CHAPTER 6
THERMOCHEMISTRY
Questions
11.
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.
12.
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.
13.
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, concentrate 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.
14.
∆H = ∆E + P∆V at constant P; from the definition of enthalpy, the difference between ∆H
and ∆E at constant P is the quantity P∆V. Thus, when a system at constant P can do pressurevolume 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.
15.
a. The ∆H value for a reaction is specific to the coefficients in the balanced equation. Because the coefficient in front of H2O is a 2, 891 kJ of heat is released when 2 mol of H2O
is produced. For 1 mol of H2O formed, 891/2 = 446 kJ of heat is released.
b. 891/2 = 446 kJ of heat released for each mol of O2 reacted.
16.
Use the coefficients in the balanced rection to determine the heat required for the various
quantities.
a. 1 mol Hg ×
90.7 kJ
= 90.7 kJ required
mol Hg
182
CHAPTER
6
THERMOCHEMISTRY
b. 1 mol O2 ×
183
90.7 kJ
= 181.4 kJ required
1/ 2 mol O2
c. When an equation is reversed, ∆Hnew = −∆Hold. When an equation is multiplied by some
integer n, then ∆Hnew = n(∆Hold).
Hg(l) + 1/2 O2(g) → HgO(s)
2Hg(l) + O2(g) → 2HgO(s)
17.
∆H = −90.7 kJ
∆H = 2(−90.7 kJ) = −181.4 kJ
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.
18.
A state function is a function whose change depends only on the initial and final states and
not on how one got from the initial to the final state. An extensive property depends on the
amount of substance. Enthalpy changes for a reaction are path-independent, but they do
depend on the quantity of reactants consumed in the reaction. Therefore, enthalpy changes
are a state function and an extensive property.
19.
The zero point for ∆H of values are elements in their standard state. All substances are measured in relationship to this zero point.
20.
No matter how insulated your thermos bottle, some heat will always escape into the
surroundings. If the temperature of the thermos bottle (the surroundings) is high, less heat
initially will escape from the coffee (the system); this results in your coffee staying hotter for
a longer period of time.
21.
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.
22.
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
23.
KE =
1 kg m 2
1
mv2; convert mass and velocity to SI units. 1 J =
2
s2
184
CHAPTER 6
Mass = 5.25 oz ×
Velocity =
THERMOCHEMISTRY
1 lb
1 kg
×
= 0.149 kg
16 oz 2.205 lb
1.0 × 10 2 mi
1h
1 min 1760 yd
1m
45 m
×
×
×
×
=
h
60 min
60 s
mi
1.094 yd
s
1
1
⎛ 45 m ⎞
× 0.149 kg × ⎜
KE = mv2 =
⎟
2
2
⎝ s ⎠
2
= 150 J
2
24.
1
1 ⎛
1 kg ⎞ ⎛⎜ 2.0 × 105 cm
1 m ⎞⎟
⎟⎟ ×
KE = mv2 = × ⎜⎜1.0 × 10 −5 g ×
= 2.0 × 10 −2 J
×
2
2 ⎝
1000 g ⎠ ⎜⎝
s
100 cm ⎟⎠
25.
1
1
⎛ 1 .0 m ⎞
KE = mv2 = × 2.0 kg × ⎜
⎟
2
2
⎝ s ⎠
2
2
1
1
⎛ 2.0 m ⎞
× 1.0 kg × ⎜
= 1.0 J; KE = mv2 =
⎟
2
2
⎝ s ⎠
= 2.0 J
The 1.0-kg object with a velocity of 2.0 m/s has the greater kinetic energy.
26.
Ball A: PE = mgz = 2.00 kg ×
9.81 m
196 kg m 2
×
10.0
m
=
= 196 J
s2
s2
At point I: All 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 ×
9.81 m
× 3.00 m = 118 J
s2
KE = Etotal − PE = 196 J − 118 J = 78 J
Heat and Work
27.
∆E = q + w = 45 kJ + (−29 kJ) = 16 kJ
28.
∆E = q + w = −125 + 104 = −21 kJ
29
a. ∆E = q + w = !47 kJ + 88 kJ = 41 kJ
b. ∆E = 82 ! 47 = 35 kJ
c. ∆E = 47 + 0 = 47 kJ
d. When the surroundings do work on the system, w > 0. This is the case for a.
30.
Step 1: ∆E1 = q + w = 72 J + 35 J = 107 J; step 2: ∆E2 = 35 J − 72 J = !37 J
∆Eoverall = ∆E1 + ∆E2 = 107 J − 37 J = 70. J
CHAPTER
31.
6
THERMOCHEMISTRY
185
∆E = 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
32.
a. ∆E = q + w = !23 J + 100. J = 77 J
b. w = !P∆V = !1.90 atm(2.80 L ! 8.30 L) = 10.5 L atm ×
101.3 J
= 1060 J
L atm
∆E = q + w = 350. J + 1060 = 1410 J
c. w = !P∆V = !1.00 atm(29.1 L ! 11.2 L) = !17.9 L atm ×
101.3 J
= !1810 J
L atm
∆E = q + w = 1037 J ! 1810 J = !770 J
33.
w = !P∆V; we need the final volume of the gas. Because T and n are constant, P1V1 = P2V2.
V2 =
V1 P1
P2
=
10.0 L(15.0 atm)
= 75.0 L
2.00 atm
w = !P∆V = !2.00 atm(75.0 L ! 10.0 L) = !130. L atm ×
101.3 J
1 kJ
×
L atm
1000 J
= !13.2 kJ = work
34.
w = !210. J = !P∆V, !210 J = !P(25 L ! 10. L), P = 14 atm
35.
In this problem, q = w = !950. J.
1 L atm
= !9.38 L atm of work done by the gases
101.3 J
− 650.
atm × (Vf ! 0.040 L), Vf ! 0.040 = 11.0 L, Vf = 11.0 L
w = !P∆V, !9.38 L atm =
760
!950. J ×
36.
∆E = q + w, !102.5 J = 52.5 J + w, w = !155.0 J ×
1 L atm
= !1.530 L atm
101.3 J
w = !P∆V, !1.530 L atm = !0.500 atm × ∆V, ∆V = 3.06 L
∆V = Vf – Vi, 3.06 L = 58.0 L ! Vi, Vi = 54.9 L = initial volume
37.
q = molar heat capacity × mol × ∆T =
20.8 J
o
C mol
× 39.1 mol × (38.0 ! 0.0)°C = 30,900 J
= 30.9 kJ
w = !P∆V = !1.00 atm × (998 L ! 876 L) = !122 L atm ×
∆E = q + w = 30.9 kJ + (!12.4 kJ) = 18.5 kJ
101.3 J
= !12,400 J = !12.4 kJ
L atm
186
38.
CHAPTER 6
THERMOCHEMISTRY
H2O(g) → H2O(l); ∆E = q + w; q = !40.66 kJ; w = !P∆V
Volume of 1 mol H2O(l) = 1.000 mol H2O(l) ×
18.02 g
1 cm 3
×
= 18.1 cm3 = 18.1 mL
mol
0.996 g
w = !P∆V = !1.00 atm × (0.0181 L ! 30.6 L) = 30.6 L atm ×
101.3 J
= 3.10 × 103 J
L atm
= 3.10 kJ
∆E = q + w = !40.66 kJ + 3.10 kJ = !37.56 kJ
Properties of Enthalpy
39.
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.
40.
One should try to cool the reaction mixture or provide some means of removing heat because
the reaction is very exothermic (heat is released). The H2SO4(aq) will get very hot and
possibly boil unless cooling is provided.
41.
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.
42.
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.
43.
4 Fe(s) + 3 O2(g) → 2 Fe2O3(s) ∆H = -1652 kJ; note that 1652 kJ of heat is released when
4 mol Fe reacts with 3 mol O2 to produce 2 mol Fe2O3.
a. 4.00 mol Fe ×
− 1652 kJ
= !1650 kJ; 1650 kJ of heat released
4 mol Fe
b. 1.00 mol Fe2O3 ×
− 1652 kJ
= !826 kJ; 826 kJ of heat released
2 mol Fe 2 O 3
CHAPTER
6
THERMOCHEMISTRY
187
c. 1.00 g Fe ×
− 1652 kJ
1 mol Fe
×
= !7.39 kJ; 7.39 kJ of heat released
55.85 g
4 mol Fe
d. 10.0 g Fe ×
1 mol O 2
1 mol Fe
= 0.179 mol Fe; 2.00 g O2 ×
= 0.0625 mol O2
55.85 g
32.00 g
0.179 mol Fe/0.0625 mol O2 = 2.86; the balanced equation requires a 4 mol Fe/3 mol O2
= 1.33 mole ratio. O2 is limiting since the actual mole Fe/mole O2 ratio is greater than the
required mole ratio.
