Transcript Chemical Equilibrium Chapter 16 Hein and Arena Version 1.1 Chapter Outline 16.1 Reversible Reactions 16.2 Rates of Reaction 16.8 Effect of Catalysts on Equilibrium 16.9 Equilibrium Constants 16.3 Chemical.

Chemical Equilibrium
Chapter 16
Hein and Arena
Version 1.1
1
Chapter Outline
16.1 Reversible Reactions
16.2 Rates of Reaction
16.8 Effect of Catalysts on
Equilibrium
16.9
16.3 Chemical Equilibrium
16.4 Le Chatelier’s Principle
Equilibrium Constants
16.10 Ion Product Constant for
Water
16.5 Effect of Concentration on
Equilibrium
16.11 Ionization Constants
16.6 Effect of Volume on
Equilibrium
16.13 Acid-Base Properties of
Salts
16.7 Effect of Temperature on
Equilibrium
16.14 Buffer Solutions:
The Control of pH
16.12 Solubility Product Constant
2
Reversible Reactions
3
reversible reaction A chemical reaction
in which the products formed react to
produce the original reactants.
4
The reaction between NO2 and N2O4
is reversible.
cooling
N2O4 is
formed
2NO2(g) → N2O4 (g)
N2O4 decomposes
when heated forming
NO2
heating
N2O4 (g) → 2NO2 (g)
5
reaction to the right
2NO2(g) → N2O4 (g)
→
reaction to the left
6
Rates of Reaction
7
chemical kinetics The study of reaction
rates and reaction mechanisms.
8
• The rate of a reaction is variable. It
depends on:
– concentrations of the reacting species
– reaction temperature
– presence or absence of catalysts
– the nature of the reactants
9
The concentration of A and B
decreases with time lowering
the rate of the forward reaction.
The concentration of C and D
increases with time increasing
the rate of the reverse reaction.
Forward reaction A + B → C + D
Reverse reaction C + D → A + B
16.2
10
Chemical Equilibrium
11
equilibrium:
a the
dynamic
state
which
At
equilibrium
concentrations
of the
chemical
equilibrium:
the in
state
in
two orthemore
processes
are
products
and
the thereactants
are
not
which
rate opposing
of
forward
reaction
taking place
at of
thethe
same
time reaction
and at the
changing.
equals
the rate
reverse
in
rate. change.
asame
chemical
12
A saturated salt solution is in equilibrium with
solid salt.
salt crystals
are dissolving
Na+ and Clare crystallizing
NaCl(s) → Na+(aq) + Cl-(aq)
→
At equilibrium the rate of salt dissolution
equals the rate of salt crystallization.
13
Le Chatelier’s Principle
14
In 1888, the French chemist Henri LeChatelier
This generalization, known as
set forth a far-reaching generalization on the
LeChatelier’s Principle, states
behavior of equilibrium systems.
If a stress or strain is applied to a system in
equilibrium, the system will respond in such a
way as to relieve that stress and restore
equilibrium under a new set of conditions.
15
Effect of Concentration
on Equilibrium
16
• For most reactions the rate of reaction
increases as reactant concentrations
increase.
• The manner in which the rate of
reaction changes with concentration
must be determined experimentally.
17
An equilibrium is disturbed when the
concentration of one or more of its
components is changed. As a result, the
concentration of all species will change
and a new equilibrium mixture will be
established.
18
The system is at equilibrium
results in C and results
D
in A and B
increases the rate of
being produced faster
being used faster than
the forward reaction
than they are used.they are produced.
A + B →C + D
→
Increasing the concentration of B
19
The system is again at equilibrium
In the new equilibrium
concentration of A has
decreased
concentrations of B, C
and D have increased
A+ B→C + D
→
After enough time has passed, the rates of the
forward and reverse reactions become equal.
20
Percent Yield
21
At The
equilibrium
forwardthe
reaction
rate
is HI
79%
formation
complete
equals
at
1.58
mol of
HI
% Yield =
x 100% = 79%
theequilibrium.
rate of HI decomposition.
