CHAPTER SIX THERMOCHEMISTRY
Transcript of 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
CHAPTER 6 THERMOCHEMISTRY 179
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.
180 CHAPTER 6 THERMOCHEMISTRY
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
CHAPTER 6 THERMOCHEMISTRY 181
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.
182 CHAPTER 6 THERMOCHEMISTRY
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
CHAPTER 6 THERMOCHEMISTRY 183
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
184 CHAPTER 6 THERMOCHEMISTRY
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:
CHAPTER 6 THERMOCHEMISTRY 185
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
186 CHAPTER 6 THERMOCHEMISTRY
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
CHAPTER 6 THERMOCHEMISTRY 187
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
188 CHAPTER 6 THERMOCHEMISTRY
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
CHAPTER 6 THERMOCHEMISTRY 189
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)
190 CHAPTER 6 THERMOCHEMISTRY
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.
CHAPTER 6 THERMOCHEMISTRY 191
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
192 CHAPTER 6 THERMOCHEMISTRY
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
CHAPTER 6 THERMOCHEMISTRY 193
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
194 CHAPTER 6 THERMOCHEMISTRY
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).
CHAPTER 6 THERMOCHEMISTRY 195
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
196 CHAPTER 6 THERMOCHEMISTRY
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
CHAPTER 6 THERMOCHEMISTRY 197
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
198 CHAPTER 6 THERMOCHEMISTRY
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
CHAPTER 6 THERMOCHEMISTRY 199
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
200 CHAPTER 6 THERMOCHEMISTRY
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)]
CHAPTER 6 THERMOCHEMISTRY 201
)]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
202 CHAPTER 6 THERMOCHEMISTRY
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
CHAPTER 6 THERMOCHEMISTRY 203
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