0.0625 mol O2 ×
44.
a. 1.00 mol H2O ×
− 1652 kJ
= !34.4 kJ; 34.4 kJ of heat released
3 mol O 2
− 572 kJ
= !286 kJ; 286 kJ of heat released
2 mol H 2 O
b. 4.03 g H2 ×
− 572 kJ
1 mol H 2
×
= !572 kJ; 572 kJ of heat released
2.016 g H 2
2 mol H 2
c. 186 g O2 ×
1 mol O 2
− 572 kJ
×
= !3320 kJ; 3320 kJ of heat released
32.00 g O 2
mol O 2
d.
n H2 =
1.0 atm × 2.0 × 108 L
PV
=
= 8.2 × 106 mol H2
0.08206 L atm
RT
× 298 K
K mol
8.2 × 106 mol H2 ×
45.
− 572 kJ
= !2.3 × 109 kJ; 2. 3 × 109 kJ of heat released
2 mol H 2 O
From Example 6.3, q = 1.3 × 108 J. Because the heat transfer process is only 60.%
100. J
efficient, the total energy required is 1.3 × 108 J ×
= 2.2 × 108 J.
60. J
Mass C3H8 = 2.2 × 108 J ×
46.
a. 1.00 g CH4 ×
1 mol C3 H 8
3
2221 × 10 J
×
44.09 g C3H 8
= 4.4 × 103 g C3H8
mol C3 H 8
1 mol CH 4
− 891 kJ
×
= !55.5 kJ
16.04 g CH 4
mol CH 4
740.
atm × 1.00 × 103 L
PV
= 39.8 mol CH4
= 760
b. n =
0.08206 L atm
RT
× 298 K
K mol
39.8 mol ×
− 891 kJ
= !3.55 × 104 kJ
mol
188
CHAPTER 6
THERMOCHEMISTRY
47.
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.
48.
∆H = ∆E + P∆V; 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
49.
Specific heat capacity is defined as the amount of heat necessary to raise the temperature of 1
gram of substance by 1 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:
4.18 J
energy = s × m × ∆T =
o
× 25.0 g × (37.0°C !15.0°C) = 2.30 × 103 J
Cg
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:
1000 J
10.7 kJ ×
energy
kJ = 140°C
=
∆T =
0.14 J
s×m
× 550. g
o
Cg
50.
a. s = specific heat capacity =
Energy = s × m × ∆T =
b. Molar heat capacity =
c. 1250 J =
0.24 J
o
Cg
0.24 J
o
Cg
0.24 J
o
Cg
0.24 J
o
Cg
=
0.24 J
since ∆T(K) = ∆T(°C).
Kg
× 150.0 g × (298 K - 273 K) = 9.0 × 102 J
×
107.9 g Ag
=
mol Ag
26 J
o
× m × (15.2°C ! 12.0°C), m =
C mol
1250
= 1.6 × 103 g Ag
0.24 × 3.2
CHAPTER
51.
6
THERMOCHEMISTRY
s = specific heat capacity =
189
q
133 J
=
= 0.890 J/ECCg
m × ∆T
5.00 g × (55.1 − 25.2) o C
From Table 6.1, the substance is solid aluminum.
52.
s=
585 J
125.6 g × (53.5 − 20.0) o C
Molar heat capacity =
53.
0.139 J
o
Cg
= 0.139 J/°C C g
×
200.6 g
27.9 J
= o
mol Hg
C mol
| Heat loss by hot water | = | heat gain by cooler water |
The magnitudes 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/°CCg = 4.18 J/KCg (∆T in °C = ∆T in K).
Heat loss by hot water = s × m × ∆T =
Heat gain by cooler water =
4.18 J
× 50.0 g × (330. K ! Tf)
Kg
4.18 J
× 30.0 g × (Tf !280. K); heat loss = heat gain, so:
Kg
209 J
125 J
× (330. K − Tf) =
× (Tf − 280. K)
K
K
6.90 × 104 ! 209Tf = 125Tf − 3.50 × 104, 334Tf = 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.
54.
Heat gained by water = heat lost by nickel = s × m × ∆T, where s = specific heat capacity.
Heat gain =
4.18 J
o
Cg
× 150.0 g × (25.0°C − 23.5°C) = 940 J
A common error in calorimetry problems is sign errors. Keeping all quantities positive helps
to eliminate sign errors.
Heat loss = 940 J =
55.
0.444 J
o
Cg
× mass × (99.8 − 25.0) °C, mass =
940
= 28 g
0.444 × 74.8
Heat loss by Al + heat loss by Fe = heat gain by water; keeping all quantities positive to
avoid sign error:
190
CHAPTER 6
THERMOCHEMISTRY
0.89 J
0.45 J
× 5.00 g Al × (100.0°C − Tf) + o
× 10.00 g Fe × (100.0 − Tf)
o
Cg
Cg
4.18 J
= o
× 97.3 g H2O × (Tf − 22.0°C)
Cg
4.5(100.0 − Tf) + 4.5(100.0 − Tf) = 407(Tf − 22.0), 450 − (4.5)Tf + 450 − (4.5)Tf
= 407Tf − 8950
416Tf = 9850, Tf = 23.7°C
56.
Heat released to water = 5.0 g H2 ×
Heat gain by water = 1.10 × 103 J =
120. J
50. J
+ 10. g methane ×
= 1.10 × 103 J
g H2
g methane
4.18 J
o
Cg
× 50.0 g × ∆T
∆T = 5.26°C, 5.26°C = Tf ! 25.0°C, Tf = 30.3°C
57.
Heat gain by water = heat loss by metal = s × m × ∆T, where s = specific heat capacity.
Heat gain =
4.18 J
o
Cg
× 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
to eliminate sign errors.
Heat loss = 2100 J = s × 150.0 g × (75.0°C - 18.3°C), s =
58.
2100 J
150.0 g × 56.7 o C
= 0.25 J/°CCg
Heat gain by water = heat loss by Cu; keeping all quantities positive helps to avoid sign
errors:
4.18 J
o
Cg
× mass × (24.9°C ! 22.3°C) =
0.20 J
o
Cg
× 110. g Cu × (82.4°C ! 24.9°C)
11(mass) = 1300, mass = 120 g H2O
59.
50.0 × 10−3 L × 0.100 mol/L = 5.00 × 10−3 mol of both AgNO3 and HCl are reacted. Thus
5.00 × 10−3 mol of AgCl will be produced because there is a 1 : 1 mole ratio between
reactants.
Heat lost by chemicals = heat gained by solution
Heat gain =
4.18 J
× 100.0 g × (23.40 − 22.60)°C = 330 J
o
Cg
CHAPTER
6
THERMOCHEMISTRY
191
Heat loss = 330 J; this is the heat evolved (exothermic reaction) when 5.00 × 10−3 mol of
AgCl is produced. So q = −330 J and ∆H (heat per mol AgCl formed) is negative with a
value of:
∆H =
− 330 J
1 kJ
×
= −66 kJ/mol
−3
5.00 × 10 mol 1000 J
Note: Sign errors are common with calorimetry problems. However, the correct sign for ∆H
can be determined easily from the ∆T data; i.e., if ∆T of the solution increases, then the
reaction is exothermic because heat was released, and if ∆T of the solution decreases, then
the reaction is endothermic because the reaction absorbed heat from the water. For
calorimetry problems, keep all quantities positive until the end of the calculation and then
decide the sign for ∆H. This will help to eliminate errors.
60.
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
We have a stoichiometric mixture. All of the NaOH and HCl will react.
0.10 L ×
1.0 mol
= 0.10 mol of HCl is neutralized by 0.10 mol NaOH.