2 mol HI
H2 + I2 combine
to form HI
HI decomposes
to form H2 + I2
700 K
H2(g) + I2(g) → 2HI(aq)
→
0.21
1 mol
0
mol
0.21
01 mol
mol
1.58
20 mol
mol
Final Concentrations in the
Equilibrium
Initial Concentrations
Concentrations
Absence of Equilibrium
22
Comparison of Equilibria
Original Equilibrium
New Equilibrium
1.00 mol H2 + 1.00 mol I2
1.00 mol H2 + 1.20 mol I2
Yield: 79% HI
Yield: 85% HI
Equilibrium mixture contains Equilibrium mixture contains
1.58 mol HI
1.70 mol HI
0.21 mol H2
0.15 mol H2
0.21 mol I2
0.35 mol I2
23
Effect of Concentration Changes
on the Chlorine Water Equilibrium
+increase
decrease Cldecrease
increase
H
H
O
O
increase
HOCl
Cl
2
32
concentrationconcentration
concentrationconcentration
concentration
Cl2(aq) + 2H2O(l) → HOCl(aq) + H3O+(aq) + Cl-(aq)
→
toright
left
Equilibrium
Equilibriumshifts
shiftsto
24
Effect of C22H
O
H33O22
Concentration Changes on pH
Add 0.200
0.100 mol NaC2H3O2
NaC2H3O2(aq) →
Na+(aq)
C2H3O2(aq)
+ C H O (aq)
Equilibrium shifts to left
HC2H3O2(aq) + H2O(l) → H3O+(aq) + C
C22H
H3O
O2-2(aq)
(aq)
→
1 L 0.100 M HC2H3O2
Equilibrium pH = 5.05
2.87
4.74
25
Effect of Volume on
Equilibrium
26
• Changes in volume significantly affect
the reaction rate only when one or more
of the reactants or products is a gas and
the reaction is run in a closed container.
• The effect of decreasing the volume is
to increase the concentrations of any
gaseous reactants or products.
27
Decrease Volume
increases CO2
concentration
CaCO3(s) → CaO(s) + CO2(g)
→
Equilibrium shifts to left
28
Increase Volume
decreases CO2
concentration
CaCO3(s) → CaO(s) + CO2(g)
→
Equilibrium shifts to right
29
In a system composed entirely of gases, a
decrease in the volume of the container will
cause the reaction and the equilibrium to
shift to the side that contains the smallest
number of molecules.
30
Decrease Volume
N2(g) + 3H2(g) → 2NH3(g)
→
1 mol
3 mol
6.02 x 1023
molecules
1.81 x 1024
molecules
2 mol
1.20 x 1024
molecules
2.41 x 1024
molecules
Equilibrium shifts to the right towards fewer molecules.
31
Decrease Volume
N2(g) + O2(g) → 2NO(g)
→
1 mol
1 mol
6.02 x 1023
molecules
6.02 x 1023
molecules
2 mol
1.20 x 1024
molecules
1.20 x 1024
molecules
Equilibrium does not shift. The number of molecules is
the same on both sides of the equation.
32
Effect of Temperature
on Equilibrium
33
Thea rate
In
reversible
of the reaction,
reaction that
the rates
absorbs
of both
heat
When the temperature of a system is
theincreased
is
forward and
to athe
greater
reverse
extent,
reactions
and the
are
raised, the rate of reaction increases.
increased byshifts
equilibrium
an increase
to favorinthat
temperature.
reaction.
34
Heat may be treated as a reactant
in endothermic reactions.
oC moles CO
At room
temperature
very
COCO
forms.
At 1000
moles
2 little
C(s) + CO2(g) + heat → 2CO(g)
→
Equilibrium shifts to right
35
Effect of Catalysts
on Equilibrium
36
A catalyst is a substance that influences
the rate of a reaction and can be
recovered essentially unchanged at the
end of the reaction.
A catalyst does not shift the equilibrium of
a reaction. It affects only the speed at
which the equilibrium is reached.
37
Energy Diagram for an Exothermic Reaction
Activation
A
catalyst
catalyst energy:
speeds
does not
up
the achange
minimum
reactionthe
energy
by energy
lowering
required
ofthe
a
38
for a reaction
activation
reaction.
energy.
to occur.
16.3
AlCl3
PCl3(l) + S(s) → PSCl3(l)
In thelittle
Very
presence
thiophosphoryl
of an aluminum
chloridechloride
is formed
catalyst
in the
absence is
reaction
ofcomplete
a catalystinbecause
a few seconds.
the reaction is so slow.
