Lecture # 7 - Chapter 6
Chemical Equilibrium
(until mid-term exam)
This
week
aA + bB
K
Titrations
=
[C]c[D]d
[A]a[B]b
Concentration
Concentration
(in Standard State)
If Keq > 1, then reaction is favored
Activity and Chemical Equilibrium
Next
week
cC + dD
Chemical Equilibrium: Review/Preview
Standard State
1. Solutes: 1 M
2. Gases: 1 bar
3. Solids/Liquids: Pure solid/liquid
Midterm Examination
e.g. [A] = [A]/(1 M) = [A] (with no units)
Note: Pure liquid/solid, [A]/[A]standard state = 1
Chemical Equilibrium
Chemical Equilibrium
cC + dD
Example:
K’
BrO3 + 2
-
Cr3+
Br-
+ 4 H2 O
+ Cr2O7
2- +
8
=
=
[Br-][Cr2O72-][H+]8
[BrO3-][Cr3+]2
K’
=
Chemical Equilibrium
AB
AB + B-
[AB]
[A+][B-]
K2
=
A+ + B- + AB + B-
AB + AB2-
A+ + 2 B- + AB
AB + AB2-
A+ + 2 B-
AB2[AB2-]
[AB][B-]
K3 = K1 K2 =
[AB2-]
[AB][B-]
Example:
AgCl (s)
Ag+ + Cl-
K1 = 1.8 x 10-10
Ag+
AgCl (aq)
K2 = 2.0 x 103
+
Cl-
What is K for AgCl (s)
AB2-
x
[BrO3-][Cr3+]2
[Br-][Cr2O72-][H+]8
Chemical Equilibrium
AgCl (s)
Ag+
Equilibrium Constants are Multiplicative:
[AB]
[A+][B-]
BrO3- + 2 Cr3+ + 4 H2O
K<< 1, so reaction is NOT favored
Reactions are Additive:
=
1/K
= 1/K = 1/1 x 1011 = 1 x 10-11
K >> 1, so reaction is favored
A+ + B-
=
Br- + Cr2O72- + 8 H+
= 1 x 1011 (at 25°C)
K1
[A]a[B]b
[C]c[D]d
H+
K’
K
aA + bB
=
[AB2-]
[A+][B-]2
+
Cl-
AgCl (s) + Ag+ + ClAgCl (s)
K = K1 K2 = (1.8 x
AgCl (aq)?
Ag+ + ClAgCl (aq)
AgCl (aq) + Ag+ + ClAgCl (aq)
10-10)(2.0
x 103) = 3.6 x 10-7
1
Chemical Equilibrium and
Thermodynamics
Chemical Equilibrium and
Thermodynamics
Enthalpy Change (ΔH): Heat absorbed or released
Enthalpy Change (ΔH): Heat absorbed or released
Exothermic (-ΔH) = Heat released
Endothermic (+ΔH) = Heat absorbed
Exothermic (-ΔH) = Heat released
Endothermic (+ΔH) = Heat absorbed
ΔHo = Standard Enthalphy Change
ΔHo < 0, then reaction favored
Entropy Change (ΔS): Degree of disorder of
reactants and products; entropy of products minus
entropy of reactants.
ΔSo = Standard Enthalphy Change
Entropy Change (ΔS): Degree of disorder of
reactants and products; entropy of products minus
entropy of reactants.
ΔSo > 0, then reaction favored
*Reactants and products in their standard states
Chemical Equilibrium and
Thermodynamics
Chemical Equilibrium and
Thermodynamics
What if ΔHo and ΔSo are BOTH negative?
Example:
Gibb’s Free Energy (ΔG)
ΔG
HCl (g)
H+ (aq) + Cl- (aq)
ΔHo = -74.85 kJ/mol (at 25°C)
ΔSo = -130.4 J/(K ⋅ mol) (at 25°C)
Chemical Equilibrium and
Thermodynamics
What if ΔHo and ΔSo are BOTH negative?
=
ΔH
-
TΔS
ΔGo =
ΔHo TΔSo
(reactants and products in standard state)
ΔGo < 0, then reaction is favored.
