Thus, when the reaction occurs to the extent indicated by the balanced chemical equation,56.4 kJ
is released. Remembering that exothermic reactions have negative values of Hrxn, we write
CuSO 4 (aq)2NaOH(aq)88nCu(OH) 2 (s)Na 2 SO 4 (aq) Hrxn56.4 kJ/mol rxn
You should now work Exercise 62(b).
EXAMPLE 15-5 Amount of Heat Versus Extent of Reaction
When aluminum metal is exposed to atmospheric oxygen (as in aluminum doors and windows),
it is oxidized to form aluminum oxide. How much heat is released by the complete oxidation
of 24.2 grams of aluminum at 25°C and 1 atm? The thermochemical equation is
4Al(s)3O 2 (g)88n2Al 2 O 3 (s) H3352 kJ/mol rxn
Plan
The thermochemical equation tells us that 3352 kJ of heat is released for every mole of reac-
tion, that is, for every 4 moles of Al that reacts. We convert 24.2 g of Al to moles, and then
calculate the number of kilojoules corresponding to that number of moles of Al, using the unit
factors
m
3
o
3
l
5
r
2
xn
kJ
and
1
4
m
m
o
o
l
l
r
A
xn
l
Solution
For 24.2 g Al,
__?kJ24.2 g Al
2
1
7
m
.0
o
g
lA
A
l
l
1
4
m
m
o
o
l
l
r
A
xn
l
m
3
o
3
l
5
r
2
xn
kJ
751 kJ
This tells us that 751 kJ of heat is released to the surroundings during the oxidation of 24.2
grams of aluminum.
You should now work Exercises 14 and 15.
Heat is released, so this is an
exothermic reaction.
15-6 Standard States and Standard Enthalpy Changes 601
STANDARD STATES AND STANDARD
ENTHALPY CHANGES
The thermodynamic standard stateof a substance is its most stable pure form under
standard pressure (one atmosphere)* and at some specific temperature (25°C or 298 K
unless otherwise specified). Examples of elements in their standard states at 25°C are
hydrogen, gaseous diatomic molecules, H 2 (g); mercury, a silver-colored liquid metal,
15-6
Problem-Solving Tip:Mole of Reaction
Remember that a thermochemical equation can imply differentnumbers of moles of
differentreactants or products. In Example 15-5 one mole of reaction also corresponds
to 3 moles of O 2 (g) and to 2 moles of Al 2 O 3 (s).
*IUPAC has changed the standard pressure from 1 atm to 1 bar. Because 1 bar is equal to 0.987 atm, the differ-
ences in thermodynamic calculations are negligible except in work of very high precision. Many tables of thermodynamic
data are still based on a standard pressure of 1 atm, so we will use it in this book.
A temperature of 25°C is 77°F.
This is slightly above typical room
temperature. Notice that these
thermodynamic “standard conditions”
are not the same as the “standard
temperature and pressure (STP)” that
we used in gas calculations involving
standard molar volume (Chapter 12).
The signtells us that heat was released,
but it would be grammatical nonsense
to say in words that “751 kJ of heat
was released.” As an analogy, suppose
you give your friend $5. Your $ is
$5, but in describing the transaction
you would not say “I gave her minus
five dollars,” but rather “I gave her five
dollars.”