L
Heat lost by chemicals = heat gained by solution
Volume of solution = 100.0 + 100.0 = 200.0 mL
Heat gain =
1.0 g ⎞
⎛
3
× ⎜ 200.0 mL ×
⎟ × (31.3 – 24.6)°C = 5.6 × 10 J = 5.6 kJ
mL ⎠
Cg ⎝
4.18 J
o
Heat loss = 5.6 kJ; this is the heat released by the neutralization of 0.10 mol HCl. Because
the temperature increased, the sign for ∆H must be negative, i.e., the reaction is exothermic.
For calorimetry problems, keep all quantities positive until the end of the calculation and then
decide the sign for ∆H.
∆H =
61.
− 5.6 kJ
= −56 kJ/mol
0.10 mol
Heat lost by solution = heat gained by KBr; mass of solution = 125 g + 10.5 g = 136 g
Note: Sign errors are common with calorimetry problems. However, the correct
∆H can easily be obtained from the ∆T data. When working calorimetry problems,
quantities positive (ignore signs). When finished, deduce the correct sign for ∆H.
problem, T decreases as KBr dissolves, so ∆H is positive; the dissolution of
endothermic (absorbs heat).
Heat lost by solution =
sign for
keep all
For this
KBr is
4.18 J
× 136 g × (24.2°C − 21.1°C) = 1800 J = heat gained by KBr
o
Cg
192
CHAPTER 6
∆H in units of J/g =
1800 J
= 170 J/g
10.5 g KBr
∆H in units of kJ/mol =
62.
THERMOCHEMISTRY
170 J
119.0 g KBr
1 kJ
×
×
= 20. kJ/mol
g KBr
mol KBr
1000 J
NH4NO3(s) → NH4+(aq) + NO3−(aq) ∆H = ?; 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 ∆T) and then deduce the correct sign for ∆H at the end
of the problem. Here, because temperature decreases as NH4NO3 dissolves, heat is absorbed
as NH4NO3 dissolves, so this is an endothermic process (∆H is positive).
Heat lost by solution =
4.18 J
× 76.6 g × (25.00 − 23.34)°C = 532 J = heat gained as
o
Cg
NH NO dissolves
4
∆H =
63.
80.05 g NH 4 NO 3
532 J
1 kJ
×
×
= 26.6 kJ/mol NH4NO3 dissolving
1.60 g NH 4 NO 3
mol NH 4 NO 3
1000 J
Because ∆H 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 ×
heat gained by solution = 8.08 × 103 J =
Tf − 25.0°C =
64.
3
1 mol CaCl 2
81.5 kJ
×
= 8.08 kJ
110.98 g CaCl 2
mol CaCl 2
4.18 J
× (125 + 11.0) g × (Tf − 25.0°C)
o
Cg
8.08 × 103
= 14.2°C, Tf = 14.2°C + 25.0°C = 39.2°C
4.18 × 136
0.100 L ×
0.500 mol HCl
= 5.00 × 10−2 mol HCl
L
0.300 L ×
0.100 mol Ba (OH ) 2
= 3.00 × 10−2 mol Ba(OH)2
L
To react with all the HCl present, 5.00 × 10−2/2 = 2.50 × 10−2 mol Ba(OH)2 is required.
Because 0.0300 mol Ba(OH)2 is present, HCl is the limiting reactant.
118 kJ
= 2.95 kJ of heat is evolved by reaction
2 mol HCl
4.18 J
Heat gained by solution = 2.95 × 103 J = o
× 400.0 g × ∆T
Cg
5.00 × 10−2 mol HCl ×
∆T = 1.76°C = Tf − Ti = Tf − 25.0°C, Tf = 26.8°C
CHAPTER
65.
6
THERMOCHEMISTRY
193
a. Heat gain by calorimeter = heat loss by CH4 = 6.79 g CH4 ×
Heat capacity of calorimeter =
340. kJ
= 31.5 kJ/°C
10.8 o C
b. Heat loss by C2H2 = heat gain by calorimeter = 16.9°C ×
∆Ecomb =
1 mol CH 4 802 kJ
×
16.04 g
mol
= 340. kJ
31. 5 kJ
= 532 kJ
o
C
− 532 kJ
26.04 g
= !1.10 × 103 kJ/mol
×
12.6 g C 2 H 2
mol C 2 H 2
Note: Because bomb calorimeters are at constant volume, qV = ∆E.
66.
First, we need to get the heat capacity of the calorimeter from the combustion of benzoic
acid. Heat lost by combustion = heat gained by calorimeter.
26.42 kJ
= 4.185 kJ
g
4.185 kJ
= 1.65 kJ/°C
Heat gain = 4.185 kJ = Ccal × ∆T, Ccal =
2.54 o C
Heat loss = 0.1584 g ×
Now we can calculate the heat of combustion of vanillin. Heat loss = heat gain.
Heat gain by calorimeter =
1.65 kJ
o
C
× 3.25°C = 5.36 kJ
Heat loss = 5.36 kJ, which is the heat evolved by combustion of the vanillin.
∆Ecomb =
− 25.2 kJ 152.14 g
− 5.36 kJ
= −25.2 kJ/g; ∆Ecomb =
×
= −3830 kJ/mol
0.2130 g
g
mol
Hess's Law
67.
Information given:
C(s) + O2(g) → CO2(g)
CO(g) + 1/2 O2(g) → CO2(g)
∆H = !393.7 kJ
∆H = !283.3 kJ
Using Hess’s law:
2 C(s) + 2 O2(g) → 2 CO2(g)
∆H1 = 2(!393.7 kJ)
2 CO2(g) → 2 CO(g) + O2(g)
∆H2 = !2(!283.3 kJ)
__________________________________________________________
∆H = ∆H1 + ∆H2 = !220.8 kJ
2 C(s) + O2(g) → 2 CO(g)
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, then the value of ∆H is multiplied by the same integer.
194
68.
CHAPTER 6
C4H4(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l)
C4H8(g) + 6 O2(g) → 4 CO2(g) + 4 H2O(l)
H2(g) + 1/2 O2(g) → H2O(l)
THERMOCHEMISTRY
∆Hcomb = !2341 kJ
∆Hcomb = !2755 kJ
∆Hcomb = !286 kJ
By convention, H2O(l) is produced when enthalpies of combustion are given, and because
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:
∆H1 = !2341 kJ
C4H4(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l)
∆H2 = ! (!2755 kJ)
4 CO2(g) + 4 H2O(l) → C4H8(g) + 6 O2(g)
2 H2(g) + O2(g) → 2 H2O(l)
∆H3 = 2(!286 kJ)
_______________________________________________________________________
C4H4(g) + 2 H2(g) → C4H8(g)
∆H = ∆H1 + ∆H2 + ∆H3 = !158 kJ
69.
2 N2(g) + 6 H2(g) → 4 NH3(g)
∆H = !4(46 kJ)
∆H = !3(-484 kJ)
6 H2O(g) → 6 H2(g) + 3 O2(g)
___________________________________________________
2 N2(g) + 6 H2O(g) → 3 O2(g) + 4 NH3(g) ∆H = 1268 kJ
No, because the reaction is very endothermic (requires a lot of heat to react), it would not be a
practical way of making ammonia because of the high energy costs required.
70.
∆H = 1/2(167.4 kJ)
ClF + 1/2 O2 → 1/2 Cl2O + 1/2 F2O
1/2 Cl2O + 3/2 F2O → ClF3 + O2
∆H = !1/2(341.4 kJ)
F2 + 1/2 O2 → F2O
∆H = 1/2(!43.4 kJ)
__________________________________________________________
∆H = !108.7 kJ
ClF(g) + F2(g) → ClF3
71.
∆H = !199 kJ
NO + O3 → NO2 + O2
∆H = !1/2(-427 kJ)
3/2 O2 → O3
∆H = !1/2(495 kJ)
O → 1/2 O2
________________________________________________
∆H = !233 kJ
NO(g) + O(g) → NO2(g)
72.
We want ∆H for N2H4(l) + O2(g) → N2(g) + 2 H2O(l). It will be easier to calculate ∆H for
the combustion of four moles of N2H4 because we will avoid fractions.
