(l) → 12 H2O(g) + 9 N2(g) + 4 CO2(g)

ΔH° =

−+

−

mol

kJ5.393mol4

mol

kJ242mol12

+

−−

mol

kJ54mol4

mol

kJ.20mol5 = 4594 kJ

73. 2 ClF3(g) + 2 NH3(g) → N2(g) + 6 HF(g) + Cl2(g) ΔH° = 1196 kJ

ΔH° = [6 HF,fHΔ

] [2 3ClF,fHΔ + 2

3NH,fHΔ ]

1196 kJ = 6 mol

−

mol

kJ271 2

3ClF,fHΔ 2 mol

−

mol

kJ46

1196 kJ = 1626 kJ 2 3ClF,fHΔ + 92 kJ,

3ClF,fHΔ = mol2

kJ)1196921626( ++− =

mol

kJ169−

74. C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(l) ΔH° = 1411.1 kJ

ΔH° = 1411.1 kJ = 2(393.5) kJ + 2(285.8) kJ 42HC,fHΔ

1411.1 kJ = 1358.6 kJ ,HΔ42HC,f

42HC,fHΔ = 52.5 kJ/mol


Energy Consumption and Sources

75. C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)

ΔH° = [2 (393.5 kJ) + 3(-286 kJ)] (278 kJ) = 1367 kJ/mol ethanol

g07.46

mol1

mol

kJ1367

−= 29.67 kJ/g

76. CH3OH(l) + 3/2 O2(g) → CO2(g) + 2 H2O(l)

ΔH° = [393.5 kJ + 2(286 kJ)] (239 kJ) = 727 kJ/mol CH3OH

g04.32

mol1

mol

kJ727

− = 22.7 kJ/g vs. 29.67 kJ/g for ethanol

Ethanol has a slightly higher fuel value than methanol.

77. C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)

ΔH° = [3(393.5 kJ) + 4(286 kJ)] [104 kJ] = 2221 kJ/mol C3H8

g09.44

mol1

mol

kJ2221

− =

mol

kJ37.50−vs. 47.7 kJ/g for octane (Sample Exercise 6.11)

The fuel values are close. An advantage of propane is that it burns more cleanly. The boiling

point of propane is 42°C. Thus, it is more difficult to store propane and there are extra

safety hazards associated with using high pressure compressed gas tanks.

78. 1 mole of C2H2(g) and 1 mole of C4H10(g) have equivalent volumes at the same T and P.

104

22

HCofvolumepercombustionofenthalpy

HCofvolumepercombustionofenthalpy=

104

22

HCofmolpercombustionofenthalpy

HCofmolpercombustionofenthalpy

104

22

HCofvolumepercombustionofenthalpy

HCofvolumepercombustionofenthalpy=

104

104

104

22

22

22

HCmol

HCg12.58

HCg

kJ5.49

HCmol

HCg04.26

HCg

kJ9.49

−

−

= 0.452

More than twice the volume of acetylene is needed to furnish the same energy as a given

volume of butane.

79. The molar volume of a gas at STP is 22.42 L (from Chapter 5).

4.19 × 106 kJ × 4

44

CHmol

CHL42.22

kJ891

CHmol1 = 1.05 × 105 L CH4


80. Mass of H2O = 1.00 gal × mL

g00.1

L

mL1000

gal

L785.3 = 3790 g H2O

Energy required (theoretical) = s × m × ΔT = Cg

J18.4o

× 3790 g × 10.0 °C = 1.58 × 105 J

For the actual (80.0% efficient) process, more than this quantity of energy is needed since

heat is always lost in any transfer of energy. The energy required is:

1.58 × 105 J = J0.80

J.100 = 1.98 × 105 J

Mass of C2H2 = 1.98 × 105 J × 22

22

3

22

HCmol

HCg04.26

J10.1300

HCmol1

= 3.97 g C2H2

Additional Exercises

81. a. 2 SO2(g) + O2(g) → 2 SO3(g) (w = PΔV); Because the volume of the piston apparatus

decreased as reactants were converted to products, w is positive (w > 0).

b. COCl2(g) → CO(g) + Cl2(g); Because the volume increased, w is negative (w < 0).

c. N2(g) + O2(g) → 2 NO(g); Because the volume did not change, no PV work is done

(w = 0).