MnO2
2KClO3(s) → 2KCl + 3O2(l)
Δ
The laboratory preparation of oxygen uses manganese
dioxide as a catalyst to increase the rate of the reaction.
39
Equilibrium Constants
40
At equilibrium the rates of the forward
and reverse reactions are equal, and the
concentrations of the reactants and
products are constant.
41
The equilibrium constant (Keq) is a value
representing the unchanging concentrations
of the reactants and the products in a
chemical reaction at equilibrium.
42
For the general reaction
aA + bB → cC + dD
→
at a given temperature
c
K eq
d
[C] [D]
=
a
b
[A] [B]
43
For the reaction
3H2 + N2 → 2NH3
→
2
K eq
[NH3 ]
=
3
[H2 ] [N2 ]
44
For the reaction
4NH3 + 3O2 → 2N2 + 6H2O
→
2
K eq
6
[N2 ] [H2O]
=
4
3
[NH3 ] [O2 ]
45
The magnitude of an equilibrium constant
indicates the extent to which the forward and
reverse reactions take place.
3H2 + I2 → 2HI3
→
2
K eq
[HI]
o
=
= 54.8 at 425 C
3
[H2 ] [I2 ] At equilibrium more product
At equilibrium more reactant
than
exists.
than reactant
product exists.
COCl2 → CO + Cl2
→
K eq
[CO][Cl2 ]
-4
o
=
= 7.6 x 10 at 400 C
[COCl2 ]
46
When the molar concentrations of all
species in an equilibrium reaction are
known, the Keq can be calculated by
substituting the concentrations into the
equilibrium constant expression.
47
Calculate the Keq for the following reaction based on
concentrations of PCl5 = 0.030 mol/L, PCl3 = 0.97
mol/L and Cl2 = 0.97 mol/L at 300oC.
PCl5(g ) → PCI3(g) + Cl2(g)
→
K eq
[PCl3 ][Cl2 ] (0.97)(0.97)
= 31
=
=
(0.030)
[PCl5 ]
48
Ion Products Constant
for Water
49
Water autoionizes to a slight degree.
H2O + H2O → H3O+ + OH-
→
or more simply
H2O → H+ + OHAt equilibrium
→
[H+] = [OH-] = 1.00 x 10-7 mol/L
The water equilibrium constant, Kw, is called the
ion product constant for water.
Kw = [H+][OH-] = 1.00 x 10-14 T = 25oC
Kw = (1 x 10-7)(1 x 10-7) = 1.00 x 10-14
50
What is the concentration of (a) H+ and (b) OH- in a
0.001 M HCl solution?
HCl is 100% ionized.
HCl → H+ + Cl-
[H+] = 1 x 10-3 M
Solve the Kw expression for [OH-].
Kw= [H+][OH-] = 1.00 x 10-14
-14
K w 1 x 10
-11
[OH ] = + =
= 1 x 10 mol/L
-3
[H ] 1 x 10
-
51
What is the pH of a 0.010 M NaOH solution?
NaOH is 100% ionized.
NaOH → Na+ + OH-
[OH-] = 1 x 10-2 M
Solve the Kw expression for [H+].
Kw= [H+][OH-] = 1.00 x 10-14
-14
K w + 1 x 10
-12
-12
[H pH
] == - log[H
x 10 ) =mol/L
12
= ] = - log(1.0
-2 = 1 x 10
[OH ] 1 x 10
+
52
What is the pH of a 0.010 M NaOH solution?
The pH may also be calculated by first
calculating the pOH.
-] = 1xx10
pOH = - log[OH-] = -[OH
log(1.0
10-2-2) =
M2
In pure water
pH + pOH = 14
Solve for pH
pH = 14 - pOH
pH = 14 – 2 = 12
53
Relationship of H+ and OHConcentrations in Water Solutions
[H+]
[OH-]
Kw
pH
pOH
1.00 x 10-2
1.00 x 10-12 1.00 x 10-14
2.00
12.00
1.00 x 10-4
1.00 x 10-10 1.00 x 10-14
4.00
10.00
2.00 x 10-6
5.00 x 10-9
1.00 x 10-14
5.70
8.30
1.00 x 10-7
1.00 x 10-7
1.00 x 10-14
7.00
7.00
1.00 x 10-9
1.00 x 10-5
1.00 x 10-14
9.00
5.00
54
16.1
Ionization Constants
55
In addition to Kw, several other ionization
constants are used.
