Chemistry Reference
In-Depth Information
EXAMPLE 14.16
Calculate
¢
H
for the complete combustion of 1.00 mol of propane,
C
3
H
8
,
a
major component of lighter fluid.
Solution
C
3
H
8
(g)
5 O
2
(g) £ 3 CO
2
(g)
4 H
2
O(
/
)
Values of
¢
H
f
are obtained from Tables 14.6 and 14.7. For the products,
¢H
f
of an element in its stan-
dard state is zero by definition.
393 kJ
1 mol CO
2
286 kJ
1 mol H
2
O
¢
H
f
3 mol CO
2
a
b
4 mol H
2
O
a
b
232
3
kJ
Be sure to distinguish between
and
For the reactants,
¢H
f
¢H
for a reaction.
105 kJ
1 mol C
3
H
8
0 kJ
1 mol O
2
¢
H
f
1 mol C
3
H
8
a
b
5 mol O
2
a
b
105 kJ
Be especially careful about the
signs when doing these
calculations!
¢
H
¢
H
f
(prod)
¢
H
f
(react)
(
232
3
kJ)
(
105 kJ)
2220 kJ
(Note that the value of of is not in Table 14.6 because we know that
of every uncombined element in its standard state is zero.)
¢
H
f
O
2
¢
H
f
Practice Problem 14.16
Calculate
¢
H
for the complete combustion
of 1.00 mol of heptane, C
7
H
16
.
We can multiply the number of kilojoules per mole by the number of moles
undergoing reaction to get the number of kilojoules for any given quantity of
reactant or product involved. Thus, for Example 14.16, if 2.50 mol of
C
3
H
8
had
been involved, the enthalpy change would have been
2220 kJ
1 mol
a
b
2.50 mol
5550 kJ
EXAMPLE 14.17
Calculate the enthalpy change for the combustion of 1.00 mol of butane (bot-
tled gas),
Table 14.7
Enthalpies of
Formation at 25
C
4
H
10
,
to give CO and
H
2
O.
Use the following equation:
Cof
Unbranched Hydrocarbons
2 C
4
H
10
(g)
9 O
2
(g) £ 8 CO(g)
10 H
2
O(
/
)
Compound
H
f
(kJ/mol)
Solution
CH
4
74.5
83.7
105
126
146
167
187
208
For the products,
C
2
H
6
110 kJ
1 mol CO
286 kJ
1 mol H
2
O
C
3
H
8
¢
H
f
8 mol CO
a
b
10 mol H
2
O
a
b
374
0
kJ
C
4
H
10
C
5
H
12
For the reactants,
C
6
H
14
C
7
H
16
126 kJ
1 mol C
4
H
10
0 kJ
1 mol O
2
C
8
H
18
¢
H
f
2 mol C
4
H
10
a
b
9 mol O
2
a
b
252 kJ
