Chemistry Reference
In-Depth Information
the following ratio is a constant:
determination of the equilibrium concentrations of the
other substances followed by substitution into the equilib-
rium constant expression to solve for
K.
We can apply the
concepts used to solve problems involving limiting quan-
tities (Section 10.4) to find the equilibrium concentrations
and use a tabulation method to make the solution process
easier. The balanced chemical equation governs the
changes
in concentrations. If initial concentrations of
some reactants and the value of
K
are given, algebraic
variables (such as
x
) are used for the equilibrium concen-
trations. We substitute these quantities in the equilibrium
constant expression and solve for
x,
the unknown equilib-
rium concentration. Approximations that involve ignoring
small concentrations only when they are added to or sub-
tracted from larger ones are often helpful (Section 18.4).
[C]
c
[D]
d
[A]
a
[B]
b
K
The equilibrium concentrations of the substances on the
right side of the chemical equation are placed in the nu-
merator and divided by the equilibrium concentrations of
the substances on the left side, all raised to their appro-
priate powers (equal to the coefficients in the balanced
equation).
Given equilibrium concentrations, the value of
K
can
be calculated by simply substituting those concentrations
in the equilibrium constant expression. A somewhat more
difficult problem gives initial concentrations of some re-
actants and one equilibrium concentration and requires
Items for Special Attention
The equilibrium concentrations, not numbers of moles, are
the factors in equilibrium constant expressions.
correspond to the coefficients in the balanced chemical
equation.
■
In equilibrium constant expressions, the concentration
terms are multiplied and divided, not added or subtracted.
The concentrations in the numerator are those of the sub-
stances on the right side of the chemical equation, and
those in the denominator are for the substances on the left
side. The exponents in the equilibrium constant expression
The balanced chemical equation governs the concentra-
tions produced and used up by the reaction (the second row
of a tabulation such as those used in the examples in Sec-
tion 18.4). We can treat concentrations in this way because
the concentration ratio of the substances in a given solution
or gas mixture is the same as the mole ratio.
■
■
Answers to Snapshot Reviews
18.1 A. In pure oxygen, because the oxygen pressure is much
higher.
18.2 A. If equilibrium is achieved in both experiments, she
should get 0.825 mol of A, just as she had remaining
in the forward reaction.
18.3 A. (a) Toward the elements. (b) Heat is used up, so the
temperature will drop.
B. All in moles per liter:
PCl
5
(g)
SO
3
(g)
E
SO
2
Cl
2
(g)
POCl
3
(g)
Initial
0.0400
0.0400
0.0000
0.0000
Change
Final
0.0325
0.0325
0.0325
0.0325
0.0075
0.0075
0.0325
0.0325
(0.0325)
2
(0.0075)
2
[SO
2
Cl
2
] [POCl
3
]
[SO
3
] [PCl
5
]
K
19
18.4 A.
K
Self-Tutorial Problems
18.1
If an equilibrium shifts to the right in response to a stress,
is that shift another stress?
(b) Only 4 mol of
Cl
2
for every 1 mol of
CH
4
can be put
into a vessel.
(c) No matter how much of each reactant is added to a
vessel, only 4 mol of and 1 mol of will react.
(d) When 4 mol of and 1 mol of are placed in a
vessel, they will react to give 4 mol of HCl.
18.2
Which of the following statements is a correct interpre-
tation of this balanced chemical equation?
Cl
2
CH
4
Cl
2
CH
4
High temp
4 Cl
2
(g)
CH
4
(g)
4 HCl(g)
CCl
4
(g)
(a) Only 4 mol of
Cl
2
and 1 mol of
CH
4
can be put into
(e)
Cl
2
and
CH
4
react in a 4 : 1 mole ratio.
a vessel.
