TMHsiung2007 160

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Chapter 06 Chemical Equilibrium
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Contents in Chapter06

• Reversible Reactions and Chemical Equilibria
• 1) Reversible Reactions and Equilibrium constant
• 2) Manipulating Equilibrium Constants
• 2. Equilibrium and Thermodynamics
• 1) Equilibrium constant vs free energy
• 2) Temperature effect in Le Châteliers
  Principle
• 3) Reaction Quotient vs. Equilibrium Constant
• Equilibrium for Precipitation Reaction
• 1) Precipitation, Solubility, and Solubility
  product
• 2) Solving Solubility Equilibrium Problems
• 3) Separation by Precipitation

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• Equilibrium for Complexation Reaction
• 1) Terms and Definitions
• 2) Effect of Complex Formation on Solubility
• 5. Equilibrium for Acid-Base Reaction
• 1) Terms and Definitions
• 2) BrØnsted-Lowry Acids and Bases
• 3) Autoprotolysis
• 4) p function
• 5) Strengths of Acids and Bases
• 6) Polyprotic acids and Bases
• 7) Ka vs Kb
• 6. Solving Equilibrium Problems with Spreadsheet

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1. Reversible Reactions and Chemical
Equilibria 1) Reversible Reactions and
Equilibrium constant
Example of expressing the equilibrium constant
(K) for the balanced chemical equation aA(aq)
bB(g) ? cC(l) dD(s)
A(aq) Standard state of (aq) is 1 M AA/1
M B(g) Standard state of (g) is 1 bar BPB/1
bar C(l) C 1 D(s) D 1
Equilibrium constant is dimensionless
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• Manipulating Equilibrium Constants
• For sum of the reactions, multiply the
  equilibrium constants
• For reverse the reaction, take 1/K
• For Multiplying reaction by n, take Kn

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Example Given the following information Rxn 1 A
B D K1 0.40 Rxn 2 A E C D F K2
0.10 Rxn 3 C E B K3 2.0 Rxn 4 F C D
B K4 5.0 What is the equilibrium constant for
the reaction 2A B C 3D Solution A B
D K1 0.40 A E C D F K2 0.10 B C
E K5 1/K3 0.50 F C D B K4 5.0 2A B
C 3D Koverall Koverall K1 x K2 x K5 x K4
0.40 x 0.10 x 0.50 x 5.0 0.10
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2. Equilibrium and Thermodynamics

• General Chemistry Revisit
• For a chemical reaction
• DG DH TDS n x F x Ecell
  (Electrochemistry discussed later)

If DS lt 0, less disorder If DS gt 0, more disorder
If DG lt 0, spontaneous If DG 0, at
equilibrium If DG gt 0, nonspontaneous
At Standard State DGo DHo TDSo Superscript
o represent standard state 25oC, 1 M or 1 bar
of the substance (Values presented on appendix of
your general chemistry text)
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• For a balanced reaction at standard state
• aA bB ? cC dD
• How to obtain ?Go?
• Way 1. Find out individual ?Go, then calculate
  overall ?Go
• ?Go (c?GoC d?GoD) (a?GoA b?GoB)
• Way 2. Find out individual ?Ho and ?So, then
  calculate overall ?Go
• ?Go ?Ho T?So

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1) Equilibrium constant vs free energy
For a balanced chemical reaction (equation) at
any moment
aA bB ? cC dD
?G ?Go RTlnQ (R 8.314 J/K, T K) At
equilibrium, ?G 0 therefore, ?Go -RTlnK
-2.303RTlogK lnK -?Go/RT or K e-?Go/RT
?Go lt 0, K gt 1 ?Go 0, K 1 ?Go gt 0, K lt 1
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2) Temperature effect in Le Châteliers Principle
K e?Go/RT e(?HoT?So)/RT
?So affected by temperature insignificantly,
therefore i) For endothermic reaction (?Ho gt
0) Increase T, increase K ii) For exothermic
reaction (?Ho lt 0) Increase T, decrease K

• Effect in Le Châteliers Principle
• For endothermic reaction Increase T, increase K,
  shift right
• For exothermic reaction Increase T, decrease K,
  shift left

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3) Reaction quotient vs Equilibrium constant
For a balanced chemical equation aA bB ? cC
dD
Quotient at any time
At equilibrium Q K (equilibrium constant)
IF Q gt K Reaction shift left IF Q K
Equilibrium IF Q lt K Reaction shift right
Le Châteliers Principle Add product, QgtK, shift
left Add reactant, QltK, shift right
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Example 1 What is ?Go for for the
reaction Ca(OH)2(s) Ca2(aq) 2OH(aq) K
6.5 105
Solution
Ans 23.9 kJ/mol
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• Example 2
• For the reaction Mg2 Cu(s) Mg(s) Cu2
• K 1092 and ?S 18 J/(Kmol).
• Under standard conditions, is ?G positive or
  negative?
• Under standard conditions, is the reaction
  endothermic or exothermic?