56
Ka
57
When acetic acid ionizes in water,
following equilibrium is established.
+
→
HC H O (aq)
H +C H O
3
→
2
2
2
3
the
22
Ka is the ionization constant for this equilibrium.
+
2
[H ][C2H3O ]
Ka =
[HC2H3O2 ]
Ka is called the acid ionization constant.
Since the concentration of water is large and does
not change appreciably, it is omitted from Ka. 58
At 25oC, a 0.100 M solution of HC2H3O2 is 1.34%
ionized and has an [H+] of 1.34 x 10-3 mol/L.
Calculate Ka for acetic acid ?
+
→
HC2H3O2(aq)
H + C2H3O22
→
Because each molecule of HC2H3O2 that
+
ionizes yields one H and one C 2H33O22 , the
concentrations of the two ions are equal.
+
-3
[H ] = [C2H3O2 ] = 1.34 x 10 mol/L
The moles of unionized acetic acid per liter are
0.100 mol/L – 0.00134 mol/L = 0.099 mol/L
59
Substitute these concentrations into the
equilibrium expression and solve for Ka.
+
2
-3
[H ] = [C2H3O ] = 1.34 x 10 mol/L
[HC2H3O2] = 0.099 mol/L
+
2
[H ][C2H3O ]
Ka =
[HC2H3O2 ]
-3
-3
(1.34x10 )(1.34x10 )
-5

= 1.8x10
(0.099)
60
What is the [H+] in a 0.50 M HC2H3O2 solution? The
ionization constant, Ka, for HC2H3O2 is 1.8 x 10-5.
The equilibrium expression and Ka for HC2H3O2
are
+
→
HC2H3O2(aq)
H + C2H3O22
→
+
2
[H ][C2H3O ]
Ka =
[HC2H3O2 ]
+
[H ] = [C
O ] LetofY HC
Because
each
that
= [H
=2[C
2H3] O
2H3molecule
2H3O2 ]
+
ionizes
one
Hequilibrium
and one C2is
H330.50
O22 , the
[HCyields
H
O
]
at
– Y.
2 3 2
+
2
concentrations of the two ions are equal.
61
Substitute these values into Ka for HC2H3O2.
Y=
[H+]
2
= [C2H3O ]
HC2H3O2 = 0.50 - Y
+
2
[H ][C2H3O ]
Ka =
[HC2H3O2 ]
2
2
2
(Y)(Y)
(Y)(Y) Y Y Y
-5
-5 -5
Ka =
= = = 1.8
= 1.8
=x1.8
10 x 10
x 10
0.50
0.50- 0.50
Y 0.50
0.50 - Y
Solve for Y2.
Assume Y is small
compared to
2
Then 0.50 – Y  0.50
0.50 -Y.
-5
Y = 0.50 x 1.8 x 10
62
Take the square root of both sides of the
2
-5
-5
-6
equation. Y = 0.50
100 x 10
0.90 x 1.8
10 x= 9.
-6
Y = 9.0 x 10
+
-3
= 3.0 x 10
[H ] = 3.0 x 10
-3
mol/L
mol/L
Making no approximation and using the
quadratic equation the answer is 2.99 x 10-3
mol/L, showing that it was justified to assume
Y was small compared to 0.5.
63
Calculate the percent ionization in a 0.50 M HC2H3O2
solution.
The ionization of a weak acid is given by
HA → H+ + A-
→
The percent ionization is given by
concentration of [H+ ] or [A - ]
x 100= percent ionized
initial concentration of [HA]
65
Calculate the percent ionization in a 0.50 M HC2H3O2
solution.