In order to predict the sign of w for a reaction, compare the coefficients of all the product

gases in the balanced equation to the coefficients of all the reactant gases. When a balanced

reaction has more mol of product gases than mol of reactant gases (as in b), the reaction will

expand in volume (ΔV positive), and the system does work on the surroundings. When a

balanced reaction has a decrease in the mol of gas from reactants to products (as in a), the

reaction will contract in volume (ΔV negative), and the surroundings will do compression

work on the system. When there is no change in the mol of gas from reactants to products (as

in c), ΔV = 0 and w = 0.

82. w = PΔV; Δn = mol gaseous products - mol gaseous reactants. Only gases can do PV work

(we ignore solids and liquids). When a balanced reaction has more mol of product gases than

mol of reactant gases (Δn positive), the reaction will expand in volume (ΔV positive) and the

system will do work on the surroundings. For example, in reaction c, Δn = 2 0 = 2 mol, and

this reaction would do expansion work against the surroundings. When a balanced reaction

has a decrease in the mol of gas from reactants to products (Δn negative), the reaction will

contract in volume (ΔV negative) and the surroundings will do compression work on the

system, e.g., reaction a where Δn = 0 1 = 1. When there is no change in the mol of gas from

reactants to products, ΔV = 0 and w = 0, e.g., reaction b where Δn = 2 2 = 0.

When ΔV > 0 (Δn > 0), then w < 0 and system does work on the surroundings (c and e).

When ΔV < 0 (Δn < 0), then w > 0 and the surroundings do work on the system (a and d).

When ΔV = 0 (Δn = 0), then w = 0 (b).


83. ΔEoverall = ΔEstep 1 + ΔEstep 2; This is a cyclic process which means that the overall initial state

and final state are the same. Since ΔE is a state function, ΔEoverall = 0 and ΔEstep 1 = ΔEstep 2.

ΔEstep 1 = q + w = 45 J + (10. J) = 35 J

ΔEstep 2 = ΔEstep 1 = 35 J = q + w, 35 J = 60 J + w, w = 25 J

84. 2 K(s) + 2 H2O(l) → 2 KOH(aq) + H2(g), ΔH° = 2(481 kJ) 2(286 kJ) = 390. kJ

5.00 g K × Kmol2

kJ.390

Kg10.39

Kmol1 − = 24.9 kJ;

24.9 kJ of heat is released upon reaction of 5.00 g K.

24,900 J = Cg

J18.4o

× (1.00 × 103 g) × ΔT, ΔT = 31000.118.4

900,24

= 5.96°C

Final temperature = 24.0 + 5.96 = 30.0°C

85. HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) ΔH = 56 kJ

0.2000 L × L

HClmol400.0= 8.00 × 210− mol HCl

0.1500 L × L

NaOHmol500.0= 7.50 × 210− mol NaOH

Because the balanced reaction requires a 1:1 mole ratio between HCl and NaOH, and because

fewer moles of NaOH are actually present as compared to HCl, NaOH is the limiting reagent.

7.50 × 210− mol NaOH × NaOHmol

kJ56−= 4.2 kJ; 4.2 kJ of heat is released.

86. Na2SO4(aq)2 + Ba(NO3)2(aq) → BaSO4(s) + 2 NaNO3(aq) ΔH = ?

1.00 L × L

mol00.2 = 2.00 mol Na2SO4; 2.00 L ×

L

mol750.0 = 1.50 mol Ba(NO3)2

The balanced equation requires a 1:1 mole ratio between Na2SO4 and Ba(NO3)2. Because we

have fewer moles of Ba(NO3)2 present, it is limiting and 1.50 mol BaSO4 will be produced

[there is a 1:1 mole ratio between Ba(NO3)2 and BaSO4].

heat gain by solution = heat loss by reaction

mass of solution = 3.00 L × mL

g00.2

L1

mol1000 = 6.00 × 103 g


heat gain by solution = Cg

J37.6o

× 6.00 × 103 g × (42.0 30.0)C = 4.59 × 105 J

Because the solution gained heat, the reaction is exothermic; q = 4.59 × 105 J for the

reaction.

H = 4

5

BaSOmol50.1

J1059.4 − = 3.06 × 105 J/mol = 306 kJ/mol

87. qsurr = qsolution + qcal; We normally assume qcal is zero (no heat gain/loss by the calorimeter).

