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1
Reversible Reactions
2
reversible reaction - a chemical reaction
in which the products formed react to
produce the original reactants.
3
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)
4
reaction to the right (forward)
2NO2(g) → N2O4 (g)
→
reaction to the left (reverse)
Ice water
Hot water
5
Rates of Reaction
6
• 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
7
Chemical Equilibrium
9
chemical equilibrium
equilibrium
A dynamic state
The in
state
which
in
At equilibrium the concentrations of the
two orthemore
which
rate opposing
of the forward
processes
reaction
are
products and the reactants are not
taking place
equals
the rate
at of
thethe
same
reverse
time reaction
and at the
in
changing.
asame
chemical
rate. change.
10
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.
11
Le Chatelier’s Principle
12
Henri LeChatelier
13
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.
14
Effect of Concentration
on Equilibrium
15
• 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.
16
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.
17
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
18
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.
19
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)
→
Equilibrium
Equilibriumshifts
shiftsto
toright
left
20
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
H33O
O2-2(aq)
(aq)
→
1 L 0.100 M HC2H3O2
Equilibrium pH = 5.05
2.87
4.74
21
Effect of Pressure
on Equilibrium
22
• Changes in pressure 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 increasing the pressure is
to increase the concentrations of any
gaseous reactants or products.
23
Increase Pressure
increases CO2
concentration
CaCO3(s) → CaO(s) + CO2(g)
→
Equilibrium shifts to left
24
Decrease Pressure
decreases CO2
concentration
CaCO3(s) → CaO(s) + CO2(g)
→
Equilibrium shifts to right
25
In a system composed entirely of gases, a
increase in the pressure of the container
will cause the reaction and the equilibrium
to shift to the side that contains the smallest
number of molecules.
26
Increase Pressure
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.
27
Increase Pressure
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.
28
Effect of Temperature
on Equilibrium
29
The rate of the reaction that absorbs heat
is increased to a greater extent, and the
equilibrium shifts to favor that reaction.
In a reversible reaction, the rates of both
When the process
temperature
is endothermic,
of a systemthe
is
the forward and the reverse reactions are
raised, the(left
forward
rate oftoreaction
right)increases.
reaction is
increased by an increase in temperature.
increased. When the process is
exothermic, the reverse (right to left)
process is increased.
30
Heat may be treated as a reactant
in endothermic reactions.
oC moles CO
At room
temperature
very
little
COCO
forms.
At 1000
moles
2 
C(s) + CO2(g) + heat → 2CO(g)
→
Equilibrium shifts to right
31
Effect of Catalysts
on Equilibrium
32
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.
https://www.youtube.com/watch?v=DP05fxR4oUk
33
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
34
for a reaction
activation
reaction.
energy.
to occur.
AlCl3
PCl3(l) + S(s) → PSCl3(l)
In the little
Very
presence
thiophosphoryl
of a catalystchloride
the reaction
is formed
is complete
in the
in
absence
a
few seconds.
of a catalyst because the reaction is so slow.
MnO2
2KClO3(s) → 2KCl + 3O2(g)
Δ
The laboratory preparation of oxygen uses manganese
dioxide as a catalyst to increase the rate of the reaction.
35
Equilibrium Constants
36
At equilibrium the rates of the forward
and reverse reactions are equal, and the
concentrations of the reactants and
products are constant.
37
The equilibrium constant (Keq) is a value
representing the unchanging concentrations
of the reactants and the products in a
chemical reaction at equilibrium.
38
For the general reaction
aA + bB → cC + dD
→
reactants
products
at a given temperature:
The concentrations of A, B, C, and D represent the equilibrium concentrations.
The brackets around [A], [B], [C], and [D] represent concentrations in Molarity.
The products are written on the top of the fraction & the reactants on the bottom.
The coefficients to balance the equation a, b, c, and d are written as exponents.
39
For the general reaction
aA + bB → cC + dD
→
at a given temperature
c
K eq
d
[C] [D]
=
a
b
[A] [B]
40
For the reaction
3H2 + N2 → 2NH3
→
2
K eq
[NH3 ]
=
3
[H2 ] [N2 ]
41
For the reaction
4NH3 + 3O2 → 2N2 + 6H2O
→
2
K eq
6
[N2 ] [H2O]
=
4
3
[NH3 ] [O2 ]
42
The magnitude of an equilibrium constant
indicates the extent to which the forward and
reverse reactions take place.
H2 + I2 → 2HI
→
2
K eq
[HI]
o
=
= 54.8 at 425 C
[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 ]
43
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.
44
Calculate the Keq for the following reaction 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 ]
45
Ionization Constants
46
In addition to Kw, several other ionization
constants are used.
47
Ka
48
When acetic acid ionizes in water,
following equilibrium is established:
+
→
HC H O
H +C
CH
HO
3
→
2
2(aq)
22 33
the
2
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. 49
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
50
Substitute these concentrations into the
equilibrium expression and solve for Ka.
