Date post: | 17-Jan-2018 |
Category: |
Documents |
Upload: | rafe-elliott |
View: | 221 times |
Download: | 0 times |
1
Chemical Equilibrium Chapter 16
Hein and Arena
Eugene PasserChemistry DepartmentBronx Community College
© John Wiley and Sons, Inc.
Version 1.1
2
Chapter Outline16.1 Reversible Reactions
16.2 Rates of Reaction
16.3 Chemical Equilibrium16.10 Ion Product Constant for Water
16.11 Ionization Constants
16.4 Le Chatelier’s Principle
16.5 Effect of Concentration on Equilibrium
16.9 Equilibrium Constants
16.12 Solubility Product Constant
16.13 Acid-Base Properties of Salts
16.6 Effect of Volume on Equilibrium
16.7 Effect of Temperature on Equilibrium
16.8 Effect of Catalysts on Equilibrium
16.14 Buffer Solutions: The Control of pH
3
Reversible ReactionsReversible Reactions
4
reversible reaction A chemical reaction in which the products formed react to produce the original reactants.
5
2NO2(g) → N2O4 (g)cooling
N2O4 (g) → 2NO2 (g)heating
The reaction between NO2 and N2O4
is reversible.N2O4 is formed
N2O4 decomposes when heated forming NO2
6
2NO2(g) N2O4 (g)→→
reaction to the right
reaction to the left
7
Rates of ReactionRates of Reaction
8
chemical kinetics The study of reaction rates and reaction mechanisms.
9
• 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
10
Forward reaction A + B → C + D
Reverse reaction C + D → A + B
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.
16.2
11
Chemical EquilibriumChemical Equilibrium
12
equilibrium: a dynamic state in which two or more opposing processes are taking place at the same time and at the same rate.
chemical equilibrium: the state in which the rate of the forward reaction equals the rate of the reverse reaction in a chemical change.
At equilibrium the concentrations of the products and the reactants are not changing.
13
NaCl(s) Na+(aq) + Cl-(aq)→→
A saturated salt solution is in equilibrium with solid salt.
salt crystalsare dissolving
Na+ and Cl- are crystallizing
At equilibrium the rate of salt dissolution equals the rate of salt crystallization.
14
Le Chatelier’s PrincipleLe Chatelier’s Principle
15
In 1888, the French chemist Henri LeChatelier set forth a far-reaching generalization on the behavior of equilibrium systems.
This generalization, known as LeChatelier’s Principle, states
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.
16
Effect of Concentration Effect of Concentration on Equilibriumon Equilibrium
17
• 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.
18
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.
19
results in A and B being used faster than they are produced.
Increasing the concentration of B
A + B C + D→→
The system is at equilibrium
results in C and D being produced faster than they are used.
increases the rate of the forward reaction
20
The system is again at equilibrium
After enough time has passed, the rates of the forward and reverse reactions become equal.
In the new equilibriumconcentration of A has decreased
concentrations of B, C and D have increased
A + B C + D→→
21
Percent Yield
22
H2(g) + I2(g) 2HI(aq)→→H2 + I2 combine
to form HIHI decomposes to form H2 + I2
700 K
Initial Concentrations
0 mol1 mol1 mol 2 mol0 mol0 mol
Final Concentrations in theAbsence of Equilibrium
1.58 mol0.21 mol0.21 mol
Equilibrium Concentrations
At equilibrium the rate of HI formation equals the rate of HI decomposition.
The forward reaction is 79% complete at equilibrium.
