CHEMISTRY 161
HW #5C
78, 80, 82, 84, 86, 88, 90, 92, 110, 112, 114, 116, 118, 120, 122, 126
5-78
The fuel value is expressed in kilojoules per gram (kJ/g), and there is no difference in the mass (in g)
of propane (for a given molar amount) whether it was in liquid or gas form. However, because energy
is used to convert liquid propane to its gaseous state, the fuel value of the gaseous propane is higher
than that of liquid propane.
5-80
mol water in 1 gallon = 128 ounces 
28.35 g 1 mol

 201 mol
1 ounce 18.02 g
6.01 kJ  
0.0753 kJ
40.67 kJ 

 
q   201 mol 
 100 C   201 mol 
   201 mol 
  10, 900 kJ

 
mol  
mol 
mol  C
5-82
4163 kJ 1 mol

 48.31 kJ/g
mol
86.18 g
48.31 kJ
(b) Heat released by 1.00 kg C6 H14  1000 g
 4.831 10 4 kJ
g
(c) Energy needed to raise 1.00 kg water from 25.0˚C to 85.0˚C:
1 mol
75.3 J
q  1000 g 

 60.0 C  2.51  10 5 J or 251 kJ
18.02 g mol  C
1g
 5.20 g
Mass of C6H14 needed to generate this energy  251 kJ 
48.31 kJ
(d) Weighted average fuel value  0.75  48.31 kJ g  0.25  48.99 kJ g
 48.48 kJ g
(a) Fuel value of C6 H14 
Mass of fuel mixture required  251 kJ 
1g
 5.18 g
48.48 kJ
5-84
The burning of carbon in oxygen (even when the stoichiometry of the reaction to produce CO is used)
produces both carbon monoxide and carbon dioxide simultaneously, so that a mixture of these
products is produced. Therefore, it is impossible to isolate the enthalpy of formation for CO.
5-86
If enthalpy were not a state function, the value we would calculate for enthalpy of a reaction via an
°
alternative pathway using Hess’s law would give us a different H rxn
value. The reaction pathway
would matter and therefore Hess’s law would be invalid.
5-88
Carbon dioxide does not appear in the overall desired equation. Arranging H f,° CO and ∆H˚comb, CO
2
chemical equations so that C and O2 are the reactants and CO is the product gives
C(s)  O 2 (g)  CO 2 (g)
H f,° CO2
CO2 (g)  CO(g)  12 O 2 (g)
C(s)  12 O 2 (g)  CO(g)
– H comb, CO
°
°
H rxn
 H f,° CO2 – H comb,
CO
5-90
Cl(g) + 2 O2 (g)  ClO(g) + O3 (g)
2 O3 (g)  3 O2 (g)
Cl(g) + O3 (g)  ClO(g) + O2 (g)
°
H rxn
 29.90 kJ
°
H rxn
 24.18 kJ
°
H rxn
 54.08 kJ
5-92
Cdiamond (s) + O2 (g)  CO2 (g)
H °  –395.4 kJ
2 CO2 (g)  2 CO(g) + O2 (g)
H °  566.0 kJ
2 CO(g)  Cgraphite (s) + CO2 (g)
Cdiamond (s)  Cgraphite (s)
H °  –172.5 kJ
H °  –1.90 kJ
5-110
The balanced chemical equation for the reaction of iron(II) oxide with oxygen to form iron(III) oxide
is
2 FeO(s) + 12 O2 (g)  Fe2 O3 (s)
4 FeO(s) + O2 (g)  2 Fe2 O3 (s)
°
:
We use the coefficients in the equation for H rxn
H rxn   2 mol Fe2 O3  –824.2 kJ/mol 
  4 mol FeO  –271.9 kJ/mol   1 mol O2  0.0 kJ/mol  
H rxn  –560.8 kJ
5-112
Moles of water converted into steam:
1 mol
 0.050 mol
18.02 g
If 0.050 mol of water requires 2.0 kJ, then one mole would require
2.0 kJ
 40 kJ
0.050 mol
0.90 g 
5-114
a)