However, if the calorimeter has a nonzero heat capacity, then some of the heat absorbed by

the endothermic reaction came from the calorimeter. If we ignore qcal, then qsurr is too small

giving a calculated H value which is less positive (smaller) than it should be.

88. The specific heat of water is 4.18 J/g°C, which is equal to 4.18 kJ/kg°C.

We have 1.00 kg of H2O, so: 1.00 kg × Ckg

J18.4o

= 4.18 kJ/°C

This is the portion of the heat capacity that can be attributed to H2O.

Total heat capacity = Ccal + OH2C , Ccal = 10.84 4.18 = 6.66 kJ/°C

89. Heat released = 1.056 g × 26.42 kJ/g = 27.90 kJ = Heat gain by water and calorimeter

Heat gain = 27.90 kJ = Ckg

J18.4o

× 0.987 kg × ΔT + C

kJ66.6o

× T

27.90 = (4.13 + 6.66) ΔT = 10.79 ΔT, ΔT = 2.586°C

2.586°C = Tf 23.32°C, Tf = 25.91°C

90. To avoid fractions, let's first calculate ΔH for the reaction:

6 FeO(s) + 6 CO(g) → 6 Fe(s) + 6 CO2(g)

6 FeO + 2 CO2 → 2 Fe3O4 + 2 CO ΔH° = 2(18 kJ)

2 Fe3O4 + CO2 → 3 Fe2O3 + CO ΔH° = (39 kJ)

3 Fe2O3 + 9 CO → 6 Fe + 9 CO2 ΔH° = 3(23 kJ)

______________________________________________________

6 FeO(s) + 6 CO(g) → 6 Fe(s) + 6 CO2(g) ΔH° = 66 kJ

So for: FeO(s) + CO(g) → Fe(s) + CO2(g) ΔH° = 6

kJ66−= 11 kJ

91. a. ΔH° = 3 mol (227 kJ/mol) 1 mol (49 kJ/mol) = 632 kJ


b. Since 3 C2H2(g) is higher in energy than C6H6(l), acetylene will release more energy per

gram when burned in air.

92. I(g) + Cl(g) → ICl(g) ΔH = (211.3 kJ)

1/2 Cl2(g) → Cl(g) ΔH = 1/2(242.3 kJ)

1/2 I2(g) → I(g) ΔH = 1/2(151.0 kJ)

1/2 I2(s) → 1/2 I2(g) ΔH = 1/2(62.8 kJ)

_______________________________________________________________

1/2 I2(s) + 1/2 Cl2(g) → ICl(g) ΔH = 16.8 kJ/mol = ICl,

ofHΔ

93. a. C2H4(g) + O3(g) → CH3CHO(g) + O2(g), ΔH° = 166 kJ [143 kJ + 52 kJ] = 361 kJ

b. O3(g) + NO(g) → NO2(g) + O2(g), ΔH° = 34 kJ [90. kJ + 143 kJ] = 199 kJ

c. SO3(g) + H2O(l) → H2SO4(aq), ΔH° = 909 kJ [-396 kJ + (286 kJ)] = 227 kJ

d. 2 NO(g) + O2(g) → 2 NO2(g), ΔH° = 2(34) kJ 2(90.) kJ = 112 kJ

Challenge Problems

94. Only when there is a volume change can PV work be done. In pathway 1 (steps 1 + 2), only

the first step does PV work (step 2 has a constant volume of 30.0 L). In pathway 2 (steps 3 +

4), only step 4 does PV work (step 3 has a constant volume of 10.0 L).

Pathway 1: w = PΔV = 2.00 atm (30.0 L 10.0 L) = -40.0 L atm × atmL

J3.101

= 4.05 × 103 J

Pathway 2: w = PΔV = 1.00 atm (30.0 L 10.0 L) = 20.0 L atm × atmL

J3.101

= 2.03 × 103 J

Note: The sign is () because the system is doing work on the surroundings (an expansion).

We get different values of work for the two pathways; both pathways have the same initial

and final states. Because w depends on the pathway, work cannot be a state function.

95. a. C12H22O11(s) + 12 O2(g) → 12 CO2(g) + 11 H2O(l)

b. A bomb calorimeter is at constant volume, so heat released = qv = ΔE:

ΔE = mol

g30.342

g46.1

kJ00.24

−= 5630 kJ/mol C12H22O11

c. PV = nRT; At constant P and T, PΔV = RTΔn where Δn = mol gaseous products mol

gaseous reactants.

