Equilibrium
Equilibrium
• Limiting reagent
• Concentrations become
constant
• Dynamic situation
• Reversible reactions
2NO2  N2O4
a
b
c
time
d
2NO2  N2O4
Equilibrium Constant (K)
represents a ratio of
the concentrations of
products to reactants
at equilibrium:
aA +bB  cC + dD
Equilibrium Constant (K)
K=
c
d
[C] [D]
a
b
[A] [B]
Equilibrium, or K,
expression
Equilibrium Constant (K)
• “[]” represents
concentration in mol/L for
(g) and (aq), only
• Each “[]” must be raised
to the power of its
coefficient
Equilibrium Constant (K)
• K < 1 indicates little
product formation
• K > 100 indicates
great amount of
product formation
Equilibrium Constant (K)
• Write the K expression
for the dimerization of
nitrogen dioxide.
Equilibrium Constant (K)
[N2O4]
K=
2
[NO2]
L/mol
What will the units of
K be in this example?
Equilibrium Constant (K)
• At 25°C, the equilibrium
concentrations of NO2 and
N2O4 are 0.0370M and
0.2315M. What is the
value of K at this T?
Equilibrium Constant (K)
[0.2315]
K=
2
[0.037]
K=0.2315mol/L
2
2
0.001369mol /L
K=169 L/mol
N2 + 3H2  2NH3
• Write the K expression
for the synthesis of
ammonia.
Equilibrium Constant (K)
2
]
[NH3
K=
3
[N2][H2]
2
2
L /mol
What will the units of
K be in this example?
Equilibrium Constant (K)
• At 300°C, the equilibrium
concentrations are:
[N2]eq= 2.59M
[H2]eq=2.77M
[NH3]eq=1.82M
What is the value of K at this
temperature?
Equilibrium Constant (K)
K=
2
[1.82]
3
[2.59][2.77]
2
2
K=3.3124mol /L
4
4
55.05mol /L
K=0.0602
2
2
L /mol
Equilibrium Constant (K)
• Small K (<1) means…
• Big K (>100) means…
• Different manner of
solving problems
Equilibrium Constant (K)
• If a reaction is reversed,
then the value of K for the
reversed reaction is the
reciprocal of K.
Equilibrium Constant (K)
• So, if the dimerization of
NO2 is reversed to be the
decomposition of N2O4…
Equilibrium Constant (K)
• K = (169
-1
L/mol)
0.00592 mol/L
or
Equilibrium Constant (K)
• At 25°C, the initial
concentration of N2O4 is
0.750M. What are the eq.
conc. of both species at this
temperature?
Equilibrium Constant (K)
• You will make an
equilibrium chart to
indicate the initial, change,
and equilibrium
concentrations.
Equilibrium Constant (K)
[N2O4]
Initial
Change
Eq.
0.750
[NO2]
0
Equilibrium Constant (K)
[N2O4]
Initial
Change
Eq.
[NO2]
0.750
0
-x
+2x
Equilibrium Constant (K)
[N2O4]
Initial
Change
Eq.
[NO2]
0.750
0
-x
+2x
0.75 - x
2x
Equilibrium Constant (K)
0.00592=
2
[2x]
[0.75 – x]
2
[2x]
0.00592=
[0.75]
4.44 x
-3
10 =
2
4x
Equilibrium Constant (K)
-3
10 =
1.11 x
0.0333= x
2
x
Equilibrium Constant (K)
5% rule…is what you
removed less than 5%
of the smaller initial
value?
Equilibrium Constant (K)
If so, then your
assumption that what
you removed was so
small it is negligible
is correct
Equilibrium Constant (K)
5% rule test:
1x1(0.0333)
X 100 =
0.75
4.44% < 5%…
assumption is good
Equilibrium Constant (K)
[N2O4]
Initial
[NO2]
0.750
0
Change
-0.0333
+2(0.0333)
Eq.
0.7167
0.0666
Equilibrium Constant (K)
Check your answer:
K=
2
[0.0666]
[0.7167]
K= 0.00619
Equilibrium Constant (K)
• At 25°C, the initial
concentration of NO2 is
0.500M. What are the eq.
conc. Of both species at this
temperature? Remember
that K = 169L/mol.
Equilibrium Constant (K)
Since K is big, lots of product
will be made. Thus, almost all
of the initial amount of reactant
will be used. You need to make
two charts for a big K
problem…Stoichiometry and
Equilibrium
Equilibrium Constant (K)
Stoichiometry Chart
[NO2]
Initial
Change
Final
0.500
[N2O4]
0
Equilibrium Constant (K)
Stoichiometry Chart
[NO2]
Initial
Change
Final
[N2O4]
0.500
0
-0.5
+0.25
Equilibrium Constant (K)
Stoichiometry Chart
[N2O4]
[NO2]
Initial
Change
Final
0.500
0
-0.5
+0.25
0
0.25
Equilibrium Constant (K)
Equilibrium Chart
[N2O4]
[NO2]
Initial
(=final)
Change
Eq.
0
0.25
Equilibrium Constant (K)
Equilibrium Chart
[N2O4]
[NO2]
Initial
Change
Eq.
0
0.25
+2x
-x
Equilibrium Constant (K)
Equilibrium Chart
[N2O4]
[NO2]
Initial
0
0.25
Change
+2x
-x
Eq.
2x
0.25-x
Equilibrium Constant (K)
[0.25 - x]
169=
2
[2 x]
[0.25]
169=
2
4x
2
676x =
0.25
Equilibrium Constant (K)
-4
10 =
3.70 x
0.0192= x
2
x
Equilibrium Constant (K)
5% rule test:
1x1(0.0192)
X 100 =
0.25
7.69% > 5%…
assumption is bad
Equilibrium Constant (K)
Since the assumption that x
was so small it was negligible
is bad, then you must re-insert x
and solve the equation with x
present.
Equilibrium Constant (K)
[0.25 - x]
169=
2
[2 x]
[0.25
x]
169=
2
4x
2
676x =
0.25 - x
Equilibrium Constant (K)
2
676x
2
ax
+ x – 0.25 = 0
+ bx + c = 0
x = -b ±
2
b
2a
– 4ac
Equilibrium Constant (K)
You will get two values of x. If
both are positive, then you will
always select the smaller one.
If one is positive and the other
negative, you will select the
positive one.
Equilibrium Constant (K)
2
1
– 4(676)(-0.25)
2(676)
x = -1 ± 1 – (-676)
1352
x = -1 ±
Equilibrium Constant (K)
x = -1 ± 677
1352
x = -1 ± 26.02
1352
Equilibrium Constant (K)
x = 25.02
= 0.0185 *accept
1352
this one
OR
x = -27.02 = -0.0200
1352
Equilibrium Constant (K)
Equilibrium Chart
[N2O4]
[NO2]
Initial
0
0.25
Change +2(0.0185)
-0.0185
Eq.
0.037
0.2315
Equilibrium Constant (K)
Check your answer:
[0.2315]
K=
2
[0.037]
K= 169L/mol