Chemical Equilibrium and
Thermodynamics
Free Energy and Equilibrium
Example:
HCl (g)
H+ (aq) + Cl- (aq)
ΔHo = -74.85 kJ/mol (at 25°C)
ΔSo = -130.4 J/(K ⋅ mol) (at 25°C)
ΔGo
=
ΔHo
-
TΔSo
K = e -ΔG
/RT
R (Gas Constant) = 8.314472 J/(K ⋅ mol)
T = Temperature (K)
= (-74.85 kJ/mol) - (298.15 K)(-130.4 J/K ⋅ mol)
= -35.97 kJ/mol
ΔGo < 0, reaction favored
2
Chemical Equilibrium and
Thermodynamics
Le Châtelier’s Principle
Example:
HCl (g)
“the direction in which the system proceeds back
to equilibrium is such that the change is partially
offset”
aA + bB
cC + dD
At equilbrium
H+ (aq) + Cl- (aq)
ΔHo = -74.85 kJ/mol (at 25°C)
ΔSo = -130.4 J/(K ⋅ mol) (at 25°C)
ΔGo
=
ΔHo
=
-35.97 kJ/mol
K=e
-ΔG /RT
-
K
TΔSo
= e -(-35.97 x 103 J/mol)/(8.314472 J/(K ⋅ mol))(298.15 K)
=e
Reaction Quotient (Q):
Q
14.51
=
Le Châtelier’s Principle
[C]c[D]d
[A]a[B]b
At ANY concentration
[C]c[D]d
[A]a[B]b
K = 2.00 x 106
At equilibrium, Q = K
Le Châtelier’s Principle
Example:
Example:
BrO3- + 2 Cr3+ + 4 H2O
K
=
Br- + Cr2O72- + 8 H+
BrO3- + 2 Cr3+ + 4 H2O
[Br-][Cr2O72-][H+]8
[BrO3-][Cr3+]2
K
=
= 1 x 1011 (at 25°C)
=
Br- + Cr2O72- + 8 H+
[Br-][Cr2O72-][H+]8
[BrO3-][Cr3+]2
= 1 x 1011 (at 25°C)
At equilibrium,
Q
=
At equilibrium,
[1.0][0.1][5.0]8
=1x
1011
[0.043][0.003]2
Q
=
[1.0][0.2][5.0]8
[0.043][0.003]2
= 2 x 1011
=K
>K
Le Châtelier’s Principle
and Free Energy (ΔG)
Le Châtelier’s Principle
and Free Energy (ΔG)
Free Energy and Equilibrium
K = e -ΔG
/RT
= e -(ΔH - TΔS)/RT
= e -ΔH/RT ⋅ e ΔS/R
If ΔH > 0, increasing T will increase e -ΔH/RT and K
(i.e. endothermic)
Endothermic Reaction:
Heat + Reactants
Products
Exothermic Reaction:
Reactants
Products + Heat
If ΔH < 0, increasing T will decrease e -ΔH/RT and K
(i.e. exothermic)
3
Solubility Product
Solubility Product
+ - = water
+
- +
- +
- +
- +
- +
Solubility Product
Solubility Product
+
+
-
- +
- +
- +- +
+
-
+
-
+
+
-
-
Solubility Product
AaBb (s)
Ksp
a A+ (aq) + b B- (aq)
=
[A+]a[B-]b
AaBb dissociates until Ksp is reached
If excess solid AB, the solution is saturated
Ion Pairs: e.g. AgCl (s) vs. AgCl (aq)
Common Ion Effect: “a salt will be less soluble if one of
its constituent ions is already present in the solution”
4
Complex Formation
= 7.9 x 10-9
K1 = [PbI]/[Pb2+][I-] = 1.0 x 102
Pb2+ + 2 I-
PbI2 (aq) β1 = [PbI2 (aq)]/[Pb2+][I-]2 = 1.4 x 103
Pb2+
PbI3-
β2 = [PbI3-]/[Pb2+][I-]3 = 8.3 x 103
PbI42-
β3 = [PbI42- ]/[Pb2+][I-]4 = 3.0 x 104
+3
I-
Pb2+ + 4 I-
Acids and Bases
What is an Acid? What is a Base?
Lewis Acids/Bases
Lewis Acid = Accepts a pair of electrons
Lewis Base = Donates a pair of electrons
:
PbI+
Pb2+ + 2 I-
e.g.
++Pb
+ : I :-
[Pb
I :]+
:
Pb2+ + I-
[Pb2+][I-]2
:
Ksp =
Ksp
:
PbI2 (s)
5
Acids and Bases
Acids and Bases
What is an Acid? What is a Base?
Brønsted-Lowry Acids/Bases
Acid = Proton Donor
Base = Proton Acceptor
e.g.
HCl (g) + NH3 (g)
(acid)
NH4+Cl - (s)
(base)
In water:
NH4+Cl - (s)
What is an Acid? What is a Base?
Brønsted-Lowry Acids/Bases
Acid = Proton Donor
Base = Proton Acceptor
(salt)
HCl (g) + H2O
H3O+ (aq) + Cl- (aq)
(acid)
(conjugate (conjugate
acid)
base)
NH4+(aq) + Cl- (aq)
HCl (g) + NH3 (g)
NH4+Cl - (s)
HCl (g) + NH3 (g)
(acid) (base)
NH4+(aq) + Cl- (aq)
(conjugate (conjugate
acid)
base)
HCl (g)
H+ (aq) + Cl- (aq)
Proton
Acids and Bases
Acids and Bases
What is an Acid? What is a Base?
