USING K VALUES IN
EQUILIBRIUM
CALCULATIONS
CH40S – UNIT 3 Equilibrium
REVIEW
When 0.800 moles of SO2 and 0.800 moles of O2 a 2.00 L container and allowed to
reach equilibrium, the equilibrium [SO3] is to be 0.300 M. Calculate the Keq value.
2 SO2
(g)
+
O2
(g)
⇋
2 SO3
(g)
REVIEW
When 0.800 moles of SO2 and 0.800 moles of O2 are placed into a 2.00 L container and
allowed to reach equilibrium, the equilibrium [SO3] is to be 0.300 M. Calculate the Keq
value.
Implies initial and not equilibrium concentrations - ICE
2SO2
I
(g)
0.400 M
+
O2
⇋
(g)
0.400M
C
-0.300 M
-0.150 M
E
0.100 M
0.250 M
x
2/
2
2SO3
(g)
0
x1/2
+0.300 M
0.300 M
Equilibrium concentrations go in the equilibrium equation!
Keq
=
[SO3]2
[SO2]2[O2]
=
(0.3)2
(0.1)2(0.25)
=
36.0
Using K Values to Find [Eqm]
¨ 
¨ 
We have just seen that we can calculate the
equilibrium concentrations of reactants and products
and the K value of a reaction as long as we know
the initial concentrations and AT LEAST ONE
EQUILIBRIUM CONCENTRATION using an ICE table.
You can also calculate equilibrium concentrations of
reactants and products when you DON’T KNOW
ANY EQUILIBRIUM CONCENTRATIONS using an ICE
table and the K VALUE of the reaction at that
temperature.
Working with K values
INITIAL CONCENTRATIONS
+
1. 
2. 
K = 25 @ this temp.
⇋
+
What will the
concentration of
be at equilibrium?
How can we
determine this?
Working with K values
INITIAL [NO2]
+
I
C
E
K = 25 @ this temp.
⇋
+
The Chemistry…
1. 
0.80 moles of H2 and 1.4 moles of S are initially put in a 4.0 L flask and allowed to
reach equilibrium. Calculate the [H2] at equilibrium.
H2(g)
+
S(s)
⇄
H2S(g)
I
0.20 M
0
C
-x
+x
E
0.20 - x
x
[H2S]
Keq
Keq= 14
x
=
=
[H2]
=
0.2 - x
14
The Math…
x
0.2 - x
+14x
1x
+
X=
14
X
1
1x
=
14(0.2 - x)
1x
=
2.8 - 14x
14x
=
2.8
15x
_____
15
= _____
2.8
15
x
= 0.19 M
+14x
The Answer!
H2(g)
+
⇄
S(s)
H2S(g)
I
0.20 M
0
C
-0.19
+0.19
E
0.20 - 0.19
0.19
[H2]
=
0.20 - x
[H2]
=
0.20 - 0.19
[H2]
=
0.01 M
Keq= 14
Approximating
10
¨ 
¨ 
Sometimes, the math gets pretty hairy when you have an
“x” involved. To simplify things, you can make the following
assumption:
If Keq is really small the reaction will not proceed to the
right very far, meaning the equilibrium concentrations will
be nearly the same as the initial concentrations of your
reactants.
0.20 – x is just about 0.20 if x is really tiny.
THE “100 RULE”
If the [initial] is at least 100x bigger than the K value for the
reaction, ignore x! Otherwise, you have to use the quadratic.
Yay! I can ignore “x”!
Carbon monoxide gas is a primary starting material in the synthesis of many organic
compounds. At 2000oC, K = 6.40 x 10-7 for the decomposition of CO2. Calculate the
equilibrium concentrations of all entities if 0.250 mol of CO2 is initially placed in a
1.000 L closed container at 2000oC.
2CO2(g)
⇄
2CO(g)
+
O2(g)
Arghh! I can’t ignore “x”!
If 0.500 mol of N2O4(g) is placed in a 1.00 L closed container at 150oC, what will the
equilibrium concentrations of all entities be? The equilibrium constant (K) for this reaction
at these conditions is 4.50.
N2O4(g)
⇄
2NO2(g)