5.2 Flashcards
The Equilibrium Constant
For a chemical equilibrium of the type:
aA + bB cC + dD
The value for K is given by:
•The constant is dimensionless and its value changes with the
temperature of the system.
•Equilibrium expressions are always written with the products as
the numerator and the reactants as the denominator. The
exponents in the expression are the same as the coefficients in the
balanced chemical equation.
TRY:
•It is used to predict the amounts of reactants and products at
equilibrium
given the amounts of starting materials.
Keg =
Write the equilibrium expression for the following
reaction.
Ng(8)
Keq =
[AB
Writing Equilibrium Expressions for
Equilibrium Reactions
+ 3H
(e) * 2NH
Rule #1: Solids NEVER appear in equilibrium expressions.
[SO.)
2) AgN0
Rule #2: Liquids and solvents NEVER appear in equilibrium
expressions.
[NH
]
[N||H
|
NH
an + H
Therefore they are CONSTANT.
Keg =
WHY? The concentration of a pure solid/liquid is
unchangeable.
+ HCI
3 (ag)
Keq =
What’s wrong with these examples?
1) C
Ha) t+ H
Keq =
[C
H] H
C
H
ao)* AgCs + HNO3 (eq)
Keg =
LAgCiJ[HN 0
]
[NH] [OH]
[NH
)
[AgNO
][HCI]
[H
0]?
[H
][0
Calculating K
from Equilibrium Concentrations
2NOC
Solution:
Place 2.00 mol of NOCI is a 1.00 L flask. At equilibrium
you find 0.66 mol/L of NO. Calculate K
Set up an ICE table of concentrations
2NOCI → 2NO
Initial conc. (mol/L)
Change in conc. (mol/L)
Equilibrium conc. (mol/L)
Example (cont’d)
•Calculate Kea
+ C
(e
2
Keq =
reverse procesS.
[NOCI]
[NOCH?
(0.66)°(0.33)
(1.34)²
=0.080
0
using the equilibrium concentrations
0.66
+
K’=
NOTE: the equilibrium constant for the forward and
reverse reactions are reciprocal values.
→K’ represents the equilibrium constant for the
-
1
d
Keq
0
= 12.5
Kea <1
.. formation of REACTANT
NOCI
REACTION) is favoured
When written in reverse:
.” Keq> 1
2NOCI
→ 2NO
.”.formation of PRODUCT (NOCI) is favoured
Homogeneous/Heterogeneous Equilibria
2NOe t C
(e)
• Homgeneous equilibria: all the reactants and
products are in the same phase.
• Heterogeneous equilibria: more than one phase exists
in a reaction mixture
Remember that solids
liquids (and solvents) DO NOT
appear in equilibrium expressions
Example: Write the equilibrium law for the dissociation of
NaOH
in water.
The Magnitude of K
eo
→ 2NOCe)
Very
small
eq
equilibrium.
NaOH * Na
Key = [Na’][OH]
eq
• When K is very large (K»1)
the equilibrium lies very
much to the right; products are favored.
Reaction proceeds
hardly at all.
• If K is very small (K«1)
the equilibrium lies very
much to the left; reactants are favored.
Keq
12.5
+ OH (aq)
(aq)
• When K is neither very large nor very small (K= 1)
neither reactants
nor products are favoured at
10-3
1
109
Appreciable concentrations
of both reactants and products
are present at equilibrium.
9-10/11
large
Reaction proceeds
nearly to completion.
Kea <1
.. formation of REACTANT
NOCI
REACTION) is favoured
When written in reverse:
.” Keq> 1
2NOCI
→ 2NO
.”.formation of PRODUCT (NOCI) is favoured
Homogeneous/Heterogeneous Equilibria
2NOe t C
(e)
• Homgeneous equilibria: all the reactants and
products are in the same phase.
• Heterogeneous equilibria: more than one phase exists
in a reaction mixture
Remember that solids
liquids (and solvents) DO NOT
appear in equilibrium expressions
Example: Write the equilibrium law for the dissociation of
NaOH
in water.
The Magnitude of K
eo
→ 2NOCe)
Very
small
eq
equilibrium.
NaOH * Na
Key = [Na’][OH]
eq
• When K is very large (K»1)
the equilibrium lies very
much to the right; products are favored.
Reaction proceeds
hardly at all.
• If K is very small (K«1)
the equilibrium lies very
much to the left; reactants are favored.
Keq
12.5
+ OH (aq)
(aq)
• When K is neither very large nor very small (K= 1)
neither reactants
nor products are favoured at
10-3
1
109
Appreciable concentrations
of both reactants and products
are present at equilibrium.
9-10/11
large
Reaction proceeds
nearly to completion.