The ionization of acetic acid is given by
+
→
HC2H3O2(aq)
H + C2H33O22
→
The percent ionization of acetic acid is given by
concentration of [H+ ] or [C2H3O-2 ]
x 100= percent ionized
initial concentration of [HC2H3O2 ]
-3 +
[H
] was previously calculated
3.0 x 10 mol/L
x 100=
0.60% percent ionized
-3
as 3.0 x 10 mol/L
0.50 mol/L
66
Ionization Constants (Ka) of
Weak Acids at 25oC
Acid
Formula
Ka
Acid
Formula
Ka
Acetic
HC2H3O2
1.8 x 10-5
Hydrocyanic
HCN
4.0 x10-10
Benzoic
Carbolic
HC7H5O2
HC6H5O
6.3 x 10-5
1.3 x 10-10
Cyanic
HCNO
2.0 x 10-4
Formic
HCHO2
1.8 x 10-4
Hypochlorous
Nitrous
Hydrofluoric
HOCl
HNO2
HF
3.5 x10-8
4.5 x10-4
6.5 x10-4
67
16.2
Solubility Product Constant
68
chemical equilibrium
equilibrium
A dynamic state
The in
state
which
in
The solubility product constant, Ksp, is
two orthemore
which
rate opposing
of the forward
processes
reaction
are
the equilibrium constant of a slightly
taking place
equals
the rate
at of
thethe
same
reverse
time reaction
and at the
in
soluble salt.
asame
chemical
rate. change.
69
Silver chloride is in equilibrium with its ions in
aqueous solution.
AgCl(s) → Ag+(aq) + Cl-(aq)
→
The equilibrium constant+is
--
Ksp = [Ag
][Cl
]]
[Ag
][Cl
The product of
Keq=and
K eq
[AgCl(s) is a constant.[AgCl(s )]
+
Rearrange
+
-
K eq x [AgCl(s)] = [Ag ][Cl ] = K sp
The The
amount
concentration
of solid
AgClof does
solid not
AgClaffect
is a
the equilibrium.
constant.
70
The solubility of AgCl in water is 1.3 x 10-5
mol/L.
AgCl(s) → Ag+(aq) + Cl-(aq)
Because each formula unit of AgCl that
dissolves yields one Ag+ and one Cl- , the
concentrations of the two ions are equal.
→
[Ag+] = [Cl-] = 1.3 x 10-5 mol/L
Ksp = [Ag+][Cl-] The Ksp has no denominator.
= (1.3 x 10-5)(1.3 x 10-5) = 1.7 x 10-10
71
When the product of the molar
concentration of the ions in solution
(each raised to its proper power) is
greater than the Ksp for that substance,
precipitation will occur. If the ion
product is less than the Ksp value no
precipitation will occur.
72
The Ksp value for lead sulfate is 1.3 x 10-8. Calculate
the solubility of PbSO4 in grams per liter.
The equilibrium equation of PbSO4 is
2+
2→
PbSO4
Pb + SO4
→
The Ksp of PbSO4 is
2+
2K sp = [Pb ][SO4 ]
Because each formula unit of PbSO4 that
dissolves yields one Pb2+ and one SO 2, the
4
concentrations of the two ions are equal.
2+
24
Let Y = [Pb ] = [SO ]
73
The Ksp value for lead sulfate is 1.3 x 10-8. Calculate
the solubility of PbSO4 in grams per liter.
Substitute Y into the Ksp equation.
2+
24
K sp = [Pb ][SO ]
K sp = (Y)(Y) = 1.3 x 10
2
-8
-8
Y = 1.3 x 10
-4
Y = 1.3 x 10 = 1.1 x 10 mol/L
-8
The solubility of PbSO4 is 1.1 x 10-4 mol/L.
74
The Ksp value for lead sulfate is 1.3 x 10-8. Calculate
the solubility of PbSO4 in grams per liter.
Convert mol/L to grams/L. The molar mass of
PbSO4 is 303.3 g/mol.
 1.1 x 10 mol   303.3 g 
-2

 = 3.3 x 10 g/L


L

  mol 
-4
The solubility of PbSO4 is 3.3 x 10-2 g/L.
75
The Common Ion Effect
76
A shift in the equilibrium position upon
An ion added to a solution already
addition of an ion already contained in
containing that ion is called a common
the solution is known as the common
ion.
ion effect.
77
The equilibrium equation of AgCl is
AgCl(s) → Ag+(aq) + Cl-(aq)
→
Upon the addition of Ag+, the
equilibrium shifts to the left, in
accordance with LeChatelier’s
Principle.
Silver nitrate dissociates in aqueous solution.
+
AgNO3 (s)  Ag (aq) + NO3 (aq)
Ag+ is common to both reactions.
78
Silver nitrate is added to a saturated AgCl solution
until the [Ag+] 0.10 M. What will be the [Cl-]
remaining in solution.
Use Ksp of AgCl to determine [Cl-].