+
-3
[H ] = [C2H3O2 ] = 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)
51
What is the [H+] & pH 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 ]
Because each molecule of HC2H3O2 that
+
ionizes yields one H and one C2 H33O22 , the
concentrations of the two ions are equal.
52
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 ]
+
2
[H ] = [C2H3O ]
Let x =
[H+]
2
= [C2H3O ]
[HC2H3O2] at equilibrium is 0.50 – x.
53
Substitute these values into Ka for HC2H3O2.
x=
[H+]
2
= [C2H3O ] HC2H3O2 = 0.50 - x
+
2
[H ][C2H3O ]
Ka =
[HC2H3O2 ]
2
(x)(x)
x
-5
Ka =
=
= 1.8 x 10
0.50 - x 0.50 - x
Assume x is small compared to
0.50 - x. Then 0.50 – x  0.50
54
Substitute these values into Ka for HC2H3O2.
x=
[H+]
2
= [C2H3O ] HC2H3O2 = 0.50 - x
+
2
[H ][C2H3O ]
Ka =
[HC2H3O2 ]
2
(x)(x)
x
-5
Ka =
=
= 1.8 x 10
0.50
0.50
2
x
-5
= 1.8 x 10
0.50
Solve for x2.
2
x = 0.50 x 1.8 x 10
-5
55
Take the square root of both sides of the
2
-5
-5
-6
equation. x == 0.90
0.50 x 1.8
10 x=10
9. 0 x 10
-6
-3
x = 9.0 x 10 = 3.0 x 10
+
[H ] = 3.0 x 10
-3
mol/L
mol/L
-3
pH = -log(3.0 x 10 ) = 2.52
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.
56
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]
57
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
3.0 x 10 mol
/ L previously calculated
x 100=
0.60% percent ionized
-3
as 3.0 x 10 mol/L
0.50 mol/L
58
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
HC7H5O2
6.3 x 10-5
Hypochlorous
HOCl
3.5 x10-8
Carbolic
HC6H5O
1.3 x 10-10
Nitrous
HNO2
4.5 x10-4
Cyanic
HCNO
2.0 x 10-4
Hydrofluoric
HF
6.5 x10-4
Formic
HCHO2
1.8 x 10-4
59
Ionic Equilibria
Problems
60
Use Ka as a measure of acid strength.
Let 1 = Hi and 5 = Lo
Acid
Ka
order
HC2H3O2 1.8  10–5
H2S
6  10–8
HF
7  10– 4
HC2Cl3O2 2  10–1
HCN
7.2  10–10
61
Use Ka as a measure of acid strength.
Let 1 = Hi and 5 = Lo
Acid
Ka
HC2H3O2 1.8  10–5
order
3
H2S
6  10–8
4
HF
7  10– 4
2
HC2Cl3O2 2  10–1
1
HCN
7.2  10–10
5
62
Find the pH in 0.1 M HCN if
Ka = 7.2  10–10
Step 1: ionize the acid
+

→
HCN
H  CN
→
Step 2: write the Ka expression
+
-
[H ][CN ]
Ka =
[HCN]
63
Step 3: let x = [H+] = [CN–]
[HCN] = 0.1 M – x ≈ 0.1 M
Step 4: substitute into Ka expression
[x][x]
10
Ka =
 7.2  10
[0.1]
Step 5: solve for x:
x = 7.2  10
2
11
x  7.2  10
11
 8.5  10
6
64
Step 6: solve for pH
6
pH  - log 8.5 10  5.07
65
Find [H+] in 0.1 M H2CO3 if
Ka = 4.0  10–17
Step 1: ionize the acid
→
H2CO3
→
2H  CO
+
-2
3
Step 2: write the Ka expression
+ 2
-2
3
[H ] [CO ]
Ka =
[H2CO3 ]
66
Step 3: let x = [H+] , [CO3–2] = ½ x
[H2CO3] = 0.1 M – ½ x ≈ 0.1 M
Step 4: substitute into Ka expression
2 x
2
3
[x] [ ] x
17
Ka =

 4.0 10
[0.1] 0.2
Step 5: solve for x:
x = 8.0  10
3
18
x  8.0  10
3
18
 2.0  10
6
67
Find the pH of a 0.001 M solution of
bombastic hydroxide (BmOH) if
Kb = 1.6  10–10
Step 1: ionize the base
+

→
BmOH
Bm  OH
→
Step 2: write the Kb expression
+
-
[Bm ][OH ]
Kb =
[BmOH]
68
Step 3: let x = [Bm+] = [OH–]
[BmOH] = 0.001 M – x ≈ 0.001 M
Step 4: substitute into Kb expression
[x][x]
10
Kb =
 1.6  10
[0.001]
Step 5: solve for x:
x = 1.6  10
2
13
x  1.6  10
13
 4.0  10
7
69
Step 6: solve for pOH
7
pOH  - log 4.0 10  6.40
Step 7: solve for pH
pH  14  6.40  7.60
70
• On the yellow “Equilibria in Aqueous
Solutions” Self-Test, you may omit
problems 11 – 16.