∆H = 9(−286 kJ)
9 H2 + 9/2 O2 → 9 H2O
∆H = −3(−317 kJ)
3 N2H4 + 3 H2O → 3 N2O + 9 H2
∆H = −1010. kJ
2 NH3 + 3 N2O → 4 N2 + 3 H2O
∆H = − (−143 kJ)
N2H4 + H2O → 2 NH3 + 1/2 O2
_____________________________________________________
∆H = −2490. kJ
4 N2H4(l) + 4 O2(g) → 4 N2(g) + 8 H2O(l)
For N2H4(l) + O2(g) → N2(g) + 2 H2O(l)
∆H =
− 2490. kJ
= −623 kJ
4
CHAPTER
6
THERMOCHEMISTRY
195
Note: By the significant figure rules, we could report this answer to four significant figures.
However, because the ∆H values given in the problem are only known to ±1 kJ, our final
answer will at best be ±1 kJ.
CaC2 →
CaO + H2O →
2 CO2 + H2O →
Ca + 1/2 O2 →
2 C + 2 O2 →
73.
Ca + 2 C
Ca(OH)2
C2H2 + 5/2 O2
CaO
2 CO2
∆H = !(!62.8 kJ)
∆H = !653.1 kJ
∆H = !(!1300. kJ)
∆H = !635.5 kJ
∆H = 2(!393.5 kJ)
_______________________________________________________________________________________________________
CaC2(s) + 2 H2O(l) → Ca(OH)2(aq) + C2H2(g)
P4O10 → P4 + 5 O2
10 PCl3 + 5 O2 → 10 Cl3PO
6 PCl5 → 6 PCl3 + 6 Cl2
P4 + 6 Cl2 → 4 PCl3
74.
∆H = !713 kJ
∆H = !(!2967.3 kJ)
∆H = 10(!285.7 kJ)
∆H = !6(!84.2 kJ)
∆H = !1225.6
__________________________________________________________________________________________
P4O10(s) + 6 PCl5(g) → 10 Cl3PO(g)
∆H = !610.1 kJ
Standard Enthalpies of Formation
75.
The change in enthalpy that accompanies the formation of 1 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 ∆H of 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)
76.
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: ∆Hcomb values assume 1 mole of compound combusted.
f.
77.
NH4Br(s) → NH4+(aq) + Br-(aq)
In general, ∆H° = 3np ∆H of , products ! 3nr ∆H of , reactants , and all elements in their standard
state have ∆H of = 0 by definition.
a. The balanced equation is 2 NH3(g) + 3 O2(g) + 2 CH4(g) → 2 HCN(g) + 6 H2O(g).
196
CHAPTER 6
THERMOCHEMISTRY
∆H° = (2 mol HCN × ∆H of , HCN + 6 mol H2O(g) × ∆H of , H 2O )
! (2 mol NH3 × ∆H of , NH 3 + 2 mol CH4 × ∆H of , CH )
4
∆H° = [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)
⎡
⎛ − 1267 kJ ⎞ ⎤
⎛ − 1433 kJ ⎞
∆H° = ⎢3 mol CaSO 4 (s)⎜
⎟⎥
⎟ + 2 mol H 3PO 4 (l)⎜
⎝ mol ⎠ ⎦
⎝ mol ⎠
⎣
⎡
⎛ − 4126 kJ ⎞
⎛ − 814 kJ ⎞ ⎤
! ⎢1 mol Ca 3 (PO 4 ) 2 (s)⎜
⎟⎥
⎟ + 3 mol H 2SO 4 (l)⎜
⎝ mol ⎠
⎝ mol ⎠ ⎦
⎣
∆H° = !6833 kJ ! (!6568 kJ) = !265 kJ
c. NH3(g) + HCl(g) → NH4Cl(s)
∆H° = (1 mol NH4Cl × ∆H of , NH 4Cl ) ! (1 mol NH3 × ∆H of , NH 3 + 1 mol HCl × ∆H of , HCl )
⎡
⎛ − 314 kJ ⎞⎤ ⎡
⎛ − 46 kJ ⎞
⎛ − 92 kJ ⎞⎤
∆H° = ⎢1 mol ⎜
⎟⎥ − ⎢1 mol ⎜
⎟⎥
⎟ + 1 mol ⎜
⎝ mol ⎠⎦ ⎣
⎝ mol ⎠
⎝ mol ⎠⎦
⎣
∆H° = !314 kJ + 138 kJ = !176 kJ
78.
a. The balanced equation is C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g).
⎡
⎛ − 278 kJ ⎞⎤
⎛ − 393.5 kJ ⎞
⎛ − 242 kJ ⎞⎤ ⎡
∆H° = ⎢2 mol ⎜
⎟⎥
⎟ + 3 mol ⎜
⎟⎥ − ⎢1 mol ⎜
⎝ mol ⎠⎦
⎝ mol ⎠
⎝ mol ⎠⎦ ⎣
⎣
∆H° = !1513 kJ ! (!278 kJ) = !1235 kJ
b. SiCl4(l) + 2 H2O(l) → SiO2(s) + 4 HCl(aq)
Because HCl(aq) is H+(aq) + Cl−(aq), ∆H of = 0 ! 167 = !167 kJ/mol.
⎡
⎡
⎛ − 687 kJ ⎞
⎛ − 286 kJ ⎞⎤
⎛ − 167 kJ ⎞
⎛ − 911 kJ ⎞⎤
∆H° = ⎢4 mol ⎜
⎟ + 2 mol ⎜
⎟⎥
⎟ + 1 mol ⎜
⎟⎥ − ⎢1 mol ⎜
⎝ mol ⎠
⎝ mol ⎠⎦
⎝ mol ⎠
⎝ mol ⎠⎦
⎣
⎣
∆H° = !1579 kJ ! (!1259 kJ) = !320. kJ
c. MgO(s) + H2O(l) → Mg(OH)2(s)
⎡
⎛ − 602 kJ ⎞
⎛ − 286 kJ ⎞⎤
⎛ − 925 kJ ⎞⎤ ⎡
∆H° = ⎢1 mol ⎜
⎟ + 1 mol⎜
⎟⎥
⎟⎥ − ⎢1 mol ⎜
⎝ mol ⎠⎦
⎝ mol ⎠
⎝ mol ⎠⎦ ⎣
⎣
∆H° = !925 kJ ! (!888 kJ) = !37 kJ
CHAPTER
79.
6
THERMOCHEMISTRY
197
a. 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g); ∆H° = 3np ∆H of , products ! 3nr ∆H of , reactants
⎡
⎛ − 46 kJ ⎞⎤
⎛ 90. kJ ⎞
⎛ − 242 kJ ⎞⎤ ⎡
∆H° = ⎢4 mol ⎜
⎟⎥ = !908 kJ
⎟ + 6 mol ⎜
⎟⎥ − ⎢4 mol ⎜
⎝ mol ⎠⎦
⎝ mol ⎠
⎝ mol ⎠⎦ ⎣
⎣
2 NO(g) + O2(g) → 2 NO2(g)
⎡
⎛ 90. kJ ⎞ ⎤
⎛ 34 kJ ⎞⎤ ⎡
∆H° = ⎢2 mol ⎜
⎟ ⎥ = !112 kJ
⎟⎥ − ⎢2 mol ⎜
⎝ mol ⎠ ⎦
⎝ mol ⎠⎦ ⎣
⎣
3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g)
⎡
⎛ − 286 kJ ⎞⎤
⎛ 34 kJ ⎞
⎛ 90. kJ ⎞ ⎤ ⎡
⎛ − 207 kJ ⎞
∆H° = ⎢2 mol ⎜
⎟⎥
⎟ + 1 mol ⎜
⎟ ⎥ − ⎢3 mol ⎜
⎟ + 1 mol ⎜
⎝ mol ⎠⎦
⎝ mol ⎠
⎝ mol ⎠ ⎦ ⎣
⎝ mol ⎠
⎣
= !140. kJ
Note: All
∆H of
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 because each step is exothermic.
80.