ΔH = ΔE + PΔV = ΔE + RTΔn


For this reaction, Δn = 12 12 = 0, so ΔH = ΔE = 5630 kJ/mol.

96. Energy needed = mol

kJ5640

OHCg30.342

OHCmol1

hr

OHCg10.20

112212

1122121122123

= 3.3 × 105 kJ/hr

Energy from sun = 1.0 kW/m2 = 1000 W/m2 = 22 ms

kJ0.1

ms

J1000=

10,000 m2 ×hr

min60

min

s60

ms

kJ0.12

= 3.6 × 107 kJ/hr

% efficiency = hourperenergyTotal

hourperusedEnergy× 100 =

kJ106.3

kJ103.37

5

= 0.92%

97. Energy used in 8.0 hours = 40. kWh = hr

s3600

s

hrkJ0.40 = 1.4 × 105 kJ

Energy from the sun in 8.0 hours = hr

min60

min

s60

ms

kJ0.12

× 8.0 hr = 2.9 × 104 kJ/m2

Only 13% of the sunlight is converted into electricity:

0.13 × (2.9 × 104 kJ/m2) × Area = 1.4 × 105 kJ, Area = 37 m2

98. a. 2 HNO3(aq) + Na2CO3(s) → 2 NaNO3(aq) + H2O(l) + CO2(g)

ΔH° = [2(467 kJ) + (286 kJ) + (393.5 kJ)] [2(207 kJ) + (1131 kJ)] = 69 kJ

2.0 × 104 gallons × mL

g42.1

qt

mL946

gal

qt4 = 1.1 × 108 g of concentrated nitric

acid solution

1.1 × 108 g solution × solutiong0.100

HNOg0.70 3 = 7.7 × 107 g HNO3

7.7 × 107 g HNO3 × 32

32

3

32

CONamol

CONag99.105

HNOmol2

CONamol1

g02.63

mol1

= 6.5 × 107 g Na2CO3

There are (7.7 × 107/63.02) mol of HNO3 from the previous calculation. There are 69 kJ

of heat evolved for every two moles of nitric acid neutralized. Combining these two

results:

7.7 × 107 g HNO3 × 33

3

HNOmol2

kJ69

HNOg02.63

HNOmol1 − = 4.2 × 107 kJ


b. They feared the heat generated by the neutralization reaction would vaporize the

unreacted nitric acid, causing widespread airborne contamination.

99. 400 kcal × kcal

kJ18.4 = 1.7 × 103 kJ 2 × 103 kJ

PE =mgz =

cm100

m1

in

cm54.2in8

s

m81.9

lb205.2

kg1lb180

2 = 160 J200 J

200 J of energy are needed to climb one step. The total number of steps to climb are:

2 × 106 J × J200

step1 = 1 × 104 steps

100. H2(g) + 1/2 O2(g) → H2O(l) ΔH° = )l(OH,

of

2HΔ = 285.8 kJ; We want the reverse reaction:

H2O(l) → H2(g) + 1/2 O2(g) ΔH° = 285.8 kJ

w = PV; Because PV = nRT, at constant T and P, PV = RTn where n = mol gaseous

products – mol gaseous reactants. Here, Δn = (1 mol H2 + 0.5 mol O2) – (0) = 1.5 mol

ΔE° = ΔH° PΔV = ΔH° ΔnRT

ΔE° = 285.8 kJ 1.50 mol × 8.3145 J/molK × 298 K ×J1000

kJ1

ΔE° = 285.8 kJ 3.72 kJ = 282.1 kJ

101. There are five parts to this problem. We need to calculate:

1. q required to heat H2O(s) from 30.C to 0C; use the specific heat capacity of H2O(s)

2. q required to convert 1 mol H2O(s) at 0C into 1 mol H2O(l) at 0C; use Hfusion

3. q required to heat H2O(l) from 0C to 100.C; use the specific heat capacity of H2O(l)

4. q required to convert 1 mol H2O(l) at 100.C into 1 mol H2O(g) at 100.C;

use Hvaporization

5. q required to heat H2O(g) from 100.C to 140.C; use the specific heat capacity of

H2O(g)