1.58 mol HI% Yield = x 100% = 79%2 mol HI
23
Original Equilibrium New Equilibrium1.00 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
Comparison of Equilibria
24
Cl2(aq) + 2H2O(l) HOCl(aq) + H3O+(aq) + Cl-(aq)→→
Effect of Concentration Changeson the Chlorine Water Equilibrium
decrease Cl2 concentration
Equilibrium shifts to left
increase H2O concentration
Equilibrium shifts to right
increase HOCl concentration
Equilibrium shifts to left
decrease H3O+ concentration
Equilibrium shifts to right
increase Cl- concentration
Equilibrium shifts to left
25
HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2(aq)→→ -2 3 2C H O ( )aq
1 L 0.100 M HC2H3O2
Equilibrium pH = 2.87
Effect of C2H3O2
Concentration Changes on pH
Add 0.100 mol NaC2H3O2
NaC2H3O2(aq) → Na+(aq) + C2H3O2(aq)-2 3 2C H O ( )aq
Equilibrium shifts to left
1 L 0.100 M HC2H3O2
Equilibrium pH = 4.74
Add 0.200 mol NaC2H3O2
1 L 0.100 M HC2H3O2
Equilibrium pH = 5.05
-2 3 2C H O
26
Effect of Volume on Effect of Volume on EquilibriumEquilibrium
27
• 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.
28
CaCO3(s) CaO(s) + CO2(g)→→increases CO2 concentration
Equilibrium shifts to left
Decrease Volume
29
CaCO3(s) CaO(s) + CO2(g)→→decreases CO2 concentration
Equilibrium shifts to right
Increase Volume
30
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.
31
N2(g) + 3H2(g) 2NH3(g)→→Equilibrium shifts to the right towards fewer molecules.
Decrease Volume
1 mol 3 mol 2 mol6.02 x 1023
molecules1.81 x 1024 molecules
1.20 x 1024
molecules
2.41 x 1024
molecules
32
N2(g) + O2(g) 2NO(g)→→Equilibrium does not shift. The number of molecules is the same on both sides of the equation.
Decrease Volume
1 mol 1 mol 2 mol6.02 x 1023
molecules6.02 x 1023
molecules1.20 x 1024
molecules
1.20 x 1024
molecules
33
Effect of Temperature Effect of Temperature on Equilibriumon Equilibrium
34
When the temperature of a system is raised, the rate of reaction increases.
In a reversible reaction, the rates of both the forward and the reverse reactions are increased by an increase in temperature.
The rate of the reaction that absorbs heat is increased to a greater extent, and the equilibrium shifts to favor that reaction.
35
Equilibrium shifts to right
At room temperature very little CO forms.
C(s) + CO2(g) + heat 2CO(g)→→At 1000oC moles CO2 moles CO
Heat may be treated as a reactant in endothermic reactions.
36
Effect of CatalystsEffect of Catalystson Equilibriumon Equilibrium
37
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.
3816.3
Energy Diagram for an Exothermic Reaction
Activation energy: the minimum energy required for a reaction to occur.A catalyst speeds up a reaction by lowering the activation energy.
A catalyst does not change the energy of a reaction.
39
PCl3(l) + S(s) → PSCl3(l)AlCl3
Very little thiophosphoryl chloride is formed in the absence of a catalyst because the reaction is so slow.In the presence of an aluminum chloride catalyst the reaction is complete in a few seconds.
MnO2
Δ 2KClO3(s) → 2KCl + 3O2(l)
The laboratory preparation of oxygen uses manganese dioxide as a catalyst to increase the rate of the reaction.
40
Equilibrium ConstantsEquilibrium Constants
41
At equilibrium the rates of the forward and reverse reactions are equal, and the concentrations of the reactants and products are constant.
42
The equilibrium constant (Keq) is a value representing the unchanging concentrations of the reactants and the products in a chemical reaction at equilibrium.
43
For the general reactionaA + bB cC + dD→→
c d
q a be[C] [D]K = [A] [B]
at a given temperature
44
For the reaction3H2 + N2 2NH3
→→
3eq
2
2
32
[NH ]K = [H ] [N ]
45
For the reaction4NH3 + 3O2 2N2
+ 6H2O→→
2 6
4 32 2
eq3 2
[N ] [H O]K = [NH ] [O ]
46
The magnitude of an equilibrium constant indicates the extent to which the forward and reverse reactions take place.