4 Na(s) + O2(g) → 2 Na2O(s)
⎛ − 416 kJ ⎞
∆H° = 2 mol ⎜
⎟ = −832 kJ
⎝ mol ⎠
2 Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g)
⎡
⎛ − 286 kJ ⎞⎤
⎛ − 470. kJ ⎞ ⎤ ⎡
∆H° = ⎢2 mol ⎜
⎟⎥ = −368 kJ
⎟ ⎥ − ⎢2 mol ⎜
⎝ mol ⎠⎦
⎝ mol ⎠ ⎦ ⎣
⎣
2Na(s) + CO2(g) → Na2O(s) + CO(g)
⎡
⎛ − 393.5 kJ ⎞⎤
⎛ − 416 kJ ⎞
⎛ − 110.5 kJ ⎞⎤ ⎡
∆H° = ⎢1 mol ⎜
⎟⎥ = −133 kJ
⎟ + 1 mol ⎜
⎟⎥ − ⎢1 mol ⎜
⎝ mol ⎠⎦
⎝ mol ⎠
⎝ mol ⎠⎦ ⎣
⎣
In Reactions 2 and 3, sodium metal reacts with the "extinguishing agent." Both reactions are
exothermic, and each reaction produces a flammable gas, H2 and CO, respectively.
81.
3 Al(s) + 3 NH4ClO4(s) → Al2O3(s) + AlCl3(s) + 3 NO(g) + 6 H2O(g)
198
CHAPTER 6
THERMOCHEMISTRY
⎡
⎛ 90. kJ ⎞
⎛ − 704 kJ ⎞
⎛ − 1676 kJ ⎞⎤
⎛ − 242 kJ ⎞
∆H° = ⎢6 mol ⎜
⎟⎥
⎟ + 3 mol ⎜
⎟ + 1 mol ⎜
⎟ + 1 mol ⎜
⎝ mol ⎠
⎝ mol ⎠
⎝ mol ⎠
⎝ mol ⎠⎦
⎣
⎡
⎛ − 295 kJ ⎞⎤
− ⎢3 mol ⎜
⎟⎥ = !2677 kJ
⎝ mol ⎠⎦
⎣
82.
5 N2O4(l) + 4 N2H3CH3(l) → 12 H2O(g) + 9 N2(g) + 4 CO2(g)
⎡
⎛ − 242 kJ ⎞
⎛ − 393.5 kJ ⎞ ⎤
∆H° = ⎢12 mol ⎜
⎟ + 4 mol ⎜
⎟⎥
⎝ mol ⎠
⎝ mol ⎠ ⎦
⎣
⎡
⎛ − 20. kJ ⎞
⎛ 54 kJ ⎞⎤
− ⎢5 mol ⎜
⎟ + 4 mol ⎜
⎟⎥ = !4594 kJ
⎝ mol ⎠
⎝ mol ⎠⎦
⎣
83.
2 ClF3(g) + 2 NH3(g) → N2(g) + 6 HF(g) + Cl2(g)
∆H° = (6 ∆H of , HF ) − (2 ∆H of ,
ClF3
+ 2∆H of ,
∆H° = −1196 kJ
NH 3 )
⎛ − 271 kJ ⎞
⎛ − 46 kJ ⎞
o
−1196 kJ = 6 mol ⎜
⎟ − 2 ∆H f , ClF3 − 2 mol ⎜
⎟
mol
⎝
⎠
⎝ mol ⎠
−1196 kJ = −1626 kJ − 2 ∆H fo, ClF3 + 92 kJ, ∆H fo, ClF3 =
84.
C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(l)
(−1626 + 92 + 1196) kJ − 169 kJ
=
mol
2 mol
∆H° = !1411.1 kJ
∆H° = !1411.1 kJ = 2(!393.5) kJ + 2(!285.8) kJ ! ∆H of , C 2 H 4
!1411.1 kJ = !1358.6 kJ ! ∆H of , C 2 H 4 , ∆H of , C 2 H 4 = 52.5 kJ/mol
Energy Consumption and Sources
85.
C(s) + H2O(g) → H2(g) + CO(g) ∆H° = −110.5 kJ − (−242 kJ) = 132 kJ
86.
CO(g) + 2 H2(g) → CH3OH(l)
87.
C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)
∆H° = −239 kJ − (−110.5 kJ) = −129 kJ
∆H° = [2(!393.5 kJ) + 3(!286 kJ)] ! (!278 kJ) = !1367 kJ/mol ethanol
1 mol
− 1367 kJ
×
= !29.67 kJ/g
mol
46.07 g
88.
CH3OH(l) + 3/2 O2(g) → CO2(g) + 2 H2O(l)
∆H° = [!393.5 kJ + 2(!286 kJ)] ! (!239 kJ) = !727 kJ/mol CH3OH
CHAPTER
6
THERMOCHEMISTRY
199
1 mol
− 727 kJ
×
= !22.7 kJ/g versus !29.67 kJ/g for ethanol
mol
32.04 g
Ethanol has a slightly higher fuel value than methanol.
89.
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)
∆H° = [3(!393.5 kJ) + 4(!286 kJ)] ! (!104 kJ) = !2221 kJ/mol C3H8
− 50.37 kJ
− 2221 kJ
1 mol
×
=
versus !47.7 kJ/g for octane (Example 6.11)
mol
mol
44.09 g
The fuel values are very 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.
90.
1 mole of C2H2(g) and 1 mole of C4H10(g) have equivalent volumes at the same T and P.
enthalpy of combustion per mol of C 2 H 2
Enthalpy of combustion per volume of C 2 H 2
=
Enthalpy of combustion per volume of C 4 H10
enthalpy of combustion per mol of C 4 H 10
Enthalpy of combustion per volume of C 2 H 2
=
Enthalpy of combustion per volume of C 4 H10
− 49.9 kJ 26.04 g C 2 H 2
×
g C2H 2
mol C 2 H 2
= 0.452
− 49.5 kJ 58.12 g C 4 H10
×
g C 4 H10
mol C 4 H10
More than twice the volume of acetylene is needed to furnish the same energy as a given
volume of butane.
91.
The molar volume of a gas at STP is 22.42 L (from Chapter 5).
4.19 × 106 kJ ×
92.
1 mol CH 4
22.42 L CH 4
×
= 1.05 × 105 L CH4
891 kJ
mol CH 4
3.785 L 1000 mL 1.00 g
×
×
= 3790 g H2O
gal
L
mL
4.18 J
× 3790 g × 10.0 °C = 1.58 × 105 J
Energy required (theoretical) = s × m × ∆T = o
Cg
Mass of H2O = 1.00 gal ×
For an 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 ×
100 . J
= 1.98 × 105 J
80 .0 J
Mass of C2H2 = 1.98 × 105 J ×
1 mol C 2 H 2
26.04 g C 2 H 2
×
= 3.97 g C2H2
3
mol C 2 H 2
1300. × 10 J
200
CHAPTER 6
THERMOCHEMISTRY
Connecting to Biochemistry
93.
Because the sign of ∆H is negative, the reaction is exothermic. Heat is evolved by the system
to the surroundings.
94.
From the problem, walking 4.0 miles consumes 400 kcal of energy.
1 lb fat ×
454 g
7.7 kcal
4 mi
1h
×
×
×
= 8.7 h = 9 h
lb
g
400 kcal 4 mi
5500 kJ
1 mol H 2 O
18.02 g H 2 O
×
×
= 4900 g = 4.9 kg H2O
h
40.6 kJ
mol
95.
2.0 h ×
96.
Heat loss by hot water = heat gain by cold water; keeping all quantities positive helps to
avoid sign errors:
4.18 J
o
Cg
mhot =
97.
× mhot × (55.0°C ! 37.0°C) =
90.0 g × 15.0 o C
o
Cg
× 90.0 g × (37.0°C ! 22.0°C)
= 75.0 g hot water needed
1.56 kJ
× 3.2°C = 5.0 kJ = heat loss by quinine
o
C
Heat loss = 5.0 kJ, which is the heat evolved (exothermic reaction) by the combustion of
0.1964 g of quinone. Because we are at constant volume, qV = ∆E.
Heat gain by calorimeter =
∆Ecomb =
98.
