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CHAPTER 14
CHEMICAL EQUILIBRIUM
14.13
c
[B]
[A]
=
K
(1)
With
K
c
=
10, products are favored at equilibrium.
Because the coefficients for both A and B are one,
we expect the concentration of B to be 10 times that of A at equilibrium.
Choice
(a)
is the best choice
with 10 B molecules and 1 A molecule.
(2)
With
K
c
=
0.10, reactants are favored at equilibrium.
Because the coefficients for both A and B are
one, we expect the concentration of A to be 10 times that of B at equilibrium.
Choice
(d)
is the best
choice with 10 A molecules and 1 B molecule.
You can calculate
K
c
in each case without knowing the volume of the container because the mole ratio
between A and B is the same.
Volume will cancel from the
K
c
expression.
Only moles of each component
are needed to calculate
K
c
.
14.14
Note that we are comparing similar reactions at equilibrium – two reactants producing one product, all with
coefficients of one in the balanced equation.
(a)
The reaction, A
+
C
U
AC has the largest equilibrium constant.
Of the three diagrams, there is the
most product present at equilibrium.
(b)
The reaction, A
+
D
U
AD has the smallest equilibrium constant.
Of the three diagrams, there is the
least amount of product present at equilibrium.
14.15
When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant
becomes the reciprocal of the original equilibrium constant.
34
11
4.17
10
−
==
=
×
×
33
'2
.
4
0
1
0
K
K
14.16
The problem states that the system is at equilibrium, so we simply substitute the equilibrium concentrations
into the equilibrium constant expression to calculate
K
c
.
Step 1:
Calculate the concentrations of the components in units of mol/L.
The molarities can be calculated
by simply dividing the number of moles by the volume of the flask.
2
2.50 mol
[H ]
0.208
12.0 L
M
5
6
2
1.35
10
mol
[S ]
1.13
10
12.0 L
−
−
×
×
M
2
8.70 mol
[H S]
0.725
12.0 L
M
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View Full DocumentCHAPTER 14:
CHEMICAL EQUILIBRIUM
377
Step 2:
Once the molarities are known,
K
c
can be found by substituting the molarities into the equilibrium
constant expression.
2
2
2
22
6
[H S]
(0.725)
[H ] [S ]
(0.208) (1.13
10
)
−
==
=
×
7
c
1.08
10
K
×
If you forget to convert moles to moles/liter, will you get a different answer?
Under what circumstances will
the two answers be the same?
14.17
Using Equation (14.5) of the text:
K
P
=
K
c
(0.0821
T
)
Δ
n
where,
Δ
n
=
2
−
3
=
−
1
and
T
=
(1273
+
273) K
=
1546 K
K
P
=
(2.24
×
10
22
)(0.0821
×
1546)
−
1
=
1.76
×
10
20
14.18
Strategy:
The relationship between
K
c
and
K
P
is given by Equation (14.5) of the text.
What is the change
in the number of moles of gases from reactant to product?
Recall that
Δ
n
=
moles of gaseous products
−
moles of gaseous reactants
What unit of temperature should we use?
Solution:
The relationship between
K
c
and
K
P
is given by Equation (14.5) of the text.
K
P
=
K
c
(0.0821
T
)
Δ
n
Rearrange the equation relating
K
P
and
K
c
, solving for
K
c
.
c
(0.0821 )
Δ
=
P
n
K
K
T
Because
T
=
623 K and
Δ
n
=
3
−
2
=
1, we have:
5
1.8
10
(0.0821)(623 K)
(0.0821 )
−
Δ
×
=
7
c
3.5
10
P
n
K
T
−
×
K
14.19
We can write the equilibrium constant expression from the balanced equation and substitute in the pressures.
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This note was uploaded on 03/02/2009 for the course CHM 2046 taught by Professor Veige/martin during the Spring '07 term at University of Florida.
 Spring '07
 veige/martin
 Chemistry, Equilibrium

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