Solution (a)
(b) ?Go ?Ho T?So ?Go is positive, and ?So is
positive, ?Ho must be positive, This reaction is
endothermic
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Example 3 For the reaction 2H2O(l) H3O(aq)
OH(aq), Kw 1.0 1014 at 25oC. If the
concentrations in a system out of equilibrium are
H3O 3.0 x 105 M and OH 2.0 x 107 M.
Will the reaction proceed to the left or to the
right to reach equilibrium?
Solution Q H3OOH 6.0 1012 Qgt K,
therefore, reaction shift to the left to reach
equilibrium
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• Equilibrium for Precipitation Reaction
• Precipitation, dissolution, and solubility
  product
• Precipitation reaction
• Example
• Ca2(aq) SO42(aq) CaSO4(s)
• ii) Dissolution
• Example
• CaSO4(s) Ca2(aq) SO42(aq)
• Solubility product
• Example For a saturated CaSO4 solution
• CaSO4(s) Ca2(aq) SO42(aq)
• Ksp Ca2SO42

Note Dissolved Ca2 in a saturated CaSO4
solution, 2/3 is Ca2, 1/3 is ion pairs form
CaSO4(aq)
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• iv) Simultaneous side reaction
• Example For a saturated Hg2Cl2 solution
• Hg2Cl2(s) Hg22(aq) 2Cl(aq) Ksp 1.2x1018
• Examples of side reaction
• a) Hydrolysis
• Hg22(aq) 2H2O(l) Hg2OH(aq) H3O(aq) K
  105.3
• Disproportionation
• 2Hg22(aq) Hg2(aq) Hg(l) K 102.1
• Common ion effect for solubility
• Common ion The same ion (common ion) of
  different substance. e.g., Cl is the common for
  AgCl and NaCl.
• Common ion effect for the solubility of salt
  Solubility of a salt (ionic compound) decreases
  when it is placed in a solution containing one of
  the salts ions.

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2) Solving Solubility Equilibrium Problems
Example 1 What is the solubility of Pb(IO3)2 in
Water ? What is its Pb2 at equilibrium? What
is its IO3 at equilibrium? (a Simple Problem)
Solution
Initial - 0 0 Change - s 2s Equilibrium - s 2s
(s)(2s)2 4s3 2.5x1013 s 4.0x105 Ans solub
ility of Pb(IO3)2 in water is 4.0x105 M at
equilibrium, Pb2 4.0x105 M, at
equilibrium, IO3 8.0x105 M
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Example 2 What is the solubility of Pb(IO3)2 in
0.10 M Pb(NO3)2 solution? What is its Pb2 at
equilibrium? What is its IO3 at equilibrium?
(Need consider common ion effect)
Solution
Initial - 0.10 0 Change - s 2s Equilibrium - 0.1
0s 2s
(0.10s)(2s)2 4s3 0.40s2 2.5x1013.. Hard
to solve!! So, Assume s ltlt 0.10, thus
(0.10)(2s)2 0.40s2 2.5x1013, s7.91x107,
Yes, s ltlt 0.10!!
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Contd
Pb2 (0.10s) 0.10 7.91x107
0.10 IO3 2s 2x(7.91x107) 1.58x106
Ans solubility of Pb(IO3)2 in water is 7.91x107
M at equilibrium, Pb2 0.10 M at
equilibrium, IO3 1.58x106 M
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Example 3 What is the solubility of Pb(IO3)2 in
1x104 M Pb(NO3)2 solution? What is its Pb2 at
equilibrium? What is its IO3 at equilibrium?
(Need consider common ion effect, more
sophisticated solving step)
Solution
Initial - 1x104 0 Change - s 2s Equilibrium -
1x104 s 2s
Assume s ltlt 1x104, thus (1x104)(2s)2
4x104s2 2.5x1013, s2.5x105, Oh, no!! s is
not ltlt 1x104!! Assumption incorrect!!
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Contd
Lets use iterative approach (method of
successive approximations).
Initial - 1x104 0 Change - s 2s Equilibrium -
1x104 s 2s
(1x104s)(2s)2 2.5x1013

s ?? 4s2 ?? s ?? 0, 2.500x109, 2.500x105 2.500x
105, 2.000x109, 2.236x105 2.236x105,
2.043x109, 2.260x105 2.260x105, 2.043x109, 2.
260x105
?????,????
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Contd
Pb2 (1x104s) 1x104 2.26x105
1.23x104 IO3 2s 2x(2.26x105) 4.52x105
Ans solubility of Pb(IO3)2 in water is 2.26x105
M at equilibrium, Pb2 1.23x104 M at
equilibrium, IO3 4.52x105 M
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3) Separation by Precipitation

• Need define complete precipitation, e.g., 99.99
  precipitated out (0.01 left in solution).