Autoprotolysis of Water
Brønsted-Lowry Acids/Bases
Acid = Proton Donor
Base = Proton Acceptor
Protic Solvents = “have a reactive H+”
Water can donate and accept a proton
Monoprotic Acids = Acids which donate one
proton (H+)
Polyprotic Acids = Acids which donate more than
one proton (H+)
(base)
H 2 O + H2 O
H3O+ + OH-
H2O
Kw
=
[H+][OH-]
H+ + OH=
1.01 x 10-14
(at 25°C)
6
Acids and Bases
Kw
=
H2O
H+ + OH-
[H+][OH-]
=
1.01 x 10-14
Acids and Bases
Kw
=
H2O
H+ + OH-
[H+][OH-]
=
(at 25°C)
What is the concentration of H+ and OH- in
pure water (at 25°C)?
1.01 x 10-14
(at 25°C)
What is the concentration of H+, if [OH-] is
1.0 x 10-10 (at 25°C)?
Varies with temperature
Kw
=
[H+][OH-]
[1.0 x 10-7][1.0 x 10-7]
=
=
1.01 x 10-14
1.01 x 10-14
Kw
=
[H+][OH-]
x ⋅ 1.0 x
10-10
=
1.01 x 10-14
=
1.01 x 10-14
x = 1.01 x 10-14 / 1.0 x 10-10 = 1.0 x 10-4
Acids and Bases
pH
Kw
=
=
[H+][OH-]
[1.0 x 10-7][1.0 x 10-7]
pH = - log (1.0 x
[1.0 x 10-4][1.0 x 10-10]
- log [H+]
=
1.01 x 10-14
=
1.01 x 10-14
10-7)
=
= 7.00
1.01 x 10-14
pH = - log (1.0 x 10-4)
= 4.00
Acids and Bases
pH
Kw
=
=
[H+][OH-]
- log [H+]
=
1.01 x 10-14
Solution is acidic, if [H+] > [OH-] and pH < 7
Solution is basic, if [H+] < [OH-] and pH > 7
7
Strengths of Acids and Bases
“Acids and bases are commonly classified as strong
or weak, depending on whether they react
‘completely’ or ‘partially’ to produce H+ or OH-”
Strong Acids/Bases Dissociate (Nearly) Completely
Strong Bases
LiOH
MgOH2
NaOH
CaOH2
KOH
SrOH2
RbOH
BaOH2
CsOH
R4NOH
Strong Acids
HCl
HBr
HI
H2SO4
HNO3
HClO4
Strengths of Acids and Bases
“Acids and bases are commonly classified as strong
or weak, depending on whether they react
‘completely’ or ‘partially’ to produce H+ or OH-”
Weak Acids/Bases Only Partially Dissociate
Weak Acids Donate
a Proton to Water
HA + H2O
H3O+ + A-
HA
Weak Acids Abstract
a Proton from Water
B + H2 O
H+ + A-
[H+][A-]
[HA]
(Acid Dissociation Constant)
Kb =
Ka =
Strengths of Acids and Bases
“Acids and bases are commonly classified as strong
or weak, depending on whether they react
‘completely’ or ‘partially’ to produce H+ or OH-”
[BH+][OH-]
[B]
(Base Hydrolysis Constant)
Carboxylic Acids/Carboxylate Ions
O
CH3
O
C
CH3
O
Weak Acids/Bases Only Partially Dissociate
C
H
e.g. Acetic Acid
Weak Acids
• Carboxylic Acids
• Ammonium Ions
• Metal Ions
BH+ + OH-
+ H+
O-
e.g. Acetate Ion
Ka = 1.75 x 10-5
Weak Bases
• Carboxylate Ions
• Amines
O
R
O
C
R
O
H
C
+ H+
O-
8
Amines/Ammonium Ions
H
:
N
H2O +
CH3
N
H
H
e.g. Methylamine
+
+ OHH
H
Methylammonium Ion
CH3
“Acids and bases are commonly classified as strong
or weak, depending on whether they react
‘completely’ or ‘partially’ to produce H+ or OH-”
HA
A- + H2O
Kb = 4.47 x 10-4
RNH2 = Primary Amine
R2NH = Secondary Amine
R3N = Tertiary Amine
Strengths of Acids and Bases
RNH3+
R2NH2+
R3NH+
H2O
H+ + A-
Ka =
[H+][A-]
[HA]
HA + OH-
Kb =
[BH+][OH-]
[B]
H+ + OH-
Kw = [H+][OH-]
Kw = KaKb
The product of Ka and Kb of conjugate acid/base is Kw
9