+
-10
K sp = [Ag ][Cl ] = 1.7 x 10
Substitute [Ag+] into the Ksp.
-10
[0.10][Cl ] = 1.7 x 10
Solve for [Cl-].
-10
1.7 x 10
-9
[Cl ] =
= 1.7x10 mol/L
[0.10]
This
In the
is an
absence
example
of AgNO
of the 3common
, [Cl-] =1.3
ionx effect.
10-5. 79
Solubility Product Constants (Ksp) at 25oC
16.3
Compound
Ksp
Compound
Ksp
AgCl
1.7 x 10-10
CaF2
3.9 x 10-11
AgBr
5 x 10-13
CuS
9 x 10-45
AgI
8.5 x 10-17
Fe(OH)3
6 x 10-38
AgC2H3O2
2 x 10-3
PbS
7x 10-29
Ag2CrO4
1.9 x 10-12
PbSO4
1.3 x 10-8
BaCrO4
8.5 x 10-11
Mn(OH)2
2.0 x 10-13
BaSO4
1.5 x 10-9
80
Acid-Base
Properties of Salts
81
hydrolysis is the term used for the general
reaction in which a water molecule is split.
82
Salts that contain the anion of a weak acid
under go hydrolysis.
The net ionic equation for the hydrolysis of
sodium acetate is
→
C2H3O (aq) + HOH(l) HC2H3O2 (aq) + OH (aq)
→
2
The water
The solution
molecule is
splits.
basic.
83
Salts that contain the cation of a weak
base under go hydrolysis.
The net ionic equation for the hydrolysis of
ammonium chloride is
+
→
NH (aq) + HOH(l ) NH3 (aq) + H3O (aq )
→
+
4
The
Thesolution
water
molecule
is acidic.
splits.
84
Salts derived from a strong acid and a strong
base do not undergo hydrolysis.
Na+ + Cl- + HOH(l ) → no reaction
85
Ionic Composition of Salts and the Nature of
the Aqueous Solutions They Form
Type of salt
Nature of
Aqueous Solution
Examples
Weak base-strong acid
Acid
NH4Cl, NH4NO3
Strong base-weak acid
Basic
NaC2H3O2
Weak base-weak acid
Depends on the salt
NH4C2H3O2, NH4NO2
Strong base-strong acid
Neutral
NaCl, KBr
86
16.4
Buffer Solutions:
The Control of pH
87
A buffer solution resists changes in pH
when diluted or when small amounts of
acid or base added.
88
A weak acid mixed with its conjugate base
form a buffer solution.
Sodium acetate when mixed with acetic acid
forms a buffer solution.
89
A weak acid mixed with its conjugate base
form a buffer solution.
If a small amount of HCl is added, the acetate
ions of the buffer will react with the H+ of the
HCl to form unionized acetic acid.
H (aq) + C2H3O (aq)  HC2H3O2 (aq)
+
2
90
A weak acid mixed with its conjugate base
form a buffer solution.
If a small amount of NaOH is added, the acetic
acid molecules of the buffer will react with the
OH- of the NaOH to form water.
OH + HC2H3O2 (aq)  H2O(l ) + C2H3O (aq)
-
2
91
Changes in pH Caused by the
Addition of HCl and NaOH
Solution
H2O (1000 mL)
pH
Change in pH
7
H2O + 0.010 mol HCl
2
5
H2O + 0.010 mol NaOH
12
5
0.10 M HC2H3O2 +
0.10 M NaC2H3O2
4.74
–
Buffer + 0.010 mol HCl
4.66
0.08
Buffer + 0.010 mol NaOH
4.83
0.09
Buffer solution (1000 mL)
92
16.5
Key Concepts
16.1 Reversible Reactions
16.2 Rates of Reaction
16.8 Effect of Catalysts on
Equilibrium
16.9
16.3 Chemical Equilibrium
16.4 Le Chatelier’s Principle
Equilibrium Constants
16.10 Ion Product Constant for
Water
16.5 Effect of Concentration on
Equilibrium
16.11 Ionization Constants
16.6 Effect of Volume on
Equilibrium
16.13 Acid-Base Properties of
Salts
16.7 Effect of Temperature on
Equilibrium
16.14 Buffer Solutions:
The Control of pH
16.12 Solubility Product Constant
93