We will sum up the heat required for all five parts and this will be the total amount of heat

required to convert 1.00 mol of H2O(s) at 30.C to H2O(g) at 140.C. (qtotal = q1 + q2 + q3 +

q4 + q5). The molar mass of H2O is 18.02 g/mol.

q1 = 2.03 J/Cg × 18.02 g × [0 – (30.)]C = 1.1 × 103 J

q2 = 1.00 mol × 6.02 × 103 J/mol = 6.02 × 103 J


q3 = 4.18 J/Cg × 18.02 g × (100. – 0)C = 7.53 × 103 J

q4 = 1.00 mol × 40.7 × 104 J/mol = 4.07 × 104 J

q5 = 2.02 J/Cg × 18.02 g × (140. – 100.) = 1.5 × 103 J

qtotal = q1 + q2 + q3 + q4 + q5 = 5.68 × 104 J = 56.9 kJ

102. When a mixture of ice and water exists, the temperature of the mixture remains at 0C until

all of the ice has melted. Because an ice water mixture exists at the end of the process, the

temperature remains at 0C. All of the energy released by the element goes to convert ice

into water. The energy required to do this is related to Hfusion = 6.02 kJ/mol (from Exercise

101).

heat loss by element = heat gain by ice cubes at 0C

heat gain = 109.5 g H2O × OHmol

kJ02.6

g02.18

OHmol1

2

2 = 36.6 kJ

specific heat of element = C)0195(g0.500

J600,36

TΔmass

qo−

= = 0.375 J/Cg

Integrative Problems

103. N2(g) + 2 O2(g) → 2 NO2(g) H = 67.7 kJ

2Nn =

K373Kmol

atmL08206.0

L250.0atm50.3

RT

PV

= = 2.86 × 210− mol N2

2On =

K373Kmol

atmL08206.0

L450.0atm50.3

RT

PV

= = 5.15 × 210− mol O2

The balanced equation requires a 2:1 O2 to N2 mole ratio. The actual mole ratio is

5.15 × 210− /2.86 × 210− = 1.80; Because the actual mole ratio < required mole ratio,

O2 in the numerator is limiting.

5.15 × 210−

mol O2 × 2

2

Omol2

NOmol2 = 5.15 ×

210− mol NO2

5.15 × 210−

mol NO2 × 2NOmol2

kJ7.67= 1.74 kJ

104. a. 4 CH3NO2(l) + 3 O2(g) → 4 CO2(g) + 2 N2(g) + 6 H2O(g)

orxnHΔ = 1288.5 kJ = [4 mol(393.5 kJ/mol) + 6 mol(242 kJ/mol)]


)]HΔ(mol4[ oNOCH,f 23

Solving: o

NOCH,f 23HΔ = 434 kJ/mol

b. Ptot = 950. torr × torr760

atm1= 1.25 atm;

22 NtotN χPP = = 1.25 atm × 0.134

= 0.168 atm

2Nn

K373Kmol

atmL08206.0

L0.15atm168.0

= = 0.0823 mol N2

0.0823 mol N2 × 2

2

Nmol1

Ng02.28 = 2.31 g N2

105. heat loss by U = heat gain by heavy water; vol of cube = (cube edge)3

mass of heavy water = 1.00 × 103 mL × mL

g11.1 = 1110 g

heat gain by heavy water = Cg

J211.4o

× 1110 g × (28.5 – 25.5)C = 1.4 × 104 J

heat loss by U = 1.4 × 104 J = Cg

J117.0o

× mass × (200.0 – 28.5)C, mass = 7.0 × 102 g U

7.0 × 102 g U × g05.19

cm1 3

= 37 cm3; cube edge = (37 cm3)1/3 = 3.3 cm

Marathon Problems

106. X → CO2(g) + H2O(l) + O2(g) + A(g) ΔH = 1893 kJ/mol (unbalanced)

To determine X, we must determine the moles of X reacted, the identity of A and the moles

of A produced. For the reaction at constant P (ΔH = q):

OH2

q− = qrxn = 4.184 J/°Cg × 1.000 × 104 g × (29.52 25.00) °C × 1 kJ/1000 J

qrxn = 189.1 kJ (carrying extra sig. figs.)

Since ΔH = 1893 kJ/mol for the decomposition reaction and since only 189.1 kJ of heat was

released for this reaction, then 189.1 kJ × 1 mol X/1893 kJ = 0.100 mol X was reacted.