3H2 + I2 2HI3→→
oeq
2 2
2
3[HI]K = 54.8= at 425 C
[H ] [I ]
COCl2 CO + Cl2→→
o2q
-4e
2
[CO][Cl ]K = 7.6 x 10= at 400 C[COCl ]
At equilibrium more product than reactant exists.At equilibrium more reactant than product exists.
47
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.
48
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)→→
3 2eq
5
[PCl ][Cl ]K = = [PCl ]
(0.97)(0.97) = 31 (0.030)
49
Ion Products Constant Ion Products Constant for Waterfor Water
50Kw = (1 x 10-7)(1 x 10-7) = 1.00 x 10-14
H2O + H2O H3O+ + OH-→→
H2O H+ + OH-→→
[H+] = [OH-] = 1.00 x 10-7 mol/L
Kw = [H+][OH-] = 1.00 x 10-14 T = 25oC
Water autoionizes to a slight degree.
or more simply
At equilibrium
The water equilibrium constant, Kw, is called the ion product constant for water.
51
What is the concentration of (a) H+ and (b) OH- in a 0.001 M HCl solution?
HCl → H+ + Cl-HCl is 100% ionized.
[H+] = 1 x 10-3 MSolve the Kw expression for [OH-].
Kw= [H+][OH-] = 1.00 x 10-14
-14
-31 x 10 = 1 x 10
-11= 1 x 10 mol/L- w+
K[OH ] = [H ]
52
What is the pH of a 0.010 M NaOH solution?
NaOH → Na+ + OH-
NaOH is 100% ionized.
[OH-] = 1 x 10-2 MSolve the Kw expression for [H+].
Kw= [H+][OH-] = 1.00 x 10-14
-14
-21 x 10 = 1 x 10
-12= 1 x 10 mol/L+ w-
K[H ] = [OH ]
pH = - log[H+] = - log(1.0 x 10-12) = 12
53
What is the pH of a 0.010 M NaOH solution?
= - log(1.0 x 10-2) = 2
The pH may also be calculated by first calculating the pOH.
pOH = - log[OH-]
pH + pOH = 14
pH = 14 - pOH
pH = 14 – 2 = 12
In pure water
Solve for pH
[OH-] = 1 x 10-2 M
54
[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
Relationship of H+ and OH- Concentrations in Water Solutions
16.1
55
Ionization ConstantsIonization Constants
56
In addition to Kw, several other ionization constants are used.
57
Ka
58
HC2H3O2(aq) H+ + C2H3O2→→ -
2 3 2C H O
When acetic acid ionizes in water, the following equilibrium is established.
Ka is the ionization constant for this equilibrium. + -
2 3 2a
2 3 2
[H ][C H O ]K = [HC H O ]
Ka is called the acid ionization constant. Since the concentration of water is large and does not change appreciably, it is omitted from Ka.
59
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+ + C2H3O2→→ -
2 3 2C H O
Because each molecule of HC2H3O2 that ionizes yields one H+ and one C2H3O2 , the concentrations of the two ions are equal.
-2 3 2C H O
+ - -32 3 2[H ] = [C H O ] = 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
60
+ -2 3 2
a2 3 2
[H ][C H O ]K = [HC H O ]
-3 -3(1.34x10 )(1.34x10 )
(0.099)-5 = 1.8x10
+ - -32 3 2[H ] = [C H O ] = 1.34 x 10 mol/L
[HC2H3O2] = 0.099 mol/L
Substitute these concentrations into the equilibrium expression and solve for Ka.
61
Because each molecule of HC2H3O2 that ionizes yields one H+ and one C2H3O2 , the concentrations of the two ions are equal.
-2 3 2C H O
What is the [H+] in a 0.50 M HC2H3O2 solution? The ionization constant, Ka, for HC2H3O2 is 1.8 x 10-5.