18.0 o C
4.18 J
− 5.0 kJ
= !25 kJ/g;
0.1964 g
∆Ecomb =
− 25 kJ 108.09 g
×
= !2700 kJ/mol
g
mol
a. C12H22O11(s) + 12 O2(g) → 12 CO2(g) + 11 H2O(l)
b. A bomb calorimeter is at constant volume, so heat released = qV = ∆E:
∆E =
− 24.00 kJ 342.30 g
×
= !5630 kJ/mol C12H22O11
1.46 g
mol
c. PV = nRT; at constant P and T, P∆V = RT∆n, where ∆n = moles of gaseous products !
moles of gaseous reactants.
∆H = ∆E + P∆V = ∆E + RT∆n
For this reaction, ∆n = 12 !12 = 0, so ∆H = ∆E = !5630 kJ/mol.
99.
C6H4(OH)2 → C6H4O2 + H2
H2O2 → H2 + O2
2 H2 + O2 → 2 H2O(g)
2 H2O(g) → 2 H2O(l)
∆H = 177.4 kJ
∆H = ! (!191.2 kJ)
∆H = 2(!241.8 kJ)
∆H = 2(!43.8 kJ)
_____________________________________________________________________________________________________
C6H4(OH)2(aq) + H2O2(aq) → C6H4O2(aq) + 2 H2O(l)
∆H = !202.6 kJ
CHAPTER
100.
6
THERMOCHEMISTRY
201
From the photosynthesis reaction, CO2(g) is used by plants to convert water into glucose and
oxygen. If the plant population is significantly reduced, not as much CO2 will be consumed
in the photosynthesis reaction. As the CO2 levels of the atmosphere increase, the greenhouse
effect due to excess CO2 in the atmosphere will become worse.
Additional Exercises
101.
a. 2 SO2(g) + O2(g) → 2 SO3(g); w = −P∆V; because the volume of the piston apparatus
decreased as reactants were converted to products (∆V < 0), w is positive (w > 0).
b. COCl2(g) → CO(g) + Cl2(g); because the volume increased (∆V > 0), w is negative
(w < 0).
c. N2(g) + O2(g) → 2 NO(g); because the volume did not change (∆V = 0), 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 moles of product gases than moles of reactant gases (as in b), the reaction
will expand in volume (∆V positive), and the system does work on the surroundings. When
a balanced reaction has a decrease in the moles of gas from reactants to products (as in a), the
reaction will contract in volume (∆V negative), and the surroundings will do compression
work on the system. When there is no change in the moles of gas from reactants to products
(as in c), ∆V = 0 and w = 0.
102.
w = −P∆V; ∆n = moles of gaseous products − moles of gaseous reactants. Only gases can do
PV work (we ignore solids and liquids). When a balanced reaction has more moles of
product gases than moles of reactant gases (∆n positive), the reaction will expand in volume
(∆V positive), and the system will do work on the surroundings. For example, in reaction c,
∆n = 2 − 0 = 2 moles, and this reaction would do expansion work against the surroundings.
When a balanced reaction has a decrease in the moles of gas from reactants to products (∆n
negative), the reaction will contract in volume (∆V negative), and the surroundings will do
compression work on the system, e.g., reaction a, where ∆n = 0 − 1 = −1. When there is no
change in the moles of gas from reactants to products, ∆V = 0 and w = 0, e.g., reaction b,
where ∆n = 2 − 2 = 0.
When ∆V > 0 (∆n > 0), then w < 0, and the system does work on the surroundings (c and e).
When ∆V < 0 (∆n < 0), then w > 0, and the surroundings do work on the system (a and d).
When ∆V = 0 (∆n = 0), then w = 0 (b).
103.
∆Eoverall = ∆Estep 1 + ∆Estep 2; this is a cyclic process, which means that the overall initial state
and final state are the same. Because ∆E is a state function, ∆Eoverall = 0 and ∆Estep 1 =
!∆Estep 2.
202
CHAPTER 6
THERMOCHEMISTRY
∆Estep 1 = q + w = 45 J + (!10. J) = 35 J
∆Estep 2 = !∆Estep 1 = !35 J = q + w, !35 J = !60 J + w, w = 25 J
104.
2 K(s) + 2 H2O(l) → 2 KOH(aq) + H2(g)
5.00 g K ×
∆H° = 2(!481 kJ) ! 2(!286 kJ) = !390. kJ
1 mol K
− 390. kJ
×
= !24.9 kJ
39.10 g K
2 mol K
24.9 kJ of heat is released on reaction of 5.00 g K.
24,900 J =
4.18 J
24,900
× (1.00 × 103 g) × ∆T, ∆T =
= 5.96°C
o
g C
4.18 × 1.00 × 103
Final temperature = 24.0 + 5.96 = 30.0°C
105.
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) ∆H = !56 kJ
0.2000 L ×
0.400 mol HCl
= 8.00 × 10 −2 mol HCl
L
0.1500 L ×
0.500 mol NaOH
= 7.50 × 10 −2 mol NaOH
L
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 with HCl, NaOH is the
limiting reagent.
7.50 × 10 −2 mol NaOH ×
106.
− 56 kJ
= !4.2 kJ; 4.2 kJ of heat is released.
mol NaOH
Na2SO4(aq) + Ba(NO3)(aq) → BaSO4(s) + 2 NaNO3(aq)
1.00 L ×
∆H = ?
2.00 mol
0.750 mol
= 2.00 mol Na2SO4; 2.00 L ×
= 1.50 mol Ba(NO3)2
L
L
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
1000 mol 2.00 g
×
= 6.00 × 103 g
1L
mL
6.37 J
× 6.00 × 103 g × (42.0 ! 30.0)EC = 4.59 × 105 J
Heat gain by solution = o
Cg
Mass of solution = 3.00 L ×
CHAPTER
6
THERMOCHEMISTRY
203
Because the solution gained heat, the reaction is exothermic; q = !4.59 × 105 J for the
reaction.
∆H =
− 4.59 × 105 J
= !3.06 × 105 J/mol = !306 kJ/mol
1.50 mol BaSO 4
107.
|qsurr| = |qsolution + qcal|; we normally assume that 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 that is less positive (smaller) than it should be.
108.
The specific heat of water is 4.18 J/°CCg, which is equal to 4.18 kJ/°CCkg.
We have 1.00 kg of H2O, so: 1.00 kg ×
4.18 J
= 4.18 kJ/°C
C kg
This is the portion of the heat capacity that can be attributed to H2O.
o
Total heat capacity = Ccal + C H 2O , Ccal = 10.84 ! 4.18 = 6.66 kJ/°C
109.
Heat released = 1.056 g × 26.42 kJ/g = 27.90 kJ = heat gain by water and calorimeter
⎛ 4.18 J
⎞
⎛ 6.66 kJ
⎞
× 0.987 kg × ∆T ⎟⎟ + ⎜ o
× ∆T ⎟
Heat gain = 27.90 kJ = ⎜⎜ o
C
⎝
⎠
⎝ C kg
⎠
27.90 = (4.13 + 6.66)∆T = (10.79)∆T, ∆T = 2.586°C
2.586°C = Tf ! 23.32°C, Tf = 25.91°C
110.
For Exercise 81, a mixture of 3 mol Al and 3 mol NH4ClO4 yields 2677 kJ of energy. The
mass of the stoichiometric reactant mixture is:
26.98 g ⎞ ⎛
117.49 g ⎞
⎛
⎜ 3 mol ×
⎟ + ⎜ 3 mol ×
⎟ = 433.41 g
mol ⎠ ⎝
mol ⎠
⎝
For 1.000 kg of fuel: 1.000 × 103 g ×
− 2677 kJ
= −6177 kJ
433.41 g
In Exercise 82, we get 4594 kJ of energy from 5 mol of N2O4 and 4 mol of N2H3CH3. The
92.02 g ⎞ ⎛
46.08 g ⎞
⎛
mass is ⎜ 5 mol ×
⎟ + ⎜ 4 mol ×
⎟ = 644.42 kJ.
mol ⎠ ⎝
mol ⎠
⎝
For 1.000 kg of fuel: 1.000 × 103 g ×
− 4594 kJ
= −7129 kJ
644.42 g
Thus we get more energy per kilogram from the N2O4/N2H3CH3 mixture.
204
111.
CHAPTER 6
1/2 D → 1/2 A + B
1/2 E + F → 1/2 A
1/2 C → 1/2 E + 3/2 D
∆H = −1/6(−403 kJ)
∆H = 1/2(−105.2 kJ)
∆H = 1/2(64.8 kJ)
F + 1/2 C → A + B + D
∆H = 47.0 kJ
THERMOCHEMISTRY
___________________________________________________________________________
112.
