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Solving Problems: A Chemistry Handbook
Chemistry: Matter and Change
171
SOLVING PROBLEMS:
A CHEMISTRY HANDBOOK
CHAPTER 17
from the expression because the concentrations of these substances
are constant values.
Example Problem 17-2
Equilibrium Constant Expressions for Heterogeneous
Equilibria
Write the equilibrium constant expression for the high-temperature
reaction of carbon dioxide and solid carbon to form carbon
monoxide.
CO 2 (g) ϩ C(s) 3 2CO(g)
Write a ratio with the concentration of the product in the numerator
and the concentrations of the reactants in the denominator. Raise
each concentration to the power equal to its coefficient in the
balanced equation.
Leave out [C] because it is a pure solid with an unchanging concentration. The result is the equilibrium constant expression.
K eq ϭ
Practice Problems
2. Write equilibrium constant expressions for the following
heterogeneous equilibria.
a. C 4 H 10 (l) 3 C 4 H 10 (g)
b. NH 4 HS(s) 3 NH 3 (g) ϩ H 2 S(g)
c. CO(g) ϩ Fe 3 O 4 (s) 3 CO 2 (g) ϩ 3FeO(s)
d. (NH 4 ) 2 CO 3 (s) 3 2NH 3 (g) ϩ CO 2 (g) ϩ H 2 O(g)
Calculating equilibrium constants The value of K eq is a constant for a given reaction at a given temperature. A K eq value greater
than 1 indicates that products are favored at equilibrium. If K eq is
less than 1, reactants are favored. The equilibrium concentrations of
the reactants and products may be used to calculate K eq , as shown in
this example problem.
▲
[CO] 2
ᎏ
[CO 2 ]
[CO] 2
ᎏᎏ
[CO 2 ][C]