5.2 Flashcards
1
Q
The Equilibrium Constant
A
2
Q
For a chemical equilibrium of the type:
A
3
Q
aA + bB cC + dD
A
4
Q
The value for K is given by:
A
5
Q
•The constant is dimensionless and its value changes with the
A
6
Q
temperature of the system.
A
7
Q
•Equilibrium expressions are always written with the products as
A
8
Q
the numerator and the reactants as the denominator. The
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9
Q
exponents in the expression are the same as the coefficients in the
A
10
Q
balanced chemical equation.
A
11
Q
TRY:
A
12
Q
•It is used to predict the amounts of reactants and products at
A
13
Q
equilibrium
A
given the amounts of starting materials.
14
Q
Keg =
A
15
Q
Write the equilibrium expression for the following
A
16
Q
reaction.
A
17
Q
Ng(8)
A
18
Q
Keq =
A
19
Q
[AB
A
20
Q
Writing Equilibrium Expressions for
A
21
Q
Equilibrium Reactions
A
22
Q
+ 3H
A
(e) * 2NH
23
Q
Rule #1: Solids NEVER appear in equilibrium expressions.
A
24
Q
[SO.)
A
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2) AgN0
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Rule #2: Liquids and solvents NEVER appear in equilibrium
27
expressions.
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[NH
]
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[N||H
|
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NH
an + H
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Therefore they are CONSTANT.
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Keg =
33
WHY? The concentration of a pure solid/liquid is
34
unchangeable.
35
+ HCI
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3 (ag)
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Keq =
38
What's wrong with these examples?
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1) C
Ha) t+ H
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Keq =
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[C
H] H
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C
H
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ao)* AgCs + HNO3 (eq)
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Keg =
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LAgCiJ[HN 0
]
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[NH] [OH]
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[NH
)
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[AgNO
][HCI]
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[H
0]?
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[H
][0
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Calculating K
from Equilibrium Concentrations
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2NOC
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Solution:
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Place 2.00 mol of NOCI is a 1.00 L flask. At equilibrium
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you find 0.66 mol/L of NO. Calculate K
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Set up an ICE table of concentrations
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2NOCI → 2NO
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Initial conc. (mol/L)
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Change in conc. (mol/L)
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Equilibrium conc. (mol/L)
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Example (cont'd)
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•Calculate Kea
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+ C
(e
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2
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Keq =
66
reverse procesS.
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[NOCI]
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[NOCH?
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(0.66)°(0.33)
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(1.34)²
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=0.080
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0
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using the equilibrium concentrations
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0.66
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+
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K'=
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NOTE: the equilibrium constant for the forward and
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reverse reactions are reciprocal values.
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→K' represents the equilibrium constant for the
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-
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1
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d
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Keq
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0
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= 12.5
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Kea <1
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.. formation of REACTANT
NOCI
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REACTION) is favoured
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When written in reverse:
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." Keq> 1
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2NOCI
→ 2NO
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.".formation of PRODUCT (NOCI) is favoured
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Homogeneous/Heterogeneous Equilibria
94
2NOe t C
(e)
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• Homgeneous equilibria: all the reactants and
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products are in the same phase.
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• Heterogeneous equilibria: more than one phase exists
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in a reaction mixture
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Remember that solids
liquids (and solvents) DO NOT
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appear in equilibrium expressions
101
Example: Write the equilibrium law for the dissociation of
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NaOH
in water.
103
The Magnitude of K
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eo
105
→ 2NOCe)
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Very
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small
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eq
109
equilibrium.
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NaOH * Na
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Key = [Na'][OH]
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eq
113
• When K is very large (K>>1)
the equilibrium lies very
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much to the right; products are favored.
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Reaction proceeds
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hardly at all.
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• If K is very small (K<<1)
the equilibrium lies very
118
much to the left; reactants are favored.
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Keq
120
12.5
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+ OH (aq)
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(aq)
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• When K is neither very large nor very small (K= 1)
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neither reactants
nor products are favoured at
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10-3
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1
127
109
128
Appreciable concentrations
129
of both reactants and products
130
are present at equilibrium.
131
9-10/11
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large
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Reaction proceeds
134
nearly to completion.
135
Kea <1
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.. formation of REACTANT
NOCI
137
REACTION) is favoured
138
When written in reverse:
139
." Keq> 1
140
2NOCI
→ 2NO
141
.".formation of PRODUCT (NOCI) is favoured
142
Homogeneous/Heterogeneous Equilibria
143
2NOe t C
(e)
144
• Homgeneous equilibria: all the reactants and
145
products are in the same phase.
146
• Heterogeneous equilibria: more than one phase exists
147
in a reaction mixture
148
Remember that solids
liquids (and solvents) DO NOT
149
appear in equilibrium expressions
150
Example: Write the equilibrium law for the dissociation of
151
NaOH
in water.
152
The Magnitude of K
153
eo
154
→ 2NOCe)
155
Very
156
small
157
eq
158
equilibrium.
159
NaOH * Na
160
Key = [Na'][OH]
161
eq
162
• When K is very large (K>>1)
the equilibrium lies very
163
much to the right; products are favored.
164
Reaction proceeds
165
hardly at all.
166
• If K is very small (K<<1)
the equilibrium lies very
167
much to the left; reactants are favored.
168
Keq
169
12.5
170
+ OH (aq)
171
(aq)
172
• When K is neither very large nor very small (K= 1)
173
neither reactants
nor products are favoured at
174
10-3
175
1
176
109
177
Appreciable concentrations
178
of both reactants and products
179
are present at equilibrium.
180
9-10/11
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large
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Reaction proceeds
183
nearly to completion.