To avoid fractions, let's first calculate ∆H for the reaction:
6 FeO(s) + 6 CO(g) → 6 Fe(s) + 6 CO2(g)
∆H° = !2(18 kJ)
6 FeO + 2 CO2 → 2 Fe3O4 + 2 CO
∆H° = ! (!39 kJ)
2 Fe3O4 + CO2 → 3 Fe2O3 + CO
3 Fe2O3 + 9 CO → 6 Fe + 9 CO2
∆H° = 3(!23 kJ)
______________________________________________________
∆H° = !66 kJ
6 FeO(s) + 6 CO(g) → 6 Fe(s) + 6 CO2(g)
So for FeO(s) + CO(g) → Fe(s) + CO2(g), ∆H° =
113.
− 66 kJ
= !11 kJ.
6
a. ∆H° = 3 mol(227 kJ/mol) ! 1 mol(49 kJ/mol) = 632 kJ
b. Because 3 C2H2(g) is higher in energy than C6H6(l), acetylene will release more energy
per gram when burned in air.
114.
I(g) + Cl(g) → ICl(g)
∆H = !(211.3 kJ)
∆H = 1/2(242.3 kJ)
1/2 Cl2(g) → Cl(g)
1/2 I2(g) → I(g)
∆H = 1/2(151.0 kJ)
1/2 I2(s) → 1/2 I2(g)
∆H = 1/2(62.8 kJ)
_______________________________________________________________
1/2 I2(s) + 1/2 Cl2(g) → ICl(g)
∆H = 16.8 kJ/mol = ∆H of , ICl
115.
a. C2H4(g) + O3(g) → CH3CHO(g) + O2(g) ∆H° = !166 kJ ! [143 kJ + 52 kJ] = !361 kJ
b. O3(g) + NO(g) → NO2(g) + O2(g) ∆H° = 34 kJ ! [90. kJ + 143 kJ] = !199 kJ
c. SO3(g) + H2O(l) → H2SO4(aq) ∆H° = !909 kJ ![!396 kJ + (!286 kJ)] = !227 kJ
d. 2 NO(g) + O2(g) → 2 NO2(g)
∆H° = 2(34) kJ ! 2(90.) kJ = !112 kJ
CHAPTER
6
THERMOCHEMISTRY
205
Challenge Problems
116.
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 = !P∆V = !2.00 atm(30.0 L ! 10.0 L) = -40.0 L atm ×
101.3 J
L atm
= !4.05 × 103 J
Pathway 2: w = !P∆V = !1.00 atm(30.0 L ! 10.0 L) = !20.0 L atm ×
101.3 J
L atm
= !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.
117.
A(l) → A(g)
∆Hvap = 30.7 kJ
w = −P∆V = −∆nRT, where ∆n = nproducts − nreactants = 1 − 0 = 1
w = −(1 mol)(8.3145 J/KCmol)(80. + 273 K) = −2940 J = −2.94 kJ
Because pressure is constant: ∆E = qp + w = ∆H + w = 30.7 kJ + (−2.94 kJ) = 27.8 kJ
118.
Energy needed =
20. × 103 g C12 H 22 O11
1 mol C12 H 22 O11
5640 kJ
×
×
= 3.3 × 105 kJ/h
h
342.3 g C12 H 22 O11
mol
Energy from sun = 1.0 kW/m2 = 1000 W/m2 =
10,000 m2 ×
1.0 kJ
sm
2
×
Percent efficiency =
119.
1000 J 1.0 kJ
=
s m2
s m2
60 s
60 min
×
= 3.6 × 107 kJ/h
min
h
energy used per hour
3.3 × 105 kJ
× 100 =
× 100 = 0.92%
total energy per hour
3.6 × 10 7 kJ
Energy used in 8.0 hours = 40. kWh =
Energy from the sun in 8.0 hours =
40. kJ h 3600 s
= 1.4 × 105 kJ
×
s
h
10. kJ 60 s 60 min
×
×
× 8.0 h = 2.9 × 104 kJ/m2
2
min
h
sm
Only 13% of the sunlight is converted into electricity:
0.13 × (2.9 × 104 kJ/m2) × area = 1.4 × 105 kJ, area = 37 m2
206
120.
CHAPTER 6
THERMOCHEMISTRY
a. 2 HNO3(aq) + Na2CO3(s) → 2 NaNO3(aq) + H2O(l) + CO2(g)
∆H° = [2(!467 kJ) + (!286 kJ) + (!393.5 kJ)] ! [2(!207 kJ) + (!1131 kJ)] = !69 kJ
2.0 × 104 gallons ×
4 qt
946 mL
1.42 g
= 1.1 × 108 g of concentrated nitric
×
×
gal
qt
mL
acid solution
1.1 × 108 g solution ×
7.7 × 107 g HNO3 ×
70.0 g HNO 3
= 7.7 × 107 g HNO3
100.0 g solution
1 mol Na 2 CO 3 105.99 g Na 2 CO 3
1 mol
×
×
63.02 g
2 mol HNO3
mol Na 2 CO 3
= 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 2 moles of nitric acid neutralized. Combining these two results:
1 mol HNO 3
− 69 kJ
7.7 × 107 g HNO3 ×
×
= !4.2 × 107 kJ
63.02 g HNO 3
2 mol HNO 3
b. They feared the heat generated by the neutralization reaction would vaporize the
unreacted nitric acid, causing widespread airborne contamination.
121.
400 kcal ×
4.18 kJ
= 1.7 × 103 kJ . 2 × 103 kJ
kcal
⎛
⎛
1 kg ⎞ 9.81 m
2.54 cm
1m ⎞
⎟⎟ ×
⎟ = 160 J . 200 J
× ⎜⎜ 8 in ×
×
PE = mgz = ⎜⎜180 lb ×
2
2.205 lb ⎠
in
100 cm ⎟⎠
s
⎝
⎝
200 J of energy is needed to climb one step. The total number of steps to climb are:
2 × 106 J ×
122.
1 step
= 1 × 104 steps
200 J
H2(g) + 1/2 O2(g) → H2O(l) ∆H° = ∆H of , H 2O ( l ) = !285.8 kJ; we want the reverse reaction:
H2O(l) → H2(g) + 1/2 O2(g) ∆H° = 285.8 kJ
w = !P∆V; because PV = nRT, at constant T and P, P∆V = RT∆n, where ∆n = moles of
gaseous products – moles of gaseous reactants. Here, ∆n = (1 mol H2 + 0.5 mol O2) – (0) =
1.5 mol.
∆E° = ∆H° !P∆V = ∆H° ! ∆nRT
⎛
1 kJ ⎞
⎟
∆E° = 285.8 kJ ! ⎜⎜1.50 mol × 8.3145 J/K • mol × 298 K ×
1000 J ⎟⎠
⎝
∆E° = 285.8 kJ ! 3.72 kJ = 282.1 kJ
CHAPTER
123.
6
THERMOCHEMISTRY
207
There are five parts to this problem. We need to calculate:
(1) q required to heat H2O(s) from !30.EC to 0EC; use the specific heat capacity of H2O(s)
(2) q required to convert 1 mol H2O(s) at 0EC into 1 mol H2O(l) at 0EC; use ∆Hfusion
(3) q required to heat H2O(l) from 0EC to 100.EC; use the specific heat capacity of H2O(l)
(4) q required to convert 1 mol H2O(l) at 100.EC into 1 mol H2O(g) at 100.EC;
use ∆Hvaporization
(5) q required to heat H2O(g) from 100.EC to 140.EC; 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.EC to H2O(g) at 140.EC.
q1 = 2.03 J/ECCg × 18.02 g × [0 – (!30.)]EC = 1.1 × 103 J
q2 = 1.00 mol × 6.02 × 103 J/mol = 6.02 × 103 J
q3 = 4.18 J/ECCg × 18.02 g × (100. – 0)EC = 7.53 × 103 J
q4 = 1.00 mol × 40.7 × 103 J/mol = 4.07 × 104 J
q5 = 2.02 J/ECCg × 18.02 g × (140. – 100.)EC = 1.5 × 103 J
qtotal = q1 + q2 + q3 + q4 + q5 = 5.69 × 104 J = 56.9 kJ
124.