HC2H3O2(aq) H+ + C2H3O2→→ -
2 3 2C H O
The equilibrium expression and Ka for HC2H3O2 are
+ -2 3 2
a2 3 2
[H ][C H O ]K = [HC H O ]
Let Y = [H+] = -2 3 2[C H O ]+ -
2 3 2[H ] = [C H O ]
[HC2H3O2] at equilibrium is 0.50 – Y.
62
Substitute these values into Ka for HC2H3O2.
+ -2 3 2
a2 3 2
[H ][C H O ]K = [HC H O ]
Y = [H+] = -2 3 2[C H O ] HC2H3O2 = 0.50 - Y
2-5
a(Y)(Y) YK = = = 1.8 x 10
0.50 - Y 0.50 - Y
2-5
a(Y)(Y) YK = = = 1.8 x 100.50 0.50
Assume Y is small compared to 0.50 -Y.Then 0.50 – Y 0.50
2-5Y = 1.8 x 10
0.50
2 -5Y = 0.50 x 1.8 x 10Solve for Y2.
63
Take the square root of both sides of the equation. 2 -5Y = 0.50 x 1.8 x 10-5= 0.90 x 10 -6= 9. 0 x 10
-6 -3Y = 9.0 x 10 = 3.0 x 10 mol/L+ -3[H ] = 3.0 x 10 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.
65
Calculate the percent ionization in a 0.50 M HC2H3O2 solution.
The percent ionization is given by
+ -concentration of [H ] or [A ] x 100= percent ionizedinitial concentration of [HA]
The ionization of a weak acid is given by
HA H+ + A- →→
66
[H+] was previously calculated as 3.0 x 10-3 mol/L
+ -2 3 2
2 3 2
concentration of [H ] or [C H O ] x 100= percent ionizedinitial concentration of [HC H O ]
Calculate the percent ionization in a 0.50 M HC2H3O2 solution.
HC2H3O2(aq) H+ + C2H3O2→→ -
2 3 2C H O
The percent ionization of acetic acid is given by
The ionization of acetic acid is given by
-33.0 x 10 mol/L x 100= 0.60% percent ionized0.50 mol/L
67
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 HOCl3.5 x10-8
Carbolic HC6H5O 1.3 x 10-10 Nitrous HNO24.5 x10-4
Cyanic HCNO 2.0 x 10-4 Hydrofluoric HF6.5 x10-4
Formic HCHO2 1.8 x 10-4
Ionization Constants (Ka) of Weak Acids at 25oC
16.2
68
Solubility Product ConstantSolubility Product Constant
69
equilibrium A dynamic state in which two or more opposing processes are taking place at the same time and at the same rate.
chemical equilibrium The state in which the rate of the forward reaction equals the rate of the reverse reaction in a chemical change.
The solubility product constant, Ksp, is the equilibrium constant of a slightly soluble salt.
70
Silver chloride is in equilibrium with its ions in aqueous solution.
AgCl(s) Ag+(aq) + Cl-(aq) →→
+ -
eq[Ag ][Cl ]K = [AgCl( )]s
The equilibrium constant is
The amount of solid AgCl does not affect the equilibrium.
The concentration of solid AgCl is a constant.
+ -eq spK x [AgCl( )] = [Ag ][Cl ] = Ks
Rearrange
The product of Keq and [AgCl(s) is a constant.
+ -spK = [Ag ][Cl ]
71
The solubility of AgCl in water is 1.3 x 10-5
mol/L.
Because each formula unit of AgCl that dissolves yields one Ag+ and one Cl-
, the concentrations of the two ions are equal.
AgCl(s) Ag+(aq) + Cl-(aq) →→Ksp = [Ag+][Cl-]
= (1.3 x 10-5)(1.3 x 10-5)
[Ag+] = [Cl-] = 1.3 x 10-5 mol/L
= 1.7 x 10-10
The Ksp has no denominator.
72
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.