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Example PbI2(s) ? Pb2(aq) 2I(aq) Ksp7.9x109
HgI2(s) ? Hg22(aq) 2I(aq) Ksp1.1x1028 Is
it possible to separating 99.99 of 0.01 M Hg22
from 0.01 M Pb2 by selectively precipitating
Hg22 with I?
Solution HgI2(s) ? Hg22(aq) 2I(aq) Initial
conc. - 0.01 0 Final conc. - 1x106 x Hg22I2
(1x106)(x)2 Ksp 1.1x1028 x I
1.0x1011 M Now, for reaction between Pb2 and
I Q Pb2I2 (0.01)(1.0x1011)2
1.0x1024 (lt Ksp of PbI2) Ans 99.99 separate
Hg22 from Pb2 is feasible
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• Equilibrium for Complexation Reaction
• 1) Terms and Definition

• Lewis acid electron pair acceptor
• Lewis base electron pair donor
• Complexation reaction The reaction produce a
  dative bond (coordinate covalent) between a Lewis
  acid (e.g., metal ions) and the Lewis base (e.g.,
  ligand which has nonbonding pairs electron).

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• Formation constant (stability constant)
• Example

• Dissociation constant
• Example

Kd 1.8x108
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• Stepwise formation constant
• Example

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• Cumulative formation constant
• Example

Cd2 NH3 ? CdNH32 ß1K1 Cd2 2NH3 ?
Cd(NH3)22 ß2K1K2 Cd2 3NH3 ?
Cd(NH3)32 ß3K1K2K3 Cd2 4NH3 ?
Cd(NH3)42 ß4K1K2K3K4
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• Effect of Complex Formation on Solubility
• i) A simple example

AgBr(s) ? Ag(aq) Br(aq) Ksp 5 x
1013 Ag(aq) 2NH3(aq) ? Ag(NH3)2(aq) ß2
1.6 x 107 What is the molar solubility of AgBr(s)
in 3.0 M NH3 ? Solution AgBr(s) 2NH3(aq) ?
Ag(NH3)2(aq) Br(aq) K Ksp x ß2 8 x 106
AgBr(s) 2NH3(aq) ? Ag(NH3)2(aq)
Br(aq) Initial
3.0 0
0 Change 2s
s s Equilibrium
3.0 2s s
s
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s 8.4x103 M.. Ans
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ii) A sophisticated example
PbI2(s) ? Pb2(aq) 2I(aq) Ksp 7.9 x
109 Pb2(aq) I(aq) ? PbI(aq) K1 1.0 x
102 Pb2(aq) 2I(aq) ? PbI2 (aq) ß2 1.4 x
103 Pb2(aq) 3I(aq) ? PbI3 (aq) ß3 8.3 x
103 Pb2(aq) 4I(aq) ? PbI42 (aq) ß4 3.0 x
104 For a PbI2(s) saturated solution, find the
concentrations Pb2(aq), PbI(aq), PbI2 (aq),
PbI3 (aq), PbI42 (aq), and total dissolved Pb
when I0.001 M.
Solution At I0.001 M Pb2Ksp/I27.9x10
3 M PbIK1Pb2 I7.9x104
M PbI2ß2Pb2 I21.1x105 M
PbI3ß3Pb2 I36.6x108
M PbI42ß4Pb2 I42.4x1010 M total
Pb8.7x103 M
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Effect of Complex Formation Total dissolved Pb
(curve with circle)
Below 0.1 M I Increase I, decrease total
dissolved Pb, Beyond 0.1 M I Increase I,
increase total dissolved Pb
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5. Equilibrium for Acid-Base Reaction 1) Terms
and Definition

• Neutralization (acid-base reaction)
• Acid Base ? Salt H2O
• Protic solvent The solvent molecule involving a
  transferable H. e.g., HOH (water), CH3COOH
  (acetic acid)
• Protic acids and bases The acids and base
  involving a transferable H. e.g., HCl, NaOH.
• Amphiprotic solvent The solvent molecules act as
  either acids or base, e.g., H2O
• HCl H2O ? H3O Cl
• base
• NH3 H2O ? NH4 OH
• acid

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2) Brfnsted-Lowry Acids and Bases (Conjugate
Acid-Base Pair) Acid A proton (H) donor Base
A proton (H) acceptor
(conjugate pair)
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• Autoprotolysis (autoionization, self-ionization)
• For amphiprotic solvent, proton transfer from
  one molecule to the other.
• HS as the symbol of solvent molecule for
  example
• HS HS ? H2S S Ks H2SS H2S2
• For water
• H2O H2O ? H3O OH Kw H3OOH
  H3O 2