When a mixture of ice and water exists, the temperature of the mixture remains at 0EC until
all of the ice has melted. Because an ice-water mixture exists at the end of the process, the
temperature remains at 0EC. 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 123).
Heat loss by element = heat gain by ice cubes at 0EC
Heat gain = 109.5 g H2O ×
1 mol H 2 O
6.02 kJ
×
= 36.6 kJ
18.02 g
mol H 2 O
Specific heat of element =
q
36,600 J
=
= 0.375 J/ECCg
mass × ∆T
500.0 g × (195 − 0) o C
208
CHAPTER 6
THERMOCHEMISTRY
Integrative Problems
125.
N2(g) + 2 O2(g) → 2 NO2(g)
∆H = 67.7 kJ
n N2 =
PV
3.50 atm × 0.250 L
=
= 2.86 × 10 −2 mol N2
0.08206 L atm
RT
× 373 K
K mol
n O2 =
PV
3.50 atm × 0.450 L
=
= 5.15 × 10 −2 mol O2
0
.
08206
L
atm
RT
× 373 K
K mol
The balanced equation requires a 2 : 1 O2 to N2 mole ratio. The actual mole ratio is
5.15 × 10−2/2.86 × 10 −2 = 1.80; Because the actual mole ratio is smaller than the required
mole ratio, O2 in the numerator is limiting.
5.15 × 10 −2 mol O2 ×
2 mol NO 2
= 5.15 × 10 −2 mol NO2
2 mol O 2
5.15 × 10 −2 mol NO2 ×
126.
67.7 kJ
= 1.74 kJ
2 mol NO 2
a. 4 CH3NO2(l) + 3 O2(g) → 4 CO2(g) + 2 N2(g) + 6 H2O(g)
∆H orxn = !1288.5 kJ = [4 mol(!393.5 kJ/mol) + 6 mol(!242 kJ/mol)] !
[4 mol ( ∆H of , CH 3 NO 2 )]
Solving: ∆H of , CH 3 NO 2 = !434 kJ/mol
1 atm
= 1.25 atm; PN 2 = Ptotal × χ N 2 = 1.25 atm × 0.134
760 torr
= 0.168 atm
0.168 atm × 15.0 L
=
= 0.0823 mol N2
0.08206 L atm
× 373 K
K mol
b. Ptotal = 950. torr ×
n N2
0.0823 mol N2 ×
127.
28.02 g N 2
= 2.31 g N2
1 mol N 2
Heat loss by U = heat gain by heavy water; volume of cube = (cube edge)3
Mass of heavy water = 1.00 × 103 mL ×
1.11 g
= 1110 g
mL
CHAPTER
6
THERMOCHEMISTRY
Heat gain by heavy water =
4.211 J
o
Heat loss by U = 1.4 × 104 J =
7.0 × 102 g U ×
Cg
× 1110 g × (28.5 – 25.5)EC = 1.4 × 104 J
0.117 J
o
209
Cg
× mass × (200.0 – 28.5)EC, mass = 7.0 × 102 g U
1 cm 3
= 37 cm3; cube edge = (37 cm3)1/3 = 3.3 cm
19.05 g
Marathon Problems
128.
X → CO2(g) + H2O(l) + O2(g) + A(g)
∆H = −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 (∆H = q):
− q H 2O = q rxn = -4.184 J/°CCg(1.000 × 104 g)(29.52 − 25.00 °C)(1 kJ/1000 J)
qrxn = −189.1 kJ (carrying extra significant figures)
Because ∆H = −1893 kJ/mol for the decomposition reaction, and because only −189.1 kJ of
heat was released for this reaction, 189.1 kJ × (1 mol X/1893 kJ) = 0.100 mol X was reacted.
Molar mass of X =
22.7 g X
= 227 g/mol
0.100 mol X
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 contains 3.00 mol CO2, 2.50 mol H2O, and 0.25 mol O2.
⎛ 44.0 g ⎞
⎛ 18.0 g ⎞
1.00 mol X = 227 g = 3.00 mol CO2 ⎜
⎟ + 2.50 mol H2O ⎜
⎟
mol
⎝
⎝ mol ⎠
⎠
⎛ 32.0 g ⎞
+ 0.25 mol O2 ⎜
⎟ + (mass of A)
⎝ mol ⎠
Mass of A in 1.00 mol X = 227 g − 132 g − 45.0 g − 8.0 g = 42 g A
To determine A, we need the moles of A produced. The total moles of gas produced can be
determined from the gas law data provided in the problem. Because H2O(l) is a product, we
need to subtract PH 2O from the total pressure.
ntotal =
PV
; Ptotal = Pgases + PH 2O ; Pgases = 778 torr − 31 torr = 747 torr
RT
210
CHAPTER 6
THERMOCHEMISTRY
⎛ 1L ⎞
⎟ = 12.0 L
V = height × area; area = πr2; V = (59.8 cm)(π)(8.00 cm)2 ⎜⎜
3 ⎟
⎝ 1000 cm ⎠
T = 273.15 + 29.52 = 302.67 K
⎛ 1 atm ⎞
⎟ (12.0 L)
747 torr⎜⎜
760 torr ⎟⎠
PV
⎝
=
= 0.475 mol = mol CO2 + mol O2 + mol A
ntotal =
0.08206 L atm
RT
(302.67 K )
K mol
Mol A = 0.475 mol total − 0.300 mol CO2 − 0.025 mol O2 = 0.150 mol A
Because 0.100 mol X reacted, 1.00 mol X would contain 1.50 mol A, which from a previous
calculation represents 42 g A.
Molar mass of A =
42 g A
= 28 g/mol
1.50 mol A
Because 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.
⎛ 1 atm ⎞
⎟⎟ (12.0 L − 0) = −12.3 L atm
b. w = −P∆V = −778 torr ⎜⎜
⎝ 760 torr ⎠
⎛ 8.3145 J/K • mol
−12.3 L atm ⎜⎜
⎝ 0.08206 L atm/K • mol
⎞
⎟⎟ = −1250 J = −1.25 kJ, w = −1.25 kJ
⎠
c. ∆E = q + w, where q = ∆H since at constant pressure. For 1 mol of X decomposed:
w = −1.25 kJ/0.100 mol = −12.5 kJ/mol
∆E = ∆H + w = −1893 kJ/mol + (−12.5 kJ/mol) = −1906 kJ/mol
∆H of for C3H5N3O9 can be estimated from standard enthalpies of formation data and
assuming ∆H rxn = ∆H orxn . For the balanced reaction given in part a, where ∆H orxn =
−1893 kJ:
−1893 kJ = (3∆H of , CO 2 + 2.5∆H of , H 2O + 0.25∆H of , O 2 + 1.5∆H of , N 2 ) − (∆H of , C3H 5 N 3O9 )
−1893 kJ = [3(−393.5) kJ + 2.5(−286) kJ + 0 + 0] − ∆H of , C3H 5 N 3O9
∆H of , C3H 5 N 3O9 = −2.5 kJ/mol = −3 kJ/mol
CHAPTER
129.
6
THERMOCHEMISTRY
211
⎛ 2 x + y/2 ⎞
CxHy + ⎜
⎟ Ο2 → x CO2 + y/2 H2O
2
⎠
⎝
[x(!393.5) + y/2 (!242)] ! ∆H oC x H y = !2044.5, !(393.5)x ! 121y ! ∆H C x H y = !2044.5
dgas =
P • MM
, where MM = average molar mass of CO2/H2O mixture
RT
0.751 g/L =
1.00 atm × MM
, MM of CO2/H2O mixture = 29.1 g/mol
0.08206 L atm
× 473 K
K mol
Let a = mol CO2 and 1.00 ! a = mol H2O (assuming 1.00 total moles of mixture)
(44.01)a + (1.00 ! a) × 18.02 = 29.1; solving: a = 0.426 mol CO2 ; mol H2O = 0.574 mol
y
y
0.574
= 2 , 2.69 = , y = (2.69)x
Thus:
x
x
0.426
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) ! ∆H oC3H8 , ∆H oC3H8 = !104 kJ/mol