73
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-
4 4PbSO Pb + SO→→
2+ 2-sp 4K = [Pb ][SO ]
The Ksp of PbSO4 is
Because each formula unit of PbSO4 that dissolves yields one Pb2+ and one , the concentrations of the two ions are equal.
2-4SO
2+ 2-4Let Y = [Pb ] = [SO ]
74
The Ksp value for lead sulfate is 1.3 x 10-8. Calculate
the solubility of PbSO4 in grams per liter.
2+ 2-sp 4K = [Pb ][SO ]
Substitute Y into the Ksp equation.
-8spK = (Y)(Y) = 1.3 x 10
2 -8Y = 1.3 x 10-8Y = 1.3 x 10
-4= 1.1 x 10 mol/L
The solubility of PbSO4 is 1.1 x 10-4 mol/L.
75
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.
-41.1 x 10 molL
-2= 3.3 x 10 g/L 303.3 g
mol
The solubility of PbSO4 is 3.3 x 10-2 g/L.
76
The Common Ion EffectThe Common Ion Effect
77
An ion added to a solution already containing that ion is called a common ion.
A shift in the equilibrium position upon addition of an ion already contained in the solution is known as the common ion effect.
78
AgCl(s) Ag+(aq) + Cl-(aq) →→Silver nitrate dissociates in aqueous solution.
+ -3 3AgNO ( ) Ag ( ) + NO ( )s aq aq
The equilibrium equation of AgCl is
Ag+ is common to both reactions.
Upon the addition of Ag+, the equilibrium shifts to the left, in accordance with LeChatelier’s Principle.
79In the absence of AgNO3, [Cl-] =1.3 x 10-5.
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-].
+ - -10spK = [Ag ][Cl ] = 1.7 x 10
Solve for [Cl-].
- -10[0.10][Cl ] = 1.7 x 10
-10- -91.7 x 10[Cl ] = = 1.7x10 mol/L
[0.10]
Substitute [Ag+] into the Ksp.
This is an example of the common ion effect.
80
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
Solubility Product Constants (Ksp) at 25oC
16.3
81
Acid-Base Acid-Base Properties of SaltsProperties of Salts
82
hydrolysis is the term used for the general reaction in which a water molecule is split.
83
→→-2 3 2 2 2
-3C H O (aq) + (l) C H O (aq) OH O+H HH ( )aq
The net ionic equation for the hydrolysis of sodium acetate is
The water molecule splits.
Salts that contain the anion of a weak acid under go hydrolysis.
The solutionis basic.
84
The net ionic equation for the hydrolysis of ammonium chloride is
The water molecule splits.
Salts that contain the cation of a weak base under go hydrolysis.
The solutionis acidic.
+3
+4 3NH ( ) + ( ) NH ( ) OH O+ ( )H Haq l aq aq→→
85
Salts derived from a strong acid and a strong base do not undergo hydrolysis.
+ -Na + Cl + ( ) no re O acH H tionl →
86
Type of salt Nature ofAqueous 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
Ionic Composition of Salts and the Nature of the Aqueous Solutions They Form
16.4
87
Buffer Solutions: Buffer Solutions: The Control of pHThe Control of pH
88
A buffer solution resists changes in pH when diluted or when small amounts of acid or base added.
89
Sodium acetate when mixed with acetic acid forms a buffer solution.
A weak acid mixed with its conjugate base form a buffer solution.
90
+ -2 3 2 2 3 2 H ( ) + C H O ( ) HC H O ( )aq aq aq
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.
A weak acid mixed with its conjugate base form a buffer solution.
91
- -2 3 2 2 2 3 2OH + HC H O ( ) H O( ) + C H O ( )aq l aq
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.
A weak acid mixed with its conjugate base form a buffer solution.
92
Solution pH Change in pHH2O (1000 mL) 7
H2O + 0.010 mol HCl 2 5
H2O + 0.010 mol NaOH 12 5
Buffer solution (1000 mL)
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
Changes in pH Caused by theAddition of HCl and NaOH
16.5
93