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At 25oC, Kw 1x1014 H3O OH 1x107, pH
pOH 7 Naming H3O (hydronium ion), OH
(hydroxide ion) Thus, Acidic aqueous solution at
25oC H3O gt 1.00 x 107, or say pH lt 7
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• P Function
• pX log(X)
• pH log(aH3O) logH3O

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5) Strengths of Acid and Bases
i) Strong Acid
ii) Strong Base
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Remark1 HF is not strong acid, HF is weak acid
because of forming ion pair between F and H3O.
Remark 2 H2SO4 Ka1 very large (complete
dissociation), Ka21.0x102
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• Weak acids
• For example

Ka Acid dissociation constant
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iv) Weak bases For example
Kb Base hydrolysis constant
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v) Miscellaneous
a) RCOOH H2O ? RCOO H3O
Carboxylic acids are weak acids, carboxylate
anios are weak bases
b) RNH2 H2O ? RNH3 OH
R2NH H2O ? R2NH2 OH R3N H2O ? R3NH OH
Amines (primary, secondary, tertiary) are weak
bases, ammonium ion (primary, secondary,
tertiary) are weak acids.
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c) Hydrolysis of hydrated metal ion
M(H2O)xn H2O ? M(H2O)x1(OH)n1 H3O thus,
metal ion act as weak acid
pKa for aqueous metal ions
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6) Polyprotic acids and Bases
i) Polyprotic acids H3PO4 for example
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ii) Polybasic (polyprotic) bases Na3PO4 for
example
PO43- H2O HPO42- OH- Kb1 HPO42- H2O
H2PO4- OH- Kb2 H2PO4- H2O H3PO4 OH- Kb3
?? Kb1Kb3 ???
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7) Ka vs Kb
i) For monoprotic acid HA H2O ? A
H3O Its conjugate base A H2O ? HA OH
Ka x Kb Kw So, Kb Kw/ Ka
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ii) For diprotic acid H2A H2O ? HA
H3O HA H2O ? A2 H3O For conjugate
base A2 H2O ? HA OH HA H2O ? H2A
OH
Ka1 x Kb2 KwKa2 x Kb1 Kw
iii) For triprotic acid Ka1 x Kb3 Kw, Ka2 x
Kb2 Kw , Ka3 x Kb1 Kw
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6. Solving Equilibrium Problems with Spreadsheet
Example 1 PbI2(s) ? Pb2(aq) 2I(aq) Ksp 7.9
x 109 Pb2(aq) I(aq) ? PbI(aq) K1 1.0 x
102 Pb2(aq) 2I(aq) ? PbI2 (aq) ß2 1.4 x
103 Pb2(aq) 3I(aq) ? PbI3 (aq) ß3 8.3 x
103 Pb2(aq) 4I(aq) ? PbI42 (aq) ß4 3.0 x
104 For a PbI2(s) saturated solution, find the
concentrations Pb2(aq), PbI(aq), PbI2 (aq),
PbI3 (aq), PbI42 (aq), and total dissolved Pb
for I from 103.5 M to 10 M.
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Calculation by Excel
Excel Ch06-1
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Example 2 For the balanced reaction I2(s)5Br2(a
q)6H2O(l) ? 2IO3(aq)10Br(aq)12H(aq)
K1x1019 For a solution containing 0.001 M
Br2(aq), 0.005 M IO3(aq), 0.02 M Br(aq), 1 M
H(aq) and excess solid I2. What are the final
concentrations of those four species when the
solution reach equilibrium?
Solution
QgtK, shift left
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I2(s)5Br2(aq)6H2O(l) ? 2IO3(aq)10Br(aq)12H(
aq)
Initial conc. excess 0.001
0.005 0.02 1 Change
2.5x x 5x
6x Final conc. excess 0.0012.5x
0.005x 0.025x 16x
Solving with Excel by Way 1 trial-and error Way
2 GOAL-SEEK
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I2(s)5Br2(aq)6H2O(l) ? 2IO3(aq)10Br(aq)12H(
aq)
Initial conc. excess 0.001
0.005 0.02 1 Change
2.5x x 5x
6x Final conc. excess 0.0012.5x
0.005x 0.025x 16x
Solving with Excel by Way 1 trial-and error Way
2 GOAL-SEEK
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Way 1 trial-and error
Excel Ch06-2
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Way 2 GOAL-SEEK
Excel Ch06-2
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Homework Problem 6-28/p.119, Due
2007/10/31 ?????????? Examples All Exercise Al
l Problems 1-11, 14-17, 22-25, 29-40, 43-51
End of Chapter06