• On the pink “Redox and
Electrochemistry: Self-Test, you may
omit all problems except 5 & 6.
71
Find the Ka for a 0.01 M HCN
solution if the pH = 6.3
Step 1: ionize the acid
+

→
HCN
H  CN
→
Step 2: write the Ka expression
+
-
[H ][CN ]
Ka =
[HCN]
72
Step 3: use pH to find [H+]
[H+] = 10–pH = 10–6.3 = 5.0  10–7
[H+] = [CN–],
[HCN] = 0.01- 5.0  10–7 ≈ 0.01
Step 4: substitute into Ka expression
7
7
[5.0 10 ][5.0 10 ]
11
Ka =
 2.5 10
[0.01]
73
Solubility Product Constant
74
The solubility product constant, Ksp, is
the equilibrium constant of a slightly
(sparingly) soluble salt.
75
Silver chloride is in equilibrium with its ions in
aqueous solution.
AgCl(s) → Ag+(aq) + Cl-(aq)
→
The equilibrium constant+is
--
K sp = [Ag
][Cl
]]
[Ag
][Cl
The product of
Keq=and
K eq
[AgCl(s) is a constant.[AgCl(s )]
+
Rearrange
+
-
Keq x [AgCl(s)] = [Ag ][Cl ] = Ksp
The The
amount
concentration
of solid
AgClof does
solid not
AgClaffect
is a
the equilibrium.
constant.
76
Ksp for Sparingly Soluble Solids
To find Ksp:
1. ionize the salt
2. write the Ksp expression
3. determine the M’s of the ions
4. substitute into the Ksp expression
77
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
78
Molar Solubility for Sparingly
Soluble Solids
1.
2.
3.
4.
5.
6.
ionize the salt
write the Ksp expression
let x equal the amount of salt that
dissolves
find concentrations of ions in terms of x
substitute in the Ksp expression
solve for x
79
The Ksp value for lead sulfate is 1.3 x 10-8. Calculate
the solubility of PbSO4 in moles 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
22+
dissolves yields one Pb and one SO 4 , the
concentrations of the two ions are equal.
2+
24
Let x = [Pb ] = [SO ]
80
The Ksp value for lead sulfate is 1.3 x 10-8. Calculate
the solubility of PbSO4 in moles per liter.
Substitute x into the Ksp equation.
2+
24
K sp = [Pb ][SO ]
K sp = (x)(x) = 1.3 x 10
2
x = 1.3 x 10
-8
x = 1.3 x 10
-8
-8
-4
= 1.1 x 10 mol/L
The solubility of PbSO4 is 1.1 x 10-4 mol/L.
81
Solubility Product Constants (Ksp) at 25oC
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
82
Practice
83
The Ksp value for silver carbonate is
6.2 x 10-12. Calculate the solubility of
Ag2CO3 in moles per liter.
+
2→
Ag2CO3
2Ag + CO3
→
+ 2
23
K sp = [Ag ] [CO ]
If x = amount of salt that dissolves,
then [Ag+] = 2x, [CO3–2] = x
84
2
K sp = [2x] [x] = 4x3
4x3 = 6.2 x 10-12
x3 = 1.6 x 10-12
3
-12
x = 1.6 x 10
x = 1.2 x 10-4 mol/L
The solubility of silver carbonate is
1.2 x 10-4 mol/L
85
The Ksp value for calcium phosphide is
1.7 x 10-40. Calculate the solubility of
Ca3P2 in moles per liter.
Solution in lecture!
86
Acid-Base
Properties of Salts
87
hydrolysis is the term used for the general
reaction in which a water molecule is split.
88
Salts that contain the anion of a weak acid
undergo 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.
89
Salts that contain the cation of a weak
base undergo 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.
90
Salts derived from a strong acid and a strong
base do not undergo hydrolysis.
Na+ + Cl- + HOH(l ) → no reaction
91
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
Acidic
NH4Cl, NH4NO3
Strong base-weak acid
Basic
NaC2H3O2
Weak base-weak acid
Close to Neutral
NH4C2H3O2, NH4NO2
Strong base-strong acid
Neutral
NaCl, KBr
92
Buffer Solutions:
The Control of pH
93
A buffer solution resists changes in pH
when diluted or when small amounts of
acid or base are added.
94
A weak acid mixed with its conjugate base
form a buffer solution.
Sodium acetate when mixed with acetic acid
forms a buffer solution.
95
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
96
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)
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HC2H3O2
NaOH
NaC2H3O2
HCl
NaC2H3O2 + H2O NaCl + HC2H3O2
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