Molar mass of X = Xmol100.0

Xg7.22= 227 g/mol


From the problem, 0.100 mol X produced 0.300 mol CO2, 0.250 mol H2O and 0.025 mol O2.

Therefore, 1.00 mol X produces 3.00 mol CO2, 2.50 mol H2O and 0.25 mol O2.

1.00 mol X = 227 g = 3.00 mol CO2

mol

g01.44+ 2.50 mol H2O

mol

g02.18

+ 0.25 mol

mol

g00.32+ (mass of A)

mass of A in 1.00 mol X = 227 g 132.0 g 45.05 g 8.00 g = 42 g A

To determine A, we need the moles of A produced. The total moles of gases produced can be

determined from the gas law data provided in the problem. Since H2O(l) is a product, we

need to subtract OH 2P (the vapor pressure of H2O) from the total pressure.

ntotal = RT

PV; Ptotal = Pgases +

OH 2

P , Pgases = 778 torr 31 torr = 747 torr

V = height × area; area = πr2; V = 59.8 cm (π) (8.00 cm)2

3cm1000

L1 = 12.0 L

T = 273.15 + 29.52 = 302.67 K

ntotal =

K67.302molK

atmL08206.0

L0.12torr760

atm1torr747

RT

PV

= = 0.475 mol = mol CO2 + mol O2 + mol A

mol A = 0.475 mol 0.300 mol CO2 0.025 mol O2 = 0.150 mol A

Since 0.100 mol X reacted, then 1.00 mol X would produce 1.50 mol A which from a

previous calculation represents 42 g A.

Molar mass of A = Amol50.1

Ag42= 28 g/mol

Since A is a gaseous element, the only element that is a gas and has this molar mass is N2(g).

Thus, A = N2(g)

a. Now we can determine the formula of X.

X → 3 CO2(g) + 2.5 H2O(l) + 0.25 O2(g) + 1.5 N2(g). For a balanced reaction, X =

C3H5N3O9, which, for your information, is nitroglycerine.

b. w = PΔV = 778 torr × torr760

atm1 × (12.0 L 0) = 12.3 L atm


12.3 L atm × atmL

J3.101= 1250 J = 1.25 kJ, w = 1.25 kJ

c. ΔE = q + w, where q = ΔH since at constant pressure. For 1 mol of X decomposed:

w = 1.25 kJ/0.100 mol = 12.5 kJ/mol; ΔE = ΔH PΔV and w = PΔV

ΔE = ΔH + w = 1893 kJ/mol + (12.5 kJ/mol) = 1906 kJ/mol

ofHΔ for C3H5N3O9 can be estimated from standard enthalpies of formation data and

assuming orxnrxn HΔHΔ = . For the balanced reaction given in part a:

orxnHΔ = 1893 kJ = ]HΔ[]HΔ5.2HΔ3[ o

ONHC,fo

OH,fo

CO,f 935322−+

1893 kJ = [3 (393.5) kJ + 2.5 (286) kJ] o

ONHC,f 9353HΔ

o

ONHC,f 9353HΔ = 2.5 kJ/mol = 3 kJ/mol

107. CxHy +

+

2

2/yx2 → x CO2 + y/2 H2O

[393.5x + y/2 (242)] o

HC yxHΔ = 2044.5, 393.5x 121y

yx HCHΔ = 2044.5

dgas = RT

M MP • where MM = average molar mass of CO2/H2O mixture

0.751 g/L =

K473molK

atmL08206.0

MMatm00.1

, MM of CO2/H2O mixture = 29.1 g/mol

Let a = mol CO2 and 1.00 a = mol H2O (assuming 1.00 total mol of mixture)

44.01 a + (1.00 a) × 18.02 = 29.1; Solving: a = 0.426 mol CO2 , mol H2O = 0.574

Thus: x

y,

x

2

y

426.0

574.0= = 2.69, y = 2.69 x

For whole numbers, multiply by three which gives y = 8, x = 3. Note that y = 16, x = 6 is

possible, along with other combinations. Because the hydrocarbon has a lower density than

Kr, the molar mass of CxHy must be less than the molar mass of Kr (83.80 g/mol). Only C3H8

works.

2044.5 = 393.5(3) 121(8) o

HCo

HC 8383HΔ,HΔ = 104 kJ/mol

chapter six thermochemistry

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