424.7 kJ  
285.8 kJ  
H rxn   1 mol HCOOH 
  1 mol H 2 O 

 
mol  
mol  


0.0 kJ  
238.7 kJ  
1 mol O 2  mol   1 mol CH 3OH  mol    471.8 kJ


(b) This reaction is exothermic because energy is produced (as a “product”) in the thermochemical
equation.
(c) The moles of methanol in 60.0 g is
1 mol
60.0 g 
 1.87 mol
32.04 g
For this number of moles, the enthalpy is
471.8 kJ
1.87 mol 
 882.3 kJ
mol
(d) We can expect that the first step that converts methanol to formaldehyde where only one oxygen
atom is present would be less exothermic than the overall conversion of methanol to formic acid, so
the enthalpy of reaction for the formation of formaldehyde from methanol is expected to be smaller
(less exothermic) than the overall enthalpy of the reaction to form formic acid.
5-116
D(s)  E(g)  F(g)
–H1
F(g)  F(s)
–H 2
D(s)  E(g)  F(s)
H 3  – H1  – H 2
5-118
NiO(s)  SO2 (g)  NiSO3 (s)
°
H rxn
 –156 kJ
S8 (s)  O2 (g)  SO2 (g)
°
H rxn
 –297 kJ
Ni(s)  12 O2 (g)  NiO(s)
°
H rxn
 –241 kJ
1
8
Ni(s)  18 S8 (s)  23 O 2 (g)  NiSO 3 (s)
°
H rxn
 –694 kJ
5-120
°
H comb
  (2 mol CO2  –393.5 kJ/mol)  (3 mol H 2 O  –241.8 kJ/mol) 
–  (1 mol C 2 H 5 OH  –277.7 kJ/mol)  (3 mol O2  0 kJ/mol) 
H
°
comb
 1234.7 kJ
5-122
All results are shown in the table below.
(a) The specific heat capacity values and molar masses are given for Au, Sn, Zn, Cu, and Fe. These
can be used to calculate the values in the fourth column. Using the calculation for Au to determine
the units gives:
197.0 g 0.125 J

 24.6 J/mol  C
1 mol
g C
(b) The average value for the fourth column values using the results for Pb, Au, Sn, Zn, Cu, and Fe is
25.5  24.6  25.5  25.4  25.2  25.7
 25.3
6
(c) Missing atomic masses for Bi and the two unknown elements in the table are found by dividing
25.3 by the specific heat capacity. The result for Bi = 210.8 g/mol, which is close to the known molar
mass in the periodic table, but that mass could also be for At or Po. For the two unknown elements,
the molar masses are calculated as 108.6 and 58.4 g/mol. This would identify these elements as
Ag (107.9 g/mol) and as either Co (58.9 g/mol) or Ni (58.7 g/mol). There is obviously some
uncertainty in assigning elements based on this application of the law of Dulong and Petit.
(d) The specific heat capacity for Pt and S are calculated by dividing 25.3 by the molar mass. See
table for results.
Element
cS[J/(g ·˚C)] }  cs
} (g/mol)
Bismuth
210.8
0.120
25.3
Lead
207.2
0.123
25.5
Gold
197.0
0.125
24.6
Platinum
195.1
0.130
25.3
Tin
118.7
0.215
25.5
Silver
108.6
0.233
25.3
Zinc
65.38
0.388
25.4
Copper
63.5
0.397
25.2
Cobalt/nickel 58.4
0.433
25.3
Iron
55.8
0.460
25.7
Sulfur
32.1
0.788
25.3
Average value in column 4
25.3
5-126
Adding the reactions gives
N 2 (g)  O 2 (g)  2NO(g)
°
H comb
 180 kJ
2NO(g)  O 2 (g)  2NO 2 (g)
°
H comb
 –112 kJ
°
N 2 (g)  O 2 (g)  2NO 2 (g)
H comb
 68 kJ
This reaction is endothermic under standard conditions.